# Checking a Proof of a Theorem

Theorem 1.2 of Bennett and Skinner (Canad. J. Math., 2004) asserts that the Diophantine equation $$xp−4yp=z2x^{p} - 4y^{p} = z^{2}$$ is unsolvable for every prime $$p≥7.p \geq 7.$$
The following is a possible proof (from an arXiv author) that Fermat’s Last Theorem is a consequence of this theorem, i.e., proof that if there exist integers $$x,y,z>0x, y, z > 0$$ such that $$(x,y)=1(x, y) = 1$$ and $$xp+yp=zp,x^{p} + y^{p} = z^{p},$$ then there exist integers $$a,b,c>0a, b, c > 0$$ such that $$(a,b)=1(a, b)=1$$ and $$ap−4bp=c2a^{p} - 4b^{p} = c^{2}$$.

Take a prime
$$p≥7.p \geq 7.$$
We will prove Fermat’s Last Theorem in the form:
Take integers $$x,y,z>0x, y, z > 0$$.
If
$$(x,y)=1,(x, y) = 1,$$
then
$$xp+yp≠zpx^{p} + y^{p} \neq z^{p}$$.

We argue by contradiction.
By the equation
$$xp+yp=zpx^{p} + y^{p} = z^{p}$$
there is a rational
$$0
such that
$$xp=rzp and yp=(1−r)zp,x^{p} = rz^{p}\ \ \mbox{and}\ \ y^{p} = (1-r)z^{p},$$
so that
$$r2−r+(xy)pz2p=0,r^{2} - r + \dfrac{(xy)^{p}}{z^{2p}} = 0,$$
and hence
$$r=1+√1−4(xy)p/z2p2 or 1−√1−4(xy)p/z2p2.r = \dfrac{1 + \sqrt{1 - 4(xy)^{p}/z^{2p}}}{2}\ \ \mbox{or}\ \ \dfrac{1 - \sqrt{1 - 4(xy)^{p}/z^{2p}}}{2}.$$
Therefore,
the difference
$$1−4(xy)p/z2p≥01 - 4(xy)^{p}/z^{2p} \geq 0$$
is to be a perfect square.
But since
$$1−4(xy)pz2p=z2p−4(xy)pz2p,1 - \dfrac{4(xy)^{p}}{z^{2p}} = \dfrac{z^{2p} - 4(xy)^{p}}{z^{2p}},$$
since
$$z2pz^{2p}$$
is a perfect square,
and since if $$z2p=4(xy)pz^{2p} = 4(xy)^{p}$$
then from the equation
$$xp+yp=zpx^{p} + y^{p} = z^{p}$$
we have
$$x=yx = y$$
so there is an integer
$$c>0c > 0$$
such that
$$z2p−4(xy)p=c2.z^{2p} - 4(xy)^{p} = c^{2}.$$
On choosing
$$a:=z2 and b:=xya := z^{2}\ \ \mbox{and}\ \ b := xy$$
we have
$$a,b>0 and ap−4bp=c2.a, b > 0\ \ \mbox{and}\ \ a^{p} - 4b^{p} = c^{2}.$$
Moreover,
because
$$(x,y)=1(x, y) = 1$$
and
$$xp+yp=zp,x^{p} + y^{p} = z^{p},$$
we have
$$(x,y)=(y,z)=(x,z)=1,(x, y) = (y, z) = (x, z) = 1,$$
whence
$$(a,b)=1.(a, b) = 1.$$
But the existence of such
$$a,b,ca, b, c$$
contradicts Theorem 1.2 of Bennett and Skinner [1].

Your proof looks OK to me. But don't rush to publish $-$ it seems to me that Theorem 1.2 (available at this link) depends on the Taniyama–Shimura–Weil conjecture (or the modularity theorem, as it should now be called), just like Wiles' proof of Fermat's Last Theorem did.
So your proof is $-$ roughly speaking $-$ using a consequence of Fermat's Last Theorem to prove Fermat's Last Theorem.