I want to prove the statement in the title. This is, how far i came:
Proof. We have $p \in \Bbb P$ and $x,y,z \in \Bbb N$ with $x^{p-1}+y^{p-1}=z^{p-1}$.
If $p=2$, we have $x+y=z$. Now if $x$ and $y$ are both even or odd at the same time, $z$ is even, so $2 \mid z$ and $2 \mid xyz$. If either $x$ or $y$, but not both at the same time, is even, we have $2 \mid xyz$ as well. So we can assume $p \geq 3$, in particular odd, in the following. Consider cases:
- Consider at first $p \mid z$, then $p \mid xyz$ trivially.
- Now suppose $p \nmid z$. By Fermat's little theorem we have $z^{p-1} \equiv 1 \mod p$. This means $x^{p-1}+y^{p-1} \equiv 1 \mod p$. Now, since $p$ is odd, $p-1$ is even. Hence we basically have $a^2+b^2 \equiv 1 \mod p$ with $a=x^k$ and $b=y^k$ for some $k \in \Bbb N$.
Here I am stuck. I have to prove $a^2+b^2 \equiv 1 \mod p \implies (p \mid a \vee p \mid b)$. I suppose that a square, $c^2$ with $c=z^k$, which is the sum of two squares, $a^2+b^2=c^2$, where neither of the two is divisible by $p$, can't be in the remainder class $\overline{1} \mod p$, as the statement for $p=3$ can be proven like this. I tried to prove this by contradiction, but did not succeed. I couldn't prove this for general $p \in \Bbb P$ with $p \geq 3$.
I would appreciate any help with this.
If none of $x,y,z$ is divisible by $p$, then by Fermat's Theorem the left-hand side is congruent to $2$ modulo $p$, but the right-hand side is congruent to $1$.