Theorem 1.2 of Bennett and Skinner (Canad. J. Math., 2004) asserts that the Diophantine equation xp−4yp=z2 is unsolvable for every prime p≥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>0 such that (x,y)=1 and xp+yp=zp, then there exist integers a,b,c>0 such that (a,b)=1 and ap−4bp=c2.Take a prime

p≥7.

We will prove Fermat’s Last Theorem in the form:

Take integers x,y,z>0.

If

(x,y)=1,

then

xp+yp≠zp.We argue by contradiction.

By the equation

xp+yp=zp

there is a rational

0<r<1

such that

xp=rzp and yp=(1−r)zp,

so that

r2−r+(xy)pz2p=0,

and hence

r=1+√1−4(xy)p/z2p2 or 1−√1−4(xy)p/z2p2.

Therefore,

the difference

1−4(xy)p/z2p≥0

is to be a perfect square.

But since

1−4(xy)pz2p=z2p−4(xy)pz2p,

since

z2p

is a perfect square,

and since if z2p=4(xy)p

then from the equation

xp+yp=zp

we have

x=y

that leads to a contradiction,

so there is an integer

c>0

such that

z2p−4(xy)p=c2.

On choosing

a:=z2 and b:=xy

we have

a,b>0 and ap−4bp=c2.

Moreover,

because

(x,y)=1

and

xp+yp=zp,

we have

(x,y)=(y,z)=(x,z)=1,

whence

(a,b)=1.

But the existence of such

a,b,c

contradicts Theorem 1.2 of Bennett and Skinner [1].

**Answer**

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.

**Attribution***Source : Link , Question Author : Megadeth , Answer Author : TonyK*