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An Approach to a Simple Proof of

Fermat’s Last Theorem

Mike Winkler

Fakult¨at f¨ur Mathematik, Ruhr-Universit¨at Bochum

mike.winkler@ruhr-uni-bochum.de

www.mikewinkler.co.nf

April 2, 2018

Abstract

Fermat’s Last Theorem states that the Diophantine equation Xn+Yn=Znhas no

non-trivial solution for any ngreater than 2. In this paper we give an approach to a

brief and simple proof of the theorem using only elementary methods.

Introduction

The only known successful proof of Fermat’s Last Theorem was given in 1994 by

Andrew Wiles [7]. Unfortunately this proof contains nearly hundred pages and can

be understood in its entirety only by some specialists. For this reason and relating to

Fermat’s famous marginal note1, many people (mostly amateurs) are still looking for

a shorter and simpler proof based on elementary methods. In this paper we give an

approach to such a proof.

To prove Fermat’s Last Theorem it suﬃces to prove it for the exponent 4 and every

odd prime exponent. A proof for the case n= 4 has already been given by Fermat

himself. Therefore we give our approach only for the prime exponents greater than 2.

1See [6] for the complete text.

Mike Winkler - An Approach to a Simple Proof of Fermat’s Last Theorem

Part One

Lemma 1. Let p, q be integers with gcd(p, q)=1, then for any odd prime nthere exists

an integer λn,p,q with gcd(p, q, λn,p,q )=1such that

(p−q)n=pn−qn−npq(p−q)λn,p,q .(1)

Proof. According to the binomial theorem, we have

(p−q)n=

n

X

k=0 n

kpn−k(−q)k

=pn−qn+

n−1

X

k=1 n

kpn−k(−q)k

=pn−qn−npq ·

n−1

X

k=1

1

nn

kpn−k−1(−q)k−1

=pn−qn−npq(p−q)·

n−2

X

k=1

1

nn−1

k+ (−1)k+1pn−k−2(−q)k−1.(2)

The term 1

nn−1

k+ (−1)k+1assumes only positive integer values for and odd prime

n. A proof can be found in Mamakani [3]. The OEIS reference for these values is

A219539 [5].

By substituting λn,p,q =Pn−2

k=1

1

nn−1

k+ (−1)k+1pn−k−2qk−1into (2) we get (1).

For n= 3 we have λ3,p,q = 1. For primes n > 3 it follows from gcd(p, q) = 1 and the

expansion of λn,p,q, given by

pn−3−n−1

2−1

npn−4q+· · · − n−1

n−3−1

npqn−4+qn−3,(3)

that gcd(p, q, λn,p,q) = 1. Because if pdivides λn,p,q then p|qn−3, and if qdivides λn,p,q

then q|pn−3. This completes the proof.

Remark 2. According to Lemma 1, for primes n≥5 the further factorisation of λn,p,q

is known. We have

(p2−pq +q2)|λn,p,q for each n≡ −1 (mod 6),

(4)

(p2−pq +q2)2|λn,p,q for each n≡1 (mod 6).

For a proof we refer the reader to [1].

Lemma 3. Let a, b, c be integers deﬁned by

a=1

z−y·

n−2

X

k=0

xn−2−k(zk+1 −yk+1),(5)

b=1

z−y·

n−1

X

k=0

xn−1−k(zk+1 −yk+1),(6)

c=1

z−y·

n

X

k=0

xn−k(zk+1 −yk+1),(7)

2

Mike Winkler - An Approach to a Simple Proof of Fermat’s Last Theorem

then the identity

zn=axy −b(x+y) + c, (8)

holds for all integers x, y, z and any nonnegative integer n.

Proof. Multiplying (5) by xgives

ax =1

z−y·

n−2

X

k=0

xn−1−k(zk+1 −yk+1).

Adding zn

−yn

z−ywe get

ax +zn−yn

z−y=1

z−y·

n−1

X

k=0

xn−1−k(zk+1 −yk+1) = b. (9)

Multiplying by (z−y) yields

ax(z−y) + zn−yn=b(z−y).(10)

Multiplying (6) by xwe have

bx =1

z−y·

n−1

X

k=0

xn−k(zk+1 −yk+1).

