junior problems - awesomemath

Junior problems

J415. Prove that for all real numbers x, y, z at least one of the numbers

23x−y + 23x−z − 2y+z+1

23y−z + 23y−x − 2z+x+1

23z−x + 23z−y − 2x+y+1

is nonnegative.

Proposed by Adrian Andreescu, Dallas, USA

Solution by the authorWe argue by contradiction assuming that all the three numbers are negative.

Then

2y+z+1 > 23x−y + 23x−z ≥ 2 ⋅ 23x−y+3x−z

2 ,

by the AM-GM Inequality. It follows that

y + z + 1 >3x − y + 3x − z

2+ 1,

implying y + z > 2x. Adding this with the analogous z + x > 2y and x + y > 2z yields a contradiction.

Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Daniel Lasaosa,Pamplona, Spain; Nikos Kalapodis, Patras, Greece; Alessandro Ventullo, Milan, Italy; Arkady Alt, San Jose,CA, USA; Jamal Gadirov, Istanbul University, Istanbul, Turkey; Ioan Viorel Codreanu, Satulung, Mara-mures, Romania; Duy Quan Tran, University of Medicine and Pharmacy at Ho Chi Minh City, Vietnam;Kousik Sett, West Bengal, Hooghly, India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Nicusor Zlo-ta, Traian Vuia Technical College, Focsani, Romania; Paul Revenant, Lycée du Parc, Lyon, France; AlbertStadler, Herrliberg, Switzerland; Stephanie Li, New York, NY, USA; Titu Zvonaru, Comănes,ti, România;Polyahedra, Polk State College, FL, USA; Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake, Kolka-ta, India; Alan Yan, Princeton Junction, NJ, USA; Ervin Ramirez, Universidad Nacional de Ingenieria,Nicaragua; Prajnanaswaroopa S., Bangalore, Karnataka, India.

Mathematical Reflections 4 (2017) 1

J416. Find all positive real numbers a and b for which

ab

ab + 1+

a2b

a2 + b+

ab2

a + b2=1

2(a + b + ab).

Proposed by Mihaela Berindeanu, Bucharest, Romania

Solution by Polyahedra, Polk State College, USABy the AM-GM inequality, ab + 1 ≥ 2

√ab, a2 + b ≥ 2a

√b, and a + b2 ≥ 2b

√a. Thus

2(a + b + ab) − 4(ab

ab + 1+

a2b

a2 + b+

ab2

a + b2) ≥ 2(a + b + ab) − 2(

√ab + a

√b + b

√a)

= (√a −

√b)2 + (

√a −

√ab)2 + (

√b −

√ab)2 ≥ 0,

with equalities if and only if a = b = 1.

Also solved by Daniel Lasaosa, Pamplona, Spain; Arkady Alt, San Jose, CA, USA; Paolo Perfetti, Uni-versità degli studi di Tor Vergata Roma, Roma, Italy; Duy Quan Tran, University of Medicine and Pharmacyat Ho Chi Minh City, Vietnam; Konstantinos Metaxas, Athens, Greece; Kousik Sett, West Bengal, Hooghly;Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Nguyen Ngoc Tu, Ha Giang, Vietnam; Nicusor Zlota,Traian Vuia Technical College, Focsani, Romania; Shuborno Das, Ryan International School, Bangalore, In-dia; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, România; Luca Ferrigno, Universitàdegli studi di Tor Vergata, Roma, Italy; Rajarshi Kanta Ghosh, Kolkata, India.


J417. Solve in positive real numbers the equation

x2 + y2

xy + 1=

√

2 −1

xy.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution by Jamal Gadirov, Istanbul University, Istanbul, TurkeyLet take square of both sides :

(x2 + y2)2

(1 + xy)2=2xy − 1

xy

After some simplifications we will getxy(x4 + y4) + 1 = 3x2y2

Since x, y positive real numbers by AM −GM inequality we have

x5y + y5x + 1 ≥ 3 3√x6y6 = 3x2y2

That is equation has solution if and only if whenever equality case holds. Thus, it has solution iff x5y = y5x =1. Now if x5y = y5x then xy(x − y)(x + y)(x2 + y2) = 0. Clearly x, y ≠ 0. Therefore there are two cases :Case 1: If x = y then x6 = 1 (roots of unity) and only real solution is x = 1 = y.Case 2: If x = −y then x6 = −1 which has no real solution.So, we conlude that there is only (x, y) = {(1,1)} solution.

Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, FL, USA; Akash SinghaRoy, Hariyana Vidya Mandir, Salt Lake, Kolkata, India; Alan Yan, Princeton Junction, NJ, USA; ErvinRamirez, Universidad Nacional de Ingenieria, Nicaragua; Konstantinos Metaxas, Athens, Greece; Luca Fer-rigno, Università degli studi di Tor Vergata, Roma, Italy; Alessandro Ventullo, Milan, Italy; Arkady Alt, SanJose, CA, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Duy Quan Tran,University of Medicine and Pharmacy at Ho Chi Minh City, Vietnam; Kousik Sett, West Bengal, Hooghly,India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Nicusor Zlota, Traian Vuia Technical College,Focsani, Romania; Paul Revenant, Lycée du Parc, Lyon, France; Shuborno Das, Ryan International School,Bangalore, India; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, România.


J418. Prove that the following inequality holds for all a, b, c ∈ [0,1]

a + b + c + 3abc ≥ 2(ab + bc + ca).

Proposed by Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution by Daniel Lasaosa, Pamplona, SpainNote that the proposed inequality is equivalent to

a(1 − b)(1 − c) + b(1 − c)(1 − a) + c(1 − a)(1 − b) ≥ 0,

clearly true since all factors in all terms are nonnegative. Note further that if c ≠ 0, then either a = 1 or b = 1,hence wlog by symmetry in the problem, we may assume that c ∈ {0,1}. If c = 1 then equality is equivalentto (1 − a)(1 − b), ie a = 1 or b = 1 but not necessarily both. On the other hand, if c = 0 equality is equivalentto a + b = 2ab. But since ab ≤ a with equality iff a = 1 or b = 0 because a, b ∈ [0,1], and similarly ab ≤ b, wehave a + b ≤ a + b, or since equality must hold, either a = b = 0 or a = b = 1. We conclude that equality holdsin the proposed inequality iff (a, b, c) = (0,0,0) or iff (a, b, c) is a permutation of (1,1, k) where k is any realin [0,1].

