I have found the following way to prove some(Wieferich's) criterion for Fermat's Last Theorem and am wondering what would be wrong. My point of doubt is calculation of the Fermat-quotients of $y,z$ being $-1$, since I found these rules on Wikipedia.
Also, should I split this in parts? I can imagine people don't feel like going through too much text.
Anyway, have fun!
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$
If:
$ \quad \quad \quad \quad x^p = y^p + z^p$,
then $p$ is Wieferich-prime.
Proof:
Consider the following congruence:
$ \quad \quad \quad \quad (x^n - y^n)/(x - y) \equiv nx^{n - 1} \pmod {x - y}$
which we can prove by induction on $n$ and in which we divide first.
Let $n = p$.
Since:
$ \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)$
$ \quad \quad \quad \quad = \gcd(x - z,p)$
$ \quad \quad \quad \quad = 1$,
it follows that:
$ \quad \quad \quad \quad x - y = r^p$,
$ \quad \quad \quad \quad (x^p - y^p)/(x - y) = s^p$,
$ \quad \quad \quad \quad x - z = t^p$,
$ \quad \quad \quad \quad (x^p - z^p)/(x - z) = u^p$,
$ \quad \quad \quad \quad rs = z$,
$ \quad \quad \quad \quad tu = y$,
for some $r,s,t,u$ with $\gcd(r,s) = \gcd(t,u) = 1$.
The following also holds for $x - z,t,u$:
$ \quad \quad \quad \quad s \equiv 1 \pmod p \implies s^p \equiv 1 \pmod {p^2}$
Now let:
$ \quad \quad \quad \quad s^p = px^{p - 1} \pmod {x - y}$
$ \quad \quad \quad \quad \implies s^p = px^{p - 1} + ar^p \equiv 1 \pmod {p^2}$, for some $a$
$ \quad \quad \quad \quad \implies s \equiv ar \equiv 1 \pmod p \implies s^p \equiv (ar)^p \equiv 1 \pmod {p^2}$
$ \quad \quad \quad \quad \implies ar^p \equiv 1/a^{p - 1} \pmod {p^2}$
$ \quad \quad \quad \quad \implies s^p = px^{p - 1} + ar^p \equiv px^{p - 1} + 1/a^{p - 1} \equiv 1 \pmod {p^2}$
$ \quad \quad \quad \quad \implies px^{p - 1} \equiv 1 - 1/a^{p - 1} \pmod {p^2}$
$ \quad \quad \quad \quad \implies p(ax)^{p - 1} \equiv a^{p - 1} - 1 \pmod {p^2}$
$ \quad \quad \quad \quad \implies q_p(a) \equiv 1 \pmod p$,
where $q_p(a)$ denotes the Fermat-quotient for $a$ modulo $p$.
So it follows that:
$ \quad \quad \quad \quad q_p(r) \equiv q_p(1/a) \equiv -q_p(a) \equiv -1 \pmod p$
Because of $q_p(s) \equiv 0 \pmod p$:
$ \quad \quad \quad \quad q_p(z) \equiv q_p(rs) \equiv q_p(r) + q_p(s) \equiv -1 + 0 \equiv -1 \pmod p$
Since the same holds for $x - z,t,u$, we now have:
$ \quad \quad \quad \quad q_p(y) \equiv q_p(z) \equiv -1 \pmod p$
From which it follows that:
$ \quad \quad \quad \quad y^{p - 1} \equiv 1 - p \pmod {p^2}$
$ \quad \quad \quad \quad z^{p - 1} \equiv 1 - p \pmod {p^2}$
$ \quad \quad \quad \quad \implies y^p \equiv y(1 - p) \pmod { p^2 }$
$ \quad \quad \quad \quad \implies z^p \equiv z(1 - p) \pmod { p^2 }$
We also note:
$ \quad \quad \quad \quad y^p \equiv (tu)^p \equiv t^p \equiv x - z \pmod {p^2}$
$ \quad \quad \quad \quad z^p \equiv (rs)^p \equiv r^p \equiv x - y \pmod {p^2}$
So we can set:
$ \quad \quad \quad \quad y(1 - p) \equiv x - z \pmod {p^2}$
$ \quad \quad \quad \quad z(1 - p) \equiv x - y \pmod {p^2}$
$ \quad \quad \quad \quad \implies (x - z)/y \equiv (x - y)/z \implies z(x - z) \equiv y(x - y) \pmod {p^2}$
$ \quad \quad \quad \quad \implies y^2 - z^2 \equiv (y + z)(y - z) \equiv x(y - z) \pmod {p^2}$
So either:
$ \quad \quad \quad \quad \implies x \equiv y + z \pmod {p^2}$
or:
$ \quad \quad \quad \quad \implies p | y - z$
Suppose $x \equiv y + z \pmod {p^2}$:
$ \quad \quad \quad \quad \implies (x - y)^p \equiv r^{p^2} \equiv r^p \equiv x - y \implies z^p \equiv z \pmod {p^2}$, contradicting:
$ \quad \quad \quad \quad z^p \equiv z(1 - p) \pmod { p^2 }$
So now we know $y \equiv z \pmod {p}$
But then:
$ \quad \quad \quad \quad y^p \equiv z^p \pmod {p^2}$
$ \quad \quad \quad \quad \implies x^p \equiv y^p + z^p \equiv 2z^p \pmod {p^2}$
Also:
$ \quad \quad \quad \quad x \equiv y + z \implies x \equiv z + z \equiv 2z \pmod p$
$ \quad \quad \quad \quad \implies x^p \equiv (2z)^p \pmod {p^2}$
We conclude:
$ \quad \quad \quad \quad x^p \equiv (2z)^p \equiv 2z^p \pmod {p^2}$
$ \quad \quad \quad \quad \implies (2z)^p - 2z^p \equiv 0 \pmod {p^2}$
$ \quad \quad \quad \quad \implies z^p(2^p - 2) \equiv 0 \pmod {p^2}$,
from which we can see $p$ must be a Wieferich-prime.