Iranian University Students Mathematics

Competitions, 1973-2007

Bamdad R. Yahaghi

School of Mathematics, Institute for Studies in Theoretical Physicsand Mathematics (IPM), P.O. Box: 19395-5746 Tehran, Iran

E-mail address: bamdad5@ipm.ir, bbaammddaadd55@gmail.com

Contents

Preface ix

Chapter 1. Problems 11.1. First Competition, University of Tehran, March 1973 11.2. Second Competition, Shiraz (former Pahlavi) University, March 1974 31.3. Third Competition, (former Jondi Shapour) University of Ahwaz,

March 1975 61.4. Fourth Competition, University of Tabriz, March 1976 81.5. Fifth Competition, Sharif (former Aryamehr) University of Technology,

March 1977 91.6. Sixth Competition, The University of Isfahan, March 1978 101.7. Seventh Competition, Ferdowsi University of Mashhad, March 1980 121.8. Eighth Competition, Shiraz University, March 1984 131.9. Ninth Competition, Tehran Teacher Training (Tarbiat Moallem)

University, September 1985 141.10. Tenth Competition, University of Sistan and Baluchestan, March

1986 161.11. Eleventh Competition, The University of Birjand, March 1987 171.12. Twelfth Competition, Guilan University , March 1988 181.13. Thirteenth Competition, University of Tehran, March 1989 201.14. Fourteenth Competition, The University of Isfahan, March 1990 211.15. Fifteenth Competition, Ferdowsi University of Mashhad, March 1991 231.16. Sixteenth Competition, Razi (Rhazes or Rasis) University of

Kermanshah, March 1992 241.17. Seventeenth Competition, Shahid Beheshti (former National)

University, March 1993 251.18. Eighteenth Competition, Sharif University of Technology, March

1994 261.19. Nineteenth Competition, University of Kerman, March 1995 271.20. Twentieth Competition, Sharif University of Technology, February

1996 281.21. Twenty First Competition, University of Tehran, March 1997 281.22. Twenty Second Competition, University of Ahwaz, March 1998 291.23. Twenty Third Competition, Sharif University of Technology, March

1999 361.24. Twenty Fourth Competition, Khajeh Nasir Toosi University of

Technology, May 2000 371.25. Twenty Fifth Competition, Imam Khomeini International University

of Qazvin, May 2001 39

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1.26. Twenty Sixth Competition, Institute for Advanced Studies in BasicSciences (IASBS) of Zanjan, May 2002 40

1.27. Twenty Seventh Competition, Bu-Ali Sina (Avecina) University ofHamedan, May 2003 41

1.28. Twenty Eighth Competition, Sharif University of Technology, May2004 43

1.29. Twenty Ninth Competition, University of Mazandaran in Babolsar,May 2005 45

1.30. Thirtieth Competition, Tafresh University, May 2006 461.31. Thirty First Competition, Ferdowsi University of Mashhad, May

2007 48

Chapter 2. Solutions 512.1. First Competition 512.2. Second Competition 562.3. Third Competition 642.4. Fourth Competition 712.5. Fifth Competition 742.6. Sixth Competition 832.7. Seventh Competition 862.8. Eighth Competition 912.9. Ninth Competition 962.10. Tenth Competition 1092.11. Eleventh Competition 1152.12. Twelfth Competition 1192.13. Thirteenth Competition 1252.14. Fourteenth Competition 1342.15. Fifteenth Competition 1432.16. Sixteenth Competition 1492.17. Seventeenth Competition 1522.18. Eighteenth Competition 1552.19. Nineteenth Competition 1602.20. Twentieth Competition 1652.21. Twenty First Competition 1722.22. Twenty Second Competition 1762.23. Twenty Third Competition 1792.24. Twenty Fourth Competition 1842.25. Twenty Fifth Competition 1962.26. Twenty Sixth Competition 2022.27. Twenty Seventh Competition 2052.28. Twenty Eighth Competition 2102.29. Twenty Ninth Competition 2152.30. Thirtieth Competition 2202.31. Thirty First Competition 226

Chapter 3. Problem Index 235First Competition 235Second Competition 235Third Competition 236

CONTENTS vii

Fourth Competition 236Fifth Competition 236Sixth Competition 237Seventh Competition 237Eighth Competition 237Ninth Competition 238Tenth Competition 238Eleventh Competition 239Twelfth Competition 239Thirteenth Competition 240Fourteenth Competition 240Fifteenth Competition 240Sixteenth Competition 241Seventeenth Competition 241Eighteenth Competition 241Nineteenth Competition 241Twentieth Competition 242Twenty First Competition 242Twenty Second Competition 242Twenty Third Competition 242Twenty Fourth Competition 243Twenty Fifth Competition 243Twenty Sixth Competition 243Twenty Seventh Competition 244Twenty Eighth Competition 244Twenty Ninth Competition 244Thirtieth Competition 245Thirty First Competition 245Classification of problems 246

Index 247

Preface

taa khod raa be tchizi nadaadi be kol-liyat,aan tchiz sa’b-o doshvaar minomaayad.

tchoon khod raa be kol-liyat be tchizi daadi,digar doshvaari namaanad.

[Until you do not devote yourself to a task totally,that task looks hard and unreachable.

As you devote yourself to a task totally,there remains no difficulty.]

–Shamsoddin Tabrizi

International mathematical competitions have gained great popularity in recentyears. They provide the first test for a young person’s mathematical prowess, andsuccess in these competitions often translates into gaining admittance to the bestresearch institutions for graduate work. For this reason some colleges and evenhigh schools have organized special classes for training mathematically talentedstudents for these competitions. The purpose of this text is to provide a selectionof mathematical problems that are not only suitable for special college level (andoccasionally even at high school level) courses designed for these competitions, butcan significantly improve the level of mathematical sophistication of students whoare intrigued by and envision a career in mathematics in general.

I was asked by the Iranian Mathematical Society (IMS) to prepare a problembook on the basis of college level competitions in Iran. Since Iranian studentshave done quite well in international competitions and their success to some extentreflects the training that they received in special courses, I decided to make the bookavailable to a wider audience. The problems (and their solutions) that are presentedin this book are from national mathematical competitions in Iran at college levelfrom 1973 to 2007. I provided my own solutions to most of the problems and alsoutilized solutions from the files of the IMS, which contained approximately about40% of all the solutions.

I should point out that I have not edited the problems of the competitions andthey are direct translations from Persian into English. Unfortunately, there weresome typos and mistakes in the original version. The errors have been corrected andexcept for trivial typographical ones, the corrections are so indicated as footnotesthroughout Chapter One. Some comments on the problems also appear in footnotes.

I am indebted, and hence express my deepest gratitude, to the colleagues whocontributed in different ways to these competitions and to this book — either asmembers of the Scientific Committee for the competitions, or proposed problemsor assisted in finding elegant solutions. Unfortunately, some of those colleagueshave left us; may they rest in peace. I am thankful to my dear friend Dr. Hossein

ix

x PREFACE

Hajiabolhassan who constantly encouraged me and helped me with the prepara-tion of the book. I would like to thank Professor Alireza Jamali, Professor EbadMahmoodian, the outgoing President of the IMS, Dr. Rashid Zaare-Nahandi, andMessrs. M. Shokoohi and M. Abdi-Zadeh and the administrative assistants at themain office of the IMS for their assistance in gaining access to the existing solutionfiles of the IMS. I am grateful to Ms. Anahita Samie for drawing the figures forthis book...

Bamdad R. Yahaghi

PREFACE xi

Acknowledgement

A number of friends and colleagues took interest in this work and their contri-butions are implicit in some elegant solutions to the problems. While inadvertentlyI may have left out the names of some, the contributions ofSaeed Akbari (the second solution of Problem 7 of 1.9.3),Kasra Alishahi (the third solution of Problem 1.b of 1.24.2),Rajendra Bhatia (the first proof of the lemma presented in Solution 3 of 2.3.1),Hossein Hajiabolhassan (Problems: 2 of 1.21.2, 2 of 1.23.2, and 6 of 1.28.1),Hossein Hajiabolhassan and the late Mojtaba Mehrabadi (Problem 2 of 1.14.2),Ramin Mohammadalikhani (the second solution of Problem 2 of 1.13.1),Ali Mohammadian (Problems: 6 of 1.27.2 and 5 of 1.25.1),Omid Naghshineh Arjmand (Problem 3.b of 1.3.1),are duly acknowledged. Needless to say I am responsible for any shortcomings orerrors.

xii PREFACE

Iranian University Students Mathematics Competitions – a historicalintroduction

The history of the Iranian Mathematics Competitions for university students isan integral part of the contemporary history of mathematics in Iran. On March 31st,1972, the first general assembly of the Iranian Mathematical Society (IMS) was heldat the former National University (now Shahid Beheshti University). Upon a mo-tion proposed by Mehdi Behzad, then president of the society –which at that timewas called the Society’s secretary– the need for a mathematics competition aimed atuniversity students was approved. Herewith a committee, consisting of HoushangAttarchi, Mehdi Behzad, Fereidoun Ghahramani, the late Mohammad-Ali Gheyni,and Bahman Vahidi, was first appointed to prepare a guideline for the competitionsand to make preparations for holding theses competitions, which were intended foruniversity students. This committee developed a guideline and required that thecompetitions must be held annually and simultaneously with the Annual IranianMathematical Society Conference. The primary goals of the competitions were todiscover, encourage, and nurture mathematical talents throughout the country andto create a friendly and scientific rivalry among students and the Iranian universi-ties. In those years, the mathematics courses in most of the universities across thecountry were heavily influenced by the traditional and old curriculum as opposed tothe modern one. Only a handful of universities, such as the University of Tehran,Sharif (then Aryamehr) University of Technology, (former Pahlavi) University ofShiraz, and Mossaheb Institute of Mathematics offered courses such as Set Theory,Modern Algebra, Modern Analysis, and Combinatorics. The secondary goal of thecompetitions was to motivate the students to view mathematics in new ways tomaintain and sustain a more diverse perspective of mathematics, thereby puttingpressure on universities to change their educational systems.

The following year, on March 30th, 1973, simultaneous with the Fourth AnnualIMS Conference held at the University of Tehran, the First National MathematicsCompetition was held with the participation of 22 students from five universitiesacross the country. Half of the questions were in the areas of Mathematical Analysisand Algebra and the rest were in other areas categorized as General, i.e., innovativequestions, Probability and Statistics, Differential Equations, and Topology. Thesecond competition took place in March 1974 at the, former Pahlavi, University ofShiraz with the participation of 51 students from eleven universities. The students’participation grew steadily in the following years so that in the thirtieth competitionmore than 180 students from 40 universities took part.

Ever since 1973, the competitions were held annually except for 1979 with clo-sure of universities because of the revolution and 1981-1983 when universities wereclosed due to the cultural revolution. Students’ and universities’ reception of thecompetitions caused a great increase in the number of participating students. Onthe other hand, when the competitions were held on the same date and locationas the annual IMS conferences, the organizational matters of the event, in particu-lar grading the student’s papers, had become so overwhelming that it would takea number of months to announce the results. This would sometimes lead to dis-content and objections, which occasionally resulted in changes to the final scores.However, it was only for the tireless and diligent efforts of the questions commit-tees in those years that the competitions survived and were kept alive despite alldifficulties and shortcomings. In view of all this, the IMS was persuaded to hold

PREFACE xiii

the competitions in a date and location different from those of the Annual IMSconferences. Consequently, from 1996-2003 the competitions took place during athree day period. Since 2004 they have been taking place in a four day period atone of the Iranian universities during the Spring of every year.

In June 1999, the IMS’ executive committee and board of trustees gatheredat Tafresh University to reconsider the bylaws and regulations of the IMS. Atthat time, a committee was established to substantially change the regulationsconcerning the National Mathematics Competitions. From 2000, the competitionsare being held according to the new regulations. It must, however, be said that someminor changes and improvements were made at some stages due to the experiencesobtained from previous competitions.

According to the current guideline of the competitions, each university or highereducational institution, can send only one team consisting of at most five studentsand a leader, possibly along with a deputy leader, to the competitions. Eachuniversity is responsible for selecting its own team, presumably through holdingpreliminary tests among its students. The contests take place in two sessions, witheach session lasting three and a half hours. The students would answer twelvequestions of which four are in the area of analysis, four in the area of algebra,and another four in other areas such as discrete mathematics, probability, numbertheory, etc, categorized as innovative questions. Attempts are made to proposeproblems whose solutions require not only mathematical knowledge but innovation.

The questions of the competitions are now designed and selected by a questionscommittee consisting of a chairperson, a supervisor, and three other people as theheads of the following sections: analysis, algebra, and innovative questions. Theheads of the three sections can each choose a vice head to assist them. The questionscommittee is appointed by the Executive Committee of the IMS for a period oftwo up to three years. The committee is responsible for designing questions andholding the competitions. It is also responsible for appointing a grading team,whose members are mostly chosen from the winners of the previous competitions.In the course of a number of months, the committee selects 24 questions which aredivided into two sets of questions, each of which has 12 questions. The contests areheld in two sessions. For each sitting, 6 questions are selected out of the 12 questionsby a jury consisting of the leaders of the participating teams. The participants areranked based on their individual scores. The top 5 individual scorers win goldmedals, the next 10 top contestants are awarded silver medals, and up to the nexttop 15 receive bronze medals. It is worth mentioning that since April 2004, i.e.,from the 28th competition on, the problems together with their proposed solutionsare posted online, on the IMS website currently only in Persian. This is done for tworeasons. Firstly, the universities can simultaneously hold the competitions for theirinterested students. Secondly, this way the problems and their proposed solutionsare preserved on the Internet permanently.

Now after thirty one competitions, it seems that most of the goals of the com-petitions have been achieved. A glance at the list of the winners of competitionsreveals that many of them have gone on to become distinguished mathematiciansand are now working at prominent universities and institutions throughout theworld. Also, a glance at the questions of these thirty one competitions shows thatthe competitions and the Iranian mathematical community, on the whole, have

xiv PREFACE

evolved to a greater level of sophistication in terms of content, quantity and qual-ity.

In 1995, the executive committee of the IMS approved the preparation of abook containing the problems and solutions to the problems of the first twentycompetitions. It took the IMS several months to collect the problems of the firsttwenty competitions. About one third of the proposed solutions of all problemswere available in the IMS files. Rashid Zaare-Nahandi informed Bamdad Yahaghiabout the decision made by the IMS and asked whether Yahaghi was interestedin preparing such book. Yahaghi showed interest and readiness and it took hima year of hard work to prepare a preliminary copy of the manuscript. Under therecommendation of the IMS, the analysis and the algebra part of the manuscriptwas then refereed by the late Karim Seddighi and Mohammad Reza Darafsheh,respectively. This was simultaneous with Yahaghi’s departure for Canada to con-tinue his PhD studies at Dalhousie University. Unfortunately, the original plan bythe IMS to publish this book within a year could not be realized. Two years ago,shortly after the author’s return to Iran, his interest in the project on which hehad spent such considerable time and energy was rekindled. He began to revise,update, and rewrite the book in both Persian and English. It must be said thatthe IMS did not have the proposed solutions to the problems of Competitions 21,23, 24, and 27. It took Yahaghi almost two years to prepare both the English andPersian versions of the book. We must point out that this book could not havecome into existence without the efforts of the colleagues who have contributed tothese competitions. This book perhaps sheds light on the time and effort that havebeen put in organizing these competitions. It is hoped that this book succeeds inintroducing these competitions to an international audience.

Following this introduction, the names of the members of the questions com-mittees1 and the date, location, and the winners of the competitions, together withtheir educational and employment affiliations, are quoted.

Bamdad R. Yahaghi and Rashid Zaare-Nahandi

1Up until 1998, the executive committee of the IMS used to appoint only one person as the

chairperson of the competition. This person was charged with the task of inviting other colleagues

to submit questions and to mark the student’s papers. As outlined before, as of 1999, the questionscommittee is appointed by the Executive Committee of the IMS. And it is this committee that is

responsible for designing questions and holding the competitions.

PREFACE xv

The First Competition, University of Tehran, March 1973Questions Committee: Mehdi Behzad.

1. Elizabeth Ebrahimzadeh, Sharif (former Aryamehr) University of Technology.PhD (University of California, Berkeley, 1984), Professor (California State Uni-

versity, Sacramento).2. Mohammdad Reza Darafsheh, University of Tehran.PhD (University of Birmingham, UK, 1978), Professor (University of Tehran,

Iran).3. Hashem Madadi-Almousavi, Ferdowsi University of Mashhad.Deceased.4. Mirebrahim Hashemi Aghdam, Sharif (former Aryamehr) University of Technol-

ogy.5. Yousef Bahrampour, (former Pahlavi) University of Shiraz.PhD (University of Oregon, USA, 1983), Professor (University of Kerman,

Iran).

The Second Competition, Shiraz (former Pahlavi) University, March1974

Questions Committee: Mehdi Behzad.

1. Mohammad Ali Najafi, Sharif (former Aryamehr) University of Technology.2. Firouz Khosroyani, University of Tehran.3. Mehdi Zekavat, (former Pahlavi) University of Shiraz.4. Hamid Hamed Akbari-Tousi, University of Tehran.5. Shahram Arshad-Riyazi, former National University.

The Third Competition, (former Jondi Shapour) University of Ahwaz,March 1975

Questions Committee: Mohammad Ali Gheyni.

1. Pirouz Vakili, Sharif (former Aryamehr) University of Technology.2. Moslem Nikfar, University of Tehran.3. Ali Asghar Babadi Margha, University of Tehran.4. Saeed Ghahramani, Sharif (former Aryamehr) University of Technology.5. Homayoun Moeen, Sharif (former Aryamehr) University of Technology.

The Fourth Competition, University of Tabriz, March 1976Questions Committee: Vahab Davarpanah.

1. Pirouz Vakili, Sharif (former Aryamehr) University of Technology.2. Homayoun Moeen, Sharif (former Aryamehr) University of Technology.3. Nasser Hosseini, (former Pahlavi) University of Shiraz.4. Shahla Marvizi, Sharif (former Aryamehr) University of Technology.5. Ali Karimi, Tarbiat Moallem (Teacher Training) University.

The Fifth Competition, Sharif (former Aryamehr) University ofTechnology, March 1977

xvi PREFACE

Questions Committee: Mohammad Reza Nouri-Moghadam.

1. Hamid Kazemi, Sharif (former Aryamehr) University of Technology.2. Hossein Masoumi Fakhar, Sharif (former Aryamehr) University of Technology.3. Masoud Khalkhali, Sharif (former Aryamehr) University of Technology.4. Ebrahim Saatchi, (former Azarabadegan) University of Tabriz.5. Safa Nourbakhash, former National University.

The Sixth Competition, The University of Isfahan, March 1978Questions Committee: Magerdich Toumanian.

1. Nasser Boroujerdian, University of Tehran.2. Hamid Kazemi, Sharif (former Aryamehr) University of Technology.3. Ali Asghar Alikhani Kouhpaee, University of Isfahan.4. Mehdi Salehi Nejad, Ferdowsi University of Mashhad.5. Ali Rejali, University of Isfahan.

The Seventh Competition, Ferdowsi University of Mashhad, March1980

Questions Committee: Akbar Hassani.

1. Fereydoun Rezakhanlou, University of Tehran.2. Mehdi Alavi Shoushtari, University of Ahwaz.3. Seyed Esmail Seyedabadi, Sharif University of Technology.4. Rajabali Kamyabigol, Ferdowsi University of Mashahd.5. Ali Asghar Jodeyri Akbarfam, University of Tabriz.

The Eighth Competition, Shiraz University, March 1984Questions Committee: Asadollah Niknam.

1. Mohammad Hassan Jahanbakht, University of Isfahan.2. Mojtaba Moniri, University of Tehran.3. Mohammad Taghi Jahandideh, Shiraz University.4. Samad Ahmadi, Tarbiat Moallem (Teacher Training) University.5. Kamal Atigh, University of Tabriz.

The Ninth Competition, Tehran Teacher Training (Tarbiat Moallem)University, September 1985

Questions Committee: Rahim Zaare-Nahandi.

1. Nasser Boroujerdian, University of Tehran.2. Mojtaba Moniri, University of Tehran.3. Majid Ashrafi, Shiraz University.4. Jamal Rouin, Tarbiat Moallem (Teacher Training) University.5. Ali Parsian, University of Tehran.

The Tenth Competition, University of Sistan and Baluchestan, March1986

PREFACE xvii

Questions Committee: Karim Seddighi.

1. Reza Jahani Nejad, University of Kashan.2. Amir Akbary Majdabadno, University of Tehran.3. Shaahin Ajoodani Namini, University of Tehran.4. Hamid Reza Farhadi, Tarbiat Moallem (Teacher Training) University.5. Masoud Amini, Ferdowsi University of Mashhad.

The Eleventh Competition, The University of Birjand, March 1987

Questions Committee: Karim Seddighi.

1. Vahid Tarokh, Sharif University of Technology.2. Masoud Amini, Ferdowsi University of Mashhad.3. Shaahin Ajoodani Namini, University of Tehran.4. Reza Karami, Isfahan University of Technology.

The Twelfth Competition, Guilan University , March 1988

Questions Committee: Mohammad Ali Shahabi.

1. Shaahin Ajoodani Namini, University of Tehran.2. Shaahin Amiri Sharifi, Sharif University of Technology.3. Gholam Hossein Eslamzadeh, Shiraz University.4. Bamdad Yahaghi, Sharif University of Technology.5. Ali Iranmanesh, Shiraz University.

The Thirteenth Competition, University of Tehran, March 1989

Questions Committee: Mohammad Ali Shahabi.

1. Kambiz Mahmoodian, University of Tehran.2. Saeed Zakeri, University of Tehran.3. Mohammad Sal Moslehian, Ferdowsi University of Mashhad.4. Bamdad Yahaghi, Sharif University of Technology.5. Behrooz Mashayekhifard, Ferdowsi University of Mashhad.

The Fourteenth Competition, The University of Isfahan, March 1990

Questions Committee: Heydar Zahed Zahedani.

1. Shaahin Amiri Sharifi, Sharif University of Technology.2. Hessam Hamidi Tehrani, Sharif University of Technology.3. Saeed Zakeri, University of Tehran.4. Hamid Mousavi, Tarbiat Moallem (Teacher Training) University.5. Shahab Shahabi, University of Tehran.

The Fifteenth Competition, Ferdowsi University of Mashhad, March1991

xviii PREFACE

Questions Committee: Mohammad Reza Darafsheh.

1. Hessam Hamidi Tehrani, Sharif University of Technology.2. Ali Rajai, Sharif University of Technology.3. Shahriar Mokhtari Sharghi, Sharif University of Technology.4. Pedram Safari, Sharif University of Technology.5. Ataollah Togha, University of Kerman.

The Sixteenth Competition, Razi (Rhazes or Rasis) University ofKermanshah, March 1992

Questions Committee: Mohammad Reza Darafsheh.

1. Shahriar Mokhtari Sharghi, Sharif University of Technology.2. Ataollah Togha, University of Kerman.3. Pedram Safari, Sharif University of Technology.4. Ali Sabetian, Shiraz University.5. Ali Rajai, Sharif University of Technology.6. Mehdi Najafikhah, Iran University of Science and Technology.

The Seventeenth Competition, Shahid Beheshti (former National)University, March 1993

Questions Committee: Mohammad Reza Darafsheh.

1. Hossein Hajiabolhassan, Sharif University of Technology.2. Payman L. Kassaei, Sharif University of Technology.3. Behrang Noohi, Sharif University of Technology.4. Arash Rastegar, Sharif University of Technology.5. Ali Dadban, University of Tehran.

The Eighteenth Competition, Sharif University of Technology, March1994

Questions Committee: Jafar Zafarani.

1. Ramin Takloo-Bighash, Sharif University of Technology.2. Kasra Rafi, Sharif University of Technology.3. Behrang Noohi, Sharif University of Technology.4. Payman L. Kassaei, Sharif University of Technology.5. Aminollah Zargarian, University of Tehran.

The Nineteenth Competition, University of Kerman, March 1995Questions Committee: Ahmad Haghani.

1. Amir Jafari, Sharif University of Technology.2. Ali Lashgari Faghani, Isfahan University of Technology.3. Fatemeh Ayatollahzadeh Shirazi, University of Tehran.4. Mohammad Reza Raoofi, Isfahan University of Technology.5. Reza Naserasr, Sharif University of Technology.

The Twentieth Competition, Sharif University of Technology, February1996

PREFACE xix

Questions Committee: Omid Ali Karamzadeh and Yahya Tabesh.

1. Keivan Mallahi-Karai, Sharif University of Technology.2. Hossein Movasati, Sharif University of Technology.3. Omid Naghshineh Arjmand, Sharif University of Technology.4. Ali Reza Amini Harandi, Isfahan University of Technology.5. Ebrahim Samei, Shahid Beheshti (former National) University.

The Twenty First Competition, University of Tehran, March 1997Questions Committee: Omid Ali Karamzadeh.

1. Kia Dalili, Sharif University of Technology.2. Maryam Mirzakhani, Sharif University of Technology.3. Ebrahim Samei, Shahid Beheshti (former National) University.4. Hadi Jorati, Sharif University of Technology.5. Hossein Abedi Andani, Isfahan University of Technology.

The Twenty Second Competition, University of Ahwaz, March 1998Questions Committee: Omid Ali Karamzadeh.

1. Maryam Mirzakhhani, Sharif University of Technology.2. Eaman Eftekhari, Sharif University of Technology.3. Payam Nasser Tayoub, University of Tehran.4. Mohammad Ahmadvand, Bu-Ali Sina (Avecina) University of Hamedan.5. Abolghassem Karimi, Shahid Beheshti (former National) University.

The Twenty Third Competition, Sharif University of Technology,March 1999

Questions Committee: Omid Ali Karamzadeh and Yahya Tabesh.

1. Hadi Salmasian, Sharif University of Technology.2. Mohsen Bahramgiri, Sharif University of Technology.3. Mohammad Javaheri, Sharif University of Technology.4. Bijan Ahmadi, Shahid Beheshti (former National) University.5. Kamal Azizi, University of Tabriz.

The Twenty Fourth Competition, Khajeh Nasir Toosi University ofTechnology, May 2000

Questions Committee: Saeed Azam, Rouzbeh Tusserkani, Ali Reza Jamali (Chair),and Hossein Mohebi.

1. Omid Amini, Sharif University of Technology.2. Kasra Alishahi, Sharif University of Technology.3. Maziar Mirrahimi, Sharif University of Technology.4. Seyed Reza Moghaddasi, Sharif University of Technology.5. Masoud Aryapour, Sharif University of Technology.

The Twenty Fifth Competition, Imam Khomeini InternationalUniversity of Qazvin, May 2001

xx PREFACE

Questions Committee: Saeed Azam, Rouzbeh Tusserkani, Ali Reza Jamali (Chair),and Hossein Mohebi.

1. Amir Mohammadi, Sharif University of Technology.2. Salman Abolfath Beigi, Sharif University of Technology.3. Hamid Reza Darbidi, Sharif University of Technology.4. Babak Amini, Shiraz University.5. Afshin Amini, Shiraz University.

The Twenty Sixth Competition, Institute for Advanced Studies inBasic Sciences (IASBS) of Zanjan, May 2002

Questions Committee: Saeed Azam, Rouzbeh Tusserkani, Ali Reza Jamali (Chair),and Hossein Mohebi.

1. Salman Abolfath Beigi, Sharif University of Technology.2. Ali Shourideh, Sharif University of Technology.3. Javad Ebrahimi Boroujeni, Sharif University of Technology.4. Amin Aminzadeh Gohari, Sharif University of Technology.5. Majid Hadian, Sharif University of Technology.

The Twenty Seventh Competition, Bu-Ali Sina (Avecina) University ofHamedan, May 2003

Questions Committee: Gholam-Hossein Eslamzadeh, Hossein Hajiabolhassan,Mohammad Reza Pournaki, and Mohammad Taghi Dibaei (Chair).

1. Mohsen Sharifi Tabar, Sharif University of Technology.2. Ali Shourideh, Sharif University of Technology.3. Mohammad Farajzadeh Tehrani, Sharif University of Technology.4. Payam Valadkhan, Sharif University of Technology.5. Hamid Hassanzadeh, Tarbiat Moallem (Teacher Training) University.6. Rahbar Rasooli, University of Tehran.7. Maryam Khosravi, Tarbiat Moallem (Teacher Training) University.8. Amir Moradifam, Iran University of Science and Technology.

The Twenty Eighth Competition, Sharif University of Technology, May2004

Questions Committee: Gholam Hossein Eslamzadeh, Hossein Hajiabolhassan,Majid Mirza-Vaziri, Abdolrasool Pourabbas, Mohammad Reza Pournaki, and RashidZaare-Nahandi (Chair).

1. Iman Setayesh, Sharif University of Technology.2. Omid Haji Mirsadeghi, Sharif University of Technology.3. Armin Morabi, Sharif University of Technology.4. Sajad Lakzian, Amir Kabir University of Technology.5. Mohammad Kazem Anvari, Ferdowsi University of Mashhad.

The Twenty Ninth Competition, University of Mazandaran inBabolsar, May 2005

PREFACE xxi

Questions Committee: Hossein Hajiabolhassan, Majid Mirza-Vaziri, MojtabaMoniri, Mohammad Reza Pournaki, Mehdi Radjabalipour (Chair), Bamdad R. Ya-haghi, Rashid Zaare-Nahandi (Supervisor), and Manouchehr Zaker.

1. Iman Setayesh, Sharif University of Technology.2. Mohammad Farajzadeh Tehrani, Sharif University of Technology.3. Mohammad Abbas Rezai, Sharif University of Technology.4. Mohammad Hossein Mousavi, Sharif University of Technology.5. Fatemeh Doroodian, Amir Kabir University of Technology.6. Mahmoud Hassanzadeh, University of Tehran.

The Thirtieth Competition, Tafresh University, May 2006Questions Committee: Hossein Hajiabolhassan, Majid Mirza-Vaziri, Mojtaba

Moniri, Mohammad Reza Pournaki, Mehdi Radjabalipour (Chair), Bamdad R. Ya-haghi, Rashid Zaare-Nahandi (Supervisor), and Manouchehr Zaker.

1. Ali Akbar Daemi, Sharif University of Technology.2. Mohammad Gharakhani, Sharif University of Technology.3. Omid Haji Mirsadeghi, Sharif University of Technology.4. Mostafa Einollahzadeh Samadi, Sharif University of Technology.5. Behzad Mehrdad, Sharif University of Technology.

The Thirty First Competition, Ferdowsi University of Mashhad, May2007

Questions Committee: Mojtaba Gheerati, Hassan Shirdareh Haghighi, OmidNaghshineh Arjmand, and Fariborz Azarpanah (Chair).

1. Ali Akbar Daemi, Sharif University of Technology.2. Nasser Talebizadeh, Sharif University of Technology.3. Erfan Salavati, Sharif University of Technology.4. Nima Ahmadipour Anari, Sharif University of Technology.5. Mohammad Gharakhani, Sharif University of Technology.

CHAPTER 1

Problems

1.1. First Competition, University of Tehran, March 1973

1.1.1. Analysis. 1. A function φ that is a pointwise limit of continuous realfunctions is called a Baire function.

(a) If f is a real function of the real variable x whose derivative exists every-where, prove that the function f ′ is a Baire function.

(b) By giving an example show that every Baire function is not necessarily thederivative of a function.

2. Let’s call the set of all n × n real matrices M. The set M is a metric space ifwe view the elements of it as n2-dimensional vectors equipped with the Euclideannorm . That is, for every matrix A = (aij), we define the norm of A as follows

||A|| =( ∑

1≤i,j≤n

a2ij

) 12 .

Prove that the set of invertible matrices is(a) open.(b) disconnected.

3. Suppose that the function f is defined on the half-line (0,+∞) by

f(x) =

0 x /∈ Q,1p+q x = p

q , p, q ∈ N, gcd(p, q) = 1.

(a) Show that the limit of this function exists at any point of (0,+∞).(b) At what points of (0,+∞) is the function f continuous? Prove your claim.

1.1.2. Algebra. 1. Let R be commutative ring with identity element.(a) Prove that an ideal P is prime if and only if

∀a ∈ R,∀b ∈ R : ab ∈ P =⇒ a ∈ P or b ∈ P.(b) Prove that every maximal ideal is a prime ideal.• Hint. In a ring R, the product of two ideals T1 and T2 is defined as ∑

finite

xy : x ∈ T1, y ∈ T2

.

An ideal M 6= R is called maximal if for any ideal T , we have

M ⊆ T =⇒ T = R.

An ideal P is prime if and only if for any two ideals T1 and T2, from T1T2 ⊆ P , itfollows that T1 ⊆ P or T2 ⊆ P .

1

2 1. PROBLEMS

2. Suppose that every element of a group G satisfies the equality x2 = e. Provethat the group G is abelian (i.e., commutative).

3. Let E and F be two isomorphic sets (that is, there is a one-to-one correspon-dence between them), and that f is a function from P(E) into P(F ) satisfying thefollowing three conditions, where P stands for the power set operation.

(a) f(∅) = ∅.(b) ∀X ∈ P(E),∀Y ∈ P(E) : f(X ∪ Y ) = f(X) ∪ f(Y ).(c) ∀X ∈ P(E) : cardX ≤ cardf(X).(cardA ≤ cardB means that there exists a one-to-one mapping from A into B.)3.1. Prove that E and f(E) are isomorphic. Moreover, if X and Y are two

finite subsets of E such that X is isomorphic to f(X) and Y is isomorphic to f(Y ),show that X ∪ Y and f(X ∪ Y ), and X ∩ Y and f(X ∩ Y ) are isomorphic sets,respectively, and that for any two such sets we have f(X ∩ Y ) = f(X) ∩ f(Y ).

3.2. Prove that if E is finite and normal, then there exists a normal subsetX0 6= ∅ of E such that for any other normal subset X of E, we have

X ∩X0 = ∅ or X0 ⊆ X.

(A subset A of E which is isomorphic to f(A) is called a normal subset of E.)

1.1.3. General. 1. A triangle with sides a, b, c such that a < b < c is given.Set

S = maxab,b

c,c

a

min

ab,b

c,c

a

.

Show that S ≥ 1.

2. Show that the decimal fraction 0.123456789101112131415 . . ., which is formedby putting consecutive positive integers next to one another, is not periodic.

3. Find two distinct real or complex numbers in such a way that each of which isthe cube of the other.

4. Let x, y be two positive real numbers with x+ y = 1. Prove that

xx + yy ≥√

2.

1.1.4. Differential Equations. Find the general solution of the followingdifferential equation

d2x

dt2+ a2x = f(t),

where a is a constant and f is a continuous real function.

1.2. SECOND COMPETITION 3

1.1.5. Probability and Statistics. Let X1, . . . , X10 be ten independent ran-

dom variables with the probability density functions fi(Xi) =e−µiµXi

i

Xi, where

µi = i (i = 1, . . . , 10) and that the codomain of the variable Xi is 0, 1, 2, . . ..Set

X =110

10∑i=1

Xi.

What is the probability that X equals one?

1.1.6. Topology. Let Q be the set of all rational numbers. For each q ∈ Q,the set ]q,+∞[ is denoted by Aq. If T is the set consisting of R and ∅ and all Aq’s,show that T is not a topology on R.

• Hint. Consider the sets of the form Aq with q >√

2 and q ∈ Q.

1.2. Second Competition, Shiraz (former Pahlavi) University, March1974

1.2.1. Analysis. 1. Let f be a real continuous function with nonnegativevalues on the closed interval [0, 1]. Set

un =(∫ 1

0

(f(x))n dx) 1

n

, M = sup0≤x≤1

f(x).

(a) Assuming that 0 < ε < 1, prove that there exist numbers α, β subject to0 ≤ α < β ≤ 1 such that M(1 − ε) ≤ f(x) ≤ M for all x belonging to the openinterval ]α, β[.

(b) Prove that limn→+∞ un = M .

2. Evaluate

D = inff∈F

(sup

0≤t≤1

∣∣1− f(t)∣∣+ ∫ 1

0

∣∣1− f(t)∣∣dt) ,

where F is the vector space of all continuous functions from [0, 1] into R whoseelements take the value zero at zero. Describe the geometrical interpretation of thenumber D.

3. Assuming that C is the Cantor set, show that

C + C = [0, 2].

• Hint. 1. If A and B are two subsets of real numbers, by definition,

A+B :=a+ b : a ∈ A, b ∈ B

.

2. The Cantor set consists of all real numbers between zero and one whoseternary expansions do not have any one.

4 1. PROBLEMS

1.2.2. Algebra. 1. Let F be a field with n elements. Prove that for everyinteger m with m ≥ 1, we have∑

x∈Fxm =

n− 1 n− 1 | m,

0 n− 1 - m.

• Hint. F ∗ = F \ 0 is a cyclic group.

2. Let A be a ring.• Definition 1. An element z ∈ A is called right quasi-regular if there is a z′ ∈ A

such that z + z′ − zz′ = 0.• Definition 2. An element z ∈ A is called left quasi-regular if there is a z′ ∈ A

such that z + z′ − z′z = 0.• Definition 3. An element z ∈ A is called quasi-regular if there is a z′ ∈ A

such that z + z′ − zz′ = z + z′ − z′z = 0.(a) An element z ∈ A is quasi regular if an only if it is right and left quasi-

regular.(b) If A has an identity element, then the identity element is not right quasi-

regular. And if x is a right quasi-regular element, then 1− x is invertible1 in A.

3. Let E be a finite-dimensional vector space over a field K and φ a homomorphismfrom the ring K[x] into the ring L(E) with the hypothesis that φ(1) is the identityelement of L(E).

(a) If P and D belong to K[x] and D is a divisor of P , prove that

ker(φ(P )

)⊃ ker

(φ(D)

).

(b) If R is another element of K[x] and D is the greatest common divisor of Pand R, prove that

ker(φ(D)

)= ker

(φ(P )

)∩ ker

(φ(R)

).

• Hint. By K[x], we mean the ring of all polynomials in one indeterminate xwith coefficients coming from K and L(E) is the ring of all endomorphisms of E.For any u belonging to L(E), keru means the kernel of the linear transformationu.

1.2.3. General. 1. Find the minimum and the maximum number of “Wednes-days occurring on the 13th of the month” which might happen in a solar year,according to the Persian calendar that is.

Remark. Recall that there are 12 months in the Persian calendar, each of thefirst six months has 31 days, each of the sixth-eleventh months has 30 days, andthe last month has 29 days except that in a leap year it has 30 days.

2. Prove that (x+ y + z + t

4

)n≤ xn + yn + zn + tn

4,

where x, y, z, t are positive real numbers and n is a natural number.

3. A particle, moving along a straight line, goes one unit of distance in one unit oftime in such a way that the speed of it at the initial point as well as the final point

1“invertible” must read “right invertible”!

1.2. SECOND COMPETITION 5

is zero. Prove that [the absolute value of] the particle acceleration at some pointon its path is greater than or equal to four.

4. We know that in Iran the plate number of every car is a five-digit numbernone of whose digits is zero. In big cities, like Tehran, in addition to a number a[Persian] letter is also used to characterize the plate, e.g., “alef”, “be”, “pe”, etc.By explaining your reasoning, find the total number of the car plates that containthe letter “dal”.

Remark. The Persian alphabet has 32 letters.

1.2.4. Probability. Let X and Y be two random variables subject to thefollowing conditions.

var(X) = var(Y ) = σ2, cov(X,Y ) = λ,

where σ and λ are given real numbers. Define the random variables U and V byU = X + Y and V = X − Y .

(a) Find the covariance of U and V .(b) Are U and V independent? If so, prove it; if not, by an example justify

your answer.

1.2.5. Topology. Consider a nonempty topological space (E, T ) and two non-empty subsets A and K of E with K ⊃ A. Suppose that B and C are two closedsubsets of the topological space (K,TK) such that

A ⊂ B ∪ C, A ∩ C 6= ∅, A ∩B 6= ∅.

Prove that if A is connected, then A ∩B ∩ C 6= ∅.• Hint. By TK we mean the induced topology which is induced by T on K.

The problem can be proved by contradiction.

1.2.6. Differential Equations. Find the general solution of the followingdifferential equation subject to the two cases |x| > 1 and |x| < 1.

3(x2 − 1)y2y′ + xy3 = x3 + x2 − x− 1.

6 1. PROBLEMS

1.3. Third Competition, (former Jondi Shapour) University of Ahwaz,March 1975

1.3.1. Analysis. 1. Let f be a real function whose domain is [a, b]. If f ′′

exists on [a, b] and is positive, then for all ξ ∈ [a, b],2 there exists a point x0 ∈ [a, b]such that

f ′(ξ) =f(a)− f(x0)

a− x0or f ′(ξ) =

f(b)− f(x0)b− x0

.

2. A sequence (an)+∞n=1 of numbers in the closed interval [0, 1] is given such that theelements of the sequence are all distinct. The goal is to find a continuous functionf from [0, 1] into [0, 1] such that f(an) = an+1. Prove that

(a) in general, the problem has no solution.(b) if the sequence is increasing or decreasing, the problem has a solution.

3. In each part of this problem, you can use the preceding parts.(a) If the complex number z is a root of the following equation with complex

coefficientsxp + c1x

p−1 + · · ·+ cp−1x+ cp = 0,then

∃ k ∈ 1, . . . , p : |z| ≤ 2 k√|ck|.

(b) The following two equations with complex coefficients are given.

xp + c1xp−1 + · · ·+ cp−1x+ cp = 0,

xp + c′1xp−1 + · · ·+ c′p−1x+ c′p = 0.

If the positive numbers K and δ satisfy the following

∀i = 1, . . . , p :(|ci| < Ki, |ci − c′i| < Kiδ

),

then for any root zj of the first equation there exists a root z′k of the second equationsuch that ∣∣zj − z′k

∣∣ < 2K p√δ.

(c) Suppose that the real numbers K and α are such that K > 0 and 0 < α ≤1. Also suppose that the complex valued functions b1, . . . , bp of a real variable tbelonging to the compact interval [t1, t2] are such that

|bi(t)| < Ki, |bi(t)− bi(t′)| < Ki|t− t′|α,

for all t, t′ ∈ [t1, t2] and each i = 1, . . . , p. Prove that if the complex valued functionf is continuous on the closed interval [t1, t2] and that it satisfies the equation

fp + b1fp−1 + · · ·+ bp−1f + bp = 0,

then ∣∣f(t2)− f(t1)∣∣ < 4pK p

√(t2 − t1)α.

• Hint. If F : I → U is a function from an interval into an open subset of thecomplex plane, the image of F lies in one of the connected components of U .

2 “ξ ∈ [a, b]” must read “ξ ∈ (a, b)”!

1.3. THIRD COMPETITION 7

1.3.2. Algebra. 1. Let M be a module on a ring K and M1 and M2 twosubmodules of M . Set

M3 = M1 +M2, M4 = M1 ∩M2.

(a) Show that the quotient modulesM3

M1and

M2

M4are isomorphic.

(b) If K is a commutative field and M1 and M2 are finite-dimensional, showthat

dimM1 + dimM2 = dimM3 + dimM4.

2. Prove that a necessary and sufficient condition for a group to be commutativeis that the following function is a homomorphism of groups.

f : G→ G, f(x) = x2.

3. Suppose that for any element y of a semigroup S with a right identity element(that is, xe = x for all x ∈ S), there exists a y with the property that yy = e.

(a) By giving an example, show that S might not be a group.(b) If the right identity element is unique, prove that S is a group.

1.3.3. General. 1. The real function F (x, y) where x and y are two realvariables is considered. Suppose that this function has the following two properties.3

∀x, y ∈ R : F (x, y) = 0 ⇐⇒ x = y,∀x, y, z ∈ R : F (y, x) ≤ F (x, z) + F (z, y).

(a) Prove that F (x, y) ≥ 0 for all x, y ∈ R.(b) Prove that F (x, y) = F (y, x) for all x, y ∈ R.

2. Without using mathematical induction, prove thatn∑

m=0

(n!

m!(n−m)!

)2

=(2n)!(n!)2

,

for all n ∈ N.

3. Two flat mirrors ∆ and ∆′ which are perpendicular to a plane P and two pointsM,N ∈ P are given. From the point M , shine a beam of light on the mirror ∆, inthe plane P , in such a way that the reflected light beam after hitting the mirror ∆′

passes through the point N .

3 “∀x, y ∈ R : F (x, y) = 0 =⇒ x = y” is redundant!

8 1. PROBLEMS

NM

Figure 1

1.4. Fourth Competition, University of Tabriz, March 1976

1.4.1. Analysis. 1. In this problem R denotes the real line and R2 the Euclid-ean plane. Prove that if f is a continuous and one-to-one function from R into R2,its inverse is not necessarily continuous.

2. Let (αn)+∞n=1 be a sequence of nonnegative real numbers such that

limn→+∞

αn = 0.

Prove that there are infinitely many number of indices n such that αm ≤ αn for allm greater than or equal to n.

1.4.2. Algebra. 1. Let A be a commutative ring and B an ideal of A. Set

R(B) =x ∈ A : ∃r ∈ N xr ∈ B

.

We call R(B) the radical of B. We call the ideal B of a commutative ring A aprime ideal if

∀a, b ∈ A, ab ∈ B =⇒ a ∈ B or b ∈ B.(a) Prove that R(B) is an ideal of A.(b) Prove that if the ideal B is prime, then so is the ideal R(B).(c) If we set A = Z, and if a is an element of Z and4

B =⟨a⟩

= aZ = aA,

prove that there are prime numbers p1, . . . , pr such that

R(B) =⟨p1p2 · · · pr

⟩.

2. Prove that in a vector space E, whose dimension is at least 2, for any two vectorsx and y which are not linearly dependent, one can find an automorphism u : E → Esuch that u(x) = x and u(y) = x+y. From this, conclude that if an endomorphismf : E → E commutes with any automorphism v (i.e., f v = v f), then for everyvector x ∈ E, the two vectors x and f(x) are linearly dependent whence f is justthe scalar product of the identity automorphism by a fixed scalar (i.e., f = λidE).

4 Obviously, we must have a 6= 0,±1.

1.5. FIFTH COMPETITION 9

3. Let R be an integral domain with identity. Suppose that every descending chainof the ideals of R terminates. (that is, for every chain of ideals of R as follows

I1 ⊇ I2 ⊇ · · · ⊇ Ik ⊇ · · · ,

there exists a positive integer n such that Im = In for all m ≥ n.) Prove that R isa field.

• Hint. Assume that x is a nonzero element in R, investigate the ideals xiR(i = 1, 2, . . .).

1.4.3. General. 1. Evaluate the following integral.

I =∫ π

2

0

sinn xcosn x+ sinn x

dx, n ∈ N.

2. We have three boxes A , B, and C, of which two are empty and one has a prizein it. We choose one of these boxes at random (say, for example B); it is plain thatthe probability that the chosen box contains the prize is 1

3 because of the threeboxes only one contains the prize. Now, suppose that of the two not-chosen boxesA and C, we open the one that does not contain the prize (say, for example A).Therefore, the prize is in one of the boxes B and C.

(a) What is the probability that the chosen box contains the prize?(b) What is the probability that the other box, i.e., C, contains the prize?5

1.5. Fifth Competition, Sharif (former Aryamehr) University ofTechnology, March 1977

1.5.1. Analysis. 1. If two functions are integrable on the interval (0, 1) inthe Riemann sense, is it true that the composition of the two functions is integrableon the interval (0, 1) in the Riemann sense? Explain your reasoning.

2. Let R be the set of all real numbers and f a continuous function from R into Rthat does not assume any value more than twice. Prove that f assumes at least avalue exactly once.

3. Let A be a subset of the real numbers R. A point p of A is called a congestionpoint of A if every neighborhood of p of the form (p−ε, p+ε) contains uncountablymany points of A. Prove that all but countably many points of A are congestionpoints of A (A is uncountable).

5 This problem is also known as the Monty Hall problem.

10 1. PROBLEMS

1.5.2. Algebra. 1. Let V be a finite-dimensional vector space over C (thecomplex numbers). If A : V → V is a linear transformation such that A 6= I,A2 6= I, A3 = I, find the eigenvalues of A. Extend this to the case where Ak = I.Do the eigenvalues of A form a group under multiplication? Find a necessary andsufficient condition for this set to form a group.

2. Let G be group and a and b two elements of it satisfying the following relations.

a 6= 1, b 6= 1, aba−1 = b2, a7 = 1,

where 1 is the identity element of the group. Find the orders of a and b.

3. Let R be a ring with identity element such that every x ∈ R satisfies thefollowing relation: x3 + 2x2 + x = 0.

(a) Prove that 2x = 0 for all x ∈ R.(b) Prove that R is a commutative ring.

1.5.3. General. 1. Let m,n ∈ N. Prove that the number(mn)!m!(n!)m

is an

integer.

2. If S = z1, . . . , zk is a subset of the complex numbers, define the set C(S) by

C(S) :=z = α1z1 + · · ·+ αkzk|αi ≥ 0, α1 + · · ·+ αk = 1

.

If f is a polynomial of degree greater than or equal to two and A and A′ denotethe sets of the roots of the equations f(z) = 0 and f ′(z) = 0, respectively, wheref ′ denotes the derivative of f , prove that

C(A) ⊃ C(A′).

3. Consider the trajectory of a billiard ball B which moves on an ellipse-shapedbilliard table ξ. Suppose that the angle of incidence is equal to the angle of reflec-tion. Assuming that the initial ball trajectory does not intersect the line segmentjoining the the two foci, show that there exists another ellipse η which is confocalwith the ellipse ξ such that the ball trajectory is always tangent to η.

• Hint. Recall that if PL and PL′ are two tangents from a point P to anellipse, then the angles FPL and F ′PL′ are equal. (F and F ′ are the foci of theellipse.)

1.6. Sixth Competition, The University of Isfahan, March 1978

1.6.1. Analysis. 1. In the xy-plane a point is called rational if both of itscoordinates are rational. Prove that if the center of a given circle in the plane isnot rational, then there are at most two rational points on the circle.

2. Suppose that in a metric space M a sequence (fn)+∞n=1 of continuous functionsis uniformly convergent to a function f . Prove that for every sequence (xn)+∞n=1

converging to a point x ∈M , we have

limn→+∞

fn(xn) = f(x).

1.6. SIXTH COMPETITION 11

3. If f is a continuous function on R and moreover∫ +∞0

∣∣f(x)∣∣dx < +∞, prove

that

limn→+∞

∫ +∞

0

∣∣f(x+1n

)− f(x)∣∣dx = 0.

1.6.2. Algebra. 1. Let G be a group with |G| = pna, where p is a prime andp and a are relatively prime. If G has subgroups A and B satisfying the followingconditions

|A| = pn, |B| = pm, 0 < m ≤ n,B * A,

prove that AB cannot be a subgroup of G.

2. Let A1, . . . , An be mutually commuting m ×m matrices such that A2i = 0 for

all 1 ≤ i ≤ n. If m < 2n, prove that A1A2 · · ·An = 0.

3. If R is a ring with identity (1 6= 0) and e and f are two commuting elements ofR such that e2 = e and f2 = f , prove that

(e− f)n = 0 =⇒ e = f,

for all positive integers n.

1.6.3. General. 1. Let n1, n2, . . . , nk be k integers of which m1,m2, . . . ,mk

is a permutation. Prove that

|n1 −m1|+ |n2 −m2|+ · · ·+ |nk −mk|

is an even integer.

2. An n × n matrix whose elements are nonnegative integers is given. If the sumof the elements on any row and any column corresponding to any nonzero elementof the matrix is at least n, prove that the sum of all elements of this matrix is noless than n2

2 .

3. Suppose 1000000 points inside a circle are given. Can one find a straight linenot passing through any of the points that divides the circle into two sections eachof which containing 500000 of the points?

12 1. PROBLEMS

1.7. Seventh Competition, Ferdowsi University of Mashhad, March1980

1.7.1. Analysis. 1. Is there a closed set S $ R2 such that for each x ∈ R2 \Sthere are exactly two points in S as the closest point of S to x?

2. For each positive rational number r (r ∈ Q+), suppose Ir is an open intervalsuch that r < s =⇒ Ir ⊂ Is. The function f on the set A =

⋃r∈Q+ Ir is defined

by

f(x) = infr : x ∈ Ir

.

Show that f is continuous.

3. Suppose that the sequence (xn)+∞n=1 is defined by

xn =n+ 12n+1

n∑k=1

2k

k, n = 1, 2, . . . .

Prove that limn→+∞ xn exists and find its value.

1.7.2. Algebra. 1. (a) If in a ring R the element e is a left identity element(i.e., ex = x,∀x ∈ R) that is unique, show that e is the identity element of this ring(i.e., ex = xe = x,∀x ∈ R).

(b) If zero is the only nilpotent element of the ring R, show that for everyidempotent element a of R we have

ax = xa, ∀x ∈ R.

• Note. The element b of a ring R is called nilpotent if there exists a positiveinteger n such that bn = 0; b is called idempotent if b2 = b.

2. Let G be a group and G1 a normal subgroup of G such that the groups GG1

and G1 are commutative. Prove that for any arbitrary subgroup H of G, there isa commutative subgroup H1 of H such that H1 is a normal subgroup of H and H

H1

is commutative.

3. An n-dimensional vector space V and a linear transformation θ : V → V aregiven. Consider the powers of θ, i.e., θ0 = 1, θ, θ2, . . .. Prove that there is a nonzerointeger s such that

V = im(θs)⊕ ker(θs).

1.8. EIGHTH COMPETITION 13

1.8. Eighth Competition, Shiraz University, March 1984

1.8.1. Analysis. 1. Let F and G be two closed subsets of Rn, where Rdenotes the real numbers endowed with its ordinary topology. Suppose that f :F → R and g : G→ R are two continuous functions which coincide on F ∩G.

(a) Prove that there is a continuous function h : F ∪G→ R such that h is anextension of both of the functions f and g.

(b) By giving an example, show that if F and G are not closed, the assertionin (a) is not correct.

2. Suppose that (an)+∞n=1 is a decreasing sequence of real numbers and that∑+∞n=1 an

is convergent. Prove that∑+∞n=1 n(an − an+1) is convergent.6

3. Let the function g be continuously differentiable on [0, 1]. Prove that

limn→+∞

∫ 1

0

xndg(x) = 0.

1.8.2. Algebra. 1. In the multiplicative group of nonsingular 2× 2 matriceswith real entries, let

a =(

2 00 1

), b =

(1 10 1

).

If H is the subgroup generated by b, show that aHa−1 is a proper subgroup of H.

2. Let R be an infinite integral domain (i.e., a commutative ring with identitywithout divisors of zero). If R has finitely many units, prove that R has infinitelymany maximal ideals. (A unit is an element which has a multiplicative inverse.)

3. Let A = (aij) be an n×n matrix over the field of real numbers such that for alli we have

∑nj=1 aij = a. If A2 = I (I is the identity matrix), find a.

1.8.3. General. 1. Without using derivative, find the minimum of the threevariable real function

f(x, y, z) = x2 + 4y2 + z2 − 6x+ 4y.

2. Five people have 719 rials altogether. If each person has an integer amount ofmoney, that the money of no two people is equal, and that the ratio of a person’smoney with respect to that of any other person with less amount of money is aninteger, determine how much money each person has.

3. A bus which must pass through the city A passes through the crossroad B of thecity with the probability of 1

3 . If the traffic-light of the crossroad is, consecutively,30 seconds red and 30 seconds green, find the average of the stop time of the busat the crossroad.

6 For a counterpart of this problem for sequences of real functions, see Problem 2 of 1.12.1.

14 1. PROBLEMS

1.9. Ninth Competition, Tehran Teacher Training (Tarbiat Moallem)University, September 1985

1.9.1. Analysis. 1. On the set of real numbers R, define the equivalencerelation ∼ as follows

a ∼ b ⇐⇒ a− b ∈ Q,where Q is the set of rational numbers. Prove that every equivalence class of ∼ isdense in R.

2. Prove that if f is a continuous and one-to-one function from R into R, then theinverse of f (from f(R) onto R) is continuous as well.

3. Let f be a continuous and increasing function from [a, b] into [a, b] and f(a) = a.Prove that if E =

x|a ≤ x ≤ b, f(x) ≥ x

, then f(E) = E.

4. Let the function f be defined on [0, 1] by

f(x) =

0 x /∈ Q,1q x ∈ Q, x = p

q , p, q ∈ N, gcd(p, q) = 1.

Prove that∫ 1

0f(x)dx exists.

5. Let f : [0, 1] → [0, 1] × [0, 1] be a function. Prove that the function f can haveany two properties of the following three properties but cannot have more than twoproperties.

Continuity, Injectivity, Surjectivity.

1.9.2. Algebra. 1. Let G be a finite group and p the smallest prime dividingthe order of G. Prove that every subgroup of index p in G is a normal subgroup ofG.

2. Let Z3 denote the field of integers mod 3. Define, explicitly, an isomorphism

from the fieldZ3[x]

(x2 + 1)onto the field

Z3[x](x2 + x+ 2)

.

3. Find the greatest common divisor of the two polynomials 4x4 − 2x2 + 1 and−3x3 + 4x2 + x+ 1 in Z7[x].

4. A quadratic extension K ⊇ F such that ch(F ) 6= 2 is given (here, ch(F ) denotesthe characteristic of the field F ). Prove that there exists an element y ∈ K suchthat y2 ∈ F and that 1, y is a basis for K over F .

5. Let F be a field with characteristic 2, V an n-dimensional vector space over F ,and T : V → V a linear transformation such that T 2 = I. Set W =

v ∈ V : Tv =

v. Prove that dim(W ) ≥ n/2. (Here, I denotes the identity transformation on

V .)

1.9. NINTH COMPETITION 15

1.9.3. General. 1. Evaluate the determinant of an n × n matrix whose di-agonal entries are all equal to r and whose off-diagonal entries are all equal toλ.

2. A husband and a wife, who are working at one of the poultry farms of thecountry, have posed a problem on the number of their chickens as follows. If theysell 75 of their chickens, their chicken food will finish twenty days after they run outof it. But if they buy another one hundred chicken, their chicken food will finishfifteen days before they run out of chicken food. Find the number of chickens inthe chicken farm.

3. Let α be a real number. Prove that there is no positive continuous functionf : [0, 1] → R such that∫ 1

0

f(x)dx = 1,∫ 1

0

xf(x)dx = α,

∫ 1

0

x2f(x)dx = α2.

4. Find the number of the solutions of the following equation in the set of positiveintegers.

x1 + x2 + · · ·+ xm = n,

where m and n are natural numbers with m < n.

5. On a square grid paper the following figure is drawn. Can you cut the paperalong the lines and divide it into two pieces in such a way that putting the piecesnext to one another forms a chessboard?

6. The start hour in a factory is 8 o’clock in the morning. A worker has estimatedthat, by car, s/he goes the distance from home to work within 10 to 20 minutes(with uniform distribution).

(a) If this worker leaves his/her home at 7 34 , what is the probability that s/he

does not get to work on time?(b) If 15 minutes is needed to have breakfast at the work place, determine the

latest time that this worker can leave home to get to work on time and with theprobability of 75% to have time to eat breakfast.

7. Let c1, . . . , cn be n real numbers. Consider the matrix (cicj)n×n. Evaluatedet(I + (cicj)

), where I is the identity matrix of size n.

16 1. PROBLEMS

1.10. Tenth Competition, University of Sistan and Baluchestan, March1986

1.10.1. Analysis. 1. The real function f satisfying the following conditionsis given.

(a) f(1) = 1.

(b) f ′(x) =1

x2 +(f(x)

)2 for all x ≥ 1.

Prove that limx→+∞ f(x) exists and is less than 1 + π4 .

2. The continuous function f : [a, b] → R satisfying the following conditions isgiven.

(a) f(a) = f(b) = 0.(b) The second derivative of f on the interval (a, b) exists [and is bounded].Show that ∫ b

a

∣∣f(x)∣∣dx ≤M

(b− a)3

12,

where M = sup∣∣f ′′(x)∣∣ : x ∈ (a, b).

• Hint. Use the auxiliary function g(t) = (t−a)(t−b)f(x)−(x−a)(x−b)f(t).

3. Prove that

limn→+∞

∫ 3

0

x2(1− x)xn

1 + x2ndx = 0.

1.10.2. Algebra. 1. Prove that the additive group of R has no maximalsubgroup.

2. Let A,B,C,D be ideals in a unital ring R such that

A+B = A+ C = A+D = R. (∗)

If M = B ∩ C ∩D, prove that A+M = R.Give an example of a unital ring R having ideals A,B,C,D which satisfy (∗).

3. Let E be an extension field of F and α ∈ E algebraic on F of degree n. If m < nand gcd(n,m!) = 1, then show that F (α) = F (αm).

4. Prove that no set of nilpotent matrices can span Mn(F ), where Mn(F ) is thevector space of all n× n matrices over the field F .

1.11. ELEVENTH COMPETITION 17

1.10.3. General. 1. Consider a test in which the probability of success (S)is equal to p and the probability of failure (F) is equal to q = 1 − p. Perform thistest two times. If the results turn out to be FS or SF , set the random variable Xto be 0 or 1, respectively. If not, do the test another two more times. Again if theresults of the two tests are FS or SF , we take the random variable X to be 0 or 1,respectively; and if not, we perform the test another two more times... Prove thatPX = 0 = PX = 1 = 1

2 .

2. Show that every closed set A in the plane with the property that A = ∅ is theboundary of an open set in the plane. (Here, A denotes the interior of the set A.)

3. Prove that every solution of the differential equation (∗) is a solution of theintegral equation (∗∗) and vice versa.

x′′ = f(t, x), x(0) = x0, x′(0) = y0, (∗)

where f(t, x) is a continuous function in a region D which contains the point (0, x0).

x(t) = x0 + y0t+∫ t

0

(t− s)f(s, x(s))ds. (∗∗)

4. Prove that the product of r consecutive natural numbers is divisible by r! and

from there conclude thatn!

r!(n− r)!is always an integer.

1.11. Eleventh Competition, The University of Birjand, March 1987

1.11.1. Analysis. 1. The function f : [0, 1] → R is differentiable and f andf ′ have no common zero. Prove that the set of zeros of f in [0, 1] is finite. (Recallthat a solution of the equation f(x) = 0 is called a zero of the function f .)

2. (a) Show that the sequence of functions fn(x) = xx···x

is pointwise convergentto a function f on [1, e

1e ] which satisfies the functional equation f(x) = xf(x).

• Hint. Prove that for all a ∈ [1, e1e ] the sequence

(fn(a)

)+∞n=1

is increasing andbounded. Then, in view of the fact that the function g(x) = ax is continuous onR, show that f(a) = af(a).

(b) Show that the function f is one-to-one. Then using the inverse of f , con-clude that f is continuous. And then, prove that the sequence (fn)+∞n=1 convergesuniformly on [1, e

1e ].

3. Suppose that for every n ∈ N, ϕn : [−1, 1] → R has second order derivative andthat ϕ′n(0) = 1. If sup

∣∣ϕ′′n(x)∣∣ : n ∈ N, x ∈ [−1, 1]< +∞, prove that

∑+∞n=1 an is

convergent, where an = 1n

∣∣ ∫ 1

−1ϕn(t) cosnπtdt

∣∣.

18 1. PROBLEMS

1.11.2. Algebra. 1. Suppose that the ringR has exactly two two-sided ideals.Prove that if there is an element u in R for which ux = x for all x ∈ R, then R isa ring with identity and u = 1R.

2. Prove that A, the field of algebraic numbers, is not a finite extension of Q.

3. Let G be a finite group with the following property: for every two elements xand y of G with x 6= e and y 6= e, where e is the identity element of G, there existsan automorphism θ ∈ Aut(G) such that y = θ(x). Prove that there is a primenumber p such that

G ∼= Zp ⊕ Zp ⊕ · · · ⊕ Zp.

1.11.3. General. 1. The matrix A is chosen at random from the set of all2 × 2 matrices with entries from Z. What is the probability that the determinantof A is an even number?

2. A real number c is given. Show that if one of the roots of the equation

x3 − 34x+ c = 0

is in the closed interval [−1, 1], then all of the roots of this equation are in theinterval [−1, 1].

3. Let X be a set with n elements and C a family of the subsets of X which satisfiesthe following conditions.

(i) If A,B ∈ C, then A ∪B ∈ C.(ii) If A ∈ C, then Ac := X \A ∈ C.(iii) ∅ ∈ C.Show that the number of the elements of C is equal to 2k where k ≤ n.

1.12. Twelfth Competition, Guilan University , March 1988

1.12.1. Analysis. 1. Let f : R → R be a continuous function. Moreover,suppose that there is an M > 0 such that for all x, y ∈ R∣∣f(x)− f(y)

∣∣ ≥M∣∣x− y

∣∣.Show that the function f is one-to-one and surjective.

2. Let (fn)+∞n=1 be a decreasing sequence of nonnegative functions on a nonvoid setS. That is,

∀n ∈ N : fn : S −→ R, fn ≥ fn+1, fn ≥ 0.

Prove that the series∑n≥1 fn converges uniformly on S if and only if the series∑

n≥1 n(fn − fn+1) does.7

7 This problem, as stated, is not correct! More precisely, for the “if part” of the problem, weneed to assume further that the sequence (fn)+∞n=1 converges uniformly to zero on S.

1.12. TWELFTH COMPETITION 19

3. Let g : [0, 1] → R be a Riemann integrable function. Prove that∫ 1

0

(∫ 1

x

g(t)dt)dx =

∫ 1

0

tg(t)dt.

1.12.2. Algebra. 1. Let G be a group of order 2p, where p is an odd prime.Prove that G has one and only one subgroup of order p. Also prove that G has psubgroups of order p, or one subgroup of order p. In the latter case, show that Gis a cyclic group.

2. Let G be an abelian group of odd order and φ a homomorphism of G of order 2.Show that every element g ∈ G can, uniquely, be written as g = xy where φ(x) = xand φ(y) = y−1.

3. Let R be a unital ring with characteristic 2 such that for every x 6= 1 and y 6= 1,we have xy2 = xy. Show that R is commutative.

4. Suppose that R is a unital ring, that every ideal of R is principal, and thatf : R→ R is a surjective homomorphism. Prove that f is an isomorphism.

5. Let A be a 3 × 3 invertible matrix over a field F such that detA = 1 andtr(A) = tr(A−1) = 0. Prove that A3 = I.

6. If α and β are two distinct roots of the equation

1 + x+x2

2!+ · · ·+ xp

p!= 0,

where p is a prime with p > 2, show that α − β and α + β and αβ are irrationalnumbers.

1.12.3. General. 1. A circle C centered at O is given and the center of asquare is on the circle. If the area of the square is not greater than half of the areaof the circle, prove that one of the vertices of the square is inside the circle.

2. 1700 people have participated in a true-false test. Knowing that 15 questionshave been given in the test, that none of the participants has answered two con-secutive questions correctly, and that all of them have answered all questions, dothere exist two equal answer sheets?

3. Evaluate1− C2

n + C4n − C6

n + · · · ,

where n ∈ N, Ckn :=n!

k!(n− k)!if k = 0, . . . , n, and Ckn := 0 otherwise.

20 1. PROBLEMS

1.13. Thirteenth Competition, University of Tehran, March 1989

1.13.1. Analysis. 1. Let f : R → R be (Riemann) integrable and f(x+ y) =f(x)+ f(y) for all x, y ∈ R. Show that there exists a number c such that f(x) = cxfor all x ∈ R.

2. Show that if an, bn are in R for all n ∈ N, (an + bn)bn 6= 0, and that both series+∞∑n=1

anbn

and+∞∑n=1

(anbn

)2 are convergent, then so is the series+∞∑n=1

anan + bn

.

3. Let (fn)+∞n=1, with fn : [0, 1] −→ R, be a sequence of differentiable functionssuch that ||f ′n||∞ ≤ 1 for all n ∈ N. Show that if for any continuous functiong : [0, 1] −→ R, we have

limn→+∞

∫ 1

0

fng = 0,

then the sequence (fn)+∞n=1 converges uniformly to zero on [0, 1].

1.13.2. Algebra. 1. Let R be a unital, commutative, and uncountable ringsuch that for every ideal 0 6= I CR, the quotient ring R/I is countable. Prove thatR is an integral domain.

2. Let R be a subring of Mn(Q). If(α 00 α

)∈ R for all α ∈ Q, (∗)

prove that every left (or right) ideal of R is finitely generated. By finding a leftideal in the ring

T =(

z q10 q2

)|z ∈ Z, q1, q2 ∈ Q

,

which is not finitely generated, show that the condition (∗) cannot be dropped.

3. Give an example of a group G which contains two elements a and b such that aand b are of order two but ab is of order infinity.

4. Let G = GL2(Z3) (the multiplicative group of all 2 × 2 invertible matrices onZ3), K = Z(G) denote the center of G, and

H =(

a b0 c

): a, b, c ∈ Z3, ac 6= 0

.

(a) Prove K ≤ H ≤ G and that the order of H is 12.(b) Prove

K =⋂x∈G

x−1Hx.

(c) ProveG

K∼= S4,

where S4 denotes the group of all permutations on four letters.

1.14. FOURTHEENTH COMPETITION 21

5. A = (aij) is an n× n matrix, where

aij =

δi,n j = 1,δi,j−1 j > 1.

Suppose that ξ is an nth root of unity in C. Set

ν(ξ) =

1ξξ2

...ξn−1

.

Prove that ν(ξ) is an eigenvector of A and find its corresponding eigenvalue.

1.13.3. General. 1. A regular n-gon is inscribed in a circle with radius one.Choose an arbitrary point on the circle and find the squares of the distances fromthe point M to the vertices of the regular n-gon. Prove that the sum of these valuesis 2n.

2. Find all polynomials f(x) with rational coefficients with the property that f(x)is irrational whenever x is irrational.

3. Let n parallel lines in the space that do not lie in a plane be given. Show thatthere are at least n distinct planes each of which passes through at least two linesof these given lines (n > 2).

1.14. Fourteenth Competition, The University of Isfahan, March 1990

1.14.1. Analysis. 1. Let f : R −→ R be twice differentiable on R \ x0 forsome x0 ∈ R. If f ′(x) < 0 < f ′′(x) on x < x0, and f ′(x) > 0 > f ′′(x) on x > x0,then f is not differentiable at x0.

2. Let g : R −→ R be a continuous function with the property that g(x) > 0 forx 6= 0 and g(0) = 0. Let f : R −→ R be a uniformly continuous and boundedfunction such that g f is integrable on R. Prove that limx→∞ f(x) = 0.

3. Let f : R −→ R be a continuous function and [a, b] and [c, d] two closed intervalssuch that [a, b] ⊂ f

([c, d]

). Prove that there exists a subinterval [r, s] of [c, d] such

that f([r, s]

)= [a, b].

22 1. PROBLEMS

1.14.2. Algebra. 1. Prove that if G is a finite p-group (p a prime number),then G′ 6= G, where G′ is the commutator (a.k.a. derived) subgroup of G.

2. Let G be a finite group and K a normal subgroup of G of order p, where p isthe smallest prime dividing |G|. Prove that K is a subgroup of Z(G), where Z(G)denotes the center of the group.

3. Give an example of a ring R having two elements which have the greatestcommon divisor in R but do not have the least common multiple in R.

4. Prove that if F is a field and n an integer greater than one, then xny+xn−1 +1is irreducible in F [x, y]. What can be said in case n = 1?

5. Let V be a finite-dimensional vector space over a field F and V1 and V2 twosubspaces of V with dimV1 = dimV2. Prove that there exists a subspace U of Vsuch that

U ⊕ V1 = U ⊕ V2.

1.14.3. General. 1. Let S be a set with n elements, and

A =A1, . . . , An

a family containing n distinct subsets of S. Show that there exists an element x ∈ Ssuch that the sets

A1 ∪ x, . . . , An ∪ x

are distinct.

2. A publisher is to exhibit 1369 titles of its published books. The books are tobe exhibited as follows. Each day 100 titles are to be placed on the exhibitiontable and that no two titles are placed on the table more than once. Determine themaximum number of the days of the publisher exhibition.

3. Find all nonzero real numbers a1, . . . , an with the following property:

n∑i=1

ami =n∑i=1

ai, m = 1, . . . , n+ 1.

1.15. FIFTEENTH COMPETITION 23

1.15. Fifteenth Competition, Ferdowsi University of Mashhad, March1991

1.15.1. Analysis. 1. The real function f is defined on [0,+∞) and is increas-ing. The function ϕ is defined on [0,+∞) by

ϕ(x) =∫ x

0

f(t)dt.

(a) Prove that for all x, y ≥ 0

ϕ(x+ y

2)≤ 1

2

(ϕ(x) + ϕ(y)

).

(b) Conclude that ϕ is convex.

2. The real function g is continuous on [0, 1] and g(0) = 0. Define the sequence(fn)+∞n=1 of functions on [0, 1] by

fn(x) =g(x)(sinx)n

1 + nx.

Prove that the sequence (fn)+∞n=1 is uniformly convergent on [0, 1].

3. A function f from (0,+∞) into (0,+∞) is defined; it is a one-to-one correspon-dence and for all x, y ∈ (0,+∞), we have

2xy ≤ xf(x) + yf−1(y).

(f−1 is the inverse of f .)(a) Show that for all x, y ∈ (0,+∞),

y − x

yf(x) ≤ f(y)− f(x) ≤ y − x

xf(y).

(b) Conclude that there exists a c ∈ R such that f(x) = cx for all x ∈ (0,+∞).

1.15.2. Algebra. 1. Let G be a group of order pαm, where p is an odd prime,a ∈ N, p - m, and that G has exactly (1 + p) Sylow p-subgroups. Show that∣∣∣ p+1⋂

i=1

Pi

∣∣∣ = pα−1,

where Pi’s are the Sylow p-subgroups of G.• Hint. [G : A ∩B] ≤ [G : A][G : B].

2. Let R be a unital ring such that every left ideal of it is also a right ideal ofR. Prove that the intersection of all prime ideals of R is equal to the set of thenilpotent elements of R.

3. Show that for every n× n matrix A, there is an n× n matrix B such that ABis an idempotent. 8

8 This problem, as stated, is trivial! To make the problem nontrivial, prove that for everyn × n matrix A with entries from a field F , there is a matrix B ∈ Mn(F ) such that AB is an

idempotent whose rank is equal to that of A (see Solution 3 of 2.15.2).

24 1. PROBLEMS

1.15.3. General. 1. Find a necessary and sufficient condition for the follow-ing property: the product of two integers is divisible by the sum of them.

2. Let f be nonnegative on [0, 1] and∫ 1

0f(x)dx = 1. Prove that∫ 1

0

(x−

∫ 1

0

uf(u)du)2

f(x)dx ≤ 14.

3. A train has n wagons. Any passenger who is boarding the train (independentof others), chooses a wagon at random to get on.

(a) What is the probability that at least one passenger gets on every wagon?(b) Using (a), evaluate the following sum.(

n

1

)1p −

(n

2

)2p +

(n

3

)3p − · · ·+ (−1)n−1

(n

n

)np,

where 1 ≤ p ≤ n.

1.16. Sixteenth Competition, Razi (Rhazes or Rasis) University ofKermanshah, March 1992

1.16.1. Analysis. 1. Let g : [0, 1] −→ R be a continuous function withg(1) = 0. If fn(x) = xng(x), show that the sequence (fn)+∞n=1 converges uniformlyon [0, 1].

2. The function f : R −→ 0, 1, ..., 9 is defined by

f(x) =a2 x 6= [x],9 x = [x],

where [x] denotes the integral part of x and a2 is the second digit of the decimalexpansion of x− [x].

(a) Prove that f is periodic and determine its period.(b) If c is the period of f , evaluate the following integral∫ c

0

xdf(x).

(If the decimal expansion of a number ends in zero, reduce the last nonzero digitby one and change all other digits on the right of it to 9.)

3. Let f : R −→ R be a uniformly continuous function. Prove that there arepositive numbers a and b such that

∣∣f(x)∣∣ ≤ a

∣∣x∣∣+ b for all x ∈ R.

1.17. SEVENTEENTH COMPETITION 25

1.16.2. Algebra. 1. Let G be a finite nonabelian group, and A,B twodistinct abelian subgroups of G such that

[G : A] = [G : B] = p,

where [G : A] and [G : B] denote the indexes of A and B in G, respectively, andp is the smallest prime dividing |G|. Prove that Inn(G) = Zp × Zp, where Inn(G)stands for the set of all inner isomorphisms of G.

2. Let R be a ring and r ∈ R be such that r − r2 is nilpotent. Prove that R has anonzero idempotent element whenever r is not nilpotent.

3. Let A = (aij) ∈Mn(Q), where aij = gcd(i, j). Is A invertible? Why?

1.17. Seventeenth Competition, Shahid Beheshti (former National)University, March 1993

1.17.1. Analysis. 1. Suppose that the function f is continuous on the inter-val [a, b] and that it is differentiable on (a, b). Also suppose that the graph of f isnot a line segment. Prove that there exists a c ∈ (a, b) such that

∣∣f ′(c)∣∣ > ∣∣∣f(b)− f(a)b− a

∣∣∣.2. If p1(x), p2(x), p3(x), p4(x) are real polynomials in the indeterminate x, provethat ∫ x

−1

p1(t)p3(t)dt∫ x

−1

p2(t)p4(t)dt−∫ x

−1

p1(t)p4(t)dt∫ x

−1

p2(t)p3(t)dt

is divisible by (x+ 1)4.

3. Let X be a metric space and f : X −→ R a continuous function. Show thatthe set x ∈ X : f(x) = 0 is an open subset of X if and only if there exists acontinuous function g : X −→ R such that f = gf2.

4. Let f : [0, 1] −→ R be a continuous function such that f(0) = f(1). Prove thatthere are two points a, b satisfying the following: 0 < a ≤ b ≤ 1, b − a = 1

2 , andf(b) = f(a).

26 1. PROBLEMS

1.17.2. Algebra. 1. Let G be a finite group and H ≤ G be such that

∀ x ∈ G(if x /∈ H =⇒ H ∩ (x−1Hx) = eG

).

Prove that [G : H] and |H| are relatively prime.

2. Let R be the ring of all n × n matrices over a field F , and R[x, y] the ring ofall polynomials in two indeterminates x, y with coefficients from R. (note: ∀a ∈R : ax = xa, ay = ya, xy = yx.) Suppose f, g ∈ R[x, y] are such that fg = 1.Determine gf .

3. Let V be a finite-dimensional vector space over a field F and T : V −→ V alinear transformation. Determine the dimension of kerT ∩ T (V ) in terms of theranks of the powers of T .

1.18. Eighteenth Competition, Sharif University of Technology, March1994

1.18.1. Analysis. 1. Let (an)+∞n=0 be a sequence of positive reals and An =a0 + · · · + an. If limn→+∞An = +∞ and lim

n→+∞

anAn

= 0, show that the radius of

convergence of the power series∑+∞n=0 anx

n is one.

2. Let f, g be continuous and periodic functions whose periods are one. Show that

limn

∫ 1

0

f(x)g(nx)dx =(∫ 1

0

f(x)dx)(∫ 1

0

g(x)dx).

3. For any real number x, write the binary expansion of x− [x] as follows

x− [x] =∞∑k=1

ak2k.

Show that the function g : R −→ R defined by g(x) = lim supn→∞

∑nj=1 aj

nmaps every

nonempty interval onto [0, 1] (i.e., for every nonempty interval I, we have g(I) =[0, 1]). Using this, construct an open map from R into R that is not continuous.

1.18.2. Algebra. 1. Let G be a finite group and K a normal subgroup of Gsuch that K as a group is simple and the square of its order does not divide theorder of G. Prove that if H is a subgroup of G such that H ∼= K, then H = K.

2. The following ring is given.

R =(

Z Q0 Q

)=(

a b0 c

): a ∈ Z, b, c ∈ Q

Prove that every ascending chain of right ideals of R is necessarily stable; that is,if

I1 ⊆ I2 ⊆ I3 ⊆ · · · , ∀j ∈ N : Ij Cr R,

then there is an n ∈ N such that In = In+1 = In+2 = · · · .

1.19. NINETEENTH COMPETITION 27

3. Let V be a finite-dimensional vector space over a field F . If Vii∈I is a set ofproper subspaces of V such that dimVi = dimVj for all i, j ∈ I and if |I| < |F |,prove that there is a proper subspace U of V such that

V = Vi ⊕ U,

for all i ∈ I.

1.19. Nineteenth Competition, University of Kerman, March 1995

1.19.1. Analysis. 1. If f : [0, 1] −→ [0, 1] is a continuous function, show thatthe equation

2x−∫ x

0

f(t)dt = 1

has only one root in [0, 1].

2. Let f : (0,+∞) −→ R be a continuously differentiable function such thatlimx→+∞

(f(x) + hf ′(x)

)= 0 for some h ∈ R+. Show that limx→+∞ f(x) =

limx→+∞ f ′(x) = 0.

3. Let A be the set of all continuous real functions on [0, 1] 9 and, for each n ∈ N,En denote the set consisting of all f ∈ A for which there is an a ∈ [0, 1] such that

|f(x)− f(a)| ≤ n|x− a|,for all x ∈ [0, 1].

(a) Show that En’s are closed sets with empty interior.(b) Conclude that there is a continuous function on [0, 1] which is nowhere

differentiable.

1.19.2. Algebra. 1. Let a group G be given. Suppose that the set

A =N C G :

G

N∼= G

is ordered by inclusion and that A has a maximal element. Prove that the trivialgroup e is the only maximal element of A.

2. Let R be a ring with identity. Set S =∑J , where J can be any minimal left

ideal of R (in case such ideals are non-existent, set S=0).(a) Prove that S is a two-sided ideal in R (i.e., S CR).(b) If S 6= 0 and the product of any two nonzero two-sided ideals of R is a

nonzero two-sided ideal of R, then S =⋂I, where I can be any nonzero two-sided

ideal of R.

3. A square matrix A = (aij)n×n over a field F is given. Suppose that there arescalars xi, yi ∈ F such that aij = xi + yj (1 ≤ i, j ≤ n). Prove that rank(A) ≤ 2.

4. Let F be a field, n ∈ N, and A ∈Mn(F ) with rank(A) = 1. Prove

det(I +A) = 1 + tr(A).

9 The set, in fact the algebra, A is assumed to be endowed with the uniform metric of A,which is induced by the uniform norm of A, usually denoted by ||.||∞

28 1. PROBLEMS

1.20. Twentieth Competition, Sharif University of Technology,February 1996

1. Let C be the field of complex numbers and(x, y), (z, t), (x′, y′), (z′, t′)

⊆ C2.

Prove that there are scalars α, β ∈ C which are not simultaneously zero such thatthe vectors θ = α(x, y)+β(z, t) and θ′ = α(x′, y′)+β(z′, t′) are linearly dependent.

2. The sequence (an)+∞n=1 of nonnegative real numbers satisfies the following prop-erty:

1 + am+n ≤ (1 + am)(1 + an),for all m,n ∈ N. Prove that the sequence (xn)+∞n=1 defined by xn = n

√1 + an is

convergent.

3. Let G be a group such that for all σ ∈ Aut(G) and for all x ∈ G we haveσ(x) = x or σ(x) = x−1. Prove that G is solvable.

4. The function f : (14 , 1) −→ R for all x ∈ ( 1

4 , 1) satisfies the following

xf(x) = f(x).

Prove that f is uniformly continuous on ( 14 , 1).

5. Let R be a commutative ring with identity and with the following properties:(a) The intersection of all of its nonzero ideals is nontrivial.(b) If x and y are zero divisors in R, then xy = 0.Prove that R has exactly one nontrivial ideal.

6. Let f , g : [0,+∞) −→ R be two functions. Suppose that f is decreasing andlimx→+∞ f(x) = 0 and that g is a periodic function such that

∫ p0g = 0, where p is

the period of g. Show that∫ +∞0

fg is convergent.