Also solved by Polyahedra, Polk State College, USA; Akash Singha Roy, Hariyana Vidya Mandir, SaltLake, Kolkata, India; Alan Yan, Princeton Junction, NJ, USA; Prajnanaswaroopa S., Bangalore, Karna-taka, India; Konstantinos Metaxas, Athens, Greece; Luca Ferrigno, Università degli studi di Tor Vergata,Roma, Italy; Rajarshi Kanta Ghosh, Kolkata, India; Arkady Alt, San Jose, CA, USA; Kevin Soto Palacios,Huarmey-Perú; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Adamopoulos Dio-nysios, 3rd High School, Pyrgos, Greece; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Duy QuanTran, University of Medicine and Pharmacy at Ho Chi Minh City, Vietnam; Kousik Sett, West Bengal,Hooghly, India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Nicusor Zlota, Traian Vuia Techni-cal College, Focsani, Romania; Paraskevi-Andrianna Maroutsou, High School of Evangeliki, Athens, Greece;Paul Revenant, Lycée du Parc, Lyon, France; Shuborno Das, Ryan International School, Bangalore, India;Albert Stadler, Herrliberg, Switzerland.


J419. Let a, b, c be positive real numbers such that abc = 1. Prove that

1

a4 + b + c4+

1

b4 + c + a4+

1

c4 + a + b4≤

3

a + b + c.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Arkady Alt, San Jose, CA, USAIn homogeneous form the original inequality becomes

∑cyc

abc

a4 + ab2c + c4≤

3

a + b + c.

Sinceabc

a4 + ab2c + c4=

b

a3

c+ b2 +

c3

a

anda3

c+c3

a≥ a2 + b2 ⇐⇒ a4 + c4 ≥ a3c + ac3 ⇐⇒

(a2 + ac + c2) (a − c)2 ≥ 0

thenb

a3

c+ b2 +

c3

a

≤b

a2 + b2 + c2

and, therefore,

∑cyc

abc

a4 + ab2c + c4≤∑cyc

b

a2 + b2 + c2=

a + b + c

a2 + b2 + c2.

And also we have

a + b + c

a2 + b2 + c2≤

3

a + b + c⇐⇒ (a + b + c)2 ≤ 3 (a2 + b2 + c2) ⇐⇒

(a − b)2 + (b − c)2 + (c − a)2 ≥ 0.

Also solved by Daniel Lasaosa, Pamplona, Spain; Nikos Kalapodis, Patras, Greece; Polyahedra, Polk StateCollege, FL, USA; Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake, Kolkata, India; KonstantinosMetaxas, Athens, Greece; Kevin Soto Palacios, Huarmey-Perú; Paolo Perfetti, Università degli studi di TorVergata Roma, Roma, Italy; Abhay Chandra, Indian Institute of Technology, New Delhi, India; NerminHodžic, Dobošnica, Bosnia and Herzegovina; Nguyen Ngoc Tu, Ha Giang, Vietnam; Nicusor Zlota, TraianVuia Technical College, Focsani, Romania; Shuborno Das, Ryan International School, Bangalore, India;Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, România.


J420. Let ABC be a triangle and let A,B,C be the magnitudes of its angles, expressed in radians. Provethat if A,B,C and cosA, cosB, cosC are geometric sequences, then the triangle is equilateral.

Proposed by Nairi Sedrakyan, Yerevan, Armenia

First solution by Joel Schlosberg, Bayside, NY, USAFirst, we show that △ABC is acute. If △ABC has a nonacute angle, then being larger than the otherangles it must be the first or last in the geometric sequence A,B,C. Then cosA ⋅ cosC, as a product withone nonpositive factor, is nonpositive; while cos2B, the square of a nonzero quantity, is positive. ThuscosA ⋅ cosC ≠ cos2B, contradicting the assumption that cosA, cosB, cosC is a geometric sequence.

Define f ∶ (0, π/2)→ R by x↦ ln cosx. Since the cosine function is strictly decreasing on (0, π/2) and thelogarithm function is strictly increasing on (0, π/2), f is strictly decreasing. Since by the AM-GM inequality√AC ≤ A+C

2 with equality iff A = C,

ln cos√AC = f (

√AC) ≥ f (

A +C

2)

with equality iff A = C. Since f ′′(x) = − sec2 x < 0, f is strictly concave, so by Jensen’s inequality

f (A +C

2) ≥

f(A) + f(C)

2= ln

√cosA cosC

with equality iff A = C. If A,B,C and cosA, cosB, cosC are geometric sequences, then ln cos√AC =

ln cosB = ln√cosA cosC, so equality in the above inequalities implies that A = C. Since B =

√AC has the

common value as well, △ABC is equilateral.

Second solution by Polyahedra, Polk State College, USASuppose that B2 = AC and cos2B = cosA cosC. Then A and C must be acute. By the AM-GM inequalityand the fact that cosx is concave and decreasing for x ∈ (0, π2 ),

cosA cosC ≤ (cosA + cosC

2)

2

≤ cos2A +C

2≤ cos2

√AC = cos2B,

with equalities if and only if A = C. Hence B = A = C.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Hariyana Vidya Mandir, SaltLake, Kolkata, India; Kousik Sett, West Bengal, Hooghly, India; Nermin Hodžic, Dobošnica, Bosnia andHerzegovina; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania.


Senior problems

S415. Letf(x) =

(2x − 1)6x

22x−1 + 32x−1.

Evaluatef (

1

2018) + f (

3

2018) +⋯ + f (

2017

2018) .


Solution by Alessandro Ventullo, Milan, ItalyObserve that

f(1 − x) =(1 − 2x)61−x

21−2x + 31−2x=

(1 − 2x)6x

22x−1 + 32x−1= −f(x),

i.e. f(x) + f(1 − x) = 0.Therefore,

1009

∑k=1

f (2k − 1

2018) =

504

∑k=1

(f (2k − 1

2018) + f (

2018 − (2k − 1)

2018)) + f (

1009

2018)

= f (1009

2018)

= f (1

2)

= 0.

