Can somebody explain to me what is wrong with the following argument for a proof of Fermat's Last Theorem?
Suppose somewhere the following theorem has been proven:
Theorem:
If:
$\quad \quad \quad \quad x^p = y^p + z^p$
$\quad \quad \quad \quad \gcd(x,y,z) = 1$
$\quad \quad \quad \quad p \gt 2$ and prime
Then:
$ \quad \quad \quad \quad x - y = r^p$,
$ \quad \quad \quad \quad (x^p - y^p)/(x - y) = s^p$,
for some $r,s$ with $\gcd(r,s) = 1$.
We now proceed to prove some corollary:
Corollary:
$\quad$ There can be no solutions to our system.
Proof:
Because $\gcd(x - y,(x^p - y^p)/(x - y)) = \gcd(x - y,z^p/(x - y) = 1$, there exist some $a,b$ such that:
$ \quad \quad \quad \quad a(x - y) + bz^p/(x - y) = 1$
$ \quad \quad \quad \quad \implies a(x - y)^2 - (x - y) + bz^p = 0$
We now use the quadratic formula to show there can be no solutions to our system by infinitely generating smaller ones.
$ \quad \quad \quad \quad D_{x - y} = 1 - 4abz^p = d^2$, for some d
$ \quad \quad \quad \quad \implies (1 - d)(1 + d) = 4abz^p$
so $d$ is odd, say $2e + 1$ for some $e$.
$ \quad \quad \quad \quad \implies -2e(2e + 2) = 4abz^p$
$ \quad \quad \quad \quad \implies -e(e + 1) = abz^p$
$ \quad \quad \quad \quad \implies e^2 + e + abz^p = 0$
$ \quad \quad \quad \quad \implies D_e = 1 - 4abz^p = f^2$, for some f.
$ \quad \quad \quad \quad \implies (1 - f)(1 + f) = 4abz^p$
so $f$ is odd, say $2g + 1$, for some $g$.
$ \quad \quad \quad \quad \implies -2g(2g + 2) = 4abz^p$
$ \quad \quad \quad \quad \implies -g(g + 1) = abz^p$
$ \quad \quad \quad \quad \implies g^2 + g + abz^p = 0$
We see we keep ending up on the equation:
$ \quad \quad \quad \quad u^2 + u + abz^p = 0$
Since we can infinitely repeat this procedure through calculation of the discriminant $D_u$, we find there is no smallest solution.
We conclude our system cannot have any solutions.