1.21. Twenty First Competition, University of Tehran, March 1997

1.21.1. Analysis. 1. A set S ⊂ R and a function f : S −→ R are given.(a) Assume that α is a limit point of the set S and that for every ε > 0 the set

x ∈ S : |f(x)| ≥ ε

is finite. Show that limx→α f(x) = 0.(b) Assume that S is compact and that for every limit point α of S, we have

limx→α f(x) = 0. Show that the setx ∈ S : |f(x)| ≥ ε

is finite for all ε > 0.

2. Let g : R −→ R be a continuous function satisfying the following

limx→+∞

(g(x+ t)− g(x)

)= 0,

for all t ∈ R.(a) Suppose that K is a compact subset of R. Show that

∀ε > 0 ∃M > 0 x > M ⇒∣∣g(x+ t)− g(x)

∣∣ < ε (∀ t ∈ K).

1.22. TWENTY SECOND COMPETITION 29

(b) Use (a) to show that

limx→+∞

(∫ x+1

x

g(u)du− g(x))

= 0, limx→+∞

g(x)x

= 0.

3. Let f : [0, 1] −→ R be a bounded function whose limit from the left exists atany point in [0, 1]. Prove that f is Riemann integrable on [0, 1].

1.21.2. Algebra. 1. Let R be a ring and A and B ideals of R with A ⊆ B.

By definition, A is said to be small in B and we write As⊆ B if for every ideal C

of R for which A + C = B, then C = B. Show that if As⊆ B and C

s⊆ D and

A ∩ C = B ∩D = 0, then A+ Cs⊆ B +D.10

2. Let G be a group and H a subgroup of G such that H ≤ Z(G), where Z(G)denotes the center of G. Prove that if [G : H] = p2, where p is prime, then thederived subgroup of G is cyclic.

3. Let A be a nonzero real n× n matrix such that

A = (aij)1≤i,j≤n, aikajk = akkaij , ∀i, j, k.Prove that

(a) tr(A) 6= 0.(b) The matrix A is symmetric.(c) The characteristic polynomial of A is equal to xn−1

(x− tr(A)

).

1.22. Twenty Second Competition, University of Ahwaz, March 1998

1.22.1. Analysis. 1. Let f : [0, 1] −→ R be a differentiable function withf(0) = f(1) − 1 = 0. Show that for each n ∈ N, there are x1, . . . , xn ∈ [0, 1] such

thatn∑i=1

1f ′(xi)

= n.

2. Let X be a metric space, f : X −→ R a continuous function, and (fn)+∞n=1 asequence of continuous nonnegative functions on X.

(a) If∑+∞n=1 fn = f2, then there exists a sequence (gn)+∞n=1 of continuous real

functions on X such that fn = gnf for all n ∈ N.(b) Prove that

∑+∞n=1 gn = f provided that

∑+∞n=1 gn is uniformly convergent

on X and that the interior of f−1(0) is the empty set.

3. Let (fn)+∞n=1 be a sequence of analytic functions from the region D ⊂ C into Csuch that fn uniformly converges to f on D. Prove that if γ is a simple closed curveinside D, then f is analytic inside and on the curve γ and that for all z0 inside γ,we have

limn

∫γ

fn(z)z − z0

dz =∫γ

f(z)z − z0

dz.

10 The hypothesis “A ∩ C = B ∩D = 0” is redundant!

30 1. PROBLEMS

1.22.2. Algebra. 1. Let G be a group, H ≤ K ≤ G, [G : K] an odd number,and [K : H] = 2. Also suppose that there is a k ∈ K of order 2 which is notconjugate to any element of H in G. Prove that G has a subgroup of index 2.

2. In a commutative ring R (which is not necessarily unital), the ideal M is calledmaximal if R2 * M and M is not contained in any ideal other than itself and R .Now, we define a subring J of R as follows. If R has no maximal ideal, set J = R.If not, set J to be the intersection of all maximal ideals of R. Prove that J has nomaximal ideal.

3. Let V be an n-dimensional vector space over a field F (n ∈ N) and S, T : V −→ Vtwo linear transformations such that the characteristic polynomial of one of themis irreducible over F . If L = TS − ST 6= 0, then the rank of L is greater than one.

1.22.3. General. 1. If limx→+∞

(b+ x

b− x

)x= 9, then b is equal to

(a) 3. (b) 3e. (c) ln 3. (d) ln 9.

2. If limx→0

12bx− sinx

∫ x

0

t2√t− a

dt exists and is nonzero, then which of the following

is admissible?

(a) a < 0, b = 12 .

(b) a = 0, b = 12 .

(c) a > 0, b = 12 .

(d) a = −1, b = −12 .

3. Which of the following is the general solution of the differential equation ex(1 +x)dx− (xex − yey)dy = 0?

(a) x2ex−y + y3 = c.(b) y2 + 2xex−y = c.(c) y2ex−y + x2 = c.(d) xex−y + 1

2y2 = 0.

4. If A and B are two ends of the diagonal of a cube with side length 3, and anant wants to travel from A to B, what is the smallest distance in meters that theant needs to travel?

(a) 3√

3. (b) 3 + 3√

3. (c) 3√

6. (d) 3√

5.

5. If f(x) =+∞∑n=1

(x+ 2)n

2n(n+ 1), then which one is the domain of f?

(a) [−4, 0).(b) [−4, 0].(c) (−4, 0].(d) None of (a), (b), and (c).

1.22. TWENTY SECOND COMPETITION 31

B

A1cm

Figure 2

6. If x3 + y3 = xy + 1 and in a neighborhood of (1, 1), y is a function of x, whichone is y′(1)?

(a) −1. (b) −23 . (c) 1. (d) 3

2 .

7. A particle, initially located in the origin, is discretely moving along the x-axisone unit to the right or one unit to the left with probability p and q = 1 − p,respectively. The probability that after 2k moves (k ≥ 5), the particle is 10 unitsfar away from the origin is

(a) Ck+52k pkqk(p10 + q10).

(b) Ck+52k pk+5qk−5.

(c) Ck+52k pk−5qk+5.

(d) Ck+52k pk−5qk−5(p10 + q10).

8. An arrow, as shown in the figure, is moving according to the following law. Ifthe arrow is put at a point B, then it jumps as much as the length of the arc AB inthe trigonometric direction. At how many point(s) one can put the arrow so thatafter 36 seconds it hits the point A?

(a) Exactly at one point.(b) Exactly at 18 points.(c) Exactly at 35 points.(d) At no point.

9. Consider the function f(x) =x2 x irrational0 x rational . The point(s) at which f

is differentiable is/are

(a) the two points x = 0, 1.(b) exactly one point.(c) rational points.(d) no point.

10. Consider the series+∞∑n=0

cos(xn)1 + n2

.

(a) The series converges only at zero.

32 1. PROBLEMS

(b) The series converges everywhere.(c) The series converges at nonzero points.(d) The series converges absolutely at one point.

11. If we consider the sequence (xn)+∞n=1 (x1 6= 0) of real numbers satisfying thefollowing recursive relation xn+1 =

xn1 + xn

, then

(a) xn → 1 as n→ +∞.(b) (xn)+∞n=1 has a subsequence converging to one.(c) (xn)+∞n=1 has a subsequence converging to zero.(d) (xn)+∞n=1 is divergent.

12. Which of the following propositions is correct?

(a) limn

∫ n0

dx1+x2 exists and is positive.

(b) limn

∫ n0

dx1+x2 exists and is negative.

(c) limn

∫ n0

dx1+x2 is divergent.

(d) f(x) = 11+x2 is not integrable on the interval [0, n].

13. If the function f : R −→ R is differentiable and its derivative is bounded, then

(a) the function f is bounded.(b) the function f is increasing.(c) the function f is uniformly continuous.(d) the function f is decreasing.

14. If the function f : [0, 1] −→ [0, 12 ] is continuous, then

(a) the graph of f intersects the line y = 2x.(b) there exists only one point x0 for which f(x0) = x0.(c) f(x) 6= 0 for all x ∈ [0, 1].(d) the equation f(x) = x has at most two solutions.

15. If f(x) =|x|

1 + |x|for all x ∈ R, then

(a) the function f is increasing.(b) f(x+ y) = f(x) + f(y).(c) f

(f(x+ y)

)≤ f

(f(x)

)+ f

(f(y)

).

(d) the function f is continuous only at zero and f(x+ y) ≤ f(x) + f(y).

16. Below is drawn the graph of the polar curve r = θ. Which one is the area ofthe shaded region?

(a)∫ π

3π6θ2dθ.

(b)∫ −π

6−π3

12θ

2dθ.

(c)∫ 5π

62π3

12θ

2dθ −∫ π

3π6

12θ

2dθ.

(d)∫ 11π

65π3

12θ

2dθ −∫ π

3π6

12θ

2dθ.

1.22. TWENTY SECOND COMPETITION 33

Figure 3

17. The value of∫ 1

0x5exdx is

(a) −44e. (b) 120 + 44e. (c) 44e. (d) e.

18. If y =∫ x2

sin x

(x+ t)dt, thendy

dxat x = 0 is equal to

(a) 5. (b) 0. (c) 2. (d) −1.

19. For what values of p, does the integral∫ 2

1

dx

x(lnx)pconverge?

(a) p = 1. (b) p > 1. (c) p < 1. (d) p = 2.

20.∫ +∞

0

e−ax − e−bx

xdx (0 < a < b) is equal to

(a) ln(ab ). (b) ln( ba ). (c) ln(ab). (d) eab.

21. Which of the following statements is not correct?

(a) A power series might converge at both of the end points of its interval ofconvergence.

(b) A power series can converge conditionally or diverge at any of the end pointsof its interval of convergence.

(c) A power series can converge absolutely at one of the end points of its intervalof convergence and converge conditionally at the other end point.

(d) A power series can converge conditionally at one of the end points of itsinterval of convergence and diverge at the other end point.

22. Using the spherical coordinates to compute the volume of the ellipsoid, theJacobian of the transformation is equal to

(a) ρ2 sin(ϕ). (b) ρ sin(ϕ). (c) abcρ2 sin(ϕ). (d) abc.

23. Letting F (x, y, z) = (x − y)−→i + (y − z)

−→j + (x − y)

−→k (−1 ≤ x, y, z,≤ 1) be

a vector field, how much is the flux of the field F through the surface of a cube ofside length two centimeters?

(a) 8 cm3. (b) 16 cm3. (c) 1 cm3. (d) 14 cm3.

34 1. PROBLEMS

24. For the vector field F (x, y, z) = f(x, y, z)−→i + g(x, y, z)

−→j + h(x, y, z)

−→k , we

have Curl(F ) = 0 but∮CF.dR 6= 0, where C : x2 + y2 = 1. Then

(a) div(F ) < 0.(b) the domain of F is connected.(c) the domain of F is simply connected.(d) the domain of F is not connected.

25. The family of orthogonal curves to the curves r = c(1 + cos θ) is

(a) r = c(1− sin θ).(b) r = c sin θ.(c) r = c(1 + sin θ).(d) r = c(1− cos θ).

26. The general solution of the differential equation y′′ − 2y′ − 3y = 64xe−x is

(a) y = c1e3x + c2e

−x − e−x(8x2 + 4x+ 1).(b) y = c1e

3x + c2e−x.

(c) y = c1e3x + c2e

−x − 8x2e−x.(d) y = c1e

3x + c2e−x − e−x(8x2 + 1).

27. Using the Gamma function for the definition of the factorial of numbers, ( 12 )!

is

a)√π. (b)

√π

2 . (c) 1√π. (d)

√π2 .

28. The equation x4 + x2 − 1 = 0 has roots as follows.

(a) Two positive roots, two negative roots.(b) One positive root, one negative root, two nonreal roots.(c) No real root.(d) Four real roots.

29. The equation f(x) = e−x − sinx = 0 has a real root in the interval [0.4, 0.8].The appropriate fixed point iteration function to find this root is

(a) g(x) = x+ e−x − sinx.(b) g(x) = − ln

(sin(x)

).

(c) g(x) = sin−1(e−x).(a) g(x) = 1

2

(2x+ e−x − sinx

).

30. In the bisection method, the number of times needed in order to obtain accuracywithin ε is equal to

(a) 1 +log b−a

ε

log 2.

(b) log(b− a)− log(ε).(c) cannot be determined.

(d) 1 +log(b− a)

log 2.

1.22. TWENTY SECOND COMPETITION 35

31. In the iteration method, Pn+1 = g(Pn) (n ≥ 0) where g(x) = x− f(x)f ′(x)

, as the

multiplicity of the root x = P increases,

(a) g′(P ) approaches zero.(b) the speed of the convergence does not change.(c) g′(P ) gets very close to one.(d) the method is not convergent.

32. Which of the following propositions is correct?

(a) If∑+∞n=1

an

an+1converges, then so does the series

∑+∞n=1 an.

(b) If∑+∞n=1 an converges, then limn nan = 0.

(c) If limn nan = 0, then the series∑+∞n=1 an is convergent.

(d) If∑+∞n=1

an+1an

converges, then so does the series∑+∞n=1 an.

33. Which one is the equation of the plane which includes the line 2y− 2 = 2x = z

and is parallel to the linex− y = 0z = 0 ?

(a) y − x = 1.(b) y = 2x.(c) z − 2y − 2x = 2.(d) y − x = −1.

34. Which of the following statements is a more complete definition of “algorithm”?

(a) “Algorithm” is a method for solving a problem.(b) “Algorithm” is a logical method together with a terminating condition for

solving a problem.(c) “Algorithm” is a logical procedure without ambiguity, which includes a

finite set of stages that are related to one another, together with a terminatingcondition.

(d) “Algorithm” just means method.

35. Every digit in octal expansion is equivalent to

(a) one digit in decimal expansion.(b) five digits in binary expansion.(c) two digits in hexadecimal expansion.(d) three digits in binary expansion.

36. A is infinite if and only if

(a) A is countable.(b) card(A \B) = card(A) for all B ⊆ A.(c) A is equivalent to any of its infinite subsets.(d) for all natural numbers n, A includes a subset which is equivalent to an

n-set.

36 1. PROBLEMS

37. Let A be an ordered set and B ⊆ C ⊆ A. Which of the following claims istrue?

(a) If sup(A) and sup(C) exist, then sup(B) ≤ sup(C).(b) If sup(B) and sup(C) exist, then sup(C) ≤ sup(B).(c) If sup(B) and sup(C) exist, then inf(B) = inf(C).(d) There is no relation between sup(B) and sup(C).

38. Which of the following is equivalent to Zorn’s Lemma?

(a) Every subset of an ordered set having an upper bound has sup.(b) There is a choice function for every nonempty family of nonempty subsets

of an arbitrary set X.(c) If card(A) ≤ card(B) ≤ card(P (A)), then card(A) = card(B) or card(B) =

card(P (A)).(d) Every bounded subset of an ordered set has a maximal element.

39. The integral∫ a

−a

sinxdxex + e−x

is equal to

(a) 2.(b) 0.(c) 2

∫ a0

sin xdxex+e−x .

(d) It does not exist.

40. If f(a+ b− x) = f(x) for all x ∈ [a, b], then∫ baxf(x)dx is equal to

(a) a−b2

∫ baf(x)dx.

(b)∫ baf(x)dx.

(c) a+b2

∫ baf(x)dx.

(c) (a+ b)∫ baf(x)dx.

1.23. Twenty Third Competition, Sharif University of Technology,March 1999

1.23.1. Analysis. 1. Let X = Cb(R) be the space of all continuous boundedfunctions g : R −→ R which is endowed with the norm ||g|| = supt∈R

∣∣g(t)∣∣. For agiven function f ∈ X , define fα : R −→ R by fα(t) = f(t+α), where α ∈ R. Proveor disprove the following propositions.

(a) Iffα : α ∈ R

is compact, then f is uniformly continuous on R.

(b) If f is uniformly continuous on R, then the setfα : α ∈ R

is compact in

X .

2. Let α ∈ R and α 6= 2kπ for all k ∈ Z. Find the sum of the series+∞∑n=1

einα

nand

justify your answer. Ditto for+∞∑n=1

sin(nα)n

and+∞∑n=1

cos(nα)n

.

1.24. TWENTY FOURTH COMPETITION 37

3. Let f, f1, f2, . . . be real continuous functions on the interval [a, b] whose deriva-tives are also continuous on [a, b]. If the sequence (fn)+∞n=1 converges to f pointwiseand the sequence (f ′n)

+∞n=1 converges pointwise to a continuous function g, show that

f ′(x) = g(x) for all x ∈ [a, b]. (Note: at the end points a and b, only the right andleft derivatives are to be considered.)

1.23.2. Algebra. 1. Let G be a nonabelian group. Prove that the group ofthe inner automorphisms of G cannot be nonabelian of order 8.11

2. Let R be a ring and H the intersection of all nonzero right ideals of R. Showthat if H 6= 0, then H is a two-sided ideal in R and that we have H2 = 0, or elseR is a division ring.

3. Let A be an n × n matrix with entries from a field F . Prove that if for everymatrix B with trace zero, we have tr(AB) = 0, then A = λI for some scalar λ inF .

1.24. Twenty Fourth Competition, Khajeh Nasir Toosi University ofTechnology, May 2000

1.24.1. First Day. 1. Let f, g : [a, b] −→ R be continuous and that g′ existson [a, b]. Show that if

(f(a)− g′(a)

)(g′(b)− f(b)

)> 0, then there exists a c ∈ (a, b)

such that f(c) = g′(c).

2. Let D be a domain in the complex plane and u a real valued harmonic functionon D. Show that if the set

A =(x0, y0) ∈ D : ∃r > 0 ∀(x, y) ∈ Br(x0, y0) ∩D ⇒ u(x, y) ≤ u(x0, y0)

is nonempty, then u is constant on D. Recall that a domain is a connected openset and that

Br(x0, y0) =(x, y) ∈ R2 :

√(x− x0)2 + (y − y0)2 < r

.

3. Let t(n) be the smallest prime factor of n for all 1 < n ∈ N. Prove thatt(n) < t(3n − 2n).

4. Let G be an infinite group with the property that for every two infinite subsetsX and Y of G, there exist x ∈ X and y ∈ Y such that xy = yx. Prove that if thecenter of G has finite index in G, then G is abelian.

5. Let M(n, d) denote the maximal number of n-tuples from 0, 1 such that everytwo of which differ at least in d components. (For instance, M(4, 3) = 2.) Provethat M(2d− 1, d) ≤ 2d.

6. A ruthless governor has made three mathematicians, who have been sentencedto death, play the following game.

11 This problem is wrong!

38 1. PROBLEMS

Two lookalike boxes, one of which contains two black marbles and a whitemarble, and the other contains two white marbles and a black marble, are provided.Each of the convicts picks a marble, which is not to be replaced, from one of theboxes at random. If the picked marble is black, the convict will be executed; if not,s/he will be set free. Assuming that every convict witnesses the choice(s) made bythe convict(s) prior to oneself and that the convicts will do the most logical effort,what is the probability that the second person survives? The probability of survivalfor the third person is higher or for the second person?

1.24.2. Second Day. 1. Let B be a nonempty bounded set in Rn with theproperty that for each pair of points x, y in B, there exists an open ball U suchthat U ⊆ B and x, y ∈ U .

(a) Prove that B is an open ball.(b) Show that if we replace Rn by a complete metric space, then the conclusion

of (a) does not necessarily hold.

2. Let f be a continuous function on the interval [a, b] such that∫ baf(t)dt 6= 0.

Show that for each k ∈ (0, 1), there exists a c ∈ (a, b) such that∫ c

a

f(t)dt = k

∫ b

a

f(t)dt.

3. Let matrices A,B,C be such that AB, BC, and ABC make sense. Denotingthe rank of any matrix P by r(P ), prove that

(a) r(BC) + r(AB) ≤ r(ABC) + r(B).(b) r(A) + r(B) ≤ r(AB) + n,where n is the number of columns of the matrix A.

4. Let R be a ring. Prove that if for any left ideal I of R, we have I2 = I, thenfor any two-sided ideal K of R and any left ideal I of R, we have I ∩ K = KI.(Remark: Note that R is not necessarily unital.)

5. In an athletic tournament, n teams have participated and any two teams haveplayed once against one another. Assuming that the game result is to be either winor lose, prove that at the end of the tournament there are two teams whose winsare equal if and only if there are three teams A,B,C such that A has won B, Bhas won C, and C has won A.

6. Two persons are playing the following game.First, the first person chooses a triple from 0, 1 (for instance, (0, 1, 0)). The

second person, who knows the triple chosen by the first person, chooses a differenttriple. Then a machine which at random generates 0 with probability 1

2 and 1 withprobability 1

2 is turned on. We write the numbers generated by the machine fromleft to right and the player whose chosen triple comes first is to win the game. Showthat no matter what choice is made by the first person, there is a choice for thesecond person so that the probability of winning the game by the second person isgreater than 1

2 .

1.25. TWENTY FIFTH COMPETITION 39

1.25. Twenty Fifth Competition, Imam Khomeini InternationalUniversity of Qazvin, May 2001

1.25.1. First Day. 1. Let G be a finite group of order n such that [G :Z(G)] = 4. Prove that 8|n. For any given natural number n such that 8|n, constructa group with the aforementioned property. (Here, Z(G) denotes the center of G.)

2. Let S, T be two linear transformations on a vector space V . Prove that if S2 = S,T 2 = T , kerT ⊆ imS, and imT ⊆ kerS, then T + S is the identity transformation.

3. Let f be a twice differentiable function such that f ′′(t) < 0 for all t ∈ R. Provethat if for two real numbers x and y we have f ′(y) + x < f(y + 1), then f(y) > x.

4. Let f : (a, b) −→ R be continuously differentiable and

limx→a+

f2(x) = 0, limx→b−

f2(x) = e− 1.

Prove that if 2f(x)f ′(x)− f2(x) ≥ 1 for all x ∈ (a, b), then 0 < b− a ≤ 1. Give anexample in which b− a = 1. (Note: f2(x) =

(f(x)

)2.)5. Seven boxes, on each of which a number from 1 to 7 is written, are at ourdisposal. Seven marbles numbered 1, seven marbles numbered 2, ..., and sevenmarbles numbered 7 are provided. We place the marbles in the boxes at randomin such a way that each box has exactly seven marbles. Play the following game.

In stage one, a marble is drawn at random from box number one (withoutreplacement). In stage i (i ≥ 2), a marble is drawn at random from the boxwhose number is identical to that of the marble drawn in stage (i − 1) (withoutreplacement). The game is stopped when it is needed to draw a marble from anempty box. Find the probability that all the marbles are drawn from all boxes.

6. Prove that the number of triangles, with integer sides, each of which havingperimeter n is equal to the number of partitions of the number n into the summands2, 3, and 4 in which the summand 3 appears at least once.

1.25.2. Second Day. 1. Let R be an integral domain and U(R) the mul-tiplicative group of the units of R. Prove that every finite subgroup of U(R) iscyclic.

2. Let p be an odd prime. Prove that every prime factor of 2p+1 which is differentfrom 3 is of the form 2kp+ 1.

3. Let (X, d) be a compact metric space and f : X −→ X a surjective function.Show that if

d(f(x), f(y)

)≥ d(x, y),

for all x, y ∈ X, then f is continuous.

4. Let f : C −→ C be a function with f(0) = 0 and such that for all z ∈ C andw ∈ 0, 1, i, we have ∣∣f(z)− f(w)

∣∣ = ∣∣z − w∣∣.

40 1. PROBLEMS

Find the function f explicitly. (i2 = −1.)

5. Let n be a an odd number and A an n×n matrix whose entries are from the set−1, 1. If the product of the entries of the ith row is shown by ai and the productof the entries of the jth column is shown by bj , prove that

n∑i=1

ai +n∑j=1

bj 6= 0.

Does the above assertion hold for an even n?

6. A society is called ideal if for every two distinct members a and b of the society,there exists a member c of it such that c is acquainted with one and only one ofa or b (We assume that acquaintance is a symmetric and nonreflexive relation.)Prove that every ideal society of n people (n ≥ 2) contains an ideal society of n− 1people.

1.26. Twenty Sixth Competition, Institute for Advanced Studies inBasic Sciences (IASBS) of Zanjan, May 2002

1.26.1. First Day. 1. Let f : [a, b] → [a, b] be a continuous function which isdifferentiable on (a, b) and f(a) = a , f(b) = b. Prove that there exist two distinctpoints x1 and x2 in (a, b) such that f ′(x1)f ′(x2) = 1.

2. Suppose that U =z ∈ C : |z| < 1

and D is an open set in C such that U ⊆ D,

and that the function f is analytic on D. Also let f(0) = 1 and that |f(z)| > 2 forall z for which |z| = 1. Prove that there exists z0 ∈ U such that f(z0) = 0.

3. Letting p be an odd prime, prove that

pp−1 + (2p− 2)p−1 ≡ 1(mod p(2p− 2)

).

4. Let R be a unital commutative ring such that every ideal of it is principal. Provethat if R has only one maximal ideal, then Rx ⊆ Ry or Ry ⊆ Rx for all x, y ∈ R.

5. A k-element cover for a set S is a collection of k nonempty subsets of S suchthat the union of it is S. An n-element cover of the set S is called minimal if non − 1 elements of it cover S. If the number of the minimal k-element covers of aset with n elements is denoted by M(n, k), prove that

M(n, n− 1) =n

2(2n − n− 1

).

6. Let X be a set with n elements and positive integers m, k be such that m ≤ kand m + k ≤ n. Denote the set of all subsets of X with m elements by Xm.Suppose that f : Xm → R is a function where R is the set of real numbers. Provethat if for every member S of Xk we have

∑T∈Xm,T⊆S

f(T ) = 0, then f ≡ 0.

1.27. TWENTY SEVENTH COMPETITION 41

1.26.2. Second Day. 1. Letr1, r2, . . .

be the set of rational numbers in

[0, 1]. For each x ∈ [0, 1], we let Ax =n ∈ N : rn ≤ x

and define the function

f : [0, 1] → R by

f(x) =∑n∈Ax

12n.

Prove that f is continuous at any irrational point of the interval [0, 1].

2. Let f : R → R be a continuous function and that(f f f

)(x) = x for all x ∈ R.

Prove that f(x) = x for all x ∈ R.

3. Let G be a group which has two distinct maximal subgroups, say, H and K.Prove that if H and K are abelian and Z(G) = e, then H ∩K = e.

4. Let A be an n× n matrix with entries in a field F . Let λ ∈ F be an eigenvalueof An corresponding to an eigenvector x such that for all a0, a1, . . . , an−1 ∈ F , if(∑n−1

i=0 aiAi)x = 0, then a0 = a1 = · · · = an−1 = 0. Prove that

(An−λI

)n = 0. 12

5. An extended die is a homogeneous cube on each of whose sides an arbitrarypositive integer is written. (These numbers are not necessarily distinct.) Can onedesign two extended dice (which are not necessarily identical) in such a way thatthe probability that the sum of the numbers on the rolled dice is the same as thoseof two ordinary dice? (That is, the probability of getting a sum of 2 is 1

36 , theprobability of getting a sum of 3 is 2

36 , etc.) In case, you answer in the affirmative,find all possible answers; and if you answer in the negative, prove your claim.

6. A matrix Mm×n = (mij) with real entries is said to be balanced, if wheneverfor all i, i′, j, j′ we have 1 ≤ i < i′ ≤ m and 1 ≤ j < j′ ≤ n, then mij + mi′j′ ≤mij′ +mi′j . Suppose that there are two rows i1 i2 such that by interchanging therows, the matrix remains balanced. Prove that for all i′1 and i′2, if i1 < i′1 < i′2 < i2,then by interchanging the rows i′1 and i′2 the matrix remains balanced.

1.27. Twenty Seventh Competition, Bu-Ali Sina (Avecina) Universityof Hamedan, May 2003

1.27.1. First Day. 1. Let n be a natural number greater than 1. Denote theset of all n × n matrices with real entries by Mn(R). Define the following metricon Mn(R). For A = (Aij) and B = (Bij),

d(A,B) = max∣∣Aij −Bij

∣∣ : i, j = 1, . . . , n.

Prove that GLn(R), the set of nonsingular matrices, is an open and disconnectedsubset of Mn(R).

2. Let f be a function that is analytic on C. Let L and M be two orthogonal linesintersecting at a point A such that f(L) = L and f(M) = M . Prove that if z1 andz2 are two complex numbers that are symmetric with respect to A, then f(z1) isthe image of f(z2) with respect to A.

12 Prove that An − λI = 0!

42 1. PROBLEMS

3. Let (an)+∞n=1 be a sequence of real numbers defined as follows

a0 = 0, a1 = b, an+1 = an

√1 + a2

n−1 + an−1

√1 + a2

n, n ≥ 1.

Determine an in terms of b.

4. Let K be a nonempty set and ∅ 6= I ⊆ K. LetAi : i ∈ I

be a family of the

subsets of K. Prove that ifi : i /∈ Ai

∈Ai : i ∈ I

∪∅,

then i ∈ Ai for all i ∈ I.

5. Let A be a 3× 2 matrix and B a 2× 3 matrix with complex entries for which

AB =

8 2 −22 5 4−2 4 5

.

Determine the matrix BA. (Hint: Compute (AB)2.)

6. Let G be a group and H a subgroup of it such that for all x ∈ G \ H and ally ∈ G, there is a u ∈ H such that y−1xy = u−1xu. Prove that H is normal in Gand G/H is abelian.

1.27.2. Second Day. 1. Let f : R → R be a differentiable function, a, b ∈ Rand a < b. If f(a) = f(b) = 0, f ′(a) > 0, and f ′(b) > 0, prove that f ′ has at leasttwo roots in the interval (a, b).

2. Prove that the interval [0, 1] cannot be written as a union of pairwise disjointclosed intervals each of which having a positive length less than one.

3. Let D be a countable subset of the Euclidean plane, i.e., R × R. Prove thatthere is a partition for D into two subsets X and Y such that any line parallel tothe x-axis intersects X in finitely many points and any line parallel to the y-axisintersects Y in finitely many points.

4. LetA1, . . . , An

be a family of finite sets and S =

⋃ni=1Ai. Suppose that a

fixed number k, 1 ≤ k ≤ n, satisfies the following conditions.(a) The union of any k elements of the family

A1, . . . , An

is equal to S.

(b) The union of any k− 1 elements of the familyA1, . . . , An

is not equal to

S.(c) |S| =

(nk−1

)Find the number of the elements of all of Ai’s.

5. Let a, b be two natural numbers such that gcd(a, b) = 1. Prove that

ordab(a+ b) = lcm[ordb(a), orda(b)]

(Here, by ordn(m), we mean the order of m modulo n.)

6. Let R and R′ be two rings whose elements are all idempotents and consider afunction f : R→ R′ that is one-to-one and onto such that f(xy) = f(x)f(y) for allx, y ∈ R. Prove that R ∼= R′.

1.28. TWENTY EIGHTH COMPETITION 43

1.28. Twenty Eighth Competition, Sharif University of Technology,May 2004

1.28.1. First Day. 1. Prove that any one-to-one and entire function is of theform f(z) = az + b, where a and b are fixed complex numbers and a 6= 0.

2. Let d1 and d2 be the Euclidean metrics on R and R × R, respectively, A =(x, y) ∈ R × R : x = 0 or y = 0

, and d = d2|A the metric induced by the

Euclidean metric of R × R on A. Prove that if f : (R, d1) → (A, d) is continuousand surjective, then f−1(0, 0) has at least three elements.13

3. Let f be an arithmetic function with the property that for every natural numbern,∑d|n f(d) = n2. (

∑d|n means sum over all positive divisors of n.) If φ is Euler’s

totient function, prove thatf(n)φ(n)

= n∏

p|n,p prime

(1 +

1p

),

for all natural numbers n > 1.

4. Let G be a subgroup of Sn with the property that for all 1 ≤ i ≤ n and1 ≤ j ≤ n, there exists σ ∈ G such that σ(i) = j. Prove that for all 1 ≤ k ≤ n,Gk ∩ Z(G) = e where Gk =

τ ∈ G : τ(k) = k

, Z(G) denotes the center of G,

and e is the identity element of G.

5. Let A be the set of all vectors of n-tuples whose entries are zero or one in sucha way that in each vector the numbers of one entries is odd. By explaining yourreasoning, determine, the maximal number of vectors from the set A such thatevery two of which share an even number of one entries.

6. Let (P,≤) be a poset (reflexive, antisymmetric, and transitive) with n elements.For every element i ∈ P , set Ui =

j : j ∈ P, j > i

and Li =

j : j ∈ P, j < i

.

Suppose that relative to every element i of P , there corresponds a real number Xi

subject to the following

Xi =

1−P

j∈LiXj

n−|Ui| if Li 6= ∅,1

n−|Ui| if Li = ∅.Prove that 0 ≤ Xi ≤ 1 for all i ∈ P .

1.28.2. Second Day. 1. Let, in Figure 4, the function f be one-to-one andcontinuous and that for every point P on the curve y = 2x2, the areas of the regionsA and B are equal to one another. Determine the function f explicitly.

2. Let f : [a, b] → (a, b) be a continuous function. Prove that for every naturalnumber n, there exist a positive number α and c ∈ (a, b) such that

f(c) + f(c+ α) + · · ·+ f(c+ nα) = (n+ 1)(c+n

2α).

13 Prove that f−1(0, 0) has infinitely many elements!

44 1. PROBLEMS

1.5

BA

P

Figure 4

3. A matrix O is called orthogonal if OOt = OtO = I, where Ot denotes thetranspose of the matrix O . Let A and B be two orthogonal square matrices withentries in C and det(A) + det(B) = 0. Can one conclude that det(A + B) = 0?Why?

4. Suppose that R is a unital ring and that there is a natural number n with theproperty that for all k ∈ n, n+ 1, n+ 2 and for all x, y ∈ R, (xy)k = xkyk. Provethat R is commutative.

5. Of three people who are suspected to have committed a murder, one is themurderer. The three people are to be interrogated by taking a test which hasfive questions. If the person being tested is innocent, the probability that theperson responds positively to any question is 0.4; and if guilty, the probability ofresponding positively is 0.8. Of these three people, one who is chosen at random isbeing interrogated. This person responds positively to four questions and negativelyto one question. What is the probability that the person is the murderer?

6. Let X be a set and r a natural number. Let Xr be the set of all subsets of Xwhich have r elements. Suppose that F is a subset of Xr with the property thatthe intersection of every k elements of F is nonempty (k ≥ 2 is a fixed number). Ifwe set

I(F ) = min|T | : T ⊆ X,T ∩A 6= ∅,∀A ∈ F

,

prove that I(F ) ≤ r−1k−1 + 1.

1.29. TWENTY NINTH COMPETITION 45

1.29. Twenty Ninth Competition, University of Mazandaran inBabolsar, May 2005

1.29.1. First Day. 1. Let f : [0, a] → R be a continuous and positive func-tion. Prove that (∫ a

0

f(x)dx)(∫ a

0

dx

f(x)

)≥ a2.

2. Let (ni)+∞i=1 be an increasing sequence (not necessarily strictly) of natural num-

bers with n1 ≥ 2 such that the series+∞∑i=1

1n1 · · ·ni

converges to a real number x.

Prove that x is rational if and only if there exists a natural number ` such thatni = n` for all i ≥ `.

3. Consider the sequence of all natural numbers whose digits consist of 1.

1, 11, 111, . . .

Prove that if the natural number m is relatively prime with respect to 30, theninfinitely many terms of the above sequence are divisible by m.

4. Let R be an arbitrary ring (not necessarily unital) that has no nonzero nilpotenttwo-sided ideal. Prove that every nonzero right ideal in R has an element whosesquare is not zero.

5. Let Z, E, and O be the set of integers, even integers, and odd integers, respec-tively. Set

X :=A ∈ P(Z) : both A ∩ E, A ∩O are infinite

,

Y :=A ∈ P(Z) : A is infinite

.

We know that there exists a one-to-one correspondence from X into Y . Exhibit alaw for a surjective function f : X → Y .

6. Let S be the k-dimensional vector space of all binary (zero and one) sequencesof length n over the field Z2. The distance between two elements X,Y of S areto be defined as the number of entries of X and Y that are different from oneanother. (More precisely, if X = (x1, . . . , xn) and Y = (y1, . . . , yn), then thedistance between X and Y is equal to the number of i’s for which xi 6= yi.) Supposethat the minimum distance between two distinct elements of S is equal to d. Provethat

d ≤ n2k−1

2k − 1.

46 1. PROBLEMS

1.29.2. Second Day. 1. Let D =z ∈ C : |z| < 1

and f : D → C be an

analytic function such that f( 1n ) ∈ R for all natural numbers n ≥ 2. Prove that for

all natural numbers n, f (n)(0) ∈ R, where f (n) denotes the nth derivative of thefunction f .

2. (i) Prove that if X is a connected metric space, then for all ε > 0 and every twopoints x, y ∈ X, there exist an n ∈ N and points x1, . . . , xn ∈ X such that x1 = x ,xn = y and that d(xi, xi+1) < ε for all i < n.

(ii) Give an example showing that the converse of the assertion of (i) does nothold.

(iii) Prove that the converse of the assertion of (i) holds provided that X iscompact.

3. Let G be a group and K a subgroup of it.

(i) Prove thatNG(K)CG(K)

is isomorphic to a subgroup of Aut(K).

(ii) Prove that if K is cyclic and K EG = G′, then K ≤ Z(G).

4. Let F be a field, Mn(F ) the set of all n × n matrices with entries in F ,A ∈ Mn(F ), and the invertible matrix P ∈ Mn(F ) be such that P−1AP is up-per triangular. Prove that every two invariant subspaces of A are comparable withrespect to inclusion if and only if there exist a λ ∈ F and a nilpotent matrixN ∈Mn(F ) with Nn−1 6= 0 such that A = λI +N .

5. Two people, named A and B, are playing a coin-flipping game as follows. Theythrow their coins, if the result of flipping the two coins turns out to be the same,A is going to win both coins; if not, B is going to take both coins. Suppose that Ahas m coins and B has n coins. On average, how many times should the game beplayed till one of the players runs out of the coins?

6. Let C be the Cantor set. Prove that C − C = [−1, 1]. (Hint: it is worthmentioning that C − C =

x − y : x, y ∈ C

and that C is equal to the set of all

numbers in [0, 1] whose ternary expansions have only 0 or 2.)

1.30. Thirtieth Competition, Tafresh University, May 2006

1.30.1. First Day. 1. Let f : [0,+∞) −→ R be a continuous function andlimx→+∞ f(x) = 1. Evaluate the following limit.

limn→+∞

∫ 2006

1385

f(nx)dx.

2. Let m ∈ N, c ∈ C, aj ∈ C, and |aj | = 1 for all 1 ≤ j ≤ m. If

limn→+∞

m∑j=1

anj = c,

then c = m and aj = 1 for all 1 ≤ j ≤ m.

1.30. THIRTIETH COMPETITION 47

3. Let R be a commutative ring with identity and a an element of R with a3−a−1 =0. Prove that if J is an ideal of R such that the quotient ring R/J has at most fourelements, then J = R.

4. Let p, q be prime numbers such that p = 2q + 1 and q ≡ 1 (mod 4). Prove that2 is a primitive root modulo p.

5. Prove that ∑|k|<

√m

(2mm+ k

)≥ 22m−1,

for all m ∈ N with m ≥ 1Hint. By the Chebyshev Inequality if X is a random variable whose mean and

variance are µ and σ2, respectively, then P (|X − µ| ≥ λσ) ≤ 1λ2 for all λ > 0.

6. A financial group has n members each of whom has a number of coins. Letk ∈ N be a fixed positive integer. A business among four people, say, p1, p2, p3,and p4, of the n people, who are chosen arbitrarily, can be done subject to thefollowing conditions:

(a)(sum of the number of coins owned by p3 and p4

)−(sum of the number

of coins owned by p1 and p2

)+ 2k > 0,

(b) each of p1 and p2 has at least k coins.If the conditions (a) and (b) are met, then a business is done as follows. Each

of p3 and p4 earns exactly k coins which are lost by each of p1 and p2. Prove thatafter a finite number of doing business, the condition (a) or (b) in the above doesnot hold for any four people of the n people.

1.30.2. Second Day. 1. Let X be a separable metric space, that is, X hasa countable dense subset. Let f : X −→ R be a function for which limx→a f(x)exists for all a ∈ X. Prove that the set of points at which f is discontinuous is atmost countable.

2. Suppose that f(z) =∑+∞n=0 anz

n defines a nonconstant analytic function, wherethe radius of the convergence of the series is equal to R > 0. Prove that the closestdistance from a zero of f to the origin is at least R|a0|

M+|a0| , where M = M(R) =sup|z|=R |f(z)|.

3. Let G be a group with the property that the order of any element of G′, thederived subgroup of G, is finite. Prove that the set of all elements of G whose ordersare finite is a subgroup of G.

4. LetK be a field, F a subfield ofK, n ∈ N, and A an F -algebraic n×nmatrix withentries in K such that rank(A) = rank(A2). Prove, firstly, that Kn = imA⊕ kerAand, secondly, that there is a polynomial f ∈ F [x] such that E := f(A) is anidempotent matrix satisfying E(x+ y) = x for all x ∈ imA and y ∈ kerA.

5. For an arbitrary subset C of the natural numbers, set C ⊕ C :=x + y|x, y ∈

C, x 6= y. Prove that there is a unique partition of the set of natural numbers into

two subsets A and B such that none of A⊕A and B ⊕B contains a prime.

48 1. PROBLEMS

Hint. Recall that by Bertrand’s conjecture (a.k.a. Bertrand’s principle) forevery natural number n there exists at least a prime number p such that n < p ≤ 2n.

6. Let C be a circle whose perimeter is equal to one and 0 < α < 12 . The distance

between two points of the circle is defined to be the length of the shortest arc joiningthe two points. Let T ⊆ C and T = I1∪· · ·∪Im, where m ∈ N and Ij ’s are disjointsarcs of the circle (1 ≤ j ≤ m). Prove that if the distance between every two pointsof T is less than or equal to α, then

m∑j=1

`(Ij) ≤ α,

where `(Ij) denotes the length of the arc Ij .

1.31. Thirty First Competition, Ferdowsi University of Mashhad, May2007

1.31.1. First Day. 1. We have colored a circle by two colors. Does therenecessarily exist a monocolored triangle inscribed in the circle which is

(a) an equilateral triangle?(b) a right triangle?(c) an isosceles triangle?

2. The Minkowski sum of two sets A,B ⊆ Rd is defined as follows

A+B =a+ b ∈ Rd|a ∈ A, b ∈ B

.

Prove that if A is bounded and B is closed, then

(A+B)′ = (A′ +B) ∪ (A+B′),

where A′ stands for the set of the limit points of the set A.

3. Suppose that there is a group G which has exactly n subgroups of index 2 (nis a natural number.). Prove that there exists a finite abelian group which hasexactly n subgroups of index 2.

4. Can one find two biased dice in such a way that the probability of getting a sumof j, for all 2 ≤ j ≤ 12, for the two simultaneously thrown dice is a number in theinterval

(233 ,

433

)?

5. Show that R2 has a dense subset whose no three points are collinear.

6. Let A be an n× n matrix with real entries. Prove that

det(A) =1n!

det

tr(A) 1 0 · · · · · · 0tr(A2) tr(A) 2 0 · · · 0

tr(A3) tr(A2) tr(A) 3. . .

......

......

. . . . . . 0tr(An−1) tr(An−2) · · · · · · tr(A) n− 1tr(An) tr(An−1) · · · · · · tr(A2) tr(A)

.

1.31. THIRTY FIRST COMPETITION 49

1.31.2. Second Day. 1. Let n be an odd natural number. Prove that thereare n consecutive natural numbers whose sum is a complete square. In this spirit,do there exist twelve natural numbers whose sum is a complete square?

2. For what values of the nonzero reals α and β the following limit exists (and isfinite)?

limx,y→0+

x2αy2β

x3α + y3β

3. Let A be a nonvoid set and An be the set of all ordered n-tuples of the elementsof A. For α = (α1, . . . , αn) and β = (β1, . . . , βn) in An, defined(α, β) = the number of corresponding components of α and β that differ with

one another.For two arbitrary elements x and y in An, prove that there is a one-to-one corre-spondence between the following two sets

C =z ∈ An : d(x, z) < d(y, z)

,

D =z ∈ An : d(y, z) < d(x, z)

.

4. Let R be a commutative ring with identity. Prove that the ring R[x] has infinitelymany maximal ideals.

5. Let x1, . . . , x2n be real numbers such that removing any of them the remainingones can be partitioned into two sets with equal sums (n ≥ 2). Prove that xi’s areall zero.

6. Let T be the union of all line segments with one end at M = (0, 1) and the otherend at a point on the x-axis having rational coordinates. In other words,

T =(tq, 1− t) ∈ R2|t ∈ [0, 1], q ∈ Q

.

(a) Assume that A,B ∈ T such that A, B, and M are not collinear. Showthat any continuous path from the point A to the point B in the set T has to passthrough the point M (by the aforementioned path, we mean a continuous functionγ : [0, 1] → T with γ(0) = A and γ(1) = B.).

(b) Prove that every continuous function from T into T has at least a fixedpoint.

CHAPTER 2

Solutions

2.1. First Competition

2.1.1. Analysis. 1. (a) For the given function f , let fn : R → R be definedby fn(x) = n

(f(x+ 1

n )− f(x)), where n ∈ N. It is plain that the sequence (fn)+∞n=1

converges pointwise to f ′ on R, finishing the proof of this part.

(b) First, we recall the following which is known as Darboux’s Theorem.

Darboux’s Theorem. The derivative of any differentiable function on aclosed interval has the intermediate value property.

Proof. Suppose that f : [a, b] −→ R is a differentiable function, f ′(a) 6= f ′(b),and λ ∈ R is between f ′(a) and f ′(b). We need to show that there exists a c ∈ (a, b)such that f ′(c) = λ. To this end, if necessary replacing f by −f , without loss ofgenerality, we may assume that f ′(a) < f ′(b). There are two cases to consider.

(i) f ′(a) < λ ≤ f(b)−f(a)b−a .

Define the function g : [a, b] −→ R by

g(x) =

f ′(a) x = a,f(x)−f(a)

x−a a < x ≤ b.

It is readily seen that g is continuous on [a, b]. We have g(a) < λ ≤ g(b). It thusfollows from the Intermediate Value Theorem that there exists an x0 ∈ (a, b] suchthat g(x0) = f(x0)−f(a)

x0−a = λ. Therefore, by the Mean Value Theorem, λ = f ′(c) forsome c ∈ (a, b), which is what we want.

(i) f(b)−f(a)b−a < λ < f ′(b).

In this case, defining the function g : [a, b] −→ R by

g(x) =

f(b)−f(x)b−x a ≤ x < b,

f ′(b) x = b,

the assertion follows in a similar fashion.

Define the function f : R → R by

f(x) =

0 x ≤ 0,1 x > 0.

51

52 2. SOLUTIONS

The function f is a Baire function because the sequence (fn)+∞n=1 of continuousfunctions, where fn : R → R is defined by

fn(x) =

0 x ≤ 0,nx 0 < x < 1

n ,1 1

n ≤ x,

converges to the function f pointwise. Yet, in view of Darboux’s Theorem , thefunction f is not the derivative of any function on R because f does not have theintermediate value property (a.k.a. Darboux’s property ).

2. (a) For X = (xij) ∈M, by definition, we have

det(X) =∑σ∈Sn

sgn(σ)x1σ1 · · ·xnσn,

where Sn denotes the symmetric group on 1, . . . , n and sgn(σ) stands for the signof the permutation σ ∈ Sn. Hence, the function det : M→ R is continuous, for fis a polynomial function in n2 variables xij where 1 ≤ i, j ≤ n. We have

U(M) = det−1(R \ 0

),

where U(M) denotes the set of invertible matrices in M. It follows that U(M) isan open subset of M because R \ 0 is an open subset of R.

(b) To see that M is disconnected, just note that

U(M) = det−1(R−) ∪ det−1

(R+)

where R− = (−∞, 0) and R+ = (0,+∞). Thus, in view of the proof of (a), theopen sets det−1

(R−) and det−1

(R+)

form a disconnectedness for U(M), finishingthe proof.

3. (a) We need the following lemma.

Lemma. Let N ∈ R+ and I a bounded interval of R. Then, in the interval Ithere are finitely many r ∈ Q whose denominators are less than or equal to N .

Proof. Without loss of generality assume that N is a natural number. Justnote that, by the Archimedean property of real numbers and the well-orderingprinciple of natural numbers, the set

⋃Nb=1

ab ∈ I : a ∈ Z is a finite set, proving

the assertion.

We claim that limx→a f(x) = 0 for all a ∈ R+. To see this, for a given ε > 0,first pick N ∈ N such that 1

N < ε. It follows from the lemma above that thereexists δ > 0 such that x = p

q ∈ Q with gcd(p, q) = 1 and 0 < |x− a| < δ imply thatq > N . So for this given ε > 0, find δ > 0 as in the above. If 0 < |x− a| < δ, thereare two cases to consider:

(i) x = pq ∈ Q with gcd(p, q) = 1. Noting that q > N , we can write

|f(x)− 0| = 1p+ q

<1q<

1N

< ε.

(ii) x /∈ Q. In this case, we have

|f(x)− 0| = 0 < ε.

From these two cases, we see that limx→a f(x) = 0, which is what we want.

2.1. FIRST COMPETITION 53

(b) In view of the claim we made in (a), it follows that f is continuous at theirrational points of (0,+∞) and is discontinuous at the rational points of (0,+∞).

2.1.2. Algebra. 1. (a) “if”: Suppose that I1, I2 C R with I1I2 ⊆ R andI1 * P . We show that I2 ⊆ P . Choosing an x0 ∈ I1 \ P and assuming that y ∈ I2is arbitrary, we have x0y ∈ I1I2 ⊆ P . Hence, y ∈ P , for x0 ∈ I1 \ P . As y wasarbitrary, we see that I2 ⊆ P , which is what we want.

“only if”: Let 〈a〉 and 〈b〉 denote the ideals generated by a and b in R, respec-tively. The ring R is commutative with identity element. So we have 〈a〉 = aR,〈b〉 = bR, and 〈a〉〈b〉 = 〈ab〉. Now from ab ∈ P , it follows that 〈a〉〈b〉 ⊆ P . But Pis prime. Therefore, 〈a〉 ⊆ P or 〈b〉 ⊆ P , implying that a ∈ P or b ∈ P , which iswhat we want.

(b) Let M be a maximal ideal in R. In view of (a), it suffices to show that Mis prime. To this end, suppose that ab ∈M , where a, b ∈ R. We need to show thata ∈M or b ∈M . If a /∈M , then M + 〈a〉 = R because M is maximal and a /∈M .In particular, we see that there are m ∈ M and x ∈ R such that 1 = m + ax,implying that b = bm+ abx ∈M , for M is an ideal and ab ∈M .

2. It follows from the hypothesis that g−1 = g for all g ∈ G. We can write

ab = (ab)−1 = b−1a−1 = ba,

for all a, b ∈ G. That is, G is abelian, which is what we want.

3. (3.1) In view of (c), we can write

cardE ≤ cardf(E) ≤ cardF = cardE.

This, together with the Schroder-Bernstein Theorem, implies that cardE = cardf(E).Now, suppose thatX and Y are two finite subsets of E such that cardX = cardf(X)and cardY = cardf(Y ). We can write

cardf(X ∪ Y ) = card(f(X) ∪ f(Y )

)= cardf(X) + cardf(Y )− card

(f(X) ∩ f(Y )

). (∗)

On the other hand,

card(X ∪ Y ) = cardX + cardY − card(X ∩ Y ). (∗∗)But card(X ∪Y ) ≤ cardf(X ∪Y ) = card

(f(X)∪ f(Y )

). So it follows from (∗) and

(∗∗) that card(f(X)∩ f(Y )

)≤ card(X ∩ Y ). As f(X ∩ Y ) ⊆ f(X)∩ f(Y ), we can

write

card(X ∩ Y ) ≤ cardf(X ∩ Y ) ≤ card(f(X) ∩ f(Y )

)≤ card(X ∩ Y ).

Therefore,cardf(X ∩ Y ) = card

(f(X) ∩ f(Y )

)= card(X ∩ Y ).

That is, X ∩ Y and f(X) ∩ f(Y ) are isomorphic sets. Now, since f(X ∩ Y ) ⊆f(X)∩ f(Y ) and these two sets are finite, it follows that f(X ∩Y ) = f(X)∩ f(Y ).This equality together with (∗) and (∗∗) implies that card(X ∪Y ) = cardf(X ∪Y ),finishing the proof of assertion in this part.

54 2. SOLUTIONS

(3.2) Set

T =card(A) : ∅ 6= A ⊆ E and card(A) = cardf(A)

.

It is obvious that the set T is a nonempty subset of N. It thus follows from thewell-ordering principle of N that the set T has an initial element, say, cardX0,where X0 is a nonempty normal subset of E. We claim that for any other normalsubset X of E, we have

X ∩X0 = ∅ or X0 ⊆ X.

Note that X0 ∩X ⊆ X0. If X0 ∩X = X0, we obtain X0 ⊆ X; if not, we show thatX0 ∩X = ∅, completing the proof. Suppose that X0 ∩X $ X0. Since X0 is finite,it follows that

card(X0 ∩X) < card(X0).

On the other hand, by (3.1), card(X0 ∩ X) = cardf(X0 ∩ X). We claim thatX0 ∩ X = ∅. Otherwise, we see that card(X0 ∩ X) ∈ T , yielding card(X0) ≤card(X0 ∩ X), for card(X0) is the initial element of T . This is a contradiction.Hence, X0 ∩X = ∅ and we are done.

2.1.3. General. 1. From a < b < c, we see thata

b< 1 ,

b

c< 1 and 1 <

c

a.

Thus,

max(ab,b

c,c

a

)=c

a, min

(ab,b

c,c

a

)6= c

a.

There are two cases to consider.(i) min

(ab,b

c,c

a

)=a

band (ii) min

(ab,b

c,c

a

)=b

c.

In both cases, in view of the above inequalities, it is easily seen that S > 1.

2. We show that the decimal fraction a = 0.123456789101112131415 . . . cannot beperiodic after any digit. To prove this by contradiction, assume that the decimalfraction a is periodic after its first k digits and that its period is l. Obviously, thenumber 10k+l occurs somewhere after the kth digit of the decimal fraction a. As10k+l has k + l zeros in its decimal expansion and that the period of a after thekth digit is assumed to be l, it follows that the digits of a after the kth digit are allzero, which is obviously impossible, settling the proof.

3. We need to solve the following system of equations in Cx = y3

y = x3 , x 6= y.

It is plain that x, y 6= −1, 0, 1, for otherwise x = y ∈ −1, 0, 1. By substitution,we obtain

x9 = x, x 6= 0 =⇒ x8 = 1 =⇒ x = ωj , 0 < j < 7, j 6= 4,

where ω = cos π4 + i sin π4 . It is now easily seen that

x = ωj

y = ω3j , 0 < j < 7, j 6= 4,

is the general solution of the above system of equations.

2.1. FIRST COMPETITION 55

4. Using the AM-GM Inequality, we can write

xx + yy ≥ 2√xxyy = 2ef(x),

where f : (0, 1) → R is defined by f(x) = x lnx+(1−x) ln(1−x). We have f ′(x) =− ln( 1

x − 1). Thus, f ′(x) < 0 on (0, 12 ) and f ′(x) > 0 on ( 1

2 , 1), implying that fattains its only local minimum and hence its absolute minimum at 1

2 . Consequently,we have

xx + yy ≥ 2ef(x) ≥ 2ef( 12 ) =

√2,

for all x, y ∈ R+ with x + y = 1 and that the equality happens if and only ifx = y = 1

2 , which is what we want.

2.1.4. Differential Equations. We know that the general solution of a non-homogeneous linear equation is y = yc + yp, where yc is the general solution of thecorresponding homogeneous equation and yp is a particular solution of the nonho-mogeneous linear equation x′′ + a2x = f(t). In view of this, first of all, it is easilyseen that yc(t) = c1 cos at + c2 sin at (t ∈ R), where c1, c2 ∈ R, is the general solu-tion of the homogeneous equation. Using the method of variation of parameters, astraightforward calculation shows that the function yp defined by

yp(t) =1a

∫ t

t0

f(x) sin a(t− x)dx (t ∈ R),

where t0 ∈ R is a fixed arbitrary real number, is a particular solution of the non-homogeneous linear equation x′′ + a2x = f(t). Therefore,

y(t) = c1 cos at+ c2 sin at+1a

∫ t

t0

f(x) sin a(t− x)dx (t ∈ R),

is the general solution of the equation x′′ + a2x = f(t).

2.1.5. Probability and Statistics. We need to calculate P(∑10

i=1Xi = 10).

To do this, let, first of all, X be a Poisson random variable with parameter λ definedby its probability density function as follows

fX(i) = p(X = i) =

e−λλi

i! i ∈ N ∪ 0,0 otherwise.

Next, we find the moment generating function of X.

mX(t) = E(etX)

=∑Xi

etXifX(Xi) =+∞∑i=0

etifX(i)

= e−λ+∞∑i=0

(λet)i

i!= eλ(et−1).

56 2. SOLUTIONS

Now, since each Xi is a Poisson random variable with parameter µi and that Xi’sare independent random variables, letting Y =

∑10i=1Xi, we can write

mY (t) = E(etP10

i=1Xi)

= E( 10∏i=1

etXi)

=10∏i=1

E(etXi

)=

10∏i=1

mXi(t) = e(P10

i=1 µi)(et−1).

Therefore, Y =∑10i=1Xi is a Poisson random variable with parameter

∑10i=1 µi,

whence

P( 10∑i=1

Xi = 10)

= fY (10) =e−

P10i=1 µi(

∑10i=1 µi)

10

10!.

But µi = i. So, we have

P(X = 1

)= P

( 10∑i=1

Xi = 10)

=e−555510

10!.

2.1.6. Topology. To prove the assertion by contradiction, suppose that T isa topology on R. We must have⋃

q>√

2q∈Q

Aq = (√

2,+∞) ∈ T,

implying that√

2 ∈ Q, which is a contradiction, settling the proof.

2.2. Second Competition

2.2.1. Analysis. 1. (a) Since f is continuous and the interval [0, 1] is com-pact, there exists an x0 ∈ [0, 1] such that

M = sup0≤x≤1

f(x) = max0≤x≤1

f(x) = f(x0).

If M = 0, there is nothing to prove. If not, as f is continuous at x0, we see that forgiven Mε > 0, there exists a neighborhood of x0, say, (α, β), where 0 ≤ α < β ≤ 1,such that

|f(x)− f(x0)| < Mε

whenever α < x < β. Noting that f(x0) = M ≥ f(x) for all x ∈ [0, 1], in view ofthe above inequality, we obtain

M(1− ε) < f(x) ≤M,

for all x ∈ (α, β), which is what we want.

(b) For given 0 < ε < 1, find 0 ≤ α < β ≤ 1 from (a). We have 0 < M(1− ε) ≤f(x) ≤M on [α, β], for f is continuous on [0, 1]. We can write

Mn(1− ε)n(β − α) ≤∫ β

α

fn ≤∫ 1

0

fn ≤Mn.

2.2. SECOND COMPETITION 57

Taking nth root, we obtain

M(1− ε)(β − α)1n ≤

(∫ 1

0

fn) 1

n

≤M. (∗)

Now taking limn and limn of both sides of the left and right inequalities above, weobtain

M(1− ε) ≤ limn

(∫ 1

0

fn) 1

n

≤ limn

(∫ 1

0

fn) 1

n

≤M.

Thus, 0 ≤ limn

(∫ 1

0fn) 1

n − limn

(∫ 1

0fn) 1

n ≤ M −M(1 − ε) = Mε. As 0 < ε < 1is arbitrary, we conclude that

limn

(∫ 1

0

fn) 1

n

= limn

(∫ 1

0

fn) 1

n

.

That is, limn

(∫ 1

0fn) 1

n

exists. Letting n→ +∞ in (∗), we obtain

M(1− ε) ≤ limn

(∫ 1

0

fn) 1

n

≤M.

Again, 0 < ε < 1 being arbitrary, we see that

limn

(∫ 1

0

fn) 1

n

= M = f(x0) = sup0≤x≤1

f(x),

finishing the proof.

2. Note first that

sup0≤t≤1

∣∣1− f(t)∣∣ ≥ ∣∣1− f(0)

∣∣ = 1,∫ 1

0

∣∣1− f(t)∣∣dt ≥ 0,

and hence

sup0≤t≤1

∣∣1− f(t)∣∣+ ∫ 1

0

∣∣1− f(t)∣∣dt ≥ 1,

for all f ∈ F . This implies that

D = inff∈F

(sup

0≤t≤1

∣∣1− f(t)∣∣+ ∫ 1

0

∣∣1− f(t)∣∣dt) ≥ 1.

We claim that D = 1. To see this, just note that for the sequence (fn)+∞n=1, wherefn : [0, 1] → R (n ∈ N) is defined by

fn(t) =nt 0 ≤ t ≤ 1

n ,1 1

n < t ≤ 1,

we have

limn

(sup

0≤t≤1

∣∣1− fn(t)∣∣+ ∫ 1

0

∣∣1− fn(t)∣∣dt) = lim

n

(1 +

12n

)= 1.

58 2. SOLUTIONS

As for the geometrical interpretation of D, we view F as a subspace of thenormed space all continuous functions on [0, 1], denoted by C[0, 1], which is equippedwith the following norm

||f || := sup0≤t≤1

∣∣f(t)∣∣+ ∫ 1

0

∣∣f(t)∣∣dt.

If 1 : [0, 1] → R is defined by 1(x) = 1 for all x ∈ [0, 1], we would then have

d(1,F) = inff∈F

||1− f ||

= inff∈F

(sup

0≤t≤1

∣∣1− f(t)∣∣+ ∫ 1

0

∣∣1− f(t)∣∣dt) = D = 1.

That is, the distance from the vector 1 ∈ C[0, 1] to the subspace F of C[0, 1] is equalto D = 1.

3. It suffices to show that [0, 2] ⊆ C +C. To this end, let x ∈ [0, 2] be arbitrary. Ifx

2= 0.x1x2 . . .

denotes the ternary expansion of x2 where xi ∈ 0, 1, 2, then we can write

x

2= 0.a1a2 . . . + 0.b1b2 . . . ,

where ai, bi ∈ 0, 1 and xi = ai + bi for all i ∈ N. So we have

x = 2(0.a1a2 . . .) + 2(0.b1b2 . . .) = 0.a′1a′2 . . . + 0.b′1b

′2 . . . := a+ b,

where a′i = 2ai and b′i = 2bi, and hence a′i, b′i ∈ 0, 2 for all i ∈ N. Therefore,

a, b ∈ C because there is no one in their ternary expansions. This finishes the proof.

2.2.2. Algebra. 1. To prove the assertion we need the following generallemma.

Lemma. Let G be a finite group with the property that the equation xn = ehas at most n solutions in G for all n ∈ N, where e is the identity element of G.Then the group G is cyclic.

Remark. In case the group G is abelian, the lemma is a quick consequence ofthe Fundamental Theorem of finite abelian groups .

First proof. Let |G| = n and C be a cyclic group with n elements so thatC = 〈a〉, where ord(a) = n. Suppose that for a divisor d of n, there is an elementgd ∈ G such that ord(gd) = d. It follows that gd, g2

d, . . . , gd−1d , gdd = e are d solutions

of the equation xd = e. As d∣∣n = |G| = |C| and C is cyclic, there is an ad ∈ C such

that ord(ad) = d. Consequently, 〈gd〉 ∼= 〈ad〉, and hence C has at least as manyelements of order d as G has. But G and C both have n elements. Thus, G musthave the same number of elements of order d as C, for all d dividing n. It followsthat G is cyclic because C has an element of order n, namely the element a. So theproof is complete.

2.2. SECOND COMPETITION 59

Second proof. Define the relation ∼ on G as follows

a ∼ b⇐⇒ ord(a) = ord(b).

It is obvious that ∼ is an equivalence relation on G. Hence, the relation ∼ partitionsG into its equivalence classes. In other words, we can write

G =k⋃i=1

[gi],

where k ∈ N is less than |G|, g1, . . . , gk is a maximal set of nonequivalent elementsof G, and [gi] denotes the equivalence class containing the element gi.

Note that if G happens to be a cyclic group of order n so that G = 〈a〉 withord(a) = n, we can write

n =∣∣G∣∣ = k∑

i=1

∣∣[gi]∣∣,where gi’s (1 ≤ i ≤ k) are as in the above and gi = aji for some 1 ≤ ji ≤ n. Letdi = gcd(ji, n). It is easily checked that

ord(ak) =ord(a)

gcd(k, ord(a)

) =n

gcd(k, n)(∗)

for all 1 ≤ k ≤ n. In view of this, we see that di 6= di′ whenever i 6= i′ andthat ak ∈ [gi] if and only if n

gcd(k,n) = nd if and only if gcd(k, n) = d if and only

if gcd(kd ,nd ) = 1. Thus,

∣∣[gi]∣∣ = φ(nd ), where φ denotes Euler’s totient function .Consequently,

n =k∑i=1

∣∣[gi]∣∣ =∑d|n

φ(nd

)=∑d|n

φ(d).

Now back to a general finite group G, letting di = ord(gi), we claim that∣∣[gi]∣∣ = φ(di). That∣∣[gi]∣∣ ≥ φ(di) follows from the hypothesis that di = ord(gi) and

that, in view of (∗), the cyclic group 〈gi〉 has exactly φ(di) generators. On the otherhand,

∣∣[gi]∣∣ ≤ φ(di), for otherwise there must exist a g ∈ G \ 〈gi〉 with ord(g) = di,which, in turn, implies that the equation xdi = e has at least di + 1 solutions,namely, g, gi, g2

i , . . . , gdii = e, which is a contradiction. Thus,

∣∣[gi]∣∣ = φ(di), andhence

n =∣∣G∣∣ = k∑

i=1

φ(di) =∑d|n

φ(d).

This, in particular, implies that G has an element of order n. Therefore, G is acyclic group of order n, which is what we want.

Corollary. Let D be a division ring and G a finite abelian subgroup of themultiplicative group of D, i.e., D∗ = D \ 0. Then, G is a cyclic group. Inparticular, the multiplicative group of any finite field is cyclic.

By the corollary above, there exists an a ∈ F ∗ such that F ∗ = 〈a〉. Lettingb = am, we can write

S :=∑x∈F

xm =n−2∑i=0

(ai)m =n−2∑i=0

bi.

60 2. SOLUTIONS

If n − 1|m, then, as an−1 = 1, we have b = 1. This together with the aboveequality implies that

∑x∈F x

m = n − 1. If n − 1 - m, then bm 6= 1. So we haveS = bS = 1 + b+ · · · bn−1, yielding (b− 1)S = 0 which, in turn, implies that S = 0,proving the assertion.

2. (a) “⇐=” This is obvious.“=⇒” Assume that z is right and left quasi-regular. Thus, there exist z′, z′′ ∈ A

such thatz + z′ − z′z = z + z′′ − zz′′ = 0. (∗)

We can write

(z + z′ − z′z)z′′ = 0 =⇒ zz′′ + z′z′′ − z′(z + z′′) = 0 =⇒ zz′′ = z′z.

This together with (∗) implies that z′ = z′′. That is, z is quasi-regular, which iswhat we want.

(b) Proceed by contradiction. The identity element of A being right quasi-regular means there exists 1′ ∈ A such that 1 + 1′ − 1′ = 0, yielding 1 = 0, whichis impossible. Likewise, the identity element is not left quasi-regular either.

As pointed out in the footnote of this problem, from the element x being rightquasi-regular one can only conclude that 1 − x is right invertible. First, let x beright quasi-regular. So there exists x′ ∈ A such that x+x′−xx′ = 0. From this, weobtain (1−x)(1−x′) = 1. That is, 1−x is right invertible. In fact, as this argumentis reversible, in a similar fashion, one can prove that the element x is right (resp.left) quasi-regular if and only if 1− x is right (resp. left) invertible. We now showthat in general from x being right quasi-regular one cannot conclude that 1− x isinvertible. To this end, in view of the aforementioned comment, it suffices to showthat in general from x being right quasi-regular one cannot conclude that x is leftquasi-regular as well. To see this, let F be a field of characteristic zero and F [x]the ring of all polynomials with coefficients from the field F . View F [x] as a vectorspace over F and use A to denote L(F [x]), the ring of all linear transformationsfrom F [x] into F [x]. Let I denote the identity transformation, D the differentiationlinear transformation, i.e. D(f0 + f1x+ · · ·+ fnx

n) = f1 + 2f2x+ · · ·+ nfnxn−1,

and I1 and I2 denote the linear transformations defined by

I1(f0 + f1x+ · · ·+ fnxn) = 1 + f0x+

f12x2 + · · ·+ fn

n+ 1xn+1,

I2(f0 + f1x+ · · ·+ fnxn) = 2 + f0x+

f12x2 + · · ·+ fn

n+ 1xn+1.

Clearly, DI1 = DI2 = I. That is, D has two right inverses, and hence D isnot invertible. This implies that I − D is right quasi-regular but it is not leftquasi-regular. Because otherwise I − (I −D) = D will become invertible, which isimpossible.

3. (a) Since D divides P , there exists Q ∈ K[x] such that P = QD. We concludethat φ(P ) = φ(Q)φ(D) because φ is a ring homomorphism. It is now obvious thatkerφ(D) ⊆ kerφ(P ), which is what we want.

(b) Suppose that D = gcd(P,R), where P,R ∈ K[x]. Since K[x] is a Euclideanring, it follows that there exist M,N ∈ K[x] such that D = MP + NR. As φ

2.2. SECOND COMPETITION 61

is a ring homomorphism, we conclude that φ(D) = φ(M)φ(P ) + φ(N)φ(R). Thisimplies kerφ(R) ∩ kerφ(P ) ⊆ kerφ(D). On the other hand, since D = gcd(P,R),in view of (a), we see that kerφ(D) ⊆ kerφ(P ) and kerφ(D) ⊆ kerφ(R), yieldingkerφ(D) ⊆ kerφ(R)∩kerφ(P ). Therefore, kerφ(D) = kerφ(R)∩kerφ(P ), finishingthe proof.

2.2.3. General. 1. If the first day up to the seventh day of the first monthof the Persian calendar, called “Farvardin”, corresponds to 1 up to 7, respectively,then the thirteenth day of the first month corresponds to 6. From this convention,we will obtain the 12-tuple

(6, 2, 5, 1, 4, 7, 3, 5, 7, 2, 4, 6)

whose ith component is corresponded to the thirteenth day of the month i of thePersian calendar (1 ≤ i ≤ 12). For instance, the number 1 being the fourthcomponent of the above 12-tuple means that the thirteenth day of the fourth monthof the Persian calendar occurs on the same day of the week as does the first day ofthe first month of the calendar. For 1 ≤ i ≤ 7, let n(i) denote the number of timesthat the number i occurs in the above 12-tuple. Obviously,

n(i) =

1 i ∈ 1, 3,2 i /∈ 1, 3.

As the first day of “Farvardin” can occur on any day of the week, we seethat “Wednesday” can correspond to any of the numbers 1, 2, . . . , 7. Therefore,in the Persian calendar, the minimum and the maximum number of “Wednesdaysoccurring on the 13th of the month” is 1 and 2, respectively.

2. Recall that a real function f on an interval I is called convex if

f(λx+ (1− λ)y

)≤ λf(x) + (1− λ)f(y),

for all x, y ∈ I and 0 ≤ λ ≤ 1. It is well-known that a function f : I → R which istwice differentiable is convex if and only if f ′′ ≥ 0 on I. From this, it follows thatthe function gn : (0,+∞) → R defined by gn(x) = xn, where n ∈ R+, is convexif and only if n ≥ 1. In particular, gn is convex whenever n ∈ N. This yieldsgn(x+y

2

)≤ gn(x)+gn(y)

2 for all n ∈ N and x, y ∈ R+. So, we can write(x+ y + z + t

4

)n= gn

(x+ y + z + t

4)≤ 1

2gn(x+ y

2)

+12gn(z + t

2)

≤ gn(x) + gn(y) + gn(z) + gn(t)4

=xn + yn + zn + tn

4,

for all n ∈ N and x, y, z, t ∈ R+, which is what we want.

3. We prove the following proposition of which the assertion is a quick consequence.Note that if the particle were moving along a straight line, then the tangentialacceleration of the motion would be equal to the acceleration of the motion alongthe straight line.

A particle moving along a curve goes s0 unit(s) of distance in t0 unit(s) oftime in such a way that the instantaneous speed of it at the initial point as well

62 2. SOLUTIONS

as the final point is zero. Prove that the absolute value of the particle tangentialacceleration at some point on its path is greater than or equal to 4s0

t20

unit of distance(unit of time)2 .

Assume that the initial and final moments of the motion are 0 and t0, respec-tively. If the equation of the motion is given by a vector function x : [0, t0] → R3,it follows that x is twice differentiable, x′(0) = x′(t0) = 0, and s(0) = 0, s(t0) = s0,where x′ = dx

dt denotes the instantaneous velocity vector and s(t) =∫ t0||x′(τ)||dτ

the length of the path at time t. It is plain that s′(0) = s′(t0) = 0. Also, recallthat the tangential acceleration of the motion, denoted by at, is, by definition, thetime derivative of the instantaneous speed, i.e., at = dv

dt = d2sdt2 = s′′(t). As s is

continuous, we see from the Intermediate Value Theorem that there is a moment0 < c < t0 at which we have s(c) = s(t0)+s(0)

2 = s(t0)2 . There are two cases to

consider.(i) 0 < c < t0

2 .In this case, using the extension of the Mean Value Theorem for second order

derivatives, we obtain a moment t1 with 0 < t1 < c such that

s(c) = s(0) + s′(0)(c− 0) + s′′(t1)(c− 0)2

2!= s′′(t1)

c2

2,

from which, we get s(t0)2 = s′′(t1) c

2

2 , and hence s′′(t1) = s(t0)c2 . This together

with the hypothesis that 0 < c < t02 implies that s′′(t1) ≥ 4s(t0)

t20> 0, yielding

|s′′(t1)| ≥ 4s(t0)t20

= 4s0t20

, which is what we want.

(ii) t02 ≤ c < t0.

Again using the extension of the Mean Value Theorem for second derivatives,we obtain a moment t1 with t0

2 < t1 < t0 such that

s(c) = s(t0) + s′(t0)(c− t0) + s′′(t1)(c− t0)2

2!,

from which, we obtain s(t0)2 = s(t0)+ s′′(t1)

(c−t0)22 , and hence s′′(t1) = −s(t0)

(c−t0)2 < 0.

This together with t02 ≤ c < t0 implies that |s′′(t1)| = s(t0)

(c−t0)2 ≥4s(t0)t20

= 4s0t20

, whichis what we want.

4. Since the digits can be selected independent of the letter, using the product ruleof combinatorics, it is obvious that the total number of the car plates that containthe letter “dal” is equal to

9× 9× 9× 9× 9× 1 = 59049.

2.2.4. Probability. (a) We can write

cov(U, V ) = E(UV )− E(U)E(V )= E

((X + Y )(X − Y )

)− E(X + Y )E(X − Y )

= E(X2 − Y 2)−((E(X))2 − (E(Y ))2

)= E(X2)− (E(X))2 −

(E(Y 2)− (E(Y ))2

)= var(X)− var(Y ) = σ2 − σ2 = 0.

2.2. SECOND COMPETITION 63

Therefore, cov(U, V ) = 0.Note: As is obvious now the condition cov(X,Y ) = λ was redundant.

(b) Not necessarily. Suppose that U, V are two random variables with cov(U, V ) =0, which are not independent. If we let X = 1

2 (U + V ) and Y = 12 (U − V ), we

would have

var(X) =14(var(U) + var(V ) + 2cov(U, V )

)=

14(var(U) + var(V )

),

var(Y ) =14(var(U) + var(V )− 2cov(U, V )

)=

14(var(U) + var(V )

).

Therefore, var(X) = var(Y ) = 14

(var(U) + var(V )

)= σ2. Note that the random

variables X + Y = U and X − Y = V are not independent by our hypothesis. Togive a concrete example, let U and V be two random variables defined by theirjoint distribution table as follows.

U V 1 2 31 0.1 0.1 0.1 0.32 0.1 0.1 0.1 0.43 0.1 0.1 0.1 0.3

0.3 0.3 0.4

It is easily verified that

UV 1 2 3 4 6 90.1 0.2 0.2 0.1 0.3 0.1

From the above table, we obtain E(UV ) = 4.2. On the other hand, we haveE(U) = 2 and E(V ) = 2.1, implying that E(UV ) = E(U)E(V ), and hencecov(U, V ) = 0. But P (U = 1, V = 1) = 0.1 and P (U = 1) = P (V = 1) = 0.3,yielding P (U = 1, V = 1) 6= P (U = 1).P (V = 1). In other words, U and V are notindependent, which is what we wanted.

2.2.5. Topology. Proceed by contradiction. As A ⊆ B∪C, we have A = (A∩B)∪(A∩C). Also, by the contradiction hypothesis, (A∩B)∩(A∩C) = A∩B∩C = ∅.But A ∩ B 6= ∅ and A ∩ C 6= ∅. Hence, it suffices to show that A ∩ B and A ∩ Care closed sets in the induced topology of A, for A would then be disconnected, acontradiction. To this end, note that B and C are closed in (K,TK). Thus, thereexist closed sets B1 and C1 in (E, T ) such that B = B1 ∩K and C = C1 ∩K, fromwhich, we obtain

A ∩B = A ∩B1, A ∩ C = A ∩ C1.

Now as B1 and C1 are closed in E, it follows that A ∩ B and A ∩ C are closed inA, which is what we want.

64 2. SOLUTIONS

2.2.6. Differential Equations. Two cases to consider.(i) |x| > 1.Rewrite the equation as

(y3)′ +x

x2 − 1y3 =

(x2 − 1)(x+ 1)x2 − 1

= x+ 1.

Multiplying both sides of the equation by

µ(x) = exp(∫

xdx

(x2 − 1)

)=√|x2 − 1| =

√x2 − 1,

we obtaind

dx

(y3√x2 − 1

)= (x+ 1)

√x2 − 1,

from which, we get

y3√x2 − 1 =

∫x√x2 − 1dx+

∫ √x2 − 1dx.

Using integration by parts and dividing both sides of the above equality by√x2 − 1,

we see that the general solution of the equation is

y3 =13(x2 − 1) +

x

2+

12√x2 − 1

ln |x+√x2 − 1|+ c√

x2 − 1,

where c is an arbitrary fixed constant.(ii) |x| < 1.In this case, note that |x2 − 1| = 1 − x2 yields µ(x) =

√1− x2. From this, a

similar argument as in (i) shows that the general solution of the equation is

y3 =−13

(x2 − 1) +x

2+

12√

1− x2arcsinx+

c√1− x2

,

where c is an arbitrary fixed constant.

2.3. Third Competition

2.3.1. Analysis. 1. Remark. It is worth mentioning that the assertion holdsunder the weaker hypothesis that f ′′ is nonnegative on (a, b) or f ′ is monotonic on[a, b]. Also in the statement of the problem “ξ ∈ [a, b]” must read “ξ ∈ (a, b)” andthat “ξ ∈ (a, b)” cannot be weakened, because for the function f : [0, 1] → R withf(x) = x2 and ξ ∈ 0, 1, there is no x0 ∈ (0, 1) such that

f ′(ξ) =f(0)− f(x0)

0− x0or f ′(ξ) =

f(1)− f(x0)1− x0

.

Since f ′′ > 0 on [a, b], it follows that f ′ is strictly increasing on [a, b]. By theMean Value Theorem, there exists a < ξ0 < b such that

f ′(ξ0) =f(b)− f(a)

b− a.

Now, if f ′(ξ) = f ′(ξ0), there is nothing to prove. If not, there are two cases toconsider.

2.3. THIRD COMPETITION 65

(i)

f ′(ξ) < f ′(ξ0) =f(b)− f(a)

b− a.

Define the function g : [a, b] → R by

g(x) =

f ′(a) x = a,f(a)−f(x)

a−x a < x ≤ b.

It is plain that

limx→a+

g(x) = limx→a+

f(a)− f(x)a− x

= f ′(a) = g(a).

Hence, g is continuous at a from the right, and therefore it is continuous on [a, b].As f ′ is strictly increasing on [a, b], we obtain f ′(a) < f ′(ξ). So we can write

g(a) = f ′(a) < f ′(ξ) < f ′(ξ0) = g(b).

Thus, it follows from the Intermediate Value Theorem that there exists an a <x0 < b such that

f ′(ξ) = g(x0) =f(a)− f(x0)

a− x0,

which is what we want.(ii)

f ′(ξ) > f ′(ξ0) =f(b)− f(a)

b− a.

The proof, which is omitted, is almost identical to that of (i) except that weneed to define the function g : [a, b] → R by

g(x) =

f(b)−f(x)b−x a ≤ x < b,

f ′(b) x = b.

2. (a) Set B =(Q ∩ [0, 1]

)\

1n : n ∈ N

. Plainly, we can write B = bn : n ∈ N.

Define the sequence (an)+∞n=1 by

an =

1n n odd,bn n even.

We claim that for this sequence (an)+∞n=1, there is no continuous function f satisfyingf(an) = an+1 for all n ∈ N. Suppose to the contrary that f : [0, 1] → [0, 1] is acontinuous function such that f(an) = an+1 for all n ∈ N. By proving that f = 0on (0, 1], and hence on [0, 1], we obtain a contradiction, proving the claim. For agiven x ∈ (0, 1], pick a subsequence (ani

)+∞i=1 such that limi→+∞ ani= x. As x 6= 0,

if necessary by passing to a subsequence, we may assume that ani /∈

1n : n ∈ N

for all i ∈ N. This implies that ni’s are all even numbers, from which, we see that(ni + 1)’s are all odd numbers and hence

f(x) = limi→+∞

f(ani) = lim

i→+∞ani+1 = lim

i→+∞

1ni + 1

= 0,

as desired.

(b) We sketch the proof for the case in which the sequence (an)+∞n=1 is increasing.In case the sequence is decreasing, the proof can be carried out in a similar fashion.

66 2. SOLUTIONS

Without loss of generality, if necessary by redefining the sequence (an)+∞n=1, we mayassume that a1 = 0. Let a∞ = limn→+∞ an. Define the function f on [0, 1] by

f(x) = an+2−an+1

an+1−an(x− an) + an+1 an ≤ x < an+1, n ∈ N,x a∞ ≤ x ≤ 1.

It is not difficult to see that f : [0, 1] → [0, 1] is continuous and that f(an) = an+1

for all n ∈ N, as desired.

3. (a) We need the following proposition.

Proposition. Let the complex number z be a root of the following equationwith complex coefficients

xp + c1xp−1 + · · ·+ cp−1x+ cp = 0,

where p ∈ N. If λk > 0 for all 1 ≤ k ≤ p and∑pk=1

1λk≤ 1, then

|z| ≤ max1≤k≤p

k√λk|ck|.

Under the hypothesis of the proposition, taking λk = 2k for all 1 ≤ k ≤ p, weobtain

|z| ≤ 2 max1≤k≤p

k√|ck|,

proving the assertion.

In order to prove the proposition, we need the following lemma.

Lemma. Let the complex numbers z1, . . . , zp (p ∈ N) be the roots of the fol-lowing equation with complex coefficients

xp + c1xp−1 + · · ·+ cp−1x+ cp = 0.

Set r0 = max1≤i≤p |zi|. If r > 0 and

rp > |c1|rp−1 + · · ·+ |cp|,

then r0 < r.