Also solved by Daniel Lasaosa, Pamplona, Spain; Nikos Kalapodis, Patras, Greece; Akash Singha Roy,Hariyana Vidya Mandir, Salt Lake, Kolkata, India; Konstantinos Metaxas, Athens, Greece; Luca Ferrigno,Universitá degli studi di Tor Vergata, Roma, Italy; Rajarshi Kanta Ghosh, Kolkata, India; Arkady Alt,San Jose, CA, USA; Kevin Soto Palacios, Huarmey-Perú; Jamal Gadirov, Istanbul University, Istanbul,Turkey; Kousik Sett, West Bengal, Hooghly, India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; PaulRevenant, Lycée du Parc, Lyon, France; Albert Stadler, Herrliberg, Switzerland; Tamoghno Kandar, Mumbai,India; Xingze Xu, Hangzhou Foreign Languages School A-level Centre, China; Titu Zvonaru, Comănes,ti,România.


S416. Let f ∶ N → {±1} be a function such that f(mn) = f(m)f(n), for all m,n ∈ N. Prove that there areinfinitely many n such that f(n) = f(n + 1).

Proposed by Oleksi Krugman, University College London, UK

Solution by Paul Revenant, Lycée du Parc, Lyon, FranceSuppose that there are only a finite number of n such that f(n) = f(n+1). Then there exists an even integerN such that:

∀n ≥ N, f(n) = −f(n + 1).

Denote ε = f(N). By induction, f(N + 2k) = ε and f(N + 2k + 1) = −ε for k ∈ N. Since N is even, the valuesof even integers greater than N are ε, and those of odd integers greater than N are −ε.

However, for all integer n, the given equality shows that f(n2) = f(n)2. Hence, f(n2) is nonnegative, andthen f(n2) = 1. Since n2 can take even and odd values greater than n, we face with a contradiction, becauseeven and odd integers greater than N shouldn’t have the same value by f .

Finally, there are infinitely many n such that f(n) = f(n + 1).

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake,Kolkata, India; Konstantinos Metaxas, Athens, Greece; Luca Ferrigno, Università degli studi di Tor Vergata,Roma, Italy; Rajarshi Kanta Ghosh, Kolkata, India; Alessandro Ventullo, Milan, Italy; Joel Schlosberg,Bayside, NY, USA; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Shuborno Das, Ryan InternationalSchool, Bangalore, India; Albert Stadler, Herrliberg, Switzerland.


S417. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that

a2

a + b+

b2

b + c+

c2

c + a≤3 (a2 + b2 + c2)

2(a + b + c).

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

First solution by Albert Stadler, Herrliberg, SwitzerlandWe note that

3(a2 + b2 + c2)

2(a + b + c)−

a2

a + b−

b2

b + c−

c2

c + a=

∑symm a4b −∑symm a

3b2

2(a + b)(b + c)(c + a)(a + b + c)≥ 0

by Muirhead’s inequality.

Second solution by Daniel Lasaosa, Pamplona, SpainAfter multiplying by the (clearly positive) product of denominators and rearranging terms, the proposedinequality is equivalent to

ab(s − c)(a − b)2 + bc(s − a)(b − c)2 + ca(s − b)(c − a)2 ≥ 0,

which is clearly true since all three terms in the LHS are nonnegative. Now, since no two of a, b, c are zero,s − a = b + c, s − b = c + a, s − c = a + b are all positive. If abc = 0, then wlog by symmetry we may considerc = 0, and the inequality becomes ab(a + b)(a − b)2 ≥ 0, where ab(a + b) > 0, and equality holds iff a = b. Ifabc ≠ 0, then ab(s − c), bc(s − a), ca(s − b) are positive, or equality holds iff a = b = c. Thus equality holds inthe proposed inequality iff either a = b = c takes a positive real value, or (a, b, c) is a permutation of (k, k,0),where k is any positive real value.

Also solved by Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake, Kolkata, India; Ervin Ramirez,Universidad Nacional de Ingenieria, Nicaragua; Prajnanaswaroopa S., Bangalore, Karnataka, India; Kon-stantinos Metaxas, Athens, Greece; Nguyen Ngoc Tu, Ha Giang, Vietnam; Arkady Alt, San Jose, CA, USA;Kevin Soto Palacios, Huarmey-Perú; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Ita-ly; Kousik Sett, West Bengal, Hooghly, India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; NguyenNgoc Tu, Ha Giang, Vietnam; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Paul Reve-nant, Lycée du Parc, Lyon, France; Titu Zvonaru, Comănes,ti, România; P.V.Swaminathan, Smart MindsAcademy, Chennai, India.


S418. Let a, b, c, d be positive real numbers such that abcd ≥ 1. Prove that

a + b

a + 1+b + c

b + 1+c + d

c + 1+d + a

d + 1≤ a + b + c + d.

Proposed by An Zhenping, Xianyang Normal University, China

Solution by the authorFrom the condition abcd ≥ 1, we can find λ ≥ 1 and x, y, z, t > 0 such that a = λx, b = λy, c = λz, d = λt andxyzt = 1. Then the given inequality becomes equivalent to

∑cyc

x + y

1 + λx≤∑

cyc

x

Now, it is sufficient to prove that∑cyc

x + y

1 + x≤∑

cyc

x.

This inequality is equivalent to

∑cyc

1 + xy

1 + x≥ 4,

Let x = nm , y = p

n , z =qp , t =

mq , where m,n, p, q > 0. Then, we have

∑cyc

1 + xy

1 + x= (m + p) (

1

m + n+

1

p + q) + (n + q) (

1

n + p+

1

q +m) ≥

≥4(m + p)

m + n + p + q+

4(n + q)

m + n + p + q= 4.


