I am wondering about correctness of the following theorem. It looks very simple to me but I can't seem to find a mistake. Hopefully anyone will take a look at it.
Theorem:
Let:
$\quad\quad p$ an odd prime,
$\quad\quad \gcd(x,y,z) = 1,$
$\quad\quad x^p = y^p + z^p,$
Then:
$\quad\quad p \equiv 1 \pmod{r^p}$ for some $r$
Proof:
$\quad\quad$ Assume $p \not | z$ and consider the following congruence:
$\quad\quad\quad\quad (x^p - y^p)/(x - y) \equiv px^{p - 1} \equiv py^{p - 1} \pmod{x - y}$
$\quad\quad$ In which division has to be done before the modulo-operation.
$\quad\quad$ It follows that:
$\quad\quad\quad\quad \gcd(x - y,(x^p - y^p)/(x - y))$
$\quad\quad\quad\quad = \gcd(x - y,px^{p - 1})$
$\quad\quad\quad\quad = \gcd(x - y,p) = 1$
$\quad\quad$ From which we conclude that:
$\quad\quad\quad\quad x - y = r^p,$
$\quad\quad\quad\quad (x^p - y^p)/(x - y) = s^p$
$\quad\quad$ for some $r,s$ with $\gcd(r,s) = 1, z = rs$
$\quad\quad \implies s^p - px^{p - 1} \equiv (x^p - y^p)/(x - y) - px^{p - 1} \equiv 0 \pmod{x - y}$
$\quad\quad$ Because:
$\quad\quad\quad\quad s^p \equiv px^{p - 1} \equiv py^{p - 1} \pmod{x - y}$
$\quad\quad x,y$ are interchangeable
$\quad\quad \implies x^p - y^p - (x - y)px^{p - 1} \equiv 0 \pmod{(x - y)^2}$
$\quad\quad \implies (x^p - px^p) - (y^p - py^p) \equiv 0 \pmod{(x - y)^2}$
$\quad\quad \implies (1 - p)(x^p - y^p) \equiv 0 \pmod{(x - y)^2}$
$\quad\quad \implies p \equiv 1 \pmod{x - y}$
$\quad\quad \implies p \equiv 1 \pmod{r^p}$