Q:

Number Theory Proof Need Logic Checked

I'm working on the following problem: 

Show that if $x^{p} + y^{p} = z^{p}$, then $p \space | \space (x + y -z)$

So far my proof looks something like this: 
Suppose $p \nmid \space (x+y-z)$ then $x^p + y^p = z^p$ shouldn't have a solution (proof by contraposition). Taking the original equation $\bmod p$ we have 
$$x^p + y^p = z^p \quad \rightarrow \quad x + y \equiv z \space \bmod p \quad \rightarrow \quad x + y - z \equiv 0 \space \bmod p \quad \rightarrow \quad p \mid (x + y -z)$$ (which we know is false). Hence the original equation holds.
So I don't think my logic is making sense, and am need of assistance if it's wrong. Any tips would be appreciated. 

A:

Hint: Fermat's little theorem states that $x^p \equiv x \pmod p$.
I'm not exactly certain how you want us to respond. It appears that you have the idea. The way that I would express it is as follows:
$x + y \equiv x^p + y^p = z^p \equiv z \pmod{p}$, hence $p \mid x+y -z$.

So, most of you know FLT as "If $x \neq 0 \pmod{p}$, then $x^{p-1} \equiv 1 \pmod{p}$. The first line is a direct corollary of this result.
If $x \equiv 0 \pmod{p}$, then $x \equiv 0 \equiv x^p \pmod{p}$. Otheriwse, $x^p \equiv x^{p-1} x \equiv x \pmod{p}$.
Hence, in all cases, $x^p \equiv p \pmod{p}$.