S419. Solve the system of equations

x(x4 − 5x2 + 5) = y

y(y4 − 5y2 + 5) = z

z(z4 − 5z2 + 5) = x


Solution by the authorFirst, we seek real solutions in the interval [−2,2]. Let x = 2 cos t, t ∈ [0, π]. Then y = 2(16 cos5 t− 20 cos3 t+5) = 2 cos 5t, z = cos 5(5t) = cos 25t, and so x = 2 cos 5(25t) = 2 cos 125t. Hence cos 125t = cos t, implying125t − t = 2k(π), k = 0,1, . . . ,124 or 125t + t = 2k′(π), k′ = 1,2, . . . ,126.

We obtain 63 + 62 distinct solutions: 2 cosk (π

62), k = 0,1, ...,62 and 2 cosk (

π

63), k′ = 1, ...,62, as 2 cos 0

and 2 cos(π) have already been found.

We do not need to look for more solutions since all 125 solutions we found, (2 coskπ

62,2 cos 5k

π

62,2 cos 25k

π

62),

k = 0,1, ...,62 and (2 coskπ

63,2 cos 5k

π

63,2 cos 25k

π

63), k = 0,1, ...,62 are distinct and the given system has

5 × 5 × 5 = 125 solutions.

Also solved by Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Ervin Ramirez, Univer-sidad Nacional de Ingenieria, Nicaragua; P.V.Swaminathan, Smart Minds Academy, Chennai, India; AngelMejía, Universidad Autónoma de Santo Domingo, Dominican Republic; Nermin Hodžic, Dobošnica, Bosniaand Herzegovina.


S420. Let T be the Toricelli point of a triangle ABC. Prove that

(AT +BT +CT )2≤ AB ⋅BC +BC ⋅CA +CA ⋅AB.

Proposed by Nguyen Viet Chung, Hanoi University of Science, Vietnam

Solution by Nikos Kalapodis, Patras, Greece

If triangle ABC contains an angle of 120○ or more then it is well-known that the Torricelli point is the vertexof this obtuse angle. So, (without loss of generality) if A ≥ 120○, then T ≡ A and the given inequality becomes(AB + CA)2 ≤ AB ⋅BC +BC ⋅ CA + CA ⋅AB or (b + c)2 ≤ ab + bc + ca. Thus, in this case, it is enough toprove that b2 + c2 + bc ≤ a(b + c) (1).Since A ≥ 120○, we have that cosA ≤ −1

2 or − cosA ≥ 12 and by the law of cosines we obtain

a2 = b2 + c2 − 2bc cosA ≥ b2 + c2 + bc. Taking into account the triangle inequality we haveb2 + c2 + bc ≤ a2 < a(b + c), so (1) is true.

If triangle ABC has no angle greater than 120○ then it is well-known that the Torricelli point is thatpoint inside triangle ABC at which ∠TAB = ∠TBC = ∠TCA = 120○. Furthermore T lies on AP where Pis the vertex of equilateral triangle PBC drawn outwardly of triangle ABC.Then AT +BT +CT = AT + TP = AP (We have that TP = BT +CT by the Ptolemy’s theorem on quadri-lateral TBPC).By the law of cosines in triangle BAP we obtain that

AP 2= a2 + c2 − 2ac cos∠ABP = a2 + c2 − 2ac cos (B + 60○) = a2 + c2 − 2ac(

1

2cosB −

√3

2sinB) =

a2 + c2 − ac cosB +√3ac sinB = a2 + c2 −

a2 + c2 − b2

2+ 2

√3S =

a2 + b2 + c2 + 4√3S

2.

Therefore we have to prove thata2 + b2 + c2 + 4

√3S

2≤ ab + bc + ca or

4√3S ≤ 2(ab+bc+ca)−a2−b2−c2. But this is the well-known Hadwiger-Finshler’s inequality and we are done.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake,Kolkata, India; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Kousik Sett, West Bengal, Hooghly,India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Nicusor Zlota, Traian Vuia Technical College,Focsani, Romania.


Undergraduate problems

U415. Prove that the polynomial P (X) = X4 + iX2 − 1 is irreducible in the ring of polynomials over Gaussintegers.

Proposed by Mircea Becheanu, University of Bucharest, Romania

Solution by Daniel Lasaosa, Pamplona, SpainLet z be a root of the polynomial. Note that

z2 =−i ±

√−1 + 4

2=−i +

√3

2, ∣z∣4 = ∣z2∣

2=1 + 3

4= 1,

or ∣z∣ = 1 for any root of the polynomial. Now, the Gauss integers with modulus 1 are 1, i,−1,−i, whosesquares are 1,−1,1,−1, clearly different from −i+

√3

2 . Or the roots of P (X) are not Gauss integers, henceP (X) does not admit any factor of the form X − z in the ring of polynomials over the Gauss integers, andif P (X) is not irreducible in this ring, then polynomials Q(X) = X2 + aX + b and R(x) = X2 + cX + d existsuch that P (X) = Q(X)R(X) and where a, b, c, d are Gauss integers. But

Q(X)R(X) =X4+ (a + c)X3

+ (ac + b + d)X2+ (bc + ad)X + bd,

or c = −a. If a = c = 0 then Q(X)R(X) = X4 + (b + d)X2 + bd, or b + d = i, bd = −1, and b, d are the roots ofr2 − ir − 1 = 0, for

r =i ±

√−1 + 4

2=i ±

√3

2,

which are clearly not Gauss integers. If c = −a ≠ 0, then b = d, and 2b − a2 = i, b2 = −1, for b = ±i, anda2 = 2b − i ∈ {i,−3i}. But if a2 = i then a = ±1+i√

2is not a Gauss integer, and if a2 = −3i then a = ±

√31−i

2 isnot a Gauss integer. The conclusion follows.

Also solved by Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake, Kolkata, India; PrajnanaswaroopaS., Bangalore, Karnataka, India; Brian Zilli, Hofstra University, Hempstead, NY, USA; Nermin Hodžic,Dobošnica, Bosnia and Herzegovina; Albert Stadler, Herrliberg, Switzerland.


U416. For any root z ∈ C of the polynomial X4 + iX2 − 1 we denote wz = z + 2z . Let f(x) = x

2 − 3. Prove that

∣(f(wz) − 1) f(wz − 1)f(wz + 1)∣

is an integer that does not depend on z.


