Wondering about correctness of this:
Theorem:
Let:
$ \quad \quad \quad \quad p$ be an odd prime
$ \quad \quad \quad \quad \gcd(x,y,z) = 1$
$ \quad \quad \quad \quad xyz \not \equiv 0 \pmod p$
$ \quad \quad \quad \quad a \equiv y/z \pmod p$
$ \quad \quad \quad \quad x^p = y^p + z^p$
Then $p$ is a Wieferich-prime to base $a$.
Proof:
Note:
$ \quad \quad \quad \quad y/z \equiv a \implies y \equiv az \pmod p$
$ \quad \quad \quad \quad \implies y^p \equiv (az)^p \pmod {p^2}$
$ \quad \quad \quad \quad \implies x^p = y^p + z^p \equiv (az)^p + z^p \equiv (a^p + 1)z^p \pmod {p^2}$
Also:
$ \quad \quad \quad \quad x \equiv y + z \equiv az + z \pmod p$
$ \quad \quad \quad \quad x^p \equiv (az + z)^p \equiv (a + 1)^pz^p \pmod {p^2}$
So:
$ \quad \quad \quad \quad x^p \equiv (a^p + 1)z^p \equiv (a + 1)^pz^p \pmod {p^2}$
$ \quad \quad \quad \quad a^p + 1 \equiv a + 1 \implies a^p \equiv a \pmod {p^2}$
So our theorem is correct.
How do you conclude in the last line that $a^p+1\equiv a+1\pmod{p^2}$? You've shown that $$(a^p+1)z^p\equiv x^p\equiv(a+1)^pz^p\pmod{p^2},$$ which implies that $a^p+1\equiv(a+1)^p\pmod{p^2}$ because $p\nmid z$. This does not immediately imply $a^p+1\equiv a+1\pmod{p^2}$, as far as I can tell.