proposed elementary proof of Fermat’s last theorem

Michael Pogorsky has offered what is said to be an elementary proof of Fermat’s last theoremMathworldPlanetmath. Is the proof correct? The intent of this entry is to show the proof up to the point at which it fails, if there is such a point. New equation numbers will be used.

proof We assume that there are positive integers a,b and c such that

an+bn=cn   (1)


We can assume without loss of generality that a,b and c are mutually coprime, so that in fact they are also pairwise coprime. The proof is split into 3 major cases: (1) n is a prime greater than 2, (2) n is divisible by a prime greater than 2, and (3) n is a power of 2.

1 n is a prime greater than 2

Write c as


for some integers k and f.



Using the binomial theorem we can write

an=f(nbn-1+12n(n-1)bn-2f++fn-1)   (2)


bn=k(nan-1+12n(n-1)an-2k++kn-1).   (3)

Lemma 1. If n is prime numberMathworldPlanetmath then n divides (nk) for 0<k<n.
The proof is easy.

Claim: gcd(f,k)=1.
Proof. A factor of f and k will also divide a and b by equations (2) and (3). But a and b are coprimeMathworldPlanetmath, so the gcd of f and k must be 1.

Now write (2) as an=fs for some integer s.

From that point in the proof the exposition is somewhat unclear so I will attempt to rearrange the steps in what seems to be a better order. First, I introduce a lemma of my own. Write (3) as bn=kt for some integer t.

Lemma 2. gcd(f,s)=nα and gcd(k,t)=nβ for some nonnegative integers α and β.
Proof. Suppose q divides gcd(f,s) where q is a prime. Then q divides a and s. We can write s=nbn-1+fT for some integer T, so that q divides s-fT and therefore q divides nbn-1. Hence q divides n or q divides bn-1. But if q divides bn-1 then q divides b, a contradictionMathworldPlanetmathPlanetmath. Hence q divides n. But n is a prime, so q=n. From this we get that gcd(f,s)=nα for some nonnegative integer α. Similarly, gcd(k,t)=nβ for some nonnegative integer β.

It is clear that at least one of α and β is zero, otherwise n divides f and k. Without loss of generality, we can assume that α=0.

The author now introduces what he calls version A and version B. I would prefer to call these Case A and Case B. But there is no claim outstanding yet, so I have to defer the case split. What seems to be the next main result is stated in the following lemma.

Lemma 3. There exist positive integers p,u,v,w such
1) a=vp,
2) if β=0 then


3) if β>0 then there is a positive integers g such that




Proof. (1) We have an=fs,where f and s are coprime. By unique factorization of integers it must be that f=vn and s=pn for some positive integers p and v. It follows that a=pv.
(2) Similarly, there are positive integers w and q such that b=wq, where wn=k. From a+k=b+f we get


and after regrouping we have


Since gcd(f,k)=1 it follows that gcd(v,w)=1, so that





u:=p-vn-1w=q-wn-1v   (4)

is an integer. Using u we can now write a=uwv+vn, b=uwv+wn, and c=uwv+vn+wn.
(3) Since α=0, we have gcd(f,s)=1. Since β>0, we can write


where τ>0 and gcd(k1,n)=1. By Lemma 1 there is a positive integer ci such that (ni)=nci for 0<i<n. We can write


where T=an-1+12(n-1)an-2k++k1nτ-1kn-2. Hence,


Claim: gcd(T,n)=1.
This follows from the fact that n divides all the terms of T except the first term. The first term is not divisible by n because k divides n and therefore n divides b and a and b are coprime.
Claim: gcd(T,k1)=1.
This follows from the fact that k divides b, so k1 divides b, and a and b are coprime. By unique factorization of integers, then, it must be that there are positive integers q, w and λ such that T=qn, k1=wn and nτ+1=λn. Since n is a prime, it follows that λ=ng for some positive integer g. Hence gn=τ+1.

It follows that bn=wnngnqn, so that b=ngwq. From a+k=b+f we get


which we can regroup to get


Since a and b are coprime, it follows that v and ngw are coprime. Hence


is an integer. It follows that


and one can now express a,b and c in terms of u,v,w.

Lemma 4. Let u be the integer of Lemma 3. There is a monic polynomialMathworldPlanetmath P with integer coefficientsMathworldPlanetmath such that
(a) P(u)=0,
(b) the sum of the roots of P is 0, and
(c) all coefficients of P are divisible by n except that when β is positive the last coefficient is not divisible by n.
Proof. We use the same cases as in Lemma 3. (1) In this case we have


The left hand side can be expanded using the binomial theorem to get a polynomialMathworldPlanetmathPlanetmath Q with coefficients that depend on v and w. For the coefficient of un-i we have to combine

=un-i(ni)(wv)n-i(-(vn+wn)i+(vn)i+(wn)i)   (5)