Solution by Paul Revenant, Lycée du Parc, Lyon, FranceLet z be a root of the polynomial X4 + iX2 − 1. After computation, we get:

∣(f(wz) − 1)f(wz − 1)f(wz + 1)∣ = ∣z6 +64

z6∣ .

Now, we remark that z6 = (z2 − i)(z4 + iz2 − 1) − i = −i, and then:

∣z6 +64

z6∣ = ∣−i −

64

i∣ = 63.

Finally, ∣(f(wz − 1)f(wz − 1)f(wz + 1)∣ is an integer that does not depend on z.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Hariyana Vidya Mandir, SaltLake, Kolkata, India; Kousik Sett, West Bengal, Hooghly, India; Nermin Hodžic, Dobošnica, Bosnia andHerzegovina.


U417. Prove that for any n ≥ 14 and for any real number x, 0 < x < π2n , the following inequality holds:

sin 2x

sinx+sin 3x

sin 2x+ . . . +

sin(n + 1)x

sinnx< 2 cotx

Proposed by Nairi Sedrakyan, Yerevan, Armenia

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, ItalyLemma:For 0 < x < π

2n ,sin(k+1)xsinkx ≤ 1 + 1

k , 0 ≤ k ≤ n

Proof of the Lemma:We need to prove

∫

kx

0cos ydy + ∫

(k+1)x

kxcos ydy ≤ (1 +

1

k)∫

kx

0cosdy

that is

∫

(k+1)x

kxcos ydy ≤

1

k∫

kx

0cos ydy

This is implied by

x cos(kx) ≤sinkx

k⇐⇒ f(x) ≐ sin(kx) − kx cos(kx) ≥ 0

f ′(x) = k cos(kx) − x cos(kx) + k2x sin(kx) = k2x sin(kx), kx ≤ π/2

Thus f(x) ≥ f(0) = 0 ∎

Based on the Lemma, it suffices to show

n

∑k=1

1 +1

k< 2

cosx

sinx, 0 < x ≤

π

2n, n ≥ 14 (1)

The concavity of cosx for 0 ≤ x ≤ π/(2n), yields

cosx ≥ 1 +2nx

π(cos

π

2n− 1)

The inequality in (1) is implied by

n

∑k=1

1 +1

k< 2

1 + 2nxπ

(cos π2n − 1)

x=2

x−4n

πsin2

π

4n

which in turn is implied by (use (sinx ≤ x)

n

∑k=1

(1 +1

k) <

4n

π−π

4n

This will be proven by induction.Let

f(n) ≐n

∑k=1

(1 +1

k) −

4n

π+π

4n, f(12) ∼ −0.1


Suppose f(n) < 0 for any 12 ≤ n ≤ r. For n = r + 1 we need to show

r

∑k=1

(1 +1

k) + 1 +

1

r + 1−4r + 4

π+

π

4r + 4< 0

By using the inductive step we come to

4r

π−π

4r+ 1 +

1

r + 1−4r + 4

π+

π

4r + 4< 0

which is equivalent to

g(n) ≐π − 4

π+

1

n + 1−π

4n+

π

4n + 4< 0 ∀ n ≥ 12

We have g(12) ∼ −0.2 and

g′(n) =π

4n2−

1

(n + 1)2−

4π

16(n + 1)2

and it easy to observe that g′(n) < 0 for n ≥ 12 so concluding the proof.

Also solved by Nermin Hodžic, Dobošnica, Bosnia and Herzegovina.


U418. Let a, b, c be positive real numbers

such that abc = 1. Prove that

√16a2 + 9 +

√16a2 + 9 +

√16a2 + 9 ≤ 1 +

14

3(a + b + c)

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, ItalyLet a = ex,

b = ey, c = ez.The condition abc = 1 becomes x + y + z = 0.

(√16e2x + 9)

′′

=32e2x(8e2x + 9)

(16e2x + 9)32

≥ 0

The convexity implies the following standard result for a convex function f(x) defined on R

f(x1) + . . . + f(xn) ≤ f(x1 + . . . + xn − (n − 1)x0) + (n − 1)f(x0)

for any x0.

By choosing x0 = 0 and clearly f(x) =√16e2x + 9 we get

√16e2x + 9 +

√16e2y + 9 +

√16e2z + 9 ≤ f(0) + 2f(0) = 5 + 10 = 15

Moreover

1 +14

3(a + b + c) ≥ 1 +

14

33(abc)

13 = 15

and we are done.

Also solved by Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake, Kolkata, India; PrajnanaswaroopaS., Bangalore, Karnataka, India; Konstantinos Metaxas, Athens, Greece; Nermin Hodžic, Dobošnica, Bosniaand Herzegovina; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Albert Stadler, Herrliberg,Switzerland.


U419. Let p > 1 be a natural number. Prove that

limn→∞

(n

∑k=1

1p√k−

p

p − 1(n

p−1p − 1)) ∈ (0,1)

Proposed by Alessandro Ventullo, Milan, Italy

Solution by Arkady Alt, San Jose, CA, USA

Let an ∶=n

∑k=1

1p√k−

p

p − 1((n + 1

p−1p ) − 1) and bn ∶=

n

∑k=1

1p√k−

p

p − 1(n

p−1p − 1) .

First we will prove that an ↑ N and bn ⇃ N..Indeed, by Mean Value Theorem there is cn ∈ (n + 1, n + 2) such that

an+1 − an =1

p√n + 1

−p

p − 1((n + 2)

p−1p − (n + 1)

p−1p ) =

1p√n + 1

−1

p√cn

>1

p√n + 1

−1

p√n + 1

= 0

and similarly, there is cn ∈ (n,n + 1)

bn − bn+1 = −1

p√n + 1

+p

p − 1((n + 1)

p−1p − n

p−1p ) =

1p√cn

−1

p√n + 1

>1

p√n + 1

−1

p√n + 1

= 0

Since, an ↑ N, bn ↓ N and an < bn , n ∈ N then an < bm for any n,m ∈ N.Indeed,an < an+m < bn+m < bm.In particular an < b1 and a1 < bn, n ∈ N.Therefore, both sequences (an )n∈N and (bn )n∈N are convergent.Let l ∶= lim

n→∞an and u ∶= lim

n→∞bn. Since lim

n→∞(bn − an) = 0 then l = u.

