zsigmondy theorem proof (1)

7/22/2019 Zsigmondy Theorem Proof (1)

1/56

Zsigmondys Theorem

Lola Thompson

Dartmouth College

August 11, 2009

Lola Thompson (Dartmouth College) Zsigmondys Theorem August 11, 2009 1 / 1
http://find/http://goback/


2/56

Introduction

Definition

o(a modp) := the multiplicative order ofa (mod p).

Recall: The multiplicative order ofa (mod p) is the smallest integer ksuch that ak 1 (mod p).

Exampleo(2 mod 5) = 4 since 21 2 (mod 5), 22 4 (mod 5), 23 3

(mod 5) and 24 1 (mod 5).

http://find/


3/56

Introduction

Theorem (Zsigmondy)

For every pair of positive integers(a, n), except n= 1 and (2,6), thereexists a prime p such that n=o(a mod p).

Lets see why the exceptional cases might not work:

http://find/


4/56

Introduction

Theorem (Zsigmondy)

For every pair of positive integers(a, n), except n= 1 and (2,6), thereexists a prime p such that n=o(a mod p).

Lets see why the exceptional cases might not work:

Ifn= 1, then 1 =o(a mod p) a1 1 (mod p). But this is only

true when a= 1.

http://find/


5/56

Introduction

Theorem (Zsigmondy)For every pair of positive integers(a, n), except n= 1 and (2,6), thereexists a prime p such that n=o (amod p).

(2, 6) is an exception means that there are no primes psuch that6 =o(2 modp), i.e. for any prime psuch that 26 1(mod p), itmust be the case that 23 1 (mod p) or 22 1 (mod p) also.

http://find/


6/56

Introduction

Theorem (Zsigmondy)For every pair of positive integers(a, n), except n= 1 and (2,6), thereexists a prime p such that n=o (amod p).

(2, 6) is an exception means that there are no primes psuch that6 =o(2 modp), i.e. for any prime psuch that 26 1(mod p), itmust be the case that 23 1 (mod p) or 22 1 (mod p) also.

The fact that (2, 6) is an exception can be proven through elementarymeans, but well get it for free in the process of proving ZsigmondysTheorem.



7/56

Outline



8/56

Cyclotomic Polynomials

Definition

We define nth cyclotomic polynomial as follows:n(x) =

primitive

nth root of 1

(x ).

n(x) has degree (n) since there are (n) primitive nth roots of unity.(Recall: If is primitive then k is primitive if and only if (k, n) = 1)

Some Other Properties:

Monic

http://find/


9/56


Definition


primitive

nth root of 1

(x ).



Monic

Irreducible

http://find/


10/56


Definition


primitive

nth root of 1

(x ).



Monic

Irreducible

In Z[x](In fact, n(x) is the minimal polynomial for over Q)



11/56


Theorem

xn 1 =d|n

d(x)

(True since xn 1 =

nth root of 1

(x ) =d|n

primitive

dth root of 1

(x )).

http://find/


12/56

The Mobius Function

The Mobius function (n) is an arithmetic function satisfying (1) = 1

andd|n

(d) = 0 for every n>1.

Example:d|2

(d) =(1) + (2) = 0.

Since (1) = 1 then it must be the case that (2) =1.

Example:d|4

(d) =(1) + (2) + (4) = 0.

Since we know that (1) + (2) = 0 then (4) = 0.

In general: (n) =

0, n=m pr, r>1

1k, n=p1p2 pk

http://find/


13/56


Theorem (Mobius Inversion Formula)

If f(n) =d|n

g(d) then g(n) =d|n

f(d) (n/d).

Since xn 1 =d|n

d(x) then n(x) =d|n

(xd 1)(n/d)

(take log of both sides, apply Mobius inversion, then undo the logs).

Example:2(x) = (x

1 1)(2/1) (x2 1)(2/2) = (x 1)1 (x2 1) =x+ 1.

http://find/


14/56


Some Examples:

1(x) =x 12(x) =x+ 13(x) =x

2 +x+ 14(x) =x

2 + 1

5(x) =x

4

+x

3

+x

2

+x+ 16(x) =x2 x+ 1

7(x) =x6 +x5 +x4 +x3 +x2 +x+ 1

8(x) =x4 + 1

In general, p(x) =xp1 +xp2 + +x+ 1.

