# The Diophantine equation $x^p - 4y^p = z^2$

If $p \geq 5$ is a prime, are there any integers $x, y, z > p$ such that $(x, y) = 1$ and $$x^{p} - 4y^{p} = z^{2}$$

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$78^3-4\times29^3=614^2$. $93^3-4\times53^3=457^2$. –  Gerry Myerson Aug 20 '14 at 1:59
Ah fantastic. Appreciated. Then I have to strengthen the lower bound. The first version of my question is for $p \geq 3,$ and that's why Gerry Myerson left the comment. –  Chou Aug 20 '14 at 2:00
Note that my earlier comment referred to an earlier version of the question. –  Gerry Myerson Aug 20 '14 at 2:02
If there were infinitely many, it would contradict the abc conjecture. –  Felipe Voloch Aug 20 '14 at 3:14
Well if there were infinitely many solutions, it would contradict a Theorem of Darmon and Granville! See math.mcgill.ca/darmon/pub/Articles/Research/12.Granville/… . –  Lucia Aug 20 '14 at 3:34

See Theorem 1.2 of the paper by Bennett and Skinner, which settles the problem for $p\ge 7$ (take there $C=1$ and $\alpha_0=2$). Note that the Bennett-Skinner results are more general. (Earlier work of Darmon and Granville (using Faltings's theorem) showed that there are only finitely many solutions; again for more general such equations.)
Finally GH from MO has kindly pointed out an earlier paper of Darmon that handles this particular equation (assuming Shimura-Taniyama) for $p=11$ or $p\ge 17$.
Actually Darmon already proved in 1993 that, assuming the Shimura-Taniyama conjecture, there are no nonzero solutions for $p\geq 17$ and for $p=11$. See Proposition 2.5 in his paper (Internat. Math. Res. Notices 1993, 263–274). –  GH from MO Aug 20 '14 at 3:48