Let c ∶= limn→∞

an = limn→∞

bn and also an < c < bn because an ↑ N, bn ↓ N andan < bn , n ∈ N.Thus, we have an < c < bn ⇐⇒

n

∑k=1

1p√k−

p

p − 1((n + 1

p−1p ) − 1) < c <

n

∑k=1

1p√k−

p

p − 1(n

p−1p − 1) ⇐⇒

p

p − 1(n

p−1p − 1) + c <

n

∑k=1

1p√k<

p

p − 1((n + 1

p−1p ) − 1) + c.

Sincep

p − 1(2

p−1p − 1) ↑ p ∈ NÓ {1} then

a1 =1

∑k=1

1p√k−

p

p − 1(((1 + 1)

p−1p ) − 1) =

1 −p

p − 1(2

p−1p − 1) ≥ 1 −

2

2 − 1(2

2−12 − 1) = 1 − 2 (

√2 − 1) = 1 −

2√2 + 1

=

√2 − 1

√2 + 1

> 0

and, therefore, a1 ≤ an < c Ô⇒ c > 0.

Also, since b1 =1

∑k=1

1p√k−

p

p − 1(1

p−1p − 1) = 1 and c < bn ≤ b1 then c < 1.

Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Luca Ferrigno,Università degli studi di Tor Vergata, Roma, Italy.


U420. Find the least length of a segment whose endpoints are on the hyperbola xy = 5 and ellipse x2

4 +4y2 = 2,

respectively.

Proposed by Titu Andreescu, USA and Oleg Mushkarov, Bulgaria

Solution by the authorsThe problem is: for real numbers a, b, c, d find the minimum of (a− c)2 + (b− d)2 given the constraints ab = 5

andc2

4+ 4d2 = 2.

We have

(a − c)2 + (b − d)2 = (2a√5− c

√5

2)

2

+ (b

√5− d

√5)

2

+ (1

5) (a − 2b)2 + 4

ab

5− (

c2

4+ 4d2) ≥ 4 − 2 = 2.

Hence the minimum is 2, achieved if and only if

a =√10, b =

√10

2, c = 4

√10

5, d =

√10

10or

a = −√10, b = −

√10

2, c = −4

√10

5, d = −

√10

10.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake,Kolkata, India; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Nermin Hodžic,Dobošnica, Bosnia and Herzegovina.


Olympiad problems

O415. Let n > 2 be an integer. An n×n square is divided into n2 unit squares. Find the maximum number ofunit squares that can be painted in such a way that every 1×3 rectangle contains at least one unpaintedunit square.

Proposed by Magauin Armanzhan, Kazakhstan and Nairi Sedrakyan, Armenia

Solution by Daniel Lasaosa, Pamplona, SpainBy "every 1 × 3 rectangle", we may understand either only those rectangles in a given orientation (ie thosecontaining the unit squares in the intersection of one row and the union of three consecutive columns), or anyrectangle of that size regardless of its orientation (ie also those containing the unit squares in the intersectionof one column and the union of three consecutive rows). Since the former is a relatively obvious proposition(see the note at the end of this solution), we will assume that the problem actually refers to the second case.

Number the rows and the columns, and denote by (i, j) the unit square in the intersection of the i-throw and the j-th column. Note that it suffices to leave ⌊n

2

3 ⌋ unit squares unpainted, more specifically thosesuch that i+ j ≡ 1 (mod 3). Clearly, any 1× 3 rectangle contains three unit squares with the same value of i(or j) and three consecutive values of j (or of i), yielding three consecutive integers as values of i+ j, one ofwhich will clearly be a multiple of three, and thus unpainted.

If n = 3k is a multiple of 3, this includes the diagonal defined by i + j = n + 1 which contains n unitsquares, and minor diagonals symmetric two by two with respect to this one, one of such diagonals containing3,6, . . . , n−3 unit squares on each side of diagonal i+ j = n+1, for a total of unpainted unit squares equal to

n(n − 3)

3+ n =

n2

3.

If n = 3k + 1 for some positive integer k, this includes minor diagonals in the same direction as diagonali + j = n + 1, with respectively 3,6, . . . , n − 1 unit squares, and on the other side of diagonal i + j = n + 1 withn − 2, n − 5, . . . ,2 unit squares, for a total of unpainted unit squares equal to

(n + 2)(n − 1)

6+n(n − 1)

6=n2 − 1

3=n2

3−1

3.

If n = 3k + 2, the same argument leads to diagonals containing 3,6, . . . , n − 2 unit squares on one side ofdiagonal i+ j = n+ 1, and diagonals containing 1,4, . . . , n− 1 unit squares on the other side of said diagonal,for a total of unpainted unit squares equal to

(n + 1)(n − 2)

6+n(n + 1)

6=n2 − 1

3=n2

3−1

3.

In all three cases, the total number of unpainted squares can be clearly described by the aforementionedexpression ⌊n

2

3 ⌋. Note now that for n = 3,4,5, this sufficient number of unpainted squares is also minimum.Indeed, in the 3 × 3 square, there must be at least one unpainted square in each line, the 4 × 4 square canbe tiled with four 1 × 3 rectangles and one 3 × 1 rectangle, leaving one unit square uncovered, and each oneof the five tiles must contain at least one unpainted square. For the 5 × 5 square, note that each row mustcontain at least one unpainted square, and for each row that contains exactly one unpainted square, it mustbe its central square. If this happens in more than two rows, there are either two adjacent rows next to oneof the edges of the square, or two rows separated by exactly one row, with exactly one unpainted squareeach, the central square in both cases. Considering the four 3 × 1 rectangles which contain the other foursquares of each one of these rows, their remaining four unit squares, which are on the same line, must be allleft unpainted. There are therefore, either at most two rows with exactly one unpainted square and at least


three rows with at least two unpainted squares each, or at least one row with at least four unpainted squaresand at least one unpainted square in each one of the other for rows, yielding at least 8 unpainted squares.