Adding zn+1−yn+1

z−ywe get

bx +zn+1 −yn+1

z−y=1

z−y·

n

X

k=0

xn−k(zk+1 −yk+1) = c. (11)

Multiplying by (z−y) we obtain

bx(z−y) + zn+1 −yn+1 =c(z−y).(12)

Now we can prove the evidence of (8). Multiplying (8) by (z−y) gives

zn(z−y) = axy(z−y)−b(x+y)(z−y) + c(z−y).

Applying (10) on the right-hand side we get

zn(z−y) = yb(z−y)−zn+yn−b(x+y)(z−y) + c(z−y)

=−bx(z−y)−yzn+yn+1 +c(z−y).

Applying (12) on the right-hand side yields

zn(z−y) = −bx(z−y)−yzn+yn+1 +bx(z−y) + zn+1 −yn+1

=zn+1 −yzn.

We obtain a true statement, which completes the proof.

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Mike Winkler - An Approach to a Simple Proof of Fermat’s Last Theorem

Theorem 4. The Diophantine equation Xn+Yn=Znhas no non-trivial solution for

any odd prime number n.

Proof. We assume that x, y, z are nonzero integers and nis an odd prime such that

xn+yn=zn.(13)

It suﬃces to consider only solutions (x, y, z) with gcd(x, y, z) = 1. Hence x, y, z are

pairwise relatively prime and exactly one of these integers is even. Applying Lemma 1

with p=z,q=x+y, we have

(z−x−y)n=zn−(x+y)n−n(x+y)z(z−x−y)λn,z,x+y.

Applying Lemma 1 with p=x,q=−y, on the right-hand side gives

(z−x−y)n=zn−xn−yn−nxy(x+y)λn,x,−y−n(x+y)z(z−x−y)λn,z,x+y.

Applying (13) on the right-hand side we obtain

(z−x−y)n=−nxy(x+y)λn,x,−y−n(x+y)z(z−x−y)λn,z,x+y,

that is

(z−x−y)n=n(x+y)xyλn,x,−y−z(z−x−y)λn,z,x+y.(14)

Because nis an odd prime we conclude from (14) that n|(z−x−y)n, hence n|(z−x−y).

Dividing (14) by n2it follows that ndivides (x+y) or xyλn,x,−y−z(z−x−y)λn,z,x+y.

From (13) we conclude that (x+y)|zn, hence if ndivides (x+y) then n|z. If n

divides xyλn,x,−y−z(z−x−y)λn,z,x+ythen n|xyλn,x,−y, because n|(z−x−y). It

follows from gcd(x, y, λn,x,−y) = 1 that ndivides one and only one of the integers x, y,

or λn,x,−y. So we have to consider two cases. Case 1: If n-λn,x,−ythen n|xyz. By

gcd(x, y, z) = 1 it follows that ndivides one and only one of the integers x, y, z.Case

2: If n|λn,x,−ythen n-xyz. This applies if and only if nis congruent 1 (mod 6) or

an exceptional prime as listed in A068209 [4]. Unfortunately, it is an open problem if

there exists a simple method of characterizing these exceptional primes.

Applying Lemma 1 with p=z,q=y, it follows from (13) that

xn

z−y= (z−y)n−1+nzyλn,z,y .(15)

Now we assume that x6=±1 and n-x. From (15) we conclude that (z−y)|xn, hence

n-(z−y). Now we consider two cases. Case 1: If xis even then y, z are odd, so

(z−y) is even. Applying Lemma 1 it follows from (3) that λn,z,y consists of an odd

number of terms, where the number of even coeﬃcients is even and the number of odd

coeﬃcients is odd. Hence, with y, z odd, λn,z,y is a sum of an odd number of odd terms

and an even number of even terms, so λn,z,y is odd. Case 2: If xis odd then y, z have

diﬀerent parity, so (z−y) is odd. Applying Lemma 1 it follows from (3) with yor z

even that each term of λn,z,y is even except (zn−3+yn−3) which is odd, so λn,z ,y is

odd. From gcd(z, y) = 1 we have gcd(z, y, z −y) = 1, hence gcd(z−y, λn,z ,y) = 1, so

gcd(z−y, zyλn,z,y ) = 1. Therefore a prime factor of xand (z−y) does not divide the

right-hand side of (15), which clearly forces that (z−y) is an n-th power. It follows

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Mike Winkler - An Approach to a Simple Proof of Fermat’s Last Theorem

that there exist integers u, v with gcd(u, v) = 1 such that x=uv and z−y=unand

v-(z−y).