For k1,pk(x) = p(xpk1 ). So pk(x) has the same number of

nonzero terms as p(x).

http://find/


15/56

Values of Cyclotomic Polynomials

Theorem

Suppose n>1. Then:

(1) n(0) = 1

(2) n(1) =

p, n=pm,m>0

1, otherwise

To prove (2): Evaluate xn1

x1 at x= 1 in 2 different ways to find

n= d|nd>1

d(1). We know that p(x) =xp1 + +x+ 1, so p(1) =p.

Moreover, pk(1) =p. By unique factorization, n=pe11 p

egg . Since

there are eidivisors ofn that are powers ofpi for each prime pi dividing nthen, from our formula above, d(1) = 1 when d is composite.

http://find/


16/56

Values of Cyclotomic Polynomials

Theorem

Suppose n>1. Then:

(3) If a>1 then (a 1)(n) p.

Proof of (3)Ifa>1 then geometry implies that a 1


17/56

Key Lemma

Lemma

Suppose that n>2 and a>1 are integers andn(a) is prime. Ifn(a)|n then n= 6 and a= 2.

Proof

Let p= n(a), where p|n.

http://find/


18/56

Key Lemma

Lemma


Proof


Ifa3 then n(a)> pby (4), which is obviously false.


K L
http://find/


19/56

Key Lemma

Lemma


Proof



Thus, a= 2 and n(2) =p.


K L
http://find/


20/56

Key Lemma

Lemma


Proof




Since n(2) =p then p|(2n 1)(since n(x) always divides x

n 1), i.e. 2n 1(mod p).


K L
http://find/


21/56

Key Lemma

Lemma


Proof




Since n(2) =p then p|(2n 1)(since n(x) always divides x

n 1), i.e. 2n 1(mod p).

So, pmust be odd.


K L
http://find/


22/56

Key Lemma

So far, we have: p= n(2), p|n, podd.


Ke Le a
http://find/


23/56

Key Lemma


Factorn=pe m where pm.


Key Lemma
http://find/


24/56

Key Lemma



Its a well-known fact from abstract algebra that ifn is as above andif is a root of n(x) over Fp then m=o().


Key Lemma
http://find/


25/56

Key Lemma




As a result, since we know that n(2) 0(mod p), thenm=o(2 mod p).


Key Lemma
http://find/


26/56

Key Lemma





Ife>1 then p= n(2) = pem(2) = m(2pe) = pm(2

pe1 ).


Key Lemma
http://find/


27/56

Key Lemma






pe1 ).

This contradicts (4), since 2pe1 2p >4.


Key Lemma
http://find/


28/56

Key Lemma






pe1 ).

This contradicts (4), since 2pe1 2p >4.

Thus, n=pm.


Key Lemma
http://find/


29/56

Key Lemma

At this point, we have deduced: p= n(2), p|n, podd, n=pmwhere pm.


Key Lemma
http://find/


30/56

Key Lemma


Were trying to show that n= 6.


Key Lemma
http://find/


31/56

Key Lemma



Now, p= pm(2) = m(2p)

m(2) > (2

p1)(m)

(2+1)(m) 2

p13 (from (3)).


Key Lemma
http://find/


32/56

Key Lemma



Now, p= pm(2) = m(2p)

m(2) > (2

p1)(m)

(2+1)(m) 2

p13 (from (3)).

But then 3p+ 1>2p, which is impossible ifp>3.


Key Lemma
http://find/


33/56

y



Now, p= pm(2) = m(2p)

m(2) > (2

p1)(m)

(2+1)(m) 2

p13 (from (3)).

But then 3p+ 1>2p, which is impossible ifp>3.

Therefore,p= 3 and m=o(2 mod 3) = 2, so n= 2 3 = 6.


Recap and Extensions
http://find/


34/56

p

We have proven the following Key Lemma:

Lemma

Suppose that n>2 and a>1 are integers andn(a) isprime. Ifn(a)|n then n= 6 and a= 2.

We can extend the Key Lemma to show that if n(a) is a divisor ofn forsome n>2 and a>1, then n= 6 and a= 2.