First proof. Let z ∈ C with |z| ≥ r be arbitrary. We can write∣∣∣∣f(z)z

∣∣∣∣ =∣∣∣1 +

c1z

+ · · ·+ cpzp

∣∣∣≥ 1−

∣∣∣c1z

∣∣∣− · · · − ∣∣∣ cpzp

∣∣∣≥ 1− |c1|

r− · · · − |cp|

rp> 0.

Thus, f(z) 6= 0, and hence r0 < r, as desired.

2.3. THIRD COMPETITION 67

Second proof. Let k(x) := xp − |c1|xp−1 − · · · − |cp|. By the hypothesis,k(r) > 0, which is equivalent to k(r)

rp = 1 −∑pi=1

|ci|ri > 0. Thus, for all x ≥ r, we

have

1−p∑i=1

|ci|xi

≥ 1−p∑i=1

|ci|ri

> 0.

Consequently,

k′(x) = (xp)′(k(x)xp

)+ xp

(k(x)xp

)′= pxp−1

(1−

p∑i=1

|ci|xi)

+ xpp∑i=1

i|ci|xi+1

> 0,

for all x ≥ r. Thus, k is strictly increasing on the interval [r,+∞). Now, let z ∈ Cbe such that |z| ≥ r. It follows that k(|z|) ≥ k(r) > 0. So, we can write

|zp + c1zp−1 + · · ·+ cp| ≥ |z|p − |c1||z|p−1 − · · · − |cp|

= k(|z|) ≥ k(r) > 0.

That is, zp+c1zp−1+· · ·+cp 6= 0 whenever |z| ≥ r. This yields r0 = max1≤i≤p |zi| <r, which is what we want.

Third proof. Define f(z) = zp + c1zp−1 + · · · + cp−1z + cp, g(z) = zp, and

h(z) = c1zp−1 + · · ·+ cp = f(z)− g(z) on C. Note that f(z) 6= 0 for all z ∈ C with

|z| = r, for otherwise

rp =∣∣zp∣∣ = ∣∣− c1z

p−1 − · · · − cp∣∣ ≤ ∣∣c1∣∣∣∣z∣∣p−1 + · · ·+

∣∣cp∣∣,from which, we obtain rp ≤ |c1|rp−1 + · · ·+ |cp|, which is a contradiction. For all zon the circle |z| = r, we have∣∣h(z)∣∣ ≤ ∣∣c1∣∣rp−1 + · · ·+

∣∣cp∣∣ < rp =∣∣g(z)∣∣.

So it follows from Rouche’s Theorem that the entire functions g(z) = zp and f(z) =h(z) + g(z) have the same number of zeros inside the circle |z| = r. Consequently,f has p zeros inside the circle |z| = r, implying that r0 < r, which is what we want.

Proof of Proposition. Set r1 = max1≤k≤pk√λk|ck|, where λk’s in R+ are

such that∑pk=1

1λk≤ 1. For all r > r1 and for each k = 1, . . . , p, we obviously have

1λk

> |ck|rk . Thus,

1 ≥p∑k=1

1λk

>

p∑k=1

|ck|rk

,

which obtainsrp > |c1|rp−1 + · · ·+ |cp|,

and hence r0 < r in view of the above lemma. As r > r1 was arbitrary, we concludethat r0 ≤ r1 = max1≤k≤p

k√λk|ck|, which is what we want.

(b) Let f(z) := zp+c1zp−1+· · ·+cp−1z+cp, g(z) := zp+c′1zp−1+· · ·+c′p−1z+c

′p,

and h(z) := g(z+ zj). First, we prove that the absolute value of the product of the

68 2. SOLUTIONS

roots of h, i.e., |h(0)|, is less than (2K)pδ. To this end, we can write∣∣h(0)∣∣ =

∣∣g(zj)∣∣ = ∣∣g(zj)− f(zj)∣∣ = ∣∣ p∑

i=1

(c′i − ci)zp−ij

∣∣≤

p∑i=1

∣∣c′i − ci∣∣∣∣zj∣∣p−i ≤ p∑

i=1

Kiδ∣∣zj∣∣p−i

≤ δ

p∑i=1

Ki(2 max1≤k≤p

k√|ck|)p−i < δ

p∑i=1

Ki(2K)p−i

= δKp

p∑i=1

2p−i = δKp(2p − 1) < (2K)pδ,

yielding∣∣h(0)

∣∣ < (2K)pδ, as desired. Now, let z′1, . . . , z′p be the roots of g(z) = 0.

It follows that z′1− zj , . . . , z′p− zj are the roots of h(z) = g(z+ zj) = 0. Obviously,there is a 1 ≤ k ≤ p such that |z′k − zj | = min1≤i≤p |z′i − zj |. Consequently,

|z′k − zj |p ≤∏

1≤i≤p

∣∣z′i − zj∣∣ = ∣∣h(0)

∣∣ < (2K)pδ,

implying that |z′k − zj | < 2K p√δ, which is what we want.

(c) We have not been able to prove the assertion. However, we prove thefollowing which can be thought of as a counterpart of the assertion at the locallevel.

Under the hypothesis of the (c) part of the problem, prove that for every t ∈[t1, t2], there exists a δ = δ(t) > 0 such that if s ∈ [t1, t2] and |s− t| < δ, then

|f(s)− f(t)| < 2K p√|s− t|α.

For a given t ∈ [t1, t2], let z1 = f(t), z2, . . . , zk (1 ≤ k ≤ p) be distinct roots ofzp+b1(t)zp−1 + · · ·+bp−1(t)z+bp(t) = 0. If k = 1, in fact f need not be continuousand yet the assertion is a quick consequence of (b). To see this, note that by (b),for the root f(s) of zp + b1(s)zp−1 + · · · + bp−1(s)z + bp(s) = 0, where s ∈ [t1, t2],there is a root of zp+ b1(t)zp−1 + · · ·+ bp−1(t)z+ bp(t) = 0, which must be the onlyroot of the equation, namely f(t), such that∣∣f(s)− f(t)

∣∣ < 2K p√|s− t|α,

as desired. So, we may without loss of generality assume that k > 1. Now, setd0 = 1

2 min1≤i<j≤k |zi − zj |. Choose a δ = δ(t) > 0 such that 2K p√|s− t|α < d0

and |f(s)− f(t)| < d0 whenever |s− t| < δ and s ∈ [t1, t2]. It follows from (b) thatfor every s ∈ [t1, t2] with |s− t| < δ, there is an is ∈ 1, . . . , k such that∣∣f(s)− zis

∣∣ < 2K p√|s− t|α < d0.

On the other hand, ∣∣f(s)− f(t)∣∣ < d0.

Consequently,∣∣z1 − zis∣∣ = ∣∣f(t)− zis

∣∣ ≤ ∣∣f(t)− f(s)∣∣+ ∣∣f(s)− zis

∣∣ < 2d0,

2.3. THIRD COMPETITION 69

implying that z1 = f(t) = zis . Thus,∣∣f(s)− f(t)∣∣ < 2K p

√|s− t|α,

whenever s ∈ [t1, t2] and |s− t| < δ, which is what we want.

2.3.2. Algebra. 1. (a) Just apply the First Isomorphism Theorem for mod-ules to the mapping ϕ : M2 → M1+M2

M1defined by ϕ(x) = x+M1, where x ∈M2.

(b) Suppose that αimi=1 is a basis for M1 ∩M2. Enlarge αimi=1 to basesαimi=1 ∪ βini=1 and αimi=1 ∪ γi

pi=1 for M1 and M2, respectively. It is easily

seen that αimi=1 ∪ βini=1 ∪ γipi=1 is a basis for M1 +M2. We can write

dimM1 + dimM2 = (m+ n) + (m+ p) = (m+ n+ p) +m

= dim(M1 +M2) + dim(M1 ∩M2)= dimM3 + dimM4,

which is what we want.

2. If G is abelian, we see that the function f : G → G defined by f(x) = x2

is a homomorphism. Conversely, if f , defined by f(x) = x2 for all x ∈ G, is ahomomorphism, then (xy)2 = x2y2 for all x, y ∈ G. Multiplying both sides by x−1

and y−1 on the left and right, respectively, we obtain xy = yx for all x, y ∈ G.That is, G is abelian, which is what we want.

3. (a) It is easily verified that the semigroup S defined by

S :=(aij) ∈M2(R) : a11 = a21 = 0, a12, a22 ∈ R \ 0

has a right identity element, namely

(0 10 1

), with respect to which every element

of S has a left inverse. And yet S is not a group, for its right identity element isnot a left identity element.

(b) Let x ∈ S be an arbitrary element with a left inverse x ∈ S so that xx = e.To prove the assertion, it suffices to show that xx = e. To this end, let x be a leftinverse of x. We have

ex = (x.x)x = x(xx) = xe = x.

By showing that xx is a right identity element of S, we conclude that xx = e,finishing the proof. Suppose that y ∈ S is arbitrary. We can write

y.(xx) = (ye)(xx) = y(ex)x = y(x.x) = ye = y.

Thus, xx = e, which is what we want.

70 2. SOLUTIONS

N’ S

N

M

R M’

Figure 5

2.3.3. General. 1. Letting z = x in the second condition, we see that

F (y, x) ≤ F (x, x) + F (x, y) = F (x, y),

for all x, y ∈ R. Therefore, F (y, x) ≤ F (x, y) for all x, y ∈ R, implying thatF (x, y) ≤ F (y, x) for all x, y ∈ R. That is, F (x, y) = F (y, x) for all x, y ∈ R. Thisproves (b). Letting y = x in the second condition, we get

0 = F (x, x) ≤ F (x, z) + F (z, x),

from which, in view of (b), we obtain 2F (x, z) ≥ 0, implying that F (x, z) ≥ 0 forall x, z ∈ R. This proves (a), completing the proof.

2. The identity (1 + x)2n = (1 + x)n(1 + x)n in R[x] together with the BinomialTheorem implies that

2n∑k=0

Ck2nxk =

( n∑k=0

Cknxk)( n∑

k=0

Cknxk),

for all n ∈ N, where Ckn =n!

k!(n− k)!is the binomial coefficient. Using the formula

for Cauchy product of two polynomials, noting that Ckn = Cn−kn , and finallyequating the coefficients of xn of both sides of the above identity, we see that

(2n)!(n!)2

= Cn2n =n∑k=0

(Ckn)2 =

n∑k=0

( n!k!(n− k)!

)2

,

proving the assertion.

3. It is plain that if from the point M , we shine a beam of light onto a point Rof the mirror ∆ so that the reflected light beam hits the mirror ∆′ at a point S,then the three points S , R, and M ′ are collinear, where M ′ is the image of M inthe mirror ∆. Hence the problem is solved as follows. Find the image of M andN in the mirrors ∆ and ∆′, respectively, to obtain the points M ′ and N ′ of theplane P . The line joining M ′ and N ′ intersects ∆ and ∆′ at the points R and S,respectively. It is now obvious that if we shine a beam of light onto the point R onthe mirror ∆, then the reflected light beam after hitting the mirror ∆′ at S wouldpass through the point N (see Figure 5).

2.4. FOURTH COMPETITION 71

(-2,0) (-1,0) (0,0) (1,0) (2,0)

Figure 6

2.4. Fourth Competition

2.4.1. Analysis. 1. Define the function f : R → R2 by

f(t) =

g(t) t < −2,(t, 0) −2 ≤ t < 2,h(t) 2 ≤ t,

where g(t) =(

−21+(t+2)2 ,

2(t+2)1+(t+2)2

)and h(t) =

(2

1+(t−2)2 ,2(t−2)

1+(t−2)2

). It is straight-

forward to see that f is continuous and one-to-one. In fact, f is a one-to-oneparametrization of the curve consisting of the upper semicircle with radius one cen-tered at (−1, 0) going from (0, 0) to (−2, 0), the line segment joining (−2, 0) and(2, 0), and the upper semicircle with radius one centered at (1, 0) going from (2, 0)to (0, 0) (see Figure 6).

We claim that the function f−1 is discontinuous at (0, 0), proving the assertion.To see this, define the sequence (an)+∞n=1 by

an =

(1n , 0)

n even,(2

1+n2 ,2n

1+n2

)n odd.

It is obvious that limn an = (0, 0) = f(0) but

f−1(an) =

1n n even,

n+ 2 n odd,

whose limit does not exist as n→ +∞. Therefore, f−1 is not continuous at (0, 0),proving the claim.

2. Set S :=n ∈ N : αn > 0

. If the set S is finite, there is nothing to prove. So

assume that S is an infinite set. For each k ∈ S with k ≥ 2, define

Tk :=N ∈ S|N ≥ k αn ≤ αk ∀n ≥ N

.

As limn αn = 0, letting ε = ak in the definition of the limit for limn αn = 0, wesee that Tk 6= ∅. Let Mk be the initial element of Tk. If Mk = k, we will haveαn ≤ αk for all n ≥ k, in which case set Nk := k. If Mk > k, set Nk to be thegreatest integer subject to k ≤ Nk < Mk and Nk ∈ S. It follows that Nk /∈ Tkand that αn = 0 for all Nk < n < Mk. Consequently, αNk

> αk > 0, for Mk

is the initial element of Tk. Now, if n ≥ Nk, then either n = Nk, in which case

72 2. SOLUTIONS

αn = αNk≤ αNk

, or Nk < n < Mk, which yields αn = 0 < αNk, or n ≥ Mk, in

which case αn ≤ αk < αNk. In other words, αn ≤ αNk

for all n ≥ Nk. On theother hand, S is an infinite set and Nk ≥ k for all k ∈ S. This shows that there arean infinite number of Nk’s such that αn ≤ αNk

for all n ≥ Nk.

2.4.2. Algebra. 1. (a) Let x, y ∈ R(B) and a ∈ A be arbitrary. It followsthat there are M,N ∈ N such that xM ∈ B and xN ∈ B. Noting that in acommutative ring the Binomial Theorem holds, (ab)n = anbn for all a, b ∈ A andn ∈ N, and that B is an ideal of A, we see that

(x− y)M+N =M+N∑i=0

CiM+Nxi(−y)M+N−i ∈ B, (ax)M = aMxM ∈ B,

yielding x− y ∈ R(B) and ax ∈ R(B). That implies R(B) is an ideal of A, whichis what we want.

(b) First proof. Let B be a prime ideal of A. To prove that R(B) is prime,suppose for given x, y ∈ A, we have xy ∈ R(B). We need to show that x ∈ R(B)or y ∈ R(B). As xy ∈ R(B), there is an n ∈ N such that xnyn = (xy)n ∈ B. ButB is prime. So we must have xn ∈ B or yn ∈ B. In other words, x ∈∈ R(B) ory ∈ R(B), as desired.

Second proof. We prove the assertion by showing that R(B) = B. Evidently,B ⊆ R(B). To show that R(B) ⊆ B, let x ∈ R(B) be arbitrary. It follows thatthere is an n ∈ N such that xn ∈ B. Now, a straightforward induction on ntogether with the hypothesis that B is prime reveals that x ∈ B. This proves thatR(B) ⊆ B, and hence R(B) = B, as desired.

(c) Remark. We must have a 6= 0,±1.Without loss of generality, assume that a > 1. Write a = pm1

1 · · · pmk

k , wheremi ∈ N and pi’s are distinct prime numbers (1 ≤ i ≤ k). We show that R(B) =〈p1p2 · · · pk〉. First, let x ∈ 〈p1p2 · · · pk〉 be arbitrary. It follows that x = p1p2 · · · pkrfor some r ∈ Z. Obviously, we can write xm1+···+mk = as ∈ B for some s ∈ Z,implying that x ∈ R(B). This yields 〈p1p2 · · · pk〉 ⊆ R(B). Next, let x ∈ R(B)be arbitrary. It follows that xn ∈ B = 〈pm1

1 · · · pmk

k 〉, form which, we obtain xn =pm11 · · · pmk

k r for some r ∈ Z. Thus, pi|xn for all 1 ≤ i ≤ k, yielding pi|x becausepi is prime. This implies p1 · · · pk|x, for pi’s are distinct primes. That is, x ∈〈p1p2 · · · pk〉, and hence R(B) ⊆ 〈p1p2 · · · pk〉. Therefore, R(B) = 〈p1p2 · · · pk〉,finishing the proof.

2. Let x, y be a linearly independent set in E, which is a vector space over a fieldF . Use Zorn’s Lemma to enlarge the independent sets x, y and x, x + y tox, y∪αii∈I and x, x+y∪βii∈I , respectively, each of which is a Hamel basisfor E. It is well-known that there exists a linear transformation u : E → E satisfyingu(x) = x, u(y) = x + y, and u(αi) = βi for all i ∈ I. The linear transformation uis invertible because it maps the elements of the Hamel basis x, y ∪ αii∈I ontothose of the Hamel basis x, x + y ∪ βii∈I . This means u is an automorphismof the vector space E. Now, suppose that a linear transformation f : E → E, alsocalled an endomorphism of the vector space E, commutes with any invertible linear

2.4. FOURTH COMPETITION 73

transformation v : E → E, which is also called an automorphism of E. We claim,first of all, that for any nonzero vector x, we have f(x) = λxx for some λx ∈ F .To prove this by contradiction, letting ux : E → E be an automorphism such thatux(x) = x and ux(f(x)) = x + f(x), we must have fux = uxf . In particular, wehave f(ux(x)) = ux(f(x)) which means f(x) = x + f(x), yielding x = 0, which isimpossible. Therefore, for any x ∈ E there exists λx ∈ F such that f(x) = λxx.Next, fix a nonzero x0 ∈ E. By showing that λx = λx0 , we prove the assertion.Without loss of generality, assume that x, x0 is linearly independent. Therefore,there is an automorphism ux : E → E such that ux(x) = x and ux(x0) = x + x0.But f commuting with ux implies that

λxx+ λx0x0 = f(x+ x0) = f(ux(x0)

)= ux

(f(x0)

)= λx0ux(x0)

= λx0x+ λx0x0,

yielding λxx = λx0x. From this, we obtain λx = λx0 because x 6= 0. So we haveshown that f(x) = λx0x for all x ∈ E, or equivalently f = λx0IE , which is what wewant, finishing the proof.

3. It suffices to show that any nonzero x ∈ R has a multiplicative inverse. To thisend, let a nonzero x ∈ R be given. It follows from the hypothesis that the followingdescending chain of the ideals of R

xR ⊇ x2R ⊇ x3R ⊇ · · ·

must terminate. That is, there exists N ∈ N such that xnR = xNR for all n ≥ N .In particular, xN+1R = xNR, implying that xN ∈ xN+1R. Hence, there existsy ∈ R such that xN = xN+1y, yielding xN (1 − xy) = 0. This, in view of the factthat R is an integral domain, implies that 1 − xy = 0, yielding xy = yx = 1. Inother words, x is invertible, which is what we want.

2.4.3. General. 1. Doing the substitution x = π2 − t, we can write

I :=∫ π

2

0

sinn xcosn x+ sinn x

dx =∫ 0

π2

sinn(π2 − t)cosn(π2 − t) + sinn(π2 − t)

(−dt)

=∫ π

2

0

cosn tsinn t+ cosn t

dt.

We have

I =12(I + I

)=

12

(∫ π2

0

sinn x+ cosn xcosn x+ sinn x

dx)

=12× π

2=π

4,

for all n ∈ R.

2. (a) Define the events a, b, c, a1, b1, c1 as follows:

a, b, c: The event that the prize is in box A, B, or C, respectively.

a1, b1, c1: The event that the box A, B, or C is opened, respectively.

74 2. SOLUTIONS

It is obvious that a, b, c is a partition of the probability space and that thedesired probability is P (b|a1). Using the Inverse Probability Theorem or Bayes’Theorem, we can write

P (b|a1) =P (b)P (a1|b)

P (a1)=

P (b)P (a1|b)P (a)P (a1|a) + P (b)P (a1|b) + P (c)P (a1|c)

=13 ×

12

13 × 0 + 1

3 ×12 + 1

3 × 1=

13.

So the probability that the chosen box, i.e. B, contains the prize is 13 .

(b) Plainly, the desired probability is P (c|a1). Again, using Bayes’ Theorem,we can write

P (c|a1) =P (c)P (a1|c)

P (a1)=

P (c)P (a1|c)P (a)P (a1|a) + P (b)P (a1|b) + P (c)P (a1|c)

=13 × 1

13 × 0 + 1

3 ×12 + 1

3 × 1=

23.

That is, the probability that the other box, i.e., C, contains the prize is 23 .

2.5. Fifth Competition

2.5.1. Analysis. 1. No, it is not. We present a counterexample on the closedinterval [0, 1] which will work on the open interval (0, 1) as well. Define the functionsf, g : [0, 1] → [0, 1] as follows

f(x) = 1

q x ∈ Q ∩ (0, 1], x = pq , p, q ∈ N, gcd(p, q) = 1,

0 x ∈ [0, 1] \ (Q ∩ (0, 1]).

g(x) =

1 x ∈ 1q |q ∈ N,

0 x /∈ 1q |q ∈ N.

Obviously,

g f(x) =

1 x ∈ Q ∩ (0, 1],0 x ∈ [0, 1] \ (Q ∩ (0, 1]).

We claim that f, g are Riemann integrable on [0, 1] but g f is not integrable on[0, 1]. In fact, it turns out that g f is not integrable on any closed interval [ε, δ],where 0 < ε < δ < 1, from which it follows that g f in not integrable on the openinterval (0, 1). In view of Solution 3 of 2.1.1, we see that

∀a ∈ (0, 1) : limx→a

f(x) = 0, limx→0+

f(x) = 0, limx→1−

f(x) = 0.

Thus, the function f is continuous at the irrational points of (0, 1) and the set ofdiscontinuity points of f , i.e., Q ∩ (0, 1), is countable. It is easily seen that g isdiscontinuous at any point of the set 1

q |q ∈ N ∪ 0 and it is continuous on theset (0, 1] \ 1

q |q ∈ N. Finally, using the fact that Q∩ (0, 1] is dense in [0, 1], we seethat g f is discontinuous at any point of the interval [0, 1]. From this point on,we present two proofs for the claim we made in the above. The first one uses somestandard theorem from analysis and the second proof is self-contained.

2.5. FIFTH COMPETITION 75

First proof. We will make use of Lebesgue’s Integrability Criterion for inte-grals in the Riemann sense which asserts that a bounded function f : [a, b] → Ris integrable in the Riemann sense if and only if the set consisting of the points atwhich f is discontinuous has measure zero. Recall that a set A ⊆ R is said to bea set of measure zero if for any ε > 0, there exists a sequence (In)+∞n=1 of intervalssuch that A ⊆

⋃+∞n=1 In and

∑+∞n=1 `(In) < ε, where `(In) := βn−αn and αn and βn

are the end-points of the interval In. It is easy to see that any finite or countablesubset of reals has measure zero and that a countable union of sets of measurezero has measure zero. In view of all that, we see that the functions f and g areRiemann integrable on [0, 1] and hence on (0, 1) but g f is not integrable (0, 1)because it is not integrable on any interval [ε, δ], where 0 < ε < δ < 1, for g f iscontinuous nowhere.

Second proof. Define the sequences (fn)+∞n=1 and (gn)+∞n=1, with fn, gn :[0, 1] → [0, 1], as follows

fn(x) = 1

q x = pq , p, q ∈ N, gcd(p, q) = 1, q ≤ n,

0 otherwise.

gn(x) =

1 x = 1q , q ∈ N, q ≤ n,

0 otherwise.

It is easily verified that the sequences (fn)+∞n=1 and (gn)+∞n=1 converge uniformly on[0, 1] to f and g, respectively. As fn and gn (n ∈ N) are discontinuous at onlyfinitely many points of the interval [0, 1], it follows that f and g are integrable inthe Riemann sense on [0, 1]. Now, we see from the following lemma that g f isnot integrable on any interval [ε, δ], where 0 < ε < δ < 1, which is what we want.

Lemma. Let f : [a, b] → R be bounded and integrable on [a, b] in the Riemannsense. Then the set of points at which f is continuous is infinite.

Proof. Since f integrable on any closed subinterval of [a, b], it suffices to provethat f is continuous at some point of the interval [a, b], for it would then followthat any subinterval of [a, b] contains a point at which f is continuous. And hence,the set of points at which f is continuous would be infinite, which is what we want.Taking ε = b−a in Riemann’s criterion for integrability on the interval I0 := [a, b],we obtain a partition P0 : a = t00 < · · · < t0n0 = b of I0 such that

U(P0, f)− L(P0, f) < b− a,

where

U(P0, f) =n0∑i=1

M0i∆t0i, L(P0, f) =n0∑i=1

m0i∆t0i, ∆t0i = t0i − t0(i−1),

M0i = supt0(i−1)≤x≤t0i

f(x), m0i = inft0(i−1)≤x≤t0i

f(x).

Hence, there exists 1 ≤ i0 ≤ n0 such that

M0i0 −m0i0 < 1,

for otherwise it would follow that U(P0, f)−L(P0, f) ≥ b− a, which is impossible.Set I1 := [a1, b1] = [t0(i0−1), t0i0 ]. As f is integrable on I1, taking ε = b1−a1

2 , asimilar argument shows that there exists a subinterval I2 := [a2, b2] of I1 such thatsupx∈I2 f(x) − infx∈I2 f(x) < 1

2 . Continuing this way, we obtain a sequence of

76 2. SOLUTIONS

nested closed interval I1 ⊃ I2 ⊃ · · · such that 0 ≤ supx∈Inf(x)− infx∈In

f(x) < 1n .

Since⋂+∞n=1 In 6= ∅ because I0 is compact, in view of the above inequality, it follows

that f is continuous at any point of⋂+∞n=1 In 6= ∅, finishing the proof.

2. Remark. Adjusting the proof below, it is not difficult to prove the followingproblem, which characterizes all continuous real functions having the property thatthey do not assume any value more than twice. Let f : R −→ R be a continuousfunction that is neither increasing nor decreasing on R and that it assumes anyvalue at most twice. Prove that one of the following statements holds.

(i) There exists an a ∈ R such that f or −f is strictly increasing on (−∞, a]and is strictly decreasing on [a,+∞).

(ii) There exist a, b ∈ R with a < b such that f or −f is strictly increasingon (−∞, a], is strictly decreasing on [a, b], and is strictly increasing on [b,+∞).Moreover, limx→−∞ f(x) ≥ limx→+∞ f(x).

If the function f is strictly increasing or strictly decreasing, there is nothing toprove. So we may assume that f is neither strictly increasing nor strictly decreasing.At this point, we need the following lemma.

Lemma. Let R be a totally ordered set with at least three elements and f :R → R a function that is neither strictly increasing nor strictly decreasing. Then,there exist x1, x2, x3 ∈ R with x1 < x2 < x3 such that f(x2) ≥ max

(f(x1), f(x3)

)or f(x2) ≤ min

(f(x1), f(x3)

).

Proof. To prove the assertion by contradiction, suppose that for all x1, x2, x3 ∈R with x1 < x2 < x3, we have f(x2) < max

(f(x1), f(x3)

)and f(x2) > min

(f(x1), f(x3)

).

Now, since max(f(x1), f(x3)

)= f(x1) or max

(f(x1), f(x3)

)= f(x3), we see that

for all x1, x2, x3 ∈ R with x1 < x2 < x3, we have

f(x1) < f(x2) < f(x3) or f(x1) > f(x2) > f(x3). (∗)

Pick a, b, c ∈ R with a < b < c. It follows that either f(a) < f(b) < f(c) orf(a) > f(b) > f(c). Suppose f(a) < f(b) < f(c). We show that f is strictlyincreasing, a contradiction. To see this, let x, y ∈ R with x < y be arbitrary. Thereare five cases to consider. (i) x < y < a, (ii) x < y = a, (iii) x < a < y, (iv)x = a < y, and (v) a < x < y. In each case, in view of (∗), it is easy to seethat f(x) < f(y), from which, we conclude that f is strictly increasing, which isa contradiction. Likewise, if f(a) > f(b) > f(c), one can see, in a similar fashion,that f is strictly decreasing which is again a contradiction. So the proof is completeby contradiction.

In view of the lemma, there are two cases to consider.(i) There are x1, x2, x3 ∈ R with x1 < x2 < x3 such that f(x2) ≥ max

(f(x1), f(x3)

).

In this case, from f(x2) ≥ max(f(x1), f(x3)) and the continuity of f , we seethat the function f attains its absolute maximum on the closed interval [x1, x3]at an internal point of the interval, say, at xM ∈ (x1, x3). By proving that fattains its only absolute maximum on R at xM , we complete the proof in this case.To this end, we first show that f(xM ) > f(x) for all x ∈ [x1, x3] with x 6= xM .Suppose to the contrary that there exists an x0 ∈ [x1, x3] with x0 6= xM suchthat f(x0) = f(xM ). Again, there are two cases to consider. (a) x1 ≤ x0 < xMand (b) xM < x0 ≤ x3. Let x1 ≤ x0 < xM and pick a x0 < t < xM . As f

2.5. FIFTH COMPETITION 77

assumes the value f(xM ) at most twice and f(xM ) is the absolute maximum of f

on [x1, x3], we see that f(t) < f(x0) = f(xM ). Set λ =f(xM )+max

(f(x3),f(t)

)2 . It

is obvious that f(t) < λ < f(x0), f(t) < λ < f(xM ), and f(x3) < λ < f(xM ).Thus, it follows from the Intermediate Value Theorem that there exist ξ1, ξ2, ξ3with x0 < ξ1 < t , t < ξ2 < xM , and xM < ξ3 < x3 such that f(ξi) = λ for eachi = 1, 2, 3, contradicting the hypothesis. Similarly, if xM < x0 ≤ x3, we obtain acontradiction. That is, we have shown that f(xM ) > f(x) for all x ∈ [x1, x3] withx 6= xM . Next, we prove that f(x) < f(xM ) for all x ∈ R with x 6= xM . Againsuppose to the contrary that there is a t ∈ R\xM such that f(t) ≥ f(xM ). Fromwhat we have proved so far, it follows that t < x1 or t > x3. If t < x1, notingthat f(x1) < f(xM ) ≤ f(t), we see that there exists an x0 ∈ R with t ≤ x0 < x1

such that f(x0) = f(xM ). Now, set λ =f(xM )+max

(f(x1),f(x3)

)2 . Again, it is

obvious that f(x1) < λ < f(x0), f(x1) < λ < f(xM ), and f(x3) < λ < f(xM ).Hence, by the Intermediate Value Theorem we obtain ξ1, ξ2, ξ3 with x0 < ξ1 < x1

, x1 < ξ2 < xM , and xM < ξ3 < x3 such that f(ξi) = λ for each i = 1, 2, 3,contradicting the hypothesis. Similarly, if x3 < t, we obtain a contradiction. Thisfinishes the proof in this case.

(ii) There are x1, x2, x3 ∈ R with x1 < x2 < x3 such that f(x2) ≤ min(f(x1), f(x3)

).

Adjusting the proof of case (i) or replacing f by −f and following the line ofargument presented in case (i), we can see that the absolute minimum of f on Roccurs in the interval (x1, x3) and that it is unique. We omit the proof for the sakeof brevity.

3. Let B =(p, q) : p, q ∈ Q, p < q

, where (p, q) denotes the open interval with

end points p and q in R. The set B is a countable basis for the ordinary topologyof R. Write B = Inn∈N, where In = (pn, qn) for some pn, qn ∈ Q, and set

T := n ∈ N : card(In ∩A) ≤ ℵ0.

Obviously, T & N because otherwise A would be countable which is impossible.Now, set

A1 =( ⋃n∈T

In)∩A =

⋃n∈T

(In ∩A).

The set A1 is countable because so is In ∩ A for all n ∈ T . By showing that anypoint x ∈ A \ A1 is a congestion point of A, we finish the proof. To prove this bycontradiction, suppose that there exists an ε0 > 0 such that (x − ε0, x + ε0) ∩ Ais countable. As B = Inn∈N is a basis for the topology of R, we see that thereexists n0 ∈ N such that x ∈ In0 ⊆ (x− ε0, x+ ε0), from which, we obtain

In0 ∩A ⊆ (x− ε0, x+ ε0) ∩A,

implying that In0 ∩ A is countable because so is (x − ε0, x + ε0) ∩ A. That is,card(In0 ∩A) ≤ ℵ0, yielding n0 ∈ T , and hence x ∈ In0 ∩A ⊆

⋃n∈T (In ∩A) = A1.

Consequently, x ∈ A1, which contradicts the hypothesis that x ∈ A \ A1. Thus, xis a congestion point of A, which is what we want.

78 2. SOLUTIONS

2.5.2. Algebra. 1. Let A : V → V be such that A 6= I , A2 6= I, A3 = I. Ifx 6= 0 is an eigenvector corresponding to the eigenvalue λ ∈ C, we can write

Ax = λx =⇒ A2x = λ2x =⇒ A3x = λ3x,

which, in view of A3 = I, implies that λ3 = 1. That is, the set of the eigenvalues ofA is a subset of the third roots of unity. Likewise, if A 6= I, . . . , Ak−1 6= I,Ak = I,then the set of the eigenvalues of A is a subset of the kth roots of unity. Theeigenvalues of A do not necessarily form a group under multiplication. To see this,let ωk = cos 2π

k + i sin 2πk , where k − 1 ∈ N, and just note that ωk 6= 1 is the only

eigenvalue of the linear transformation ωkIn, where In : Cn → Cn is the identitytransformation. Obviously, ωk does not form a group under multiplication. Asfor a necessary and sufficient condition for the set of eigenvalues of A to form agroup, nothing interesting can be said. Having said that, it is easily seen that theset of eigenvalues of A forms a group under multiplication, in fact a cyclic subgroupof the multiplicative group of the kth roots of unity, if and only if the set of theeigenvalues of A is closed under multiplication.

2. We have ord(a) = 7 because a7 = 1, a 6= 1, and 7 is prime. Induction on ktogether with the relations ab = b2a and a = b2ab−1 yields

ak = b2k

akb−1,

for all k ∈ N. In particular, a7 = b27a7b−1 together with a7 = 1 implies that

b27−1 = 1. That is, b127 = 1. But 127 is prime and b 6= 1. Hence, ord(b) = 127.

Therefore, ord(a) = 7 and ord(b) = 127, which is what we want.

3. (a) We can write

(−x)3 + 2(−x)2 + (−x) = 0 =⇒ −x3 + 2x2 − x = 0,

for all x ∈ R. On the other hand, x3 + 2x2 + x = 0 for all x ∈ R. Adding up thetwo equalities, we obtain 4x2 = 0 for all x ∈ R. We can also write

(2x)3 + 2(2x)2 + (2x) = 0 =⇒ 8x3 + 8x2 + 2x = 0,

for all x ∈ R. But 8x3 = 8x2 = 0 because 4x2 = 0 for all x ∈ R. This togetherwith the above yields 2x = 0 for all x ∈ R, which is what we want.

(b) Using (a), we can write

(x+ 1)3 + 2(x+ 1)2 + (x+ 1) = 0 =⇒ x2 + x = 0,

for all x ∈ R. Consequently, x2 = −x = x because 2x = 0. Hence, x2 = x forall x ∈ R. Therefore, R is a Boolean ring, which is commutative by the followingargument.

(x+ y)2 = x+ y =⇒ x2 + y2 + xy + yx = x2 + y2

=⇒ xy = −yx = yx,

for all x, y ∈ R, which is what we want.

2.5. FIFTH COMPETITION 79

2.5.3. General. 1. We need the following lemma

Lemma. The product of k consecutive integers is divisible by k!.Proof. It suffices to show that the product of k natural numbers is divisible by

k!. To this end, let k, n ∈ N with k ≤ n be given. To show that n(n−1) · · · (n−k+1)

is divisible by k!, it is enough to prove that Ckn :=n!

k!(n− k)!is an integer because

n(n− 1) · · · (n− k + 1) = k!Ckn.

By proving the polynomial identity below, which is also known as the BinomialTheorem

(1 + x)n =n∑k=0

Cknxk, (∗)

we conclude that Ckn is an integer, finishing the proof. Recall that, by Taylor’sFormula, for two polynomial functions p, q : R → R of degree n, we have p = qif and only if p(k)(0) = q(k)(0) for each k = 0, 1, . . . , n, where p(k), q(k) denote thekth derivative of p and q, respectively, and the zeroth derivative of a function, bydefinition, is just the function. Also, it is readily verified that

((a+ x)k

)(i) =

k(k − 1) · · · (k − (i− 1))(a+ x)k−i 1 ≤ i < k,k! i = k,0 i > k,

where a ∈ R is a constant. Use p and q to denote the left and right hand side of(∗). Obviously, p(0) = q(0). We can write

p(k)(0) =((1 + x)n

)(k)(0) = n(n− 1) · · · (n− k + 1)(1 + 0)n−k = k!Ckn,

q(k)(0) =( n∑i=0

Cinxi)(k) = Ckn(xk)(k) = Cknk!,

for each k = 1, . . . , n. So we have shown that p(k)(0) = q(k)(0) for each k =0, 1, . . . , n, implying that p = q, which is what we want.

Now to prove the assertion, note first that

(mn)! =m−1∏k=0

n∏j=1

(kn+ j)

.

Butn∏j=1

(kn+ j) =n−1∏j=1

(kn+ j)× n(k + 1).

It follows from the lemma that∏n−1j=1 (kn + j) is divisible by (n − 1)!, whence∏n

j=1(kn+ j) is divisible by (n− 1)!× n(k + 1) = n!(k + 1). This implies (mn)! =∏m−1k=0

(∏nj=1(kn+ j)

)is divisible by

∏m−1k=0 n!(k+1) = (n!)mm!. Thus, the number

(mn)!m!(n!)m

is an integer, which is what we want.

2. The assertion is known as the Gauss-Lucas Theorem . As is shown in whatfollows, it suffices to prove that if the roots of a polynomial p lie in a closed half

80 2. SOLUTIONS

x’ x

h

zH’

Figure 7

plane, then so do those of p′. For a subset S of C, the closed convex hull of S,denoted by the symbol C(S), is said to be the intersection of all closed convexsubsets of C which include S, i.e.,

C(S) :=⋂

C : S ⊆ C,C is closed and convex.

It is straightforward to see that if S = zini=1, where n ∈ N, then C(S) = C(S).We claim that

C(S) = C(S) =⋂

H : S ⊆ H,H is a closed half plane.

Set C1 :=⋂

H : S ⊆ H,H is a closed half-plane. As C1 is closed and convex, it

follows that C(S) = C(S) ⊆ C1. We establish our claim by showing that C(S)c

=C(S)c ⊆ Cc1. To see this, let x /∈ C(S) = C(S) be arbitrary. As C(S) is closed andx is compact, it follows that there exists an x′ ∈ C(S) such that

|x− x′| = infy∈C(S)

|x− y|.

Now, consider the perpendicular bisector ∆ of the line segment joining x and x′.Let H ′ be the closed half-plane containing the point x′ ∈ S. The half plane H ′

includes S because if, contrary to this, there exists a point z ∈ S in the other halfplane, then in the triangle x′xz the perpendicular foot h of the altitude passingthrough the vertex x belongs S, for x′, z ∈ S, and moreover |x−h| < |x−x′|, whichis a contradiction. Therefore, the closed half plane H ′ includes S, implying thatC1 ⊆ C(S), which is what we want.

To complete the proof we need the following lemma.

Lemma. If the roots of a polynomial with complex coefficients lie in a closedhalf plane, then so do those of the derivative of the polynomial.

Proof. Suppose z1, . . . , zn are the roots of a polynomial p so that p(z) =c(z − z1) · · · (z − zn), where c ∈ C. Also suppose that the closed half plane H =z ∈ C : Im z−a

b ≤ 0 includes the roots of p. We show that H includes the roots ofp′, the derivative of p. If z = zi, for some 1 ≤ i ≤ n, is a root of p of multiplicity

2.5. FIFTH COMPETITION 81

greater than one, then zi is a root of p′ as well, in which case there is nothing toprove. So, without loss of generality, we may assume that

p′(z)p(z)

=1

z − z1+ · · ·+ 1

z − zn.

Now, suppose z /∈ H. We complete the proof by showing that z cannot be a rootof p′. To this end, as z /∈ H and zi ∈ H, we obtain

Imz − zib

= Imz − a

b− Im

zi − a

b> 0,

for all 1 ≤ i ≤ n. It follows that Im bz−zi

< 0 for all 1 ≤ i ≤ n. From this, we obtain

Imbp′(z)p(z)

=n∑i=1

Imb

z − zi< 0,

implying that p′(z) 6= 0, proving the lemma.

Now, assume that f is a polynomial of degree greater than or equal to 2 andA and A′ denote the sets of the roots of the equations f(z) = 0 and f ′(z) = 0,respectively. Suppose that H is an arbitrary closed half plane including A. Itfollows from the lemma that H includes A′. This yields

A′ ⊆⋂

H : A ⊆ H,H is a closed half plane

= C(A).

But since C(A) is a closed convex set which includes A′, we see that

C(A′) = C(A′) ⊆ C(A),

which is what we want.

3. Remark. The lemma below together with the proof following the lemma showsthat if the initial ball trajectory intersects the line segment joining the two foci, thenthere exists a hyperbola confocal with the original ellipse to which all trajectorysegments are tangent.

We need the following lemma.

Lemma. (i) Let F and F ′ be two points and l a line in a plane not intersectingthe line segment F ′F (resp. intersecting the line segment FF ′ between its end pointsexcept the perpendicular bisector of F ′F ). Then, there exists a unique ellipse (resp.hyperbola) whose foci are F ′ and F and to which the line l is tangent.

(ii) Let F and F ′ be two points and Pm and Pn two half-lines through a pointP in the plane such that the half-lines do not intersect the line segment F ′F (resp.intersect FF ′ between its end points, i.e., between F and F ′). Then, there exists aunique ellipse (resp. hyperbola) whose foci are F ′ and F and to which the half-linesPm and Pn are tangent if and only if ∠FPm = ∠F ′Pn.

Proof. (i) Since the foci, i.e., F and F ′, of the desired ellipse (resp. hyperbola)are fixed, it suffices to show that the sum (resp. the difference) of the focal radii ofany point of the ellipse (resp. hyperbola) is uniquely determined by the given linel. To this end, suppose that the line l is tangent to an ellipse (resp. a hyperbola)whose foci are F and F ′. By the optical property of ellipses (resp. hyperbolas) ,the tangent point is obtained as follows. Reflect the focus F in the line l to get F1.Then, join F1 and F ′ to intersect the line l at the tangent point, say, A ∈ l. It isplain that AF+AF ′ = F1F

′ = F2F (resp. |AF−AF ′| = F1F′ = F2F ), where F2 is

82 2. SOLUTIONS

l

F

F’

F

A

1

FF’

F1

A

l

Figure 8

Figure 9

the reflection of F ′ across the line l. In other words, the sum (resp. the difference)of the focal radii of the point A, and hence any point, of the desired ellipse (resp.hyperbola) is F1F

′ = F2F , which is uniquely determined by the given line l. Thisproves (i).

(ii) The uniqueness follows from (i). Let F1 and F2 be the reflection of F andF ′ across the half-lines Pm and Pn, respectively.

First, suppose there exists an ellipse (resp. a hyperbola) whose foci are F ′ andF and to which the half-lines Pm and Pn are tangent. It is easily seen that the sum(resp. the difference) of the focal radii of any point of the ellipse (resp. hyperbola)is equal to F2F = F1F

′. Now, the triangles 4F2PF and 4F ′PF1 are congruentbecause F2F = F1F

′, F2P = F ′P , and PF = PF1. This, in particular, implies∠F2PF = ∠F ′PF1, yielding

∠F ′PF + 2∠F ′Pn = ∠F ′PF + 2∠FPm

2.6. SIXTH COMPETITION 83

(resp.∠F ′PF − 2∠F ′Pn = ∠F ′PF − 2∠FPm).

Thus, ∠F ′Pn = ∠FPm, which, in turn, yields

∠FPn = ∠F ′PF + ∠F ′Pn = ∠F ′PF + ∠FPm = ∠F ′Pm,

(resp.

∠FPn = ∠F ′PF − ∠F ′Pn = ∠F ′PF − ∠FPm = ∠F ′Pm, )

as desired.Next, suppose ∠FPn = ∠F ′Pm. We can write

∠F2PF = ∠F ′PF + 2∠FPn = ∠F ′PF + 2∠F ′Pm = ∠F1PF′,

(resp.

∠F2PF = ∠F ′PF − 2∠FPn = ∠F ′PF − 2∠F ′Pm = ∠F1PF′).

That is, ∠F2PF = ∠F1PF′. It thus follows that the triangles4F2PF and4F ′PF1

are congruent because F2P = F ′P , ∠F2PF = ∠F1PF′, and PF = PF1. In

particular, we must have F2F = F1F′. In other words, the two ellipses (resp.

hyperbolas) whose foci are F and F ′ and are tangent to the half-lines Pm and Pncoincide, proving the “if part” of the assertion.

Let F and F ′ be the foci of the ellipse ξ and PQ,QR,RS be three consecutivesegments of the ball trajectory. It is well known that in any ellipse the focal radii ofany point on the ellipse form equal angles with the tangent at the point. In otherwords, the normal and tangent lines at any point on an ellipse bisect the anglesbetween the focal radii of the point. From this, it follows that if a billiard segmentintersects the interior of the line segment F ′F , so does all other billiard segments;if a billiard segment passes through one of the foci, then so does all other billiardsegments; and finally, if a billiard segment does not intersect the line segment F ′F ,then neither does any other billiard segment. Now, with P,Q,R, S as in the above,it follows that ∠F ′QP = ∠FQR and ∠F ′RQ = ∠FRS. This together with thelemma implies that there are two ellipses ηQ and ηR confocal with ξ each of whichis tangent to QR. We see from part (i) of the lemma that ηQ = ηR for any billiardsegments QR. That is, there is an ellipse η confocal with the ellipse ξ such thatthe ball trajectory is always tangent to η, which is what we want.

2.6. Sixth Competition

2.6.1. Analysis. 1. To prove the assertion by contradiction, suppose thatthere are three rational points A = (x1, y1) , B = (x2, y2), and C = (x3, y3) onthe circle. Since A,B,C ∈ Q2, it follows that the equations of the perpendicularbisectors of the line segments AB and BC are of the form a1x + b1y = c1 anda2x+ b2y = c2, respectively, where ai, bi, ci ∈ Q (i = 1, 2) and a1b2 − b1a2 6= 0. Itis obvious that the center (x, y) of the circle is the solution of

a1x+ b1y = c1a2x+ b2y = c2

,

which must be rational because ai, bi, ci ∈ Q (i = 1, 2). This is a contradiction,proving the assertion, which is what we want.

84 2. SOLUTIONS

2. As the sequence (fn)+∞n=1 is uniformly convergent to f on M , for a given ε > 0,there exists a natural number N1 such that for all n ≥ N1 and x ∈M we have

d(fn(x), f(x)

)<

ε

6.

Now, since the function fN1 is continuous at the point x ∈M , for the given ε > 0,there exists a δ > 0 such that for all y ∈M with d(x, y) < δ we have

d(fN1(y), fN1(x)

)<

ε

3.

From limn xn = x, we see that for δ > 0 obtained from the above, there exists anatural number N2 such that for all n ≥ N2

d(xn, x) < δ.

Let N = max(N1, N2). For the given ε > 0 and for all n ≥ N , we can write

d(fn(xn), f(x)

)≤ d

(fn(xn), f(xn)

)+ d(f(xn), fN1(xn)

)+d(fN1(xn), fN1(x)

)+ d(fN1(x), f(x)

)<

ε

6+ε

6+ε

3+ε

3= ε.

That is, limn→+∞ fn(xn) = f(x), which is what we want.

3. Since∫ +∞0

∣∣f(x)∣∣dx < +∞, we see that for given ε > 0, there exists a natural

number N1 such that for all M ≥ N1∫ +∞

M

∣∣f(x)∣∣dx <

ε

3.

The function f is continuous on the compact interval [0, N1 + 1]. Hence, it isuniformly continuous on [0, N1 +1]. Therefore, for given ε > 0, there exists a δ > 0such that for all x, y ∈ [0, N1 + 1] with |x− y| < δ, we have∣∣f(x)− f(y)

∣∣ < ε

3N1.

Now, for a given ε > 0, find N1 > 0 and δ > 0 as in the above and pick N ∈ N suchthat 1

N < δ. For all n ≥ N , we can write∫ +∞

0

∣∣f(x+1n

)− f(x)∣∣dx =

∫ N1

0

∣∣f(x+1n

)− f(x)∣∣dx+

∫ +∞

N1

∣∣f(x+1n

)− f(x)∣∣dx

<ε

3N1(N1 − 0) +

∫ +∞

N1

∣∣f(x+1n

)∣∣dx+

∫ +∞

N1

∣∣f(x)∣∣dx

<ε

3+ε

3+ε

3= ε.

That is, ∣∣∣∣∫ +∞

0

∣∣f(x+1n

)− f(x)∣∣dx− 0

∣∣∣∣ < ε,

for al n ≥ N . Therefore, limn→+∞∫ +∞0

∣∣f(x+ 1n )− f(x)

∣∣dx = 0, which is what wewant.

2.6. SIXTH COMPETITION 85

2.6.2. Algebra. 1. Proceed by contradiction. Suppose that AB is a subgroupof G. Using Lagrange’s Theorem , we see that |AB| = pnb, where b ∈ N and b|a.Since gcd(p, a) = 1 and b|a, gcd(p, b) = 1. On the other hand, we can write

|AB| =|A| × |B||A ∩B|

.

But A∩B $ B because B * A. Hence, |A∩B| < |B|. Now, in view of Lagrange’sTheorem, let |A ∩B| = pi for some 0 ≤ i < m. We can write

pnb =pn.pm

pi=⇒ b = pm−i,

implying that gcd(p, b) = p, which is a contradiction. Therefore, AB cannot be asubgroup of G, finishing the proof.

2. We prove the assertion for matrices over a general division ring ∆. Induct on m.If m = 1, the assertion is easily seen to hold. Assuming that the assertion holds formatrices of size less than m, we prove the assertion for matrices of size m. To thisend, note first that r := rank(A1) ≤ dim kerA1 because A2

1 = 0. Now, from theRank-Nullity Theorem, we see that r ≤ m

2 . Let K := kerA1. Since Ai’s commute,K is invariant under all Ai’s (1 ≤ i ≤ n). Therefore, after a similarity, one canwrite

Ai =(Bi Ci0 Di

),

where Bi ∈ Mm−r(∆) with B1 = 0, Ci ∈ M(m−r)×r(∆), and Di ∈ Mr(∆) for all1 ≤ i ≤ n. We have m < 2n, yielding r ≤ m

2 < 2n−1. As Ai’s commute, so doDi’s in Mr(∆) (1 ≤ i ≤ n). Hence, from the induction hypothesis, we see thatD2 · · ·Dn = 0, which, in turn, implies that

A2 · · ·An =(B C0 0

), (∗)

for some appropriate matrices B ∈Mm−r(∆) and C ∈M(m−r)×r(∆). On the otherhand,

A1 =(

0 C1

0 D1

). (∗∗)

It now easily follows from (∗) and (∗∗) that

A1A2 · · ·An = A1(A2 · · ·An−1An) = 0,

which is what we want.

3. Since e2 = e, f2 = f , and ef = fe, using the Binomial Theorem, we see thatfor all n ∈ N, there exists a Kn ∈ N such that (e − f)n = e + (−1)nf + Knef .Suppose (e − f)n = 0. Without loss of generality, we may assume that n is even.It follows that

(e− f)n = e+ f +Knef = 0.Multiplying both sides of the above by e and f , respectively, and noting that e2 = e,f2 = f , and ef = fe, we obtain

e+ ef +Knef = 0 = f + ef +Knef,

yielding e = f = −(1 +Kn)ef , which proves the assertion.

86 2. SOLUTIONS

2.6.3. General. 1. Noting that (−1)|n| = (−1)n for all n ∈ N, we can write

(−1)|n1−m1|+···+|nk−mk| = (−1)|n1−m1| × · · · × (−1)|nk−mk|

= (−1)n1−m1 × · · · × (−1)nk−mk

= (−1)(n1+···+nk)−(m1+···+mk) = (−1)0 = 1,

implying that |n1 −m1|+ · · ·+ |nk −mk| is an even number.

2. Proceed by induction on n. If n = 1, the assertion is easy. Assuming that theassertion holds for n×n matrices, we prove it for (n+1)×(n+1) matrices. Supposethat A is a (n+ 1)× (n+ 1) matrix such that the sum of the elements on any rowand any column corresponding to any nonzero element of A is at least n + 1. Asthe entries of A are all nonnegative integers, if the entries are all nonzero, then thesum of the entries of the matrix A is at least (n+ 1)2 which is no less than (n+1)2

2 .Now, assume that for some 1 ≤ i, j ≤ n+ 1, aij = 0. Let Aij be the n× n matrixwhich is obtained from eliminating row i and column j of the matrix A. Obviously,the sum of the elements on any row and any column corresponding to any nonzeroelement of the matrix Aij is at least n. So it follows from the induction hypothesisthat the sum of the entries of Aij is no less than n2

2 . Noting that the sum of theentries of row i and column j of A is at least n + 1, we see that the sum of theentries of the matrix A is at least

n2

2+ n+ 1 =

(n+ 1)2 + 12

>(n+ 1)2

2,

which is what we want, finishing the proof.

3. The assertion is a special case of the following: Suppose k + m points insidea circle are given. Then, there exists a straight line not passing through any ofthe points that divides the circle into two sections one of which containing k pointsand the other the remaining m points. To see this, note that the number of linespassing through a pair of points from these k +m points is finite; in fact it is lessthan or equal to

(k+m

2

). As the number of directions determined by these lines is

finite, there is a line d whose direction is not parallel to any of the lines determinedby any pair of points from the given points. Let AA′ be a diameter of the circlewhose direction is perpendicular to the line d. Project the given points on the linesegment AA′ to get k+m distinct points. Going from A to A′, if P is the kth pointand Q the (k+1)st point, then any line perpendicular to AA′ at any point betweenP and Q does not pass through any of the given points inside the circle and yetit divides the circle into two sections one of which contains k points, including thepoint whose image is P , and the other contains the remaining m points.

2.7. Seventh Competition

2.7.1. Analysis. 1. First of all, the hypothesis that S is closed is redundantbecause it follows from the other hypothesis of the problem that such an S is closed.The assertion is a consequence of the following. Let n ∈ N. Then there is no propersubset S of Rn with the property that for all x ∈ Rn \ S there exists nx ∈ N withnx > 1 such that there are exactly nx points in S as the closest point of S to x. To

2.7. SEVENTH COMPETITION 87

prove this by contradiction, pick a point x ∈ Rn \S. It follows from the hypothesisthat there exists an nx ∈ N with nx > 1 and nx points s1, . . . , snx in S such that

infs∈S

||x− s|| = ||x− si|| = δx,

for all 1 ≤ i ≤ nx. It is obvious that there is no point of S in Bδx(x), the open ball

centered at x with radius δx. Let y = x+s12 and δ = δx

2 . Plainly, the only point ofS in Bδ(y), the closed ball centered at y with radius δ, is s1. As Bδ(y) ⊆ Bδx

(x),it follows that for the point y ∈ Rn \ S there exists exactly one point s1 ∈ S suchthat

infs∈S

||x− s|| = ||x− s1||,

which is a contradiction, finishing the proof.

2. We need the following lemma.

Lemma. Let (Ir)r∈Q+ and A and f be as in the statement of the problem.Then,

(i) for all x ∈ A with x ∈ Ir, we have f(x) ≤ r.(ii) for all x ∈ A with x /∈ Ir, we have f(x) ≥ r.Proof. Set Sx = r : x ∈ Ir. We have f(x) = inf Sx. To prove (i), note that

if x ∈ Ir, then Ir ⊆ Is whenever r < s where s ∈ Q, from which, we obtain x ∈ Isfor all s ∈ Q with r < s. Consequently, Sx includes all rational numbers greaterthan r, whence f(x) = inf Sx ≤ r.

To prove (ii), note that if x /∈ Ir, then Is ⊆ Ir whenever s < r where s ∈ Q.This implies x /∈ Is for all s ∈ Q with s < r. Therefore, Sx contains no rationalnumber less than r, whence f(x) = inf Sx ≥ r, proving the lemma.

We now prove that f is continuous. Let x0 be an arbitrary point of A. It sufficesto prove that for any open interval (c, d) with f(x0) ∈ (c, d) there exists an openneighborhood N around x0 such that f(N) ⊆ (c, d). To this end, as c < f(x0) < d,pick rational numbers p and q such that c < p < f(x0) < q < d. We claim thatthe open set N = Iq \ Ip contains x0 and moreover f(N) ⊆ [p, q] ⊆ (c, d), finishingthe proof. Firstly, x0 ∈ N because, in view of the lemma, x0 ∈ Iq and x0 /∈ Ip.Secondly, f(N) ⊆ [p, q] ⊆ (c, d). To see this, let x ∈ N = Iq \ Ip be arbitrary. Itfollows that x ∈ Iq ⊆ Iq, which, in view of the lemma, implies f(x) ≤ q. On theother hand, since x /∈ Ip, we conclude from the lemma that f(x) ≥ p. That is, wehave proved f(x) ∈ [p, q] for all x ∈ N . Consequently, f(N) ⊆ [p, q] ⊆ (c, d), whichis what we want.

3. First solution: To find the limit, we need the following lemma whose first andsecond parts are known as Stolz’s First and Second Theorems on limits.

Lemma. (i) Let (pn)+∞n=1 be a sequence of reals whose elements are eventuallynonnegative, sn = p1 + · · · + pn, and limn sn = +∞. If (an)+∞n=1 is a sequence ofreal numbers such that limn an ∈ R ∪ ±∞, then

limn

p1a1 + · · ·+ pnanp1 + · · ·+ pn

= limnan.

88 2. SOLUTIONS

(ii) Let (xn)+∞n=1 and (yn)+∞n=1 be sequences of real numbers with limn yn = +∞and such that (yn)+∞n=1 is eventually increasing. If limn

xn−xn−1yn−yn−1

∈ R ∪ ±∞, then

limn

xnyn

= limn

xn − xn−1

yn − yn−1.

Proof. (i) There are two cases to consider.(a) limn an = ±∞.As p1(−a1)+···+pn(−an)

p1+···+pn= −p1a1+···+pnan

p1+···+pn, it suffices to prove the assertion for

the case when limn an = +∞. To this end, it follows from the hypothesis that forgiven M > 0, there exists an N1 > 0 such that an > 2M whenever n > N1. LetAN1 = p1a1 + · · ·+ pN1aN1 . We can write

p1a1 + · · ·+ pnanp1 + · · ·+ pn

=AN1

sn+pN1+1aN1+1 + · · ·+ pnan

sn

>AN1

sn+ 2M

sn − sN1

sn

But limnAN1sn

= 0 and limnsn−sN1sn

= 1. Thus, there exists N2 > 0 such thatAN1sn

> −M3 and sn−sN1

sn> 2

3 whenever n > N2. Letting N = max(N1, N2), for alln > N , we have

p1a1 + · · ·+ pnanp1 + · · ·+ pn

>−M

3+

4M3

= M.

As M > 0 is arbitrary, this means limnp1a1+···+pnan

p1+···+pn= +∞, proving the assertion

in this case.(b) limn an = a ∈ R.First, let limn an = 0. It follows that for given ε > 0 there exists N1 > 0 such

that |an| < ε2 whenever n > N1. Let AN1 be as in (a). For all n > N1, we can write∣∣∣p1a1 + · · ·+ pnanp1 + · · ·+ pn

∣∣∣ ≤ ∣∣AN1

∣∣sn

+ε

2× sn − sN1

sn≤∣∣AN1

∣∣sn

+ε

2.

Now, as limnAN1sn

= 0, we see that there exists an N2 > 0 such that∣∣AN1

∣∣sn

< ε2

whenever n > N2. Letting N = max(N1, N2), for all n > N , we have∣∣∣p1a1 + · · ·+ pnanp1 + · · ·+ pn

∣∣∣ < ε

2+ε

2= ε.

That is, limnp1a1+···+pnan

p1+···+pn= 0, which is what we want.

Next, let limn an = a 6= 0. As limn(an − a) = 0, from what we just proved, itfollows that

limn

(p1a1 + · · ·+ pnanp1 + · · ·+ pn

− a

)= lim

n

p1(a1 − a) + · · ·+ pn(an − a)p1 + · · ·+ pn

= limn

(an − a) = 0.

This implies that limnp1a1+···+pnan

p1+···+pn= a, finishing the proof.

(ii) Define the two sequences (pn)+∞n=1 and (an)+∞n=1, inductively, as follows

pn = yn − yn−1, p1 = y1, (n ≥ 2),xn − xn−1 = pnan, x1 = p1a1, (n ≥ 2).

2.7. SEVENTH COMPETITION 89

Firstly, pn’s are eventually nonnegative. Secondly, yn = p1 + · · · + pn → +∞ asn→ +∞. Thirdly, an = xn−xn−1

yn−yn−1and p1a1+· · ·+pnan = xn, and p1+· · ·+pn = yn.

Therefore, (i) applies, proving the assertion.

Now, note that

xn =21

1 + · · ·+ 2n

n2n+1

n+1

=Xn

Yn.

First of all, it is easily seen that the sequence (Yn)+∞n=1, where Yn = 2n+1

n+1 for alln ∈ N, is increasing. Secondly, limn Yn = +∞. Thirdly, we can write

limn

Xn −Xn−1

Yn − Yn−1= lim

n

2n

n2n+1

n+1 −2n

n

= limn

12nn+1 − 1

= 1.

Therefore, it follows from the lemma that limn xn = limnXn

Ynexists, and moreover

limnxn = lim

n

Xn

Yn= lim

n

Xn −Xn−1

Yn − Yn−1= 1,

which is what we want.

Second solution: It is readily checked that

xn+1 =n+ 22n+2

n+1∑k=1

2k

k=

n+ 22(n+ 1)

(xn + 1

),

from which, we obtain

xn+2 − xn+1 =(n+ 2)2(xn+1 − xn)− xn+1 − 1

2(n+ 1)(n+ 2),

for all n ∈ N. As xn > 0 for all n ∈ N, in view of the above equality, we see thatxn+2−xn+1 < 0 whenever xn+1−xn ≤ 0. But a straightforward calculation revealsthat x3 = x4 = 5

3 . Thus, by an easy induction on n ≥ 3, we see that the sequence(xn)+∞n=1 is decreasing for n ≥ 3. From this, together with the fact that xn > 0 forall n ∈ N, we conclude that limn→+∞ xn exists, and hence limn→+∞ xn = L forsome L ∈ R. Consequently,

L = limn→+∞

xn+1 = limn→+∞

n+ 22(n+ 1)

(xn + 1

)=

12(L+ 1) =⇒ L = 1.

Therefore, limn→+∞ xn = 1, which is what we want.

2.7.2. Algebra. 1. (a) Let x0 ∈ R be arbitrary. We show that x0e = x0. Tothis end, note that for all x ∈ R, we can write

(x0 + e− x0e)x = x0x+ ex− x0(ex) = x0x+ x− x0x = x.

That is, (x0 +e−x0e)x = x for all x ∈ R. As the left identity element is unique, weobtain x0 + e− x0e = e, implying x0e = x0. This means e a right identity elementas well. Consequently, e is the identity element of R.

(b) Let a be an idempotent of R, i.e. a2 = a, and x ∈ R be arbitrary. Itis straightforward to see that (axa − ax)2 = 0. That is, axa − ax is a nilpotentelement of R, and hence axa−ax = 0, by the hypothesis. Likewise, (axa−xa)2 = 0,

90 2. SOLUTIONS

implying axa− xa = 0 for all x ∈ R. So we have shown that ax = xa = axa for allx ∈ R, which is what we want.

2. First solution: Let H be an arbitrary subgroup of G. Set H1 = H ∩G1. Weshow that H1 is commutative, H1 E H, and that H

H1is commutative. First, H1

is commutative because H1 ⊆ G1 and G1 is commutative. Next, H1 ≤ H becauseH1 = H ∩ G1 and H and G1 are subgroups of G. That H1 E H follows fromG1 E G and H1 = H ∩ G1. Finally, to show that H

H1is commutative, using the

Second Isomorphism Theorem for groups, we can writeH

H1=

H

H ∩G1

∼=HG1

G1.

But HG1 ≤ G. Hence, HG1G1

≤ GG1

. Consequently, HG1G1

is commutative because sois G

G1. This, in view of the above isomorphism, completes the proof.

Second solution: Use G′ to denote the derived subgroup of G. Since GG1

isabelian, it follows that G′ ≤ G1. Now for a given subgroup H of G, set

H1 = H ′ :=⟨h1h2h

−11 h−1

2 : h1, h2 ∈ H⟩.

It is plain that H1 is a normal subgroup of H and that HH1

is abelian because H1 isthe derived subgroup of H. To see that H1 is abelian, just note that H1 ≤ G′ ≤ G1

and that G1 is abelian by the hypothesis. This completes the proof.

3. For the given θ : V → V , it is obvious that

ker θ ⊆ ker θ2 ⊆ · · ·is an ascending chain of the subspaces of V . As V is finite-dimensional, it followsthat there exists an s ∈ N such that ker θs = ker θs+1.

We claim that if ker θs = ker θs+1, then ker θs = ker θs+k for all k ∈ N. Weprove this by induction on k. If k = 1, there is nothing to prove. Suppose theassertion holds for k. To prove the assertion for k + 1, note first that ker θs ⊆ker θs+k+1. Next, let x ∈ ker θs+k+1 be arbitrary. We obtain θs+k(θx) = 0. Thatis, θx ∈ ker θs+k, which, in view of the induction hypothesis, yields θx ∈ ker θs.This implies θs+1x = 0 which means x ∈ ker θs+1. But ker θs+1 = ker θs. Thus,x ∈ ker θs. As x ∈ ker θs+k+1 was arbitrary, we see that ker θs+k+1 ⊆ ker θs.Therefore, ker θs+k+1 = ker θs, proving the claim by induction on k.

We now show that V = im(θs)⊕ ker(θs), where s is as in the above. Note thatby the Rank-Nullity Theorem

dim im(θs) + dim ker(θs) = dimV.

On the other hand, by Problem 1(a) of 1.3.2, we can write

dim(im(θs) + ker(θs)) = dim im(θs) + dim ker(θs)− dim(im(θs) ∩ ker(θs)

),

which, in view of the preceding equality, implies V = im(θs)⊕ker(θs) as soon as weshow that im(θs)∩ ker(θs) = 0. To see this, let x ∈ im(θs)∩ ker(θs) be arbitrary.It follows that there exists x1 ∈ V such that x = θsx1, from which, we obtainθs+sx1 = 0 because x ∈ ker θs. On the other hand, by the claim we made in theabove, we have ker θs+s = ker θs, yielding x1 ∈ ker θs. In other words, x = θsx1 = 0.Since x ∈ im(θs) ∩ ker(θs) was arbitrary, we conclude that im(θs) ∩ ker(θs) = 0,which is what we want. Therefore, V = im(θs)⊕ ker(θs), completing the proof.

2.8. EIGHTH COMPETITION 91

2.8. Eighth Competition

2.8.1. Analysis. 1. (a) Define the function h : F ∪G→ R by

h(x) =f(x) x ∈ F,g(x) x ∈ G.

As f(x) = g(x) on F ∩G, the function h is well-defined. It is obvious that h is anextension of f and g to F ∪ G. To prove that h is continuous, it suffices to showthat h−1(C) is a closed subset of F ∪G whenever C is a closed subset of R. To thisend, note that for any closed subset C of R, we can write

h−1(C) = f−1(C) ∪ g−1(C).

Now, as f and g are continuous and C is a closed subset of R, we see that f−1(C)and g−1(C) are closed subsets of F and G, respectively. On the other hand, F andG are closed in F ∪G. Hence, f−1(C) and g−1(C) are closed subsets of F ∪G. Thistogether with the above equality implies that h−1(C) is a closed subset of F ∪ G,which is what we want.

(b) For given n ∈ N, let F = B1(0) :=x ∈ Rn : ||x|| < 1

and G = F c =

x ∈

Rn : ||x|| ≥ 1. Obviously, F ∩G = ∅ and F ∪G = Rn. Now, define the functions

f : F → R and g : G → R by f(x) = 0 and g(x) = 1 whenever x ∈ F and x ∈ G,respectively. For this f and g, we show that the conclusion of (a) does not hold.Suppose to the contrary that h : Rn = F ∪ G → R is an extension of f and g toRn = F ∪G. We must have

h(x) =

0 x ∈ F,1 x ∈ G,

which is not continuous on Rn = F ∪ G, for otherwise h(Rn) = 0, 1, which isdisconnected whereas Rn is connected. This leads to a contradiction, proving theassertion.

2. Remark. It is worth mentioning that using a proof almost identical to theproof below one can show that if (an)+∞n=1 is a decreasing sequence of real numberswith limn an = 0, then

∑+∞n=1 an is convergent if and only if

∑+∞n=1 n(an − an+1) is

convergent. Also, see Problem 2 of 1.12.1.

We have limn an = 0 because∑+∞n=1 an is convergent. Since (an)+∞n=1 is decreasing,

it follows that an ≥ an+k for all n, k ∈ N. Letting k → +∞, we obtain an ≥ 0 forall n ∈ N. Now, as

∑+∞n=1 an is convergent, it follows that it is Cauchy. Thus, for

given ε > 0, there exists an N > 0 such that∣∣ n∑k=m

ak∣∣ < ε

2,

for all n ≥ m ≥ N . Let m = [n2 ] > N . We obtain n > m ≥ N . This together withthe fact that an’s are nonnegative implies

n∑k=[ n

2 ]

ak <ε

2.

92 2. SOLUTIONS

As an’s are decreasing, we have (n− [n2 ] + 1)an ≤∑nk=[ n

2 ] ak, yielding

nan2

≤ nan2

+(n2− [

n

2] + 1

)an

=(n− [

n

2] + 1

)an ≤

n∑k=[ n

2 ]

ak

<ε

2,

for all n > 2N + 2. Consequently, nan < ε for all n > 2(N + 1). That is,limn nan = 0. We can write

+∞∑n=1

n(an − an+1) =+∞∑n=1

((nan − (n+ 1)an+1

)+ an+1

).

Since limn nan = 0, we see that the telescopic series∑+∞n=1

(nan − (n + 1)an+1

)converges to a1. On the other hand,

∑+∞n=1 an+1 is convergent. Therefore, the

series∑+∞n=1 n(an − an+1) converges and we have

+∞∑n=1

n(an − an+1) =+∞∑n=1

((nan − (n+ 1)an+1

)+ an+1

)=

+∞∑n=1

(nan − (n+ 1)an+1

)+

+∞∑n=1

an+1

= a1 ++∞∑n=1

an+1,

finishing the proof.

3. First solution: As the function g is continuously differentiable on [0, 1], itfollows that g is of bounded variation on [0, 1]. We show that g can be written asthe difference of two increasing functions g1 and g2 each of which is continuouslydifferentiable on [0, 1]. To see this, let g′ = f , where f is continuous on [0, 1]. Wecan write f = f+ − f−, where f+ = max(f, 0) = f

2 + |f |2 and f− = max(−f, 0) =

−f2 + |f |

2 . Obviously, f+ and f− are nonnegative continuous functions on [0, 1].Using the Second Fundamental Theorem of Calculus, we can write

g(x) = g(0) +∫ x

0

f =(g(0) +

∫ x

0

f+)−∫ x

0

f− = g1(x)− g2(x),

where g1(x) = g(0) +∫ x0f+ and g2(x) =

∫ x0f−. Applying the First Fundamental

Theorem of Calculus, we obtain

g′1(x) = f+(x) ≥ 0, g′2(x) = f−(x) ≥ 0,

for all x ∈ [0, 1]. Therefore, g = g1 − g2, where g1 and g2 are continuously differen-tiable and increasing on [0, 1]. We can write∫ 1

0

xndg =∫ 1

0

xnd(g1 − g2) =∫ 1

0

xndg1 −∫ 1

0

xndg2.

2.8. EIGHTH COMPETITION 93

Now, as g1 and g2 are continuously differentiable on [0, 1], we have∫ 1

0

xndg =∫ 1

0

xng′1(x)dx−∫ 1

0

xng′2(x)dx =∫ 1

0

xn(g′1(x)− g′2(x)

)dx.

Hence, ∫ 1

0

xndg =∫ 1

0

xng′(x)dx.

Now, as g′ is continuous on [0, 1], there is an M > 0 such that |g′| ≤ M on [0, 1],from which, we see that∣∣∣∣∫ 1

0

xndg

∣∣∣∣ =∣∣∣∣∫ 1

0

xng′(x)dx∣∣∣∣ ≤ ∫ 1

0

xn∣∣g′(x)∣∣dx

≤ M

∫ 1

0

xndx =M

n+ 1,

implying that limn

∫ 1

0xndg = 0 because limn

Mn+1 = 0.

Second solution: As g is continuously differentiable on [0, 1], it is of boundedvariation on [0, 1]. Hence, there exists increasing functions g1 and g2 on [0, 1] suchthat g = g1 − g2. By the definition of the Riemann-Stieltjes integrals with respectto integrands of bounded variations, we have∫ 1

0

xndg =∫ 1

0

xndg1 −∫ 1

0

xndg2.

Using integration by parts, we obtain∫ 1

0

xndgi = xngi(x)∣∣∣10−∫ 1

0

gi(x)dxn = gi(1)−∫ 1

0

ngi(x)xn−1dx,

for each i = 1, 2. So we have∫ 1

0

xndg = g(1)−∫ 1

0

ng(x)xn−1dx =∫ 1

0

n(g(1)− g(x)

)xn−1dx.

Now, for given ε > 0, as g is continuous at 1, there exists a 0 < δ < 1 such that|g(x)− g(1)| < ε

2 whenever 1− δ < x < 1. We can write∣∣∣∣∫ 1

0

xndg

∣∣∣∣ =∣∣∣∣∫ 1

0

n(g(1)− g(x)

)xn−1dx

∣∣∣∣ ≤ ∫ 1

0

n∣∣g(1)− g(x)

∣∣xn−1dx

=∫ 1−δ

0

n∣∣g(1)− g(x)

∣∣xn−1dx+∫ 1

1−δn∣∣g(1)− g(x)

∣∣xn−1dx.

By the continuity of g on [0, 1], there exists M > 0 such that |g| ≤ M on [0, 1].With this in mind, we have∣∣∣∣∫ 1

0

xndg

∣∣∣∣ ≤ 2Mn

∫ 1−δ

0

xn−1dx+nε

2

∫ 1

1−δxn−1dx

= 2M(1− δ)n +ε

2(1− (1− δ)n

)< 2M(1− δ)n +

ε

2.

Consequently, ∣∣∣∣∫ 1

0

xndg

∣∣∣∣ < 2M(1− δ)n +ε

2.

94 2. SOLUTIONS

Now, limn(1− δ)n = 0 because 0 < 1− δ < 1. Hence, there exists N > 0 such that(1− δ)n < ε

4M for all n ≥ N . Therefore,∣∣∣∣∫ 1

0

xndg

∣∣∣∣ < 2M × ε

4M+ε

2= ε,

for all n ≥ N . This means limn

∫ 1

0xndg(x) = 0, which is what we want.

2.8.2. Algebra. 1. Note that b = I + N , where I is the identity matrix,N = b− I, and N2 = 0. From this, using the Binomial Theorem and noting thatb−1 = I −N , it is easily verified that

bn =(

1 n0 1

),

for all n ∈ Z. Therefore,

H = bn : n ∈ Z =(

1 n0 1

): n ∈ Z

.

To find aHa−1, note that a−1 =(

12 00 1

). We can write

abna−1 =(

2 00 1

)(1 n0 1

)(12 00 1

)=(

1 2n0 1

).

Therefore,

aHa−1 =(

1 2n0 1

): n ∈ Z

.

This implies that aHa−1 H. That is, aHa−1 is a proper subgroup of H, whichis what we want.

2. We need the “only if part” of the following standard lemma.

Lemma. Let R be a commutative ring with identity. Then, x ∈ J(R) :=⋂m C R : m is maximal in R

if and only if 1−xy is a unit in R for all y ∈ R.

Proof. “⇐=” Let x ∈ J(R). Suppose to the contrary that there exists ay0 ∈ R such that 1 − xy0 is not a unit. As 1 − xy0 is not invertible, a standardargument using Zorn’s Lemma shows that there exists a maximal ideal m0 suchthat 1 − xy0 ∈ m0. On the other hand, since x ∈ J(R), x ∈ m0, and hencexy0 ∈ m0. So, we obtain 1 = (1 − xy0) + xy0 ∈ m0, yielding m0 = R, which is acontradiction. Thus the assertion follows by contradiction.

“=⇒” Again, we proceed by contradiction. Suppose that there exists a maximalideal m0 such that x /∈ m0. As m0 $ m0 + xR and m0 is maximal, we see thatm0 + xR = R. Hence, there exist a ∈ m0 and y0 ∈ R such that 1 = a + xy0.This together with the hypothesis implies that a = 1 − xy0 ∈ m0 is a unit in R,which, in turn, implies m0 = R, a contradiction. Therefore, the assertion followsby contradiction.

To prove the assertion by contradiction, suppose that R has finitely manymaximal ideals m1, . . . ,mn for some n ∈ N. Firstly, 0 $ mi for each i = 1, . . . , nbecause otherwise 0 and R are the only ideal of R, implying that R is a field,

2.8. EIGHTH COMPETITION 95

which is a contradiction, for R would then have infinitely many units. On the otherhand, we have

n⋂i=1

mi ⊇ m1 · · ·mn.

We also have m1 · · ·mn 6= 0 because 0 $ mi for each i = 1, . . . , n and R isan integral domain. Now, pick 0 6= a ∈ m1 · · ·mn ⊆

⋂ni=1mi =

⋂m C R :

m is maximal in R. It follows from the lemma that 1−ax is a unit in R for x ∈ R.But, as R is an integral domain, 1 − ax 6= 1 − ay whenever x 6= y. This togetherwith the fact that R is an infinite integral domain implies that R has infinitelymany units, which is a contradiction. So the assertion follows by contradiction.

3. Let A = (aij) ∈ Mn(R). Use the symbol (A2)ij to denote the ij entry of thematrix A2. Using A2 = I, we can write

n∑i=1

n∑j=1

(A2)ij =n∑i=1

n∑j=1

(I)ij = n.

On the other hand,n∑i=1

n∑j=1

(A2)ij =n∑i=1

n∑j=1

n∑k=1

aikakj =n∑i=1

n∑k=1

aik

n∑j=1

akj

=n∑i=1

n∑k=1

aika = a

n∑i=1

n∑k=1

aik = a

n∑i=1

a = na2.

Therefore, na2 = n, yielding a = ±1. Note that both cases a = 1 or a = −1 canhappen because A = ±I yields a = ±1, respectively.

2.8.3. General. 1. Note that f(x, y, z) = (x− 3)2 + 4(y+ 12 ) + z2 − 10. This

yields f(x, y, z) ≥ −10 for all x, y, z ∈ R and that f(x, y, z) = −10 if and only ifx = 3 , y = − 1

2 , and z = 0. Therefore, the function f assumes it absolute minimumat (3,− 1

2 , 0), which is what we want.

2. Let x1 < · · · < x5 be the amount of money owned by these five people inincreasing order. We have x1

∣∣x2, x2

∣∣x3, x3

∣∣x4, and x4

∣∣x5. Hence, there existnatural numbers ki (1 ≤ i ≤ 4) such that xi+1 = kixi. Using the hypothesis, wecan write

x1(1 + k1 + k1k2 + k1k2k3 + k1k2k3k4) = 719.But 719 is prime. So, we obtain

x1 = 1, 1 + k1 + k1k2 + k1k2k3 + k1k2k3k4 = 719.

Consequently, k1(1+k2 +k2k3 +k2k3k4) = 718 = 2×359. As 2 and 359 are primes,we must have

k1 = 2, 1 + k2 + k2k3 + k2k3k4 = 359, or,k1 = 1, 1 + k2 + k2k3 + k2k3k4 = 718, or,k1 = 359, 1 + k2 + k2k3 + k2k3k4 = 2, or,k1 = 718, 1 + k2 + k2k3 + k2k3k4 = 1.

The last three cases are easily seen to be refuted. Thus,

k1 = 2, 1 + k2 + k2k3 + k2k3k4 = 359.

96 2. SOLUTIONS

Simplifying, we obtain k2(1 + k3 + k3k4) = 358 = 2 × 179. Since 2 and 179 areprimes, we see that

k2 = 2, 1 + k3 + k3k4 = 179, or,k2 = 1, 1 + k3 + k3k4 = 358, or,k2 = 179, 1 + k3 + k3k4 = 2, or,k2 = 358, 1 + k3 + k3k4 = 1.

Once again, refuting the last three cases, we obtain k2 = 2, 1 + k3 + k3k4 = 179,which yields k3(1 + k4) = 178 = 2× 89. Likewise, noting that 2 and 89 are primes,we get

k3 = 2, 1 + k4 = 89, or,k3 = 1, 1 + k4 = 178, or,k3 = 89, 1 + k4 = 2, or,k3 = 178, 1 + k4 = 1.

Again, refuting the last three cases, we obtain k3 = 2, 1 + k4 = 89, which yieldsk4 = 88. Therefore, the only solution of x1 + · · ·+x5 = 719 subject to the imposedconditions is x1 = 1, x2 = k1x1 = 2, x3 = k2x2 = 4, x4 = k3x3 = 8, andx5 = k4x4 = 704. That is, these five people in increasing order have

x1 = 1, x2 = 2, x3 = 4, x4 = 8, x5 = 704,

amounts of money, which is what we want.

3. Without loss of generality, we may assume that starting with every minute, thetraffic-light of the crossroad B is 30 seconds red and 30 seconds green, respectively.If t denotes the tth second of a particular minute at the crossroad B, then the stoptime function of the bus at the crossroad, denoted by f , on that particular minute,is

f(t) =

30− t 0 ≤ t ≤ 30,0 30 ≤ t ≤ 60.

Let µf denote the average value of the function f , which means the average of thestop time at the crossroad B. By definition, we have

µf =160

∫ 60

0

f(t)dt =160

∫ 30

0

(30− t)dt = 7.5.

But the bus passes through the crossroad B of the city with the probability of13 . Therefore, the average of the stop time of the bus at the crossroad is equal to13µf = 7.5

3 = 2.5, which is what we want.

2.9. Ninth Competition

2.9.1. Analysis. 1. Suppose that C is an equivalence class of the equivalencerelation ∼. Pick an x0 ∈ C. It is plain that

C =x0 + r|r ∈ Q

.

It suffices to show that C ∩ (a, b) 6= ∅ for all open intervals (a, b) of R. To thisend, for a given open interval (a, b), as Q is dense in R, we see that there exists anr ∈ Q ∩ (a − x0, b − x0). This yields x0 + r ∈ C ∩ (a, b), implying C ∩ (a, b) 6= ∅,which is what we want.