Applying (13) on the right-hand side of (10) we obtain ax(z−y) + xn=b(z−y),

which gives

xn= (b−ax)(z−y).(16)

Substituting x=uv and z−y=uninto (16) gives (uv)n= (b−auv)un, hence

vn=b−auv =b−ax. (17)

It follows from (17) that v|b, and consequently from (8) and gcd(x, y, z) = 1 that v-c.

According to Lemma 3 we can show by a similar proof that xn=ayz −b(y+z) + c,

which yields with v|x,v|b,v-cand gcd(x, y, z) = 1 that v-a. With gcd(u, v)=1

and v|b,v-ait follows from (17) that v2-b. Substituting z−y=uninto (12) we

obtain bxun+zn+1 −yn+1 =cun, which gives

zn+1 −yn+1

c−bx =un.(18)

...

Remark 5. The terms from (5)–(7) represent special cases of the trinomial expansion

of (x+y+z)nwith the peculiarity that all trinomial coeﬃcients given by n

i,j,k=n!

i!j!k!

were set equal to 1. Let x, y, z be integers, then for any nonnegative integer nwe have

1

z−y·

n

X

k=0

xn−k(zk+1 −yk+1) = X

i+j+k=n

xiyjzk,

where i, j, k are all nonnegative integers such that i+j+k=n.

Remark 6. We can rewrite the terms from (5)–(7) as fractions. From (5) we obtain

a=z

z−y·

n−2

X

k=0

xn−2−kzk−y

z−y·

n−2

X

k=0

xn−2−kyk

=z

z−y·zn−1−xn−1

z−x−y

z−y·yn−1−xn−1

y−x

=xn(z−y) + yn(x−z) + zn(y−x)

(z−y)(x−z)(x−y)

=xn(z−y) + yn(x−z) + zn(y−x)

x2(z−y) + y2(x−z) + z2(y−x).

5

Mike Winkler - An Approach to a Simple Proof of Fermat’s Last Theorem

In a similar way, we may show that

b=xn+1(z−y) + yn+1 (x−z) + zn+1(y−x)

x2(z−y) + y2(x−z) + z2(y−x),

c=xn+2(z−y) + yn+2 (x−z) + zn+2(y−x)

x2(z−y) + y2(x−z) + z2(y−x).

Table 1 gives an overview on all possible integer values for a, b, c from (5)–(7) for any

nonnegative integer n.

n a b c

0 0 0 1

1 0 1 ∈Z

2 1 ∈Z∈N

odd ≥3∈Z∈N∈Z

even ≥4∈N∈Z∈N

Table 1: Possible values for a, b, c.

Part Two

Acknowledgements

The author wishes to express his thanks to Andreas Fillipi. Without his contribution

in a mathematics forum, I probably would never have worked on this topic again [2].

References

[1] Brown, Kenneth S.: Sums of Powers in Terms of Symmetric Functions, web docu-

ment. (mathpages.com/home/kmath097.htm)

[2] Fillipi, Andreas: Beitrag im Forum Elementare Zahlentheorie, February 9, 2015,

www.matheplanet.de. (tinyurl.com/y8jh4dvr)

[3] K. Mamakani, F. Ruskey, New roses: simple symmetric Venn diagrams with 11 and

13 curves, Disc. Comp. Geom., 52 (2014), pp. 71–87, Lemma 2.

[4] The On-Line Encyclopedia of Integer Sequences: Considering the congruence

(x+1)ˆp-xˆp=1 (mod pˆ2) sequence gives values of p of the form 3k-1 such there

exist nontrivial solutions (x other than 0 or -1 modulo p), A068209.

[5] The On-Line Encyclopedia of Integer Sequences: T(n,k) is the number of k-points

on the left side of a crosscut of simple symmetric n-Venn diagram, A219539.

[6] Wikipedia, Fermat’s Last Theorem, 2.2 Fermat’s conjecture.

(tinyurl.com/y9hwwdb5)

[7] Wiles, Andrew: Modular Elliptic Curves and Fermat’s last theorem, Annals of Math-

ematics 142 (1995), pp. 443-551.

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