Fermats conjecture requires divisibility by n squared

Question

Assuming a, b and c to be positive integers and n a prime greater than 2, a necessary sub-condition for $a^n + b^n = c^n$ to be true (however no such integers exist according to Andrew Wiles proof) is that $x^{n-1} -1$ is divisible by $n^2$ for x = a, b or c as well as for x = 2. I have not been able to find a proof for the last (x=2) condition. Can anybody please help. Jens

Answer 1

It is "natural" that you have not been able to prove what you say for $x =2$ and in addition this has been proved only when the prime $n$ does not divide $abc$ (the so-called first FLT case). The proof is very technical and made by Wieferich more than a century ago, based on works by Kummer.

Theorem.-(Wieferich, 1909) If $a^n+b^n=c^n$ with the prime $n$ does not divide $abc$ then $2^{n-1}\equiv 1\pmod {n^2}$.

Hence if the prime $n$ does not divide $abc$ and satisfies $2^{n-1}\equiv 1\pmod {n^2}$ then $a^n+b^n=c^n$ is possible maybe for $n$ (we know now it is impossible). It was only in $1913$ that Meissner found the prime $n=1093$ such that $2^{1092}-1$ is divisible by $1093^2$ and in $1922$ that Beeger discovered the second example $n=3511$. This proved that the first case of FLT is true for all prime less than $3511$ (I add this in case be of interest for you. You can read for more Ribenboim who is a classic on this topic).

Answer 2

To provide some insight I submit the following: Let p, q and r be positive integers. It is not hard to prove that a, b and must have the form c-b = p^n, c-a = q^n and a+b = r^n. Further, manipulating these expressions and using Fermats little theorem (FLT), we obtain the forms a = p^n + 2kpqrn^2, b= q^n + 2kpqrn^2 and c = r^n - 2kpqr*n^2, where k is some integer. Raising any form to the power of n and manipulating this with the unraised form using FLT will result in the stated sub-conditions for x = a, b or c. But what about x = 2.