2.9. NINTH COMPETITION 97

2. First, we claim that the function f is strictly increasing or strictly decreas-ing. To prove this by contradiction, suppose that f is neither strictly increasingnor strictly decreasing. By the lemma presented in Solution 2 of 2.5.1, there ex-ist x1, x2, x3 ∈ R with x1 < x2 < x3 such that f(x2) ≥ max

(f(x1), f(x3)

)or

f(x2) ≤ min(f(x1), f(x3)

). As f is one-to-one and x1 < x2 < x3, we see that

f(x2) > max(f(x1), f(x3)

)or f(x2) < min

(f(x1), f(x3)

). First, suppose f(x2) >

max(f(x1), f(x3)

). Letting λ =

f(x2)+max(f(x1),f(x3)

)2 , we see that f(x1) < λ <

f(x2) and f(x3) < λ < f(x2). Hence, by the Intermediate Value Theorem,there exist x1 < ξ1 < x2 and x2 < ξ2 < x3 such that f(ξ1) = λ = f(ξ2).As f is one-to-one, this yields ξ1 = ξ2, which is a contradiction. Likewise, if

f(x2) < min(f(x1), f(x3)

), considering λ =

f(x2)+min(f(x1),f(x3)

)2 , we obtain a

contradiction, proving the claim. Thus, the function f is either strictly increasingor strictly decreasing. We prove the assertion in the case when f is strictly increas-ing. The case when f is strictly decreasing can be done similarly or by replacing f by−f and repeating the above argument. To prove that f−1 is continuous, it sufficesto show that f is an open map. To this end, note first that f(−∞) = limx→−∞ f(x)and f(+∞) = limx→+∞ f(x) exist, where f(±∞) ∈ R ∪ ±∞, and that, in viewof the Intermediate Value Theorem, f(R) =

(f(−∞), f(+∞)

). Analogously, if

(a, b) is an open interval in R, we obtain f(a, b) =(f(a), f(b)

). Now, suppose that

G is an open subset of R. It follows that there exists a family (ai, bi)i∈I of openintervals of R such that G =

⋃i∈I(ai, bi). We can write

f(G) = f(⋃i∈I

(ai, bi))

=⋃i∈I

f(ai, bi) =⋃i∈I

(f(ai), f(bi)

).

But(f(ai), f(bi)

)is an open interval, hence an open subset, of f(R) =

(f(−∞), f(+∞)

).

This, in view of the above equality, implies that f(G) is an open subset of f(R).Therefore, f is an open map, which is what we want.

3. To show f(E) = E, assume first that x ∈ E is arbitrary. We need to showthat x ∈ f(E). To see this, note that f(x) ≥ x. If f(x) = x, there is nothingto prove. If f(x) > x, we have f(a) = a ≤ x < f(x). It thus follows from theIntermediate Value Theorem that there exists a c ∈ [a, x) such that f(c) = x,implying x ∈ f [a, b]. By showing c ∈ E, we conclude that x ∈ f(E), which iswhat we want. If c /∈ E, then f(c) < c, from which, as f is increasing, we see thatf(f(c)

)< f(c). But f(c) = x, which obtains f(x) < x. This is a contradiction,

yielding x ∈ f(E), implying E ⊆ f(E).Next, suppose that y = f(x) ∈ f(E), where x ∈ E, is arbitrary. As x ∈ E, we

have f(x) ≥ x, from which, as f is increasing, we obtain f(f(x)

)≥ f(x). That is,

y ∈ [a, b] and f(y) ≥ y, which means y = f(x) ∈ E. In other words, f(E) ⊆ E,proving the assertion.

4. Remark. The function f , as defined, in not well-defined at 0. Since the valueof f at any point has no affect on its integrability, we redefine f : [0, 1] → R by

f(x) = 1

q x = pq , p, q ∈ N, gcd(a, b) = 1,

0 otherwise.

98 2. SOLUTIONS

First solution: Just as explained in Solution 1 of 2.5.1, the sequence (fn)+∞n=1,with fn : [0, 1] → [0, 1], defined by

fn(x) = 1

q x = pq , p, q ∈ N, gcd(p, q) = 1, q ≤ n,

0 otherwise,

converges uniformly to f on [0, 1]. As fn (n ∈ N) is zero everywhere but at finitelymany points of the interval [0, 1], it follows that fn’s are integrable on [0, 1]. Hence,f is Riemann integrable on [0, 1], and moreover∫ 1

0

f(x)dx = limn

∫ 1

0

fn(x)dx = 0.

Second solution: As shown in Solution 3 of 2.1.1, we have limx→a f(x) = 0 forall a ∈ (0, 1), limx→0+ f(x) = 0, and limx→1− f(x) = 0. Thus, the function f iscontinuous on the set [0, 1] \

(Q∩ (0, 1]

)and it is discontinuous on Q∩ (0, 1], which

is a countable set. From this point on, we present two proofs for the integrabilityof f on [0, 1].

First proof. Just note that the set of points at which f is discontinuous iscountable and hence has measure zero. Therefore, using the Lebesgue’s IntegrabilityCriterion for Riemann integrals, we see that f is Riemann integrable, which is whatwe want.

Second proof. Using Lebesgue’s Number Lemma, we directly prove that f isRiemann integrable on [0, 1]. Recall that Lebesgue’s Number Lemma asserts thatif an open cover of a compact metric space X is given, then there exists a numberδ > 0, called a Lebesgue number for the cover, such that every subset of X ofdiameter less than δ is contained in some element of the cover. We use Riemann’scriterion for integrability to show that f is integrable. We need to show that forgiven ε > 0, there exists a partition P ∈ P[0, 1], the set of all partitions of [0, 1],such that

U(P, f)− L(P, f) =n∑i=1

(Mi −mi)∆xi < ε,

where P : 0 = x0 < · · · < xn = 1, Mi = supxi−1≤x≤xif(x), mi = infxi−1≤x≤xi

f(x),and ∆xi = xi − xi−1. To this end, for given ε > 0, choose N ∈ N such that1

2N < ε2 and let Q ∩ (0, 1] = ri+∞i=1 . Set Ni =

(ri − 1

2N+i , ri + 12N+i

)∩ [0, 1].

Obviously,∑+∞i=1 l(Ni) ≤

12N < ε

2 , where l(Ni) denotes the length of the intervalNi. On the other hand, Q ∩ (0, 1] = ri+∞i=1 ⊆

⋃+∞i=1 Ni. Therefore, for any

x ∈ C := [0, 1]\⋃+∞i=1 Ni, the function f is continuous at x. Consequently, for given

ε > 0, there exists an open interval Nx containing x such that

MNx(f)−mNx

(f) <ε

2,

where MNx(f) = supt∈Nx

f(t) and mNx(f) = inft∈Nx

f(t). It is obvious thatNi+∞i=1 ∪ Nxx∈C is an open cover for the compact interval [0, 1]. Let δ > 0 be aLebesgue number for this open cover. Choose P ∈ P[0, 1] such that the length ofits subintervals are all less than δ. That is, if

P : 0 = x0 < · · · < xn = 1,

2.9. NINTH COMPETITION 99

then ∆xi = xi − xi−1 < δ for all 1 ≤ i ≤ n. Therefore, for all 1 ≤ i ≤ n, we have[xi−1, xi] ⊆ Nj for some j ∈ N or [xi−1, xi] ⊆ Nx for some x ∈ C. Letting

A =i ∈ N : 1 ≤ i ≤ n,∃j ∈ N [xi−1, xi] ⊆ Nj

,

B =i ∈ N : 1 ≤ i ≤ n,∃x ∈ C [xi−1, xi] ⊆ Nx

,

we can write

U(P, f)− L(P, f) =n∑i=1

(Mi −mi)∆xi

≤∑i∈A

(Mi −mi)∆xi +∑i∈B

(Mi −mi)∆xi

≤∑i∈A

1×∆xi +∑i∈B

ε

2∆xi ≤

+∞∑i=1

l(Ni) +ε

2

n∑i=1

∆xi

≤ 12N

+ε

2< ε.

This means, U(P, f)− L(P, f) < ε, which is what we want.To find the value of the integral, it is obvious that

L(P, f) =n∑i=1

mi∆xi = 0,

for all P ∈ P[0, 1]. Since f is integrable, we have∫ 1

0

f(x)dx = supP∈P[a,b]

L(P, f) = 0.

5. First, we show that no function f : [0, 1] → [0, 1] × [0, 1] can have the threeproperties, namely, continuity, injectivity, and surjectivity. Suppose to the contrarythat f is a continuous and one-to-one function from [0, 1] onto [0, 1] × [0, 1]. Asf is invertible and [0, 1] is compact, it follows that f is a closed map, and hencef−1 : [0, 1] × [0, 1] → [0, 1] is a continuous function. Let f( 1

2 ) = (a, b) for somea, b ∈ [a, b]. Obviously, f−1(a, b) = 1

2 . Letting C = [0, 1]× [0, 1] \ (a, b), we have

f−1(C) = [0, 1] \ 12.

Since C is connected and f−1 is continuous, it follows that [0, 1]\ 12 is connected,

which is a contradiction. Therefore, the assertion follows by contradiction.We now prove that f can have any two properties of the aforementioned prop-

erties. There are three cases to consider.

(i) f : [0, 1] → [0, 1]× [0, 1] can be continuous and one-to-one.

The function f : [0, 1] → [0, 1]× [0, 1] defined by f(t) = (t, 0) is continuous andone-to-one.

(ii) f : [0, 1] → [0, 1]× [0, 1] can be continuous and onto.

100 2. SOLUTIONS

Such a function exists. We present the well-known example due to Schoenberg(1938). First, define the function φ on [0, 2] by

φ(t) =

0 0 ≤ t ≤ 1

3 or 53 ≤ t ≤ 2,

3t− 1 13 ≤ t ≤ 2

3 ,1 2

3 ≤ t ≤ 43 ,

−3t+ 5 43 ≤ t ≤ 5

3 .

Then, extend φ to R via φ(t + 2) = φ(t). It is plain that φ : R → R is 2-periodicand that 0 ≤ φ(t) ≤ 1 for all t ∈ R. Now, define f1, f2 : R → R by

f1(t) =+∞∑n=1

φ(32n−2t)2n

, f2(t) =+∞∑n=1

φ(32n−1t)2n

.

As 0 ≤ φ ≤ 1 on R, we see that the series defining f1 and f2 converge uniformlyand absolutely on R, and hence f1 and f2 are continuous on R and moreover, 0 ≤f1, f2 ≤ 1 on R because 0 ≤ φ ≤ 1 on R. Thus, the function f : [0, 1] → [0, 1]× [0, 1]defined by f(t) = (f1(t), f2(t)) is well-defined and continuous. To prove that fis onto, let (a, b) ∈ [0, 1] × [0, 1] be arbitrary. We need to show that there existsa c ∈ [0, 1] such that f(c) = (a, b). To this end, write the binary expansions ofa, b ∈ [0, 1] as follows

a =+∞∑n=1

an2n, b =

+∞∑n=1

bn2n,

where an, bn = 0 or 1 for all n ∈ N. Let

c = 2+∞∑n=1

cn3n,

where

cn =an+1

2n odd,

bn2

n even.

Thus, cn = 0 or 1 for all n ∈ N. Consequently, as 2∑+∞n=1

13n = 1, we obtain

0 ≤ c ≤ 1. We claim thatφ(3kc) = ck+1, (∗)

for all k ∈ Z with k ≥ 0. Suppose that (∗) holds. We would then have

φ(32n−2c) = c2n−1 = an, φ(32n−1c) = c2n = bn,

for all n ∈ N. This easily yields f(c) = (f1(c), f2(c)) = (a, b), which is what wewant. Now, to prove (∗), suppose k ∈ Z with k ≥ 0. We can write

3kc = 2k∑

n=1

cn3n−k

+ 2+∞∑

n=k+1

cn3n−k

= 2Nk + dk,

where Nk =∑kn=1 3k−ncn ∈ N and dk = 2

∑+∞n=k+1

cn

3n−k . As φ is periodic and itsperiod is 2, we have

φ(3kc) = φ(dk).If ck+1 = 0, then

0 ≤ dk ≤ 2+∞∑n=2

13n

=13,

2.9. NINTH COMPETITION 101

in which case, φ(dk) = 0, whence φ(3kc) = ck+1 = 0. If ck+1 = 1, then 23 ≤ dk ≤ 1,

in which case, φ(dk) = 1, whence φ(3kc) = ck+1 = 1. That is,

φ(3kc) = ck+1,

for all k ∈ Z with k ≥ 0, proving the claim and hence finishing the proof.

(iii) f : [0, 1] → [0, 1]× [0, 1] can be one-to-one and onto.

Such a function exists. First, define f : [0, 1] → [0, 1]× [0, 1] by

f(t) = (0.t1t3t5 . . . , 0.t2t4t6 . . .),

where 0.t1t2t3 . . ., with ti = 0 or 1 for all i ∈ N, denotes the binary expansion ofthe number t. We set the convention that in the binary expansion of a numbert the number 1 is not allowed to be repeated for all the digits after any digitexcept when t = 1, in which case its binary expansion, by definition, is set to be0.111 . . .. It is obvious that f is onto. Next, as mentioned in (i), the functionf : [0, 1] → [0, 1] × [0, 1] defined by f(t) = (t, 0) is (continuous and) one-to-one.It thus follows from the Schroder-Bernstein Theorem that there is a function f :[0, 1] → [0, 1]× [0, 1], which is one-to-one and onto, as desired.

2.9.2. Algebra. 1. First solution: Proceed by contradiction. Suppose thatthere exists a nonnormal subgroup H of index p in G so that |G| = p|H|, where pis the smallest prime dividing the order of G. It follows that there is a g ∈ G suchthat

K = Hg := g−1Hg = g−1hg : h ∈ H 6= H.

It is obvious that |K| = |H|. Also, we have |H ∩ K| < |H| = |K|, for otherwise|H ∩K| = |H| = |K|, yielding H ∩K = H = K, which is a contradiction. Now, weclaim that |K|

|H∩K| ≥ p. To see this, let q be a prime dividing |K||H∩K| . It follows that

q divides |K|, and hence |G|. This obtains p ≤ q ≤ |K||H∩K| because p is the smallest

prime dividing |G|. On the other hand,

|G| ≥ |HK| = |H| |K||H ∩K|

≥ |H|p = |G|,

implying that HK = G. Now, let g = hk for some h ∈ H, k ∈ K. We can write

Hg = Hhk = (Hh)k = Hk.

Thus,G = Gk = (HK)k = HkKk = HgK = KK = K,

which is a contradiction. So, the assertion follows by way of contradiction.

Second solution: Let H be a subgroup of G of index p in G, where p is thesmallest prime dividing the order of G. First, we claim that if x /∈ H, then xi /∈ Hfor all i ∈ N with 1 ≤ i ≤ p− 1. By way of contradiction, suppose that there existsthe smallest j ∈ N with 1 < j ≤ p − 1 such that xj ∈ H. It follows that xi /∈ Hfor all 1 ≤ i < j. Set n = |G| and m = ord(x). We have m|n. As j < p and p isthe smallest prime that divides n, it follows that m is not divisible by j. Hence,

102 2. SOLUTIONS

there are numbers r, q ∈ N with 0 < r < j such that m = jq + r. Noting thatm = ord(x), we can write

e = xm = xjq+r = (xj)qxr,

yielding (xj)qxr ∈ H. But xj ∈ H, from which, we obtain xr ∈ H, a contradictionbecause 0 < r < j. Consequently, xi /∈ H for all 1 ≤ i ≤ p− 1, as we claimed.

Now, to prove the assertion, we proceed by contradiction again. Suppose thatthere exist g ∈ G and h ∈ H such that ghg−1 /∈ H. This yields g /∈ H, which,in turn together with the above claim, implies that gi /∈ H for all i ∈ N with1 ≤ i ≤ p− 1. On the other hand, giH 6= gjH whenever 1 ≤ i < j ≤ p− 1 becauseotherwise gj = gih′ for h′ ∈ H, implying gj−i = h′ ∈ H, which is a contradiction,for 1 ≤ j − i ≤ p− 1. Therefore, H, gH, . . . , gp−1H are p distinct left cosets of H.Consequently,

G

H=H, gH, . . . , gp−1H

.

Let g1 = ghg−1. Likewise, as g1 /∈ H, we conclude that H, g1H, . . . , gp−11 H are p

distinct left cosets of H, implyingG

H=H, g1H, . . . , g

p−11 H

.

Therefore, gH = gr1H for some 1 ≤ r ≤ p−1, and hence g = gr1h1 for some h1 ∈ H.On the other hand, gr1 = ghrg−1, from which, we obtain g = ghrg−1h1. This yieldsg = h1h

r, implying g ∈ H, which is a contradiction. Therefore, H is a normalsubgroup of G, finishing the proof.

Third solution: Let H be a subgroup of G of index p in G, where p is the smallestprime dividing the order of G. Use L to denote the set of all left cosets of H inG, which has p elements by the hypothesis. For each g ∈ G define the mappingτg : L −→ L by τg(xH) = gxH. In other words, G acts on L by multiplication fromthe left. It is readily verified that τg’s (g ∈ G) give rise to a group homomorphismφ : G −→ S(L) ∼= Sp which is defined by φ(g) = τg, where S(L) denotes the groupof all permutations of L which is isomorphic to Sp because |L| = p (here, as isusual, Sp denotes the symmetric group of degree p, whose order is p!). Obviously,K := kerφ ⊆ H. We prove the assertion by showing that H = K. By the FirstIsomorphism Theorem for groups, G

K is isomorphic to a subgroup of Sp implyingthat |GK |

∣∣p!. On the other hand, every divisor of |GK | divides |G|. Also, of the primedivisors of p! only p divides |G| because p is the smallest prime dividing the orderof G. Thus, |GK | = p or |GK | = 1. The latter is impossible because |GK | ≥ |GH | = p.Therefore, |GK | = p = |GH |, implying that |H| = |K|. This obtains H = K becauseK ⊆ H, completing the proof.

Fourth solution: By a standard result from Galois theory, e.g., see PropositionV.2.16 of “Algebra” by T.W. Hungerford, there are fields E,K such that K is anextension of E and that Gal(K/E) ∼= G. In view of this, and two other standard re-sults from Galois theory, e.g., see Theorem V.2.5 and Corollary V.3.15 of “Algebra”by T.W. Hungerford, it suffices to prove the following.

Under the hypothesis of the problem, let K be an extension field of E such thatGal(K/E) ∼= G. If F ⊆ K is an extension field of E such that [F : E] = p, then Fis normal over E.

2.9. NINTH COMPETITION 103

To prove this, note first that every element of K is separable over E, i.e., theminimal polynomial of every element of K over E splits into distinct linear factors,for K is Galois over E. Therefore, F is a finite separable extension of E and hence,by the Primitive Element Theorem (see Proposition V.6.15 of “Algebra” by T.W.Hungerford, there is an α ∈ F such that F = E(α). Let g ∈ E[x] be an arbitraryirreducible polynomial that has a root, say h(α) for some h ∈ E[x], in F = E(α).It follows that the splitting field of g is contained in K, for h(α) ∈ F ⊆ K and Kis Galois, and hence normal, over E. Note that deg(g) = [E

(h(α)

): E] ≤ [E(α) :

E] = p. We need to show that g splits into linear factors in E(α). Let β 6= h(α) be aroot of g in the splitting field of g, which, as we just saw, is contained inK. It sufficesto prove that β ∈ F = E(α). To this end, we note that [F (β) : F ] < deg(g) ≤ pbecause g(β) = 0, g

(h(α)

)= 0, and h(α) ∈ F . This yields [F (β) : F ] = 1 because p

is the smallest prime dividing |G| and [F (β) : F ]∣∣|G|, for β ∈ K and [K : E] = |G|

since Gal(K/E) ∼= G. Consequently, β ∈ F , and hence F is normal over E, asdesired.

2. We determine all isomorphisms from the fieldZ3[x]

(x2 + 1)onto the field

Z3[x](x2 + x+ 2)

.

Since f : Z3[x]〈x2+1〉 →

Z3[x]〈x2+x+2〉 is an isomorphism of fields, we see that f(a+〈x2+1〉) =

a + 〈x2 + x + 2〉 for all a ∈ Z3. Thus, to determine f , we only need to determinef(x+ 〈x2 + 1〉). To this end, suppose that f(x+ 〈x2 + 1〉) = ax+ b+ 〈x2 + x+ 2〉for some a, b ∈ Z3. We must have(

f(x+ 〈x2 + 1〉

))2

= f(x2 + 〈x2 + 1〉

)= f

(− 1 + 〈x2 + 1〉

),

implying that (ax + b)2 + 1 ∈ 〈x2 + x + 2〉. But the polynomial (ax + b)2 + 1 =a2x2 + 2abx + b2 + 1 is divisible by x2 + x + 2 in Z3[x] if and only if so is thepolynomial (2ab− a2)x+(a2 + b2 +1), which is obtained from a2x2 +2abx+ b2 +1by letting x2 = −x − 2 = −x + 1, if and only if a and b satisfy the equationsa(2b − a) = 0 and a2 + b2 + 1 = 0. It is readily seen that a = 2, b = 1 anda = 1, b = 2 are the only solutions of the above equations in Z3. Therefore, anisomorphism f : Z3[x]

〈x2+1〉 →Z3[x]

〈x2+x+2〉 must be given by

f(ax+ b+ 〈x2 + 1〉

)= a(2x+ 1) + b+ 〈x2 + x+ 2〉,

or byf(ax+ b+ 〈x2 + 1〉

)= a(x+ 2) + b+ 〈x2 + x+ 2〉.

That is, either

f(ax+ b+ 〈x2 + 1〉

)= 2ax+ (a+ b) + 〈x2 + x+ 2〉,

orf(ax+ b+ 〈x2 + 1〉

)= ax+ (2a+ b) + 〈x2 + x+ 2〉,

which are both easily seen to be monomorphisms of rings. It follows that such an f

is an isomorphism from the fieldZ3[x]

(x2 + 1)onto the field

Z3[x](x2 + x+ 2)

because they

both have 9 elements.

3. Note that the ring Z7[x] is a Euclidean ring because Z7 is a field. With that inmind, we can use the Euclidean algorithm to find the greatest common divisor of

104 2. SOLUTIONS

the two polynomials 4x4 − 2x2 + 1 and −3x3 + 4x2 + x+ 1 in Z7[x]. We have

4x4 − 2x2 + 1 = (−3x3 + 4x2 + x+ 1)(x− 1) + x2 + 2,−3x3 + 4x2 + x+ 1 = (x2 + 2)(−3x+ 4).

Therefore,gcd(4x4 − 2x2 + 1,−3x3 + 4x2 + x+ 1) = x2 + 2,

which is what we want.

4. Let 1, y1, where y1 ∈ K, be a basis for the vector space K over F . If y21 ∈ F ,

there is nothing to prove. Suppose y21 /∈ F . As y2

1 ∈ K, it follows that there existscalars a, b ∈ F with b 6= 0 such that y2

1 = a+ by1. We have

y21 − by1 = b.

On the other hand, since ch(F ) 6= 2, we have 2−1 = 12 ∈ F , and hence we can write

y21 − by1 = (y1 −

b

2)2 − (

b

2)2,

whence

(y1 −b

2)2 = a+ (

b

2)2 ∈ F.

Let y = y1 − b2 . It is easily verified that 1, y is a basis for K over F and

y2 = a+ ( b2 )2 ∈ F , which is what we want.

5. Let S = T − I. Obviously, W = kerS. On the other hand, by the Rank-NullityTheorem, we have

dim kerS + dimSV = dimV = n.

We claim that SV ⊆ kerS. To see this, let y = Sx ∈ SV , where x ∈ V , bearbitrary. We have

Sy = S2x = (T − I)2x = (T 2 − 2T + I)x = (2I − 2T )x = 0,

because T 2 = I and ch(F ) = 2. This means, y ∈ kerS, proving the claim. Now,since SV ⊆ kerS, we can write

n = dim kerS + dimSV ≤ 2 dim kerS,

yielding dimW = dim kerS ≥ n2 , which is what we want.

2.9.3. General. 1. Let

A =

r λ λ · · · λ λλ r λ · · · λ λ...

. . . . . . . . ....

......

.... . . . . . . . .

...λ λ · · · λ r λλ λ · · · · · · λ r

.

2.9. NINTH COMPETITION 105

Subtracting the first row from all the other rows and then adding the jth column(2 ≤ j ≤ n) to the first column, we obtain an upper triangular matrix whosedeterminant is the product of the entries on its main diagonal. Thus, we can write

detA = det

r λ λ · · · λ λλ− r r − λ 0 · · · 0 0λ− r 0 r − λ 0 · · · 0

......

. . . . . . . . ....

λ− r 0 · · · 0 r − λ 0λ− r 0 · · · · · · 0 r − λ

= det

r + (n− 1)λ λ λ · · · λ λ0 r − λ 0 · · · 0 00 0 r − λ 0 · · · 0...

.... . . . . . . . .

...0 0 · · · 0 r − λ 00 0 · · · · · · 0 r − λ

=

(r + (n− 1)λ

)(r − λ

)n−1.

That is, detA =(r + (n− 1)λ

)(r − λ

)n−1, which is what we want.

2. Let x denote the number of chickens, a the amount of food a chicken consumesper day, and t the number of the days before the farmers run out of chicken food.The amount of food in the chicken farm is equal to

(t+ 20)(x− 75)a = (t− 15)(x+ 100)a = txa.

Simplifying, we obtain(t+ 20)(x− 75) = (t− 15)(x+ 100)(t− 15)(x+ 100) = tx

=⇒x = 5t.20t− 3x− 300 = 0.

Substituting x = 5t into the second equation yields x = 300, which impliest = 60. Therefore, there are x = 300 chickens in the farm.

3. To prove the assertion by contradiction, suppose that such a function f exists.We can write∫ 1

0

(x− α)2f(x)dx =∫ 1

0

x2f(x)dx− 2α∫ 1

0

xf(x)dx+ α2

∫ 1

0

f(x)dx

= α2 − 2α2 + α2 = 0.

That is,∫ 1

0(x − α)2f(x) = 0. As (x − α)2f(x) is continuous and nonnegative on

[0, 1], it follows that (x− α)2f(x) = 0 for all x ∈ [0, 1]. This implies f = 0 on [0, 1]except possibly at x = α. Now, from the continuity of f , we see that f = 0 on[0, 1], a contradiction. Therefore, no such function f exists, finishing the proof.

4. First solution: Consider the points with integer coordinates in the closedinterval [0, n] of the real line. To any solution (x1, . . . , xm) of the equation

x1 + x2 + · · ·+ xm = n

in N, there corresponds m−1 points p1 < · · · < pm−1 from the set in−1i=1 as follows

p1 = x1, p2 = x1 + x2, . . . , pm−1 = x1 + · · ·+ xm−1.

106 2. SOLUTIONS

Conversely, for any m − 1 points p1 < · · · < pm−1 from the set in−1i=1 , a solution

(x1, . . . , xm) of the equation

x1 + x2 + · · ·+ xm = n

is determined as follows

x1 = p1, x2 = p2 − p1, . . . , xm−1 = pm−1 − pm−2, xm = n− pm−1.

Therefore, the number of the solutions of the equation x1 + x2 + · · ·+ xm = n in Nis equal to the number of ways of choosing m− 1 points pim−1

i=1 from n− 1 pointsin−1

i=1 , which, as is well-known, is equal to(n−1m−1

).

Second solution: We only briefly elaborate on this solution. It is not difficult tosee that the number of the solutions of the equation

x1 + x2 + · · ·+ xm = n

in N is equal to the coefficient of xn in the expansion of

(x+ x2 + · · ·+ xn−m)m

or in that of

(x+ x2 + x3 + · · · )m = (x

1− x)m = xm

+∞∑n=0

(m+ n− 1m− 1

)xn,

which is easily seen to be(n−1m−1

).

5. Yes, just cut the paper along the heavy lines.

6. Since the time needed to get to work from home obeys the uniform distribution,it follows that the density probability function is equal to

f(x) =

0 x < 10110 10 ≤ x ≤ 200 x > 20

.

(a) As the worker needs 15 minutes to get to work on time, the desired proba-bility is

P (x > 15) =∫ 20

15

dx

10=

12.

2.9. NINTH COMPETITION 107

(b) First solution. Noting that it takes the worker at least 10 minutes to getto work, let x0 > 10 denote the time needed for the worker to get to work on timeand with the probability of 75% to have time to eat breakfast. We must have

P (10 < x < x0) =∫ min(x0,20)

10

dx

10= 0.75 ⇐⇒ min(x0, 20) = 17.5

⇐⇒ x0 = 17.5.

Therefore, the latest time that this worker can leave home to get to work on time andwith the probability of 75% to have time to eat breakfast is equal to 7, 45′−17.5′ =7, 27.5′ = 7, 27′, 30′′.

Second solution. Suppose the worker heads off for work m minutes before 8o’clock in the morning. In order for the worker to get to work on time and havetime to eat breakfast, we must have m−15 ≥ 10, implying m ≥ 25. Assuming thatit takes the worker x minutes to get to work, the worker has m−x minutes to havebreakfast. Thus, to make sure that the m−x minutes, which is the remaining timebefore the start hour of the factory, is enough for the worker to have breakfast withthe probability of 75%, we must have

P (m− x ≥ 15) = 0.75 ⇐⇒ P (x ≤ m− 15) = 0.75

⇐⇒∫ min(m−15,20)

10

dx

10= 0.75

⇐⇒ min(m− 15, 20) = 17.5⇐⇒ m = 32.5.

Therefore, if the worker heads off for work 32.5 minutes before 8 a.m., i.e., at8, 00′ − 32.5′ = 7, 27.5′ = 7, 27′, 30′′, s/he will get to work on time and with theprobability of 75% have time to eat breakfast.

7. First solution: We solve the problem on a general field F . Let n ∈ N,A = (cicj) ∈Mn(F ), and In the identity matrix of size n. To evaluate det(In+A),we may, without loss of generality, assume that ci 6= 0 for all 1 ≤ i ≤ n becauseif ci0 = 0 for some i0 ∈ 1, . . . , n, then det(In + A) = det(In−1 + A′) whereA′ = (c′ic

′j) ∈ Mn−1(F ) with c′i ∈ c1, . . . , cn \ ci0, which has the same general

form of A = (cicj) ∈Mn(F ). Use the basic properties of the determinant functionand perform the following operations: factor out ci of row i and of column i foreach i = 1, . . . , n; then subtract the first row from all the other rows; then multiplycolumn i by c2i (1 ≤ i ≤ n); and finally add row i to row 1 for each i = 2, . . . , n, toobtain an upper triangular whose determinant is equal to that of In + A. We can

108 2. SOLUTIONS

write

det(In +A) = c21 · · · c2n det

1 + 1c21

1 · · · 1 11 1 + 1

c22· · · 1 1

......

. . ....

...1 1 · · · 1 + 1

c2n−11

1 1 · · · 1 1 + 1c2n

= c21 · · · c2n det

1 + 1c21

1 · · · 1 1− 1c21

1c22

· · · 0 0...

.... . .

......

− 1c21

0 · · · 1c2n−1

0

− 1c21

0 · · · 0 1c2n

= det

1 + c21 c22 · · · c2n−1 c2n−1 1 · · · 0 0...

.... . .

......

−1 0 · · · 1 0−1 0 · · · 0 1

= det

1 + c21 + · · ·+ c2n c22 · · · c2n−1 c2n

0 1 · · · 0 0...

. . . . . ....

...0 0 · · · 1 00 0 · · · 0 1

= (1 + c21 + · · ·+ c2n)× 1× · · · × 1 = 1 + c21 + · · ·+ c2n.

Therefore, det(In +A) = 1 + c21 + · · ·+ c2n, which is what we want.

Second solution: Recall that if A ∈ Mn(F ), λi’s (1 ≤ i ≤ n) are the eigenvaluesof A in the algebraic closure of F , and f(x) = det(xIn − A) is the characteristicpolynomial of A so that

f(x) = xn + fn−1xn−1 + · · ·+ f0,

then, fn−1 = −∑ni=1 λi, f0 = (−1)nλ1 · · ·λn,

∑ni=1 λi = tr(A), and λ1 · · ·λn =

det(A). Also, if α ∈ F and g(x) = det(xIn − (A + αIn)

)is the characteristic

polynomial of A+ αIn, then

g(x) = det((x− α)In −A

)= f(x− α).

With all that in mind, note first that the rank of A is 1 because row i is ci(c1, . . . , cn)which is a multiple of the fixed vector (c1, . . . , cn). It follows from the Rank-NullityTheorem that dim kerA = n− 1, and hence 0 is an eigenvalue of A of multiplicityn− 1. Let λ be the only other eigenvalue of A. We must have

f(x) = det(xIn −A) =n∏i=1

(x− λi) = xn−1(x− λ) = xn − λxn−1.

2.10. TENTH COMPETITION 109

On the other hand,n∑i=1

λi = λ = tr(A) =n∑i=1

c2i ,

implying λ =∑ni=1 c

2i . If g denotes the characteristic polynomial of In + A, we

have

g(x) = f(x− 1)= (x− 1)n − λ(x− 1)n−1

= xn + gn−1xn−1 + · · ·+ g1x+ (−1)n(1 + λ).

Recall that(−1)n(1 + λ) = (−1)n det(In +A),

whence

det(In +A) = 1 + λ = 1 + tr(A) = 1 +n∑i=1

c2i ,

which is what we want.

2.10. Tenth Competition

2.10.1. Analysis. 1. From (b), we see that f ′(x) > 0 for all x ≥ 1, implyingthat f is strictly increasing on [1,+∞) and hence f(t) > f(1) = 1 for all t > 1.From this, we obtain

f ′(t) =1

t2 + (f(t))2<

11 + t2

,

for all t > 1. It thus follows that∫ +∞1

f ′(t)dt <∫ +∞1

dt1+t2 = π

4 because f ′ iscontinuous and f ′(t) < 1

1+t2 on [1,+∞). Using the Second Fundamental Theoremof Calculus , we can write

f(x) = 1 +∫ x

1

f ′(t)dt < 1 +∫ x

1

dt

1 + t2< 1 +

∫ +∞

1

dt

1 + t2= 1 +

π

4.

Therefore, f is bounded from the above. This together with the hypothesis that fis increasing implies that limx→+∞ f(x) exists and that

limx→+∞

f(x) = 1 +∫ +∞

1

f ′(t)dt < 1 +∫ +∞

1

11 + t2

= 1 +π

4,

implying limx→+∞ f(x) < 1 + π4 , which is what we want.

2. For an x ∈ (a, b), define g : [a, b] → R by

g(t) = (t− a)(t− b)f(x)− (x− a)(x− b)f(t).

Plainly, g(a) = g(x) = g(b) = 0. From this, applying Rolle’s Theorem twice to gand g′, respectively, we obtain a c ∈ (a, b) such that g′′(c) = 0. We can write

g′′(c) = 2f(x)− (x− a)(x− b)f ′′(c) = 0,

which implies ∣∣f(x)∣∣ ≤ M(x− a)(b− x)

2,

110 2. SOLUTIONS

for all x ∈ [a, b]. Taking integrals of both sides of the above yields∫ b

a

∣∣f(x)∣∣dx ≤ M

2

∫ b

a

(− x2 + (a+ b)x− ab

)dx = M

(b− a)3

12.

That is,∫ ba

∣∣f(x)∣∣dx ≤M (b−a)3

12 , which is what we want.

3. First solution: We can write∣∣∣∣∫ 3

0

x2(1− x)xn

1 + x2ndx

∣∣∣∣ ≤∫ 3

0

x2|1− x|xn

1 + x2ndx

=∫ 1

0

x2|1− x|xn

1 + x2ndx+

∫ 3

1

x2|1− x|xn

1 + x2ndx

≤∫ 1

0

xndx+ 18∫ 3

1

xn

x2ndx

=1

n+ 1+

18n− 1

− 183n−1(n− 1)

.

It thus follows from the Squeeze Lemma that

limn→+∞

∫ 3

0

x2(1− x)xn

1 + x2ndx = 0,

which is what we want.

Second solution: In view of a standard theorem from classical analysis, it sufficesto show that the sequence

(x2(1−x)xn

1+x2n

)+∞n=1

uniformly converges to the zero functionon [0, 3]. Note first that ∣∣∣∣x2(1− x)xn

1 + x2n

∣∣∣∣ ≤ |1− x|, (∗)

for all x ∈ [0, 3]. To see this, we have xn+2 ≤ 1 on [0, 1], implying xn+2 ≤ 1 + x2n

on [0, 1]. Also if n ≥ 2, then xn+2 ≤ x2n on [1, 3], which implies xn+2 ≤ 1 + x2n

on [1, 3]. Thus, xn+2 ≤ 1 + x2n on [0, 3] of which the above inequality is a quickconsequence. Now, let 0 < ε < 1 be given. It follows from (∗) that∣∣∣∣x2(1− x)xn

1 + x2n

∣∣∣∣ < ε,

for all n ≥ 2 and x ∈ (1− ε, 1 + ε). On the interval [0, 1− ε], for all n > 2 we canwrite ∣∣∣∣x2(1− x)xn

1 + x2n

∣∣∣∣ ≤ xn+2

1 + x2n≤ xn+2 ≤ (1− ε)n+2.

As 0 < 1− ε < 1, we have limn(1− ε)n+2 = 0. Hence, there exists an N1 > 1 suchthat (1− ε)n+2 < ε for all n ≥ N1. On the interval [1 + ε, 3], we have∣∣∣∣x2(1− x)xn

1 + x2n

∣∣∣∣ < 2xn+2

1 + x2n=

2x2

x−n + xn<

2xn−2

≤ 2(1 + ε)n−2

,

for all n ≥ 2. As limn2

(1+ε)n−2 = 0, there exists an N2 > 1 such that 2(1+ε)n−2 < ε

for all n ≥ N2. Letting N = max(N1, N2), for all x ∈ [0, 3] and n ≥ N we have∣∣∣∣x2(1− x)xn

1 + x2n

∣∣∣∣ < ε.

2.10. TENTH COMPETITION 111

That is, the sequence(x2(1−x)xn

1+x2n

)+∞n=1

uniformly converges to the zero function on[0, 3], yielding

limn→+∞

∫ 3

0

x2(1− x)xn

1 + x2ndx =

∫ 3

0

limn→+∞

x2(1− x)xn

1 + x2ndx = 0,

which is what we want.

2.10.2. Algebra. 1. We prove the following lemma of which the assertion isa quick consequence.

Lemma. Let V be a left (resp. right) vector space over a division ring D whosecharacteristic is zero. Then the additive group of V has no maximal subgroup.

Proof. To prove the assertion by contradiction, suppose that M is a maximalsubgroup of the additive group of V . As M ≤ V , V is abelian, and M is maximal, itfollows that V

M is a simple abelian group. Consequently, VM is a finite cyclic group,

implying that there exists a prime number p such that VM = Zp, and hence M is of

finite index in V . By proving that (V,+) has no proper subgroup of finite index, weobtain a contradiction, proving the assertion. Suppose to the contrary that thereexists a proper subgroup H of the additive group of V such that [V : H] = n ∈ N.As | VH | = n, it follows from Lagrange’s Theorem that nx ∈ H for all x ∈ V . On theother hand, x = n( xn ) and x

n ∈ V for all x ∈ V , implying x ∈ H for all x ∈ V . Thisyields V = H, a contradiction. Thus, V has no proper subgroup of finite index,which is what we want, finishing the proof.

To prove the assertion, just let V = D = R in the lemma.

2. Since the ring R is unital, it follows from (∗) that there exist elements a, a′, a′′ ∈A, b ∈ B, c ∈ C, and d ∈ D such that

a+ b = 1, a′ + c = 1, a′′ + d = 1.

Multiplying the above equalities and simplifying, we see that there is a u ∈ A suchthat

1 = (a+ b)(a′ + c)(a′′ + d) = u+ bcd,

implying 1 = u+ bcd ∈ A+M because bcd ∈M . This obviously yields A+M = R,which is what we want.

Let R = Z, A = 2Z, B = 3Z, C = 5Z, and D = 7Z. It is easily verified thatM = B ∩ C ∩D = 105Z and that R and its ideals A,B,C,D satisfy (∗).

3. We solve the problem under the weaker hypothesis that gcd(n, c) = 1, wherec = lcm(1, 2, . . . ,m− 1). It is obvious that F ⊆ F (αm) ⊆ F (α). So we can write

[F (α) : F ] = [F (α) : F (αm)][F (αm) : F ]. (∗)As [F (α) : F ] = n, it follows that there exists a polynomial P ∈ F [x] of degree nsuch that

P (α) = p0 + p1α+ · · ·+ pnαn = 0. (∗′)

Using the division algorithm, dividing k by m for each k = 0, . . . , n, and collectingterms appropriately in (∗′), we see that there exists a polynomial Q ∈ F (αm)[x]with deg(Q) ≤ m− 1 such that

P (α) = Q(α) = q0 + q1α+ · · ·+ qm−1αm−1 = 0.

112 2. SOLUTIONS

This clearly shows that [F (α) : F (αm)] ≤ m − 1, implying that [F (α) : F (αm)]divides lcm(1, 2, . . . ,m − 1) = c. But as, in view of (∗), [F (α) : F (αm)] divides nand gcd(n, c) = 1, we conclude that [F (α) : F (αm)] = 1, whence F (α) = F (αm),yielding [F (αm) : F ] = [F (α) : F ]. This, in turn, implies F (α) = F (αm), which iswhat we want.

4. Recall that the trace of any nilpotent matrix N ∈ Mn(F ) is zero. Withthat in mind, suppose by way of contradiction that there exist nilpotent matri-ces N1, . . . , Nk ∈ Mn(F ) (k ∈ N) that span Mn(F ). In particular, for the matrixE11 ∈ Mn(F ), where E11 denote the matrix with 1 in the 11 place and zero else-where, there are scalars c1, . . . , ck ∈ F such that

E11 = c1N1 + · · ·+ ckNk,

which implies

1 = tr(E11) = c1tr(N1) + · · ·+ cktr(Nk) = 0,

a contradiction, proving the assertion.

2.10.3. General. 1. As the tests are independent, we can write

P(X = 0

)= P (FS) + P (SSFS) + P (FFFS) + P (SSSSFS)

+P (SSFFFS) + P (FFSSFS) + P (FFFFFS) + · · ·= qp+ p2qp+ q2qp+ p4qp+ p2q2qp+ p2p2qp+ q4qp+ · · ·= qp

(1 + (p2 + q2) + (p2 + q2)2 + · · ·

)=

qp

1− p2 − q2=

1412

=12.

That is, P(X = 0

)= 1

2 . Similarly, one can show that P(X = 1

)= 1

2 .

2. We prove the assertion for the more general case where A is a subset of atopological space. With that in mind, let A be a closed subset of a topologicalspace X such that A = ∅. We prove the assertion by showing that A = ∂A = ∂Ac,where ∂A denotes the boundary of A. Recall that by definition ∂A = A ∩ Ac. So,we can write

∂Ac = Ac ∩Acc = Ac ∩A = ∂A.

That is, ∂A = ∂Ac. On the other hand,

∂A = A \A = A \ ∅ = A = A.

So we have A = ∂A = ∂Ac, where Ac is an open set because A is a closed set,which is what we want.

3. Let x(t) be a solution of (∗). We have

x′′ = f(t, x), x(0) = x0, x′(0) = y0.

2.10. TENTH COMPETITION 113

As f is continuous, using the First Fundamental Theorem of Calculus and changingthe order of integration, we can write

x′(t) = x′(0) +∫ t

0

f(s, x(s)

)ds =⇒ x(t) = x(0) + y0t+

∫ t

0

(∫ τ

0

f(s, x(s)

)ds

)dτ

=⇒ x(t) = x0 + y0t+∫ t

0

(∫ t

s

dτ

)f(s, x(s)

)ds

=⇒ x(t) = x0 + y0t+∫ t

0

(t− s)f(s, x(s)

)ds,

which means x(t) is a solution of (∗∗).Now, let x(t) be a solution of (∗∗). We can write

x(t) = x0 + y0t+∫ t

0

(t− s)f(s, x(s)

)ds,

implying that

x(t) = x0 + y0t+ t

∫ t

0

f(s, x(s)

)ds−

∫ t

0

sf(s, x(s)

)ds.

Again, using the First Fundamental Theorem of Calculus, we can write

x′(t) = y0 +∫ t

0

f(s, x(s)

)ds+ tf

(t, x(t)

)− tf

(t, x(t)

),

which obtains

x′(t) = y0 +∫ t

0

f(s, x(s)

)ds.

Taking derivative of both sides of the above yields

x′′(t) = f(t, x(t)

).

It is obvious that x(0) = x0 and x′(0) = y0. Therefore, x(t) satisfies (∗), which iswhat we want.

4. First Solution: See Solution 1 of 2.5.3.

Second solution: Just as in the first solution, we settle the problem by provingthe polynomial identity

(1 + x)n =n∑k=0

Cknxk, (∗)

114 2. SOLUTIONS

where Ckn =n!

k!(n− k)!. Proceed by induction on n. If n = 1, the assertion is easy.

Suppose that (∗) holds for n. To prove that (∗) holds for n+ 1, we can write

(1 + x)n+1 = (1 + x)n(1 + x) = (1 + x)n∑k=0

Cknxk

=n∑k=0

Cknxk +

n∑k=0

Cknxk+1

= 1 +n∑k=1

Cknxk +

n∑k=1

Ck−1n xk + xn+1

= 1 +n∑k=1

(Ckn + Ck−1n )xk + xn+1.

A straightforward calculation shows that Ckn+Ck−1n = Ckn+1, from which, we easily

obtain

(1 + x)n+1 =n∑k=0

Ckn+1xk,

proving the induction assertion, which is what we want.

Third Solution: For n, k ∈ N with k ≤ n, use P kn to denote n(n−1) · · · (n−k+1).

We have P kn = k!Ckn for each n, k ∈ N with k ≤ n, where Ckn =n!

k!(n− k)!. Thus,

to prove the assertion, we need to show that P kn is divisible by k! for all n, k ∈ Nwith k ≤ n. We proceed by induction on n. If n = 1, there is nothing to prove.Suppose that P kn is divisible by k! for all k ≤ n. We need to show that P kn+1 isdivisible by k! for all k ≤ n + 1. If k = n + 1, then P kn+1 = k!, in which case, theassertion is trivial. If k ≤ n, we can write

P kn+1 = (n+ 1)n(n− 1) · · · (n+ 1− k + 1)= (n− k + 1 + k)n(n− 1) · · · (n− k + 2)

= kP k−1n + P kn .

By the induction hypothesis, P k−1n and P kn are divisible by (k− 1)! and k!, respec-

tively. Thus, P kn+1 = kP k−1n + P kn is divisible by k!, which is what we want.

Fourth solution: As we know, it suffices to prove that Ckn is an integer. To thisend, we need, first of all, the following lemma, which is due to Legendre (1808).

Lemma. Let p a prime number and n ∈ N. Then, the largest power of pdividing n! is equal to

pP+∞

i=1

hn

pi

i,

where [.] denotes the integer part function.Proof. As n! = 1 × 2 × · · · × n, without loss of generality, we may assume

that p ≤ n. Among the consecutive integers 1, . . . , n those that are divisible by pare

p, 2p, . . . , m1p,

2.11. ELEVENTH COMPETITION 115

where m1 =[np

]. So, there are

[np

]such integers. Likewise, there are

[npi

]consec-

utive integers from 1 up to n that are divisible by pi (note that if pi > n, then thenumber of integers between and including 1 and n that are divisible by pi is zero).Note that any multiple of pk which is not a multiple of pk+1 contributes as much ask to the largest power of p that divides n!. Therefore, the largest power of p thatdivides n! is equal to

pP+∞

i=1

hn

pi

i,

proving the lemma.

Now to prove the assertion, note first that

[a+ b] ≥ [a] + [b],

for all a, b ∈ R. Thus, we can write[n

pi

]=[n− k + k

pi

]≥[n− k

pi

]+[k

pi

],

for all i ∈ N. Adding up these inequalities, we obtain+∞∑i=1

[n

pi

]≥

+∞∑i=1

[n− k

pi

]+

+∞∑i=1

[k

pi

].

This means that the largest power of any prime p that divides n! is divisible by thelargest power of p that divides k!(n− k)!, proving the assertion.

2.11. Eleventh Competition

2.11.1. Analysis. 1. To prove the assertion by contradiction, suppose thatthe set of zeros of f , i.e.,

Z :=x ∈ [0, 1]|f(x) = 0

= f−1(0)

is infinite. As the interval [0, 1] is compact, it follows that there exists a sequence(xn)+∞n=1 of distinct points in Z such that limn xn = x0 for some x0 ∈ [0, 1]. Thisimplies f(x0) = 0 because f is continuous. But the function f is differentiable atx0. So we can write

f ′(x0) = limn

f(xn)− f(x0)xn − x0

= limn

0xn − x0

= 0.

That is, f ′(x0) = 0. On the other hand, f(x0) = 0, implying that x0 is a commonzero of f and f ′, which is in contradiction with the hypothesis of the problem.Thus, Z is finite, which is what we want.

2. (a) To show that the sequence (fn)+∞n=1 is pointwise convergent to a function f on[1, e

1e ], it suffices to prove that the sequence

(fn(a)

)+∞n=1

is increasing and boundedfrom the above for all a ∈ [1, e

1e ]. To this end, letting a ∈ [1, e

1e ] be arbitrary, we

prove these assertions by induction on n. We have

f1(a) = a, fn+1(a) = afn(a) ∀n ∈ N.

116 2. SOLUTIONS

First, by induction on n, we show that 1 ≤ fn(a) < e for all n ∈ N. If n = 1, then1 ≤ f1(a) = a ≤ e

1e < e. Assuming that 1 ≤ fn(a) < e, we can write

1 ≤ fn+1(a) = afn(a) < (e1e )e = e.

This proves 1 ≤ fn(a) < e for all n ∈ N by induction. We note that the functiong : R → R defined by g(x) = ax is increasing whenever a > 1. To prove that thesequence

(fn(a)

)+∞n=1

is increasing, again we use induction on n. If n = 1, we canwrite

f2(a) = g(a) > g(1) = f1(a).

Now assuming that fn+1(a) > fn(a), we have

fn+2(a) = g(fn+1(a)

)> g(fn(a)

)= fn+1(a).

This proves that the sequence(fn(a)

)+∞n=1

is increasing. Thus, the sequence (fn)+∞n=1

is pointwise convergent to a function f : [1, e1e ] → R. In view of the continuity of

g, we can write

f(a) = limnfn+1(a) = lim

ng(fn(a)

)= g(limnfn(a)

)= g(f(a)

)= af(a),

implying that f(a) = af(a) and 1 ≤ f(a) ≤ e for all a ∈ [1, e1e ], as desired.

(b) Define the function g : [1, e] → [1, e1e ] by g(x) = e

ln xx . By inspecting the

derivative of g, one can easily verify that the function g : [1, e] → [1, e1e ] is one-to-

one and onto. Moreover, g is continuous and so is its inverse g−1 : [1, e1e ] → [1, e];

in fact g and, hence its inverse, g−1 are differentiable. From f(x) = xf(x), weeasily obtain g(f(x)) = x for all x ∈ [1, e

1e ]. This shows that f = g−1. Thus, f is

one-to-one and continuous because so is g−1.Now, the functional sequence (fn)+∞n=1 of continuous functions converges point-

wise to the continuous function f on the compact interval [1, e1e ] and moreover

fn(x) < fn+1(x) for all x ∈ [1, e1e ]. It thus follows from Dini’s Theorem that the

sequence (fn)+∞n=1 converges uniformly to the function f on [1, e1e ], which is what

we want.

3. SetM = sup

|φ′′n(x)| : n ∈ N, x ∈ [−1, 1]

.

It follows from the Mean Value Theorem that there exists a c between 0 and xsuch that

φ′n(x)− φ′n(0) = xφ′′n(c),

from which, we obtain ∣∣φ′n(x)∣∣ =∣∣1 + xφ′′n(c)

∣∣ ≤M + 1,

for all n ∈ N and x ∈ [−1, 1]. Using integration by parts, we get∣∣∣∣∫ 1

−1

φn(t) cosnπtdt∣∣∣∣ ≤ 1

nπ

∫ 1

−1

∣∣φ′n(t) sinnπt∣∣dt

≤ M + 1nπ

∫ 1

−1

∣∣ sinnπt∣∣dt =M + 1nπ2

,

2.11. ELEVENTH COMPETITION 117

implying that

an =1n

∣∣∣∣∫ 1

−1

φn(t) cosnπtdt∣∣∣∣ ≤ M + 1

n2π2,

for all n ∈ N. This, in view of the Limit Comparison Test for series, implies that∑+∞n=1 an is convergent, which is what we want.

2.11.2. Algebra. 1. Set

I =xu− x : x ∈ R

.

It is obvious that I is a two-sided ideal of R. It thus follows from the hypothesisthat I = 0 or I = R. If I = 0, we see that xu = x, yielding xu = ux = x forall x ∈ R, which means u = 1R. By showing that the case I = R is impossible, wefinish the proof. Suppose to the contrary that I = R. Thus, there exists an x0 ∈ Rsuch that u = x0u− x0. As u2 = u, we can write

u = u2 = (x0u− x0)u = x0u− x0u = 0,

and hencex = ux = 0x = 0,

for all x ∈ R. This means R = 0, implying that R has only one ideal, which is acontradiction. Therefore, I = 0, finishing the proof.

2. To prove the assertion by contradiction, suppose that A is a finite extension ofQ so that [A : Q] = n for some n ∈ N. Pick a prime number p, e.g., p = 2, and notethat, by Eisenstein’s Criterion, the monic polynomial xn+1 − p is irreducible overZ and hence over Q. If α is one of the roots of the equation xn+1 − p = 0, we have

Q ⊆ Q(α) ⊆ A.On the other hand, [A : Q] = n < n+1 = [Q(α) : Q], which is obviously impossible.Thus, A is not a finite extension of Q, which is what we want.

3. First we need to recall that if G is a finite p-group, where p is a prime number,then the center of G, denoted by Z(G), is nontrivial, i.e., |Z(G)| > 1. To provethis, first recall that the class equation of G is

|G| = |Z(G)|+k∑i=1

[G : CG(xi)],

where x1, . . . , xk is a maximal set of nonconjugate elements of G \ Z(G) andCG(xi) is the centralizer of xi in G (1 ≤ i ≤ k). Note that since xi’s are in G\Z(G),CG(xi)’s are proper subgroups of G, implying that whose orders are all powers of p,and hence so are [G : CG(xi)]’s because G is a finite p-group. Consequently, fromthe class equation of G, in view of |G| = pn for some n ∈ N, we conclude that pdivides |Z(G)|, yielding |Z(G)| > 1, as desired.

We now prove the assertion. It follows from the hypothesis that any two el-ements of G which are different from the identity share the same orders. Nowsuppose that p is a prime dividing |G|. It follows from Cauchy’s Theorem thatthere exists an x ∈ G whose order is p. Hence, all of the elements of G but e haveorder p, whence the order of G is a power of p. That is, G is a p-group. Conse-quently, the center of G is nontrivial, i.e., Z(G) 6= e. Since α

(Z(G)

)= Z(G) for

118 2. SOLUTIONS

all α ∈ Aut(G), from the hypothesis, we see that Z(G) = G. That is, G is abelian.Therefore, by the Fundamental Theorem of finite abelian groups, we obtain

G ∼= Zp ⊕ · · · ⊕ Zp,

which is what we want.

2.11.3. General. 1. Note that for an A ∈ M2(Z), the evenness of det(A)is equivalent to saying that det(A′) = 0, where A′ ∈ M2(Z2) is the matrix whoseentries are obtained from those of A viewed as elements of Z2. With that in mind,the desired probability is equal to the probability that a randomly chosen matrixA′ ∈M2(Z2) has determinant zero. Obviously, |M2(Z2)| = 24. On the other hand,the matrices in M2(Z2) with zero determinants are as follows(

0 00 0

),

(0 10 1

),

(1 01 0

),

(0 01 1

),

(1 10 0

),

(1 11 1

),

(0 01 0

),

(1 00 0

),

(0 10 0

),

(0 00 1

).

Therefore, the desired probability is equal to 1016 = 5

8 , which is what we want.

2. Plainly, the equation is equivalent to the following

4x3 − 3x = p,

where p = −4c. Now, letting x = cos z, where z ∈ C, the above equation becomesequivalent to the following equation

cos 3z = p. (∗)

As the original equation has a root x0 in the closed interval [−1, 1], it follows thatp ∈ [−1, 1], whence there exists z0 ∈ R such that x0 = cos z0. It is now obviousthat z0, z0 + 2π

3 , and z0 + 4π3 satisfy (∗). Thus, the corresponding x’s which are

cos z0, cos(z0+ 2π3 ), cos(z0+ 4π

3 ) ∈ [−1, 1] are all of the roots of the original equation,proving the assertion.

3. In view of the hypotheses of the problem, we see that C, endowed with thesymmetric difference of sets, denoted by ∆, which is defined by

A∆B = (A \B) ∪ (B \A)

forms a subgroup of P(X), where P(X) denotes the power set of X which obviouslyforms a group under ∆ itself. As is well-known, |P(X)| = 2n, where n = |X|. Thus,it follows from Lagrange’s Theorem that |C| divides |P(X)| = 2n, yielding |C| = 2k

for some k ≤ n, which is what we want.

2.12. TWELFTH COMPETITION 119

2.12. Twelfth Competition

2.12.1. Analysis. 1. First Solution: Suppose that f(x) = f(y) for somex, y ∈ R. It follows that M |x − y| ≤ 0, implying x = y. That is, f is one-to-one.Now, as f is continuous and one-to-one, we see from Problem 2 of 1.9.1 that f isstrictly monotonic. Without loss of generality, we may assume that f is strictlyincreasing. It now easily follows from the hypothesis that limx→−∞ f(x) = −∞ andlimx→+∞ f(x) = +∞. This, in view of the Intermediate Value Theorem, impliesthat f is onto, which is what we want.

Second Solution: Just as we saw in the first solution, the function f is one-to-one and strictly monotonic, and hence the inverse of f , i.e., f−1 : f(R) → R iscontinuous. This implies that f is both an open and closed map. Now, since R isboth open and closed in R, we see that so is f(R), yielding f(R) = R because R isconnected. This proves the assertion.

2. “=⇒” Since∑+∞n=1 fn converges uniformly on S, it is uniformly Cauchy on S,

and hence for given ε > 0, there exists an N > 0 such that∣∣ n∑k=m

fk∣∣ < ε

2,

for all m,n ≥ N . Now, letting m =[n2

]> N , it follows that n > m ≥ N , from

which, in view of the fact that (fn)+∞n=1 is a decreasing sequence of nonnegativefunctions on S, we see that

n∑k=[n

2 ]fk <

ε

2

=⇒ nfn2

≤ (n−[n2

]+ 1)fn ≤

n∑k=[n

2 ]fk <

ε

2,

for all n ≥ 2N + 1. This obviously implies nfn < ε for all n ≥ 2N + 1. That is, thesequence (nfn)+∞n=1 converges uniformly to zero on S. On the other hand,

+∞∑n=1

n(fn − fn+1) =+∞∑n=1

((nfn − (n+ 1)fn+1

)+ fn+1

). (∗)

But (n + 1)fn+1 converges uniformly to zero on S. Hence, the telescopic series∑+∞n=1

(nfn − (n+ 1)fn+1

)converges uniformly to f1 on S. This together with the

hypothesis and (∗) implies that∑+∞n=1 n(fn−fn+1) converges uniformly to

∑+∞n=1 fn

on S, which is what we want.

“⇐=” Note that as pointed out in the footnote of the problem in Section 1.12.1,for this implication we need to assume further that the sequence (fn)+∞n=1 convergesuniformly to zero on the set S. That is because for the numerical sequence (fn)+∞n=1,where fn = 2+ 1

n2 , the series∑+∞n=1 n(fn−fn+1) converges but

∑+∞n=1 fn is divergent.

120 2. SOLUTIONS

t=1t

x=1

x

Figure 10

Now, since (fn)+∞n=1 converges uniformly to zero on S, it follows that the tele-scopic series

∑+∞n=1(fn − fn+1) converges uniformly to f1 on S. We can write

nfn = n

+∞∑k=n

(fk − fk+1) =+∞∑k=n

n(fk − fk+1) ≤+∞∑k=n

k(fk − fk+1), (∗∗)

for all n ∈ N. By the hypothesis, the series∑+∞n=1 n(fn−fn+1) is uniformly Cauchy

on S. This, in view of (∗∗), easily implies that the sequence (nfn)+∞n=1 convergesuniformly to zero on S. Thus, the telescopic series

∑+∞n=1

(nfn − (n + 1)fn+1

)converges uniformly to f1 on S. We can write

+∞∑n=1

fn+1 =+∞∑n=1

(n(fn − fn+1)−

(nfn − (n+ 1)fn+1

)),

implying that∑+∞n=1 fn+1 converges uniformly on S, for so do the series

∑+∞n=1 n(fn−

fn+1) and∑+∞n=1

(nfn − (n + 1)fn+1

)on S. Therefore,

∑+∞n=1 fn+1, and hence∑+∞

n=1 fn, converges uniformly on S. Moreover,

+∞∑n=1

fn =+∞∑n=1

n(fn − fn+1).

So the proof is complete.

3. First solution: Using integration by parts, we can write∫ 1

0

(∫ 1

x

g(t)dt)dx = x

∫ 1

x

g(t)dt∣∣∣∣1x=0

−∫ 1

0

xd

(∫ 1

x

g(t)dt)

= 0 +∫ 1

0

xg(x)dx

=∫ 1

0

xg(x)dx,

which is what we want.

2.12. TWELFTH COMPETITION 121

Second solution: Changing the order of integration, we can write∫ 1

x=0

(∫ 1

t=x

g(t)dt)dx =

∫ 1

t=0

(∫ t

x=0

g(t)dx)dt

=∫ 1

t=0

g(t)(∫ t

x=0

dx

)dt

=∫ 1

0

tg(t)dt,

which is what we want.

2.12.2. Algebra. 1. By Sylow’s Third Theorem, the number of subgroupsof order p is equal to 1+ kp for some k ∈ N∪0 and that 1+kp divides |G|. Thatis, 1 + kp|2p. This yields 1 + kp = 1, implying k = 0. Thus, G has only one Sylowp-subgroup. Therefore, G has only one subgroup of order p.

Using Sylow’s Third Theorem again, we see that the number of subgroups oforder 2 is equal to 1+2k for some k ∈ N∪0 and that 1+2k divides |G|. That is,1 + 2k|2p, from which we obtain 1 + 2k = 1 or 1 + 2k = p. Therefore, G has eitherone and only one subgroup of order 2 or it has p subgroups of order 2.

Now, suppose that G has only one subgroup of order 2. Let H and K be theonly Sylow 2-subgroup and Sylow p-subgroup, respectively. We have |H| = 2 and|K| = p. This implies |H ∩ K| = 1. In other words, H ∩ K = e. Pick h ∈ Hand k ∈ K such that ord(h) = 2 and ord(k) = p. As H ∩ K = e, we see thathk = kh because hkh−1k−1 ∈ H ∩ K, and hence ord(hk) = 2p, for ord(h) = 2,ord(k) = p, and gcd(2, p) = 1 (for a more detailed proof of ord(hk) = 2p, see thelemma presented in Solution 1 of 2.25.2). This implies that the cyclic subgroupgenerated by hk is G. Therefore, G is a cyclic group, which is what we want.

2. Note first that G has no elements of even order, in particular of order 2, becauseG is of odd order. We shall show that the map ψ : G→ G defined by ψ(g) = g2 isan automorphism of G. To see that ψ is one-to-one, suppose ψ(a) = ψ(b) for somea, b ∈ G. It easily follows that (ab−1)2 = e, where e is the identity element of G.But ord(ab−1) 6= 2. So ab−1 = e, yielding a = b. That is, ψ is one-to-one. It nowfollows that ψ is an automorphism of G because G is finite and abelian. Thus, foran arbitrary g ∈ G, there exists an a ∈ G such that g = a2. Set x = aφ(a) andy = aφ(a−1). Noting that G is abelian, we can write

xy =(aφ(a)

)(aφ(a−1)

)= a2φ(aa−1) = a2 = g.

On the other hand,

φ(x) = φ(aφ(a)

)= φ(a)φ2(a) = aφ(a) = x,

φ(y) = φ(aφ(a−1)

)= φ(a)φ2(a−1) = a−1φ(a) = y−1,

as desired. To show that such x, y are unique, suppose that for g ∈ G, thereare x, y, z, t ∈ G such that g = xy = zt, φ(x) = x, φ(z) = z, φ(y) = y−1, and

122 2. SOLUTIONS

φ(t) = t−1. We have

φ(g) = φ(xy) = φ(zt) =⇒ φ(x)φ(y) = φ(z)φ(t)=⇒ xy−1 = zt−1 =⇒ xyy−2 = zt−1

=⇒ zty−2 = zt−1 =⇒ y2 = t2

=⇒ ψ(y) = ψ(t) =⇒ y = t,

which, in view of xy = zt, yields x = z. Therefore, any element g ∈ G can, uniquely,be written as g = xy, where φ(x) = x and φ(y) = y−1, finishing the proof.

3. As every Boolean ring is commutative, it suffices to show that R is a Booleanring. To this end, pick x ∈ R such that x 6= 0, 1. As ch(R) = 2, there existsx1 ∈ R \ 0, 1 such that x = 1 + x1. It follows from the hypothesis that xy = xy2

for all y ∈ R. In other words, (1 + x1)y = (1 + x1)y2, from which, in view ofx1y = x1y

2, we obtain y = y2 for all y ∈ R. That is, R is a Boolean ring, and hencecommutative, which is what we want.

4. First, we claim that every ascending chain of ideals of R terminates. To thisend, suppose

I1 ⊆ I2 ⊆ · · ·

is an ascending chain of ideals of R. Set I =⋃+∞n=1 In. It is easily seen that I is an

ideal of R, and hence there exists an x0 ∈ R such that I = 〈x0〉 = x0R. As x ∈ I,it follows that there exists an N ∈ N such that x0 ∈ IN . We can write

IN ⊆ I =+∞⋃k=1

Ik = x0R ⊆ IN =⇒ IN =+∞⋃k=1

Ik

=⇒ ∀n ∈ N : In ⊆ IN .

On the other hand, IN ⊆ In for all n ≥ N , from which, in view of the aboveinclusion, we obtain IN = In for all n ≥ N . This proves the claim. Now, supposethat f : R→ R is a surjective homomorphism. To see that f is one-to-one, we notethat

ker f ⊆ ker f2 ⊆ ker f3 ⊆ · · ·

is an ascending chain of ideals of R. It follows from the claim that there exists anN ∈ N such that ker fn = ker fN for all n ≥ N . In particular, ker fN+1 = ker fN .Suppose x ∈ ker f is arbitrary so that f(x) = 0. As f is onto, so is fn for alln ∈ N, and hence there exists an x0 ∈ R such that x = fN (x0). Consequently,fN+1(x0) = 0, yielding x0 ∈ ker fN+1 = ker fN . Therefore, x = fN (x0) = 0,implying ker f = 0. That is, f is one-to-one, which is what we want.

5. First solution: Let λi’s (1 ≤ i ≤ 3) be the eigenvalues of the matrix A in thealgebraic closure of F . As A is invertible, λ−1

i ’s (1 ≤ i ≤ 3) are the eigenvalues ofA−1. If f is the characteristic polynomial of A, we have

f = (x− λ1)(x− λ2)(x− λ3)= x3 − (λ1 + λ2 + λ3)x2 + (λ1λ2 + λ1λ3 + λ2λ3)x− λ1λ2λ3.

2.12. TWELFTH COMPETITION 123

By the hypothesis,

tr(A) = λ1 + λ2 + λ3 = 0,

tr(A−1) = λ−11 + λ−1

2 + λ−13 = 0,

det(A) = λ1λ2λ3 = 1.

On the other hand, λ1λ2 + λ1λ3 + λ2λ3 = λ−11 + λ−1

2 + λ−13 = 0. So, we must have

f(x) = x3 − 1. Now, by the Cayley-Hamilton Theorem, A3 = I, which is what wewant.

Second solution: Let λi’s (1 ≤ i ≤ 3) be as in the first solution. We see thatλi’s (1 ≤ i ≤ 3) are distinct because λ1 + λ2 + λ3 = 0 = λ−1

1 + λ−12 + λ−1

3 andλ1λ2λ3 = 1. With that in mind, suppose f(x) = x3+ax2+bx+c is the characteristicpolynomial of A. We have a = tr(A) = 0 and c = det(A) = −1. By the Cayley-Hamilton Theorem, A3 + bA − I = 0, yielding (A−1)3 − b(A−1)2 − I = 0. As λi’s(1 ≤ i ≤ 3) are distinct, the characteristic polynomial of A−1, denoted by g, isequal to g(x) = x3 − bx2 − 1. But b = tr(A−1) = 0. Therefore, (A−1)3 − I = 0,implying A3 = I, which is what we want.

6. It is obvious that α and β are roots of the polynomial

f(x) = p! + p!x+p!2!x2 + · · ·+ xp,

which is irreducible over Q by Eisenstein’s Criterion. Thus, f is the minimal poly-nomial of α and β over Q. We prove the assertion by way of contradiction.

First suppose that α−β = r ∈ Q. We have f(β+r) = f(α) = 0. That is, β is aroot of the polynomial f(x+r) whose coefficients are in Q. But f(x) is the minimalpolynomial of β over Q. It follows that f(x)

∣∣f(x+r). As f(x) and f(x+r) are bothmonic polynomials of the same degree, we obtain f(x + r) = f(x). Consequently,if α1, . . . , αp are all of the roots of f(x) = 0, then so are α1 + r, . . . , αp + r. Inparticular, we must have

(α1 + r) + · · ·+ (αp + r) = α1 + · · ·+ αp =⇒ pr = 0=⇒ r = 0=⇒ α = β,

which is a contradiction. Thus, α− β = r /∈ Q, as desired.Now , assume that α+ β = r ∈ Q. We have −f(r− β) = f(α) = 0. That is, β

is a root of the polynomial −f(r − x) whose coefficients are in Q. But f(x) is theminimal polynomial of β over Q. Thus, f(x)

∣∣ − f(r − x). As f(x) and −f(r − x)are both monic polynomials of the same degree, we obtain −f(r − x) = f(x).Consequently, letting x = r

2 , we obtain −f( r2 ) = f( r2 ), which yields f( r2 ) = 0.Hence, f(x) is divisible by x− r

2 , which is in contradiction with the fact that f(x)is irreducible over Q. Therefore, α+ β = r /∈ Q, as desired.

Next, suppose that αβ = r ∈ Q. Obviously, α, β 6= 0. We can write f( rβ ) =f(α). It thus follows that β is a root of the polynomial xp

p! f( rx ) whose coefficientsare in Q. Again, we obtain f(x)

∣∣xp

p! f( rx ), from which, as f(x) and xp

p! f( rx ) areboth monic polynomials of the same degree, we see that xp

p! f( rx ) = f(x). Thereare two cases to consider: (i) r > 0 and (ii) r < 0. If r > 0, substituting x =

√r

into xp

p! f( rx ) = f(x), we obtain f(√r) = 0. If

√r ∈ Q, then f(x) is divisible by

124 2. SOLUTIONS

A

R

O

B

Mr

Figure 11

x−√r, which is in contradiction with the irreducibility of f over Q. And if

√r /∈ Q,

then x2 − r would be the minimal polynomial of√r over Q. From this, in view

of f(√r) = 0, we see that x2 − r

∣∣f(x), implying that f is not irreducible over Q,a contradiction again. If r < 0, substituting x = i

√−r into xp

p! f( rx ) = f(x), weobtain f(i

√−r) = 0. It follows that x2 +r is the minimal polynomial of

√r over Q.

From this, in view of f(i√−r) = 0, we see that x2 + r

∣∣f(x), implying that f is notirreducible over Q, which is again a contradiction. Therefore, αβ = r /∈ Q, whichis what we want. It might be worth mentioning that one can show, in a similarfashion, that α

β /∈ Q.

2.12.3. General. 1. Let R and a, respectively, be the radius of the circle Ccentered at O and the side length of the square whose center M is on the circle.From the hypothesis, we see that a2 < πR2

2 , whence a√

2 ≤ R√π < 2R. That is,

the diameter of the square is smaller than that of the circle.

It follows that the vertices of the square, with center M , are on the circlecentered at M with radius r = a

√2

2 < R. Let A and B be the points at which thesetwo circle intersect one another. Consider two isosceles triangles OAM and OBM .We have AM = MB < OA = OB = OM . It is plain that ∠AMO > 60 and∠BMO > 60, and hence ∠AMB > 120. Therefore, the arc AB from the circlecentered at M is greater than 120, and hence some vertex of the square must beon this arc which itself lies insides the circle with radius R centered at O. Thisproves the assertion.

2. The answer is: yes, it does. To see this, we count the number of the desiredanswer sheets in a test in which n questions are given. Suppose that there are nquestions in the test. Use 1 and 0 to denote a false and a true answer, respectively,and An to denote the number of desired answer sheets. Then to any answer sheet,there corresponds an n-tuple of 0’s and 1’s. Let’s call an answer sheet or n-tupleadmissible if there are no consecutive 1’s in it. First, for a given k (0 ≤ k ≤ n), wecount the number of admissible n-tuples each of which contains k ones and use Nkto denote this number. To do so, note that in any such admissible n-tuple, thereare k ones and n − k zeros and that we want to insert a zero between any twoones unless the zero occurs in the 1st position or the nth position. With that inmind, there are n− k− 1 + 2 = n− k+ 1 positions and of those we need to choosek positions to each of which assign a one. Thus, the number Nk is equal to the

2.13. THIRTEENTH COMPETITION 125

number of ways of choosing k positions out of n− k+ 1 positions which is equal to(n−k+1

k

). Obviously, k ≤ p :=

[n+1

2

]and that we have

An =p∑k=0

Nk =p∑k=0

(n− k + 1

k

)=

(n+ 1

0

)+(n

1

)+ · · ·+

(n− p+ 1

p

).

It thus follows that the number of desired answer sheets in the test in which 15questions are given is equal to

A15 =(

160

)+(

151

)+ · · ·+

(88

)= 1597.

As 1700 people have participated in the test, there are necessarily two equal answersheets.

3. We have

(1 + i)n =n∑k=0

Cknin−k,

where i ∈ C with i2 = −1. On the other hand, 1 + i =√

2(cos π4 + i sin π

4

), and

hence

(1 + i)n = 2n2(cos

nπ

4+ i sin

nπ

4).

So we haven∑k=0

Cknin−k = 2

n2(cos

nπ

4+ i sin

nπ

4),

whence(C0n − C2

n + C4n − · · ·

)+ i(C1n − C3

n + C5n − · · ·

)= 2

n2 cos

nπ

4+ i2

n2 sin

nπ

4,

form which, we obtain

C0n − C2

n + C4n − · · · = 2

n2 cos

nπ

4,

C1n − C3

n + C5n − · · · = 2

n2 sin

nπ

4,

which is what we want.

2.13. Thirteenth Competition

2.13.1. Analysis. 1. First solution: As f : R → R is integrable on anyclosed interval, for all x, y ∈ R, we can write∫ x

0

f(t+ y)dt =∫ x

0

f(t)dt+∫ x

0

f(y)dt,

126 2. SOLUTIONS

which easily yields

xf(y) =∫ x

0

f(t+ y)dt−∫ x

0

f(t)dt

=∫ x+y

y

f(t)dt−∫ x

0

f(t)dt

=∫ x+y

0

f(t)dt−∫ x

0

f(t)dt−∫ y

0

f(t)dt,

and hence xf(y) = yf(x) for all x, y ∈ R. Therefore, f(x)x = f(1)

1 = f(1) for allx ∈ R with x 6= 0. This implies f(x) = f(1)x for all x ∈ R, as desired.

Second solution: As f : R → R is integrable, it follows from the lemma presentedin Solution 1 of 2.5.1 that the function f is continuous at infinitely many pointsof any closed interval of R. On the other hand, f(x + y) = f(x) + f(y) for allx, y ∈ R and hence f(r) = rf(1) for all r ∈ Q. From this, it is easily seen that f iscontinuous on R and that f(x) = xf(1) for all x ∈ R, which is what we want.

2. First solution: We need the following well-known lemma which is known asDedekind’s Extension of Abel’s Theorem.

Lemma. Let (an)+∞n=1 and (bn)+∞n=1 be sequences of real numbers. If the series∑+∞n=1 an and

∑+∞n=1 |bn − bn+1| converge, then so does the series

∑+∞n=1 anbn.

Proof. It is easily verified that

an+1bn+1 + · · ·+ an+pbn+p = (sn+1 − sn)(bn+1 − bn+2) + · · ·+(sn+p−1 − sn)(bn+p−1 − bn+p) + (sn+p − sn)bn+p,

where sn =∑nk=1 ak and n, p ∈ N. Note that as

∑+∞n=1 |bn − bn+1| converges, so

does the telescopic series∑+∞n=1(bn − bn+1), from which, we see that limn bn exists,

and hence the sequence (bn)+∞n=1 is bounded. Now, the above equality, together withthe facts that (sn)+∞n=1 and

(∑nk=1 |bk − bk+1|

)+∞n=1

are Cauchy and that (bn)+∞n=1 isbounded, implies that (

∑nk=1 akbk)

+∞n=1 is Cauchy, and hence the series

∑+∞n=1 anbn

is convergent, which is what we want.

To prove the assertion, letting cn = an

bn, we can write

anan + bn

=cn − c2n1− c2n

=(cn − c2n

)(1 +

c2n1− c2n

).

Let An = cn − c2n and Bn = c2n1−c2n

. We have

anan + bn

= An +AnBn.

As∑+∞n=1 c

2n is convergent, limn c

2n = 0, and hence there exists an N ∈ N such that

c2n <12 for all n ≥ N . It follows that for all n ≥ N we have

0 ≤ Bn =c2n

1− c2n< 2c2n,

2.13. THIRTEENTH COMPETITION 127

from which, in view of the Comparison Test, we see that∑+∞n=1Bn is convergent.

Note that ∣∣Bn −Bn+1

∣∣ ≤ ∣∣Bn∣∣+ ∣∣Bn+1

∣∣ = Bn +Bn+1,

for all n ≥ N . Again the Comparison Test together with the above inequality, inview of the convergence of

∑+∞n=1Bn, implies that

∑+∞n=1 |Bn−Bn+1| is convergent.

On the other hand, it follows from the hypothesis that∑+∞n=1An is convergent. Now,

using Dedekind’s Extension of Abel’s Theorem, we see that∑+∞n=1AnBn converges.

This together with the convergence of∑+∞n=1An implies that

∑+∞n=1(An + AnBn)

converges. Thus, so does the series∑+∞n=1

an

an+bn, which is what we want.

Second solution: We can writean

an + bn=

cn + c2n − c2n1 + cn

= cn −c2n

1 + cn,

where cn = an

bn. As limn cn = 0, it follows that there exists an N ∈ N such that

c2n1+cn

> 0 for all n ≥ N . This together with the convergence of∑+∞n=1 c

2n, in view

of the Limit Comparison Test, implies that the series∑+∞n=1

c2n1+cn

converges. Fromthe above equality and the hypothesis that

∑+∞n=1 cn is convergent, we see that the

series∑+∞n=1

an

an+bnis convergent and that we have

+∞∑n=1

anan + bn

=+∞∑n=1

cn −+∞∑n=1

c2n1 + cn

,

finishing the proof.

3. First solution: To prove the assertion by contradiction, suppose that thesequence (fn)+∞n=1 does not converge uniformly to zero on [0, 1]. It follows thatthere are an ε > 0, a sequence (nk)+∞k=1, with nk ≥ k, of natural numbers, and asequence (xk)+∞k=1 in [0, 1] such that |fnk

(xk)| ≥ ε for all k ∈ N. If necessary, bypassing to a subsequence of (xk)+∞k=1 and replacing fnk

by −fnk, without loss of

generality, we may assume that fnk(xk) > 0 for all k ∈ N and that there is an

x0 ∈ [0, 1] such that limk xk = x0. We can write

fnk(xk)− fnk

(x0) =∫ xk

x0

f ′nk(t)dt,

from which, in view of ||f ′n||∞ ≤ 1 for all n ∈ N, we obtain∣∣fnk(xk)− fnk

(x0)∣∣ ≤ ∣∣xk − x0

∣∣,for all k ∈ N. Let K1 ∈ N be such that |xk −x0| < ε

2 for all k ≥ K1. We must have∣∣fnk(x0)

∣∣ ≥ ∣∣fnk(xk)

∣∣− ∣∣fnk(xk)− fnk

(x0)∣∣ ≥ ε− ε

2=ε

2,

for all k ≥ K1. Now, from this and

fnk(x)− fnk

(x0) =∫ x

x0

f ′nk(t)dt,

in a similar fashion, we see that ∣∣fnk(x)∣∣ ≥ ε

4,

128 2. SOLUTIONS

for all x ∈ [0, 1] with |x−x0| < ε4 and k ≥ K2, where K2 is such that |xk −x0| < ε

4for all k ≥ K2. Since fnk

’s (k ≥ K2) are continuous functions, we conclude thatfnk

(x) > 0, and hence,fnk

(x) ≥ ε

4,

for all x ∈ [0, 1] with |x− x0| < ε4 and k ≥ K2, for otherwise, by the Intermediate

Value Theorem, there must exist zk ∈ [0, 1] with |zk − x0| < ε4 and k ≥ K2 such

that fnk(zk) = 0, which is impossible. Let [a, b] = [0, 1] ∩ [x0 − ε

4 , x0 + ε4 ]. Now

define g : [0, 1] → R by

g(x) =

0 0 ≤ x ≤ a,

3b−a (x− a) a ≤ x ≤ a+ b−a

3 ,

1 a+ b−a3 ≤ x ≤ a+ 2 b−a3 ,

−3b−a (x− b) a+ 2 b−a3 ≤ x ≤ b,

0 b ≤ x ≤ 1.

The function g is continuous and nonnegative on [0, 1]. For all k ≥ K2, we canwrite ∫ 1

0

fnkg =

∫ a+ b−a3

a

fnkg +

∫ a+2 b−a3

a+ b−a3

fnkg +

∫ b

a+2 b−a3

fnkg

≥∫ a+2 b−a

3

a+ b−a3

fnk≥ (b− a)ε

12,

contradicting the hypothesis that limn

∫ 1

0fnk

g = 0. Therefore, the assertion followsby contradiction.

Second solution: It suffices to show that every subsequence (gk)+∞k=1 of (fn)+∞n=1,where gk = fnk

(k ∈ N), in turn, has a subsequence (hj)+∞j=1, where hj = gkj

(j ∈ N), which converges uniformly to zero on [0, 1]. We first show that the sequence(fn)+∞n=1 is uniformly bounded on [0, 1]. To this end, noting that fn : [0, 1] → R and||f ′n||∞ ≤ 1 for all n ∈ N, from the Mean Value Theorem, we see that∣∣fn(x)∣∣ ≤ 1 +

∣∣fn(0)∣∣,

for all x ∈ [0, 1] and n ∈ N. We claim that the sequence(fn(0)

)+∞n=1

is bounded.

Suppose to the contrary that there exists a subsequence(fnk

(0))+∞n=1

of(fn(0)

)+∞n=1

such that limk fnk(0) = +∞ or limk fnk

(0) = −∞. If necessary, replacing fn by−fn, we may assume that limk fnk

(0) = +∞. Thus, there exists a K ∈ N suchthat fnk

(0) ≥ 2 for all k ≥ K. The sequence (fn)+∞n=1 is uniformly equicontinuouson [0, 1] because ||f ′n||∞ ≤ 1 for all n ∈ N. It follows that the sequence (fnk

)+∞n=1 isuniformly equicontinuous on [0, 1] as well. So choosing ε = 1, we obtain 0 < 2δ < 1such that ∣∣fnk

(x)− fnk(0)∣∣ < 1,

whenever 0 < x < 2δ and k ∈ N. This easily yields

fnk(x) ≥ 2− 1 = 1,

for all 0 < x < 2δ and k ∈ N with k ≥ K. Now, define g : [0, 1] → R by

g(x) =

1 0 ≤ x ≤ δ,−xδ + 2 δ ≤ x ≤ 2δ,0 2δ < x ≤ 1.

