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Theorem 1.2 of Bennett and Skinner (Canad. J. Math., 2004) asserts that the Diophantine equation $x^{p} - 4y^{p} = z^{2}$ is unsolvable for every prime $p \geq 7.$ 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 $x, y, z > 0$ such that $(x, y) = 1$ and $x^{p} + y^{p} = z^{p},$ then there exist integers $a, b, c > 0$ such that $(a, b)=1$ and $a^{p} - 4b^{p} = c^{2}$.

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

We argue by contradiction. By the equation $x^{p} + y^{p} = z^{p}$ there is a rational $0 < r < 1$ such that $$x^{p} = rz^{p}\ \ \mbox{and}\ \ y^{p} = (1-r)z^{p},$$ so that $$r^{2} - r + \dfrac{(xy)^{p}}{z^{2p}} = 0,$$ and hence $$r = \dfrac{1 + \sqrt{1 - 4(xy)^{p}/z^{2p}}}{2}\ \ \mbox{or}\ \ \dfrac{1 - \sqrt{1 - 4(xy)^{p}/z^{2p}}}{2}.$$ Therefore, the difference $1 - 4(xy)^{p}/z^{2p} \geq 0$ is to be a perfect square. But since $$1 - \dfrac{4(xy)^{p}}{z^{2p}} = \dfrac{z^{2p} - 4(xy)^{p}}{z^{2p}},$$ since $z^{2p}$ is a perfect square, and since if $z^{2p} = 4(xy)^{p}$ then from the equation $x^{p} + y^{p} = z^{p}$ we have $x = y$ that leads to a contradiction, so there is an integer $c > 0$ such that $$z^{2p} - 4(xy)^{p} = c^{2}.$$ On choosing $$a := z^{2}\ \ \mbox{and}\ \ b := xy$$ we have $$a, b > 0\ \ \mbox{and}\ \ a^{p} - 4b^{p} = c^{2}.$$ Moreover, because $(x, y) = 1$ and $x^{p} + y^{p} = z^{p},$ 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].

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    $\begingroup$ $$z^{2p} - 4(xy)^{p} = (x^p+y^p)^2- 4(xy)^{p}= (x^p-y^p)^2$$ so your $c=x^p-y^p$ ;) $\endgroup$
    – N. S.
    Nov 23, 2014 at 23:59
  • $\begingroup$ Are $a$ and $c$ relatively prime? $\endgroup$
    – David
    Nov 24, 2014 at 0:04
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    $\begingroup$ @Genomeme: "the answer to your question is no": What question? Whose question? $\endgroup$
    – TonyK
    Nov 24, 2014 at 0:17
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    $\begingroup$ Good point. I suppose a question was not asked. My point was that the proof may not in fact be a valid proof. $\endgroup$ Nov 24, 2014 at 13:43
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    $\begingroup$ To highlight the excellent comment of @N.S.: the whole proof in the question can be simplified to: “as $$(x^p + y^p)^2 - 4(xy)^p = (x^p - y^p)^2$$ (a purely algebraic identity), we have $$x^p + y^p = z^p \implies (z^2)^p - 4(xy)^p = (x^p - y^p)^2$$ which is a solution to $a^p - 4b^p = c^2$.” $\endgroup$ Apr 18, 2017 at 6:28

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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.

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