If the result holds for an n × n square, note that an (n + 3) × (n + 3) square may be partitioned into onen×n square, which by hypothesis of induction contains at least ⌊n

2

3 ⌋ unpainted squares, n+3 rectangles 3×1and n rectangles 1 × 3, for a minimum total number of unpained squares equal to

⌊n2

3⌋ + (n + 3) + n = ⌊

n2

3+ 2n + 3⌋ = ⌊

n2 + 6n + 9

3⌋ = ⌊

(n + 3)2

3⌋ ,

and the result holds for all n ≥ 3. Thus the maximum number of painted squares is

n2 − ⌊n2

3⌋ = ⌈

2n2

3⌉ .

Note: For the case where rectangles 1 × 3 of only one orientation are considered (wlog those orientedalong a row), it suffices to see that at least ⌊n

3⌋ unit squares must be left unpainted in each row, otherwise

partitioning the row in ⌊n3⌋ rectangles 1×3 and a remainder of 0, 1 or 2 unit squares in the end, by Dirichlet’s

pigeonhole principle, at least one of these rectangles would not have an unpainted unit square. On theother hand, numbering the columns from 1 to n and leaving unpainted all squares in the columns which aremultiples of 3 clearly solves the problem. It follows that the maximum number of painted squares in this(significantly easier at least at first sight) case is n2 − n ⌊n

3⌋.

Also solved by Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Joel Schlosberg, Bayside, NY, USA.


O416. Let a, b, c real numbers such that a2 + b2 + c2 − abc = 4. Find the minimum of

(ab − c)(bc − a)(ca − b) and all triples (a, b, c) for which the minimum is attained.

Proposed by Titu Andreescu,University of Texas at Dallas,USA

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy(ab − c)(bc − a)(ca − b) = (abc)2 − abc(a2 + b2 + c2) + (a2b2 + b2c2 + c2a2) − abc

We change variablesa + b + c = 3u, ab + bc + ca = 3v2, abc = w3 so we get w3 = 9u2 − 6v2 − 4 and

(ab − c)(bc − a)(ca − b) = −9(v2)2 + v2(30 − 36u) − 45u2 + 54u3 − 24u + 20

This is a concave parabola whose minimum is attained at oneof the extreme points of the variable v2. Thetheory states that once fixed the values of u and wthe extreme values of v occur whena = b (or cyclic) so we set a = b and a2 = 2 + c froma2 + b2 + c2 − abc = 4. Clearly c ≥ −2.

(ab − c)(bc − a)(ca − b)∣a=b=±

√2+c

= 2(2 + c)(c − 1)2 ≥ 0

Thus the minimum is zero and is attained when

(a, b, c) = (0,0,−2), (√3,

√3,1), (−

√3,−

√3,1)

and symmetric.

Also solved by Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake, Kolkata, India; Nermin Hodžic,Dobošnica, Bosnia and Herzegovina; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania.


O417. Let x1, x2, . . . , xn be real numbers such that x12 + x22 +⋯ + xn2 ≤ 1. Prove that

∣x1∣ + ∣x2∣ +⋯ + ∣xn∣ ≤√n(1 +

1

n) + n

n−12 x1x2⋯xn.

When does equality occur?

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution by Joel Schlosberg, Bayside, NY, USABy the arithmetic mean-quadratic mean inequality,

∣x1∣ +⋯ + ∣xn∣ ≤ n

√x21 +⋯ + x2n

n≤√n.

By the AM-GM inequality,

nn−12 ∣x1⋯xn∣ =

1√n

(n n√

x21⋯x2n)

n/2

≤1

√n

(x21 +⋯ + x2n)n/2

=1

√n.

Therefore,

∣x1∣ +⋯ + ∣xn∣ ≤√n +

1√n− n

n−12 ∣x1⋯xn∣ ≤

√n +

1√n+ n

n−12 x1⋯xn =

√n(1 +

1

n) + n

n−12 x1⋯xn.

Unless ∣x1∣ = ∣x2∣ = ⋯ = ∣xn∣ and equivalently x21 = ⋯ = x2n, inequality is strict in the AM-QM and AM-GMinequalities, and thus in the desired inequality.

Let the common absolute value be x. Since nx2 = x21 +⋯ + x2n ≤ 1,

∣x1∣ +⋯ + ∣xn∣ + nn−12 x1⋯xn = nx ± n

n−12 xn ≤

√n +

1√n,

with strict inequality unless x = 1√n. For x = 1√

n, equality implies that

n√n=√n +

1√n+ n

n−12 x1⋯xn

x1⋯xn = −(1

√n)

n

and therefore that an odd number of x1, . . . , xn are negative.Therefore, an odd number of x1, . . . , xn are − 1√

nand all the rest are 1√

n. Conversely, any such values for

x1, . . . , xn yield equality, so those are exactly the conditions when equality occurs.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake,Kolkata, India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Paul Revenant, Lycée du Parc, Lyon,France; Albert Stadler, Herrliberg, Switzerland.


O418. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that

a5

c3 + 1+

b5

a3 + 1+

c5

b3 + 1≥3

2.

Proposed by Konstantinos Metaxas, Athens, Greece

Solution by Daniel Lasaosa, Pamplona, SpainNote first that since 1 = a2+b2+c2

3 ≤ a3+b3+c3

3 because of the inequality between quadratic and cubic means, theproposed inequality is weaker than

6a5

4c3 + a3 + b3+

6b5

4a3 + b3 + c3+

6c5

4b3 + c3 + a3≥ a2 + b2 + c2,

both being equivalent iff a = b = c = 1. Now, this new inequality may rewrite, after multiplying both sides bythe product of denominators and rearranging terms, as follows:

∑cyca3 (3a8 + b8 − 4a6b2) +∑

cycb3 (3b8 + a8 − 4a2b6)+

+3∑cyca3b3 (3a5 + 2b5 − 5a3b2) +

3

7∑cyc

(33a8b3 + 6b8c3 + 10c8a3 − 49a6b3c2)+

+3

7∑cyc

(30a8b3 + b8c3 + 18c8a3 − 49a6b2c3)+

+∑cyc

(5a11 + 5b11 + 29a8b3 + 2a3b8 + 3a5b6 − 44a7b4)

+44

37∑cyc

(19a7b4 + 5b7c4 + 13c7a4 − 37a5b3c3)+

+4 (a11 + b11 + c11 − a3b3c3 (a2 + b2 + c2)) ≥ 0.

