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# An Approach to a Simple Proof of Fermat's Last Theorem

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## Abstract

Fermat's Last Theorem states that the Diophantine equation X^n + Y^n = Z^n has no non-trivial solution for any n greater than 2. In this paper we give an approach to a brief and simple proof of the theorem using only elementary methods.
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 . 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  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
(pq)n=pnqnnpq(pq)λn,p,q .(1)
Proof. According to the binomial theorem, we have
(pq)n=
n
X
k=0 n
kpnk(q)k
=pnqn+
n1
X
k=1 n
kpnk(q)k
=pnqnnpq ·
n1
X
k=1
1
nn
kpnk1(q)k1
=pnqnnpq(pq)·
n2
X
k=1
1
nn1
k+ (1)k+1pnk2(q)k1.(2)
The term 1
nn1
k+ (1)k+1assumes only positive integer values for and odd prime
n. A proof can be found in Mamakani . The OEIS reference for these values is
A219539 .
By substituting λn,p,q =Pn2
k=1
1
nn1
k+ (1)k+1pnk2qk1into (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
pn3n1
21
npn4q+· · · − n1
n31
npqn4+qn3,(3)
that gcd(p, q, λn,p,q) = 1. Because if pdivides λn,p,q then p|qn3, and if qdivides λn,p,q
then q|pn3. This completes the proof.
Remark 2. According to Lemma 1, for primes n5 the further factorisation of λn,p,q
is known. We have
(p2pq +q2)|λn,p,q for each n≡ −1 (mod 6),
(4)
(p2pq +q2)2|λn,p,q for each n1 (mod 6).
For a proof we refer the reader to .
Lemma 3. Let a, b, c be integers deﬁned by
a=1
zy·
n2
X
k=0
xn2k(zk+1 yk+1),(5)
b=1
zy·
n1
X
k=0
xn1k(zk+1 yk+1),(6)
c=1
zy·
n
X
k=0
xnk(zk+1 yk+1),(7)
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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
zy·
n2
X
k=0
xn1k(zk+1 yk+1).
yn
zywe get
ax +znyn
zy=1
zy·
n1
X
k=0
xn1k(zk+1 yk+1) = b. (9)
Multiplying by (zy) yields
ax(zy) + znyn=b(zy).(10)
Multiplying (6) by xwe have
bx =1
zy·
n1
X
k=0
xnk(zk+1 yk+1).
zywe get
bx +zn+1 yn+1
zy=1
zy·
n
X
k=0
xnk(zk+1 yk+1) = c. (11)
Multiplying by (zy) we obtain
bx(zy) + zn+1 yn+1 =c(zy).(12)
Now we can prove the evidence of (8). Multiplying (8) by (zy) gives
zn(zy) = axy(zy)b(x+y)(zy) + c(zy).
Applying (10) on the right-hand side we get
zn(zy) = yb(zy)zn+ynb(x+y)(zy) + c(zy)
=bx(zy)yzn+yn+1 +c(zy).
Applying (12) on the right-hand side yields
zn(zy) = bx(zy)yzn+yn+1 +bx(zy) + 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
(zxy)n=zn(x+y)nn(x+y)z(zxy)λn,z,x+y.
Applying Lemma 1 with p=x,q=y, on the right-hand side gives
(zxy)n=znxnynnxy(x+y)λn,x,yn(x+y)z(zxy)λn,z,x+y.
Applying (13) on the right-hand side we obtain
(zxy)n=nxy(x+y)λn,x,yn(x+y)z(zxy)λn,z,x+y,
that is
(zxy)n=n(x+y)xyλn,x,yz(zxy)λn,z,x+y.(14)
Because nis an odd prime we conclude from (14) that n|(zxy)n, hence n|(zxy).
Dividing (14) by n2it follows that ndivides (x+y) or xyλn,x,yz(zxy)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,yz(zxy)λn,z,x+ythen n|xyλn,x,y, because n|(zxy). 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 . 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
zy= (zy)n1+nzyλn,z,y .(15)
Now we assume that x6=±1 and n-x. From (15) we conclude that (zy)|xn, hence
n-(zy). Now we consider two cases. Case 1: If xis even then y, z are odd, so
(zy) 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 (zy) is odd. Applying Lemma 1 it follows from (3) with yor z
even that each term of λn,z,y is even except (zn3+yn3) which is odd, so λn,z ,y is
odd. From gcd(z, y) = 1 we have gcd(z, y, z y) = 1, hence gcd(zy, λn,z ,y) = 1, so
gcd(zy, zyλn,z,y ) = 1. Therefore a prime factor of xand (zy) does not divide the
right-hand side of (15), which clearly forces that (zy) 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 zy=unand
v-(zy).
Applying (13) on the right-hand side of (10) we obtain ax(zy) + xn=b(zy),
which gives
xn= (bax)(zy).(16)
Substituting x=uv and zy=uninto (16) gives (uv)n= (bauv)un, hence
vn=bauv =bax. (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 zy=uninto (12) we
obtain bxun+zn+1 yn+1 =cun, which gives
zn+1 yn+1
cbx =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
zy·
n
X
k=0
xnk(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
zy·
n2
X
k=0
xn2kzky
zy·
n2
X
k=0
xn2kyk
=z
zy·zn1xn1
zxy
zy·yn1xn1
yx
=xn(zy) + yn(xz) + zn(yx)
(zy)(xz)(xy)
=xn(zy) + yn(xz) + zn(yx)
x2(zy) + y2(xz) + z2(yx).
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Mike Winkler - An Approach to a Simple Proof of Fermat’s Last Theorem
In a similar way, we may show that
b=xn+1(zy) + yn+1 (xz) + zn+1(yx)
x2(zy) + y2(xz) + z2(yx),
c=xn+2(zy) + yn+2 (xz) + zn+2(yx)
x2(zy) + y2(xz) + z2(yx).
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 ZN
odd 3ZNZ
even 4NZN
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 .
References
 Brown, Kenneth S.: Sums of Powers in Terms of Symmetric Functions, web docu-
ment. (mathpages.com/home/kmath097.htm)
 Fillipi, Andreas: Beitrag im Forum Elementare Zahlentheorie, February 9, 2015,
www.matheplanet.de. (tinyurl.com/y8jh4dvr)
 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.
 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.
 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.
 Wikipedia, Fermat’s Last Theorem, 2.2 Fermat’s conjecture.
(tinyurl.com/y9hwwdb5)
 Wiles, Andrew: Modular Elliptic Curves and Fermat’s last theorem, Annals of Math-
ematics 142 (1995), pp. 443-551.
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