2.13. THIRTEENTH COMPETITION 129

Obviously, g is continuous and nonnegative on [0, 1]. So it follows from the hypoth-esis that limn

∫ 1

0fng = 0, from which, we obtain

limk

∫ 1

0

fnkg = 0.

Note that as g ≥ 0 on [δ, 2δ], for all k ≥ K, we have∫ 1

0

fnkg =

∫ 2δ

0

fnkg

=∫ δ

0

fnk+∫ 2δ

δ

fnkg

≥ 1× (δ − 0) = δ.

That is,∫ 1

0fnk

g ≥ δ for all k ≥ K, which is in contradiction with limk

∫ 1

0fnk

g = 0and 0 < δ. Therefore,

(fn(0)

)+∞n=1

is bounded and hence so is (fn)+∞n=1 because|fn(x)| ≤ 1 + |fn(0)| for all x ∈ [0, 1] and n ∈ N.

Now suppose (gk)+∞k=1, with gk = fnk(k ∈ N), is an arbitrary subsequence of

(fn)+∞n=1. The sequence (gk)+∞k=1 is uniformly bounded and equicontinuous on [0, 1]because so is (fn)+∞n=1 on [0, 1]. It thus follows from Arzela’s Theorem that thesequence (gk)+∞k=1 has a subsequence (hj)+∞j=1, with hj = gkj

(j ∈ N), such that(hj)+∞j=1 converges uniformly to a function h : [0, 1] → R on [0, 1]. The function his continuous because it is a uniform limit of continuous functions. This togetherwith the hypothesis yields limj

∫ 1

0hjh = 0. But it is plain that (hjh)+∞j=1 converges

uniformly to h2 on [0, 1]. So, we can write∫ 1

0

h2 =∫ 1

0

limjhjh = lim

j

∫ 1

0

hjh = 0,

implying∫ 1

0h2 = 0, which, in turn, implies h = 0 because h is continuous on

[0, 1]. That is, we have shown that every subsequence of (fn)+∞n=1, in turn, has asubsequence converging uniformly to zero on [0, 1]. This proves the assertion.

2.13.2. Algebra. 1. Note first that every nonzero ideal I of R is uncountable.To see this, as R

I is countable, we have

R

I=an + I : an ∈ R

.

It follows that R =⋃+∞n=1(an + I), implying that I is uncountable because other-

wise R would be countable, which is impossible. Now, to prove the assertion bycontradiction, suppose that 0 6= a ∈ R is a divisor of zero in R. Define

I :=r ∈ R : ar = 0

.

It is plain that I is a nonzero ideal of R and hence it is uncountable. Also note thataR is a nonzero ideal of R. Define the map f : aR → R

I by f(ax) = x + I. Themap f is obviously onto. Suppose that f(ax1) = f(ax2). We have x1 + I = x2 + I,implying that x1 − x2 ∈ I, whence a(x1 − x2) = 0. That is, ax1 = ax2 whichmeans f is one-to-one. Therefore, f is a one-to-one correspondence between theuncountable set aR and the countable set R

I , a contradiction. Thus, R has nodivisor of zero, and hence R is an integral domain, which is what we want.

130 2. SOLUTIONS

2. Let S = M2(Q). We have QI2 ≤ R ≤ S and dimQ R ≤ dimQ S ≤ 4, where I2denotes the 2×2 identity matrix. Suppose that I is a nonzero left ideal of R. As Rincludes QI2, the left ideal I can be viewed as a vector space over Q and we havedimQ I ≤ 4. Thus, there are xi ∈ I (1 ≤ i ≤ k ≤ 4) such that I = Qx1 + · · ·+ Qxk.This together with QI2 ≤ R implies that I = Rx1 + · · · + Rxk. In other words,I is finitely generated. Likewise, one can see that every right ideal of R is finitelygenerated as well. To show that the condition (∗) cannot be dropped, define

J =(

0 G0 0

),

where G = 〈 12i 〉+∞i=1 ≤ Q is the additive group generated by 1

2i ’s in Q. It is easilyverified that J is a left ideal of the ring T which is not finitely generated, which iswhat we want.

3. First solution: Let G = GL2(Q) be the multiplicative group of all 2 × 2invertible matrices over the field Q and

a =(

1 10 −1

), b =

(−1 10 1

).

It is easily seen that a, b ∈ G, a2 = b2 = I2, and ab = −I2 + 2N , where

I2 =(

1 00 1

), N =

(0 10 0

).

We have(ab)n = (−1)nI2 + (−1)n−12nN 6= I2,

for all n ∈ N. That is, ab has order infinity and yet both a and b have order two,which is what we want.

Second solution: Let G be the group of isometries of the Euclidean plane undercomposition of isometries. Let A,B ∈ R2 be two distinct points in the plane anda, b the half turns around the points A and B, respectively. If e denotes the identityisometry, we have a2 = b2 = e. On the other hand, it is easy to see that ab is atranslation isometry along the vector 2

−−→AB, and hence ab has order infinity because

it is a translation. This is what we want, finishing the proof.

4. (a) It is well-known that

|GLn(Zp)| = (pn − 1)(pn − p) · · · (pn − pn−1).

To see this, note that a matrix A ∈ GLn(Zp) if and only if the columns of A arelinearly independent. In view of this, there are pn−1 choices for the first column ofan arbitrary element, say A, of GLn(Zp) (the zero column vector being excluded).For an 1 ≤ i < n, assuming that the first i columns of an arbitrary element, say A,of GLn(Zp) are chosen, there are exactly pn− pi choices for the (i+ 1)st column ofA, because the (i + 1)st column of A cannot be a linear combination of the first icolumns of A. Therefore, by the product rule of combinatorics, GLn(Zp) has exactly(pn−1)(pn−p) · · · (pn−pn−1) elements, as desired. Thus, |G| = (32−1)(32−3) = 48.As K = Z(G) includes only scalar matrices, we see that K = I2, 2I2, where I2 isthe identity matrix, and hence |K| = 2. It is now obvious that K ≤ H ≤ G. Tocalculate |H|, we note that a, b, c ∈ Z3 can be chosen independent of one another

2.13. THIRTEENTH COMPETITION 131

and that a, c can be 1 or 2 and b can be 0, 1, 2. Thus, using the product rule ofcombinatorics, we have

|H| = 2× 2× 3 = 12.

(b) Let N =⋂x∈G x

−1Hx. First, we claim that N is the largest subgroup ofH that is normal in G. We present two proofs for the claim.

First proof. Firstly, N ⊆ e−1He = H. Secondly, N is a subgroup of Gbecause it is an intersection of subgroups of G, namely, x−1Hx’s where x ∈ G. Tosee that N is a normal subgroup of G, suppose that a ∈ N and g ∈ G are arbitrary.As a ∈ N =

⋂x∈G x

−1Hx and xg−1 ∈ G, we see that a ∈ gx−1Hxg−1, yieldingg−1ag ∈ x−1Hx. It follows that g−1ag ∈

⋂x∈G x

−1Hx = N . That is, N is anormal subgroup of G. Now, suppose that N1 ⊆ H is a normal subgroup of G. Wecan write

N1 = x−1N1x ⊆ x−1Hx,

for all x ∈ G, implying that N1 ⊆⋂x∈G x

−1Hx = N , as desired.

Second proof. We see from (a) that [G : H] = 4. Consequently, if X is usedto denote the set of left cosets of H in G, then there exist g1 = e, g2, g3, g4 ∈ Hsuch that

X =g1H, . . . , g4H

.

Observe that G acts on X via left multiplication. That is, g(giH) = ggiH ∈X, where g ∈ G. Now, for a fixed g ∈ G, define the map `g : X → X by`g(giH) = ggiH. It is easily verified that `g is a one-to-one map which is ontobecause |X| = 4 < ∞. In other words, `g defines a permutation on X, i.e., `g ∈S(X) ∼= S4, where S4 denotes the symmetric group of degree four. Now, define themap φ : G→ S(X) ∼= S4 by φ(g) = `g. As G acts on X via left multiplication, wesee that `g1g2 = `g1`g2 for all g1, g2 ∈ G. This means

φ(g1g2) = φ(g1)φ(g2),

for all g1, g2 ∈ G. Thus, φ is a homomorphism of groups. We claim that

kerφ = N =⋂x∈G

x−1Hx ⊆ e−1He = H,

from which, we see that N is a normal subgroup of G which is contained in H. Toprove this last claim, suppose g ∈ kerφ. We have φ(g) = `g = `e, where e is theidentity element of G. It follows that `g(x−1H) = `e(x−1H) for all x ∈ G. Thatis, gx−1H = x−1H, yielding xgx−1 ∈ H, and hence g ∈ x−1Hx for all x ∈ G. Thisimplies g ∈

⋂x∈G x

−1Hx = N . Conversely, if g ∈⋂x∈G x

−1Hx = N , we concludethat g ∈ kerφ. Therefore, kerφ =

⋂x∈G x

−1Hx = N , proving the claim.

We now show that N = K. To this end, let n ∈ N be arbitrary. As N ⊆ H,

we have n =(a1 b10 c1

), where a1, b1, c1 ∈ Z3 and a1c1 6= 0. Suppose g =(

a bc d

)∈ G, where a, b, c, d ∈ Z3 with ad − bc 6= 0, is arbitrary. As N is

normal in G, we must have h = g−1ng ∈ N ⊆ H. This implies that h21 =−aca1−c2b1+acc1

ad−bc = 0, where h21 denotes the 21 entry of the matrix h. Simplifying,

132 2. SOLUTIONS

we obtain ac(c1 − a1) + c2(−b1) = 0 for all a, b, c, d ∈ Z3 with ad − bc 6= 0. Thiseasily yields a1 = c1 and b1 = 0, proving that n ∈ K, which is what we want.

(c) We have shown that the map φ : G→ S(X) ∼= S4 defined by φ(g) = `g is ahomomorphism of groups and that

kerφ =⋂x∈G

x−1Hx = N = K.

In view of the First Isomorphism Theorem for groups, we can write

G

kerφ∼= im(φ) ≤ S4.

That is, GK∼= im(φ) ≤ S4. On the other hand, |im(φ)| = [G : K] = 24 = |S4|.

Thus, im(φ) ∼= S4, and hence GK∼= S4, which is what we want.

5. Let(Aν(ξ)

)i1

(1 ≤ i ≤ n) denote the i1 entry of the column matrix Aν(ξ). Wehave

(Aν(ξ)

)i1

=n∑k=1

aik(ν(ξ)

)k1

= ai1(ν(ξ)

)11

+n∑k=2

aik(ν(ξ)

)k1

= δi,n +n∑k=2

δi,k−1ξk−1 =

1 i = n,ξi i < n.

In other words,

Aν(ξ) =

ξξ2

...ξn−1

1

=

ξξ2

...ξn−1

ξn

= ξ

1ξξ2

...ξn−1

= ξν(ξ).

That is, ν(ξ) is an eigenvector of A whose corresponding eigenvalue is ξ.

2.13.3. General. 1. Let M be an arbitrary point on the unit circle in thecomplex plane centered at the origin and A1, . . . , An denote the vertices of a regularn-gon which is inscribed in the circle. We have

M = exp iθ = cos θ + i sin θ,Aj = exp iθj = cos θj + i sin θj ,

2.13. THIRTEENTH COMPETITION 133

where θ, θ1 ∈ R, and θj = θ1 + (j − 1) 2πn (1 ≤ j ≤ n). We can write

n∑j=1

MA2j =

n∑j=1

∣∣ exp iθ − exp iθj∣∣2

=n∑j=1

(2− exp i(θ − θj)− exp i(θj − θ)

)= 2n− exp i(θ − θ1)

n∑j=1

(exp−i2π

n

)j−1 − exp i(θ1 − θ)n∑j=1

(exp i

2πn

)j−1

= 2n− exp i(θ − θ1)

(exp−i2πn

)n − 1exp−i2πn − 1

− exp i(θ1 − θ)

(exp i 2πn

)n − 1exp i 2πn − 1

= 2n− exp i(θ − θ1)× 0− exp i(θ1 − θ)× 0 = 2n.

That is,∑nj=1MA2

j = 2n, which is what we want.

2. Letf(x) = anx

n + an−1xn−1 + · · ·+ a1x+ a0,

where ai ∈ Q (0 ≤ i ≤ n) and an > 0, be a polynomial with the property that f(x)is irrational whenever x is irrational. Let p be a prime and

g(x) = lcm(a1, . . . , an)f(x)− lcm(a1, . . . , an)a0 − p

= gnxn + gn−1x

n−1 + · · ·+ g1x− p,

where gi = lcm(a1, . . . , an)ai (1 ≤ i ≤ n). As the leading coefficient of g is positiveand g(0) = −p < 0, it follows from the Intermediate Value Theorem that g hasa positive root which must be rational because, in view of the above equality, thepolynomial g has the property that g(x) is irrational whenever x is irrational. Nowby the Rational Root Theorem, the positive root is of the form 1

d or pd for some

d dividing gn. If n > 1, it is obvious that by choosing p large enough none of thenumbers 1

d and pd , where d divides gn, can be a root of g(x) = 0. Therefore, n ≤ 1,

yielding f(x) = ax+ b for some a, b ∈ Q with a 6= 0.

3. To prove the assertion, we need the following theorem which is known asSylvester’s problem.

Theorem. Let S be a finite set of the points of a plane such that the linepassing through any two points of S contains a third point of S. Then, all of thepoints of S lie on a straight line.

Proof. To prove the assertion by contradiction, suppose that all of the pointsof S are not collinear. As S is finite, there exist A,B,C ∈ S such that the smallestheight of the triangle ABC is minimal among all triangles with vertices from S.Without loss of generality, assume that the altitude AH drawn from the vertex Aperpendicular to the base BC is the smallest height of the triangle ABC.

If H = B or H = C, then the altitude drawn from H to the base opposite toit is smaller than AH, which is impossible. If not, then H is between B and C andthere is a third point D of S on the line BC. Thus, if necessary by renaming thepoints B, C, and D, we may assume that B and C are both on the left or the rightof AH such that BH < CH. Assume, with no loss of generality, that B and C areon the right of AH such that BH < CH. It is now easily seen that the altitude

134 2. SOLUTIONS

A A

BC=H B=H H B

H’B’

CC

A

Figure 12

drawn from B to AC is smaller than AH, a contradiction. Thus, all of the pointsof S are collinear, finishing the proof.

Remark. Motivated by Sylvester’s problem, we pose the following problem.Let S be a finite set of the points of a plane containing no three collinear pointssuch that the circle passing through any three points of S contains a fourth pointof S. Then, all of the points of S lie on a circle. Hint. Perform an inversion withcenter at one of the given points and any radius, say, radius one.

First, we show that there exists a plane which contains only two lines of thegiven n parallel lines. To this end, intersect the n parallel lines with a plane toobtain n points of intersection on the plane. It follows from the above theoremthat in the plane there is a line which contains exactly two points of the n points.Now, this line and two lines of the given n parallel lines form a plane which passesthrough only two lines of the n parallel lines. With this in mind, we prove theassertion by induction on n. If n = 3, the assertion is obvious. Assuming that theassertion holds for n− 1 parallel lines, to prove the assertion for n parallel lines, asexplained in the above find a plane that contains exactly two lines of the given nlines. Now, one of these two lines and the n − 2 remaining lines are not coplanar.It thus follows from the induction hypothesis that there exist at least n−1 distinctplanes each of which includes at least two lines of these n − 1 lines. These n − 1distinct planes together with the plane that contains only two lines of the given nlines form n planes satisfying the desired property, which is what we want.

2.14. Fourteenth Competition

2.14.1. Analysis. 1. To prove the assertion by contradiction, suppose thatf ′(x0) exists. As f ′(x) < 0 on (−∞, x0), f is strictly decreasing on (−∞, x0],from which, we obtain f(x) > f(x0) for all x < x0. Likewise, since f ′(x) > 0 on(x0,+∞), we see that f(x) > f(x0) for all x > x0. Thus, f assumes it absoluteminimum, and hence local minimum, at x0, yielding f ′(x0) = 0. Now, f ′′(x) < 0on (x0,+∞) implies f ′ is strictly decreasing on [x0,+∞), which, in turn, impliesf ′(x) < f ′(x0) = 0 for all x > x0, which is in contradiction with the hypothesis.This shows that f ′(x0) does not exist, which is what we want.

2. We need the following lemma.

2.14. FOURTEENTH COMPETITION 135

Lemma. Let g : R → R be a uniformly continuous and nonnegative functionon R such that

∫ +∞−∞ g <∞. Then

limx→−∞

g(x) = 0, limx→+∞

g(x) = 0.

Proof. We prove that limx→+∞ g(x) = 0. One can prove limx→−∞ g(x) = 0in a similar fashion. Proceed by way of contradiction. Thus, there exists an ε0 > 0such that for all n ∈ N, there is an xn > n for which

g(xn) =∣∣g(xn)∣∣ ≥ ε0.

From the hypothesis that g is uniformly continuous on R, we see that for ε02 > 0,

there exists a 0 < δ0 < 1 such that∣∣g(x)− g(y)∣∣ < ε0

2,

for all x, y ∈ R with |x − y| < δ0. In particular, for all i ∈ N and x ∈ R with|x− xi| < δ0, we have ∣∣g(x)− g(xi)

∣∣ < ε02,

from which, in view of |g(xi)| − |g(x)| ≤ |g(x)− g(xi)|, we obtain

g(x) =∣∣g(x)∣∣ ≥ ∣∣g(xi)∣∣− ε0

2≥∣∣g(xi)∣∣

2=g(xi)

2≥ ε0

2.

Thus, g(x) ≥ ε02 on (xi − δ0, xi + δ0) for all i ∈ N, yielding∫ xi+δ0

xi−δ0g ≥ 2δ0

ε02

= ε0δ0.

On the other hand,∫ xi+δ0−∞ g −

∫ xi−δ0−∞ g =

∫ xi+δ0xi−δ0 g. Thus, letting i → +∞ in the

above, in view of the hypothesis, we obtain 0 ≥ ε0δ0, which is a contradiction.Therefore,

limx→+∞

g(x) = 0,

which is what we want.

First solution: Suppose by way of contradiction that there exists an ε0 > 0 suchthat for all n ∈ N there is an xn ∈ R with |xn| > n for which |f(xn)| ≥ ε0. As f isuniformly continuous on R, for ε0

2 > 0 there exists a δ0 > 0 such that∣∣f(x)− f(y)∣∣ < ε0

2,

whenever x, y ∈ R and |x − y| < δ0. In particular, for all i ∈ N and |x − xi| < δ0,we have ∣∣f(x)− f(xi)

∣∣ < ε02.

But ∣∣f(xi)∣∣− ∣∣f(x)

∣∣ ≤ ∣∣f(x)− f(xi)∣∣

from which, we see that∣∣f(x)∣∣ ≥ ∣∣f(xi)

∣∣− ε02≥ |f(xi)|

2≥ ε0

2. (∗)

On the other hand, f is bounded on R. So there exists an M > 0 such that |f(x)| ≤M for all x ∈ R. By the continuity of g on R, g is continuous on the compact

136 2. SOLUTIONS

interval [−M,M ], and hence it is uniformly continuous on [−M,M ]. Consequently,for given ε > 0, there exists a δ1 > 0 such that∣∣g(x)− g(y)

∣∣ < ε, (∗′)whenever x, y ∈ [−M,M ] and |x− y| < δ1. It follows from the uniform continuityof f on R that for the above δ1 > 0, which was obtained from the given ε > 0,there exists a δ > 0 such that ∣∣f(x)− f(y)

∣∣ < δ1,

whenever x, y ∈ R and |x− y| < δ. Now, for all x, y ∈ R with |x− y| < δ, we have|f(x) − f(y)| < δ1. This together with the fact that f(x), f(y) ∈ [−M,M ] for allx, y ∈ R with |x− y| < δ, in view of (∗′), implies∣∣g(f(x))− g(f(y))

∣∣ < ε.

That is, g f is uniformly continuous on R. On the other hand, g f is integrableon R. It thus follows from the lemma that

limx→−∞

g f(x) = 0, limx→+∞

g f(x) = 0.

Now, setm = inf

g(x)|x ∈ [−M,−ε0

2] ∪ [

ε02,M ]

.

As g(x) > 0 whenever x 6= 0 and that g is continuous on the compact set [−M,− ε02 ]∪

[ ε02 ,M ], we see that there exists a c ∈ [−M,− ε02 ]∪ [ ε02 ,M ] such that m = g(c), and

hence m = g(c) > 0. Now, from limx→∞ g f(x) = 0, it follows that for m > 0,there exists an N > 0 such that

g(f(x)

)=∣∣g(f(x)

)− 0∣∣ < m,

whenever x ∈ R and |x| > N . On the other hand, we have limi |xi| = +∞, andhence for the above N > 0, there exists an I > 0 such that |xi| > N wheneveri ≥ I. In particular, |xI | > N , implying g

(f(xI)

)< m. But, by (∗), we have ε0

2 <

f(xI) ≤ M , yielding g(f(xI)

)≥ m, a contradiction. Therefore, limx→∞ f(x) = 0,

which is what we want.

Second solution: Again we proceed by contradiction and continue up to (∗) andset

m = infg(x)|x ∈ [−M,−ε0

2] ∪ [

ε02,M ]

.

As we saw in the first solution, there exists a c ∈ [−M,− ε02 ] ∪ [ ε02 ,M ] such that

m = g(c), and hence m = g(c) > 0. It follows from (∗) that for all i ∈ N andx ∈ R such that x ∈ (xi − δ0, xi + δ0), we have M ≥ |f(x)| ≥ ε0

2 , implyingf(x) ∈ [−M,− ε0

2 ] ∪ [ ε02 ,M ], whence g(f(x)

)≥ m > 0. Thus, we can write∫ xi+δ0

xi−δ0g f ≥ m(xi − δ0 − xi + δ0) = 2δ0m (∗′′)

On the other hand,∫ xi+δ0

xi−δ0g f =

∫ xi+δ0

−∞g f −

∫ xi−δ0

−∞g f.

Thus, letting i → +∞ in (∗′′), in view of the hypothesis, we obtain 0 ≥ 2δ0m,which is a contradiction. Therefore, limx→∞ f(x) = 0, which is what we want.

2.14. FOURTEENTH COMPETITION 137

3. Define the sets A and B as follows

A :=x ∈ [c, d] : f(x) = a

= g−1

(a),

B :=x ∈ [c, d] : f(x) = b

= g−1

(b),

where g = f |[c,d], which is continuous on [c, d] because so is f on [a, b]. It followsthat A = g−1(a) and B = g−1(b) are closed, and hence compact, subsets ofthe compact interval [c, d]. Thus, there exist r ∈ A and s ∈ B such that∣∣s− r

∣∣ = inf∣∣x− y

∣∣ : x ∈ A, y ∈ B.It is plain that r, s ∈ [c, d] and r 6= s. Assume, without loss of generality, thatr < s. By showing that f([r, s]) = [a, b], we settle the proof. To this end, notefirst that f(r) = a and f(s) = b. Next, suppose that λ ∈ [a, b] is arbitrary. Wehave f(r) = a ≤ λ ≤ b = f(s). From this and the Intermediate Value Theorem,we see that there exists a c ∈ [r, s] such that f(c) = λ. That is, λ ∈ f([r, s]),

implying [a, b] ⊆ f([r, s]). Now, suppose that λ = f(c) ∈ f([r, s]), where c ∈ [r, s],is arbitrary. We need to show that λ ∈ [a, b]. Suppose to the contrary that λ /∈ [a, b].This implies λ < a or λ > b. If λ < a, we can write

f(c) = λ < a < b = f(s).

From this, in view of the Intermediate Value Theorem, we see that there existsr1 ∈ (c, s) such that f(r1) = a. On the other hand, r ≤ c < r1 < s, f(r) = f(r1) =a, f(s) = b, and |s− r1| < |s− r|, which is in contradiction with our choice of r ands. This implies a ≤ λ. Likewise, one can see that λ ≤ b. Thus, a ≤ λ ≤ b, yieldingf([r, s]) ⊆ [a, b] because λ ∈ f([r, s]) was arbitrary. Therefore, f([r, s]) = [a, b],which is what we want.

2.14.2. Algebra. 1. First solution: As G is a finite p-group, we have |G| =pn for some n ∈ N. It now follows from Sylow’s First Theorem that G has a normalsubgroup H such that |H| = pn−1. We have |G|

|H| = p. Thus, GH is a cyclic, and

hence an ableian, group because p is a prime number. Since GH is abelian, we obtain

G′ ⊆ H $ G, implying that G′ 6= G, as desired.

Second solution: Recall that a group is solvable if and only if for some n ∈ N∪0,we have G(n) = e, where G(0) = G and G(n+1) =

(G(n)

)′ (n ∈ N ∪ 0). It iswell-known that every finite p-group is solvable. With all that in mind, proceedby contradiction. Suppose G′ = G. It follows that G(n) = G for all n ∈ N ∪ 0.Consequently, G(n) 6= e for all n ∈ N ∪ 0, and hence G cannot be solvable,which is in contradiction with the fact that every finite p-group is solvable. Thus,G′ 6= G, which is what we want.

2. Let p be the smallest prime dividing |G| and K a normal subgroup of G with pelements. As p is prime, it follows that K is cyclic so that K = 〈a〉 for some a ∈ Gwith ord(a) = p. Obviously, it suffices to prove that a ∈ Z(G). To this end, supposethat g ∈ G is arbitrary. It follows that g−1ag ∈ K because K is a normal subgroupof G. Hence, there is an integer j with 0 ≤ j ≤ p−1 such that g−1ag = aj . Firstly,j 6= 0 because otherwise a = e, which is impossible, for ord(a) = p. So 0 < j ≤ p−1and we can write

aj = g−1ag =⇒ (aj)j = g−1ajg,

138 2. SOLUTIONS

whence aj2

= g−2ag2. By an easy induction on n, we see that ajn

= g−nagn forall n ∈ N. In particular, if n = p− 1, we obtain aj

p−1= g−(p−1)agp−1. Now, since

0 < j ≤ p− 1 and p is prime, it follows that gcd(j, p− 1) = 1. This together withFermat’s Little Theorem yields jp−1 p

≡ 1. That is, there exists an integer k suchthat jp−1 = 1 + kp. As ord(a) = p, we have

(gp−1)−1agp−1 = ajp−1

= a1+kp = a(ap)k = aek = a,

which yields(gp−1)−1agp−1 = a.

In other words, agp−1 = gp−1a for all g ∈ G. To finish the proof, we need toshow that the map ϕ : G → G defined by ϕ(g) = gp−1 is onto. As |G| is finite, itsuffices to show that the map ϕ is one-to-one. To this end, suppose ϕ(g1) = ϕ(g2)for some g1, g2 ∈ G. We have gp−1

1 = gp−12 and that gcd(p − 1, |G|) = 1, for p is

the smallest prime that divides |G|. It follows that there exist r, s ∈ Z such thatr(p− 1) + s|G| = 1. Now, noting that g|G| = e for all g ∈ G, we can write

g1 = gr(p−1)+s|G|1 =

(g(p−1)1

)r(g|G|1

)s =(g(p−1)2

)r(g|G|2

)s = gr(p−1)+s|G|2 = g2.

That is, g1 = g2. In other words, ϕ : G→ G is one-to-one, and hence onto, finishingthe proof.

3. Recall that if R is a commutative ring and a, b ∈ R are nonzero elements ofit, then the greatest common divisor of a and b in R, denoted by gcd(a, b), is anonzero element d ∈ R satisfying the following. (i) d|a and d|b, that is, there existk1, k2 ∈ R such that a = k1d and b = k2d. (ii) If d′|a and d′|b for some 0 6= d′ ∈ R,then d′|d. Also, the least common multiple of a and b, denoted by lcm(a, b), is anonzero element c ∈ R satisfying the following. (i) a|c and b|c. (ii) If a|c′ and b|c′for some 0 6= c′ ∈ R, then c|c′.

We claim that in Z6, the ring of integers mod 6, the greatest common divisorof 2 and 3 is 1, whereas 2 and 3 have no common multiple. First, it is obvious that1|2 and 1|3. Next, suppose d′|2 and d′|3. It follows that d′|3 − 2 = 1. Hence, 1 isa greatest common divisor of 2 and 3. Use contradiction to prove that 2 and 3 hasno common multiple. Suppose to the contrary that for 0 6= c ∈ Z6 we have 2|c and3|c. It follows that there exist k1, k2 ∈ Z6 such that c = 2k1 and c = 3k2. We canwrite

3c = 6k1 = 0, 2c = 6k2 = 0,

whence c = 3c − 2c = 0 − 0 = 0. That is, c = 0, which is impossible. This provesthe claim by contradiction.

4. First, if n ∈ N and n > 1, we show that xny+ xn−1 + 1 is irreducible in F [x, y].Suppose to the contrary that there exist f, g ∈ F [x, y] such that xny+xn−1+1 = fg.As degy(xny+xn−1+1) = 1, in view of the above equality, without loss of generality,we may assume that degy(f) = 0 and degy(g) = 1. Consequently, there exist a non-invertible f0 ∈ F [x] and g0, g1 ∈ F [x] such that f = f0 and g = g0 + yg1. We canwrite

xny + xn−1 + 1 = f0g0 + f0g1y,

yielding f0g1 = xn and f0g0 = xn−1 + 1. As f0 is not invertible, there exists 1 <j0 < n such that f0 = xj0 and g1 = xn−j0 . From this, we obtain g0xj0 = xn−1 + 1,

2.14. FOURTEENTH COMPETITION 139

which is impossible because j0 > 1 and n > 1. Thus, the polynomial xny+xn−1 +1is irreducible in F [x, y] whenever n > 1.

Next, if n = 1, we show that xy + 1 + 1 = xy + 2 is irreducible in F [x, y] ifand only if ch(F ) 6= 2. To this end, note first that if ch(F ) = 2, then xy + 2 = xyis reducible in F [x, y]. Now, suppose that xy + 2 is reducible in F [x, y]. We showthat ch(F ) = 2. Suppose to the contrary that ch(F ) 6= 2. It follows that 2 has aninverse in F . Since xy+ 2 is reducible in F [x, y], just as we saw in the above, thereexist a non-invertible f0 ∈ F [x] and g0, g1 ∈ F [x] such that f = f0, g = g0 + yg1,and xy + 2 = fg. We must have

xy + 2 = f0g0 + f0g1y,

from which, we obtain f0g0 = 2 and f0g1 = x. Now, since 2 is invertible in F , itfollows that so is f0 in F [x], a contradiction. Thus, ch(F ) = 2, which is what wewant.

5. We may assume that V1 * V2 and V2 * V1 because otherwise, in view ofdimV1 = dimV2, we see that V1 = V2, in which case the assertion is easy to prove.More precisely, pick a basis αimi=1 for V1 = V2, where m = dimV1, and enlarge itto a basis αimi=1 ∪ βini=1 for V , where m+ n = dimV . It is plain that

U ⊕ V1 = U ⊕ V2 = V,

where U = 〈β1, . . . , βm〉 is the vector subspace spanned by βi’s (1 ≤ i ≤ n). Weprove the assertion by induction on dimV − dimV1 + 1. If dimV − dimV1 + 1 = 1,then dimV1 = dimV2 = dimV . Hence, for U = 0, we obviously have U ⊕ V1 =U ⊕ V2 = V . Assuming that the assertion holds whenever dimV − dimV1 + 1 = n,we prove it whenever dimV − dimV1 + 1 = n+ 1. As V1 * V2 and V2 * V1, we seethat V1 ∪ V2 $ V , and hence there exists a vector β ∈ V \ (V1 ∪ V2). Set

V ′1 = V1 ⊕ 〈β〉, V ′

2 = V2 ⊕ 〈β〉.We have dimV ′

1 = dimV ′2 = dimV1 + 1, yielding dimV − dimV ′

1 + 1 = dimV −dimV1 = n. It thus follows from the induction hypothesis that there exists asubspace U1 such that

U1 ⊕ V ′1 = U1 ⊕ V ′

2 = V.

Letting U = U1 ⊕ 〈β〉, we obviously obtain

U ⊕ V1 = U ⊕ V2 = V,

proving the induction assertion, which is what we want.

2.14.3. General. 1. First solution: Let A = A1, . . . , An be a familycontaining n distinct subsets of the set S with n elements. Define the edge-labeledgraph G with vertices in A as follows. For 1 ≤ i, j ≤ n with i 6= j, connect thedistinct vertices Ai and Aj with an edge labeled s ∈ S whenever Ai = Aj ∪ s orAj = Ai ∪ s. Now, define the relation ∼ on A as follows. For 1 ≤ i, j ≤ n, wewrite Ai ∼ Aj whenever i = j or there exists a path in the graph G joining thevertices Ai and Aj of the graph. It is easily verified that ∼ is an equivalence relationon A and that the equivalence classes of the relation ∼ are connected componentsof the graph G. Let Giki=1 (1 ≤ k ≤ n) denote the connected components of G.Also, let Ti (1 ≤ i ≤ k) be a maximal tree of Gi (recall that a tree is connectedgraph not having any cycle). That is, Ti is a subgraph of Gi which is a tree and that

140 2. SOLUTIONS

it is maximal as a tree in Gi (from this, we immediately see that Ti is a spanningtree of Gi, i.e., Ti a subgraph of Gi that contains all the vertices of Gi and is atree). For a subgraph H of G, use L(H) to denote the set of labels of the edgesof H. We claim that L(Ti) = L(Gi) for all 1 ≤ i ≤ k. Obviously, it suffices toshow that L(Gi) ⊆ L(Ti). To this end, let s ∈ L(Gi) be arbitrary. It follows thats ∈ S is the label of an edge AiAj of the subgraph Gi. If the edge AiAj is alreadyin the tree Ti, there is nothing to prove. If not, as Ti is a maximal tree of Gi, wesee that the graph obtain by adding the edge AiAj to the tree Ti is not a tree.It thus follows that the edge AiAj participates in a cycle whose all edges but theedge AiAj come from the tree Ti. Now, s ∈ S being the label of AiAj , we see thatAi = Aj ∪ s or Aj = Ai ∪ s. By symmetry, we may assume without loss ofgenerality, that Ai = Aj ∪ s. Note that s /∈ Aj , for i 6= j. Now, going from Ajto Ai along the other edges of the cycle, in which the edge AiAj participates, andnoting that s /∈ Aj but s ∈ Ai, we see that there must be an edge AkAl ∈ Ti, where1 ≤ k, l ≤ n, having s as its label, because otherwise s /∈ Ai, which is impossible.Thus, s ∈ L(Ti), whence L(Gi) ⊆ L(Ti), implying L(Gi) = L(Ti), as desired. It isplain that L(Ti) ≤ ε(Ti), where ε(Ti) denotes the number of the edges of Ti. Onthe other hand, since Ti is a tree, by a standard theorem from the theory of graphs,we have

ε(Ti) = ν(Ti)− 1,

where ν(Ti) denotes the number of vertices of Ti. The number k being the numberof connected components of G, we can write

L(G) =k∑i=1

L(Gi) ≤k∑i=1

ε(Ti) =k∑i=1

ν(Ti)− k,

implying

L(G) ≤k∑i=1

ν(Ti)− k = ν(G)− k,

where ν(G) denotes the number of vertices of G. This implies L(G) ≤ ν(G)− k <ν(G). Thus, there exists an s0 ∈ S such that s0 /∈ L(G). It is now plain that thesets

A1 ∪ s0, . . . , An ∪ s0are distinct, for otherwise we must have Ai = Aj ∪s0 or Aj = Ai ∪s0 for some1 ≤ i, j ≤ n with i 6= j, yielding s0 ∈ L(G), which is impossible. This completesthe proof of the assertion.

Second solution: This solution is taken from Linear Algebra Methods in Com-binatorics by L. Babai and P. Frankl. Without loss of generality, assume thatS = 1, . . . , n. It suffices to show that there exists an element x ∈ S such that thesets

A1 \ x, . . . , An \ xare all distinct because so will then be the sets

A1 ∪ x, . . . , An ∪ x.

To see this, just note that for some x ∈ S and 1 ≤ i, j ≤ n, Ai ∪ x = Aj ∪ xif and only if Ai \ x = Aj \ x. Define the n × n matrix M = (mij) with

2.14. FOURTEENTH COMPETITION 141

mij ∈ 0, 1 (1 ≤ i, j ≤ n) as follows

mij =

1 j ∈ Ai,0 j /∈ Ai.

By the hypothesis, the rows ofM are all distinct. We need to show that this remainsto be the case after omitting an appropriate column from M . There are two casesto consider. (i) detM = 0; and (ii) detM 6= 0. If detM = 0, there is a columnof M , say, column j for some 1 ≤ j ≤ n, which is linearly dependent on the othercolumns of M . We see that after deleting column j of M , the remaining rows areall distinct. Suppose on the contrary that this is not the case. That is, the matrixobtained from deleting column j of M has two equal rows, say rows i1 and i2 forsome 1 ≤ i1 < i2 ≤ n. It follows that rows i1 and i2 of M are equal as well becausecolumn j of M depends linearly on the other columns of M . This is a contradiction,proving the assertion in this case. If detM 6= 0, if necessary by interchanging tworows of M , we may assume that the first row of M is a row with the minimumnumber of ones. Note that there might be several rows with the minimum numberof ones. Expanding detM by the first row, we see that m1j detM1j 6= 0, whereM1j is the matrix obtained by eliminating row 1 and column j from the matrixM . Consequently, m1j = 1 and no two rows of M1j are the same. It thus followsthat deleting column j leaves no two equal rows. Suppose on the contrary that tworows of the matrix obtained from deleting column j of M are equal. Since no tworows of M1j are the same, we see that for some 1 < i ≤ n, row 1 and row i of thematrix obtained from deleting column j of M must be equal. This easily impliesthat mij = 0, for otherwise rows 1 and i of M would be equal, which is impossible.Thus, the number of ones of row i of M is less than that of row 1, which is acontradiction. Therefore, no two rows of the matrix obtained from deleting columnj of M are equal, which is what we want.

2. We prove that the answer is 14 days. It is obvious that the maximum number ofthe exhibition days is attained provided that there is exactly one common book forany two days of the exhibition. With that in mind, first, we obtain the number ofbooks needed to exhibit the books for k days in such a way that 100 books are tobe exhibited in each day and that the exhibited books of any two days have exactlyone book in common. Then, we find the maximum number of the days subjectto the condition that the number of the books is less than or equal to 1369. Indoing so, suppose that we would like to run the exhibition for k days. First, choosek − 1 books (b1i )

ki=2 for the first day of the exhibition and set aside the book b1j to

be exhibited for a second time in the jth day. Then, choose k − 2 books (b2i )ki=3,

from the books not already chosen, for the second day and set aside the book b2j(3 ≤ j ≤ k) to be exhibited for a second time in the jth day. Then, choose k − 3books (b3i )

ki=4, from the books not already chosen, for the third day and set aside

the book b3j (4 ≤ j ≤ k) to be exhibited for a second time in the jth day. Continuethis way to finally choose one book bk−1

k , from the books not already chosen, forthe (k − 1)st day and set aside the book bk−1

k to be exhibited for a second time inthe kth day. So far, for each day we have k− 1 books to be exhibited and that theexhibited books of any two days, say day i and day j where i < j, have exactly onebook in common, namely, the book bij . In order to exhibit 100 books per day, foreach day we need another 100 − (k − 1) = 101 − k books. So the number of the

142 2. SOLUTIONS

books needed is

(k − 1) + (k − 2) + · · ·+ 2 + 1 + k(101− k) =k(k − 1)

2+ k(101− k)

=k(201− k)

2.

Thus, for k = 14, the number of the books needed is 14(201−14)2 = 1309 and for

k = 15 the number of the books needed is 15(201−15)2 = 1395. Since we have only

1369 books, it follows that the maximum number of the days of the exhibition is14, finishing the proof.

3. The assertion is a quick consequence of the following lemma.

Lemma. Let F be a field and with ch(F ) = 0 or > n, K an extension of F ,and λ1, . . . , λn ⊆ K. Also, let m ∈ N ∪ 0 and c ∈ F . If

λk1 + · · ·+ λkn = ck−m(λm1 + · · ·+ λmn ),

for each k = m,m+ 1, . . . ,m+ n, then λi = 0 or c for all 1 ≤ i ≤ n.Proof. We prove the assertion by induction on n. If n = 1, the proof is

obvious. Suppose that the assertion holds for all k < n. Let λ1, . . . , λn ⊆ K bea given subset as described in the lemma. There are two cases to consider.

(i) First, suppose c = 0 or λm1 + · · ·+ λmn = 0. Then

λk1 + · · ·+ λkn = 0,

for each k = m + 1, . . . ,m + n. We claim that λi = 0 for all 1 ≤ i ≤ n. Suppose,by contradiction, that this is not the case. Let µi ∈ K (1 ≤ i ≤ l) be nonzero anddistinct such that µ1, . . . , µl ∪ 0 = λ1, . . . , λn ∪ 0. It follows that thereexist n1, . . . , nl ∈ N with n1 + · · ·+ nl ≤ n such that

n1µk1 + · · ·+ nlµ

kl = λk1 + · · ·+ λkn = 0,

for each k = m+ 1, . . . ,m+ n. In other words, µm+11 · · · µm+1

l...

. . ....

µm+n1 · · · µm+n

l

n1

...nl

=

0...0

,

implying that

µm+11 · · ·µm+1

l det

1 · · · 1...

. . ....

µn−11 · · · µn−1

l

= 0.

This in turn, in view of Vandermonde’s determinant formula, implies µi = 0 forsome 1 ≤ i ≤ l or µi = µj for some 1 ≤ i < j ≤ l, a contradiction in any event.Thus, λi = 0 for all 1 ≤ i ≤ n, which is what we want.

(ii) Next, suppose c 6= 0, and hence λm1 + · · ·+ λmn 6= 0. For each k = 1, . . . , n,let Sk denote the elementary symmetric polynomial in λ1, . . . , λn of degree k, i.e.,S1 = λ1 + · · ·+ λn, . . . , Sn = λ1 · · ·λn. Obviously, we can write

xn − S1xn−1 + · · ·+ (−1)nSn = (x− λ1) · · · (x− λn), (∗)

from which, we obtain

λn+mi − S1λ

n+m−1i + · · ·+ (−1)nSnλmi = 0,

2.15. FIFTEENTH COMPETITION 143

for all 1 ≤ i ≤ n. Adding up these equations, we getn∑i=1

λn+mi − S1

n∑i=1

λn+m−1i + · · ·+ (−1)nSn

n∑i=1

λmi = 0.

It now follows from the hypothesis that

cnn∑i=1

λmi − S1cn−1

n∑i=1

λmi + · · ·+ (−1)nSnn∑i=1

λmi

=n∑i=1

λn+mi − S1

n∑i=1

λn+m−1i + · · ·+ (−1)nSn

n∑i=1

λmi = 0.

Thus, ( n∑i=1

λmi)(cn − S1c

n−1 + · · ·+ (−1)nSn)

= 0,

yieldingcn − S1c

n−1 + · · ·+ (−1)nSn = 0.

This together with (∗) implies that c = λi for some 1 ≤ i ≤ n. It is now plain thatthe induction hypothesis can be applied to the set λ1, . . . , λn \ c, completingthe proof.

Now, to prove the assertion, just let F = K = R and c = m = 1 in the lemmaabove.

2.15. Fifteenth Competition

2.15.1. Analysis. 1. (a) Letting 0 ≤ x < y, we can write

2ϕ(x+ y

2

)= 2

∫ x+y2

0

f =∫ x

0

f +∫ x+y

2

0

f +∫ x+y

2

x

f.

As f is increasing on [0,+∞) andy − x

2is positive, we obtain∫ x+y

2

x

f ≤∫ x+y

2

x

f(t+

y − x

2)dt =

∫ y

x+y2

f.

So, we have

2ϕ(x+ y

2

)=

∫ x

0

f +∫ x+y

2

0

f +∫ x+y

2

x

f

≤∫ x

0

f +∫ x+y

2

0

f +∫ y

x+y2

f

=∫ x

0

f +∫ y

0

f

= ϕ(x) + ϕ(y).

This proves (a).

144 2. SOLUTIONS

(b) Note first that f is integrable on any closed and bounded subinterval of[0,+∞) because f is increasing. Consequently, ϕ is continuous on [0,+∞). On theother hand, it easily follows from (a) that

ϕ(λx+ (1− λ)y

)≤ λϕ(x) + (1− λ)ϕ(y), (∗)

where λ = k2k , 1 ≤ k ≤ 2n, and k, n ∈ N. Now, as the set

k2k : k, n ∈ N, 1 ≤ k ≤

2n

is dense in [0, 1], in view of the continuity of ϕ, we see that (∗) holds for allλ ∈ [0, 1]. In other words, ϕ is convex, which is what we want.

2. It is worth mentioning that the hypothesis g(0) = 0 is redundant. The functiong is continuous on the closed interval [0, 1], so it is bounded on [0, 1]. Hence, thereexists an M > 0 such that |g(x)| ≤ M for all x ∈ [0, 1]. Noting that 0 ≤ sinx ≤ 1for all x ∈ [0, 1], we can write

|fn(x)| =∣∣∣∣g(x)(sinx)n1 + nx

∣∣∣∣ ≤ M sinx1 + nx

≤ M

n.

From this, we conclude that the sequence (fn)+∞n=1 uniformly converges to the zerofunction on [0, 1]. So the assertion follows.

3. (a) For all x, z ∈ (0,+∞), we have

2xz ≤ xf(x) + zf−1(z).

Letting z = f(y), we obtain

2xf(y) ≤ xf(x) + yf(y),

for all x, y ∈ (0,+∞). We can write

xf(y)− xf(x) ≤ yf(y)− xf(y) =⇒ f(y)− f(x) ≤ y − x

xf(y),

for all x, y ∈ (0,+∞). Interchanging x, y in the above, we obtain

f(x)− f(y) ≤ x− y

yf(x) =⇒ f(y)− f(x) ≥ y − x

yf(x)

for all x, y ∈ (0,+∞). This together with the preceding inequality proves (a).

(b) First, we prove that f is increasing on (0,+∞). To this end, we have

2xf(y) ≤ xf(x) + yf(y),

2yf(x) ≤ yf(y) + xf(x),yielding

2(xf(y) + yf(x)

)≤ 2

(xf(x) + yf(y)

),

for all x, y ∈ (0,+∞). Consequently,

(y − x)(f(y)− f(x)

)≥ 0,

for all x, y ∈ (0,+∞). This shows that f is increasing on (0,+∞). It thus followsthat f has limit from the left and limit from the right at any point and that the twolimits at any point coincide because f is increasing and onto. Thus, f is continuouson (0,+∞). From (a), we have

y − x

yf(x) ≤ f(y)− f(x) ≤ y − x

xf(y),

2.15. FIFTEENTH COMPETITION 145

for all x, y ∈ (0,+∞). Dividing by y − x where x, y ∈ (0,+∞) with x < y, we canwrite

f(x)y

≤ f(y)− f(x)y − x

≤ f(y)x

.

Letting y → x+ in the above, we obtain

f ′(x+) =f(x)x

,

where f ′(x+) denotes the right derivative of f at x. Likewise, dividing by y − xwhere x, y ∈ (0,+∞) with x > y, we can write

f(x)y

≥ f(y)− f(x)y − x

≥ f(y)x

,

from which by letting y → x−, we obtain

f ′(x−) =f(x)x

,

where f ′(x−) denotes the left derivative of f at x. That is, f ′(x) exists for allx ∈ (0,+∞) and moreover f ′(x) = f(x)

x . This implies xf ′(x)− f(x) = 0 and henceddx

(f(x)x

)= xf ′(x)−f(x)

x2 = 0 for all x ∈ (0,+∞). Therefore, there exists a c ∈ R

such that f(x)x = c, yielding f(x) = cx for all x ∈ (0,+∞), which is what we want.

2.15.2. Algebra. 1. Let Ω =P1, . . . , Pp+1

denote the set of all (distinct)

Sylow p-subgroups of G. The group G acts on Ω by conjugation. The kernel ofthe action is

H :=g ∈ G| ∀i = 1, . . . , p+ 1 : g−1Pig = Pi

=p+1⋂i=1

NG(Pi),

where NG(Pi) denotes the normalizer of Pi in G. Recall that for a subset S of G,the normalizer of S in G, denoted by NG(S), is defined by

NG(S) =g ∈ G : g−1Sg = S

.

We note that by Sylow’s Second Theorem H is a normal subgroup of G. Fromthe First Isomorphism Theorem for groups, we see that the group G

H is isomorphicto a subgroup of the symmetric group Sp+1, and since p

∣∣(p+ 1)! but p2 -(p+ 1)!,

it follows that pα∣∣ |H| or pα−1

∣∣ |H|. Note that the subgroup H ∩ Pi is a Sylow

p-subgroup of H for each i = 1, . . . , p + 1, for∣∣HPi

∣∣∣∣Pi

∣∣ =∣∣H∣∣∣∣H∩Pi

∣∣ . Now, as H ∩ Piand H ∩ Pj are Sylow p-subgroups of H, by Sylow’s Second Theorem, they areconjugate in H, and hence there exists an h ∈ H such that h−1(H ∩Pi)h = H ∩Pj .On the other hand, h ∈ H, implies h ∈ NG(Pi) and hence

H ∩ Pi = (h−1Hh) ∩ (h−1Pih) = h−1(H ∩ Pi)h = H ∩ Pj .

That is,H ∩ Pi = H ∩ Pj ,

146 2. SOLUTIONS

for all 1 ≤ i, j ≤ p + 1. From this, we obtain H ∩ Pi ⊆ Pj for all 1 ≤ i, j ≤ p + 1,yielding

H ∩ Pi ⊆p+1⋂j=1

Pj ,

for all 1 ≤ i ≤ p+ 1. On the other hand,p+1⋂i=1

Pi ⊆ H ∩ Pi,

for all 1 ≤ i ≤ p+ 1. Therefore,p+1⋂j=1

Pj = H ∩ Pi,

for all 1 ≤ i ≤ p+ 1. This implies that⋂p+1j=1 Pj is a Sylow p-subgroup of H. Now,

if pα∣∣ |H|, then

∣∣⋂p+1j=1 Pj

∣∣ = pα = |Pi|, and hence⋂p+1j=1 Pj = Pi, implying that

Pi = Pj , a contradiction. Thus, pα−1∣∣ |H|. Consequently, a Sylow p-subgroup of

H must be of order pα−1, and hence |⋂p+1i=1 Pi| = pα−1, which is what we want.

2. First, we claim that if P is a prime ideal of R and ab ∈ P for some a, b ∈ R,then a ∈ P or b ∈ P . To see this, from ab ∈ P , we get Rab ⊆ P . Now, as the leftideal Ra is also a right ideal of R, we see that RaR ⊆ Ra, implying RaRb ⊆ Rab.This yields RaRb ⊆ P , from which, we obtain Ra ⊆ P or Rb ⊆ P , for P is a primeideal of R. This, in turn, implies a ∈ P or b ∈ P because R is unital.

Now to prove the assertion, first, suppose that x ∈ R is nilpotent. So thereexists an n ∈ N such that xn = 0. We have xn = 0 ∈ P for all prime ideals of R. Inview of the above claim, we see that x ∈ P for all prime ideals of R. In other words,x ∈

⋂P : P C R,P is prime

. Conversely, suppose that x ∈ R and that xn 6= 0

for all n ∈ N. In other words, for the multiplicative set S := xn : n ∈ N, we haveS ∩ 0 = ∅. A standard argument using Zorn’s Lemma shows that there existsa prime ideal P such that P ∩ S = ∅. This yields x /∈ P . In other words, we haveproved that if x ∈

⋂P : P C R,P is prime

, then x must be nilpotent. Therefore,

the intersection of all prime ideals of R is equal to the set of the nilpotent elementsof R, which is what we want.

3. If we do not require the matrix AB to be a nonzero idempotent, then theassertion is trivial. Just let B = 0. Then, AB = 0 is an idempotent, settling theproof. However, we state and prove the following nontrivial problem.

Let D be a division ring, n ∈ N, and A ∈Mn(D). Then, there exists a matrixB ∈Mn(D) such that AB is an idempotent whose rank is equal to that of A.

Let Dn denote the right vector space of all n × 1 column vectors with entriesin D; that is, the addition x + y is defined componentwise and the multiplicationof the scalar λ ∈ D into the vector x = (xi)ni=1 ∈ Dn is defined by xλ := (xiλ)ni=1.The members of Mn(D) can be viewed as linear transformations acting on the leftof Dn via the usual matrix multiplication; that is, we can write Mn(D) = L(Dn),where L(Dn) is the ring of all right linear transformations acting on the left ofDn. For x ∈ Dn and f ∈ (Dn)′, where (Dn)′ = Dn denotes the dual of Dn

which is the left vector space of all 1× n row vectors with entries in D, define therank-one linear transformation x ⊗ f ∈ L(Dn) by (x ⊗ f)(y) := xf(y). Choose

2.15. FIFTEENTH COMPETITION 147

yi ∈ Dn (1 ≤ i ≤ r) such that Ayi1≤i≤r is a basis for the range of A. Setxi := Aiyi and enlarge xi1≤i≤r to a basis B ∪ xi1≤i≤r for Dn, where the setB is linearly independent. Now, let fi1≤i≤r be a dual subset with respect toB ∪ xi1≤i≤r so that 〈B〉 ⊆ ker fi and fi(xj) = δij for each i, j = 1, . . . , r. LetB = y1⊗f1+· · ·+yr⊗fr. Since 〈B〉 ⊆ ker fi and fi(xj) = δij for each i, j = 1, . . . , r,we easily see that AB = A(y1 ⊗ f1 + · · ·+ yr ⊗ fr) = x1 ⊗ f1 + · · ·+ xr ⊗ fr is anidempotent whose rank is r = rank(A), proving the assertion.

2.15.3. General. 1. We prove that a necessary and sufficient condition forthe product of two integers a, b to be divisible by their sum is that there existintegers x, y, z such that x, y are relatively prime and that

a = x(x+ y)z, b = y(x+ y)z.

Sufficiency is easy. To prove necessity, from a+ b∣∣ab, we see that

ab = m(a+ b),

for some m ∈ Z. If one of a or b is zero, say, a = 0, then letting x = 0, y = 1,z = b proves the assertion. So without loss of generality, assume that a and b arenonzero. Set d = gcd(a, b) and

x =a

d, y =

b

d.

It is obvious that x and y are relatively prime. From ab = m(a + b), we obtaindxy = m(x+y), implying x+y

∣∣dxy. But xy and x+y are relatively prime becauseso are x and y. Thus, x + y

∣∣d, and hence there exists an integer z such thatd = (x+ y)z. Form this, we obtain

a = x(x+ y)z, b = y(x+ y)z,

which is what we want.

2. Set β =∫ 1

0xf(x)dx. We prove that∫ 1

0

(x− β)2f(x)dx ≤∫ 1

0

(x− α)2f(x)dx, (∗)

for all α ∈ R. To this end, using∫ 1

0f(x)dx = 1 and

∫ 1

0xf(x)dx = β, we can write∫ 1

0

(x− α)2f(x)dx =∫ 1

0

((x− β) + (β − α)

)2f(x)dx

=∫ 1

0

(x− β)2f(x)dx+ (β − α)2 + 2(β − α)β − 2(β − α)β

=∫ 1

0

(x− β)2f(x)dx+ (β − α)2

≥∫ 1

0

(x− β)2f(x)dx.

As (x− 12 )2 ≤ 1

4 on [0, 1], we can write∫ 1

0

(x− 1

2)2f(x)dx ≤

∫ 1

0

14f(x)dx =

14.

148 2. SOLUTIONS

Now, in view of (∗), we have∫ 1

0

(x− β)2f(x)dx ≤∫ 1

0

(x− 1

2)2f(x)dx ≤ 1

4,

which is what we want.

3. (a) The desired probability is the ratio of the number of “favorable cases” to thenumber of “total cases”. The number of “total cases” is obviously np. To calculatethe number of “favorable cases”, let Ai be the event that no one gets on the ithwagon. It follows that the number of “disfavorable cases” is equal to |A1∪· · ·∪An|.Using the inclusion-exclusion principle from combinatorics, we can write

|A1 ∪ · · · ∪An| =∑i

|Ai| −∑i,j

|Ai ∩Aj |+ · · ·+ (−1)n−1|A1 ∩ · · · ∩An|

=∑i

(n− 1)p −∑i,j

(n− 2)p + · · ·+ (−1)n−10p

=(n

1

)(n− 1)p −

(n

2

)(n− 2)p + · · ·+ (−1)n−2

(n

n− 1

)1p.

Thus, the number of “favorable cases” is equal to

np −(n

1

)(n− 1)p +

(n

2

)(n− 2)p + · · ·+ (−1)n−1

(n

n− 1

),

and hence the desired probability is

1np

(np −

(n

1

)(n− 1)p +

(n

2

)(n− 2)p + · · ·+ (−1)n−1

(n

n− 1

)),

which is what we want.

(b) First, if p < n, the probability calculated in (a) is obviously zero, implyingthat the number of “favorable cases” is zero. This, in view of

(nr

)=(nn−r), yields(

n

1

)1p −

(n

2

)2p +

(n

3

)3p − · · ·+ (−1)n−1

(n

n

)np = 0.

Next, if p = n, then the number of wagons is equal to that of the passengers. In thiscase, in order not to have any empty wagon, we must have one passenger in eachwagon. Thus, the number of “favorable cases” is equal n!. On the other hand, ifwe let p = n in the above formula, which we obtained for the number of “favorablecases”, and equate these two values, we obtain(

p

1

)1p −

(p

2

)2p +

(p

3

)3p − · · ·+ (−1)p−1

(p

p

)pp = (−1)p−1p!,

finishing the proof.

2.16. SIXTEENTH COMPETITION 149

2.16. Sixteenth Competition

2.16.1. Analysis. 1. As g is continuous on the compact interval [0, 1], itfollows that there exists an M > 0 such that

∣∣g(x)∣∣ ≤ M for all x ∈ [0, 1]. Fromthe continuity of g at 1 and g(1) = 0, we see that for given ε > 0, there exists a0 < δ < 1 such that ∣∣g(x)∣∣ < ε,

whenever x ∈ [0, 1] and 0 < 1 − x < δ. On the other hand, for given ε > 0, thereexists an N ∈ N such that (1− δ)nM < ε for all n ≥ N . It thus follows that for agiven ε > 0, for all x ∈ [0, 1] and n ≥ N , we have∣∣fn(x)− 0

∣∣ = ∣∣xn∣∣∣∣g(x)∣∣ ≤ max(

(1− δ)nM, sup1−δ≤x≤1

∣∣g(x)∣∣) < ε,

proving the assertion.

2. (a) Let x = [x] + 0.a1a2a3 . . .. We have

x+ 0.1 = [x] + a0.a′1a2a3 . . . ,

where a0 = 0 or 1 and a′1 ∈ 0, 1, . . . , 9 and hence

f(x) = f(x+ 0.1) = a2.

Hence, 0.1 is a period for f . To prove that 0.1 is the period of f , suppose to thecontrary that 0 < α < 0.1 is a period of f . If we let x1 = 0.1 − (0.1)α, we have0.09 < x1 < 0.1. Consequently, x1 = 0.09α1α2 . . ., where all of αi’s are not equal to9. It follows that f(x1) = 9. On the other hand, if we let x2 = α+x1 = 0.1+(0.9)α,we have

0.0999 . . . = 0.1 < x2 < 0.19 = 0.18999 . . . .

Thus, x2 = 0.1β1β2 . . ., where β1 ≤ 8. Hence, f(x2) = β2 ≤ 8, whence f(x1) 6=f(x2), which is a contradiction. This shows that α cannot be the period of f .

(b) Using integration by parts, we can write∫ c

0

xdf(x) =∫ 0.1

0

xdf(x) = 0.1f(0.1)−∫ 0.1

0

f(x)dx

= (0.1).9−9∑k=0

k1

100= 0.45,

which is what we want.

3. We prove the assertion in the normed linear space setting. Let X be a realor complex normed linear space and f : X → X a uniformly continuous function.Then, there exists a, b ∈ R+ such that

||f(x)|| ≤ a||x||+ b,

for all x ∈ X, where ||.|| denotes the norm of X. As f is uniformly continuous onX,for ε = 1, there exists δ0 > 0 such that ||f(x)− f(y)|| < 1 whenever x, y ∈ Rn and||x−y|| ≤ δ0. We show that for a = 1

δ0and b = |f(0)|+1, we have ||f(x)|| ≤ a||x||+b

150 2. SOLUTIONS

for all x ∈ X. To this end, for x ∈ X, set N =[‖x‖δ0

], where bracket stands for the

integer part function. We can write

||f(x)|| ≤ ||f(0)||+ ||f(x)− f(0)||≤ ||f

(0)||

+N∑k=1

∥∥∥∥f(kδ0 x

||x||)− f

((k − 1)δ0

x

||x||)∥∥∥∥+

∥∥∥∥f(x)− f(Nδ0

x

||x||)∥∥∥∥ .

From N =[||x||δ0

], we obtain

N ≤ ‖x‖δ0

< N + 1,

yielding ∥∥∥∥x−Nδ0x

||x||

∥∥∥∥ <‖x‖N + 1

< δ0,∥∥∥∥kδ0 x

||x||− (k − 1)δ0

x

||x||

∥∥∥∥ =∥∥∥∥ δ0‖x‖

x

∥∥∥∥ = δ0 ≤ δ0,

for all 1 ≤ k ≤ N . So, for all x ∈ X, we can write

‖f(x)‖ ≤ ‖f(0)‖+N + 1 ≤ 1δ0‖x‖+ (‖f(0)‖+ 1),

whence||f(x)|| ≤ a||x||+ b,

for all x ∈ X, where a, b are as in the above, which is what we want.

2.16.2. Algebra. 1. First, recall that Inn(G) ∼= GZ(G) , where Z(G) denotes

the center of the group G. To see this, it is easily checked that the map f :G −→ Aut(G) defined by fg(x) = gxg−1 is a homomorphism of groups and thatker f = Z(G). It thus follows from the First Isomorphism Theorem for groupsthat Inn(G) ∼= G

Z(G) . Next, we need to recall that if p is a prime, then everygroup G of order p2 is abelian. To prove this by contradiction, suppose that Gis not abelian. Note first that |Z(G)| > 1 because G is a p-group (see Solution 3of 2.11.2). Consequently, |Z(G)| = p. But then

∣∣ GZ(G)

∣∣ = p, implying that GZ(G) is

cyclic and hence G is abelian, a contradiction. Thus, G is abelian, as desired.We now prove the assertion. To this end, from the hypothesis that [G : A] =

[G : B] = p, it follows that |A| = |B|. And since p is the smallest prime that divides|G|, from Problem 1 of 2.9.2, we see that A and B are both normal in G. It thusfollows that AB ≤ G. By the Second Isomorphism Theorem for groups, we haveABA = B

A∩B . So we can write

[G : AB] =|G||AB|

=[G : A]

[B : A ∩B]=

p

[B : A ∩B].

If [B : A∩B] = 1, we obtain A = B, which is a contradiction. Thus, [B : A∩B] = p,and hence G = AB. We now prove that A∩B = Z(G). To this end, for x ∈ A∩B,

2.16. SIXTEENTH COMPETITION 151

as is usual, use CG(x) to denote the centralizer of the element x of G. As A and Bare abelian, we have A,B ≤ CG(x), and hence

p = [G : A] = [G : CG(x)][CG(x) : A],

p = [G : B] = [G : CG(x)][CG(x) : B],

from which, we see that CG(x) = G. That is, x ∈ Z(G), and hence A ∩B ≤ Z(G).Now, as

[G : A ∩B] = [G : B][B : A ∩B] = [G : B][AB : A] = [G : B][G : A] = p2,

we conclude that [G : Z(G)] = 1 or p or p2 because [G : Z(G)]∣∣[G : A ∩ B]. As G

is nonabelian, the cases [G : Z(G)] = 1 or p are impossible. Thus, [G : Z(G)] = p2,from which, in view of [G : A ∩B] = p2, we see that A ∩B = Z(G). Consequently,∣∣Inn(G)

∣∣ =∣∣ GZ(G)

∣∣ = p2, and hence Inn(G) is abelian. This together with theFundamental Theorem of finite abelian groups implies Inn(G) ∼= G

Z(G)∼= Zp2

or Zp ⊕ Zp. From Inn(G) ∼= GZ(G)

∼= Zp2 , it follows that G is abelian, which isimpossible. Therefore, Inn(G) ∼= Zp ⊕ Zp, which is what we want.

2. There exists an n ∈ N such that (r2− r)n = 0, for r2− r is nilpotent. It is plainthat there exists a g ∈ Z[x] such that

0 = (r2 − r)n = rn − rn+1g(r),

whence rn = rn+1g(r). Setting f(x) = xng(x)n, we have

f(r)2 = r2ng(r)2n,

and hence we can write

rn = rn+1g(r) = rg(r)rn = rg(r)rn+1g(r) = rn+2g(r)2 = r2g(r)2rn

= r2g(r)2rn+1g(r) = rn+3g(r)3 = · · · = rn+ng(r)n = r2ng(r)n.

That is, rn = r2ng(r)n, yielding

f(r)2 = r2ng(r)2n = r2ng(r)ng(r)n = rng(r)n = f(r).

If f(r) = 0, then

0 = rnf(r) = rnrng(r)n = r2ng(r)n = rn,

implying rn = 0, which contradicts the hypothesis that r is not nilpotent. There-fore, f(r) is a nonzero idempotent element of R, which is what we want.

3. Remark. Adjusting the proof below one can prove that

det(lcm(i, j)

)= n!f(1) · · · f(n),

where f(n) =∑d|n dµ(d) and µ denotes the Mobius function. Hint. Use lcm(i, j) =

ijgcd(i,j) .

We show that the matrix A is invertible. To this end, define the matrices B = (bij)and C = (cij) in Mn(Q) as follows

bij =ϕ(i) i = j0 i 6= j

, cij =

1 j|i0 j - i ,

152 2. SOLUTIONS

where ϕ is the Euler’s totient function. Let Ct denote the transpose of C. Wehave

(CBCt)ij =n∑k=1

cik(BCt)kj =n∑k=1

cik

n∑k′=1

bkk′cjk′

=n∑k=1

cikϕ(k)cjk =∑

k|i, k|j

ϕ(k)cikcjk =∑

k|gcd(i,j)

ϕ(k)

= gcd(i, j) = aij = (A)ij ,

for all 1 ≤ i, j ≤ n (for a proof of∑d|n φ(d) = n for all n ∈ N, see Solution 1 of

2.2.2). Consequently, A = CBCt. So we can write

detA = detCBCt = detC detB detCt

= 1.detB.1 = detB = ϕ(1)ϕ(2) · · ·ϕ(n),

implying that detA 6= 0. Thus, A is invertible, which is what we want.

2.17. Seventeenth Competition

2.17.1. Analysis. 1. To prove the assertion by contradiction, suppose that|f ′(x)| ≤

∣∣ f(b)−f(a)b−a

∣∣ for all x ∈ (a, b). There are two cases to consider.

(i) f(b)−f(a)b−a ≥ 0.

Define the function g : [a, b] → R by g(x) = f(x)− f(b)−f(a)b−a (x−a). Now, from

the contradiction hypothesis, we see that

g′(x) = f ′(x)− f(b)− f(a)b− a

≤ 0,

for all x ∈ (a, b). Thus, g is non-increasing on [a, b]. On the other hand, wehave g(a) = f(a) = g(b), and hence f(a) = g(b) ≤ g(x) ≤ g(a) = f(a) for allx ∈ [a, b]. Therefore, g(x) = f(a) for all x ∈ [a, b]. In other words, f(x) =f(a) + f(b)−f(a)

b−a (x− a) for all x ∈ [a, b]. That is, the graph of f is a line segment,which is a contradiction. This proves the assertion in this case.

(ii) f(b)−f(a)b−a ≤ 0.

Adjusting the above proof or replacing f by −f and repeating the above argu-ment, one can prove the assertion in this case as well.

2. Let

p(x) =∫ x

−1

p1p3

∫ x

−1

p2p4 −∫ x

−1

p1p4

∫ x

−1

p2p3.

It is plain that p ∈ R[x] and p(−1) = 0. Form this, it follows that p is divisible byx+ 1. We can write

p′(x) = p1(x)p3(x)∫ x

−1

p2p4 + p2(x)p4(x)∫ x

−1

p1p3

−p1(x)p4(x)∫ x

−1

p2p3 − p2(x)p3(x)∫ x

−1

p1p4,

2.17. SEVENTEENTH COMPETITION 153

from which, we obtain p′(−1) = 0. Thus, p(x) is divisible by (x + 1)2. Takingderivative of the both sides of the above equality yields

p′′(x) = (p1p3)′(x)∫ x

−1

p2p4 + (p1p3p2p4)(x) + (p2p4)′(x)∫ x

−1

p1p3

+(p2p4p1p3)(x)− (p1p4)′(x)∫ x

−1

p2p3 − (p1p4p2p3)(x)

+(p2p3)′(x)∫ x

−1

p1p4 − (p2p3p1p4)(x),

implying p′′(−1) = 0, and hence p(x) is divisible by (x+ 1)3. Finally, we can write

p′′′(x) = (p1p3)′′(x)∫ x

−1

p2p4 +((p1p3)′p2p4

)(x) + (p2p4)′′(x)

∫ x

−1

p1p3

+((p2p4)′p1p3

)(x)− (p1p4)′′(x)

∫ x

−1

p2p3 −((p1p4)′p2p3

)(x)

−(p2p3)′′(x)∫ x

−1

p1p4 −((p2p3)′p1p4

)(x),

which obtains

p′′′(−1) =((p1p3)′p2p4 + (p2p4)′p1p3

)(−1)−

((p1p4)′p2p3 + (p2p3)′p1p4

)(−1)

= (p1p2p3p4)′(−1)− (p1p2p3p4)′(−1) = 0.

Therefore, p(x) is divisible by (x+ 1)4, which is what we want.

3. “ =⇒” Suppose that Z(f) := x ∈ X : f(x) = 0 = f−1(0) is an open subsetof X. It follows that Z(f) is both open and close, and hence so is Z(f)c = X \Z(f).Now, since X = Z(f)∪Z(f)c and Z(f) and Z(f)c are both open sets, the functiong : X → R defined by

g(x) = 1

f(x) x ∈ Z(f)c

0 x ∈ Z(f)

is continuous on X and furthermore satisfies f = gf2 on X. This is what we want.

“ ⇐=” Suppose that there exists a continuous function g : X → R such thatf = gf2. To show that Z(f) := x ∈ X : f(x) = 0 = f−1(0) is an open subsetof X, we prove that Z(f)c := X \ Z(f) is closed. It suffices to show that for anysequence (xn)+∞n=1, with xn ∈ Z(f)c, we have x∞ = limn xn ∈ Z(f)c. To this end,first note that f(xn) = g(xn)f(xn)2 for all n ∈ N. As f(xn) 6= 0 for all n ∈ N, wesee that f(xn)g(xn) = 1 for all n ∈ N. Now, in view of the continuity of f and g,letting n→ +∞, we obtain f(x∞)g(x∞) = 1, yielding f(x∞) 6= 0. In other words,x∞ ∈ Z(f)c, which is what we want.

4. Define the function g : [0, 12 ] → R by g(x) = f(x+ 1

2 )− f(x). We have

g(0) = f(12)− f(0) = −

(f(1)− f(

12))

= −g(12),

yielding g(0)g( 12 ) = −g(0)2 ≤ 0. It thus follows from the Intermediate Value

Theorem that there exists a c ∈ [0, 12 ] such that g(c) = 0. That is, f(c+ 1

2 ) = f(c).Letting a = c and b = c+ 1

2 , we have b− a = 12 and f(a) = f(a), which is what we

want.

154 2. SOLUTIONS

2.17.2. Algebra. 1. For a, b ∈ G, define a ∼ b if and only if h1ah2 = b forsome h1, h2 ∈ H. It is readily verify that ∼ is indeed an equivalence relation onG. Also, for any x ∈ G, we have [x] = HxH, where [x] denotes the equivalenceclass of x. With all that in mind, let e denote the identity element of G andx1 = e, x1, . . . , xn be a maximal set of nonequivalent elements of G. It followsfrom the hypothesis that n > 1. We have

G =n⋃i=1

HxiH.

This yields ∣∣G∣∣ = n∑i=1

∣∣H∣∣∣∣x−1i Hxi

∣∣∣∣H ∩ (x−1i Hxi)

∣∣ =n∑i=1

|H|2∣∣H ∩ (x−1i Hxi)

∣∣ ,from which, together with the hypothesis, we obtain∣∣G∣∣ = ∣∣H∣∣+ n∑

i=2

∣∣H∣∣2 =∣∣H∣∣+ (n− 1)

∣∣H∣∣2.This implies [G : H] = |G|

|H| = 1+(n−1)|H|. In other words, [G : H]−(n−1)|H| = 1,which easily yields gcd([G : H], |H|) = 1, which is what we want.

2. The assertion is a special case of the following. Let R be a unital ring with theproperty that ab = 1 implies ba = 1 whenever a, b ∈ R. Then, the ring R[x] hasthe same property, i.e., if a, b ∈ R[x] and ab = 1, then ba = 1. If D is a divisionring and n ∈ N, it follows from the Rank-Nullity Theorem that for A,B ∈Mn(D),we have AB = In if and only if BA = In, where In denotes the identity matrix.Thus, the ring R = Mn(F ) satisfies the above property. As a matter of fact, itcan be shown that any left or right Noetherian ring satisfies the above property.To prove the above more general assertion, suppose that for f, g ∈ R[x] we havefg = 1. We prove that gf = 1. Suppose to the contrary that gf 6= 1. Notefirst that (gf)2 = g(fg)f = gf . Assuming that f = a0 + a1x + · · · + anx

n andg = b0 + b1x + · · · + bmx

m, we obtain a0b0 = 1 because fg = 1. Thus, b0a0 = 1.So we can write gf = 1 + cxk + · · · , where k is the least exponent of x such thatc 6= 0. Now, from (gf)2 = gf , we see that

(1 + cxk + · · · )2 = 1 + cxk + · · · ,

from which, we obtain 2c = c, yielding c = 0, which is a contradiction. Thusgf = 1, which is what we want.

3. We prove the counterpart of the problem for left (resp. right) finite-dimensionalvector spaces over a division ring D. So assume that V is a left (resp. right)vector space over a division ring D and T : V −→ V a left (resp. right) lineartransformation on V . As T 2V ⊆ TV , the linear transformation S : TV → TVdefined by S(Tx) = T 2x defines a left (resp. right) linear transformation on theleft (resp. right) vector space TV . By the Rank-Nullity Theorem, we can write

dimTV = dimS(TV ) + dim kerS.

As S(TV ) = T 2V and kerS = kerT ∩ TV , we see that

dim(kerT ∩ TV ) = dimTV − dimT 2V.

2.18. EIGHTEENTH COMPETITION 155

Therefore,dim(kerT ∩ TV ) = rank(T )− rank(T 2),

as desired.

2.18. Eighteenth Competition

2.18.1. Analysis. 1. We have an = An − An−1. This together with thehypothesis easily yields limn

An−1An

= 1, and hence limnAn

An−1= 1 because An > 0

for all n ∈ Z with n ≥ 0. From this, it follows that the radius of convergence of thepower series

∑+∞n=0Anx

n is 1. In particular, the series absolutely converges for allx ∈ (−1, 1). Now, as a0 = A0 and an < An for all n ∈ N, it follows that

0 ≤ an∣∣x∣∣n ≤ An

∣∣x∣∣n,for all n ∈ N ∪ 0 and x ∈ (−1, 1). Thus, by the Comparison Test,

∑+∞n=0 an|x|n

converges for all x ∈ (−1, 1) because so does∑+∞n=0An|x|n for all x ∈ (−1, 1). So, if

R is the radius of convergence of the series∑+∞n=0 anx

n, then R ≥ 1. But R > 1 isimpossible, for otherwise the series

∑+∞n=0 an must be convergent, which contradicts

the hypothesis that limnAn = +∞. Therefore, R = 1, which is what we want.

2. We prove the assertion under the weaker hypothesis that the functions f and gare Riemann integrable on [0, 1]. In the following integrals, perform the substitu-tions nx = t and s = t− k + 1, respectively, to obtain∫ 1

0

f(x)g(nx)dx =∫ n

0

1nf( tn

)g(t)dt

=n∑k=1

∫ k

k−1

1nf( tn

)g(t)dt

=n∑k=1

∫ k

k−1

1nf( sn

+k − 1n

)g(s+ (k − 1)

)ds.

But the period of g is one. So we can write∫ 1

0

f(x)g(nx)dx =∫ 1

0

fn(s)g(s)ds,

where fn(s) = 1n

∑nk=1 f

(sn + k−1

n

). Suppose we have proved that the sequence

(fn)+∞n=1 uniformly converges to∫ 1

0f on [0, 1]. Since g is bounded on [0, 1], we see

that the sequence (fng)+∞n=1 uniformly converges to g∫ 1

0f on [0, 1]. Hence, we can

write

limn

∫ 1

0

f(x)g(nx)dx = limn

∫ 1

0

fn(s)g(s)ds

=∫ 1

0

limnfn(s)g(s)ds =

∫ 1

0

(∫ 1

0

f

)g(s)ds

=(∫ 1

0

f

)(∫ 1

0

g

),

156 2. SOLUTIONS

proving the assertion. It remains to show that the sequence (fn)+∞n=1 uniformlyconverges to

∫ 1

0f on [0, 1]. To this end, as f is Riemann integrable, it follows from

Riemann’s criterion for integrability that for given ε > 0, there exists a δ > 0 suchthat for any partition P ∈ P[0, 1], where P[0, 1] denotes the set of all partitions ofthe interval [0, 1], with ||P || < δ, we have

U(P, f)− L(P, f) =n∑i=1

(Mi −mi)∆xi < ε,

where

P : x0 = 0 < x1 < · · · < xn = 1, ||P || = max1≤i≤n ∆xi

U(P, f) =∑ni=1Mi∆xi, L(P, f) =

∑ni=1mi∆xi,

mi = infx∈[xi−1,xi] f(x), Mi = supx∈[xi−1,xi] f(x).

We also know that if U(P, f) − L(P, f) < ε for some P ∈ P[0, 1], then for anyxi−1 ≤ ξi ≤ xi, we have ∣∣∣∣∣

n∑i=1

f(ξi)∆xi −∫ 1

0

f

∣∣∣∣∣ < ε.

Note that fn(s) corresponds to the Riemann sum with respect to the uniformpartition of [0, 1] with n subintervals of the same length, i.e., ∆xi = 1

n for all1 ≤ i ≤ n, and the mid points ξi = s

n + i−1n . In view of this, for given ε > 0, find

δ > 0 from the above and let N ∈ N such that 1N < δ. Now, for each n ≥ N , let Pn

be the uniform partition of [0, 1] with n subintervals of the same length. We have||Pn|| = 1

n <1N < δ. It thus follows that∣∣∣∣fn(s)− ∫ 1

0

f

∣∣∣∣ =

∣∣∣∣∣ 1nn∑i=1

f( sn

+i− 1n

)−∫ 1

0

f

∣∣∣∣∣=

∣∣∣∣∣n∑i=1

f(ξi)∆xi −∫ 1

0

f

∣∣∣∣∣≤ U(Pn, f)− L(Pn, f) < ε.

That is, ∣∣∣∣fn(s)− ∫ 1

0

f

∣∣∣∣ < ε,

for all s ∈ [0, 1] and n ≥ N . In other words, the sequence (fn)+∞n=1 uniformlyconverges to

∫ 1

0f on [0, 1], which is what we want.

3. We need the following lemma.

Lemma. If α ∈ [0, 1], then there exists a sequence (ai)+∞i=1 with ai ∈ 0, 1

(i ∈ N) such that limn→+∞

∑ni=1 ain

= α.

2.18. EIGHTEENTH COMPETITION 157

Proof. If α = 0, let ai = 0 for all i ∈ N. If not, let

ai =

1 if i = bmα c for some m ∈ N,0 otherwise,

where b.c stands for the integer part function. We have∑ni=1 ain

=1n

cardm ∈ N :

⌊mα

⌋≤ n

=

1n

cardm ∈ N : m < α(n+ 1)

=bα(n+ 1)c

n.

This obtains

limn

∑ni=1 ain

= limn

bα(n+ 1)cn

= α,

as desired.

To prove the assertion, note first that if a sequence (ai)+∞i=1 has the property

that limn→+∞

∑ni=1 ain

= α, then the property remains intact under changing a finite

number of the terms of (ai)+∞i=1 . Now, let I be a nonempty open interval of R sothat I = (x−r, x+r) for some x ∈ I and r > 0. Clearly, g(I) ⊆ [0, 1]. Let α ∈ [0, 1]

be arbitrary. Choose n ∈ N such that12n

< ε. And choose x′ ∈ R such that the

first n digits of its binary expansion are the same of those of x and from its (n+1)stdigit onward its digits are equal to ai’s. That is,

x′ = 0.x1 . . . xna1a2a3 . . . ,

where x = 0.x1 . . . xnxn+1 . . .. It is obvious that x′ ∈ I and that g(x′) = α.Consequently, [0, 1] ⊆ g(I), and hence g(I) = [0, 1], as desired.

For the rest, note first that any open subset G of R includes an open interval I,implying that g(G) ⊇ g(I) = [0, 1]. This implies g(G) = [0, 1] because g(G) ⊆ [0, 1].That is, for any open subset G of R, we have g(G) = [0, 1]. Now, define the functionh : R → R by h = f g, where the function f : [0, 1] → R is defined by

f(x) =x 0 < x < 1,12 x ∈ 0, 1.

It is obvious that for any open subset G of R, we have

h(G) = f(g(G)

)= f

([0, 1]

)= (0, 1).

In other words, h is an open map. By proving that h has no limit at any point ofR, we show that h is not continuous on R. Suppose to the contrary that there existx0, ` ∈ R such that

limx→x0

h(x) = `.

It follows that for every ε > 0, there exists a δ > 0 such that

`− ε < h(x) < `+ ε,

whenever x ∈ (x0− δ, x0 + δ). We have 0 ≤ l ≤ 1 because 0 < h < 1. Consequently,we can choose ε > 0 such that (` − ε, ` + ε) ∩ (0, 1) $ (0, 1). For this ε > 0, findδ > 0 from the above. Thus, for all x ∈ (x0 − δ, x0 + δ), we have

h(x) ∈ (`− ε, `+ ε). (∗)

158 2. SOLUTIONS

Letting G = (x0 − δ, x0 + δ), we obtain

h(G) = f(g(G)

)= (0, 1),

from which, in view of (∗), we see that

(0, 1) ⊆ (`− ε, `+ ε),

implying (`− ε, `+ ε)∩ (0, 1) = (0, 1), contradicting our choice of ε > 0. Therefore,the function h has no limit at any point of R, which is what we want.

2.18.2. Algebra. 1. The subgroupH∩K is normal inK becauseK is normalin G. Since K is simple, we have H ∩ K = K or H ∩ K = e, where e is theidentity element of G. If H ∩K = K, then K ⊆ H, implying that K = H because|G| <∞ and H ∼= K. To finish the proof, we show that H ∩K 6= e. Suppose tothe contrary that H ∩K = e. As K is normal in G, we have HK = KH, whenceHK ≤ G, and hence |HK|

∣∣ |G|. But∣∣HK∣∣ = ∣∣H∣∣∣∣K∣∣∣∣H ∩K∣∣ =

∣∣K∣∣2.In other words, |K|2

∣∣ |G|, contradicting the hypothesis that the square of the orderof K does not divide that of G. Thus, H ∩K 6= e, which is what we want.