Clearly if i=1 this coefficient is 0. If i=0 the coefficient is (wv)n. For the other terms we can write them as


so that the coefficient is divisible by (wv)n. The coefficient is also divisible by n if 1in by Lemma 1. So we set P:=Q/(wv)n to get the conclusionMathworldPlanetmath for case (1).
(2) We proceed as in case (1). The left side of the equation


can be expanded by the binomial theorem to get a polynomial Q with coefficients that depend on v and w. It is clear that the leading term is (nguwv)n. For the coefficient of un-i we have to combine

=un-i(ni)(ngwv)n-i((vn)i+(ngn-1wn)i-(vn+ngn-1wn)i.   (6)

This form makes it clear that the coefficient is 0 if i=1 and divisible by n if 1i<n. If i=0, the coefficient is (ngwv)n. Equation (6) is equal to


which shows that ngnwnvn divides each coefficient. Set P:=Q/(ngwv)n to get the conclusion for case (2).

Lemma 5. The polynomial P of Lemma 4 has exactly one positive root.
Proof. By (5) and (6) the coefficients of P are negative except for the leading coefficient. So there is exactly one sign change and by Descartes’s rule of signs there is exactly one positive root.

Definition. For each real root ui of P we can define a, b and c. (For example, a=uiwv+vn and so on.) We say that a root ui is acceptable if the resulting a,b,c are all positive integers.

Lemma 6. The only acceptable root of P is u and u>0.
Proof. Suppose that ui is a nonpositive acceptable root. Then a,b,c are all positive and in case (1) we have


while in case (2) we have




which is a contradiction. Since u is acceptable, it must be that u>0.

The following lemma 7 is incorrect.
Lemma 7. n does not divide a+b.
Proof. We use the cases of Lemma 3. (1) We write




It is known that the common divisorMathworldPlanetmathPlanetmath if a+b and Q is n and that if ns||a+b then ns+1||an+bn. Hence, we can write




where gcd(n,δ)=1, gcd(n,γ)=1 and gcd(δ,γ)=1. From cn=an+bn=(a+b)Q=ns+1δγ we get s=n-1, n||c and

a+b=nn-1δ.   (7)

Since n divides a+b and c we have n divides 2c-(a+b)=vn+wn. It is also known that

vn+wn=(v+w)[(v+w)n-1-nvw(vn-1++wn-1)]   (8)

so that n divides v+w. But then from (8) again, n2 divides vn+wn. Now from (7) we have n2 divides a+b. Hence


is divisible by n2 and this is a contradiction.
(2) In this case


so that if n divides a+b then n divides vn and therefore n divides v. From a=vp we get then n divides a and therefore n divides b=a+b-a. But a and b are coprime, so we have a contradiction.

Since Lemma 7 is incorrect, Lemma 8 is also incorrect.
Lemma 8. There are positive integers up and cp such that a+b=upn and c=upcp.
Proof. We can write




It is an old result first attributed to Nicolas Malebranche (1638-1715) that if x and y are coprime and d is a prime divisorMathworldPlanetmathPlanetmath of x+y and Q then d divides z. I will give a proof of this here. Define




Then d divides Q2. Define


In general


and each Qn is divisible by d. Hence,


will be divisible by d. But d does not divide y (since otherwise it would also divide x, which would contradict that x and y are coprime). Hence, d divides z. Using this result, we can say that if


then a+b and Q are coprime. Because if d is a prime divisor of each then d divides n, so d=n and then n divides a+b, which contradicts Lemma 7. By unique factorization there are positive integers up and cp such that




Then cn=(a+b)Q=upncpn. Hence c=upcp.

Lemma 9. Let p,u,v,w be as in Lemma 3. Let cp be as in Lemma 8. Suppose there are positive integers hand q such that


Then one of the following possibilities holds:
(a) h=hkc,q=qkc for some integers hk and qk;
(b) h=q=jcp for some integer j;
(c) h=jwn(n-1),q=jvn(n-1) for some integer j;
(d) h=jb,q=jwn for some integer j;
(e) h=jvn,q=-jwn for some integer j;
(f) h=jvn,q=j(2uwv+wn+2vn) for some integer j;
(g) h=j(2uwv+2wn+vn),q=jwn for some integer j.
Proof. At this point I think the proof is incomplete since he does not prove the result, but rather verifies that each of the possible solutions is indeed a solution. Later on,he needs to know that these are the only solutions.

2 n is divisible by a prime greater than 2

If n=mz where z is a prime greater than 2, then


and we can apply the results of section 1 to conclude that no such z can exist.

3 n is a power of 2

It is known that if n=4 then Fermat’s Last Theorem is true. For example, see [1]. So if n=2t,t3, then we can write


which contradicts the theorem for n=4.


  • 1 G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, page 191.
Title proposed elementary proof of Fermat’s last theorem
Canonical name ProposedElementaryProofOfFermatsLastTheorem
Date of creation 2013-03-22 17:36:55
Last modified on 2013-03-22 17:36:55
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 78
Author Mathprof (13753)
Entry type Proof
Classification msc 11D41