# Question about (probably false) elementary proof of Fermat's Last Theorem

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.

• Why is $x-y=r^p$? Is $p$ a prime here? – Dietrich Burde Jun 28 '17 at 12:30
• This couldn't be it as it would not fit in the margin. – Mikhail Katz Jun 28 '17 at 12:31
• @Niels How did you get from $a(x−y)+bz^p/(x−y)=1$ to $a(x−y)^2−(x−y)+bz^p=0$? – Toby Mak Jun 28 '17 at 12:32
• You want $p = n$? – Dirk Jun 28 '17 at 12:33
• @Dietrich: yes $p$ is prime. – user451710 Jun 28 '17 at 12:33

You will have $f = d$. Thus, your process will not be decreasing and it is totally ok (and not a contradiction) that it can be repeated infinitely.
When you write $d = 2e+1$, $\implies e = \dfrac{d-1}{2}$
On the other hand, $e = \dfrac{-1 \pm f}{2}$
So that, if $e$ is positive, $d=f$ and so on, all integers obtained are equal.