2. It suffices to show that any right ideal of R is finitely generated. To this end,let I be a right ideal in R. If I = 0, the assertion is trivial. Suppose I 6= 0. As isusual, use Eij to denote the 2 × 2 matrix whose ij entry is 1 and zero elsewhere.There are two cases to consider.

(i) There exists A = a0E11 + b0E12 + c0E22 ∈ I such that a0 6= 0.Using the well-ordering principle of natural numbers, let am be the least pos-

itive integer a for which there are b, c ∈ Q such that aE11 + bE12 + cE22 ∈ I. Itis easily verified that if aE11 + bE12 + cE22 ∈ I for some a ∈ Z and b, c ∈ Q,then am divides a. We claim that I is generated by amE11, E12, E22. SupposeamE11 + bE12 + cE21 ∈ I for some b, c ∈ Q. It follows that

(amE11 + bE12 + cE22)E12 = amE12 ∈ I,

implying

(amE12)( 1am

E22

)= E12 ∈ I.

That is, E12 ∈ I. We can write

(amE11 + bE12 + cE22)E11 = amE11 ∈ I,

yielding amE11 ∈ I. Now, as amE11, E12 ∈ I, we see that if A = kamE11 +bE12 + cE22 ∈ I for some k ∈ Z and b, c ∈ Q, then cE22 ∈ I. If c = 0 for allA = kamE11 + bE12 + cE22 ∈ I, then I is generated by amE11, E12. Otherwise,just as we saw in the above, cE22 ∈ I for some nonzero c ∈ Q, yielding

(cE22)(1cE22

)= E22 ∈ I.

That is, E22 ∈ I, in which case the right ideal I is generated by amE11, E12, E22.This proves the assertion in this case.

(ii) Every A ∈ I is of the form A = bE12 + cE22, where b, c ∈ Q.

2.18. EIGHTEENTH COMPETITION 159

Suppose A0 = b0E12+c0E22 ∈ I, where b0, c0 ∈ Q are such that (b0, c0) 6= (0, 0).Set J = QA0. It is easily verified that J is a right ideal of R. If I = J , then I isgenerated by A0. If not, then there exist b1, c1 ∈ Q such that b1c0 − b0c1 6= 0 andb1E12 + c1E22 ∈ I. It follows that (b0c1 − b1c0)E12 = (b0E12 + c0E22)(b1E12 +c1E22)− (b1E12 + c1E22)(b0E12 + c0E22) ∈ I, and hence

(b0c1 − b1c0)E12

( 1b0c1 − b1c0

E22

)= E12 ∈ I.

That is, E12 ∈ I. So we have

E12(b0E22) = b0E12 ∈ I.

From this, we obtain c0E22 ∈ I because b0E12 + c0E22 ∈ I. If c0 = 0 wheneverb0E12 + c0E22 ∈ I, then I is generated by E12. If not, then c0E22 ∈ I for somenonzero c0 ∈ Q. This implies E22 = (c0E22)( 1

c0E22) = E22 ∈ I. Thus, I is

generated by E12, E22 in this case. So in any event, I is finitely generated.We now prove that every ascending chain of right ideals of R necessarily termi-

nates. To this end, let (In)+∞n=1 be an ascending sequence of right ideals of R. SetI = ∪+∞

n=1In. As (In)+∞n=1 is an ascending sequence of right ideals of R, it is easilyverified that I is a right ideal of R. It follows from the above that there are at mostthree, not necessarily distinct, elements A1, A2, A3 ∈ I such that I = 〈A1, A2, A3〉.Since I = ∪+∞

n=1In, there exist n1, n2, n3 ∈ N such that Ai ∈ Inifor each i = 1, 2, 3.

Letting N = max(n1, n2, n3), we have Ai ∈ In for all n ≥ N and i = 1, 2, 3. Thisimplies

In ⊆ I =⟨A1, A2, A3

⟩⊆ In,

for all n ≥ N . Therefore, In = I for all n ≥ N . In other words, In = IN for alln ≥ N , which is what we want.

3. We need the following lemma.

Lemma. Let V be a finite-dimensional vector space over a field F and Vii∈Ia family of proper subspaces of V such that |I| < |F |. Then,⋃

i∈IVi $ V.

Proof. Without loss of generality, we may assume that Vii∈I is a familyof distinct proper subspaces of V . With that in mind, we prove the assertion byinduction on dimV . If dimV = 1, we must have dimVi = 0, implying Vi = 0 forall i ∈ I, in which case the assertion is trivial. Assuming that the assertion holdsfor any vector space V with dimV < k, we prove the assertion for any vector spaceV with dimV = k. Let Vi (i ∈ I) be as in the lemma. Set

S =

dimVi|i ∈ I.

Plainly, S ⊆ N and S is nonempty and bounded from above by k = dimV . Itfollows that S has a terminal element. That is, there exists i0 ∈ I such that

dimVi ≤ dimVi0 ,

for all i ∈ I. Set W = Vi0 , J = I \ i0, and finally Wj = Vj ∩ Vi0 . Firstly, for allj ∈ J we have Wj $ Vi0 . Suppose to the contrary that Wj = Vj∩Vi0 = Vi0 for somej ∈ J , then Vi0 ⊆ Vj , implying dimVi0 ≤ dimVj . On the other hand, dimVj ≤dimVi0 , yielding Vi0 = Vj which is impossible because j 6= i0. Thus, Wj $ W = Vi0

160 2. SOLUTIONS

for all j ∈ J . We have dimW = dimVi0 < dimV = k and |J | ≤ |I| < |F |. Soit follows from the induction hypothesis that

⋃j∈JWj $ W = Vi0 . Consequently,

there exists a vector v0 ∈ Vi0 such that v0 /∈ Wj = Vj ∩ Vi0 for all j ∈ J . That is,v0 /∈ Vj for all j ∈ I with j 6= i0. On the other hand, since Vi0 $ V , there exists avector v1 ∈ V \ Vi0 . We claim that there exists an f0 ∈ F such that v1 + f0v0 /∈ Vifor all i ∈ I. To prove this by contradiction, suppose that for each f ∈ F thereexists an if ∈ I such that v1 + fv0 ∈ Vif . Note that if f, f ′ ∈ F and f 6= f ′, thenv1 + fv0 6= v1 + f ′v0. Also note that the hypothesis |I| < |F | implies that thereexist f, f ′ ∈ F with f 6= f ′ such that if = if ′ . Firstly, we observe that if 6= i0, forotherwise v1 + fv0 ∈ Vi0 , yielding v1 ∈ Vi0 , because v0 ∈ Vi0 , which is impossible.Secondly, from v1 + fv0, v1 + f ′v0 ∈ Vif = Vif′ , we obtain (f − f ′)v0 ∈ Vif , yieldingv0 ∈ Vif , which, in turn, implies if = i0, contradicting if 6= i0 as observed inthe above. That is, we obtain a contradiction in any event. Thus, there exists anf0 ∈ F such that v1 + f0v0 /∈ Vi for all i ∈ I. In other words, v1 + f0v0 /∈

⋃i∈I Vi,

and hence⋃i∈I Vi $ V , which is what we want.

We now use the lemma to prove the assertion. From this point on, the proof isalmost identical to that of Problem 5 of 1.14.2.

Fix i0 ∈ I. We prove the assertion by induction on dimV −dimVi0 . If dimV −dimVi0 = 1, then for U = u, where u ∈ V \(

⋃i∈I Vi), we obviously have Vi⊕U =

V for all i ∈ I. Assuming that the assertion holds whenever dimV −dimVi0 = n, weprove it whenever dimV −dimVi0 = n+1. As, in view of the lemma,

⋃i∈I Vi $ V ,

there exists a vector β ∈ V \ (⋃i∈I Vi). Set

V ′i = Vi ⊕ 〈β〉.

We have dimV ′i = dimV ′

j = dimVi0 + 1 for all i, j ∈ I, yielding dimV − dimV ′i0

=dimV −dimV ′

i = n for all i ∈ I. It thus follows from the induction hypothesis thatthere exists a subspace U ′ such that

V ′i ⊕ U ′ = V,

for all i ∈ I. Letting U = 〈β〉 ⊕ U ′, we obviously obtain

Vi ⊕ U = V,

for all i ∈ I, proving the induction assertion, which is what we want.

2.19. Nineteenth Competition

2.19.1. Analysis. 1. Define the function g : [0, 1] → R by

g(x) = 2x− 1−∫ x

0

f.

It suffices to show that g has only one zero on the interval [0, 1]. As f is continuouson [0, 1], g is differentiable on (0, 1). Using the First Fundamental Theorem ofCalculus, we can write

g′(x) = 2− f(x),for all x ∈ (0, 1). But 0 ≤ f(x) ≤ 1 for all x ∈ (0, 1). In particular, we must haveg′(x) = 2 − f(x) ≥ 1 > 0 for all x ∈ (0, 1), implying that g is strictly increasingon [0, 1]. Therefore, g has at most one zero on [0, 1]. With that in mind, note thatg(0) = −1 < 0 ≤ 1 −

∫ 1

0f = g(1). To see the second inequality, just note that

2.19. NINETEENTH COMPETITION 161

from f ≤ 1 on [0, 1], it follows that∫ 1

0f ≤ 1, yielding g(1) = 1−

∫ 1

0f ≥ 0. It thus

follows from the Intermediate Value Theorem that there exists a c ∈ (0, 1] suchthat g(c) = 0. This completes the proof.

2. First Solution: Note that limx→+∞ exh = +∞ whenever h > 0. With that in

mind, using L’Hopital’s rule, we can write

limx→+∞

f(x) = limx→+∞

exh f(x)e

xh

= limx→+∞

(e

xh f(x)

)′(e

xh

)′= lim

x→+∞

1he

xh f(x) + e

xh f ′(x)

1he

xh

= limx→+∞

(f(x) + hf ′(x)

)= 0,

implying limx→+∞ f(x) = 0. This together with

limx→+∞

(f(x) + hf ′(x)

)= 0,

where h ∈ R+, yields limx→+∞ f ′(x) = 0, which is what we want. (We note that thehypothesis that f is continuously differentiable is redundant. The differentiabilityof f is enough for the assertion to be true.)

Second solution: It follows from the hypothesis that for given ε > 0, there existsan M = M(ε) > 0 such that ∣∣f(x) + hf ′(x)

∣∣ < ε

2,

whenever x ≥ M . This easily implies∣∣ ddx

(e

xh f(x)

)∣∣ < ε2he

xh whenever x ≥ M . We

can write ∣∣∣e xh f(x)− e

Mh f(M)

∣∣∣ =∣∣∣ ∫ x

M

d

dt

(e

th f(t)

)dt∣∣∣

≤ ε

2h

∫ x

M

eth dt =

ε

2(e

xh − e

Mh

),

for all x ≥M . Consequently, for all x ≥M ,∣∣f(x)− eM−x

h f(M)∣∣ < ε

2(1− e

M−xh

),

which obtains ∣∣f(x)∣∣ ≤

∣∣f(x)− eM−x

h f(M)∣∣+ ∣∣eM−x

h f(M)∣∣

≤ ε

2(1− e

M−xh

)+ e

M−xh

∣∣f(M)∣∣

<ε

2+ e

M−xh

∣∣f(M)∣∣,

for all x ≥M . Consequently,

lim supx→+∞

f(x) ≤ lim supx→+∞

(ε2

+ eM−x

h

∣∣f(M)∣∣) =

ε

2,

for all ε > 0. Thus, lim supx→+∞ f(x) ≤ 0, and hence limx→+∞ f(x) = 0.As we pointed out in the first solution, this together with the hypothesis thatlimx→+∞

(f(x) + hf ′(x)

)= 0, where h ∈ R+, yields limx→+∞ f ′(x) = 0, finishing

the proof.

162 2. SOLUTIONS

3. Recall that A equipped with the uniform metric, induced by the uniform normof A, denoted by ||.||∞, which is defined by

d(f, g) = ||f − g||∞ = supx∈[0,1]

∣∣f(x)− g(x)∣∣,

is a complete normed space or a real Banach space. Also recall that by Problem 2of 1.6.1 the following lemma holds.

Lemma. Let X and Y be metric spaces and f, fn : X → Y continuous func-tions such that the sequence (fn)n∈N uniformly converges to f on X. Also letx0, xn ∈ X be such that the sequence (xn)n∈N converges to x0. Then, limn fn(xn) =f(x0).

With all that in mind, we now prove the assertion.

(a) LetN ∈ N be arbitrary. To show that EN is closed, suppose that f : [0, 1] →R is a limit point of EN with respect to the topology induced by the uniform normof A. We show that f ∈ EN . To this end, as f is a limit point of EN , we see thatthere exists a sequence (fn)n∈N in EN such that fn → f in A as n→ +∞. In otherwords, fn’s uniformly converge to f on [0, 1]. Now, from fn ∈ EN for each n ∈ N,it follows that there exists an an ∈ [0, 1] such that∣∣fn(x)− fn(an)

∣∣ ≤ N∣∣x− an

∣∣,for all x ∈ [0, 1]. It is plain that the sequence (an)n∈N in the compact interval[0, 1] has a subsequence (ank

)k∈N converging to some a0 ∈ [0, 1]. Let bk = ankand

gk = fnkfor all k ∈ N. We can write∣∣gk(x)− gk(bk)

∣∣ ≤ N∣∣x− bk

∣∣,for all x ∈ [0, 1] and k ∈ N. Now, letting k → +∞ and using the above lemma, wesee that ∣∣f(x)− f(a)

∣∣ ≤ N∣∣x− a

∣∣,for all x ∈ [0, 1]. That is, f ∈ EN , and hence EN is closed for all N ∈ N. To showthat the interior of EN is empty for all N ∈ N, we prove that the set A\EN is densein A. Suppose that f ∈ A is arbitrary and ε > 0 is given. From the continuityof f on the compact interval [0, 1], we see that f is uniformly continuous on [0, 1].Thus, there exists a δ > 0 such that∣∣f(x)− f(y)

∣∣ < ε

2,

whenever x, y ∈ [0, 1] and |x−y| < δ. Choose n ∈ N such that 1n < δ and set xi = i

n ,where 1 ≤ i ≤ n. By constructing a function g satisfying ||f−g||∞ < ε and g /∈ EN ,we show that A\EN is dense in A, finishing the proof. To construct g, it suffices todo so on any subinterval [xi−1, xi] (1 ≤ i ≤ n) in such a way that g is (piecewise)continuous, ||f − g||∞ < ε, and g /∈ EN . Here is a description of the graph of g onthe interval [xi−1, xi]. Let g be any function on [xi−1, xi] (1 ≤ i ≤ n) whose graphis a polygonal line in the rectangle

[xi−1, xi

]×[f(xi−1) − ε

2 , f(xi) + ε2

]joining

the two points(xi−1, f(xi−1)

)and

(xi, f(xi)

)satisfying the following properties.

(i) The consecutive vertices of any segment of the polygonal line lie on the linesy = f(xi−1) − ε

2 and y = f(xi) + ε2 , respectively. (ii) The absolute value of the

slope of any segment formed by any two consecutive vertices of the polygonal lineis greater than N . It is plain that for any function g : [0, 1] → R as described inthe above, we have ||f − g||∞ < ε and g /∈ EN , which is what we want.

2.19. NINETEENTH COMPETITION 163

(b) As A is a complete metric space with respect to the metric induced by theuniform norm of A, it follows from the Baire Category Theorem that A cannot bewritten as a countable union of nowhere dense subsets of A. Recall that is subsetof a topological space X is called nowhere dense if the complement of its closure isdense in X. It thus follows from (a) that EN is nowhere dense for all N ∈ N, andhence, by the Baire Category Theorem,

⋃N∈N EN $ A. Define the set E as follows

E :=f ∈ A|∃a ∈ [0, 1] f ∈ Da

=

⋃a∈[0,1]

Da,

where Da, with a ∈ (0, 1), is the set of all functions that are differentiable ata ∈ (0, 1); the set D0 (resp. D1) is defined to be the set of functions that are right(resp. left) differentiable at 0 (resp. 1). We claim that E ⊆

⋃N∈N EN . To prove

this, suppose f ∈ E is given. It follows that there exists a ∈ [0, 1] such that f ∈ Da.Define the function g : [0, 1] → R by

g(x) =

f(x)−f(a)x−a x ∈ [0, 1] \ a,f ′(a) x = a.

We have limx→a g(x) = limx→af(x)−f(a)

x−a = f ′(a) = g(a) for all a ∈ (0, 1). Likewise,limx→0+ g(x) = g(0) and limx→1− g(x) = g(1). Therefore, the function g is con-tinuous on the compact interval [0, 1], and hence it is bounded on [0, 1]. So thereexists an M > 0 such that ∣∣g(x)∣∣ ≤ M,

for all x ∈ [0, 1]. In other words,∣∣f(x)− f(a)∣∣ ≤ M

∣∣x− a∣∣,

for all x ∈ [0, 1]. Consequently, if we let N ∈ N be such that N ≥M , then∣∣f(x)− f(a)∣∣ ≤ N

∣∣x− a∣∣,

for all x ∈ [0, 1]. That is, f ∈ EN , yielding f ∈⋃N∈N EN , proving the claim.

Now, to prove the assertion, recall that E ⊆⋃N∈N EN $ A. This implies that

there exists a function f ∈ A \ E. In other words, the function f : [0, 1] → R iscontinuous and yet it is nowhere differentiable because f /∈ E. This proves (b).

2.19.2. Algebra. 1. To prove the assertion by contradiction, suppose thatN0 6= e is a maximal element of A. We have G

N0∼= G. It follows that there exists

an isomorphism φ : G → GN0

. Since N0 6= e is a normal subgroup of G and φ

is an isomorphism, we see that φ(N0) 6= e is a normal subgroup of GN0

. Fromthis, we obtain a normal subgroup M0 % N0 of G such that φ(N0) = M0

N0. Thus,

φ|N0 : N0 → M0N0

is an isomorphism of groups, yielding N0∼= M0

N0. Now, the Third

Isomorphism Theorem for groups together with GN0

∼= G and N0∼= M0

N0implies

thatG

M0

∼=GN0M0N0

∼=G

N0

∼= G.

In other words, GM0

∼= G. But N0 $ M0, contradicting the hypothesis that N0 is amaximal element of A. Thus, the assertion follows by contradiction, which is whatwe want.

164 2. SOLUTIONS

2. (a) Note first that for any nonzero a ∈ R and any minimal left ideal J of R,the left ideal Ja is also a minimal left ideal of R. To prove this by contradiction,suppose that Ja is not a minimal left ideal of R for some nonzero element a ∈ Rand minimal left ideal J of R. It follows that there exists a nonzero left ideal I ofR such that I $ Ja. Define

J1 :=j ∈ J |ja ∈ I

.

It is plain that J1 is a left ideal of R, J1 $ J , because I $ Ja, and that J1 6= 0,because I 6= 0. This is in contradiction with J being a minimal left ideal of R. Thus,Ja is a minimal left ideal of R whenever J is a minimal left ideal of R and a ∈ Ris nonzero. To prove the assertion, if R has no minimal left ideal, the assertion istrivial because, by definition, S = 0. If not, as S =

∑J : J is a minimal left ideal

of R, we see S is a left ideal in R. To see that S is a right ideal of R as well, letr ∈ R and s ∈ S be arbitrary. We need to show that sr ∈ S. Since s ∈ S, it followsthat there are minimal left ideals J1, . . . , Jn, where n ∈ N, and jk ∈ Jk (1 ≤ k ≤ n)such that

s = j1 + · · ·+ jn,

yielding

sr = j1r + · · ·+ jnr.

But jkr ∈ Jkr and Jkr is a minimal left ideal of R for all 1 ≤ k ≤ n. It thus followsthat sr ∈

∑J : J is a minimal left ideal of R = S. That is, S is a right ideal of

R as well, and hence an ideal of R, as desired.

(b) It follows from (a) that ⋂ICR

I ⊆ S,

for the two sided ideal S appears in the intersection. To prove the inclusion in theopposite direction, it suffices to show that for any minimal left ideal J of R, wehave

J ⊆⋂ICR

I.

To this end, assuming that J and I are, respectively, an arbitrary minimal left idealand an arbitrary two-sided ideal of R, it is enough to show that J ⊆ I. To see this,note first that

IJ ⊆ J,

for J is a left ideal of R. Secondly, IJ 6= 0 because otherwise I(JR) = (IJ)R = 0,where I 6= 0 and JR 6= 0 are two-sided ideals of R, contradicting the hypothesis.Thus, IJ 6= 0, and hence

0 6= IJ ⊆ J.

That is, the nonzero left ideal IJ is contained in the minimal left ideal J . Thisimplies IJ = J . Now, since I is a two-sided ideal of R, we see that IJ ⊆ I ∩ J ,whence J ⊆ I ∩ J , yielding J ⊆ I, which is what we want.

2.20. TWENTIETH COMPETITION 165

3. Recall that the rank of a matrix is equal to its column rank as well as its rowrank. With that in mind, let Aj denote the jth column of the matrix A. We have

Aj =

x1 + yjx2 + yj

...xn + yj

=

x1

x2

...xn

+ yj

11...1

= α+ yjβ,

where α =

x1

x2

...xn

and β =

11...1

. That is, Aj ∈ 〈α, β〉, yielding 〈A1, . . . , An〉 ⊆

〈α, β〉. Thus, rank(A) = dim〈A1, . . . , An〉 ≤ dim〈α, β〉 ≤ 2. Therefore, rank(A) ≤2, as desired.

4. First solution: Let A ∈ Mn(F ) be a rank one n × n matrix over F . Sincerank(A) = 1, it follows from the Rank-Nullity Theorem that the rank and nullityof the linear transformation T : Mn×1(F ) → Mn×1(F ) defined by TX = AX, are,respectively, equal to 1 and n−1. Let αin−1

i=1 be a basis for kerT . Enlarge αin−1i=1

to a basis B′ = αini=1 for Mn×1(F ), where αn is any vector in Mn×1(F )\〈αi〉n−1i=1 ,

where 〈αi〉n−1i=1 denotes the vector subspace spanned by α1, . . . , αn−1. It is plain

that the matrix of T with respect to the basis B′, denoted by B, has the form

B = (bij)n×n =

0 · · · 0 b10 · · · 0 b2... · · ·

......

0 · · · 0 bn

.

On the other hand, the matrix of T with respect to the standard basis of Mn×1(F )is equal to A. It thus follows that there exists an invertible matrix P ∈ Mn(F )such that B = P−1AP , whence A = PBP−1. We can write

det(I +A) = det(P (I +B)P−1) = det(I +B)

= det

1 · · · 0 b10 · · · 0 b2... · · ·

......

0 · · · 0 1 + bn

= 1 + bn = 1 + tr(B) = 1 + tr(A).

That is, det(I +A) = 1 + tr(A), which is what we want.

Second solution: A proof identical to that of the second solution of Problem 7 of1.9.3 shows that det(I +A) = 1 + tr(A), which is what we want.

2.20. Twentieth Competition

1. If (x, y), (z, t) or (x′, y′), (z′, t′) is linearly dependent, then there exist α, β ∈C, which are not simultaneously zero, such that

θ = α(x, y) + β(z, t) = (0, 0) or θ′ = α(x′, y′) + β(z′, t′) = (0, 0).

166 2. SOLUTIONS

In this case, the assertion is trivial because θ = 0 or θ′ = 0. So suppose that(x, y), (z, t) and (x′, y′), (z′, t′) are linearly independent. Set

A =(x yz t

), A′ =

(x′ y′

z′ t′

).

The matrices A and A′ are invertible because their rows are linearly independent.It follows that the matrix A−1A′ ∈ M2(C) is also invertible. Now, suppose thatλ ∈ C is a nonzero eigenvalue of the invertible matrix A−1A′. We see that thematrix λI2−A−1A′ is not invertible. Hence, neither is λA−A′ = A(λI2−A−1A′).This implies that the rows of the noninvertible matrix

λA−A′ = λ

(x yz t

)−(x′ y′

z′ t′

)are linearly dependent. Thus, there exist α, β ∈ C, which are not simultaneouslyzero, such that

α(λx− x′, λy − y′) + β(λz − z′, λt− t′) = (0, 0).

A straightforward calculation reveals that

λθ − θ′ = α(λx− x′, λy − y′) + β(λz − z′, λt− t′) = (0, 0),

where θ = α(x, y)+β(z, t) and θ′ = α(x′, y′)+β(z′, t′). In other words, the vectorsθ and θ′ are linearly dependent, which is what we want.

2. Set xn = ln(1 + an). It follows from the hypothesis that

xm+n ≤ xm + xn,

for all m,n ∈ N. We prove that limnxn

n exists. To see this, letting α = infn∈Nxn

n ∈R, we show that limn

xn

n = α. To this end, let ε > 0 be given. Since α = infn∈Nxn

n ,there exists a positive integer N1 = N1(ε) such that

α ≤ xN1

N1≤ α+

ε

2.

Now, for any n ≥ N1, we can write n = N1q+ r, where q ∈ N and r ∈ N∪0 with0 ≤ r < N1. In view of the hypothesis, we can write

α− ε < α ≤ xnn

≤ qxN1 + rx1

N1q + r≤ qxN1

N1q+rx1

n

<xN1

N1+N1x1

n≤ α+

ε

2+N1

nx1.

Now, pick N2 ∈ N such that N2 > 2εN1x1. Letting N = max(N1, N2), for all

n ≥ N , we have

α− ε <xnn< α+

ε

2+N1

nx1 < α+ ε.

That is, ∣∣∣xnn− α

∣∣∣ < ε,

for all n ≥ N . In other words, limn

xnn

= α, implying that limnn√

1 + an = eα,which is what we want.

2.20. TWENTIETH COMPETITION 167

3. It follows from the hypothesis that σ2 = id for all σ ∈ aut(G). Consequently,σ−1 = σ for all σ ∈ aut(G), from which we obtain

σ1σ2 = (σ1σ2)−1 = σ−12 σ−1

1 = σ2σ1,

for all σ1, σ2 ∈ aut(G). In other words, aut(G) is abelian. On the other hand, justas we saw in Solution 1 of 2.16.2 using the First Isomorphism Theorem for groups,we have G

Z(G)∼= In(G), where In(G) denotes the set of all inner automorphisms of

G, which is a normal subgroup of aut(G). It follows that In(G) is commutative andhence so is G

Z(G) . This yields G′ ⊆ Z(G), where G′ denotes the derived subgroup(a.k.a. the commutator subgroup) of G. From this, we see that G′′ = e, implyingthat G is solvable, which is what we want.

4. Note first that f(x) > 0 for all x ∈ ( 14 , 1) because f(x) = xf(x) for all such x’s.

Next, since f(x) > 0, we see that 0 < ( 14 )f(x) < xf(x) < 1f(x) = 1, implying that

0 < f(x) < 1 for all x ∈ ( 14 , 1). Now, noting that xf(x) = f(x) and f(x) > 0 for

all x ∈ ( 14 , 1), and taking ln, we obtain f(x) lnx = ln f(x) for all x ∈ ( 1

4 , 1). Thisimplies

x = eln f(x)

f(x) ,

for all x ∈ ( 14 , 1). Define g : (0, 1) → (0, 1) by g(x) = e

ln xx . The function g is

differentiable and we have g′(x) = 1−ln xx2 e

ln xx . Thus, g′(x) > 0 for all 0 < x < 1,

and hence g is strictly increasing on (0, 1). The function g : (0, 1) → (0, 1) issurjective because limx→0+ g(x) = 0 and limx→1− g(x) = 1. From this, it followsthat g has an inverse, say, g−1 : (0, 1) → (0, 1), and that g−1 is differentiableand strictly increasing on (0, 1). In particular, g−1 is continuous on (0, 1). Also,it is easily verified that g( 1

2 ) = 14 . With all that in mind, define the function

g : [0, 1] → [0, 1] by

g(x) =

0 x = 0,g(x) 0 < x < 1,

1 x = 1.It is now plain that the function g : [0, 1] → [0, 1] is surjective, one-to-one andcontinuous and that its inverse g−1 : [0, 1] → [0, 1] is also continuous, and henceuniformly continuous, because [0, 1] is a compact interval. Noting 0 < f(x) < 1 forall x ∈ ( 1

4 , 1), we can write

g(f(x)

)= g(f(x)

)= x,

from which, we obtain f(x) = g−1(x) for all x ∈ ( 14 , 1). In other words, f =

g−1|( 14 ,1)

: (14 , 1) → ( 1

2 , 1). From this, it follows that f : ( 14 , 1) → ( 1

2 , 1) is uniformlycontinuous because so is g−1 : [0, 1] → [0, 1], which is what we want.

5. Let I 6= 0 and Z, respectively, denote the intersection of all nonzero ideals ofR and the set of all zero divisors of R including zero. By showing that Z is amaximal ideal of R and that Z = I, we settle the proof, for, from this, it followsthat Z = I is the only nontrivial ideal of R. First, we show that I ⊆ Z. Supposeto the contrary there exists a nonzero x0 ∈ I such that x0 /∈ Z. We see that x2

0 6= 0because x0 is not a zero divisor of R. As R is unital and I is the intersection ofall nonzero ideals of R, we obtain I ⊆ x0R 6= 0 and I ⊆ x2

0R 6= 0. On the otherhand, x0R ⊆ I and x2

0R ⊆ I, for x0 ∈ I. It thus follows that I = x0R = x20R.

In particular, we must have x0 ∈ x20R, which yields x0 = x2

0r for some r ∈ R.

168 2. SOLUTIONS

This obtains x0(1 − x0r) = 0, which, in turn, implies 1 = x0r ∈ I, for x0 /∈ Z, acontradiction. Thus, 0 6= I ⊆ Z, and hence Z 6= 0.

Next, we show that Z is a maximal ideal of R. To this end, let x ∈ Z \ 0and Ann(x) = r ∈ R : rx = 0. Note that Z ⊆ Ann(x) because Z2 = 0. Onthe other hand, Ann(x) ⊆ Z whenever x 6= 0. It thus follows that Ann(x) = Zwhenever x ∈ Z \ 0. Now, let 0 6= x0 ∈ I ⊆ Z = Ann(x) be arbitrary. As we sawin the above Rx0 = I. It is plain that the map φ : R→ Rx0 defined by φ(r) = rx0

is an epimorphism of the left modules R and Rx0. From the First IsomorphismTheorem for modules, we see that R

kerφ∼= Rx0, implying that Rx0

∼= RAnn(x)

∼= RZ

because kerφ = Ann(x) = Z. We now show that Rx0 is a simple left module,i.e., it has no nontrivial submodules. To this end, suppose that M 6= 0 is a leftsubmodule of Rx0 = I. Choose a nonzero y0 ∈ M . As M ≤ Rx0 = I, there existsan r0 ∈ R such that y0 = r0x0. We have y0R ⊆ I because x0 ∈ I. On the otherhand, I ⊆ y0R, for R is unital and I is the intersection of all nonzero ideals of R.Therefore, I = y0R. Consequently, we have M ⊆ I and I = y0R ⊆ M becausey0 ∈ M , yielding M = I = y0R. This proves that Rx0 = I is simple as a leftmodule. This together with Rx0

∼= RZ implies that Z is a maximal left ideal, and

hence a maximal ideal of R because R is commutative.Finally, we show that Z ⊆ I. To this end, suppose that 0 6= x0 ∈ Z is

arbitrary. We have x0R 6= 0, from which, we obtain 0 6= I ⊆ x0R. Now,choose an arbitrary nonzero element y0 ∈ I. It follows that there exist r0 ∈ R suchthat r0x0 = y0 ∈ I ⊆ Z. Note that r0 /∈ Z, for otherwise y0 = r0x0 = 0 which isimpossible. If r0 = 1, then x0 = y0 ∈ I yielding Z ⊆ I, as desired. If r0 6= 1, weobtain r0R+Z = R because Z is a maximal ideal and r0 /∈ Z. Consequently, thereexist r1 ∈ R and x1 ∈ Z such that r0r1 + x1 = 1, whence 1 − r1r0 = x1 ∈ Z. Onthe other hand, x0 ∈ Z. Since Z2 = 0, we see that x0(1− r1r0) = 0. This yieldsx0 = x0r0r1 = y0r1, which, in turn, implies x0 ∈ I because y0 ∈ I. That is, x0 ∈ I,and hence Z ⊆ I, for x0 was arbitrary. Therefore, Z = I and Z is a maximal idealof R, proving the assertion, which is what we want.

6. First solution: We need the following lemmas.

Lemma 1 (Cauchy’s criterion for the convergence of improper inte-grals). Let f, α : [a,+∞) → R be two functions such that α is increasing on[a,+∞) and f is integrable with respect to α in the sense of Riemann-Stieltjes, i.e.,f ∈ R(α), on any closed interval [a, x], where x ∈ R with x > a. Then, a necessaryand sufficient condition for the convergence of

∫ +∞a

fdα is that for every ε > 0,there exists an x0 > a such that ∣∣∣∣∫ x2

x1

fdα

∣∣∣∣ < ε,

whenever x2 > x1 > x0. As the lemma is standard, we omit its proof. The interested reader may consult

standard books on analysis to see a proof of the lemma. The following lemma isalso standard. We however include a proof for the reader’s convenience.

Lemma 2 (Second Mean Value Theorem for integrals). Let g :[a, b] → R be a Riemann integrable function on [a, b].

2.20. TWENTIETH COMPETITION 169

(a) If f : [a, b] → R is a nonnegative and decreasing function on [a, b], thenthere exists a number ξ ∈ [a, b] such that∫ b

a

fg = f(a)∫ ξ

a

g.

(b) If f : [a, b] → R is a nonnegative and increasing function on [a, b], thenthere exists a number η ∈ [a, b] such that∫ b

a

fg = f(b)∫ b

η

g.

Proof. (a) Set M = supx∈[a,b] |g(x)|. Obviously, g(x)+M ≥ 0 for all x ∈ [a, b].With that in mind, suppose that P ∈ P[a, b] is a partition of the closed interval[a, b] which is given by

P : x0 = a < x1 < · · · < xn = b.

We can write∫ b

a

f(g +M) =n∑i=1

∫ xi

xi−1

f(g +M) ≤n∑i=1

f(xi−1)∫ xi

xi−1

(g +M)

=n∑i=1

f(xi−1)∫ xi

xi−1

g +Mn∑i=1

f(xi−1)(xi − xi−1).

Now, let G : [a, b] → R be defined by G(x) =∫ xag. As G(a) = 0 and f is

nonnegative and decreasing on [a, b], we haven∑i=1

f(xi−1)∫ xi

xi−1

g =n∑i=1

f(xi−1)(G(xi)−G(xi−1)

)=

n∑i=1

f(xi−1)G(xi)−n−1∑i=1

f(xi)G(xi)

=n∑i=1

(f(xi−1)− f(xi)

)G(xi) + f(b)G(b)

≤n∑i=1

(f(xi−1)− f(xi)

)supx∈[a,b]

G(x) + f(b) supx∈[a,b]

G(x)

=(f(a)− f(b)

)supx∈[a,b]

G(x) + f(b) supx∈[a,b]

G(x)

= f(a) supx∈[a,b]

G(x).

So we can write∫ b

a

f(g +M) ≤ f(a) supx∈[a,b]

G(x) +Mn∑i=1

f(xi−1)(xi − xi−1).

On the other hand, by a result due to Darboux, we have

lim||P ||→0

n∑i=1

f(xi−1)(xi − xi−1) =∫ b

a

f,

170 2. SOLUTIONS

where ||P || = max1≤i≤n(xi − xi−1). So letting ||P || → 0 in the above inequality,we obtain ∫ b

a

f(g +M) ≤ f(a) supx∈[a,b]

G(x) +M

∫ b

a

f,

implying ∫ b

a

fg ≤ f(a) supx∈[a,b]

G(x).

Replacing g by −g in the above inequality, we obtain∫ b

a

f(−g) ≤ f(a) supx∈[a,b]

−G(x),

yielding ∫ b

a

fg ≥ f(a) infx∈[a,b]

G(x).

These inequalities together with the facts that f(a) ≥ 0 and that G is continuous,in view of the Intermediate Value Theorem, imply that there exists a numberξ ∈ [a, b] such that ∫ b

a

fg = f(a)G(ξ) = f(a)∫ ξ

a

g,

which is what we want.(b) A proof similar to that of (a) settles (b). However, here is a slick proof of

(b) using (a). Define functions f1, g1 : [0, b− a] → R as follows

f1(x) = f(b− x), g1(x) = g(b− x).

It is plain that f1 and g1 satisfy the hypotheses of (a). It thus follows from (a) thatthere exists ξ ∈ [0, b− a] such that∫ b−a

0

f1g1 = f1(0)∫ ξ

0

g1.

That is,∫ b−a0

f(b− x)g(b− x)dx = f(b)∫ ξ0g(b− x)dx. Performing the substitution

b− x = t, we easily obtain ∫ b

a

fg = f(b)∫ b

η

g,

where η = b− ξ, as desired.

Now, to prove the assertion, let M =∫ p0|g|. If M = 0, the assertion is trivial

because we would have∫ bafg = 0 for all b > a, from which, the assertion trivially

follows. So suppose M > 0. As limx→+∞ f(x) = 0 and f is decreasing on [0,+∞),we see that f(x) ≥ 0 on [0,+∞) and that for given ε > 0, there exists an x0 > 0such that

f(x) =∣∣f(x)

∣∣ < ε

2M,

whenever x ≥ x0. It now follows from Lemma 2 that for all x0 < x1 < x2, thereexists ξ ∈ [x1, x2] such that∣∣∣∣∫ x2

x1

fg

∣∣∣∣ =∣∣∣∣∣f(x1)

∫ ξ

x1

g

∣∣∣∣∣ = f(x1)

∣∣∣∣∣∫ ξ

x1

g

∣∣∣∣∣ .

2.20. TWENTIETH COMPETITION 171

Since g is periodic with period p, so is |g|. And since∫ p

0

g = 0,

we see that ∫ jp

ip

g = 0,∫ ip

(i−1)p

∣∣g∣∣ = ∫ p

0

∣∣g∣∣,for all i, j ∈ N. Now, set T = kp : k ∈ N∩ [x1, ξ]. It is plain that T is a finite set.For x0 < x1 < x2, there are two cases to consider. (i) T = ∅. (ii) T 6= ∅.

First, if T = ∅, then we have [x1, ξ] ⊆ [(i − 1)p, ip] for some i ∈ N. So we canwrite ∣∣∣∣∫ x2

x1

fg

∣∣∣∣ = f(x1)

∣∣∣∣∣∫ ξ

x1

g

∣∣∣∣∣ ≤ f(x1)∫ ξ

x1

∣∣g∣∣≤ ε

2M

∫ ip

(i−1)p

∣∣g∣∣ = ε

2M

∫ p

0

∣∣g∣∣ = ε

2MM =

ε

2< ε.

That is, ∣∣∣∣∫ x2

x1

fg

∣∣∣∣ < ε.

Next, if T 6= ∅, let ip and jp be the closest points of T to x1 and ξ, respectively.Note that i, j ∈ N and i ≤ j. We can write∣∣∣∣∫ x2

x1

fg

∣∣∣∣ = f(x1)

∣∣∣∣∣∫ ξ

x1

g

∣∣∣∣∣ = f(x1)

∣∣∣∣∣∫ ip

x1

g +∫ jp

ip

g +∫ ξ

jp

g

∣∣∣∣∣= f(x1)

∣∣∣∣∣∫ ip

x1

g +∫ ξ

jp

g

∣∣∣∣∣ ≤ ε

2M

(∫ ip

x1

∣∣g∣∣+ ∫ ξ

jp

∣∣g∣∣) .But [x1, ip] ⊆ [(i− 1)p, ip] and [jp, ξ] ⊆ [jp, (j + 1)p]. So we can write∣∣∣∣∫ x2

x1

fg

∣∣∣∣ ≤ ε

2M

(∫ ip

(i−1)p

∣∣g∣∣+ ∫ (j+1)p

jp

∣∣g∣∣)

=ε

2M

(∫ p

0

∣∣g∣∣+ ∫ p

0

∣∣g∣∣) =ε

2M(M +M) = ε.

That is, ∣∣∣∣∫ x2

x1

fg

∣∣∣∣ ≤ ε.

Thus, in any case, ∣∣∣∣∫ x2

x1

fg

∣∣∣∣ < 2ε,

and hence from Lemma 1, it follows that∫ +∞0

fg converges, which is what we want.

Second solution: Let G : [0,+∞) → R be defined by G(x) =∫ x0g. It follows

from the hypothesis that∫ x+px

g = 0, and hence

G(x) =∫ x

np

g =∫ x−np

0

g,

172 2. SOLUTIONS

where n =⌊xp

⌋and b.c denotes the integer part function. Consequently,

∣∣G(x)∣∣ ≤∫ p

0

∣∣g∣∣ = M for all x ≥ 0. Thus, the function G is bounded. Now, we prove that theassertion holds true under this hypothesis as well. That is, the assertion will remaintrue if in the problem one replaces “g is a p-periodic function such that

∫ p0g = 0” by

“the function G : [0,+∞) → R, defined by G(x) =∫ x0g, is bounded”. It is worth

mentioning that the idea of this proof is very much like that of Dirichlet’s Test forconvergent series. Just as in the first solution, it suffices to show that for givenε > 0 there is an N > 0 such that

∣∣∫ yxfg∣∣ < ε whenever x, y ∈ R with N < x < y.

By the First Fundamental Theorem of Calculus, G′ = g. So we can write∣∣∣∣∫ y

x

fg

∣∣∣∣ =∣∣∣∣∫ y

x

fG′∣∣∣∣ = ∣∣∣∣∫ y

x

fdG

∣∣∣∣ = ∣∣∣∣fG∣∣∣yx−∫ y

x

Gdf

∣∣∣∣≤ |f(y)G(y)− f(x)G(x)|+

∣∣∣∣∫ y

x

Gd(−f)∣∣∣∣ ≤ 2Mf(x) +

∣∣∣∣∫ y

x

Gd(−f)∣∣∣∣

≤ 2Mf(x) +M

∫ y

x

d(−f) = 2Mf(x) +M(f(x)− f(y)

)≤ 3Mf(x).

But limx→+∞ f(x) = 0. Thus, there is an N > 0 such that f(x) < ε3M whenever

x > N . Therefore, ∣∣∣∣∫ y

x

fg

∣∣∣∣ < ε,

whenever N < x < y, and hence∫ +∞0

fg converges, as desired.

2.21. Twenty First Competition

2.21.1. Analysis. 1. (a) Suppose α is a limit point of the set S such that forevery ε > 0 the set x ∈ S : |f(x)| ≥ ε is finite. For given ε > 0, let

δ = min∣∣α− x

∣∣ : ∣∣f(x)∣∣ ≥ ε

.

It is easily verified that |f(x)| < ε whenever 0 < |x − α| < δ. This proves theassertion.

(b) To prove the assertion by contradiction, suppose there exist an ε > 0 andan infinite sequence (xi)+∞i=1 of S such that |f(xi)| ≥ ε for all i ∈ N. Since the setS is compact, the subset xi+∞i=1 has a limit point in S, say, α ∈ S. It is obviousthat limx→α f(x) 6= 0, a contradiction. Thus, the assertion follows.

2. (a) To prove the assertion by contradiction, suppose that there exists an ε0 > 0such that to each n ∈ N, there corresponds an xn ∈ R with xn > n and a tn ∈ Ksatisfying ∣∣g(xn + tn)− g(xn)

∣∣ ≥ ε0,

for all n ∈ N. Since K is compact, if necessary, by passing to a subsequence of(tn)+∞n=1, we may assume that limn tn = t0 for some t0 ∈ K. Define f : [0,+∞) → Rby f(x) = g(x + t0) − g(x). As limx→+∞ f(x) = 0 and g is continuous, we easilysee that f is uniformly continuous on [0,+∞). Thus, for given ε0 > 0, there existsa δ0 > 0 such that ∣∣g(x+ t0)− g(y + t0)

∣∣ = ∣∣f(x)− f(y)∣∣ < ε0

2,

2.21. TWENTY FIRST COMPETITION 173

whenever x, y ∈ [0,+∞) and |x − y| < δ0. On the other hand, it follows from thehypothesis that for given ε0 > 0, there exists an M > 0 such that∣∣g(x+ t0)− g(x)

∣∣ < ε02,

whenever x > M . Now, since K is compact, and hence bounded, and limn tn = t0,choose n ∈ N large enough so that xn + tn − t0 > 0, xn > M , and |tn − t0| < δ0,where M > 0 and δ0 are as in the above. We can write

ε0 ≤∣∣g(xn + tn)− g(xn)

∣∣≤

∣∣g((xn + tn − t0) + t0)− g(xn + t0)

∣∣+ ∣∣g(xn + t0)− g(xn)∣∣

<ε02

+ε02

= ε0,

implying ε0 < ε0. So the assertion follows by way of contradiction.

(b) To prove limx→+∞

(∫ x+1

xg(u)du− g(x)

)= 0, note first that by the Mean

Value Theorem for integrals, for all x ∈ R, there exists a tx ∈ [0, 1] such that∫ x+1

x

g(u)du = g(x+ tx).

Now, letting K = [0, 1] in (a), for given ε > 0, find M > 0 from (a). We have∣∣∣∣∫ x+1

x

g(u)du− g(x)∣∣∣∣ = ∣∣g(x+ tx)− g(x)

∣∣ < ε,

whenever x > M . That is, limx→+∞

(∫ x+1

xg(u)du− g(x)

)= 0, as desired.

To prove limx→+∞g(x)x = 0, it suffices to show that limx→+∞

g(x)[x] = 0, where

[.] denotes the integer part function, for limx→+∞[x]x = 1. To this end, note that

we can writeg(x)[x]

=g([x] + tx

)− g([x])

[x]+g([x])

[x],

where tx = x − [x] ∈ [0, 1) for all x ∈ R with x > 1. We need to show that

limx→+∞g([x]+tx

)−g([x])

[x] = limx→+∞g([x])

[x] = 0. To this end, letting K = [0, 1] in(a), for given ε > 0, there exists an M > 0 such that∣∣g(x+ t)− g(x)

∣∣ < ε,

whenever x > M and t ∈ [0, 1]. In particular, if x > M +1, then [x] > M , implyingthat ∣∣g([x] + tx)− g([x]

)∣∣ < ε,

because tx ∈ [0, 1]. Thus, limx→+∞(g(x)− g([x])

)= 0, and hence

limx→+∞

g([x] + tx

)− g([x])

[x]= 0.

To prove limx→+∞g([x])

[x] = 0, it suffices to show that limn→+∞g(n)n = 0. To see

this, note that part (ii) of the lemma presented in Solution 3 of 2.7.1, which isknown as Stolz’s Second Theorem, together with the hypothesis yields

limn→+∞

g(n)n

= limn→+∞

g(n)− g(n− 1)n− (n− 1)

= limn→+∞

(g(n)− g(n− 1)

)= 0,

174 2. SOLUTIONS

as desired. Therefore, limx→+∞g(x)x = 0, which is what we want.

3. We need the following lemma which is (essentially) due to Leo M. Levine (1977).

Lemma. Let f : (a, b) → R be a bounded function. Let D, L, and R denotethe set of points at which f is discontinuous, has a left hand limit, and has a righthand limit, respectively. Then, D ∩ (L ∪R) is countable.

Proof. Define ω : (a, b) → [0,+∞) by

ω(x) = limδ→0+

(sup f

([x− δ, x+ δ]

)− inf f

([x− δ, x+ δ]

)).

It is readily verified that x ∈ D if and only if ω(x) > 0. This implies D =⋃n∈N Dn,

where Dn = x ∈ (a, b) : ω(x) > 1n, and hence

D ∩ (L ∪R) =( ⋃n∈N

(Dn ∩ L))⋃( ⋃

n∈N(Dn ∩R)

).

Thus, to establish the lemma, it suffices to show that Dn ∩ L and Dn ∩ R arecountable for all n ∈ N. To this end, suppose x0 ∈ Dn ∩L (resp. x0 ∈ Dn ∩R). Asx0 ∈ L (resp. x0 ∈ R), there exists a δ > 0 such that |f(x) − f(x−0 )| < 1

2n (resp.|f(x)− f(x+

0 )| < 12n ) whenever x ∈ (x0 − δ, x0) (resp. x ∈ (x0, x0 + δ)). From this,

we obtain ∣∣f(x1)− f(x2)∣∣ < 1

n,

whenever x1, x2 ∈ (x0− δ, x0) (resp. x1, x2 ∈ (x0, x0 + δ)). Therefore, ω(x) ≤ 1n for

all x ∈ (x0 − δ, x0) (resp. x ∈ (x0, x0 + δ)), implying x /∈ Dn. It thus follows thatany point of Dn ∩ L (resp. Dn ∩ R) is the right (resp. left) end point of an openinterval which contains no point of Dn ∩L (resp. Dn ∩R). But these intervals areobviously disjoint and hence form a countable set. So Dn ∩ L (resp. Dn ∩ R) iscountable for all n ∈ N, as desired.

It is now obvious that the lemma together with Lebesgue’s Integrability Cri-terion for Riemann integrals implies that a bounded function f : [a, b] → R isintegrable in the Riemann sense if and only if [a, b] \ (L ∪ R) is a set of measurezero. This clearly proves the assertion. It is also worth mentioning that, in viewof the lemma, one can mimic the second proof presented for Solution 4 of 2.9.1 togive a direct proof of the assertion. We omit the details for the sake of brevity.

2.21.2. Algebra. 1. It is easily checked that if I, J,K are ideals of R with

J ⊆ I, then I ∩ (J + K) = J + I ∩ K. Let As⊆ B and C

s⊆ D. We prove that

A+Cs⊆ B+D. To see this, suppose E is an ideal of R such that A+C+E = B+D.

We need to show that E = B +D. To this end, we can write

B ∩ (A+ C + E) = B ∩ (B +D).

But A ⊆ B and B ⊆ B +D. So, we have

A+B ∩ (C + E) = B,

implying that B ∩ (C + E) = B because As⊆ B. Thus, B ⊆ C + E. Likewise,

from D ∩ (A + C + E) = D ∩ (B +D), we obtain D ⊆ A + E. Note that A ⊆ Band B ⊆ C + E. Thus, A ⊆ C + E, and hence B + D = A + C + E = C + E.

2.21. TWENTY FIRST COMPETITION 175

Now, from C + E = B + D, we obtain D ∩ (C + E) = D ∩ (B + D), yielding

C+(D∩E) = D, which, in turn in view of Cs⊆ D, implies D∩E = D. Therefore,

D ⊆ E. Analogously, since C ⊆ D ⊆ A+E, we have B+D = C+A+E = A+E,

yielding B∩(A+E) = B∩(B+D). Hence, A+B∩E = B, which in view of As⊆ B,

implies B∩E = B. Therefore, B ⊆ E. It follows that B+D ⊆ E+E = E ⊆ B+D.In other words, E = B +D, as desired.

2. First we need to recall that if p is a prime, then every group G of order p2 isabelian (for a proof see Solution 1 of 2.16.2).

We now prove the assertion. From H ≤ Z(G) ≤ G, we see that

p2 = [G : H] = [G : Z(G)][Z(G) : H],

implying that [G : Z(G)] = 1 or p or p2. If [G : Z(G)] = 1, then G = Z(G). Inother words, G is abelian and hence G′ = e, which is cyclic. If [G : Z(G)] = p,then G

Z(G) is cyclic, implying that G is abelian. This again yields G′ = e, whichis cyclic. Finally, if [G : Z(G)] = p2, then G

Z(G) is abelian because every group oforder p2 is abelian whenever p is a prime. It thus follows from the FundamentalTheorem of finite abelian groups that G

Z(G) is either cyclic or there are x, y ∈ G

such that GZ(G) = 〈xZ(G), yZ(G)〉. Just as we saw in the above, if G

Z(G) is cyclic,then G is abelian, in which case the assertion trivially holds. So assume thatG

Z(G) = 〈xZ(G), yZ(G)〉 for some x, y ∈ G. It follows that there is a z ∈ Z(G) suchthat xy = yxz, for G

Z(G) is abelian. Note that z = x−1y−1xy ∈ Z(G). By provingthat G′ = 〈x−1y−1xy〉 = zk : k ∈ Z, we complete the proof. A straightforwardinduction on i+ j reveals that

xiyj = yjxizij ,

for all i, j ∈ N. Also, as yx = xyz−1 and y−1x = xy−1z, inducting on i + j, weobtain

yjxi = xiyjz−ij , y−jxi = xiy−jzij ,

for all i, j ∈ N. Again, as GZ(G) = 〈xZ(G), yZ(G)〉 is abelian, for all g ∈ G there are

i, j ∈ N and zg ∈ Z(G) such that g = xiyjzg. Therefore, for given g, g′ ∈ G, thereare i, j, i′, j′ ∈ N and zg, zg′ ∈ Z(G) such that g = xiyjzg and g′ = xi

′yj

′zg′ . We

can write

g−1g′−1gg′ =(y−jx−i

)(y−j

′x−i

′)(xiyj

)(xi

′yj

′)=

(z−ijx−iy−j

)(y−j

′x−i

′)(xiyj

)(yj

′xi

′zi′j′)

= x−iy−(j+j′)x−i′xi(yj+j

′xi

′)zi′j′−ij

= x−iy−(j+j′)x−i′xixi

′yj+j

′z−(j+j′)i′zi

′j′−ij

= x−i(y−(j+j′)xi

)yj+j

′z−j(i+i

′)

= x−ixiy−(j+j′)yj+j′zi(j+j

′)z−j(i+i′)

= zij′−i′j .

In other words, g−1g′−1gg′ = (x−1y−1xy)k, where k = ij′ − i′j ∈ Z. Consequently,G′ is the cyclic group generated by z = x−1y−1xy ∈ Z(G), completing the proof.

176 2. SOLUTIONS

3. (a) Let At ∈ Mn(R) denote the transpose of the matrix A ∈ Mn(R). For all1 ≤ i, j ≤ n, use (AAt)ij to denote the ij entry of the matrix AAt. It follows fromthe hypothesis that

(AAt)ij = tr(A)aij ,

for all 1 ≤ i, j ≤ n. It is now obvious that tr(A) 6= 0, for otherwise (AAt)ij = 0for all 1 ≤ i, j ≤ n, yielding tr(AAt) = 0, which obtains A = 0, contradicting thehypothesis. Thus, tr(A) 6= 0, which is what we want.

(b) In view of (a), we have

A =1

tr(A)AAt.

Since AAt is symmetric, so is A, as desired.

(c) In view of the second proof presented for Solution 7 of 2.9.3, it suffices toshow that rank(A) = 1. To this end, suppose A ∈ Mn(R) satisfies aikajk = akkaijfor all 1 ≤ i, j, k ≤ n. For 1 ≤ j ≤ n, let us use Aj to denote the jth column of thematrix A. As tr(A) 6= 0, there exists a 1 ≤ k ≤ n such that akk 6= 0. It thus followsfrom the hypothesis that for all 1 ≤ j ≤ n, we have Aj = bjkAk, where bjk = ajk

akk.

This obviously proves that rank(A) = 1, which is what we want.

2.22. Twenty Second Competition

2.22.1. Analysis. 1. By the Mean Value Theorem, there exists a c ∈ (0, 1)such that f ′(c) = f(1)−f(0)

1−0 = 1. If f ′(x) ≥ 1 (resp. f ′(x) ≤ 1) for all x ∈ (0, 1), wesee that f(x) = x for all x ∈ [0, 1], in which case the assertion is trivial. To see this,suppose to the contrary that there exists c ∈ (0, 1) such that f(c) > c or f(c) < c.Suppose f(c) > c. It follows from the Mean Value Theorem that there exists d ∈(c, 1) (resp. d ∈ (0, c)) such that f ′(d) = f(1)−f(c)

1−c < 1 (resp. f ′(d) = f(c)−f(0)c−0 > 1),

a contradiction in any event. Likewise, if f(c) < c, we obtain a contradiction. Thus,there exist a, b ∈ (0, 1) with a < c < b such that (f ′(a)−1)(f ′(b)−1) < 0. It followsfrom Darboux’s Theorem that the range of f ′, denoted by R, includes the closedinterval [m,M ], where m = min

(f ′(a), f ′(b)

)< 1 < M = max

(f ′(a), f ′(b)

). It

thus follows that [1 − ε, 1 + ε] ⊆ R, where ε = min(1 −m,M − 1). It is obviousthat for all y1 ∈ [1− ε, 1+ ε], with y1 = f ′(x1) < 1 for some x1 ∈ [0, 1], there existsy2 ∈ [1− ε, 1 + ε], with y2 = f ′(x2) > 1 for some x2 ∈ [0, 1], such that

1f ′(x1)

+1

f ′(x2)= 2. (∗)

This obviously proves the assertion when n ∈ N is even. If n ∈ N is odd, use (∗)and the fact that 1

f ′(c) = 1, where c is as in the above, to prove the assertion.

2. (a) We have fn ≤∑+∞i=1 fi = f2 for all n ∈ N because fn’s are all nonnegative.

Define gn : X −→ R by

gn(x) =

0 x ∈ f−1(0),

fn(x)f(x) x ∈ X \ f−1(0).

2.22. TWENTY SECOND COMPETITION 177

It is obvious that fn = gnf for all n ∈ N. It remains to show that gn is continuousfor all n ∈ N. To this end, note that gn is continuous on the open set X \ f−1(0)because so are f and fn on X \ f−1(0) (n ∈ N). Suppose that x0 ∈ X \ f−1(0)is arbitrary. It follows that for given ε > 0, there exists δ > 0 such that∣∣∣∣fn(x)f(x)

− fn(x0)f(x0)

∣∣∣∣ < ε,

whenever x ∈ X \f−1(0) and |x−x0| < δ. Since X \f−1(0) is an open set andx0 ∈ X \ f−1(0), choose δ small enough such that x ∈ X \ f−1(0) whenever|x− x0| < δ. So we have ∣∣gn(x)− gn(x0)

∣∣ < ε,

whenever |x−x0| < δ. That is, gn is continuous at x0. Next, suppose x0 ∈ f−1(0)is arbitrary. As f is continuous, we see that for given ε > 0, there exists a δ > 0such that ∣∣f(x)

∣∣ = ∣∣f(x)− f(x0)∣∣ < ε,

whenever |x− x0| < δ. Now if |x− x0| < δ, we can write

∣∣gn(x)− g(x0)∣∣ = ∣∣gn(x)∣∣ = 0 < ε x ∈ f−1(0)∣∣ fn(x)

f(x)

∣∣ ≤ ∣∣f(x)∣∣ < ε x ∈ X \ f−1(0) .

In other words, |gn(x)−g(x0)| < ε whenever |x−x0| < δ. That is, gn is continuousat x0, as desired.

(b) Suppose that∑+∞n=1 gn converges uniformly on X and that the interior of

f−1(0) is empty. It follows that∑+∞n=1 gn is a continuous function on X and

that X \ f−1(0) is a dense subset of X. We see from (a) that∑+∞n=1 gnf = f2,

yielding f(f −

∑+∞n=1 gn

)= 0. This easily implies f(x)−

∑+∞n=1 gn(x) = 0 whenever

x ∈ X \ f−1(0). That is, the continuous function f −∑+∞n=1 gn vanishes on

the dense subset X \ f−1(0). Thus, f −

∑+∞n=1 gn vanishes on X, and hence

f =∑+∞n=1 gn, which is what we want.

3. Note first that f is continuous on D because the sequence (fn)+∞n=1 uniformlyconverges to f on D as n tends to infinity and that fn’s are all analytic, andhence continuous. Thus, in view of Morera’s Theorem, it suffices to show that∫γ0f(z)dz = 0, where γ0 is any simple closed curve inside γ. We have∫

γ0

f(z)dz = limn

∫γ0

fn(z)dz = 0,

because (fn)+∞n=1 uniformly converges to f on γ0 and fn’s are analytic. Assumingthat z0 is inside γ, we see from Cauchy’s Integral Formula that

fn(z0) =1

2πi

∫γ

fn(z)z − z0

dz, f(z0) =1

2πi

∫γ

f(z)z − z0

dz,

proving the assertion because fn(z0) → f(z0) as n→ +∞.

178 2. SOLUTIONS

2.22.2. Algebra. 1. Suppose [G : K] = 2n − 1 for some n ∈ N. We have[G : H] = [G : K][K : H] = 2(2n− 1). The group G acts on Ω = Hg : g ∈ G, theset of all right cosets of H in G, by multiplication from the right. That is, the actionis given by the map ϕ : G×Ω → Ω defined by ϕ(a,Hg) = Hga. Recall that ϕ givesrise to a group homomorphism ψ : G → S(Ω) defined by ψ(a) = ϕ(a, .) : Ω → Ω,where S(Ω) denotes the group of the permutations of Ω, i.e., the set of all one-to-one maps from Ω onto Ω, which forms a group under composition of maps. Thekernel of the action, by definition, is kerψ =

⋂g∈G g

−1Hg. It follows from the FirstIsomorphism Theorem for groups that G

kerψ∼= im(ψ) ≤ S(Ω) ∼= S2(2n−1). That

is, Gkerψ is isomorphic to a subgroup of the symmetric group of degree 2(2n − 1).

Since k ∈ K has order 2, we see that for k kerψ ∈ Gkerψ , we have (k kerψ)2 =

k2 kerψ = kerψ. Thus, the order of k kerψ, as an element of Gkerψ , is equal to 1

or 2. But k /∈ kerψ because otherwise, in particular, k ∈ H, which is impossible.Consequently, ord(k kerψ) = 2. Note that G

kerψ acts on Ω via (a kerψ,Hg) → Hga

and that the element (k kerψ, .) ∈ S(Ω) ∼= S2(2n−1) has no fixed point, for otherwiseHgk = Hg for some g ∈ G, yielding gkg−1 ∈ H, contradicting the hypothesis.Therefore, the element (k kerψ, .) ∈ S(Ω) corresponds to an odd permutation ofS(Ω) because, in view of (k kerψ, .)2 = id ∈ S(Ω), it decomposes into the productof 2n−1 transpositions or 2-cycles. Thus, the subgroup of G

kerψ , say, Jkerψ for some

J ⊇ kerψ, which consists of all elements that correspond to even permutations ofS(Ω) forms a subgroup of G

kerψ of index 2. Since [G : kerψ] = [G : J ][J : kerψ], wecan write

[G : J ] =[ G

kerψ:

J

kerψ

]= 2,

which is what we want.

2. If J = R or J = 0, the assertion is trivial. So suppose that J is a nontrivialideal of R and J has a maximal ideal, say, M . By obtaining a contradiction,we settle the proof. Let f : J → J

M , defined by f(x) = x + M , be the naturalhomomorphism from J onto J

M . As M is a maximal ideal of J , it follows that JM

has no nontrivial ideal. This implies that JM is a field or it is a finite ring with zero

multiplication. The latter is impossible because it would mean (x+M)(y+M) = Mfor all x, y ∈ J , yielding J2 ⊆M , which is impossible. Thus, J

M is a field. Let e+Mdenote the identity element of the field J

M , where e ∈ J \M . Define f∗ : R→ JM by

f∗(x) = f(xe+M). It is readily seen that f∗ is a homomorphism of rings. So, wesee from the First Isomorphism Theorem for rings that R

ker f∗∼= J

M . Consequently,ker f∗ is a maximal ideal of R, for J

M is field. Thus, J ⊆ ker f∗. This, in particular,implies f∗(e) = M . On the other hand, f∗(e) = f(e2 +M) = (e +M)2 = e +M .So, we must have M = e+M , yielding e ∈M , which is impossible. The assertionthus follows by contradiction.

3. To prove the assertion by contradiction, suppose that rank(L) = 1 so thatLV = 〈α〉 for some α ∈ V . With no loss of generality, we may assume that thecharacteristic polynomial of T is irreducible over F . We claim that

T kα ∈ kerL,

2.23. TWENTY THIRD COMPETITION 179

for each k = 0, 1, 2, . . .. As the rank of LT k is at most one and

tr(LT k) = tr(TST k − ST k+1) = 0,

for each k = 0, 1, 2, . . ., it follows that LT k is nilpotent. And hence its eigenvaluesare all zero. But LT kα ∈ LV = 〈α〉 for all nonnegative integers k. That is, α is aneigenvector of LT k, and hence LT kα = 0 for all nonnegative integers k. Now, set

W =⟨T kα : k = 0, 1, 2, . . .

⟩.

It is plain that W is invariant under T and that W is a nontrivial subspace ofV because α is nonzero and W ⊆ kerL. It thus follows that the characteristicpolynomial of T |W , the restriction of T to the invariant subspace W , divides theminimal polynomial of T , which is equal to the characteristic polynomial of T ,for the characteristic polynomial of T is irreducible over F . This contradicts thehypothesis that the characteristic polynomial of T is irreducible. Thus, the assertionfollows by contradiction.

2.22.3. General.a b c d a b c d a b c d

1. F 15. F 29. F2. F 16. F 30. F3. F 17. F 31. F4. F 18. F 32. F5. F 19. F 33. F6. F 20. F 34. F7. F 21. F 35. F8. F 22. F 36. F9. F 23. F 37. F10. F 24. F 38. F11. F 25. F 39. F12. F 26. F 40. F13. F 27. F 41.14. F 28. F 42.

2.23. Twenty Third Competition

2.23.1. Analysis. 1. (a) We prove the assertion by way of contradiction.Suppose that the set fα : α ∈ R is compact with respect to the uniform normof Cb(R) and yet f is not uniformly continuous on R. It follows that there areε > 0 and sequences (xn)+∞n=1 and (yn)+∞n=1 in R such that limn xn = limn yn = ±∞,limn(xn − yn) = 0, and |f(xn) − f(yn)| ≥ ε for all n ∈ N. By the hypothesis,the sequence (gn)+∞n=1, where gn = fxn , has a subsequence, say (gnk

)+∞k=1, whichconverges uniformly to fα on R for some α ∈ R. It follows that limk gnk

(ynk−xnk

) =fα(0). On the other hand, limk gnk

(0) = fα(0). Consequently, limk

(gnk

(ynk−

xnk)− gnk

(0))

= 0. But∣∣gnk(ynk

− xnk)− gnk

(0)∣∣ = ∣∣f(ynk

)− f(xnk)∣∣ ≥ ε,

180 2. SOLUTIONS

for all k ∈ N, which is in contradiction with

limk

(gnk

(ynk− xnk

)− gnk(0))

= 0.

Thus, f is uniformly continuous on R, as desired.

(b) We disprove the proposition by showing that for the function f : R −→ Rdefined by f(x) = e−x

2, the sequence fα : α ∈ R is not compact with respect to

the uniform norm of Cb(R). Note first that f is bounded and uniformly continuouson R because limx→±∞ f(x) = 0. To disprove the proposition by contradiction,suppose that fα : α ∈ R is compact with respect to the uniform norm of Cb(R). Itfollows that the sequence (fn)+∞n=1 has a subsequence, say (fnk

)+∞k=1, which convergesuniformly on R to fα for some α ∈ R. Thus, 0 = limk e

−(x+nk)2 = e−(x+α)2 ,implying that e−(x+α)2 = 0 for all x ∈ R, which is obviously impossible. So theproposition is disproved by way of contradiction.

2. First solution: First we need to recall Abel’s Continuity Theorem whichasserts that if

∑+∞n=0 an is a convergent series of complex numbers, then

limr→1−

+∞∑n=0

anrn =

+∞∑n=0

an.

Without loss of generality, if necessary by adding or subtracting multiples of2π, we may assume that 0 < α < 2π. Set

Sn(α) =n∑k=1

eikα

k.

We have

S′n(α) = in∑k=1

eikα = iei(n+1)α − eiα

eiα − 1.

Consequently,

Sn(α)− Sn(π) = i

∫ α

π

ei(n+1)t

eit − 1dt− i

∫ α

π

(1 +

1eit − 1

)dt

=∫ α

π

1eit − 1

d(ei(n+1)t

n+ 1

)− i(α− π)− i

∫ α

π

(cos t− 1)− i sin t2(1− cos t)

dt

=1

n+ 1ei(n+1)t

eit − 1

∣∣∣απ

+i

n+ 1

∫ α

π

ei(n+2)t

(eit − 1)2dt− i(α− π) +

i

2

∫ α

π

dt

−∫ α

π

1sin t

2

d(

sint

2

)=

1n+ 1

ei(n+1)α

eiα − 1+

(−1)n

2(n+ 1)+

i

n+ 1

∫ α

π

ei(n+2)t

(eit − 1)2dt− i(α− π)

2

− ln sinα

2.

2.23. TWENTY THIRD COMPETITION 181

So we can write

Sn(α)−(− ln 2 sin

α

2+ i

π − α

2)

=(ln 2 + Sn(π)

)+

1n+ 1

ei(n+1)α

eiα − 1

+(−1)n

2(n+ 1)+

i

n+ 1

∫ α

π

ei(n+2)t

(eit − 1)2dt.

On the other hand, we have

limnSn(π) = −

+∞∑k=1

(−1)k−1

k= − ln 2.

To see this, just apply Abel’s Continuity Theorem to the power series∑+∞n=1

(−1)n−1

n xn =ln(1 + x) and note that the alternating series

∑+∞n=1

(−1)n

n converges by Leibniz’sTheorem. Also, it is plain that

limn

(1

n+ 1ei(n+1)α

eiα − 1+

(−1)n

2(n+ 1)

)= 0.

Finally, ∣∣∣∣ i

n+ 1

∫ α

π

ei(n+2)t

(eit − 1)2dt

∣∣∣∣ ≤ 1n+ 1

∣∣∣∣∣∫ α

π

∣∣ei(n+2)t∣∣∣∣eit − 1∣∣2 dt

∣∣∣∣∣=

1n+ 1

∣∣∣∣∣∫ α

π

1∣∣eit − 1∣∣2 dt

∣∣∣∣∣ ,for all n ∈ N and limn

1n+1 = 0, implying that

limn

i

n+ 1

∫ α

π

ei(n+2)t

(eit − 1)2dt = 0.

Consequently,

limn

(Sn(α)−

(− ln 2 sin

α

2+ i

π − α

2))

= 0.

In other words,+∞∑n=1

einα

n= − ln 2 sin

α

2+ i

π − α

2.

Now, equating the real and imaginary parts of the above equality obtains+∞∑n=1

cos(nα)n

= − ln 2 sinα

2,

+∞∑n=1

sin(nα)n

=π − α

2,

which is what we want.

Second solution: For this solution, first we need to recall Dirichlet’s Theoremwhich asserts that if (an)+∞n=1 and (bn)+∞n=1 are sequences of complex and real num-bers, respectively, such that the sequence of partial sums of (an)+∞n=1 is bounded

182 2. SOLUTIONS

and that (bn)+∞n=1 is monotonic and tends to zero as n → +∞, then∑+∞n=1 anbn

converges. Note that, as we saw in the first solution, we haven∑k=1

eikα =ei(n+1)α − eiα

eiα − 1,

implying that ∣∣ n∑k=1

eikα∣∣ ≤ 2∣∣eiα − 1

∣∣ ,for all n ∈ N. On the other hand, the sequence ( 1

n )+∞n=1 is decreasing and tends tozero as n→ +∞. Thus, by Dirichlet’s Theorem, the series

+∞∑n=1

einα

n

converges whenever α ∈ R \ 2kπ : k ∈ Z.Again, without loss of generality, we may assume that 0 < α < 2π. Let

z = reiα, where 0 ≤ r < 1. Note that 1 − z = (1 − r cosα) + i(−r sinα). We canwrite

− ln(1− z) =+∞∑n=1

zn

n=

+∞∑n=1

einα

nrn,

where ln denotes the principal value of the natural logarithm, which is defined byln z = ln |z|+ i arg z, where arg z = arctan Im(z)

Re(z) ∈ (−π2 ,

π2 ). On the other hand,

− ln(1− z) = −(ln |1− z|+ i arg(1− z)

)= −

(ln√

1− 2r cosα+ r2 + i arctan−r sinα

1− r cosα

).

Letting r → 1−, in view of Abel’s Continuity Theorem, we obtain+∞∑n=1

einα

n= − ln

√2− 2 cosα− i arctan

− sinα1− cosα

= − ln√

4 sin2 α

2+ i arctan cot(

α

2)

= − ln 2 sinα

2+ i

π − α

2.

Thus,+∞∑n=1

einα

n= − ln 2 sin

α

2+ i

π − α

2.

Equating the real and imaginary parts of the above equality, we obtain+∞∑n=1

cos(nα)n

= − ln 2 sinα

2,

+∞∑n=1

sin(nα)n

=π − α

2,

which is what we want.

2.23. TWENTY THIRD COMPETITION 183

3. It is worth mentioning that the hypothesis that f ′ and f ′n’s are continuous isredundant. Let x ∈ [a, b) be arbitrary. In view of the definition of the derivative,the Mean Value Theorem, and the continuity of g, we can write

f ′(x) = limn→+∞

n(f(x+

1n

)− f(x))

= limn→+∞

limk→+∞

n(fk(x+

1n

)− fk(x))

= limn→+∞

limk→+∞

f ′k(xn)

= limn→+∞

g(xn) = g(x),

where x < xn < x + 1n . Thus, f ′(x) = g(x) for all x ∈ [a, b). If x = b, just

write f ′(b) = limn→+∞−n(f(b− 1

n )− f(b))

and repeat the above argument to getf ′(b) = g(b). So the proof is complete.

2.23.2. Algebra. 1. This problem is wrong! For n ∈ N, use Dn to denotethe dihedral group of order 2n. A presentation of Dn is⟨

a, b : an = e, b2 = e, ak 6= e (0 < k < n), ab = ba−1⟩,

and henceDn =aibj : 0 ≤ i < n, j = 0, 1

. We show thatD8 is a counterexample.

First, recall that Inn(G) ∼= GZ(G) , where Z(G) denotes the center of the group G

(see Solution 1 of 2.16.2). Therefore, Inn(D8) ∼= D8Z(D8)

. To show that D8 is acounterexample, it suffices to prove that D2n

Z(D2n)∼= Dn for all n ∈ N. To this end,

just note that Z(D2n) =e, an

, where a is the element of order 2n in the above

presentation of D2n, and that

D2n

Z(D2n)=

AiBj : A = aZ(D2n), B = bZ(D2n), 0 ≤ i < n, j = 0, 1

∼= Dn,

for one can easily verify that An = Z(D2n) = B2, AB = BA−1, and that Ak 6=Z(D2n) for all 0 < k < n.

2. First, we prove that H is a two-sided ideal of R by showing that for every r ∈ Rand x ∈ H, there is a t ∈ R such that rx = xt. To this end, note first that rR is aright ideal of R. If rR = 0, there is nothing to prove. If rR 6= 0, then x ∈ rR,and hence there is a y ∈ R such that x = ry. Likewise, considering the right idealyR, we obtain a t ∈ R such that x = yt. So we can write rx = (ry)t = xt. Thatis, for r ∈ R and x ∈ H, there is a t ∈ R such that rx = xt, as desired. So far,we have actually shown that the intersection of all nonzero right ideals of R is thesame of those of the nonzero left ideals of R.

Next, suppose that H2 6= 0. We prove that R is a division ring. To this end,we first show that R has no zero divisors. Suppose to the contrary that there arenonzero elements x, y ∈ R such that xy = 0. As H2 6= 0, there are a, b ∈ H suchthat ab 6= 0. For a, b ∈ R, just as we saw in the above, there are s, t ∈ R such thatxs = a and yt = b because a, b ∈ H. Again, for s ∈ R and b ∈ H, there is a u ∈ Rsuch that sb = bu. We can write

xy = 0 =⇒ xyt = 0 =⇒ xb = 0 =⇒ xbu = 0 =⇒ xsb = 0 =⇒ ab = 0,

184 2. SOLUTIONS

a contradiction. Thus, R has no zero divisors. To show that R has a multiplicativeidentity, choose a nonzero a ∈ H and note that a ∈ aR, implying that a = ae forsome e ∈ R. Since R has no zero divisors and a(ea−a) = 0, we obtain ea = a. Nowfor an arbitrary x ∈ R, we have (xe− x)a = a(ex− x) = 0, from which, we obtainxe = ex = x. That is, e is the multiplicative identity of R. For the rest, since His an ideal of R, we need to prove that every nonzero element of H is invertible.To this end, for a nonzero a ∈ H, note that a2 6= 0, and hence a = a2t for somet ∈ R. It follows that a(e − at) = 0, which, in turn, yields at = e. This impliesthat ta(e− ta) = 0, from which, we obtain e = ta. Thus e = ta = at. That is, a isinvertible, which is what we want.

3. Let A = (aij) ∈ Mn(F ). Use Eij and In to denote the matrix whose ij entryis 1 and zero elsewhere and the identity matrix, respectively. If 1 ≤ i, j ≤ nwith i 6= j, then tr(Eij) = 0. By the hypothesis, aji = tr(AEij) = 0 for all1 ≤ i, j ≤ n with i 6= j. Thus, A is a diagonal matrix. For 1 ≤ i ≤ n, letBi = diag(1, 0, . . . , 0)−diag(δ1i, δ2i, . . . , δni), where δji denotes the Kronecker delta.Since tr(Bi) = 0, the hypothesis implies aii − a11 = tr(ABi) = 0 for all 1 ≤ i ≤ n.Therefore, A = a11In, as desired.

2.24. Twenty Fourth Competition

2.24.1. First Day. 1. Define F : [a, b] −→ R by F (x) =∫ xaf(t)dt. By

the First Fundamental Theorem of Calculus, F ′(x) = f(x) for all x ∈ [a, b]. Leth = F − g. Obviously, h is differentiable on [a, b] and we have h′ = f − g′. Itfollows from the hypothesis that h′(a)h′(b) < 0. This together with Darboux’sTheorem implies that there exists a c ∈ (a, b) such that h′(c) = 0. In other words,f(c) = g′(c), which is what we want.

2. Note first that all real harmonic functions satisfy the Mean Value Property.That is, if D is a domain such that Br0(a0) ⊆ D for some a0 = (x0, y0) ∈ D andr0 > 0 and u : D −→ R is a harmonic function, then

u(a0) =12π

∫ 2π

0

u(a0 + r0eiθ)dθ.

To see this, let B be an open ball such that Br0(a0) ⊆ B ⊆ D and f a holomorphicfunction on the simply connected domain B such that u|B = Ref . It easily followsfrom Cauchy’s Integral Formula that

f(a0) =12π

∫ 2π

0

f(a0 + r0eiθ)dθ.

Taking the real part of the both sides of the above equation proves the counterpartof it with f replaced by u, as desired. Next, we need the following version of theMaximum Modulus Theorem for harmonic functions, which in fact holds true forall functions satisfying the Mean Value Property.

Theorem. Let D be a domain and u : D −→ R be a harmonic function. Ifthere is a point a0 = (x0, y0) ∈ D such that u(z) ≤ u(a0) for all z ∈ D, thenu(z) = u(a0) for all z ∈ D.

2.24. TWENTY FOURTH COMPETITION 185

Proof. SetL = z ∈ D : u(z) = u(a0).

It is plain that L is a nonempty closed subset of D. By proving that L is an opensubset of D as well, in view of the connectedness of D, we see that L = D, provingthe assertion. To this end, suppose to the contrary that there is a z0 ∈ L whichis not an interior point of L. Choose r > 0 such that Br(z0) ⊆ D. It followsthat there is a b0 ∈ Br(z0) such that u(b0) < u(a0) = u(z0). And hence, from thecontinuity of u, we see that u(z) < u(z0) for all z in a small enough neighborhoodof b0. Letting r0 = |z0 − b0| and b0 = z0 + r0e

iθ0 for some 0 ≤ θ0 ≤ 2π, there thusexists a proper subinterval I of [0, 2π] such that θ0 ∈ I and u(z0 + r0e

iθ) < u(z0)for all θ ∈ I. This, in view of the Mean Value Property of harmonic functions,yields

u(z0) =12π

∫ 2π

0

u(z0 + r0eiθ)dθ < u(z0),

which is a contradiction. Thus, L is open, completing the proof.

To prove the assertion, let the set A be as in the statement of the problem.By the Maximum Modulus Theorem for harmonic functions, the set A is open.The assertion follows as soon as we prove that A is closed because D is connectedand A is nonempty by the hypothesis. To prove that A is closed, suppose that(an)+∞n=1 is a sequence of the elements of A such that limn an = a for some a ∈ D.We need to show that a ∈ A. To this end, choose r > 0 such that Br(a) ⊆ D.It follows that there is an n ∈ N such that an ∈ Br(z0). As an ∈ A, thereis an rn > 0 such that Brn

(an) ⊆ Br(a) ⊆ D and that u(z) ≤ u(an) for allz ∈ Brn

(an). Since Br(a) is simply connected, there is a holomorphic functionf on Br(a) such that u|Br(a) = Ref . By the Maximum Modulus Theorem forharmonic functions, Ref |Brn (an) = Ref(an), which, in view of the Cauchy-RiemannEquations, yields f |Brn (an) = f(an). Since f is holomorphic on Br(a), it followsthat f |Br(a) = f(an) = f(a), which, in turn, implies u(z) = u(a) for all z ∈ Br(a).Therefore, a ∈ A, which is what we want, completing the proof.

3. First solution: Let p = t(3n − 2n). In particular, p∣∣3n − 2n. In other words,

3np≡ 2n. Obviously, gcd(6, p) = 1. And hence, 2 has an inverse modulo p. That is,

there is an a ∈ N with 1 < a < p such that 2ap≡ 1. So we can write

3nanp≡ 2nan

p≡ (2a)n

p≡ 1,

implying (3a)np≡ 1. Use m to denote the order of 3a modulo p, i.e., m is the

least positive integer satisfying (3a)mp≡ 1. It follows that m

∣∣n. By Fermat’s Little

Theorem, (3a)p−1 p≡ 1, which obtains m

∣∣p−1. Consequently, m∣∣gcd(n, p−1). Note

that m > 1, for otherwise 3ap≡ 1, yielding 3

p≡ 2, which is impossible. This implies

that any prime divisor of m, say q, is a divisor of n as well and that q ≤ p− 1 < p.Thus, t(n) ≤ q < p = t(3n − 2n), as desired.

Second solution: Let m be the least positive integer such that 3mp≡ 2m. Ob-

viously, 2m3m 6p≡ 0. Note that if 3k

p≡ 2k for some k ∈ N, then m

∣∣k. To see this,

186 2. SOLUTIONS

letting k = qm+ r, where r + 1, q ∈ N and 0 ≤ r < k, we can write

3rp≡ 3k−qm

p≡ 3k(3m)−q

p≡ 2k(2m)−q

p≡ 2k−mq = 2r,

implying that 3rp≡ 2r. Thus, r = 0, yielding k = qm, i.e., m

∣∣k. This together with

the hypothesis implies that m∣∣n. On the other hand, as 6 6

p≡ 0, in view of Fermat’s

Little Theorem, we can write

3p−1 p≡ 1

p≡ 2p−1,

yielding m∣∣p − 1. Consequently, m

∣∣gcd(n, p − 1), which, as we saw in the firstsolution, entails t(n) < t(3n − 2n), as desired.

4. Use Z to denote the center of G. It follows from the hypothesis that there existn ∈ N and gi ∈ G (1 ≤ i ≤ n) such that

G

Z=g1Z, . . . , gnZ

.

It is obvious that giZ’s (1 ≤ i ≤ n) are all infinite subsets of G. It suffices toshow that gigj = gjgi for all 1 ≤ i, j ≤ n. To this end, for given 1 ≤ i, j ≤ nwith i 6= j, it follows from the hypothesis that there exist zi, zj ∈ Z such that(gizi)(gjzj) = (gjzj)(gizi). This easily yields (gigj)(zizj) = (gjgi)(zizj), which, inturn, implies gigj = gjgi for all 1 ≤ i, j ≤ n, completing the proof.

5. First solution: First, we show that if there are m n-tuples from the set 0, 1in such a way that every two of which differ at least in d components, then

mn

2≥ (m− 1)d.