Now, all terms in the LHS except for the last one are nonnegative by use of the weighted AM-GM inequality,whereas the last term is nonnegative by use of the power-mean inequality and the AM-GM inequality. Thelast term is zero iff a = b = c = 1, and this condition is sufficient for equality in all other conditions, henceequality holds in the proposed inequality iff a = b = c = 1.

Also solved by Luke Robitaille, Euless, Texas, USA; Akash Singha Roy, Hariyana Vidya Mandir, SaltLake, Kolkata, India; Arkady Alt, San Jose, CA, USA; Kevin Soto Palacios, Huarmey-Perú; Paolo Perfetti,Università degli studi di Tor Vergata Roma, Roma, Italy; Nermin Hodžic, Dobošnica, Bosnia and Herzegovi-na; Nguyen Ngoc Tu, Ha Giang, Vietnam; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania;Albert Stadler, Herrliberg, Switzerland.


O419. Let x1, x2, . . . , xn be real numbers in the interval (0, π2 ). Prove that

1

n2(tanx1x1

+⋯ +tanxnxn

)2

≤tan2 x1 +⋯ + tan2 xn

x12 +⋯ + xn2.

Proposed by Mircea Becheanu, University of Bucharest, Romania

Solution by Arkady Alt, San Jose, CA, USA

First note thattanx

xincrease in (0, π/2) .

Indeed(tanx

x)′

=x − sinx ⋅ cosx

x2 cos2 x=

(x − sinx) + sinx (1 − cosx)

x2 cos2 x> 0

Then, because n-tiples (x21, ..., x2n) and (

tan2 x1x21

, ...,tan2 xnx2n

) agreed in order, that is

sign (x2i − x2j) = sign

⎛

⎝

tan2 xix2i

−tan2 xj

x2j

⎞

⎠for any i, j ∈ {1,2, ..., n} , by Chebishev’s Inequality

n

∑k=1

tan2 xk =n

∑k=1

x2k ⋅tan2 x k

x2k≥

n

∑k=1

x2k ⋅ (1

n

n

∑k=1

tan2 x k

x2k) .

Also, by Quadratic Mean-Arithmetic Mean Inequality

1

n

n

∑k=1

tan2 x k

x2k≥ (

1

n

n

∑k=1

tanxkxk

)

2

.

Thus,

∑nk=1 tan

2 xkn

∑k=1

x2k

≥

∑nk=1 x

2k ⋅ (

1

n∑nk=1

tan2 x k

x2k)

∑nk=1 x

2k

≥1

n2(n

∑k=1

tanxkxk

)

2

.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Hariyana Vidya Mandir, Salt Lake,Kolkata, India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Albert Stadler, Herrliberg, Switzerland.


O420. Let n ≥ 2 and let A = {1,4, . . . , n2} be the set of the first n nonzero perfect squares. A subset B of Ais called Sidon if whenever a + b = c + d for a, b, c, d ∈ B, we have {a, b} = {c, d}. Prove that A containsa Sidon subset of size at least Cn1/2 for some absolute constant C > 0.Can the exponent 1/2 be improved?

Proposed by Cosmin Pohoata, Caltech, USA

Solution by the authorTo show that A = {1,4, . . . , n2} contains a Sidon subset of size at least Cn1/2 for some absolute constantC > 0, we invoke a theorem from

V. S. Konyagin, An estimate of L1−norm of an exponential sum, The Theory of Approximations ofFunctions and Operators. Abstracts of Papers of the International Conference Dedicated to Stechkin’s 80thAnniversay [in Russian]. Ekaterinburg (2000), 88–89.

If E be the number of solutions to a2 + b2 = c2 + d2 with a, b, c, d all in A and such that {a, b} ≠ {c, d},Konyagin’s theorem says that E ≤ cn5/2 for an absolute constant c > 0. Using this result, we can argue asfollows. Let p ∈ [0,1] and let A′ be a random subset of A obtained by considering each element of A withprobability p (independently). Such a random set A′ may have solutions to a2 + b2 = c2 + d2 with a, b, c, d allin A and such that {a, b} ≠ {c, d}, but it can’t have too many. In particular, from Konyagin’s theorem A′

has at most cn5/2 such 4-tuples. A 4-tuple may be so that all a, b, c, d are pairwise distinct or a = c, in whichcase b = −d (or vice versa), or a = b, in which case c, d satisfy c2 + d2 = 2a2. In the latter case, c and d maybe distinct or c = d, in which case c = −

√2a. The number of distinct triplets c, d, a so that c2 + d2 = 2a2 is

upper bounded by n2/√log ∣A∣ from a classical theorem of Landau, and the number of the other solutions

solutions is at most dn for some constant d, which is negligible for the argument to follow.With this information, we define set A′′ by taking the random A′ and removing one element from each

4-tuple for which a2 + b2 = c2 + d2. By construction, this set A′′ won’t have any solutions to a2 + b2 = c2 + d2.The expected size of A′′ is equal to

E[∣A′′∣] ≫ np − n5/2p4 −

n2√logn

p3.

Taking p 1100n

−1/2, we get that E[∣A′′∣] ≫ Cn1/2 for some absolute constant C > 0. By pidgeonhole, thismeans that there must be some Sidon subset A′′ of size at least Cn1/2. One can improve on this result, ifwe use a better bound on E. Such bounds are available in the recent literature.

Note that if we avoid Konyagin’s theorem, then one can show the existence of a Sidon subset of sizeCn1/3, but using Landau’s theorem to bound E by n3/

√logn. It is also worth mentioning that any set of n

real numbers contains a Sidon subset of size Cn1/2 for some absolute constant C > 0 (not just the set of thefirst n perfect squares), but the proof is a bit more complicated.


junior problems - awesomemath

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