Good Pairs, Bad Pairs
http://find/


35/56

,

DefinitionLet a, nZ+, a>1. The pair (a, n) is good ifn=o(a mod p) for someprime p.

Lemma (Good Pairs Condition)

(a, n) is good if and only if there is a prime p such that p|(an 1) butp(an/q 1) for every prime factor q|n.

Example32 1 uses the same primes as 31 1, so (3, 2) is bad.


Good Pairs, Bad Pairs
http://find/


36/56

Lemma

(a, 1) is bad when a= 2.(a, 2) is bad when a= 2m 1 for some m>1.All other pairs(a, 2k) are good.

Example22 1 = 3, 23 1 = 7, 26 1 = 63 = 32 7. Thus, (2, 6) is bad.


Zsigmondys Theorem
http://find/


37/56

Theorem (Zsigmondy)

If n2, the only bad pair(a, n) is(2, 6).

[In other words, there exists a prime psuch that n=o(a mod p) for every

pair (a, n) except (2, 6)]

Proof Outline Suppose (a, n) is bad and n>2. We will translate thisinto a problem about cyclotomic polynomials and use the Key Lemma to

derive a contradiction unless a= 2 and n= 6.


Two More Lemmas
http://find/


38/56

In order to prove Zsigmondys Theorem, we will need the following two

lemmas:

Lemma (1)

If xn 1 = n(x) n(x) then n(x) = d|nd


39/56

Theorem (Zsigmondy)


Proof

Pick an odd prime factor p|n(a).


Proving Zsigmondys Theorem
http://find/


40/56

Theorem (Zsigmondy)


Proof


Suppose that (a, n) is bad, so that k=o(a mod p) is a proper divisor

ofn.


http://find/


41/56

Theorem (Zsigmondy)


Proof



ofn.

Let xn 1 = n(x) n(x).


http://find/


42/56

Theorem (Zsigmondy)


Proof



ofn.


From Lemma (1), (ak 1)| n(a), so p|(an 1) also.


http://find/


43/56

Theorem (Zsigmondy)


Proof



ofn.



Sop2 is a factor of n(a) n(a) =an 1.


http://find/


44/56

Theorem (Zsigmondy)


Proof



ofn.



Sop2 is a factor of n(a) n(a) =an 1.

By Fermats little Theorem, ap1 1(mod p), so k |p 1, hencek


45/56

From Lemma (2), we know that ifk |n and ak 1(mod p), then ifk


46/56



47/56



48/56

Weve already shown that p is odd and p|n and n(a) =pt.


Finishing Up
http://find/


49/56


Ift>2 then p2 divides an1

an/p1, since (an/p 1)| n(a).


Finishing Up
http://goforward/http://find/http://goback/


50/56




We can use the exponent law to derive a contradiction to thestatement that p2 | a

n1an/p1

. Thus, n(a) =p.


Finishing Up
http://find/


51/56





n1an/p1

. Thus, n(a) =p.

Now, ifa3 then n(a)>p, which we know is false.


Finishing Up
http://find/


52/56





n1an/p1

. Thus, n(a) =p.


Hence, we must have a= 2. By the Key Lemma, n= 6.


Finishing Up
http://find/


53/56





n1an/p1

. Thus, n(a) =p.



Therefore, (2, 6) is the only bad pair.


Finishing Up
http://find/


54/56





n1an/p1

. Thus, n(a) =p.



Therefore, (2, 6) is the only bad pair.

Weve proven Zsigmondys Theorem!


A Special Case of Zsigmondys Theorem
http://find/


55/56

A special case of Zsigmondys Theorem states the problem in terms ofMersenne numbers:

Consider the kth Mersenne number Mk= 2k 1. Then, each ofM2,M3,M4,... has a prime factor that does not occur as a factor of anearlier member of the sequence EXCEPT for M6.


Acknowledgements
http://find/


56/56

Thanks to Dr. Dan Shapiro, whose expository notes on the subject wereinvaluable in writing this talk.

Bibliography:

1. Introduction to the Theory of Numbersby Harold N. Shapiro

2. Abstract Algebra by David S. Dummit and Richard M. Foote

http://find/

zsigmondy theorem proof (1)

Documents