To this end, the given n-tuples form the following m× n matrix a11 · · · a1n

... · · ·...

am1 · · · amn

.

whose row i is the ith n-tuple (1 ≤ i ≤ m). Denote by A the total number of differ-ences between any two rows of these m rows. In other words, A =

∑1≤i<j≤m dij ,

where dij is the number of components in which row i and row j differ. As dij ≥ dfor all 1 ≤ i < j ≤ m, we obtain A ≥

(m2

)d. To find an upper bound for A, fix a

column, say column j, and use cj to denote the number of differences that these mvector can mutually have in their jth component. It is obvious that A =

∑1≤j≤n cj .

If there are xj zeros in column j, then cj = xj(m−xj) ≤ m2 (m− m

2 ) = m2

4 becausethe polynomial function f(x) = x(m − x) assumes its maximum at x = m

2 . Itfollows that A =

∑1≤j≤n cj ≤

∑1≤j≤n

m2

4 = m2n4 . Consequently,

m2n

4≥ A ≥

(m

2

)d =

m(m− 1)2

d,

from which, we obtainmn

2≥ (m− 1)d,

as desired.

2.24. TWENTY FOURTH COMPETITION 187

Now, to prove the assertion, let m = M(2d − 1, d) and note that n = 2d − 1.In view of the above inequality, we have

m(2d− 1)2

≥ (m− 1)d,

yielding2md−m ≥ 2md− 2d =⇒ m = M(2d− 1, d) ≤ 2d,

which is what we want.

Second solution: First we need to recall Turan’s Theorem from Graph Theory.For the sake of completeness, we quote a simple proof of it which is due to WilliamStaton.

Theorem (Turan, 1941). Graphs with n vertices containing no Kr have nomore than (r−2)n2

2r−2 edges, for r ≥ 2.

Proof (William Staton). Induct on r. If r = 2, the result is obvious. Nowif the statement is true for Kr-free graphs it must be shown that Kr+1-free graphshave no more than (r−1)n2

2r edges. Let G be such a graph, and let x be the numberof vertices in a largest Kr-free induced subgraph of G. Since the neighbors of anyvertex induce a Kr-free subgraph, no vertex of G has degree exceeding x. Let A bea largest induced Kr-free subgraph of G. By induction, there are at most (r−2)x2

(2r−2)

edges in A. Each edge of G not in A is incident with at least one of the n − xvertices not in A, so summing the degrees of these vertices counts each such edgeat least once. Hence there are at most x(n − x) such edges and so G has at most(r−2)x2

(2r−2) + x(n− x) edges. Since

r − 22r − 2

(x2) + x(n− x) =r − 12r

n2 − r

2r − 2

(x− (r − 1)n

r

)2

,

the result follows.

To prove the assertion by contradiction, suppose that M(2d−1, d) > 2d so thatthere are 2d+1 (2d−1)-tuples any two of which differ in at least d components. LetG be an edge-labeled graph as follows. The set of vertices is the 2d+1 (2d−1)-tuples.For any two vertices, if the two vertices differ in their k component (1 ≤ k ≤ 2d−1),then connect the two vertices with an edge with label k. Consequently, there areat least d labeled edges connecting any two vertices of the graph G and hence thegraph G has at least

(2d+1

2

)× d = (2d + 1)d2 edges. Since 2d − 1 numbers are

assigned to the edges and there are at least (2d+ 1)d2 edges, we see that there is alabel 1 ≤ k0 ≤ 2d−1 which is assigned to at least

[(2d+1)d2

2d−1

]+1 edges of the graph,

where [.] denotes the integer part function. Now consider the induced graph onthese edges to each of which k0 is assigned. This induced graph is indeed a simplegraph having 2d+ 1 vertices and

[(2d+1)d2

2d−1

]+ 1 edges. Note that[

(2d+ 1)d2

2d− 1

]+ 1 >

(2d+ 1)d2

2d− 1> d2 + d =

[(2d+ 1)2

4

].

It thus follows from Turan’s Theorem that the induced graph has a K3, i.e., atriangle. This means that there are three vertices any two of which differ in their

188 2. SOLUTIONS

k0 component. This is a contradiction because the components are either 0 or 1,completing the proof.

6. First we need the following simple lemma.

Lemma. Let two lookalike boxes contain w1 white marbles and b1 black mar-bles, and w2 white marbles and b2 black marbles, respectively. The probability ofpicking a white marble from one of the boxes is equal to 1

2

(w1

w1+b1+ w2

w2+b2

).

Proof. Use A to denote the event that one picks a marble from the box thatcontains w1 white marbles and b1 black marbles, and use B to denote the eventthat one picks a marble from the other box. If W denotes the event of pickinga white marble from one of the boxes, as A,B is a partition of the probabilityspace, we can write

P (W ) = P((A ∩W ) ∪ (B ∩W )

)= P (A)P (W |A) + P (B)P (W |B)

=12× w1

w1 + b1+

12× w2

w2 + b2=

12( w1

w1 + b1+

w2

w2 + b2

),

proving the lemma.

Suppose that the first person has picked a marble from one of the boxes, say,box b. Use b′ to denote the other box. For each i = 1, 2, 3, define the followingevents

Wi (resp. W ′i ): the event that the ith person picks a white marble from b (resp.

b′).Bi (resp. B′

i): the event that the ith person picks a black marble from b (resp.b′).

By the lemma above, we have

P (W1) =12( 11 + 2

+2

1 + 2)

=12.

Likewise, P (B1) = 12 .

Use B011 and B001 (resp. B′011 and B′

001) to denote the events that box b (resp.b′) contains one black marble and two white marbles and two black marbles andone white marble, respectively.

First, suppose that the first person has picked a white marble from b. UsingBayes’ Theorem, we can write

P (B011|W1) =P (B011)P (W1|B011)

P (W1)=

12 ×

23

12

=23,

P (B001|W1) =P (B001)P (W1|B001)

P (W1)=

12 ×

13

12

=13.

And hence, P (B′011|W1) = 1 − P (B011|W1) = 1 − 2

3 = 13 and P (B′

001|W1) =1−P (B001|W1) = 1− 1

3 = 23 . Let U = B011|W1, V = B001|W1, U ′ = B′

011|W1, andV ′ = B′

001|W1. We have P (U) = P (V ′) = 23 and P (V ) = P (U ′) = 1

3 . Noting thatU, V is a partition of the probability space, we have

P (W2) = P (U)P (W2|U) + P (V )P (W2|V ) =23× 1

2+

13× 0 =

13.

2.24. TWENTY FOURTH COMPETITION 189

That is, assuming that the first person has picked a white marble from b, theprobability that the second person picks a white marble from b is 1

3 . Analogously,as U ′, V ′ is a partition of the probability space, we can write

P (W ′2) = P (U ′)P (W ′

2|U ′) + P (V ′)P (W ′2|V ′) =

13× 2

3+

23× 1

3=

49.

In other words, assuming that the first person has picked a white marble from b,the probability that the second person picks a white marble from b′ is 4

9 , which isgreater than 1

3 . So, in this case, the second person should pick a marble from theother box, i.e., b′.

Now, assuming that the first person has picked a black marble from b, again us-ing Bayes’ Theorem, we will have P (B011|B1) = 1

3 and P (B001|B1) = 23 , and hence

P (B′011|B1) = 1 − P (B011|B1) = 1 − 1

3 = 23 and P (B′

001|B1) = 1 − P (B001|B1) =1− 2

3 = 13 . So, letting S = B011|B1, T = B001|B1, S′ = B′

011|B1, and T ′ = B′001|B1,

we can write

P (W2) = P (S)P (W2|S) + P (T )P (W2|T ) =13× 1 +

23× 1

2=

23,

P (W ′2) = P (S′)P (W ′

2|S′) + P (T ′)P (W ′2|T ′) =

23× 2

3+

13× 1

3=

59.

Since P (W2) = 23 is greater than P (W ′

2) = 59 , the second person should pick a

marble from box b. Therefore, the probability of survival for the second person is

12× 4

9+

12× 2

3=

59,

for, by the above lemma, the first person picks a white (resp. black) marble withthe probability of 1

2 .To investigate the probability of survival for the third person, there are two

cases to consider.(i) the first person survives.Without loss of generality, we may assume that the first person has picked a

white marble from box b. In view of what we showed in the above, there are twosubcases to consider. (a) The second person picks a white marble from box b′; and(b) The second person picks a black marble from box b′. In case (a), since the firstand second person have picked white marbles from b and b′, respectively, the boxesdo not make any difference for the third person to pick a marble from. Thus, by thelemma above, the third person picks a white marble from one of the boxes with theprobability of 1

4 . So if we use S1 to denote the event that the third person survivesin this case, we will have

P (S1) =12× 4

9× 1

4=

118.

In case (b), letting C = W1 ∩B′2, in view of Bayes’ Theorem, we can write

P (B011|C) =P (C|B011)P (B011)

P (B011)P (C|B011) + P (B001)P (C|B001)

=23 ×

23 ×

12

12 ×

23 ×

23 + 1

2 ×13 ×

13

=45.

190 2. SOLUTIONS

Thus, P (B001|C) = 1 − P (B011|C) = 15 . Consequently, the probability that the

third person picks a white marble from b is equal to

P (B001|C)P (W3

∣∣B001|C) + P (B011|C)P (W3

∣∣B011|C) =15× 0 +

45× 1

2=

25.

Likewise, the probability that the third person picks a white marble from b′ is equalto

P (B001|C)P (W ′3

∣∣B001|C) + P (B011|C)P (W ′3

∣∣B011|C) =15× 1 +

45× 1

2=

35.

Thus, the third person should pick a marble from b′, which is the box from whichthe second person picked a marble. So, the probability of survival for the thirdperson in this case, denoted by P (S2), is equal to

P (S2) =12× 5

9× 3

5=

16.

(ii) the first person does not survive.Again, we may assume that the first person has picked a black marble from

box b. There are two subcases to consider. (a) The second person picks a whitemarble from box b; and (b) The second person picks a black marble from box b. Incase (a), letting C = B1 ∩W2, in view of Bayes’ Theorem, we can write

P (B001|C) =P (C|B001)P (B001)

P (B001)P (C|B001) + P (B011)P (C|B011)

=23 ×

12 ×

12

12 ×

23 ×

12 + 1

2 ×13 × 1

=12.

Thus, the boxes do not make any difference for the third person to pick a marblefrom. So the probability of picking a white marble becomes 2

4 = 12 . And hence, if

S3 denotes the event that the third person survives in this case, we will have

P (S3) =12× 2

3× 1

2=

16.

Finally, if the first and second person both pick black marbles from b, then thethird person can survive by picking the remaining white marble from box b. So theprobability of survival in this case, which is denoted by P (S4), is equal to

P (S4) =12× 1

3× 1 =

16.

Consequently, the probability of survival for this person, denoted by P (S), isequal to

P (S) = P (S1) + · · ·+ P (S4) =118

+16

+16

+16

=59.

That is, the probability of survival for the third person is the same as that of thesecond person.

2.24. TWENTY FOURTH COMPETITION 191

2.24.2. Second Day. 1. (a) Use B to denote the closure of B with respectto the Euclidean norm of Rn. As B is compact because it is bounded and closed,there is a positive real r such that

diam(B) := sup||a− b|| : a, b ∈ B

= 2r, .

As diam(B) = diam(B), in view of the hypothesis, we see that for all n ∈ N thereare xn, yn ∈ B and an open ball Brn(zn), the ball centered at zn with radius rn,such that xn, yn ∈ Brn

(zn) ⊆ B and ||xn − yn|| > 2r − 1n . Since zn’s are in the

compact set B, if necessary by passing to a subsequence of (zn)+∞n=1, we may assumethat there is a z ∈ B such that limn zn = z. We claim that B = Br(z), the open ballwith radius r centered at z, proving the assertion. First, we prove that Br(z) ⊆ B.To this end, let x ∈ Br(z) be arbitrary. We can write ||x − z|| = r − ε for someε > 0. As limn

1n = 0, there is an N ∈ N such that 1

N < ε and ||z − zN || < ε2 . We

have

||x− zN || ≤ ||x− z||+ ||z − zN || < r − ε+ε

2

= r − ε

2< r − 1

2N<||xN − yN ||

2≤ rN ,

implying that x ∈ BrN(zN ) ⊆ B. In other words, x ∈ B, and hence Br(z) ⊆ B.

To prove that B ⊆ Br(z), we argue by contradiction. Suppose that there is a pointx ∈ B \Br(z) so that ||x− z|| > r. Join the two points x, z to intersect the closedball Br(z) at two antipodal points y, y′ ∈ Br(z) so that y and y′ are, respectively,on and off the line segment that joins x and z. It is obvious that

||x− y′|| = ||x− z||+ ||z − y′|| = ||x− z||+ r > 2r = diam(B) = diam(B),

which is a contradiction because x, y′ ∈ B. Therefore, B ⊆ Br(z), and henceB = Br(z), completing the proof.

(b) First solution: Let X = (C[0, 1], ||.||∞) be the normed space of all real valuedcontinuous functions on the closed interval [0, 1] equipped with the uniform norm.As is well-known, X is a Banach space, and hence in particular a complete metricspace. Let Y = (C[0, 1], d) be the metric space equipped with the metric d which isdefined on C[0, 1] by

d(f, g) :=||f − g||∞

1 + ||f − g||∞,

where ||f−g||∞ = supx∈[0,1] |f(x)−g(x)|. It is readily verified that Y is a completemetric space. We claim that the set B = f ∈ C[0, 1] : f(x) > 0 is a boundedsubset of Y with the property that for each pair of points x, y in B, there existsan open ball U such that U ⊆ B and x, y ∈ U and yet B is not an open ball inY . Plainly, Y is bounded, and hence so is B. Now, suppose f, g ∈ B are arbitrary.Set M = supx∈[0,1] max

(f(x), g(x)

)and m = infx∈[0,1] min

(f(x), g(x)

). We have

0 < m,M ∈ R because f and g are continuous on the compact interval [0, 1].Define t : [0,+∞) → [0,+∞) by t(x) = x

1+x . First, we show that f, g ∈ Br(h) ⊆ B,

where r = t(M−m2

2 ) and h ∈ B is defined by h(x) = M+m2 for all x ∈ [0, 1]. To

see f ∈ Br(h), we need to show that d(f, h) < r. To this end, note first thatm ≤ f(x) ≤ M implying that 3m

4 < f(x) < M + m4 for all x ∈ [0, 1], from which,

192 2. SOLUTIONS

we easily obtain ||f − h||∞ <M−m

22 . But t is strictly increasing on [0,+∞). This

yields

d(f, h) = t(||f − h||∞

)< t(M − m

2

2)

= r,

as desired. Likewise, we see that g ∈ Br(h). Next, we need to show that Br(h) ⊆ B.To this end, let k ∈ Br(h) be arbitrary. It follows that

d(k, h) = t(||k − h||∞

)< r = t

(M − m2

2),

which yields

−M − m

2

2< k(x)− M +m

2<M − m

2

2,

for all x ∈ [0, 1] because t is strictly increasing on [0,+∞). This obtains

0 <3m4

< k(x) < M +m

4,

for all x ∈ [0, 1], implying that k ∈ B, which is what we want. It remains to provethat B is not a an open ball. Suppose to the contrary that B = Bt(s)(l), for somes > 0 and l ∈ B. Now, let k ∈ B = Bt(s)(l) be arbitrary. We see that

d(k, l) = t(||k − l||∞

)< t(s),

which obtains−s < k(x)− l(x) < s,

for all x ∈ [0, 1]. Consequently, supx∈[0,1] k(x) < s+M0, where M0 = supx∈[0,1] l(x).This is a contradiction because k ∈ B is arbitrary. Therefore, B is a bounded subsetof the complete metric space Y = (C[0, 1], d) with the property that for each pairof points f, g in B there exists an open ball Br(h) such that f, g ∈ Br(h) ⊆ B andyet B is not an open ball in Y , proving the claim.

Second solution: A simpler proof similar to that of first solution shows that theopen interval (0,+∞) of the complete metric space (R, d), where d(x, y) := |x−y|

1+|x−y|and |.| denotes the absolute value function, is a counterexample to (a) when Rn isreplaced by the complete metric space (R, d). We omit the details for the sake ofbrevity.

Third solution: Let X = A1, . . . , A5, where A1 = (1, 0, 0), A2 = (0, 1, 0),A3 = −A1, A4 = −A2, and A5 = (0, 0, 4

3 ). Note that if we use d and s to,respectively, denote the diameter and side length of the square whose vertices areA1, A2, A3, A4, we have d = 2 and s =

√2. Also note that s < A5Ai = 5

3 < d foreach i = 1, 2, 3, 4. It is readily checked that X with respect to the Euclidean metricof R3 is a complete metric space. Set B = A1, . . . , A4. It is now easily verified

that B is a bounded subset of the complete metric space X with the property thatfor each pair of points x, y in B, there exists an open ball Br(z), where r = s+A5A1

2 ,such that x, y ∈ Br(z) ⊆ B and yet B is not an open ball in X, which is what wewant.

2. First solution: It might be worth noting that the hypothesis that f is continu-ous is redundant. To prove the assertion, it is enough to assume that f is integrable

2.24. TWENTY FOURTH COMPETITION 193

on [a, b]. Without loss of generality, we may assume that∫ baf(t)dt > 0. For given

k ∈ (0, 1), define the continuous function gk : [a, b] −→ R by

gk(x) =∫ x

a

f(t)dt− k

∫ b

a

f(t)dt.

We have gk(a) = −k∫ baf(t)dt < 0 and gk(b) = (1− k)

∫ baf(t)dt > 0. Thus, by the

Intermediate Value Theorem, there is a ck ∈ (a, b) such that g(ck) = 0, provingthe assertion.

Second solution: Define g : [a, b] −→ R by

g(x) =

∫ xaf(t)dt∫ b

af(t)dt

.

Plainly, g is continuous on [a, b] and g(a) = 0 < 1 = g(b). Thus, by the IntermediateValue Theorem, there exists a ck ∈ (a, b) such that g(ck) = k, implying

∫ ck

af(t)dt−

k∫ baf(t)dt = 0, as desired.

3. (a) The inequality is known as the Frobenius Inequality. We prove the assertionover division rings. LetD be a division ring and A ∈Mm×n(D), B ∈Mn×p(D), C ∈Mp×q(D), where m,n, p, q ∈ N. View A (resp. B, C) as a linear transformationsacting on the left of Dn (resp. Dp, Dq), the right vector space of all n × 1 (resp.p× 1, q × 1) column vectors over D. Let A1 = A|B(Dp). We can write

dim kerA1|BC(Dq) ≤ dim kerA1 = dimB(Dp)− dimA1

(B(Dp)

),

which, in view of A1

(B(Dp)

)= AB(Dp), yields

dim kerA1|BC(Dq) ≤ r(B)− r(AB).

On the other hand,

dim kerA1|BC(Dq) = dimBC(Dq)− dimA1

(BC(Dq)

)= r(BC)− dimABC(Dq) = r(BC)− r(ABC).

Therefore,r(BC)− r(ABC) ≤ r(B)− r(AB),

implyingr(BC) + r(AB) ≤ r(ABC) + r(B),

as desired.

(b) To prove the assertion which is known as the Sylvester Inequality, just setA = A, B = In, and C = B in (a).

4. Let I and K be a two-sided ideal and a left ideal of the ring R, respectively.It is plain that KI ⊆ I ∩K. To prove I ∩K ⊆ KI, choose an arbitrary elementx ∈ I ∩K. It suffices to show that x ∈ KI. To this end, by the hypothesis, we have(I ∩K)2 = I ∩K because I ∩K is a left ideal of R. It follows that x ∈ (I ∩K)2.And hence there are n ∈ N and xi, yi ∈ I ∩K (1 ≤ i ≤ n) such that x =

∑ni=1 xiyi.

But xiyi ∈ KI for all 1 ≤ i ≤ n. Therefore, x ∈ KI, which is what we want,completing the proof.

194 2. SOLUTIONS

5. First solution: First, suppose that the number of wins of no two teams areequal. As there are n teams and any team has played against all of the other n− 1teams, we see that for each i = 0, . . . , n − 1, there is exactly one team, which wecall the ith team, whose wins is equal to i. It is obvious that the (n−1)st team haswon all the other n− 1 teams, the (n− 2)nd team losses to the (n− 1)st team andwins all the remaining n−2 teams, and so on and so forth. For the ith team, wherei ∈ 0, 1, . . . , n − 1, let’s call i to be the label of the team. Thus, for any threeteams, the team with the smallest label has lost to the other two teams. Therefore,there are no three teams A,B,C such that A wins B, B wins C, and C wins A.

Next, suppose that there are no such three teams. Define the relation ≥ on theteams as follows. For two teams A and B, we write A ≥ B if and only if A = Bor A wins B. Since there are no such three teams, in view of the hypothesis, it iseasily verified that the relation ≥ is a linear order on the set of the participatingteams. So if we put the teams in the decreasing order, say T1 > T2 > · · · > Tn,then Ti, the ith team, has won exactly n − i + 1 teams, namely, Ti+1 > · · · > Tn.Consequently, no two teams have scored the same number of wins, which is whatwe want.

Second solution: First, suppose that there are two teams A and B that havescored k wins, where k ∈ 0, . . . , n−1. Without loss of generality, we may assumethat A wins B. We claim that there is a team C such that A wins B, B wins C,and C wins A. To see this, as B has won k teams, one of them, say C, must havewon A, for otherwise A must have won k + 1 teams, the k teams lost to B plus Bitself, which is impossible. Thus, such team C exists, settling the implication.

Next, suppose that there are three teams A , B, and C such that A has wonB, B has won C, and C has won A. We prove the assertion by induction on n, thenumber of participating teams. If n = 3, the assertion is trivial. Assuming that theassertion holds for n − 1, to prove the assertion for n, argue by contradiction. Sono two teams have scored the same number of wins. Consequently, there is exactlyone team, say D, that has scored n− 1 wins, for n teams have participated in thegame. Now, D is not one of A, B, or C because D has won them all. Exclude Dand consider the game between the remaining n − 1 teams which include A,B,C.It follows from the induction hypothesis that there are two teams in the remainingn− 1 teams whose wins are equal. But the two teams have both lost to D. Thus,the two teams have scored the same number of wins in the original game with nteams, which is a contradiction. So the assertion follows.

6. We convert this problem into the following coin-flipping game of which theproblem is a special case. We then present a proof of the counterpart of the assertionfor the coin-flipping game, which we have taken from “Concrete Mathematics”, abook by Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. Here is thecounterpart of the problem.

Two persons, A and B, are playing the following coin-flipping game.First, A chooses a pattern of length ` (` ≥ 3) of heads and tails (for instance,

HTH, where H stands for heads and T for tails). Then B, who knows the patternchosen by A, chooses a different pattern of the same length of heads and tails. Thena fair coin is flipped until one of the patterns is first obtained, in which case thethe player whose pattern occurs first is to win the game. Show that no matter what

2.24. TWENTY FOURTH COMPETITION 195

choice is made by A, there is a choice for B so that the probability of winning thegame by B is greater than 1

2 .

Denote, respectively, by SA and SB the sum of A’s and B’s winning positions.Use N to denote the sum of the patterns each of which does not contain anyoccurrences of the patterns A and B chosen by A and B, respectively. For instance,if A chooses A = HTH and B chooses B = TTH, we have

SA =

HTH +HHTH + THTH +HHHTH + TTHTH + THHTH + · · · ,SB =

TTH +HTTH + TTTH +HHTTH + TTTTH + THTTH +HTTTH + · · · ,N =

1 +H + T +HH + TT + TH +HT +HHH +HHT + THT + TTT + · · · .Obviously, if we set H = T = 1

2 , the resulting values for SA and SB , respectively,become the probability that A and B wins the game. We have

1 +N (H + T ) = N + SA + SB ,

NA = SA∑k=1

A(`−k)δA(k),A(k)+ SB

∑k=1

A(`−k)δB(k),A(k),

NB = SA∑k=1

B(`−k)δA(k),B(k)+ SB

∑k=1

B(k−`)δB(k),B(k),

where δα,β denotes the Kronecker delta, A(k) and A(k) (resp. B(k) and B(k)) denotethe last and the first k characters of A (resp. B). To see the first equality, just notethat every term on the left side of it either ends with A, or B, or it does not endwith either of A and B meaning that the term belongs to N ; conversely, every termon the right of the first equality, is either empty or it belongs to NH or NT . Thesecond equality holds because every term on the left either completes a term of SAin such a way that the last k characters of A coincides with its first k charactersfor some 1 ≤ k ≤ `, or a term of SB in such a way that the last k characters ofB equals the first k characters of A for some 1 ≤ k ≤ `; and conversely becauseevery term on the right belongs to the left. Analogously, the third equality holds.As noted in the above, by setting H = T = 1

2 , we obtain the wining probabilitiesfor A and B, which we denote by P (A) and P (B), respectively. It follows from thefirst equality above that P (A) + P (B) = 1. Let

A:A =∑k=1

2k−1δA(k),A(k),B:A =

∑k=1

2k−1δB(k),A(k),

A:B =∑k=1

2k−1δA(k),B(k),B:B =

∑k=1

2k−1δB(k),B(k).

Using the second and third equalities, we can write

N = 2(P (A)(A:A) + P (B)(B:A)

),

N = 2(P (A)(A:B) + P (B)(B:B)

),

196 2. SOLUTIONS

which obtainsP (A)P (B)

=B:B − B:AA:A−A:B

.

We now claim that if A chooses the pattern A = τ1τ2 · · · τ`, then B has a bet-ter chance of winning the game by choosing B = τ ′2τ1 · · · τ`−1, where τ ′2 is theheads/tails opposite of τ2. It suffices to show that P (A) < P (B). Suppose thecontrary, implying that

B:B − B:A ≥ A:A−A:B. (∗)Note that A:A ≥ 2`−1 and B:B < 2`−1 + 2`−3, and B:A ≥ 2`−2, for A(`) = A(`),B(`) = B(`) but B(`−2) 6= B(`−2), and B(`) 6= A(`) but B(`−1) = B(`−1). It followsthat B:B − B:A < 2`−1 + 2`−3 − 2`−2. Since A(`) 6= B(`) and A(`−1) 6= B(`−1), weconclude A(`−2) = B(`−2), for otherwise A:B ≤ 2`−3, from which, we obtain

A:A−A:B ≥ 2`−1 − 2`−3 ≥ 2`−1 + 2`−3 − 2`−2 > B:B − B:A,which is impossible. Consequently, A(`−2) = B(`−2), yielding τ ′2 = τ3, τ1 = τ4, τ2 =τ5, τ3 = τ6, . . . , τ`−3 = τ`. But then, A:A ≥ 2`−1 +2`−4 + · · · , A:B ≤ 2`−3 +2`−6 +· · · , B:A ≥ 2`−2 + 2`−5 + · · · , and B:B < 2`−1 + 2`−4 + · · · , implying that

A:A−A:B ≥ (2`−1 − 2`−3) + (2`−4 − 2`−6) + · · ·> (2`−1 − 2`−2) + (2`−4 − 2`−5) + · · ·> B:B − B:A.

In other words, A:A−A:B > B:B−B:A, which is in contradiction with (∗). There-fore, P (A) < P (B), proving the assertion.

2.25. Twenty Fifth Competition

2.25.1. First Day. 1. As |G| = n and [G : Z(G)] = 4, we have

Z(G) $ CG(x) $ G,

for all x ∈ G \ Z(G), where CG(x) := g ∈ G : xg = gx denotes the centralizer ofthe element x in G. We can write

4 = [G : Z(G)] = [G : CG(x)][CG(x) : Z(G)],

which easily implies [G : CG(x)] = 2. It follows that the size of any conjugacy classof the elements of G\Z(G) is equal to 2, and hence 2

∣∣∣∣G\Z(G)∣∣. On the other hand,∣∣G\Z(G)

∣∣ = 3n4 . Thus, 2

∣∣ 3n4 , implying 8

∣∣3n, which, in turn, yields 8∣∣n. For a given

n ∈ N satisfying 8∣∣n, set G := Q8 ×Z n

8, where Q8 denotes the quaternionic group

with 8 elements. It is obvious that Z(G) = −1, 1 × Z n8, yielding [G : Z(G)] = 4,

as desired.

2. From T 2 = T , we see that V = kerT ⊕ imT . Now, let α ∈ V be arbitrary. Itfollows that there are β ∈ kerT and γ ∈ V such that α = β + Tγ. In view of thehypotheses, we can write

(T + S)α = (T + S)(β + Tγ) = Tβ + T 2γ + Sβ + STγ = Tγ + Sβ.

Since β ∈ kerT , by the hypothesis, there is a δ ∈ V such that β = Sδ. So we canwrite

(T + S)α = Tγ + Sβ = Tγ + S2δ = Tγ + Sδ = β + Tγ = α.

2.25. TWENTY FIFTH COMPETITION 197

Thus, T + S is the identity transformation, for α ∈ V was arbitrary, finishing theproof.

3. By the Mean Value Theorem, there is a c ∈ (y, y + 1) such that

f(y + 1)− f(y) = f ′(c).

As f ′′(t) < 0 for all t ∈ R, f ′ is strictly decreasing on R. Thus, f ′(y) > f ′(c)because y < c. So we can write

f(y + 1)− f(y) = f ′(c) < f ′(y) < f(y + 1)− x,

yielding x < f(y), as desired.

4. Define g : (a, b) −→ R by g(x) = ln(1 + f2(x)

)− x. We have

g′(x) =2f(x)f ′(x)1 + f2(x)

− 1 =2f(x)f ′(x)− f2(x)− 1

1 + f2(x)≥ 0,

for all x ∈ (a, b). Thus, g is nondecreasing on (a, b), implying that−a = limx→a+ g(x) ≤limy→b− g(y) = 1− b. Therefore, −a ≤ 1− b, yielding b− a ≤ 1. As for an examplefor which b− a = 1, just let a = 0 = b− 1 and f(x) =

√ex − 1 on (0, 1).

5. Note first that the game stops exactly when the 7th numbered one marble isdrawn from a box. To see this, it is obvious that if the 7th numbered one marbleis drawn from a box, the the game stops because one needs to draw a marble frombox one which is empty. Conversely, if the game stops when a marble numbered iis drawn from a box, then i = 1. Because otherwise the box numbered i must beempty for some i ≥ 2, which means seven marbles numbered i must have alreadybeen drawn from the ith box, implying that there are 8 marbles numbered i, acontradiction.

Now consider the following extended game. Assume that, after a stop in theoriginal game, the game is continued by choosing the box which is not empty yetand whose number is minimal among all nonempty boxes, and that the game iscontinued in this manner until all the marbles are drawn from all boxes. In thisextended game, to each permutation of the 49 marbles, there corresponds a round ofthe game. Conversely, to each round of the game, there corresponds a permutationof the 49 marbles. Also, a round of the original game continues until all the marblesare drawn from all boxes only when the number on the last drawn marble, i.e., the49th drawn marble, reads one, in which case the original game and the extendedgame are the same. It is now obvious that the probability of the event that all themarbles are drawn from all boxes in the original game is equal to that of that sameevent in the extended game which is equal to

48!6!7!6

49!7!7

=17,

which is what we want.

6. Let 0 < c ≤ b ≤ a with a, b, c ∈ N and a + b + c = n for some n ∈ N, be theside lengths of a desired triangle. The following gives an algorithm for finding apartition of the number n into the summands 2, 3, and 4 in which the summand 3appears at least once. As the following algorithm is reversible, the assertion follows.We have a < b+ c, e.g., 7 < 6 + 3. We explain the algorithm in three stages.

198 2. SOLUTIONS

Stage (i): To create summands of 3, subtract one from the three side lengths,i.e., a, b, c, one at a time and continue this for c+ b− a times so that the inequalitya < b+ c becomes an equality, e.g.,

7 < 6 + 3 → 6 < 5 + 2 → 5 = 4 + 1.

Stage (ii): To create summands of 4, in the equality obtained in stage (i),subtract two units and one unit, one at a time, from the left hand side and righthand side of the equality, respectively. Continue in this manner for a− b times tilla zero appears in the equality , e.g.,

5 = 4 + 1 → 3 = 3 + 0.

Stage (iii): To create summands of 2, in the last equality obtained in stage (ii),subtract one unit at a time from each nonzero number of the equality and continuein this way for b− c times till all numbers become zero, e.g.,

3 = 3 + 0 → 2 = 2 + 0 → 1 = 1 + 0 → 0 = 0 + 0.

So, the corresponding partition for a triangle with side lengths 0 < a < b < cand a + b + c = n is given by n = 4(a − b) + 3(b + c − a) + 2(b − c), e.g., 16 =4 + 3 + 3 + 2 + 2 + 2.

2.25.2. Second Day. 1. The assertion is a quick consequence of the followingproposition.

Let R be a unital ring without zero divisors, i.e., ∀a, b ∈ R : ab = 0 ⇒ a = 0or b = 0, and U(R) denote the multiplicative group of the units of R. Then, everyfinite abelian subgroup of U(R) is cyclic.

We present two proofs for this proposition.First proof. Let G be an abelian subgroup of U(R) and Z denote the prime

ring of R, i.e., Z = k1 : k ∈ Z. It follows from the hypothesis that

Z[G] := n∑i=1

kigi : n ∈ N, ki ∈ Z, gi ∈ G (1 ≤ i ≤ n)

is an integral domain and that G is a subgroup of the multiplicative group of theunits of Z[G]. So without loss of generality, we may assume that R is an integraldomain. To prove the assertion, we need the following lemma.

Lemma. (i) Let G be a group and g ∈ G with ord(g) = n1 · · ·nk, wherek ∈ N and nj’s (1 ≤ j ≤ k) are pairwise relatively prime. Then, there existunique g1, . . . , gk ∈ G such that g = g1 · · · gk, gigj = gjgi, and ord(gi) = ni foreach i, j = 1, . . . , k. Conversely, if for g ∈ G, there are g1, . . . , gk ∈ G satisfyingg = g1 · · · gk, gigj = gjgi and ord(gi) = ni for each i, j = 1, . . . , k such that nj’s(1 ≤ j ≤ k) are pairwise relatively prime, then ord(g) = n1 · · ·nk.

(ii) Let G be a group and g1, . . . , gk ∈ G be of finite order such that gigj = gjgifor each i, j = 1, . . . , k. Then there is a g ∈ G such that ord(g) = lcm

(ord(g1), . . . , ord(gk)

).

Proof. (i) First, recall that if g ∈ G is such that ord(g) = n for some n ∈ N,then ord(gi) = n

gcd(i,n) for all i ∈ Z. Now, set mi =∏ki6=j=1 nj . As nj ’s (1 ≤ j ≤ k)

are pairwise relatively prime, we see that gcd(m1, . . . ,mk) = 1 and hence there arecj ’s (1 ≤ j ≤ k) in Z such that

∑kj=1 cjmj = 1. We note that gcd(ci, ni) = 1

for each i = 1, . . . , k. To see this, it suffices to show that no prime p divides

2.25. TWENTY FIFTH COMPETITION 199

gcd(ci, ni). Suppose to the contrary that there is a prime p that divides gcd(ci, ni)for some 1 ≤ i ≤ k. It follows that p

∣∣ci and p∣∣ni and hence p

∣∣ci and p∣∣mj for all

j ∈ 1, . . . , k \ i. Consequently, p∣∣∑k

j=1 cjmj = 1, which is impossible. Thus,gcd(ci, ni) = 1 for each i = 1, . . . , k. Now, letting gi = gcimi , we can write

ord(gi) = ord(gcimi) =ord(g)

gcd(cimi, ord(g))=

minigcd(cimi,mini)

=ni

gcd(ci, ni)= ni.

Also, we can write

g1 · · · gk = gc1m1 · · · gckmk = gPk

j=1 cjmj = g.

Finally, it is obvious that gj ’s (1 ≤ j ≤ k) commute. To see that gj ’s (1 ≤ j ≤ k) areunique, suppose that there are g′1, . . . , g

′k ∈ G such that g = g′1 · · · g′k, g′ig′j = g′jg

′i,

and ord(g′i) = ni for each i, j = 1, . . . , k. We can write

gcjmj = (g′1 · · · g′k)cjmj = g′cjmj

1 · · · g′cjmj

k .

But g′mj

i = e whenever i 6= j because ord(g′i) = ni and mj =∏kj 6=i=1 ni. Thus,

gcjmj = g′cjmj

j = g′1−Pk

j 6=i=1 cimi

j = g′j ,

for all 1 ≤ j ≤ k, as desired.As for the converse, suppose that ord(g) = m. We need to show that m =

n1 · · ·nk. Note first that m∣∣n1 · · ·nk because gn1···nk = (g1 · · · gk)n1···nk = e. It

thus suffices to show that n1 · · ·nk∣∣m. To this end, letting 1 ≤ i ≤ k be arbitrary

but fixed, we can write

gm = e =⇒ (g1 · · · gk)m = e =⇒ gmi =k∏

i6=j=1

g−mj .

As gnii = e, we get (gmi )ni = e, from which, we obtain ord(gmi )

∣∣ni. On the otherhand,

(gmi )Qk

i6=j=1 nj =( k∏i6=l=1

g−ml)Qk

i6=j=1 nj =k∏

i6=l=1

(gQk

i6=j=1 nj

l

)−m = e,

implying that ord(gmi )∣∣∏k

i6=j=1 nj . Thus, ord(gmi )∣∣gcd

(ni,∏ki6=j=1 nj

)= 1, which

yields ord(gmi ) = 1. This obtains gmi = e for all 1 ≤ i ≤ k. Hence, ni∣∣m for all

1 ≤ i ≤ k. It thus follows that n1 · · ·nk∣∣m because nj ’s (1 ≤ j ≤ k) are pairwise

relatively prime. Therefore, m = ord(g) = n1 · · ·nk, which is what we want.

(ii) It is plain that there exists an l ∈ N and primes p1, . . . , pl such that

ord(gi) = pmi11 · · · pmil

l ,

where mij ’s are nonnegative integers (1 ≤ i ≤ k, 1 ≤ j ≤ l). For 1 ≤ i ≤ k, we seefrom (i) that there are gi1, . . . , gil ∈ G such that gi = gi1 · · · gil, that gij ’s (1 ≤ i ≤k, 1 ≤ j ≤ l) all commute, and that ord(gij) = p

mij

j . It is obvious that for each

200 2. SOLUTIONS

1 ≤ j ≤ l, there is an 1 ≤ ij ≤ k such that ord(gijj) = maxord(g1j), . . . , ord(gkj)

.

Let g′ =∏lj=1 gijj . It thus follows from (i) that

ord(g′) =l∏

j=1

pmax

m1j ,...,mkj

j = lcm

(ord(g1), . . . , ord(gk)

),

which is what we want.

Note that U(R), and hence G, is commutative. Let a ∈ G be such that itsorder, say m, is maximal among the elements of G. Let b be an arbitrary elementof G. Since G is commutative, by part (ii) of the above lemma, there is a c ∈ Gwhose order is equal to the least common multiple of ord(a) and ord(b). It followsthat ord(c) = ord(a) because ord(a) is maximal. Thus, ord(b)

∣∣ord(c) = ord(a) = m,yielding bm = 1. In other words, every element of G is a root of f = xm−1 ∈ R[x].As R is an integral domain, f has at most m roots, and hence |G| ≤ m. Therefore,|G| = m = ord(a). That is, G is cyclic, as desired.

Second proof. Just as we saw in the first proof, we may assume that R is anintegral domain. Consequently, the equation xn = 1 has at most n solutions in Gfor all n ∈ N. It thus follows from the lemma presented in Solution 1 of 2.2.2 thatG is cyclic, as desired.

2. Let q be a prime factor of 2p+1 which is different from 3. We have q∣∣2p+1, and

hence q∣∣22p − 1, yielding 22p q

≡ 1. Use t = ordq(2) to denote the order of 2 moduloq. It follows that t divides 2p. By showing that t 6= 1, 2, p, we conclude that t = 2p.

Firstly, t 6= 1 because 21q

6≡ 1. Secondly, t 6= 2 because 22q

6≡ 1, for q 6= 3. Thirdly,t 6= p because otherwise 2p

q≡ 1, which, in view of q

∣∣2p + 1, obtains q∣∣2, which is

impossible, for q is odd. Thus, t = 2p. Now, since t = ordq(2)∣∣ |Z∗q | = q − 1, where

Z∗q = Zq \ 0, we have 2p∣∣q − 1 or q − 1 = 2kp for some k ∈ N, which is what we

want.

3. It is plain that f is a one-to-one function from X onto X. Thus, f is invertible.Use f−1 to denote the inverse of f . It follows from the hypothesis that

d(f−1(x), f−1(y)

)≤ d(x, y),

for all x, y ∈ X. Thus, the function f−1 : X −→ X is a Lipschitz function, andhence it is (uniformly) continuous. Now, since f−1 is continuous and X is compact,we see that f−1 is a closed map, implying that f = (f−1)−1 is continuous, whichis what we want.

4. Let α = f(1) and β = f(i). By the hypothesis, |f(z)| = |z|, |f(z)−α| = |z− 1|,and |f(z)− β| = |z − i| for all z ∈ C. In particular, by substituting z = 1, i in theabove equalities, we obtain

|α| = |β| = 1, |α− β| =√

2.

We can write

α2 + β2 = α2ββ + β2αα = αβ(αβ + βα)= αβ

(αα+ ββ − (α− β)(α− β)

)= αβ

(|α|2 + |β|2 − |α− β|2

)= αβ(1 + 1− 2) = 0,

2.25. TWENTY FIFTH COMPETITION 201

yielding β = εα, where ε = ±i. Simplify |f(z)−α| = |z−1| and |f(z)−β| = |z− i|to, respectively, obtain αf(z) + αf(z) = z + z and αf(z)− αf(z) = −εiz + εiz forall z ∈ C. Adding up these two equalities, we obtain

2αf(z) = (1− εi)z + (1 + εi)z,

which, in view of α = f(1) and ε = ±i, yields f(z) = f(1)z for all z ∈ C orf(z) = f(1)z for all z ∈ C, as desired.

5. Let n = 2k− 1 for some k ∈ N and Kn denote the n× n matrix whose elementsare all 1. For a given n× n matrix A with entries in −1, 1, let’s call

C(A) :=n∑i=1

ai +n∑j=1

bj

the content of the matrix A, where ai’s and bj ’s are as in the statement of theproblem. We have

C(Kn) = n+ n = 4k − 2.

It is plain that any n× n matrix A, whose entries are in −1, 1, can be obtainedfrom Kn within a finite number of steps so that at each step a fixed entry, say,aij = 1, of Kn is changed to −aij = −1. After each step, the number two will beadded to or will be subtracted from each of the two sums in the definition of thecontent. Thus, the content remains intact or it will be increased or decreased byfour at each step. Therefore, the content of any n × n matrix A will always be ofthe form 4k′ − 2, for some k′ ∈ Z, which is never zero. This proves the first part ofthe assertion. As for the case when n is even, the assertion does not hold becausefor the matrix A defined by

A =(Km −Km

Km Km

),

where m = n2 , we have C(A) = 0.

6. For a member a ∈ S, use A(a) to denote the set of all of the acquaintancesof a. It follows from the hypothesis that A(a) 6= A(b) whenever a 6= b. Proceedby contradiction. Let x ∈ S be a member for which A(x) is maximal among allmembers of S. Since S is ideal but S \ x is not ideal, there are two distinctmembers p, q ∈ S \ x such that

A(p) = A(q) \ x, x ∈ A(q). (∗)

Likewise, since S is ideal but S \ p is not ideal, there are two distinct membersr, s ∈ S \ p such that

A(r) = A(s) \ p, p ∈ A(s). (∗∗)

We claim that r = x. By (∗∗), p and s are acquainted. This, in view of (∗), impliesthat s 6= x and that s and q are acquainted. But since, by (∗∗), r is acquaintedwith all of the acquaintances of s but p, we see that r and q are acquainted. Thisimplies r = x because r is not acquainted with p. Thus, A(x) = A(s) \ p. Thatis, the number of the acquaintances of s is greater than those of x by one, which isa contradiction, proving the assertion.

202 2. SOLUTIONS

2.26. Twenty Sixth Competition

2.26.1. First Day. 1. First solution: Define g : [a, b] −→ R by g(x) =f(x)+x− (a+ b). The function g is continuous on [a, b] and differentiable on (a, b).Moreover, g(a) = a − b < 0 and g(b) = b − a > 0. So, by the Intermediate ValueTheorem, there is a c ∈ (a, b) such that g(c) = 0, which means f(c) = a + b − c.It now follows from the Mean Value Theorem that there are x1, x2 ∈ (a, b) withx1 6= x2 such that

f ′(x1) =f(a)− f(c)

a− c=c− b

a− c, f ′(x2) =

f(c)− f(b)c− b

=a− c

c− b,

yielding f ′(x1)f ′(x2) = 1, proving the assertion.

Second solution: Let g = f f : [a, b] −→ R. We have g(a) = a and g(b) = b. Bythe Mean Value Theorem, there is a c ∈ (a, b) such that

1 =g(b)− g(a)b− a

= f ′(c)f ′(f(c)),

yielding f ′(c)f ′(f(c)) = 1. There are two cases to consider. If f(c) 6= c, then theassertion is proved by letting x1 = c and x2 = f(c). If f(c) = c, then using theMean Value Theorem for f on the intervals [a, c] and [c, b], respectively, we see thatthere are x1 ∈ (a, c) and x2 ∈ (c, b) such that

f ′(x1) =f(c)− f(a)

c− a= 1, f ′(x2) =

f(b)− f(c)b− c

= 1,

yielding f ′(x1) = f ′(x2) = 1, and hence f ′(x1)f ′(x2) = 1, proving the assertion.

2. Suppose to the contrary that f has no zero in U . Define g : U −→ C byg(z) = 1

f(z) . It follows that g is analytic on U . Moreover, g(0) = 1 and |g(z)| ≤ 12 on

|z| = 1. This contradicts the Maximum Modulus Principle, proving the assertion.

3. Since gcd(p, 2p−2) = 1, it suffices to prove the congruence modulo p and modulo2p − 2. Note that, by Fermat’s Little Theorem, (2p − 2)p−1 p

≡ 1. This, togetherwith pp−1 p

≡ 0, yields pp−1 + (2p− 2)p−1 p≡ 1. Also, note that

pp−1 − 1 = (p− 1)(pp−2 + · · ·+ p+ 1) = 2k(p− 1),

which implies pp−1 2p−2≡ 1. This, together with (2p − 2)p−1 2p−2

≡ 0, yields pp−1 +

(2p− 2)p−1 2p−2≡ 1. Therefore, pp−1 + (2p− 2)p−1

p(2p−2)≡ 1, as desired.

4. We need to show that for all x, y ∈ R, x divides y or y divides x. To provethis by contradiction, suppose that there are x, y ∈ R such that x - y and y - x. Itfollows from the hypothesis that there is a nonzero z ∈ R such that Rx+Ry = Rz.Thus, there are r, s ∈ R such that x = rz and y = sz. The elements r, s ∈ R are notunits because otherwise x

∣∣y or y∣∣x, which is not possible. Therefore, the elements

r and s belong to M , the only maximal ideal of R. From Rx + Ry = Rz, we seethat there are a, b ∈ R such that z = ax + by. This, together with x = rz andy = sz, yields z = (ar + bs)z, implying

(1 − (ar + bs)

)z = 0. On the other hand,

1− (ar+ bs) is a unit in R, for otherwise 1− (ar+ bs) ∈M , yielding 1 ∈M , whichis a contradiction. Therefore, z = 0, a contradiction. So the assertion follows.

2.26. TWENTY SIXTH COMPETITION 203

5. For the sake of brevity, we say a k-cover for the set S to mean a k-element coverfor the set S. It is plain that any element of a minimal cover of S is nonempty andthat it includes exactly one element of S by itself. Therefore, a minimal (n − 1)-cover for the set S is either a partition of the set S, or there exists exactly oneelement x ∈ S which belongs to two or more elements of the (n− 1)-cover. In thefirst case, the number of such (n − 1)-covers of S, i.e., such partitions of S into(n − 1) subsets, is equal to

(n2

). In the latter case, there are

(n1

)ways to choose

x ∈ S and 2n−1−(n1

)= 2n−1−n ways to put x in the elements of the (n−1)-cover

so that x is put in at least two elements of the cover. So, we must have

M(n, n− 1) =(n

2

)+(n

1

)(2n−1 − n

)=n

2(2n − n− 1),

which is what we want.

6. Let g(R,S) =∑

f(Y ) : Y ∈ Xm, R ⊆ Y ⊆ S. We prove the following more

general proposition. Under the hypotheses of the problem, g(R,S) = 0 wheneverR ⊆ S and

∣∣S\R∣∣ ≥ k. The assertion is a consequence of the above proposition. Tosee this, note that for every T ∈ Xm, there is an S ∈ Xm+k such that T ⊆ S.This implies f(T ) = g(S, T ) = 0, as desired. We prove the above proposition byinduction on |R|. If |R| = 0, assuming that |S| ≥ k, we have(

|S|k

)g(∅, S) =

∑T∈Xk,T⊆S

g(∅, T ) = 0,

for, by the hypothesis, g(∅, T ) =∑Y ∈Xm,Y⊆T f(Y ) = 0 for all T ∈ Xk such

that T ⊆ S. On the other hand, the following recurrence equation holds.

g(R,S) = g(R \ α, S

)− g(R \ α, S \ α

),

for all α ∈ R. Therefore, the assertion follows by induction.

2.26.2. Second Day. 1. Let a ∈ [0, 1] \ Q be arbitrary. We show that f iscontinuous at a. To this end, for a given ε > 0, choose k ∈ N such that

∑+∞i=k

12i < ε

and letδ = min

∣∣a− r1∣∣, . . . , ∣∣a− rk

∣∣.Note that δ > 0. If a− δ < x ≤ a, we can write∣∣f(x)− f(a)

∣∣ = ∑i∈n∈N:x<rn≤a

12i≤

+∞∑i=k+1

12i< ε.

Analogously, if a ≤ x < a+ δ, then∣∣f(x)− f(a)∣∣ = ∑

i∈n∈N:a<rn≤x

12i≤

+∞∑i=k+1

12i< ε.

That is, limx→a

(f(x) − f(a)

)= 0. Thus, f is continuous at a, which is what we

want.

2. It easily follows from the hypothesis that f is one-to-one. Since f is continuous,we conclude that f is strictly increasing or decreasing on R. But f is not strictlydecreasing, for otherwise f f would be strictly increasing and hence f f f = I

204 2. SOLUTIONS

would be strictly decreasing, which is impossible. Thus, f is strictly increasing onR. Now, to prove the assertion by contradiction, suppose that there is an a ∈ Rsuch that f(a) 6= a. Two cases to consider. If a < f(a), we see that f(a) < f(f(a)),yielding f(a) < f

(f(a)

)< f

(f(f(a)

))= a, a contradiction. Likewise, if a > f(a),

we obtain a contradiction. So, the assertion follows by way of contradiction.

3. Suppose by way of contradiction that there is an x ∈ H ∩ K \ e, where edenotes the identity element of G. Since H is abelian and Z(G) = e, we haveH ⊆ CG(x) & G, where CG(x) denotes the centralizer of the element x in G. Thus,H = CG(x) because H is maximal. Likewise, we see that K = CG(x). Therefore,H = K, which is a contradiction. So the assertion follows.

4. We prove the assertion by showing that An = λI, where I denotes the identitymatrix of size n. Let Fn denote the n-dimensional vector space of all n× 1 columnvectors with entries in F . By the hypothesis, the vectors x,Ax, . . . , An−1x arelinearly independent and hence they form a basis for Fn. We can write

An(Aix) = Ai(Anx) = λAix,

implying(An − λI)(Aix) = 0,

for all 0 ≤ i ≤ n − 1. This yields An − λI = 0, and hence An = λI, becausex,Ax, . . . , An−1x forms a basis for Fn.

5. Let (a1, . . . , a6) and (b1, . . . , b6) be the sequences of the numbers appearing ontwo extended dice. Let f(x) = xa1 + · · ·+xa6 and g(x) = xb1 + · · ·+xb6 . It followsfrom the hypothesis that

f(x)g(x) = (x1 + · · ·+ x6)2 = x2(1 + x)2(1 + x+ x2)2(1− x+ x2)2.

As f(0) = g(0) = 0, the factor x must appear in the factorizations of both of fand g. Since f(1) = g(1) = 6, the factors 1 + x and 1 + x+ x2 must appear in thefactorizations of both of f and g as well. Therefore, the only choices which do notlead to two ordinary dice are as follows

f(x) = x(1 + x)(1 + x+ x2) = x+ 2x2 + 2x3 + x4,

g(x) = x(1 + x)(1 + x+ x2)(1− x+ x2)2 = x+ x3 + x4 + x5 + x6 + x8.

That is, aside from the ordinary dice, the sequences of the numbers on the desiredextended dice are (1, 2, 2, 3, 3, 4) and (1, 3, 4, 5, 6, 8).

6. In view of the definition of a balanced matrix and the hypothesis that thebalanceness remains intact under interchanging the rows, it follows that for allj1 < j2 and i1 < i < i2, we have

ci1j1 + cij2 ≤ ci1j2 + cij1cij1 + ci1j2 ≤ cij2 + ci1j1

⇐⇒ ci1j1 + cij2 = cij2 + cij1

⇐⇒ ci1j1 + cij1 = ci1j2 + cij2 .

That is, the difference of two rows i and i′ is a constant vector whenever i1 ≤ i ≤i′ ≤ i2. Conversely, it follows from the above relations that if the difference of tworows is a constant vector, then the balanceness remains intact under interchangingthe two rows.

2.27. TWENTY SEVENTH COMPETITION 205

2.27. Twenty Seventh Competition

2.27.1. First Day. 1. First solution: See Problem 2 of 2.1.1.

Second solution: Since all norms on the n2-dimensional normed space Mn(R) areequivalent, it suffices to show that GLn(R) is open and disconnected with respect tothe operator norm ofMn(R). To this end, let A ∈ GLn(R) be arbitrary. By showingthat B ∈ GLn(R) whenever ||B −A|| < ||A−1||−1, where ||.|| denotes the operatornorm of Mn(R), we prove that GLn(R) is open. It is plain that B ∈ GLn(R) if andonly if BA−1 ∈ GLn(R). But ||I −BA−1|| ≤ ||A−B|| ||A−1|| < 1. Thus,

(BA−1)−1 =(I − (I −BA−1)

)−1 =+∞∑i=1

(I −BA−1)n.

Consequently, BA−1 ∈ GLn(R), implying B ∈ GLn(R), as desired. To show thatGLn(R) is disconnected, suppose to the contrary that it is connected. Since GLn(R)is open and connected, it is path connected. Choose A,B ∈ GLn(R) with det(A) =−det(B) = 1. It follows that there is a continuous function φ : [0, 1] −→ GLn(R)with φ(0) = A and φ(1) = B. Since det : Mn(R) −→ R is continuous, so isdet(φ) : [0, 1] −→ R. But det

(φ(0)

)= det(A) = 1 and det

(φ(1)

)= det(B) = −1.

It thus follows from the Intermediate Value Theorem that there is a c ∈ (0, 1)such that det

(φ(c)

)= 0. This implies φ(c) /∈ GLn(R), which is a contradiction.

Therefore, GLn(R) is disconnected, proving the assertion.

2. Let 0 ≤ θ < π be the angle of inclination of the line L, i.e., θ is the angle betweenL and the x-axis. Define the function g : C −→ C by g(z) =

(f(eiθz+A)−A

)e−iθ.

It is plain that g is analytic on C, g takes the x-axis to the x-axis and the y-axisto the y-axis, and that g(0) = 0. We can write

g(z) =+∞∑n=1

anzn,

where an = g(n)(0)n! . Set

g(z) = g(z) =+∞∑n=1

anzn,

Since g maps the x-axis into the x-axis, we obtain

(g − g)(z) =+∞∑n=1

(an − an)zn = 0,

whenever z ∈ R. It thus follows from the Uniqueness Theorem for analytic functionsthat g = g because the zeros of the analytic function g − g has a limit point in C.Therefore, an = an which, in turn, implies an ∈ R for all n ∈ N. Define h : C −→ Cby

h(z) = ig(iz) =+∞∑n=1

anin+1zn.

Since g maps the x-axis into the x-axis and the y-axis into the y-axis, it followsthat the analytic function h maps the x-axis into the x-axis. Hence, just as we saw

206 2. SOLUTIONS

in the above, we see that anin+1 ∈ R for all n ∈ N. This yields a2n = 0 for alln ∈ N. Therefore,

g(z) =+∞∑n=1

a2n−1z2n−1,

where a2n−1 ∈ R for all n ∈ N. Obviously, g(−z) = −g(z) for all z ∈ C. That is, ifz1, z2 ∈ C are symmetric with respect to 0, then so are g(z1) and g(z2) with respectto 0. Now, suppose z1 and z2 are symmetric with respect to A. So, z1 + z2 = 2A.We can write

f(z2) = eiθg(e−iθ(z2 −A)

)+A

= eiθg(− e−iθ(z1 −A)

)+A = −eiθg

(e−iθ(z1 −A)

)+A

= −f(z1) + 2A,

implying f(z1)+ f(z2) = 2A. Thus, f(z1) and f(z2) are symmetric with respect toA, finishing the proof.

3. Note first that sinh : R −→ R defined by sinh(x) = ex−e−x

2 is one-to-one andonto. Thus, for each n ∈ N, there is a unique αn ∈ R such that an = sinh(αn). Wecan write

sinh(αn+1) = sinh(αn)√

1 + sinh2(αn−1) + sinh(αn−1)√

1 + sinh2(αn)

= sinh(αn) cosh(αn−1) + sinh(αn−1) cosh(αn)= sinh(αn−1 + αn),

implying αn+1 = αn−1 + αn for each n ∈ N. The characteristic polynomial ofthis recurrence equation is x2 − x − 1 = 0, yielding x = 1±

√5

2 . Thus, αn =A(

1+√

52

)n + B(

1−√

52

)n for some A,B ∈ R, where n ∈ N ∪ 0. To determine Aand B, we have

0 = sinh−1(0) = α0 = A+B,

sinh−1(b) = A(1 +

√5

2)

+B(1−√5

2),

implying A = −B = sinh−1(b)√5

. So we have

sinh−1(an) = αn =sinh−1(b)√

5

((1 +√

52

)n − (1−√52

)n),

yielding

an = sinh

(sinh−1(b)√

5

((1 +√

52

)n − (1−√52

)n)),

for all n ∈ N, which is what we want.

4. To prove the assertion by contradiction, suppose that there is an i0 ∈ I such thati0 /∈ Ai0 . It follows that i : i /∈ Ai 6= ∅, from which, in view of the hypothesis, wesee that i : i /∈ Ai = Ai1 for some i1 ∈ I. Two cases to consider. If i1 ∈ Ai1 , theni1 /∈ i : i /∈ Ai = Ai1 , a contradiction. If i1 /∈ Ai1 , then i1 ∈ i : i /∈ Ai = Ai1 , acontradiction again. Therefore, i ∈ Ai for all i ∈ I, as desired.

2.27. TWENTY SEVENTH COMPETITION 207

5. It is easily verified that (AB)2 = 9(AB). We can write

(BA)3 = B(AB)2A = 9B(AB)A = 9(BA)2,

yielding (BA)2(BA − 9I2) = 0. If we use f to denote the minimal polynomial ofBA, we see that f divides x2(x − 9). As the degree of f is less than or equal to2, there are four cases to consider. If f = x, then BA = 0, yielding AB = 0, forA(BA)B = (AB)2 = 9(AB), which is impossible. If f = x2, then (BA)2 = 0,implying tr(AB) = tr(BA) = 0, which is again impossible. If f = x(x − 9), thentr(AB) = tr(BA) = 9, which is impossible. If f = x− 9, we obtain BA− 9I2 = 0.Thus, BA = 9I2, which is what we want.

6. We first show that H is normal in G. To prove this by contradiction, supposethat there exist g ∈ G and h ∈ H such that g−1hg /∈ H. Since g−1hg ∈ G \ H,it follows from the hypothesis that there is a u ∈ H such that g(g−1hg)g−1 =u−1(g−1hg)u, from which, we obtain g−1hg = uhu−1 ∈ H, a contradiction. Thus,H is normal in G. To prove that G/H is abelian, it suffices to show that g1g2H =g2g1H, or equivalently, g−1

2 g−11 g2g1 ∈ H for all g1, g2 ∈ G. If g2 ∈ H, we see

that g−12 (g−1

1 g2g1) ∈ H because H is normal in G. So we may, with no loss ofgenerality, assume that g2 /∈ H. Hence, for g1 ∈ G, there is a u ∈ H such thatg−11 g2g1 = u−1g2u. This implies g−1

2 g−11 g2g1 = (g−1

2 u−1g2)u ∈ H, for H is normalin G. It thus follows that G/H is abelian, as desired.

2.27.2. Second Day. 1. First solution: Since f ′(a) > 0, f ′(b) > 0, andf(a) = f(b) = 0, there is a δ > 0 such that a + δ < b − δ, f(x) > 0 whenevera < x < a + δ, and that f(x) < 0 whenever b − δ < x < b. It follows thatf(a+ δ) ≥ 0 and f(b− δ) ≤ 0. So by the Intermediate Value Theorem, there is ac ∈ [a+ δ, b− δ] such that f(c) = 0. We have f(a) = f(c) = f(b) = 0. Applying theRolle’s Theorem, we obtain c1 ∈ (a, c) and c2 ∈ (c, b) such that f ′(c1) = f ′(c2) = 0,as desired.

Second solution: Recall that by Darboux’s Theorem, the derivative functionhas the intermediate value property. If there is a c ∈ (a, b) such that f ′(c) < 0,then f ′(a)f(c) < 0 and f ′(c)f ′(b) < 0. Hence, by Darboux’s Theorem, there arec1 ∈ (a, c) and c2 ∈ (c, b) such that f ′(c1) = f ′(c2) = 0, proving the assertion inthis case. If f ′(x) ≥ 0 for all x ∈ (a, b), then f is increasing on [a, b]. This implies0 = f(a) ≤ f(x) ≤ f(b) = 0, whence f(x) = 0 for all x ∈ [a, b], in which case theassertion is trivial. So the proof is complete.

2. First we need to recall the following lemma from theory of metric spaces. Alsorecall that a subset of a metric space is said to be perfect if it is closed and has noisolated points.

Lemma. Every nonempty perfect subset of a complete metric space is uncount-able.

Proof. Let P be a nonempty perfect subset of a complete metric space X.As P is closed, P equipped with the metric X induces on P forms a completemetric space itself. Now to prove the assertion by contradiction, suppose that Pis countable so that P = xi+∞i=1 , where xi ∈ X for all i ∈ N. As P is perfect,each singleton xi (i ∈ N) is a nowhere dense subset of P , implying that P is a

208 2. SOLUTIONS

countable union of nowhere dense subsets, namely xi’s where i ∈ N. This is incontradiction with the Baire Category Theorem which asserts that every completemetric space is of second category in itself, i.e., the space cannot be written as acountable union of nowhere dense subsets of it. Thus, P is uncountable, which iswhat we want.

To prove the assertion by contradiction, suppose that [0, 1] is written as aunion of mutually disjoint closed intervals each of which having a positive lengthless than one. There are a countable number of the intervals participating in theunion because they are mutually disjoint. So we may assume that

[0, 1] =+∞⋃n=1

In,

where In = [an, bn] for some an, bn ∈ R with 0 < bn − an < 1 and Im ∩ In = ∅whenever m,n ∈ N and m 6= n. Set

P =an+∞n=1

∪bn+∞n=1

.

The set P is closed. Because if x ∈ [0, 1] \P , then x ∈ (an, bn) for some n ∈ N, andhence x is an interior point of [0, 1] \P . We now show that any point x ∈ P \ 0, 1is a limit point of P . Note that 0 and 1 are isolated points of P . If x = an > 0for some n ∈ N, then for every 0 < δ < an, there exists an m ∈ N such that(an − δ, an) ∩ Im 6= ∅. Since In ∩ Im = ∅, we see that bm ∈ (an − δ, an). Thus,x = an is a limit point of P , as claimed. Likewise, if x = bn < 1 for n ∈ N, then xis a limit point of P . Therefore, P \0, 1 is a nonempty perfect subset of R, whichis a complete metric space endowed with the ordinary metric of R. By the lemmaabove, the set P \ 0, 1 is uncountable, and hence so is P , which is obviously acontradiction. This completes the proof.

3. As D is countable, there are a countable number of lines which are all parallelto the x-axis so that each of which interests D. Use `xi

Nxi=1 to denote these lines,

where `xi: x = xi and Nx ∈ N ∪ +∞. Likewise, there are a countable number

of lines, denoted by `yjNy

j=1, where `yj: y = yj and Ny ∈ N ∪ +∞, which are

all parallel to the y-axis so that each of which intersects D. It is obvious thatD ⊆ A×B, where A = xiNx

i=1 and B = yjNy

j=1. Set

Γ = D ∩(xi, yj) : j ≤ i

,

∆ = D ∩(xi, yj) : j > i

.

As D ⊆ A×B, Γ,∆ is a partition of D into two subsets. Now, any line parallelto the x-axis whose equation is given by x = a, where a ∈ R, intersects Γ at nopoint or at most at the i points (xi, y1), . . . , (xi, yi) depending on whether a /∈ Aor a = xi for some 1 ≤ i ≤ Nx < +∞ or for some i ∈ N if Nx = +∞. Analogously,any line parallel the y-axis whose equation is given by y = b, where b ∈ R, intersects∆ at no point or at most at the j − 1 points (x1, yj), . . . , (xj−1, yj) depending onwhether b /∈ B or b = yj for some 1 ≤ j ≤ Ny < +∞ or for some j ∈ N if Ny = +∞.So the assertion follows.

4. For a subset I ⊆ Nn := 1, . . . , n, set BI = (⋃i∈I Ai)

c = S\(⋃i∈I Ai). Also, let

K := I ⊆ Nn : |I| = k−1. We claim that if I, J ∈ K, then BI ∩BJ = ∅ wheneverI 6= J . To see this, note first that BI ∩ BJ = BI∪J . Since I, J ∈ K and I 6= J ,

2.27. TWENTY SEVENTH COMPETITION 209

we see that |I ∪ J | ≥ k, which, in view of the hypothesis, yields⋃i∈I∪J Ai = S,

and hence BI ∩ BJ = BI∪J = S \ S = ∅. It is now plain that the number of BI ’s,where I ∈ K, is equal to

(nk−1

)and that such BI ’s form a partition of S into

(nk−1

)subsets. In other words,

S =⋃

BI : I ∈ K, (∗)

where⋃

stands for the disjoint union. Thus, |BI | = 1 for all I ∈ K, because thenumber of such BI ’s is equal to

(nk−1

)= |S| and that they are all nonempty. Now,

fix an arbitrary i0 ∈ I. Note that K is a disjoint union of K0 := I ∈ K : i0 ∈ Iand K′0 := I ∈ K : i0 /∈ I. It is plain that |K0| =

(n−1k−2

)and that if x ∈ BI for

some I ∈ K0 (resp. I ∈ K′0), then x /∈ Ai0 (resp. x ∈ Ai0). This together with (∗),in view of the fact that |BI | = 1 for all I ∈ K, implies that

|Ai0 | =(

n

k − 1

)−(n− 1k − 2

)=(n− 1k − 1

).

In other words, Ai =(n−1k−1

)for all i ∈ Nn, which is what we want.

5. Let k = ordab(a + b), m = ordb(a), n = orda(b), and [m,n] = lcm(m,n). We

have amb≡ 1 and bn

a≡ 1. Also, [m,n] = mq = nq′ for some q, q′ ∈ N. So we canwrite

(a+ b)[m,n] a≡ b[m,n] a≡ bnq′ a≡ 1.

Analogously,

(a+ b)[m,n] b≡ a[m,n] b≡ amqb≡ 1.

Since gcd(a, b) = 1, it follows that (a+b)[m,n] ab≡ 1, which implies ordab(a+b) ≤[m,n] = lcm[ordb(a), orda(b)]. On the other hand, (a + b)k

ab≡ 1. So, in particular,

(a + b)ka≡ 1 and (a + b)k

b≡ 1, which, in turn, yields bka≡ (a + b)k

a≡ 1 and

akb≡ (a+b)k

b≡ 1. Thus, m∣∣k and n

∣∣k, and hence lcm[ordb(a), orda(b)] = [m,n]∣∣k =

ordab(a+ b). Therefore,

ordab(a+ b) = lcm[ordb(a), orda(b)],

as desired.

6. The rings R and R′ are both abelian and have characteristic 2 because they areBoolean rings. Since f is onto and 0 ∈ R′, there is an x ∈ R such that f(x) = 0.We can write

f(0) = f(0x) = f(0)f(x) = f(0)0 = 0.

That is, f(0) = 0. Now, let a, b ∈ R be arbitrary. By the surjectivity of f , there isan x ∈ R such that f(a + b) − f(a) − f(b) = f(x). We need to show that x = 0.We can write

f(ax)(f(a+ b)− f(a)− f(b)

)= f(ax)f(x) = f(ax),

implying thatf(ax+ abx)− f(ax)− f(abx) = f(ax),

and hencef(ax+ abx) = f(abx).

210 2. SOLUTIONS

Since f is one-to-one, we obtain ax + abx = abx, implying that ax = 0. Likewise,we see that bx = 0. Now, in view of ax = bx = 0, we can write

f(x)(f(a+ b)− f(a)− f(b)

)= f(x)f(x) = f(x),

which yieldsf(ax+ bx)− f(ax)− f(bx) = f(x).

Hence, f(x) = 0 = f(0), from which, in view of the injectivity of f , we see thatx = 0, which is what we want.

2.28. Twenty Eighth Competition

2.28.1. First Day. 1. Let f(z) =∑+∞n=0 cnz

n. If f is a polynomial, then,by the Fundamental Theorem of Algebra, we have f(z) = c0 + c1z because f isone-to-one. If not, then f( 1

z ) would have an essential singularity at zero. Thus, byPicard’s Great Theorem, f( 1

z ) is not one-to-one in any neighborhood of zero, andhence neither is f , a contradiction. This proves the assertion.

2. We prove the assertion by showing that f−1((0, 0)

)is not bounded, and

hence has infinitely many points. As f is surjective, f−1((0, 0)

)is nonempty.

Suppose to the contrary that f−1((0, 0)

)is bounded. Thus, f−1

((0, 0)

)is

a nonvoid closed bounded subset of R. Consequently, there exist a, b ∈ R suchthat a = min f−1

((0, 0)

)< b = max f−1

((0, 0)

). Let A1 = 0 × (0,+∞),

A2 = 0 × (−∞, 0), A2 = (0,+∞) × 0, and A4 = (−∞, 0) × 0. As f iscontinuous and [a, b] is a compact subset of R, we see that f

([a, b]

)is a bounded

subset of A. Now, since f is continuous and f(x) 6= 0 for all x ∈ (−∞, a)∪ (b,+∞),there are i1, i2 ∈ 1, 2, 3, 4, not necessarily distinct, such that f(−∞, a) ⊆ Ai1and f(b,+∞) ⊂ Ai2 . Therefore, f(R) ∩ Ai is a bounded subset Ai for each i ∈1, 2, 3, 4\i1, i2, which is a contradiction because f is surjective. So the assertionfollows.

3. By the hypothesis and the Mobius Inversion Formula, we have

f(n) =∑d|n

µ(d)(n

d)2 = n2

∑d|n

µ(d)d2

= n2g(n),

for all n ∈ N, where g(n) =∑d|n

µ(d)d2 . The arithmetic function g is multiplicative

because µ(d)d2 is a multiplicative function of d. If p is prime and α is a positive

integer, we can write

g(pα) =∑d|pα

µ(d)d2

=µ(1)

1+µ(p)p2

= 1− 1p2.

Thus, if n =∏p|n p

αp , then

g(n) = g(∏p|n

pαp) =∏p|n

g(pαp) =∏p|n

(1− 1

p2

)

=∏p|n

(1− 1

p

)∏p|n

(1 +

1p

)=φ(n)n

∏p|n

(1 +

1p

).

2.28. TWENTY EIGHTH COMPETITION 211

Consequently,

f(n)φ(n)

=n2g(n)φ(n)

=n2

φ(n)φ(n)n

∏p|n

(1 +

1p

)= n

∏p|n

(1 +

1p

),

for all n ∈ N with n > 1, as desired.

4. Let 1 ≤ i, k ≤ n be arbitrary and τ ∈ Gk∩Z(G). By the hypothesis, there existsa σ ∈ G such that σ(k) = i. We can write

λ ∈ Gi ⇐⇒ λ(i) = i ⇐⇒ λ(σ(k)

)= σ(k)

⇐⇒ σ−1λσ(k) = k ⇐⇒ σ−1λσ ∈ Gk⇐⇒ λ ∈ σGkσ−1.

Thus, Gi = σGkσ−1, and hence στσ−1 ∈ Gi because τ ∈ Gk. On the other hand,

τ ∈ Z(G) implies τ = στσ−1 ∈ Gi. It follows that τ ∈ ∩ni=1Gi = e because i wasarbitrary. This proves Gk ∩ Z(G) = e, which is what we want.

5. We prove that the maximum number of such vectors is n. To this end, letthe vectors v1, . . . , vk ∈ A be such that any two of which share an even numberof one entries. View the elements of A as vectors in the n-dimensional vectorspace Zn2 , where Z2 = 0, 1 is the field of integers modulo 2. We claim thatthe set v1, . . . , vk is a linearly independent subset of Zn2 . To see this, supposec1v1 + · · · + ckvk = 0 for some c1, . . . , ck ∈ Z2. If “.” denotes the usual innerproduct on Zn2 , for all 1 ≤ i ≤ n, we have

ci = vi.(c1v1 + · · ·+ ckvk) = 0,

proving the claim. Thus, k ≤ n. On the other hand, if ei is the vector with 1 inthe ith place and zero elsewhere, then any two elements of the set e1, . . . , en, thestandard basis of Zn2 , share an even number of one entries. Therefore, the maximumnumber of such vectors is n, as desired.

6. Let P = p1, . . . , pn. As is usual, use Sym(P ) to denote the set of all per-mutations on (the poset) P . Let’s view Sym(P ) as a probability space. Choose apermutation σ on 1, . . . , n at random which, in turn, gives rise to a permutationsσ ∈ Sym(P ) at random so that sσ(pi) = pσi for all 1 ≤ i ≤ n. For any 1 ≤ i ≤ n,there exists a 1 ≤ j ≤ n such that σj = i, which is equivalent to pi = sσ(pj) = pσj .With all that in mind, for an element pi ∈ P for which pi = sσ(pj) = pσj for some1 ≤ j ≤ n, let Fpi

be the event that the permutation sσ ∈ Sym(P ) satisfies thefollowing

∀p ∈ P \ Upi: p = sσ(pk) = pσk =⇒ σj ≤ σk.

In other words, if

sσ =(

p1 p2 · · · pnpσ1 pσ2 · · · pσn

),

then sσ ∈ Fpi if and only if the element pi precedes all of the elements of P \Upi butpi in the second row of the above representation of sσ. Set Epi

:= Fpi

⋂(⋂p∈Lpi

F cp ),where F cp := Sym(P ) \ Fp.

We note that if pi and pi′ are distinct, then the events Epi and Epi′ cannot occursimultaneously, and hence they are disjoint. Let’s prove this. First, assume thatfor pi, pi′ ∈ P , Epi

, and hence Fpi, occurs, and that pi and pi′ are not comparable.

Then, in particular pi′ 6> pi yielding pi′ ∈ P \Upi, which, in turn, implies σj < σj′,

212 2. SOLUTIONS

where i = σj and i′ = σj′. Consequently, Fpi′ , and hence Epi′ , cannot occur, forotherwise, we must have σj′ < σj because pi 6> pi′ , which is a contradiction. Thatis, Fpi and Fpi′ , and hence Epi and Epi′ , cannot occur simultaneously wheneverpi and pi′ are not comparable. Next, if the elements pi, pi′ ∈ P are comparable,then it is plain that the events Epi

and Epi′ cannot occur simultaneously. Thus,the events Epi

and Epi′ are disjoint whenever pi, pi′ ∈ P are distinct.We now use induction on |Lpi | to prove Xpi = P (Epi). If |Lpi | = 0, i.e., if pi

is a minimal element of P , then Epi = Fpi and sσ ∈ Fpi with sσ(pj) = pi if andonly if σj ≤ σk whenever p = s(pk) = pσk ∈ P \ Upi

for some 1 ≤ k ≤ n. As∣∣P \ Upi

∣∣ = n−∣∣Upi

∣∣, in this case, we can write

P (Epi) = P (Fpi

) =(|Upi |)!

(n

n−|Upi|)(n− |Upi | − 1)!

n!=

1n− |Upi |

.

Now, assuming that Xp = P (Ep) for all p ∈ Lpi , we prove Xpi = P (Epi). We claimthat ⋂

p∈Lpi

F cp =⋂

p∈Lpi

Ecp,

for all pi ∈ P . To see this, first, let sσ ∈⋂p∈Lpi

F cp be arbitrary. We showsσ ∈

⋂p∈Lpi

Ecp. Suppose the contrary, implying that there is a q ∈ Lpisuch that

sσ 6∈ Ecq . It follows that sσ ∈ Eq, and hence sσ ∈ Fq. This is in contradictionwith sσ ∈

⋂p∈Lpi

F cp . Thus,⋂p∈Lpi

F cp ⊆⋂p∈Lpi

Ecp. Next, letting sσ ∈⋂p∈Lpi

Ecpbe arbitrary, we show sσ ∈

⋂p∈Lpi

F cp . Again suppose the contrary, implying thatthere is a minimal q ∈ Lpi so that sσ 6∈ F cq but sσ ∈ F cp for all p ∈ Lq. It followsthat sσ ∈ Fq but sσ 6∈ Fp for all p ∈ Lq. Consequently, sσ ∈ Eq = Fq

⋂(⋂p∈Lq

F cp ),which is a contradiction. Therefore,

⋂p∈Lpi

Ecp ⊆⋂p∈Lpi

F cp , proving the claim.Now, as Ep’s (p ∈ Lpi) are disjoint, in view of the claim, we can write

P( ⋂p∈Lpi

F cp)

= P( ⋂p∈Lpi

Ecp)

= 1− P( ⋃p∈Lpi

Ep)

= 1−∑p∈Lpi

Xp.

On the other hand, the events Fpi and⋂p∈Lpi

F cp are independent because theposition of pi in the permutation has no affect on Fp’s (p ∈ Lpi

). Thus,

P (Epi) = P((⋂

p∈Lpi

F cp )⋂Fpi

)= P

( ⋂p∈Lpi

F cp)P (Fpi

) =1−

∑p∈Lpi

Xp

n− |Upi|

.

Therefore, Xpi = P (Epi) for all pi ∈ P , and hence 0 ≤ Xpi ≤ 1 for all pi ∈ P ,which is what we want.

2.28.2. Second Day. 1. Let g = f−1. It follows from the hypothesis thatfor all x, y with y = 2x2, we have

A =∫ x

0

(2t2 − t2)dt =13x3 =

√2

12y

32 =

B =∫ y

0

(√ t

2− g(t)

)dt =

√2

3y

32 −

∫ y

0

g(t)dt,

2.28. TWENTY EIGHTH COMPETITION 213

1.5

BA

P

Figure 13

implying ∫ y

0

g(t)dt =√

24y

32 .

Taking derivative of both sides yields

x = g(y) =3√

28y

12 =⇒ x2 =

932y =⇒ y =

329x2.

Thus, y = f(x) = 329 x

2, which is what we want.

2. Set h(x) = f(x) − x. We have h(a) > 0 and h(b) < 0. Since h is continuous,there exists a δ > 0 such that h(a + x) > 0 and h(b − x) < 0 whenever x ∈ (0, δ).Set α = δ

n+1 and

g(x) = h(x) + h(x+ α) + · · ·+ h(x+ nα)=

(f(x) + · · ·+ f(x+ nα)

)−(x+ · · ·+ (x+ nα)

).

It is readily seen that g(a) > 0 and g(b − nα) < 0. By the Intermediate ValueTheorem, there exists a c ∈ (a, b− nα) such that g(c) = 0. This implies

f(c) + f(c+ α) + · · ·+ f(c+ nα) = c+ (c+ α) + · · ·+ (c+ nα)

= (n+ 1)c+n(n+ 1)

2α = (n+ 1)(c+

n

2α),

which is what we want.

3. The answer is yes. To see this, noting that det(A+B) = det(At +Bt), we canwrite

det(A) det(A+B) = det(A) det(At +Bt) = det(I +ABt)= det(I +BAt) = det(B) det(At +Bt)= det(B) det(A+B).

214 2. SOLUTIONS

Consequently, in view of det(A) + det(B) = 0, we have

det(A) det(A+B) = det(B) det(A+B) = −det(A) det(A+B),

yielding 2 det(A) det(A+B) = 0. But det(A) = ±1 because A is orthogonal. Thus,det(A+B) = 0, as desired.

4. Let x, y ∈ R be arbitrary. We can write

xn+2yn+2 = (xy)n+2 = (xy)n+1xy = xn+1yn+1xy,

yieldingxn+1(xyn+1 − yn+1x)y = 0.

As the above equality holds for all x, y ∈ R, changing x to x+ 1 in the equality, weobtain

(1 + x)n+1(xyn+1 − yn+1x)y = 0,or ((

n+ 10

)+(n+ 1

1

)x+ · · ·+

(n+ 1n+ 1

)xn+1

)(xyn+1 − yn+1x)y = 0.

Multiplying both sides of the equality by xn and using the aforementioned equality,we get

xn(xyn+1 − yn+1x)y = 0.Analogously, changing x to x+ 1 in this equality, we see that

xn−1(xyn+1 − yn+1x)y = 0.

Continuing in this way, we finally obtain

(xyn+1 − yn+1x)y = 0. (∗)On the other hand, by the hypothesis, we have

xn+1yn+1 = (xy)n+1 = (xy)nxy = xnynxy,

yieldingxn(xyn − ynx)y = 0.

Likewise, from this, we obtain

(xyn − ynx)y = 0.

Multiplying both sides of the above by y from the left, we get

yxyn+1 − yn+1xy = 0,

from which, in view of (∗), we see that

(yx− xy)yn+1 = 0.

Once again, preforming an argument involving changing y to y+1 and multiplyingboth sides of the obtained equality by an appropriate power of y, we eventuallyconclude that yx − xy = 0. That is, xy = yx for all x, y ∈ R, which is what wewant.

5. Suppose that a person is chosen at random. Define the following events.A (resp. B) := the event that the person is guilty (resp. innocent).+ (resp. −) := the event that the person responds positively (resp. negatively).E := the event that the person responds positively to four questions and neg-

atively to one question.

2.29. TWENTY NINTH COMPETITION 215

It follows from the hypothesis that P (A) = 13 , P (B) = 2

3 , P (+|A) = 0.8,P (+|B) = 0.4, P (−|A) = 0.2, and P (−|B) = 0.6. Using Bayes’ Theorem, we seethat the desired probability is equal to

P (A|E) =P (A ∩ E)P (E)

=P (A)P (E|A)

P (A)P (E|A) + P (B)P (E|B)

=13

(54

)(0.8)4(0.2)

13

(54

)(0.8)4(0.2) + 2

3

(54

)(0.4)4(0.6)

=816.

Thus, P (A|E) = 816 is the desired probability.

6. We claim that for each j = 1, . . . , k, there are subsets A1, . . . , Aj satisfying thefollowing inequalities.

1 ≤ |A1 ∩ · · · ∩Aj | ≤ r − (j − 1)(I(F )− 1) (∗).Obviously, the assertion follows from (∗) by letting j = k. We prove (∗) by inductionon j. If j = 1, the assertion is trivial; in fact, any element A1 of F satisfies (∗) whenj = 1. Suppose that (∗) holds for j, where j ≤ k − 1. We prove the assertion forj + 1. To this end, it follows from the hypothesis that A1 ∩ · · · ∩Aj has nonemptyintersection with any element of F . This implies |A1 ∩ · · · ∩ Aj | ≥ I(F ). Let S bea subset of A1 ∩ · · · ∩ Aj with I(F )− 1 elements. Thus, there exists an Aj+1 ∈ Fsuch that S ∩Aj+1 = ∅. So we can write

1 ≤ |A1 ∩ · · · ∩Aj ∩Aj+1| ≤ |(A1 ∩ · · · ∩Aj) ∩ Sc|= |(A1 ∩ · · · ∩Aj) \ S| = |A1 ∩ · · · ∩Aj | −

(I(F )− 1

)≤ r − (j − 1)

(I(F )− 1

)− (I(F )− 1) = r − j

(I(F )− 1

),

proving the induction assertion, finishing the proof.

2.29. Twenty Ninth Competition

2.29.1. First Day. 1. First solution: We can write

I :=∫ a

0

f(x)dx∫ a

0

dx

f(x)

=∫ a

0

∫ a

0

f(x)1

f(y)dxdy =

∫ a

0

∫ a

0

f(y)1

f(x)dxdy,

from which, we obtain

2I = I + I =∫ a

0

∫ a

0

(f(x)f(y)

+f(y)f(x)

)dxdy

=∫ a

0

∫ a

0

f(x)2 + f(y)2

f(x)f(y)dxdy.

On the other hand, obviously, we have f(x)2+f(y)2

f(x)f(y) ≥ 2 for all 0 ≤ x, y ≤ a. Thus,

2I ≥∫ a

0

∫ a

0

2dxdy = 2a2,

yieldingI ≥ a2,

as desired.

216 2. SOLUTIONS

Second solution: By the Cauchy-Schwarz Inequality, we can write

a2 =

∣∣∣∣∣∫ a

0

√f(x)× 1√

f(x)dx

∣∣∣∣∣2

≤(∫ a

0

f(x)dx)(∫ a

0

dx

f(x)

),

which is what we want.

2. The “if part” is trivial. We prove the “only if part” of the assertion by con-tradiction. To this end, suppose x = a

b for some a, b ∈ N. It follows from thecontradiction hypothesis that there exists a k ∈ N such that nk+1 − 1 > b. Set

xk =k∑i=1

1n1 · · ·ni

.

We can writea

b− xk = x− xk =

1n1 · · ·nk+1

+1

n1 · · ·nk+1nk+2+ · · ·

<1

n1 · · ·nk

(1

nk+1+

1n2k+1

+1

n3k+1

+ · · ·)

=1

n1 · · ·nk1

nk+1 − 1.

Consequently,

0 <a

bn1 · · ·nk − xkn1 · · ·nk <

1nk+1 − 1

,

whence

0 < an1 · · ·nk − bxkn1 · · ·nk <b

nk+1 − 1< 1,

which is a contradiction because an1 · · ·nk − bxkn1 · · ·nk ∈ N. Thus, the assertionfollows by contradiction.

3. Note that 10φ(m) ≡ 1 (mod m) for all positive integers m satisfying gcd(m, 10) =1, where φ denotes Euler’s totient function. It follows that m

∣∣10kφ(m) − 1 when-ever k,m ∈ N with gcd(m, 10) = 1. Moreover, if gcd(m, 3) = 1, then

m

∣∣∣∣10kφ(m) − 110− 1

= 1 + 10 + · · ·+ 10kφ(m)−1 .

Thus, if gcd(m, 30) = 1, then m∣∣ kφ(m) times︷ ︸︸ ︷

11 . . . 1 for all k ∈ N, proving the assertion.

4. Proceed by way of contradiction. Suppose I is a nonzero right ideal in R whoseelements all have square zero. It follows that (x + y)2 = 0, and hence xy = −yxfor all x, y ∈ R. Thus, (xI)(xI) = 0, from which, we see that

(RxI)(RxI) ⊆ (Rx)I(xI) = R(xI)(xI) = 0,

for all x ∈ R. In other words, the two-sided ideal RxI is nilpotent, which, in viewof the hypothesis, yields RxI = 0. This shows that xI is contained in the leftannihilator of R in R, denoted by l.ann(R). But l.ann(R) is nilpotent. Thus,xI ⊆ l.ann(R) = 0, yielding xI = 0 for all x ∈ R. This easily implies I2 =0, from which, we obtain (RI)(RI) ⊆ RI2 = 0. But RI is a two-sided idealof R. So it follows from the hypothesis that RI = 0. This implies that I iscontained in the right annihilator of R in R, denoted by r.ann(R). But r.ann(R) is

2.29. TWENTY NINTH COMPETITION 217

obviously a nilpotent two-sided ideal. Thus, I ⊆ l.ann(R) = 0, yielding I = 0,a contradiction. This proves the assertion.

5. Define the function f : X −→ Y by

f(A) =k ∈ Z : 2k ∈ A

.

It suffices to show that for any B ∈ Y , there exists a A ∈ X such that f(A) = B.To this end, for given B ∈ Y , set

A :=2k : k ∈ B

.

It is obvious that A ∈ X and f(A) = B, as desired.

6. Note first that since S ≤ Zn2 is a vector space over Z2, d is equal to the minimumof the number of ones occurring in the nonzero elements of S. It is plain that S has2k elements. Use a matrix A ∈ M(2k−1)×n(Z2) to denote the nonzero elements ofS, so that every nonzero element of S corresponds to a row of A. We may assumethat the first k rows of A form a basis for S. Denote this k × n submatrix of A byB. It is plain that the number of one entries of the matrix A is at least d(2k − 1).By showing that this number is at most n2k−1, we conclude that d(2k−1) ≤ n2k−1,proving the assertion. We claim that the number of ones in any column of A iseither 0 or 2k−1. Since every row of A is a linear combination of the rows of B, thenumber of ones in any column of A is computed as follows. First, if there exists no“one” in column i (1 ≤ i ≤ n) of B, then neither does exist in the same column ofA. Next, let t be the number of ones in column i (1 ≤ i ≤ n) of B. Since any linearcombination of the rows of B gives rise to a row of A, and since the coefficientscome from Z2, it follows that the number of ones in column i of A, denoted by 1i,is equal to the number of all subsets of 1, 2, . . . , n with k elements each of whichcontains an odd number of one entries from column i of B. Thus, 1i is equal to thenumber of all odd numbered subsets of a set with t elements times the number ofall subsets of a set with k − t elements. In other words,

1i = 2t−12k−t = 2k−1.

Now, as A has k columns, the number of all ones is at most n2k−1, finishing theproof.

2.29.2. Second Day. 1. First solution: Let f(z) =∑+∞n=0 anz

n. Set g(z) =f(z). It is plain that

g(z) =+∞∑n=0

anzn,

and that g is analytic on D. It follows that f − g is analytic on D and moreover(f − g)( 1

n ) = f( 1n )− g( 1

n ) = 0 for all n ≥ 2. Hence, the set of zeros of the analyticfunction f −g has a limit point in D. It thus follows from the Uniqueness Theoremfor analytic functions that f(z) = g(z) for all z ∈ D, whence an = an for all n ∈ N.Therefore, an ∈ R for all n ∈ N. This, in view of an = f(n)(0)

n! , yields f (n)(0) ∈ Rfor all n ∈ N, as desired.

Second solution: Let f(z) =∑+∞n=0 anz

n, where an = f(n)(0)n! for all n ∈ N ∪ 0.

It suffices to show that an ∈ R for all n ∈ Z with n ≥ 0. We prove this by induction

218 2. SOLUTIONS

on n. If n = 0, the assertion is trivial because a0 = f(0) = limn f( 1n ). Assuming

that a0, . . . , an−1 ∈ R, to prove an ∈ R, set

g(z) =+∞∑i=0

ai+nzi =

1zn

(f(z)−

∑n−1i=0 aiz

i)

z 6= 0,an z = 0.

It is readily verified that g is an analytic function satisfying g( 1n ) ∈ R for all natural

numbers n ≥ 2. Thus, an = g(0) = limn g( 1n ) ∈ R, which is what we want.

2. (i) The assertion can be stated as follows. For a given ε > 0, every point of thespace X can be reached from any other point of the space within a finite number ofsteps each of which having a length less than ε. With this in mind, for given ε > 0,fix an x ∈ X and set

Cε(x) :=y ∈ X : ∃x1 = x, x2, . . . , xn = y ∈ X ∀i < n : d(xi, xi+1) < ε

.

To prove the assertion, it suffices to show that Cε(x) = X. This, in view of thehypothesis that X is connected, follows as soon as we show that Cε(x) is a clopensubset, i.e., both a closed and open subset, of X. First, suppose z ∈ Cε(x). Itfollows that there exists a sequence (yn)+∞n=1 in Cε(x) such that limn yn = z. Thus,there exists an N ∈ N such that d(yn, z) < ε whenever n ≥ N . On the other hand,since yn ∈ Cε(x), we have Bε(yn) := t ∈ X : d(t, yn) < ε ⊆ Cε(x). Consequently,z ∈ Cε(x) because z ∈ Bε(yn). Thus, Cε(x) is a closed subset of X. Next, supposey ∈ Cε(x) is arbitrary. It follows that Bε(y) ⊆ Cε(x). This implies that Cε(x) isan open subset of X. Therefore, Cε(x) = X because X is connected.

(ii) Plainly, X = [0, 1) ∪ (1, 2] is an example showing the converse of (i) doesnot necessarily hold in general.

(iii) To prove the assertion by contradiction, suppose X is disconnected. Itfollows that there is a clopen subset M of X. Since X is compact and both M andM c = X \M are closed subsets of X, they both are compact subsets of X. ButM ∩M c = ∅. Hence, d(M,M c) := infd(x, y) : x ∈ M,y ∈ M c > 0. Pick x ∈ Mand y ∈ M c. It is now obvious that x cannot be reached from y within a finitenumber of steps each of which having a length less than ε = d(M,M c). Because,otherwise ε = d(M,M c) ≤ d(xi, xi+1) < ε, where xi ∈ M and xi+1 ∈ M c are theconsecutive foot steps in a walk from x to y with a finite number of steps each ofwhich having a length less than ε. This is a contradiction. Thus, the assertionfollows.

3. (i) For given g ∈ G, define ϕg : K −→ K by ϕg(x) = g−1xg. Since g ∈ NG(K),the map ϕg is well-defined because g−1xg ∈ K for all x ∈ K. This implies ϕg ∈Aut(K). In other words, the map ϕ : NG(K) −→ Aut(K) defined by ϕ(g) = ϕg iswell-defined. It is easily verified that ϕ is a homomorphism of groups and that

kerϕ =g ∈ NG(K)| ϕ(g) = id

=

g ∈ NG(K)| ∀x ∈ K : g−1xg = x

=

g ∈ NG(K)| ∀x ∈ K : gx = xg

= CG(K).

Now, the assertion is a quick consequence of the First Isomorphism Theorem forgroups.

2.29. TWENTY NINTH COMPETITION 219

(ii) By (i), in view of the hypothesis that KEG, the group GCG(K) is isomorphic

to a subgroup of Aut(K). Since K is cyclic, Aut(K) is abelian, and hence so isG

CG(K) , whence G′ ≤ CG(K). It follows that G ≤ CG(K) ≤ G because G′ = G.Thus, G = CG(K), implying K ≤ Z(G), which is what we want.

4. To prove the “if part”, it suffices to show that any two invariant subspaces ofany nilpotent transformation N on Fn with Nn−1 6= 0 are comparable with respectto inclusion. Note first that Nn = 0 and that there exists a vector α ∈ Fn suchthat Nn−1α 6= 0 because Nn−1 6= 0. It follows that the set Nn−1α, . . . , Nα, α isa basis for Fn. To see this, suppose c1Nn−1α+ · · ·+ cn−1Nα+ cnα = 0 for someci ∈ F (1 ≤ i ≤ n). Taking Nn−1 of both sides of this equality yields cnNn−1α = 0,implying cn = 0. Thus, c1Nn−1α+ · · ·+ cn−1Nα = 0. Now, taking Nn−2 of bothsides of the equality yields cn−1 = 0. Continuing in this manner, we obtain ci = 0for all 1 ≤ i ≤ n. Thus, the set Nn−1α, . . . , Nα, α is a basis for Fn becausedimFn = n. Set

M0 =⟨0⟩

= 0, Mi =⟨Nn−1α, . . . , Nn−iα

⟩.

We haveM0 = 0 < M1 =

⟨Nn−1α

⟩< · · · < Mn = Fn.

It is plain that Mi is an invariant subspace of N for all 1 ≤ i ≤ n. This shows thatany nilpotent transformation with Nn−1 6= 0 is triangularizable. Now, supposethat M is an invariant subspace for N . We prove the assertion by showing thatM = Mj for some 1 ≤ j ≤ n. To this end, let 1 ≤ j ≤ n be the greatest naturalnumber for which there exists a nonzero vector x ∈M such that

x = c1Nn−1α+ · · ·+ cjN

n−jα, (∗)

where cj 6= 0. We show that M = Mj . Taking N j−1 of both sides of (∗), weget cjNn−1α = N j−1x ∈ M, whence Nn−1α ∈ M because cj 6= 0. Now, usingNn−1α ∈M and taking N j−2 of both sides of (∗), we see that Nn−2α ∈M. Con-tinuing this way, we conclude thatNn−jα ∈M. Thus,Mj = 〈Nn−1α, . . . , Nn−jα〉 ⊆M. On the other hand, M ⊆Mj because Nn−jα ∈ M. Therefore, M = Mj , asdesired.

To prove the “only if part”, note first that the matrix A can have at mostone eigenvalue. Because, otherwise the corresponding eigenspaces of two distincteigenvalues are not comparable with respect to inclusion, which is in contradictionwith the hypothesis. So, with no loss of generality, we may assume that zero isthe only eigenvalue of A, for A − λI and A share the same lattice of invariantsubspaces. In other words, we may assume that A is nilpotent. We claim thatthere exists a nonzero vector α ∈ Fn such that the set An−1α, . . . , Aα, α is abasis for Fn. Suppose to the contrary that the set An−1α, . . . , Aα, α is not abasis for Fn. Let M := 〈An−1α, . . . , Aα, α〉. It follows that the subspace Mis a nontrivial invariant subspace of A. Pick β ∈ Fn \ M. It follows from thecontradiction hypothesis that the subspace M′ := 〈An−1β, . . . , Aβ, β〉 is also anontrivial invariant subspace of A. Obviously, the subspaces M and M′ are notcomparable, a contradiction. Therefore, there exists a nonzero vector α ∈ Fn

such that the set An−1α, . . . , Aα, α is a basis for Fn. Since A is assumed to benilpotent, we see that An = 0 but An−1 6= 0 because An−1α 6= 0. This completesthe proof.

220 2. SOLUTIONS

5. Use A(m,n) to denote the average number of times that the players should playtill one of them runs out of the coins. If the game is played once, A wins with theprobability of 1

2 changing the number of the coins of A and B to m+ 1 and n− 1,respectively; and, likewise, B wins with the probability of 1

2 changing the numberof the coins of A and B to m− 1 and n+ 1, respectively. This yields the followingrecurrence relation on m,n ∈ N.

A(m,n) = 1 +12A(m− 1, n+ 1) +

12A(m+ 1, n− 1)

The boundary conditions are A(0, n) = A(m, 0) = 0. To solve this boundaryrecurrence equation, view A as a function of one variable, say n. It follows fromthe recurrence equation that on the line m+ n = c, where c is a constant, we have

A(m− 1, n+ 1) +A(m+ 1, n− 1)− 2A(m,n) = −2. (∗)

Thus, A(m,n) must be a quadratic function. As A(0, n) = A(m, 0) = 0, we obtainA(m,n) = cmn. Substituting this into (∗), we obtain

−2c = c(m− 1)(n+ 1) + c(m+ 1)(n− 1)− 2cmn = −2,

implying c = 1. Consequently, A(m,n) = mn. Therefore, on average, the game isplayed as many as the product of the coins owned by A and B altogether.

6. It suffices to show that [0, 1] ⊆ C −C. To this end, let y ∈ [0, 1] be arbitrary. If

1− y

2= 0.y1y2 . . .

denotes the ternary expansion of 1−y2 , where yi ∈ 0, 1, 2, then we can write

1− y

2= 0.a1a2 . . .+ 0.b1b2 . . . ,

where ai, bi ∈ 0, 1 and yi = ai + bi for all i ∈ N. So we have

1− y = 2(0.a1a2 . . .) + 2(0.b1b2 . . .),

implying

y = 0.a′1a′2 · · · + 0.b′1b

′2 · · · := a+ b,

where a′i = 2− 2ai and b′i = 2bi, and hence a′i, b′i ∈ 0, 2 for all i ∈ N. Therefore,

a, b ∈ C because there is no one in their ternary expansions. This finishes the proof.

2.30. Thirtieth Competition

2.30.1. First Day. 1. First solution: It suffices to evaluate the followinglimit

limt→+∞

∫ 2006

1385

f(tx)dx.

Setting tx = u, we have

limt→+∞

∫ 2006

1385

f(tx)dx = limt→+∞

1t

∫ 2006t

1385t

f(u)du.

2.30. THIRTIETH COMPETITION 221

As limt→+∞ t = +∞, using L’Hopital’s rule, we can write

limt→+∞

∫ 2006

1385

f(tx)dx = limt→+∞

(2006f(2006t)− 1385f(1385t)

)= 621.

Thus,

limn→+∞

∫ 2006

1385

f(nx)dx = 621,

as desired.

Second solution: We can write

limn→+∞

∫ 2006

1385

f(nx)dx = limn→+∞

(∫ 2006

1385

(f(nx)− 1

)dx+ 621

)= 621,

provided that limn→+∞∫ 2006

1385

(f(nx) − 1

)dx = 0. It follows from the hypothesis

that for given ε > 0, there is an N > 0 such that |f(x)− 1| < ε621 whenever x > N .

Thus, for all n > N1385 and x ∈ [1385, 2006], we have∣∣f(nx)− 1

∣∣ < ε.

This yields ∣∣∣ ∫ 2006

1385

(f(nx)− 1

)dx∣∣∣ ≤ ∫ 2006

1385

∣∣f(nx)− 1∣∣dx ≤ ε,

for all n > N1385 . That is, limn→+∞

∫ 2006

1385

(f(nx)− 1

)dx = 0, settling the assertion.

2. We need the following lemma.

Lemma. Let απ 6∈ R \ Q, where π = 3.14159 . . .. Then, the set einα : n ∈ N

is dense in the unit circle.Proof. Use T to denote the unit circle. Set S = einα : n ∈ N. Plainly, the

set S consists of distinct points, for απ 6∈ R\Q, and it is a multiplicative semigroup.

As T is compact, we see that S has a limit point in T. Thus, there is a strictlyincreasing sequence (nk)+∞k=1 of natural numbers such that limk e

inkα exists. Thisimplies that limk e

imkα = 1, where mk = nk+1 − nk. Now, let eit : t ∈ (a, b) bean arbitrary open arc of the unit circle, where a, b ∈ R and a < b. It follows thatthere is an ` ∈ N such that eimkα ∈ eit : t ∈ (−δ, δ) for all k ≥ `, where δ = b−a

2 .Consequently, einm`α : n ∈ N ∩ eit : t ∈ (a, b) 6= ∅. This shows that S is densein T, which is what we want.

Remark. A proof similar to that of the above lemma shows that every additivesubgroup of R is either dense or isolated, i.e., it has no limit points. A quickconsequence of this is the following. Let α ∈ R\Q. Then, the set

m+nα : m,n ∈

Z

is dense in R.

First, we claim that there is a sequence (kp)+∞p=1 of natural numbers such that

limp→+∞ akp

j = 1 for all 1 ≤ j ≤ m. Proceed by induction on m. If m = 1, thenlimn→+∞ an1 = c. Firstly, c 6= 0 because |c| = limn→+∞ |a1|n = 1. Secondly, byshowing that a1 = 1, we prove the claim in this case. From c = limn→+∞ an1 =limn→+∞ an+1

1 = ca1, we obtain c(1− a1) = 0, which yields a1 = 1 because c 6= 0.Assume that the assertion holds for m−1. It follows from the induction hypothesisthat there is a sequence (k`)+∞`=1 such that lim`→+∞ ak`

j = 1 for all 1 ≤ j ≤ m− 1.

222 2. SOLUTIONS

Without loss of generality, if necessary by passing to a subsequence of (k`)+∞`=1 , wecan assume that lim`→+∞ ak`

j = 1 for all 1 ≤ j ≤ m − 1 and lim`→+∞ ak`m = bm,

where bm ∈ C and |bm| = 1. Set

A =(ak1 , . . . , a

km−1, a

km) : k ∈ N

.

Let A′ be the set of limit points of A. It is obvious that (1, . . . , 1, bkm) ∈ A′ for allk ∈ N. As |bm| = 1, it is easily seen from the lemma above that there exists a sub-sequence (kr)+∞r=1 such that limr→+∞ bkr

m = 1. Consequently, (1, . . . , 1, 1) ∈ (A′)′ ⊆A′. Thus, there is a subsequence (kp)+∞p=1 such that limp→+∞(akp

1 , . . . , akp

m−1, akpm ) =

(1, . . . , 1, 1), and hence limp→+∞ akp

j = 1 for all 1 ≤ j ≤ m, proving the claim. Itthus follows that

limp→+∞

m∑j=1

akp+1j = c = lim

p→+∞

m∑j=1

akp

j =m∑j=1

1 = m,

yielding∑mj=1 aj = m because limp→+∞ a

kp

j = 1 for all 1 ≤ j ≤ m. Now, we canwrite

m∑j=1

Re(aj) = Re( m∑j=1

aj)

= m,

yielding∑mj=1 Re(aj) = m. This implies Re(aj) = 1, for Re(aj) ≤ 1 for all 1 ≤ j ≤

m. As the aj ’s are on the unit circle, we see that aj = 1 for all 1 ≤ j ≤ m, whichis what we want.

3. It is worth noting that the hypothesis that R is commutative is redundant.Consider the following five elements of R/J

J, 1 + J, a+ J, a2 + J, a3 + J.

It follows from the hypothesis that at least two of the above elements are equal.Thus, one of the following 10 =

(52

)cases will occur

J = 1 + J, J = a+ J, J = a2 + J, J = a3 + J,1 + J = a+ J, 1 + J = a2 + J, 1 + J = a3 + J,a+ J = a2 + J, a+ J = a3 + J, a2 + J = a3 + J.

Consequently, at least one of the following elements of R belongs to J

1, a, a2, a3, a− 1, a2 − 1, a3 − 1 = a, a2 − a, a3 − a, a3 − a2.

In view of the identity a(a − 1)(a + 1) = 1, it is readily verified that the aboveelements are invertible elements of R, and hence J = R, as desired.

4. Since p is an odd prime, by Fermat’s Little Theorem, we have 2p−1 ≡ 1 (mod p).Let s be the smallest natural number for which 2s ≡ 1 (mod p). It follows thats∣∣p − 1 = 2q, and hence s ∈ 1, 2, q, 2q, for q is prime. To prove the assertion by

contradiction, assume that s < 2q. If s = 1, then p∣∣1, which is a contradiction.

If s = 2, we obtain p = 3, yielding q = 1, which is a contradiction. Finally, ifs = q, then the equation x2 ≡ 2p (mod p) is solvable for x in Z. Thus

(2q

p

)= 1,

where(ap

), with a, p ∈ Z, denotes the Legendre symbol, which is defined to be ±1

2.30. THIRTIETH COMPETITION 223

depending on whether or not the equation x2 ≡ a (mod p) is solvable for x in Z.On the other hand,(2q

p

)=(2p

)q =(− 1) q(p2−1)

8 =(− 1) q2(q+1)

2 = −1,

which is a contradiction. Note that the last equality in the above is implied byq ≡ 1 (mod 4). Therefore, we obtain a contradiction in any event. So the assertionfollows.

5. Let X = X1 + · · · + X2m, where Xi’s (1 ≤ i ≤ 2m) are independent randomvariables assuming the values zero and one with the probability of 1

2 . It is plainthat

µ = µX = m, σ = σ2X =

m

2.

Using the Chebyshev Inequality with λ =√

2, we have

P(|X −m| ≥

√m)≤ 1

2,

implying

P(|X −m| <

√m)≥ 1

2.

On the other hand,

12≤ P

(|X −m| <

√m)

=∑

|k|<√m

(2mm+ k

)(12)2m

,

yielding ∑|k|<

√m

(2mm+ k

)≥ 22m−1,

which is what we want.

6. Use p1, . . . , pn to denote the members of the financial group. Suppose that thenumber of coins owned by person pi after doing business for j times is aij , wherei = 1, . . . , n and j ∈ N. Without loss of generality, assume that the (j + 1)stbusiness is done among p1, p2, p3, and p4. It follows that

a3j + a4j − a1j − a2j + 2k > 0.

Multiplying both sides of the above inequality by 2k, we easily conclude that

(a3j + k)2 + (a4j + k)2 + (a1j − k)2 + (a2j − k)2 > a21j + a2

2j + a23j + a2

4j .

Consequently,n∑i=1

a2i(j+1) >

n∑i=1

a2ij .

On the other hand, the sum of the number of coins owned by p1, . . . , pn is a fixednumber. Therefore, a finite number of business can be done among these people,proving the assertion.

224 2. SOLUTIONS

2.30.2. Second Day. 1. First solution: As X is separable, X has a count-able base, say B = Bi+∞i=1 , e.g., the family of all open balls with centers at thepoints of a countable dense subset of X and with positive rational radii is easilyseen to be a countable base for X. Let D1 = a ∈ X : limx→a f(x) < f(a)and D2 = a ∈ X : limx→a f(x) > f(a). It suffices to show that D1 ∪ D2 isat most countable. To this end, for a ∈ D1 (resp. a ∈ D2) choose an ra ∈ Qwith limx→a f(x) < ra < f(a) (resp. limx→a f(x) > ra > f(a)) and Bia ∈ B witha ∈ Bia such that f(x) < ra (resp. f(x) > ra) for all x ∈ Bia \ a. Definea function ι1 : D1 −→ Q × B (resp. ι2 : D2 −→ Q × B) by ι1(a) = (ra, Bia)(resp. ι2(a) = (ra, Bia)). If (ra, Bia) = (rb, Bib) for some a, b ∈ X, then a = b,for otherwise f(x) < ra = rb (resp. f(x) > ra = rb) whenever x ∈ Bia \ a. ButBia = Bib . Thus, f(b) < ra = rb < f(b) (resp. f(b) > ra = rb > f(b)), which is acontradiction. This proves that ι1 (resp. ι2) is one-to-one. Therefore, D1 and D2

are countable, and hence so is D1 ∪D2, which is what we want.

Second solution: For a given f : X −→ R, use D and L to denote the the set ofpoints at which f is discontinuous and has a limit, respectively. By showing thatD ∩ L is at most countable, we prove the assertion. Define ω : X −→ [0,+∞) by

ω(x) = limδ→0+

(sup f

(Bδ(x)

)− inf f

(Bδ(x)

)),

where Bδ(x) = y ∈ X : d(y, x) < δ denotes the open ball with center at xand radius δ > 0. Note that D =

⋃+∞n=1Dn, where Dn = x ∈ X : ω(x) > 1

n.Consequently, D ∩ L =

⋃+∞n=1(Dn ∩ L). It thus suffices to prove that Dn ∩ L is

at most countable for all n ∈ N. To this end, for a given a ∈ Dn ∩ L, we havelimx→a f(x) = la for some la ∈ R. It follows that there exists a δna > 0 such that|f(x)− la| < 1

2n whenever x ∈ Bδna(a) \ a. This implies |f(x)− f(y)| < 1

n for allx, y ∈ Bδna

(a) \ a. Hence, ω(x) ≤ 1n for all x ∈ Bδna

(a) \ a. In other words,x /∈ Dn for all x ∈ Bδna

(a) \ a. This implies that Dn ∩ L is at most countablefor all n ∈ N, for otherwise Dn ∩ L would have (uncountably many) limit pointsbecause X is separable. This obviously is in contradiction with x /∈ Dn for allx ∈ Bδna

(a) \ a and a ∈ Dn ∩ L, finishing the proof.

2. First solution: If M = +∞, the assertion is trivial. So, we may, without lossof generality, assume that M < +∞. Let D = z ∈ C : |z| < 1 denote the unit diskaround zero. Define g : D −→ C by g(z) = f(Rz)−a0

M+|a0| . Plainly, g is analytic on D andg(0) = 0. By the Maximum Modulus Principle, |f(Rz)| ≤ sup|z|=R |f(z)| = M forall z ∈ D. It follows that∣∣g(z)∣∣ ≤ ∣∣f(Rz)

∣∣+ ∣∣a0

∣∣M +

∣∣a0

∣∣ ≤M +

∣∣a0

∣∣M +

∣∣a0

∣∣ = 1,

for all z ∈ D. Thus, by the Schwarz Lemma, |g(z)| ≤ |z| for all z ∈ D. Now,if f(z0) = 0 for some z0 ∈ C, then z0

R ∈ D because |z0| < R. Consequently,|g( z0R )| ≤ | z0R |. But∣∣g(z0

R

)∣∣ = ∣∣∣f(R z0R

)− a0

M +∣∣a0

∣∣ ∣∣∣ = ∣∣a0

∣∣M +

∣∣a0

∣∣ ≤∣∣z0∣∣R,

yielding |z0| ≥ R|a0|M+|a0| , finishing the proof.

2.30. THIRTIETH COMPETITION 225

Second solution: Again, we may, without loss of generality, assume that M <

+∞. Let 0 < r < R be arbitrary. By the Cauchy Inequality, |an| ≤ M(r)rn for all

n ∈ N. In view of the Maximum Modulus Principle, we have M(r) ≤ M for allr < R, and hence we can write∣∣f(z)− a0

∣∣ ≤ +∞∑n=1

∣∣an∣∣∣∣z∣∣n ≤M(r)+∞∑n=1

( ∣∣z∣∣r

)n =M∣∣z∣∣

r −∣∣z∣∣ ,

for all z ∈ C with |z| < R. On the other hand,

M∣∣z∣∣

r −∣∣z∣∣ < ∣∣a0

∣∣⇐⇒ |z| < r|a0|M + |a0|

.

Consequently, f(z) 6= 0 whenever |z| < r|a0|M+|a0| . Since r < R is arbitrary, we see

that f(z) 6= 0 whenever |z| < R|a0|M+|a0| . This proves the assertion.

3. Let H be the set of all elements of G whose orders are finite. Obviously, H isnot empty because 1 ∈ H. Let x, y ∈ H be arbitrary. It follows that there is ann ∈ N such that xn = yn = 1. Consequently,

(xy−1)nG′ = (xy−1G′)n =((xG′)(y−1G′)

)n= (xG′)n(y−1G′)n = (xnG′)(y−nG′) = G′,

where the equality((xG′)(y−1G′)

)n = (xG′)n(y−1G′)n follows from the hypothesisthat G

G′ is abelian. Thus, (xy−1)n ∈ G′, from which, we obtain an m ∈ N such that(xy−1)nm = 1. That is, xy−1 ∈ H, and hence H ≤ G, as desired.

4. We prove the following, of which the assertion is a quick consequence. Let Dbe a division ring and F a subfield of its center. If A ∈ Mn(D) is algebraic overthe subfield F and rank(A) = rank(A2), then there is an idempotent E in the F -algebra generated by A which projects Dn onto the range of A along ker(A), i.e.,Dn = im(A)⊕ ker(A) and E(x+ y) = x, where x ∈ im(A) and y ∈ ker(A).

As is usual, the members of Mn(D) can be viewed as linear transformationsacting on the left of Dn via the usual matrix multiplication. To show that Dn =im(A)⊕ ker(A), note that by the Rank-Nullity Theorem, we have dim

(im(A)

)+

dim(ker(A)

)= dim(Dn) = n. On the other hand,

dim(im(A) + ker(A)

)= dim(im(A)) + dim(ker(A))− dim

(im(A) ∩ ker(A)

).

Thus, to prove Dn = im(A)⊕ker(A), it suffices to show that im(A)∩ker(A) = 0.To this end, let B := A|im(A) : im(A) −→ im(A). We can write

dim(im(A2)

)= rank(A2) = rank(A)

= dim(im(A)

)= dim

(ker(B)

)+ dim

(im(B)

)= dim

(im(A) ∩ ker(A)

)+ dim

(im(A2)

),

yielding dim(im(A) ∩ ker(A)

)= 0, whence im(A) ∩ ker(A) = 0.

For the rest, let f(A) = Am + am−1Am−1 + · · · + a1A + a0I = 0, where m is

minimal and a0, a1, ..., am−1 ∈ F . If a0 6= 0, then we have nothing to prove. Ifa0 = 0, then (Am−1 + am−1A

m−2 + · · · + a1I)A = 0. We claim that a1 6= 0. Ifnot, then (Am−2 + am−1A

m−3 + · · · + a2I)A2 = 0. Now, since im(A) = im(A2)because rank(A) = rank(A2), it follows that (Am−2 +am−1A

m−3 + · · ·+a2I)A = 0,

226 2. SOLUTIONS

a contradiction. So a1 6= 0. Set G = a−11 Am−1 + a−1

1 am−1Am−2 + · · ·+ a−1

1 a2A+ Iand note that Dn = ker(A) ⊕ im(A). Then E = I − G is the desired idempotent.Because for each x = Ax′ ∈ im(A) and y ∈ ker(A), we can write

E(x+ y) = Ex+ Ey

=(Ax′ − a−1

1 f(A)x′)−(a−11 Am−1 + a−1

1 am−1Am−2 + · · ·+ a−1

1 a2A)y

= (Ax′ − 0)− 0 = Ax′ = x.

So the proof is complete.

5. It is plain that the sets of odd and even numbers, denoted by O and E, respec-tively, is a partition of the natural numbers such that neither O ⊕ O nor E ⊕ Econtains any prime. To prove the uniqueness, suppose that N = A ∪ B and thatneither A ⊕ A nor B ⊕ B contains any prime. Without loss of generality, assumethat 1 ∈ A. By showing that A = O and B = E, we finish the proof. Proceed byinduction. As 1 ∈ A and 1 + 2 = 3, 2 ∈ B. Assume that

1, 3, . . . , 2k − 1⊆ A,

and 2, 4, . . . , 2k

⊆ B.

It follows from Bertrand’s conjecture that for n = 2k + 1, there is a prime p suchthat

2k + 1 < p ≤ 4k + 2.We have 2k + 1 < p ≤ 4k + 2 because k ≥ 1. But p − (2k + 1) is an even numberand 0 < p− (2k + 1) ≤ 2k. Hence, p− (2k + 1) ∈ B, whence 2k + 1 ∈ A. Likewise,from 0 < p − (2k + 2) ≤ 2k + 1, we see that 2k + 2 ∈ B. So the proof is completeby way of induction.

6. Let T denote the closure of T . It is plain that the distance between any twopoints of T is less than or equal to α. So, we may, with no loss of generality,assume that T is closed. Let x, y ∈ T be such that the distance between x andy is maximal. Use −→xy to denote the shortest arc joining x and y in the clockwisedirection. Construct T ′ from T as follows. If z ∈ T ∩ −→xy, then z ∈ T ′. If z ∈ Tand z /∈ −→xy, then z′ ∈ T ′, where z′ denotes the antipodal point of z. In this case,we have z′ /∈ T , for otherwise the distance between z and z′ is equal to one halfwhereas 0 < α < 1

2 . Also, z′ ∈ −→xy, because otherwise the distance between z and xor z and y will be greater than the distance between x and y, which is impossible.Thus, T ′ ⊆ −→xy. On the other hand, the sum of the lengths of the arcs whose unionis T is equal to the sum of the lengths of the arcs whose union is T ′. This obviouslyproves the assertion.

2.31. Thirty First Competition

2.31.1. First Day. 1. (a) and (b) The answer is no. Color the circle by twocolors, say black (b) and white (w), as shown in the figure. That is, (b) and (w) are,respectively, assigned to the upper and lower half-open semicircles. It is obviousthat any equilateral triangle inscribed in the circle has a white vertex and two black

2.31. THIRTY FIRST COMPETITION 227

Figure 14

Figure 15

vertices or vise versa. Also, the hypotenuse of any right triangle inscribed in thecircle is a diameter of the circle whose vertices are of the two colors.

(c) The answer is yes. Proceed by contradiction. Assume that there is acoloring of the circle for which there is no monocolored isosceles triangle inscribedin the circle. First choose a monocolored chord, say with white ends, which is nota diameter. Use w1w2 to denote the chord. The points on the circle that togetherwith this chord form an isosceles triangle lie on the perpendicular bisector of w1w2,which is a diameter of the circle. By the contradiction hypothesis, the two endsof this diameter are both black. Use b1b2 to denote this diameter. Likewise, thetwo ends of the diameter perpendicular to b1b2 must be both white. Use w3w4 todenote this diameter. Now consider the two chords w1w3 and w2w4. It follows thatthe ends of the diameters perpendicular to w1w3 and w2w4, respectively, must beall black. Use b3b4 and b5b6 to denote these diameters. It thus follows that theblack-colored triangles b2b4b5, b2b3b6, b1b4b5, and b1b3b6 are all isosceles. This is acontradiction, finishing the proof.

2. Using the Sequence Lemma, it is readily seen that

A′ +B ⊆ (A+B)′, A+B′ ⊆ (A+B)′,

implying that (A′+B)∪(A+B′) ⊆ (A+B)′. To see (A+B)′ ⊆ (A′+B)∪(A+B′),let x ∈ (A + B)′ be arbitrary. By the Sequence Lemma, there exists a sequence(an + bn)+∞n=1 with distinct terms such that x = limn(an + bn), where the sequences

228 2. SOLUTIONS

(an)+∞n=1 and (bn)+∞n=1 are in A and B, respectively. As A is bounded, if necessary,by passing to a subsequence, we may assume that there is an a ∈ A := A ∪ A′such that a = limn an. It follows that limn bn = limn

((an + bn) − an

)= x − a,

and hence b := x − a ∈ B because B is closed. There are two cases to consider.If a ∈ A′, then x = a + b ∈ (A′ + B) ⊆ (A′ + B) ∪ (A + B′), which is what wewant. If a ∈ A \ A′, then, if necessary by discarding a finite number of indices,we may assume that an = a for all n ∈ N. This implies that bn’s (n ∈ N) are alldistinct. Moreover, limn bn = x − a ∈ B′. Thus, x = a + (x − a) ∈ (A + B′) ⊆(A′+B)∪ (A+B′), as desired. So, in any case, x ∈ (A′+B)∪ (A+B′). Therefore,(A+B)′ ⊆ (A′ +B) ∪ (A+B′). This completes the proof.

3. Let H1, . . . ,Hn be all of the subgroups each of which has index 2 in G. It thusfollows that each Hi (i = 1, . . . , n) is normal in G, and hence so is H1 ∩ · · · ∩Hn inG. Also, H1 ∩ · · · ∩Hn is of finite index in G because so is every Hi (i = 1, . . . , n)in G. Set A := G

H1∩···∩Hn. We claim that A is the desired group. Suppose that B is

a subgroup of A of index 2. It follows that B := KH1∩···∩Hn

, where H1 ∩ · · · ∩Hn ⊆K E G. By the Third Isomorphism Theorem for groups , we have

A

B=

GH1∩···∩Hn

KH1∩···∩Hn

∼=G

K.

Thus, the index of K in G is 2. Therefore, K = Hi for some 1 ≤ i ≤ n. Conse-quently, A has exactly n subgroup of index 2 which are

H1

H1 ∩ · · · ∩Hn,

H2

H1 ∩ · · · ∩Hn, . . . ,

Hn

H1 ∩ · · · ∩Hn.

(Note that as Hi’s are mutually disjoint, so are the aforementioned groups.) Itremains to show that A is abelian. For each i = 1, . . . , n, the group G

Hiis abelian

because Hi has index 2 in G. Thus, G′ ⊆ Hi for all 1 ≤ i ≤ n, from which weobtain G′ ⊆ H1 ∩ · · · ∩Hn. Therefore, A is abelian, as desired.

4. No, one cannot find such two dice. To see this, proceed by contradiction.Consider two such dice A and B. For each i = 1, . . . , 6, let Ai and Bi, re-spectively, denote the probability of getting i for the two dice A and B. Foreach j = 2, . . . , 12, denote by Pj , the probability of getting a sum of j. Con-sequently, Pj =

∑jk=1AkBj−k. We have P2 = A1B1, P12 = A6B6, and P7 =

A1B6 +A2B5 +A3B4 +A4B3 +A5B2 +A6B1. So, we can write

P7 ≥ A1B6 +A6B1 ≥ 2√A1B6A6B1 = 2

√A1B1A6B6

= 2√P2P12 ≥ 2

√233.233

=433,

which is impossible because Pi ∈(

233 ,

433

)for all 2 ≤ i ≤ 12.

5. Note that the set of all balls whose centers are rational points of R2 and whoseradii are positive rationals is countable. Use B1, B2, B3, . . . to denote this set.Choose a set xi+∞i=1 such that xi ∈ Bi for all i ∈ N. As the set D := xi+∞i=1 isevidently dense in the plane, it suffices to show that D can be so chosen that nothree points of D are collinear. To this end, use induction on n to choose xn in sucha way that for all i < j < n, the points xi, xj , xn are not collinear. This obviouslycan be done because no ball is a union of a finite number of lines.

2.31. THIRTY FIRST COMPETITION 229

6. Let p(x) = xn + a1xn−1 + · · ·+ an be the characteristic polynomial of A and let

λ1, . . . , λn be the eigenvalues of A, which are the roots of p(x) = 0. As, for eachi = 1, . . . , n, the polynomial p(x) is divisible by x− λi, we can write

p(x)x− λi

= xn−1 + (λi + a1)xn−2 + · · ·+ (λn−1i + a1λ

n−2i + · · ·+ an−2λi + an−1).

We note that p′(x) = p(x)x−λ1

+ · · ·+ p(x)x−λn

. It thus follows that

p′(x) = nxn−1 +(tr(A) + na1

)xn−2 + · · ·

+(tr(An−1) + a1tr(An−2) + · · ·+ an−2tr(A) + nan−1

).

On the other hand, p′(x) = nxn−1 + (n− 1)a1xn−2 + · · ·+ an−1. Thus,

a1 = −tr(A)a1tr(A) + 2a2 = −tr(A2)

a1tr(A2) + a2tr(A) + 3a3 = −tr(A3)... =

...a1tr(An−2) + · · ·+ an−2tr(A) + (n− 1)an−1 = −tr(An−1)

a1tr(An−1) + · · ·+ an−1tr(A) + nan = −tr(An)

Note that the last equality in the above is obtained from p(λ1) + · · · + p(λn) = 0.It now follows from Cramer’s Rule that

an =

det

1 0 0 · · · · · · −tr(A)tr(A) 2 0 · · · · · · −tr(A2)tr(A2) tr(A) 3 0 · · · −tr(A3)

......

. . . . . ....

...tr(An−2) tr(An−3) · · · tr(A) n− 1 −tr(An−1)tr(An−1) tr(An−2) · · · · · · tr(A) −tr(An)

det

1 0 0 · · · · · · 0tr(A) 2 0 · · · · · · 0tr(A2) tr(A) 3 0 · · · 0

......

. . . . . . . . ....

tr(An−2) tr(An−3) · · · tr(A) n− 1 0tr(An−1) tr(An−2) tr(An−3) · · · tr(A) n

.

Note that the denominator of the fraction above is equal to n!. Thus, in view ofan = (−1)n detA, the assertion follows by substitution.

2.31.2. Second Day. 1. For the first part, just note that(n− 12

+ 1)

+(n− 1

2+ 2)

+ · · ·+(n− 1

2+ n

)= n2.

For the rest, suppose that m, k are positive integers such that (m+ 1) + (m+ 2) +· · · + (m + 12) = k2. It follows that 12m + 78 = k2. Consequently, n is even, andhence k = 2t for some t ∈ N. By substitution, we obtain 2t2 − 6m = 39, which isimpossible because the left hand side is an even integer.

230 2. SOLUTIONS

2. Performing the substitution u = xα and y = yβ and depending on the signs ofα and β, we need to investigate the limit of

u2v2

u3 + v3

as u and v tend to 0 or +∞. There are four cases to consider.(a) α > 0, β > 0.In this case, we need to investigate limu,v→0+

u2v2

u3+v3 . But by the SandwichTheorem, this limit exists and equals zero because

0 ≤ u2v2

u3 + v3=((uv

) 32 +

( vu

) 32)−1√

uv ≤√uv

2,

for all u, v > 0.(b) α > 0, β < 0.In this case, we need to investigate limu→0+,v→+∞

u2v2

u3+v3 . Again, by the Sand-wich Theorem, the limit exists in this case and equals zero because

0 ≤ u2v2

u3 + v3=

u2

u3

v2 + v≤ u2

v,

for all u, v > 0.(c) α < 0, β > 0.Just as in (b), one can readily see that limu→+∞,v→0+

u2v2

u3+v3 = 0.(d) α < 0, β < 0.By showing that there is a path on which the limit is infinity, we conclude that

the limit does not exist in this case. To this end, assuming that u = v, we have

limu,v→+∞,u=v

u2v2

u3 + v3= limu→+∞

u

2= +∞,

as desired.Alternatively, the cases (a), (b), and (c) above could be proven using the fol-

lowing inequalities

0 ≤ u2v2

u3 + v3≤ min

(u2

v,v2

u

)≤ max(u, v),

for all u, v > 0.

3. Let x = (x1, . . . , xn) and y = (y1, . . . , yn). Define the function f : An → An byf(z) = z′, where z = (z1, . . . , zn), z′ = (z′1, . . . , z

′n), and

z′j =

xj zj = xj 6= yj ,yj zj = yj 6= xj ,zj otherwise,

for all 1 ≤ j ≤ n. It is readily seen that

d(z, x) = d(f(z), y) = d(z′, y), d(z, y) = d(f(z), x) = d(z′, x).

Therefore, f is a one-to-one correspondence between the sets C and D, which iswhat we want.

4. First, we prove that “if F is a field, then the ring F [x] has infinitely manymaximal ideals.” To see this, noting that the ideal generated by an irreduciblemonic polynomial is a maximal ideal and that such ideals are distinct whenever

2.31. THIRTY FIRST COMPETITION 231

the generating polynomials are, it suffices to show that there are infinitely manyirreducible monic polynomials in F [x]. To this end, motivated by Euclid’s proof ofthe infinitude of primes, suppose by contradiction that for some n ∈ N, there areonly n irreducible monic polynomials in F [x], say f1, . . . , fn. But the polynomialf1 · · · fn + 1 must have an irreducible monic divisor. This implies that some fi(1 ≤ i ≤ n) divides f1 · · · fn + 1. Consequently, fi must divide 1, which is acontradiction. Thus, there are infinitely many irreducible monic polynomials inF [x].

Now, to prove the assertion, let M be a maximal ideal of R. By the precedingparagraph, the ring R

M [x] has infinitely many maximal ideals. Since the mappingφ : R[x] → R

M [x], defined by f(anxn + · · · + a1x + a0) = (an + M)xn + · · · +(a1 +M)x+(a0 +M) is an epimorphism, i.e., a surjective homomorphism, of rings,the ring R[x] has has infinitely many maximal ideals as well. This is because forevery maximal ideal I of R

M [x], φ−1(I) is also a maximal ideal of R[x] and thatφ−1(I) 6= φ−1(J) whenever I 6= J . This completes the proof.

5. First solution: The hypothesis can be rephrased as follows. We have AX = 0,where

A =

0 ±1 · · · ±1

±1. . . . . .

......

. . . . . . ±1±1 · · · ±1 0

2n×2n

, X =

x1

...

...x2n

2n×1

, 0 =

0......0

2n×1

.

That is, the diagonal entries of A are all zero and the off diagonal entries of A areall either 1 or −1. It suffices to prove that A is invertible. Since the determinantof A is an integer, we prove the assertion by showing that detA

2≡ 1. We can write

detA = det

0 ±1 · · · ±1

±1. . . . . .

......

. . . . . . ±1±1 · · · ±1 0

2≡ det

0 1 · · · 1

1. . . . . .

......

. . . . . . 11 · · · 1 0

= the number of derangements of 2n elements

= 2n!( 1

0!− 1

1!+

12!− · · ·+ (−1)2n

1(2n)!

)2≡ 1.

Recall that a derangement on n elements is a permutation on n elements hav-ing no fixed points. Also, a standard argument employing the inclusion-exclusionprinciple shows that the number of derangements on n elements is equal to

n!n∑k=0

(−1)k

k!.

Second solution: We prove the following, of which the assertion is a quick conse-quence. Let G be an abelian torsion free group, i.e., the identity is the only elementof the group having finite order. Let n, k ∈ N with n > k ≥ 2 and xi’s (1 ≤ i ≤ n)

232 2. SOLUTIONS

be elements of G such that removing any of them the remaining ones can be par-titioned into k subsets with equal sums. Prove that xi’s are all equal. Therefore,either xi’s are all zero or they are all equal to a nonzero element, in which case

nk≡ 1.

Since the subgroup generated by xi’s (1 ≤ i ≤ n) is finitely generated, inview of the Fundamental Theorem of Finitely Generated Abelian Groups, we may,without loss of generality, assume that G = Z. Also, if necessary, by adding anappropriate positive integer to all xi’s (1 ≤ i ≤ n), we may assume that xi ∈ Nfor each i = 1, . . . , n. First from the hypothesis, we easily obtain xj

k≡∑ni=1 xi for

each j = 1, . . . , n. Next, write xi’s in the basis k. Plainly, there is an N ∈ N suchthat for all 1 ≤ j ≤ n, we can write

xj =N∑i=0

aijki,

where aij ∈ N with 0 ≤ aij < k and 0 ≤ i ≤ N . Note that xjk≡∑ni=1 xi for each

j = 1, . . . , n obtains a0j = a01 for each j = 1, . . . , n. We use induction on N toprove that that xi’s are all equal. If N = 0, the assertion is easy because in thiscase xj = a0j = a01 = x1 for all 1 ≤ j ≤ n, as desired. Assuming that the assertionholds for N , we prove it for N + 1. To this end, noting that a0j = a01 for eachj = 1, . . . , n, define

yj :=xj − a01

k=N+1∑i=1

aijki−1,

where 1 ≤ j ≤ n. It readily follows from the hypothesis that removing any of theyi’s (1 ≤ i ≤ n) the remaining ones can be partitioned into k subsets with equalsums. Thus, the induction hypothesis applies to yi’s (1 ≤ i ≤ n), and hence yi’sare all equal. Therefore, so are xi’s (1 ≤ i ≤ n), which is what we want.

6. (a) Let A = (t1q1, 1 − t1) and B = (t2q2, 1 − t2), where t1, t2 ∈ (0, 1] andq1, q2 ∈ Q are distinct. Let r be an irrational number between q1, q2 ∈ Q. Itfollows that the line ` joining M and (r, 0) partitions the plane into the line ` andtwo open half-planes, say P1 and P2. It is obvious that T \ M ⊂ P1 ∪ P2 andthat A,B 6⊂ P1 whereas A,B ⊂ T \ M ⊂ P1 ∪ P2. It thus follows that anycontinuous path from A to B in the set T must intersect the line ` at the point Mbecause every other point of the line ` does not belong to the set T . This provesthe assertion.

(b) Let f : T → T be a continuous function and f(M) = M ′ = (t0q0, 1 − t0)for some t0 ∈ [0, 1] and q0 ∈ Q. If t0 = 0, there is nothing to prove, for thepoint M would be a fixed point of f . If not, define the function g : [0, 1] → T byg(t) = f(tt0q0, 1− tt0). Plainly, g is continuous. Define the set D as follows

D =

(tt0q0, 1− tt0) ∈ T : t ∈ [0, 1]

=t(t0q0, 1− t0) + (1− t)(0, 1) : t ∈ [0, 1]

.

It is plain that D is the line segment joining the points M = (0, 1) and M ′ =(t0q0, 1 − t0). Note that D is homeomorphic to the compact interval [0, 1]. Weprove the assertion by showing that f has a fixed point in D. If f takes no pointof D to the point M , the assertion is evident. That is because it would then followfrom (a) that f(D) is contained in D, and hence f has a fixed point in D, for D

2.31. THIRTY FIRST COMPETITION 233

is homeomorphic to the compact interval [0, 1]. So assume that there is a t ∈ [0, 1]such that g(t) = M . Define

tm = inft ∈ [0, 1] : g(t) = M

.

Since g is continuous and g(0) = f(M) 6= M , we see that tm > 0 and moreoverg(tm) = M . Also, it follows from (a) that the image of [0, tm] under g is containedin D. Use π2 : R2 → R to denote the projection on the y-axis. We can write

π2

(g(0)

)= 1− t0 < 1− 0t0 = 1

andπ2

(g(tm)

)= π2(M) = 1 > 1− tmt0.

Consequently, there is a t1 ∈ (0, tm) such that π2

(g(t1)

)= 1−t1t0. Since g(t1) ∈ D,

we obtainf(t1t0q0, 1− t1t0) = (t1t0q0, 1− t1t0),

which means f has a fixed point, as desired.

CHAPTER 3

Problem Index

First Competition

Analysis.Problem 1. (routine) Real analysis. Derivatives. Baire functions.Problem 2. Matrix analysis.Problem 3. Real analysis. Limits. Continuity.

Algebra.Problem 1. (routine) Ring theory. Commutative rings with identity. Prime

ideals. Maximal ideals.Problem 2. (routine) Group theory. Abelian groups.Problem 3. Baby set theory. Isomorphic sets.

General.Problem 1. (routine) Triangles.Problem 2. (routine) Decimal expansion.Problem 3. (routine) Complex numbers.Problem 4. (routine) Real numbers. Inequalities.

Differential Equations.Problem 1. (routine) Second order differential equations.

Probability and Statistics.Problem 1. (routine) Independent random variables.

Topology.Problem 1. (routine) Point-set topology of real numbers. (Ir)rational numbers.

Second Competition

Analysis.Problem 1. Real analysis. Continuous functions. Limit of sequences.Problem 2. (routine) Real analysis. Integral norm of continuous functions.Problem 3. Point-set topology of real numbers. Cantor set.

Algebra.Problem 1. (routine) Finite fields.Problem 2. (routine) Ring theory. (Right/Left) Quasi-regular elements.Problem 3. (routine) Homomorphism of rings. Polynomial rings in one variable.

General.Problem 1. (routine) Arithmetic. Congruences.Problem 2. (routine) Real numbers. Inequalities.

235

236 3. PROBLEM INDEX

Problem 3. Real analysis. Second order derivatives. Mechanics. Particle accel-eration.

Problem 4. (routine) Counting. The product rule of combinatorics.

Probability and Statistics.Problem 1. (routine) Covariance of random variables.

Topology.Problem 1. (routine) Point-set topology. Connectedness.

Differential Equations.Problem 1. (routine) First order differential equations.

Third Competition

Analysis.Problem 1. Real analysis. Derivatives.Problem 2. (routine) Real analysis. Continuous functions shifting forward se-

quences of real numbers.Problem 3(a). Roots of complex polynomials.Problem 3(b). Continuity of the roots of complex polynomials.Problem 3(c). Continuous algebraic functions. Holder functions.

Algebra.Problem 1. (routine) Modules. Dimension of vector spaces.Problem 2. (routine) Group theory. Homomorphisms characterizing abelian

groups.Problem 3. (routine) Semigroups. Groups.

General.Problem 1. (routine) Real functions of two variables.Problem 2. (routine) Binomial coefficients.Problem 3. Plane geometry.

Fourth Competition

Analysis.Problem 1. Real analysis. Continuity. Injectivity.Problem 2. Sequences of nonnegative real numbers.

Algebra.Problem 1. (routine) Ring theory. Commutative rings with identity. Radicals

of ideal. Prime ideals.Problem 2. (routine) Linear algebra. Center of the algebra of all linear trans-

formations.Problem 3. Ring theory. Artinian integral domains are fields.

General.Problem 1. Integral calculus.Problem 2. Conditional probability.

Fifth Competition

Analysis.

EIGHTH COMPETITION 237

Problem 1. Real analysis. Composition of Riemann integrable functions.Problem 2. Real analysis. Continuous functions not assuming any value more

than twice.Problem 3. Point-set topology of real numbers. Congestions points.

Algebra.Problem 1. (routine) Linear algebra. Eigenvalues of certain linear transforma-

tions.Problem 2. Group theory. Groups having two elements satisfying certain rela-

tions.Problem 3. Ring theory. Rings whose elements satisfy certain polynomial equa-

tions.

General.Problem 1. Number theory. Divisibility.Problem 2. Roots of complex polynomials and those of their derivatives. Closed

convex hulls.Problem 3. Plane geometry. Conic sections. Optics.

Sixth Competition

Analysis.Problem 1. (routine) Analytic geometry. Rational points.Problem 2. (routine) Metric spaces. Uniform convergence of sequences of con-

tinuous functions.Problem 3. Real analysis. Limits of improper integrals.

Algebra.Problem 1. Group theory. Subgroups.Problem 2. Matrices.Problem 3. Ring theory. Rings.

General.Problem 1. Integer numbers.Problem 2. Integer matrices.Problem 3. (routine) Plane geometry.

Seventh Competition

Analysis.Problem 1. Point set topology of real numbers.Problem 2. Real analysis. Continuity.Problem 3. Limits of sequences of of real numbers.

Algebra.Problem 1. Ring theory. Nilpotent elements. Idempotent elements.Problem 2. Group theory. Normal subgroups.Problem 3. Linear algebra. Linear transformations.

Eighth Competition

Analysis.Problem 1. (routine) Real analysis. Continuous extension.

238 3. PROBLEM INDEX

Problem 2. Convergence of series of real numbers.Problem 3. (routine) Riemann-Stieltjes integral. Convergence of sequences of

Riemann-Stieltjes integrals.

Algebra.Problem 1. (routine) Real matrices. The general linear group of order 2.Problem 2. Ring Theory. Infinite integral domains.Problem 3. (routine) Real matrices.

General.Problem 1. (routine) Minimum without using derivative.Problem 2. (routine) Arithmetic.Problem 3. Probability. Average function value.

Ninth Competition

Analysis.Problem 1. (routine) Point-set topology of real numbers. Equivalence relation.Problem 2. Real analysis. Continuity of inverse functions.Problem 3. (routine) Real analysis. Increasing continuous function.Problem 4. Real analysis. Riemann integrable function.Problem 5. Real analysis. Continuity. Injectivity. Surjectivity. Square filling

curves.

Algebra.Problem 1. Group theory. Finite groups. Subgroups of prime indexes. Normal

subgroups.Problem 2. Field of integers mod 3, Z3. Polynomial ring with coefficients in

Z3. Finite fields.Problem 3. (routine) Polynomial ring with coefficients in Z7. Greatest common

divisor.Problem 4. Field theory. Quadratic extensions.Problem 5. Linear algebra. Linear transformations.

General.Problem 1. Determinants.Problem 2. (routine) Algebraic equations.Problem 3. Real analysis. Continuous functions. Integrals.Problem 4. Diophantine equations. Number of solutions.Problem 5. Puzzles.Problem 6. Probability.Problem 7. Determinants.

Tenth Competition

Analysis.Problem 1. Real analysis. Derivatives. Limits.Problem 2. Real analysis. Higher derivatives. Definite integrals.Problem 3. (routine) Real analysis. Limits of definite integrals. Convergence

of function sequences.

Algebra.

TWELFTH COMPETITION 239

Problem 1. Group theory. Additive group of real numbers. Maximal sub-groups.

Problem 2. (routine) Ring theory. Ideals.Problem 3. Field theory. Field extensions.Problem 4. (routine) Matrix theory over general fields. Nilpotent matrices.

General.Problem 1. Probability. Random variables.Problem 2. Point-set topology of the plane.Problem 3. (routine) Differential and integral calculus. Differential equations.

Integral equations.Problem 4. (routine) Arithmetic. Divisibility.

Eleventh Competition

Analysis.Problem 1. Real analysis. Derivatives.Problem 2. Real analysis. Convergence of function sequences. Functional equa-

tions. Uniform convergence.Problem 3. Real analysis. Higher order derivatives. Convergence of series.

Algebra.Problem 1. Ring theory. Ideals. (Right) identity elements.Problem 2. Field theory. Algebraic numbers. Finite field extensions.Problem 3. Group theory. Finite groups.

General.Problem 1. Probability. Determinants.Problem 2. Real numbers. Algebraic equations.Problem 3. Baby set theory. Group theory.

Twelfth Competition

Analysis.Problem 1. Real analysis. Continuity. Injectivity. Surjectivity.Problem 2. Real analysis. Uniform convergence of series.Problem 3. (routine) Integral calculus. Definite integrals.

Algebra.Problem 1. Group theory. Finite groups. Sylow theory.Problem 2. Group theory. Finite abelian groups. Homomorphisms.Problem 3. Ring theory.Problem 4. Ring theory. Principal ideals.Problem 5. Matrix theory over general fields. Determinant. Trace.Problem 6. Algebraic numbers.

General.Problem 1. Plane Geometry.Problem 2. Combinatorics. Pigeonhole principal.Problem 3. Binomial coefficients.

240 3. PROBLEM INDEX

Thirteenth Competition

Analysis.Problem 1. Real analysis. Riemann integrable functions. Additive functions.Problem 2. Convergence of series of real numbers.Problem 3. Real analysis. Convergence of function sequences. Definite inte-

grals.

Algebra.Problem 1. Ring theory. Integral domains.Problem 2. Ring theory. Right (Left) ideals.Problem 3. Group theory.Problem 4. Group theory. Permutations.Problem 5. (routine) Matrix theory. Eigenvectors. Eigenvalues.

General.Problem 1. (routine) Plane geometry.Problem 2. Rational polynomials.Problem 3. Space geometry.

Fourteenth Competition

Analysis.Problem 1. Real analysis. Differentiability.Problem 2. Real analysis. Uniform continuity. Limits.Problem 3. Real analysis. Continuous functions.

Algebra.Problem 1. Group theory. Finite p-groups. Derived subgroups.Problem 2. Group theory. Finite groups. Normal subgroups. Center of groups.Problem 3. Ring theory. Greatest commons divisors. Least common multiples.Problem 4. (routine) Ring theory. Polynomial rings of two variables. Irre-

ducibility.Problem 5. Finite-dimensional vector spaces. Direct sums.

General.Problem 1. Combinatorics.Problem 2. Combinatorics. Counting.Problem 3. Real numbers.

Fifteenth Competition

Analysis.Problem 1. Real analysis. Convexity.Problem 2. (routine) Real analysis. Continuous functions. Uniform conver-

gence of function sequences.Problem 3. Real analysis. Real functions.

Algebra.Problem 1. Group theory. Finite groups. Sylow p-subgroups.Problem 2. Ring theory. (Right/Left) ideals. Prime ideals. Nilpotent elements.Problem 3. (routine, as stated, but its nontrivial counterpart is nonroutine!)

NINETEENTH COMPETITION 241

General.Problem 1. Elementary number theory.Problem 2. Integral calculus.Problem 3. Combinatorics. Probability. Binomial coefficients.

Sixteenth Competition

Analysis.Problem 1. (routine) Real analysis. Uniform convergence.Problem 2. Real analysis.Problem 3. Real analysis. Uniform continuity.

Algebra.Problem 1. Group theory. Finite abelian groups. Inner automorphisms.Problem 2. Ring theory. Nilpotent elements. Idempotent elements.Problem 3. Matrix theory. Number theory.

Seventeenth Competition

Analysis.Problem 1. Real analysis. Derivatives.Problem 2. (routine) Differential and integral calculus. Real polynomials. Di-

visibility.Problem 3. Real continuous functions on metric spaces.Problem 4. Real analysis. Continuity.

Algebra.Problem 1. Group theory. Finite groups.Problem 2. Ring theory. Polynomial rings in two variables with coefficients

from matrix rings over fields.Problem 3. Linear algebra. Linear transformations. Rank. Nullity.

Eighteenth Competition

Analysis.Problem 1. Real analysis. Power series.Problem 2. Real analysis. Periodic functions. Limits of definite integrals.Problem 3. Real analysis. Open maps.

Algebra.Problem 1. Group theory. Finite groups. Normal subgroups.Problem 2. Ring theory. Ascending chains of right ideals.Problem 3. Vector spaces. Direct sums.

Nineteenth Competition

Analysis.Problem 1. (routine) Real analysis. Differential and integral calculus. Zeros of

continuous functions.Problem 2. Real analysis. Limits.Problem 3. Real analysis. Continuous nowhere differentiable functions.

Algebra.Problem 1. Group theory. Normal subgroups. Posets.

242 3. PROBLEM INDEX

Problem 2. Ring theory. Minimal left ideals. Ideals.Problem 3. (routine) Matrix theory. Rank.Problem 4. Matrix theory. Rank. Determinant. Trace.

Twentieth Competition

Problem 1. Vector spaces. Complex numbers. Linear independence.Problem 2. Real analysis. Sequences of real numbers. Limits.Problem 3. Group theory. Automorphism groups. Solvability.Problem 4. Real analysis. Functional equations. Uniform continuity.Problem 5. Ring theory. Commutative rings with identity. Zero divisors.Problem 6. Real analysis. Periodic functions. Convergence of improper inte-

grals.

Twenty First Competition

Analysis.Problem 1. Real analysis. Limits.Problem 2. Real analysis. Continuous functions. Limits. Integrals.Problem 3. Real analysis. One-sided limits. Riemann integrable functions.

Algebra.Problem 1. Ring theory. Small ideals.Problem 2. Group theory. Center of groups. Derived subgroups.Problem 3. Matrix theory. Trace. Symmetric matrices. Characteristic polyno-

mials.

Twenty Second Competition

Analysis.Problem 1. Real analysis. Derivatives.Problem 2. Real continuous functions on metric spaces. Series. Uniform con-

vergence.Problem 3. Complex analysis. Sequences of analytic functions. Integrals.

Algebra.Problem 1. Group theory. Group index.Problem 2. Ring theory. Commutative rings. Maximal ideals.Problem 3. Linear algebra. Characteristic polynomials. Rank.

General.This section has forty routine multiple choice problems.

Twenty Third Competition

Analysis.Problem 1. Real analysis. Continuous bounded functions. Uniform norm.

Compactness. Uniform continuity.Problem 2. Real analysis. Trigonometric series.Problem 3. Real analysis. Convergence of function sequences. Differentiability.

Algebra.Problem 1. Group theory. Inner automorphism groups. Nonabelian groups.Problem 2. Ring theory. (Right) Ideals. Division rings.

TWENTY SIXTH COMPETITION 243

Problem 3. (routine) Matrix theory. Zero trace.

Twenty Fourth Competition

First Day.Problem 1. Real analysis. Derivatives.Problem 2. Real analysis. Real valued harmonic functions.Problem 3. Number theory. Primes. Divisibility. Congruences.Problem 4. (routine) Group theory. Infinite abelian groups.Problem 5. Combinatorics. Counting.Problem 6. Probability. Conditional probability.

Second Day.Problem 1. Euclidean spaces. Metric spaces. Open balls.Problem 2. (routine) Real analysis. Differential and integral calculus. Contin-

uous functions.Problem 3. Matrix theory. Rank inequalities.Problem 4. (routine) Ring theory. (Left) ideals.Problem 5. Combinatorics.Problem 6. Combinatorics. Game theory.

Twenty Fifth Competition

First Day.Problem 1. Group theory. Finite groups. Group centers.Problem 2. Linear algebra. Idempotents. Image. Kernel.Problem 3. Real analysis. Higher derivatives.Problem 4. Real analysis. Derivatives.Problem 5. Probability.Problem 6. Combinatorics. Triangles with integer sides. Partitions of numbers.

Second Day.Problem 1. Ring theory. Integral domains. Multiplicative group of units.Problem 2. Number theory. Prime numbers. Divisibility. Congruences.Problem 3. Continuous functions on metric spaces.Problem 4. Functions of one complex variable.Problem 5. Combinatorics. Matrices with −1, 1 entries.Problem 6. Finite sets. Relations.

Twenty Sixth Competition

First Day.Problem 1. Real analysis. Continuous functions. Derivatives.Problem 2. Complex analysis. Analytic functions.Problem 3. (routine) Number theory. Primes. Congruences.Problem 4. Ring theory. Commutative rings with identity. Maximal ideals.Problem 5. Combinatorics.Problem 6. Combinatorics. Finite sets.

Second Day.Problem 1. Real analysis. Continuous functions.Problem 2. Real analysis. Continuous functions.

244 3. PROBLEM INDEX

Problem 3. Group theory. Maximal subgroups.Problem 4. Matrix theory. Eigenvalues. Eigenvectors.Problem 5. Combinatorics. Probability.Problem 6. Discrete mathematics.

Twenty Seventh Competition

First Day.Problem 1. Matrix analysis.Problem 2. Complex analysis. Analytic functions.Problem 3. Sequences of real numbers. Recurrence relations.Problem 4. Set theory.Problem 5. Matrices. Product of matrices.Problem 6. Group theory. Normal subgroups.

Second Day.Problem 1. Real analysis. Derivatives.Problem 2. Point set topology of real numbers.Problem 3. Analytic geometry.Problem 4. Combinatorics. Finite sets.Problem 5. Number theory. Congruences. Divisibility.Problem 6. Ring theory. Boolean rings. Isomorphisms.

Twenty Eighth Competition

First Day.Problem 1. Complex analysis.Problem 2. Euclidean spaces. Continuous functions.Problem 3. Number theory. Arithmetic functions. Euler’s totient function.Problem 4. Group theory. Symmetric groups. Group center.Problem 5. Vector spaces over Z2.Problem 6. Posets.

Second Day.Problem 1. (routine) Integral calculus.Problem 2. Real analysis. Continuous functions.Problem 3. Complex matrices. Orthogonal matrices. Determinants.Problem 4. Ring theory. Unital rings whose elements satisfy certain relations.Problem 5. Probability. Conditional probability.Problem 6. Combinatorics. Set theory.

Twenty Ninth Competition

First Day.Problem 1. (routine) Integral calculus.Problem 2. Rational numbers. Series of rational numbers.Problem 3. (routine) Number theory. Divisibility.Problem 4. Ring theory. Nilpotent ideals. Right ideals.Problem 5. (routine) Baby set theory.Problem 6. Combinatorics.

Second Day.

THIRTY FIRST COMPETITION 245

Problem 1. Complex analysis. Analytic functions.Problem 2. Connected metric spaces. Compact metric spaces.Problem 3. Group theory. Normalizer subgroup. Centralizer subgroup. Auto-

morphism group. Cyclic groups. Derived subgroups. Group center.Problem 4. Matrix theory. Triangular matrices. Invariant subspaces. Nilpotent

matrices.Problem 5. Probability. Expectation values.Problem 6. Point-set topology of real numbers. Cantor set.

Thirtieth Competition

First Day.Problem 1. (routine) Integral calculus.Problem 2. Limits of sequences of complex numbers.Problem 3. Ring theory.Problem 4. Number theory. Primes. Congruences. Divisibility.Problem 5. Binomial coefficients.Problem 6. Natural numbers.

Second Day.Problem 1. Real functions on separable metric spaces. Points of discontinuity.Problem 2. Complex analysis. Analytic functions. Power series.Problem 3. Group theory. Derived subgroups. Elements of finite order.Problem 4. Matrix theory. Rank. Image. Kernel. Direct sum. Idempotents.Problem 5. Number theory. Primes.Problem 6. Point-set topology. Circles. Arc lengths.

Thirty First Competition

First Day.Problem 1. Plane geometry. Triangles. Coloring circles by two colors.Problem 2. Point-set topology of Rn. Minkowski sums. Limit points.Problem 3. Group theory. Subgroups of index two. Finite abelian groups.Problem 4. Probability. Dice throwing.Problem 5. Point-set topology of R2. Collinearity.Problem 6. Matrix theory. Determinant. Trace.

Second Day.Problem 1. Number theory. Complete squares.Problem 2. Calculus of two real variables. Limits.Problem 3. Set theory. One-to-one correspondence.Problem 4. Ring theory. Polynomial rings over commutative rings with iden-

tity. Maximal ideals.Problem 5. Real numbers. Partitions of sets with equal sums.Problem 6. Metric spaces. Fixed point theory.

246 3. PROBLEM INDEX

Classification of problems

• Algebra,Ring theory,

Index

Abel’s Continuity Theorem, 180, 182

Acceleration, 5Algorithm, 35

AM-GM Inequality, 55

Algebraic, 16, 47Algebraic numbers, 18

Archimedean property of real numbers, 52

Area, 19, 32, 43Arzela’s Theorem, 129

Automorphism, 18

Baire Category Theorem, 163, 208

Baire function, 1, 52

Banach space, 162, 191Bayes’ Theorem, 74, 188-190, 215

Bertrand’s conjecture, 48, 226

Bertrand’s principle, 48Binomial Theorem, 70, 72, 79, 85, 94

Bisection method, 34

Boolean ring, 78, 122, 209

Cantor set, 3, 46

Cauchy Inequality, 225Cauchy product of two polynomials, 70

Cauchy-Riemann Equations, 185

Cauchy-Schwarz Inequality, 216Cauchy’s criterion for the convergence of

improper integrals, 168

Cauchy’s Integral Formula, 177, 184Cauchy’s Theorem, 117

Cayley-Hamilton Theorem, 123

Centralizer, 117, 151, 196, 204Chebyshev Inequality, 47, 223Clopen subset, 218Chain

ascending, 26, 90, 122, 122, 159

descending, 9, 73Characteristic, 14, 19

Circle, 10, 11, 19, 21, 48Class equation, 117Closed convex hull, 80Commutator subgroup, 22, 167

Compact, 6, 28, 36Comparison Test, 127, 155Complete metric space, 38

Complete square, 49Congestion point, 9, 77

Connected, 5, 34, 37, 46

Connected component, 6

Convergent, 13, 17, 20, 28, 35pointwise, 17

uniformly, 10, 23, 29

Darboux’s property, 52

Darboux’s Theorem, 51, 52, 169, 176, 184,

207Dedekind’s Extension of Abel’s Theorem,

126, 127

Dense, 14, 47, 48Derangement, 231

Derivative, 1, 10, 13, 16, 17, 32, 37, 46

Derived subgroup, 22, 29, 47, 90, 167Determinant, 15, 18, 105, 107, 118, 231

Dihedral group, 183Dini’s Theorem, 116

Dirichlet’s Theorem, 181

Disconnected, 1, 41Division algorithm, 111

Division ring, 37, 59, 85, 111, 146, 154, 193,

225Domain, 37

Eigenvalue, 10, 21, 41Eigenvector, 21, 41

Eisenstein’s Criterion, 117, 123

Elementidempotent, 12, 25, 42

identity, 4, 12

invertible, 4left identity, 12left quasi-regular, 4

maximal, 27nilpotent, 12

right quasi-regular, 4

right identity, 7quasi-regular, 4

Ellipse, 10, 81Endomorphism, 4, 8Equation

differential, 2, 5, 17, 30, 34functional, 17integral, 17

Euclideanalgorithm, 103

247

248 INDEX

metric, 43, 192

norm, 1

plane, 8, 42, 130

ring, 60, 103

Euclid’s proof of the infinitude of primes,

231

Euler’s totient function, 43, 59, 152, 216

Expansion,

binary, 26, 35

decimal, 24, 35

hexadecimal, 35

octal, 35

ternary, 3, 46

Extension, 13, 14, 16, 18

Fermat’s Little Theorem, 138, 185, 202, 222

Field, 4, 7, 9, 13, 14, 16, 18, 19, 22, 23, 26,27, 28, 30, 37, 41, 45, 46, 47

First Isomorphism Theorem

for groups, 102, 132, 145, 150, 167, 178,

218

for modules, 69, 168

for rings, 178

Flux, 33

Frobenius Inequality, 193

Function

analytic, 29, 40, 41, 46, 47

arithmetic, 43

auxiliary, 16

bounded, 21, 29, 32

choice, 36

complex valued, 6

continuous, 1, 3, 6, 6, 8, 9, 10, 12, 13, 14,

15, 16, 17, 17, 18, 20, 21, 24, 25, 26, 27,

28, 29, 32, 37, 39, 40, 41, 43, 45, 46,49

continuous bounded, 36

continuously differentiable, 13, 27, 39

convex, 23

decreasing, 28, 32

differentiable, 17, 20, 21, 25, 29, 32, 39,

40, 42

entire, 43

harmonic, 37

increasing, 14, 23, 32

integral part, 24

nonnegative, 3, 24

nowhere differentiable, 27

one-to-one, 8, 14, 17, 18, 42, 43

periodic, 24, 26, 28

real, 1, 2, 6, 7, 16, 23

surjective, 18, 39, 43, 45

uniformly continuous, 21, 24, 28, 32, 36

Fundamental Theorem of Algebra, 210

Fundamental Theorems of Calculus,

The First, 92, 113, 160, 184

The Second, 92, 109

Fundamental Theorem of finite(ly

generated) abelian groups, 58, 118,

151, 175, 232

Gamma function, 34

Gauss-Lucas Theorem, 79

Greatest common divisor, 14, 22

Group

abelian, 2, 19, 37, 42, 48

commutative, 2, 19, 37

cyclic, 4, 19, 58, 59, 111, 121

finite, 14, 18, 22, 26, 26, 39

finite abelian, 48

finite nonabelian, 25

finite p-, 22

infinite, 37

nonabelian, 25, 37

of inner automorphisms, 25, 37, 150, 167

of permutations, 20, 102, 178, 211

of units, 198

p, 22, 117, 137, 150

symmetric, 52, 102, 131, 145, 178

quaternionic, 196

simple, 26

simple abelian, 111

solvable, 28, 137, 167

Hamel basis, 72

Hyperbola, 81

Ideal

left, 20, 23, 38

maximal, 1, 13, 30, 40, 49

minimal left, 27

nilpotent, 45

prime, 1, 8, 23

principal, 19, 40

right, 20, 23, 26, 37, 45

small, 29

two-sided, 18, 27, 37, 38, 45

Inclusion, 27, 46

Inclusion-exclusion principle, 148, 231

Instantaneous

speed, 61

velocity (vector), 62

Integral, 9, 24, 33, 36

Integral domain, 9, 13, 20, 39

Integration by parts, 64, 93, 116, 120, 149

Intermediate value property, 51, 207

Intermediate Value Theorem, 51, 62, 65,77, 97, 119, 128, 133, 137, 137, 153,

161, 170, 193, 202, 205, 207, 213

Invariant subspace(s), 46

Inverse Probability Theorem, 74

Isomorphic sets, 2

Iteration method, 35

Jacobian, 33

INDEX 249

Labeled graph, 139, 187

Lagrange’s Theorem, 85, 111, 118

Largest prime power dividing a factorial,

114

Least common multiple, 22

Lebesgue number, 98, 98

Lebesgue’s Integrability Criterion for

Riemann integrals, 75, 98, 174

Lebesgue’s Number Lemma, 98

Left annihilator, 216

Legendre symbol, 222

Leibniz’s Theorem, 181

L’Hopital’s rule, 161, 221

Limit, 1, 29, 46, 49

Limit Comparison Test, 117, 127

Limit point, 28, 48

Line, 4, 11, 21, 35, 41, 42

Linearly dependent, 8, 28

Linear transformation, 4, 10, 12, 14, 26, 30,

39

Matrix, 1, 11, 13, 15, 15, 18, 19, 21, 23, 27,29, 37, 38, 40, 41, 42, 44, 46, 47, 48

Maximal subgroup, 16

Maximum Modulus Principle, 202, 224

Maximum Modulus Theorem for harmonic

functions, 184, 185

Mean Value Property, 184

Mean Value Theorem, 51, 64, 116, 128,176, 183, 197, 202

Mean Value Theorem for integrals, 173

Mean Value Theorem for second order

derivatives, 62, 62

Metric space

compact, 39, 46,

complete, 38

connected, 46

separable, 47

Minkowski sum, 48

Module, 7

Morera’s Theorem, 177

Mobius

function, 151

Inversion Formula, 210

Nilpotent matrix, 16, 46

Noetherian ring, 154

Normalizer, 145

Normal subgroup, 12, 14, 22, 26, 42

Normed linear space, 149

Nowhere dense subset, 163, 207

Odd prime, 19, 23, 39, 40

One-to-one correspondence, 2, 23, 45, 49

Open map, 26

Operator norm, 205

Optical property of

ellipses, 81

hyperbolas, 81

Orthogonal matrix, 44

Partition, 39, 42, 47

Perfect subset, 207

Permutation, 11, 20

Persian alphabet, 5

Persian calendar, 4

Picard’s Great Theorem, 210

Plane, 7, 21, 35

Poisson random variable, 55, 56

Polynomial, 10, 14, 21, 25, 26, 47

characteristic, 29, 30, 108, 122, 178, 206,

229

minimal, 123, 179, 207

elementary symmetric, 142

Poset, 43

Prime ring, 198

Primitive root, 47

Probability, 3, 9, 13, 15, 17, 18, 24, 31, 38,39, 41, 44, 48

Radical of an ideal, 8

Radius of convergence, 26, 47

Random variables, 3, 5, 17, 47

Rank, 23, 26, 27, 30, 38, 47

Rank-Nullity Theorem, 85, 90, 104, 108,

154, 165, 225

Rational Root Theorem, 133

Regular n-gon, 21

Riemann integrable, 9, 19, 20, 21, 29

Riemann’s criterion for integrability, 75, 98,

156

Right annihilator, 216

Ring

commutative, 1, 8, 10, 19, 20, 28, 30, 40,44, 47, 49

division, 37, 59, 85, 111, 146, 154, 193,225

noetherian, 154

uncountable,

unital, 16, 19, 20, 23, 40, 44

with identity, 1, 9, 10, 11, 12, 18, 27, 28,47, 49

Rolle’s Theorem, 109, 207

Root, 6, 10, 18, 19, 21, 27, 34, 35, 42

Rouche’s Theorem, 67

Sandwich Theorem, 230

Schroder-Bernstein Theorem, 53, 101

Schwarz Lemma, 224

Second Isomorphism Theorem for groups,90, 150

Second Mean Value Theorem for integrals,168

Semigroup, 7

Sequence, 6, 8, 10, 12, 17, 20, 23, 24, 26,

28, 29, 32, 37, 42, 45

decreasing, 6, 13, 18

250 INDEX

increasing, 6, 17, 45

Series, 18, 20, 31, 35, 36, 45

power, 26, 33, 47Set

of measure zero, 75

partially ordered, a.k.a. poset, 43power, 2

totally ordered, 76

Simple left module, 168Speed, 4

Square, 19

Squeeze Lemma, 110Stolz’s Theorems on limits

the first theorem, 87the second theorem, 87, 173

Sylow p-subgroup(s), 23, 121, 145

Sylow’s Theoremsthe first theorem, 137

the second theorem, 145

the third theorem, 121Sylvester Inequality, 193

Sylvester’s problem, 133

Symmetric difference of sets, 118

Tangential acceleration, 62

Taylor’s Formula, 79Third Isomorphism Theorem for groups,

163, 228

Tree, 139Triangle, 2, 39, 48

Topological space, 5

Torsion free group, 231Turan’s Theorem, 187, 187

Uniqueness Theorem for analytic functions,205, 217

Vandermonde’s determinant formula, 142Variation of parameters, 55

Vector field, 33, 34

Vector space, 3, 4, 8, 10, 12, 14, 16, 22, 26,30, 39, 45

Well-ordering principle of natural numbers,52, 54, 158

Zero divisor(s), 28Zorn’s Lemma, 36, 72, 94, 146