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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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    $\begingroup$ $78^3-4\times29^3=614^2$. $93^3-4\times53^3=457^2$. $\endgroup$ Aug 20, 2014 at 1:59
  • $\begingroup$ 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. $\endgroup$
    – Yes
    Aug 20, 2014 at 2:00
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    $\begingroup$ Note that my earlier comment referred to an earlier version of the question. $\endgroup$ Aug 20, 2014 at 2:02
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    $\begingroup$ If there were infinitely many, it would contradict the abc conjecture. $\endgroup$ Aug 20, 2014 at 3:14
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    $\begingroup$ 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/… . $\endgroup$
    – Lucia
    Aug 20, 2014 at 3:34

1 Answer 1

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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 $p\ge 17$.

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    $\begingroup$ 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). $\endgroup$
    – GH from MO
    Aug 20, 2014 at 3:48
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    $\begingroup$ @GHfromMO: Thanks for pointing that out! I'll also add a link to Darmon's paper. $\endgroup$
    – Lucia
    Aug 20, 2014 at 3:51
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    $\begingroup$ And very similar result were proved by Ivorra, see Acta Arith. 108 (2003), 327–338. $\endgroup$
    – GH from MO
    Aug 20, 2014 at 3:52
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    $\begingroup$ And also by Siksek, see J. Théor. Nombres Bordeaux 15 (2003), 839–846. $\endgroup$
    – GH from MO
    Aug 20, 2014 at 3:55
  • $\begingroup$ Thanks so much. I wish I can vote you up more than once. $\endgroup$
    – Yes
    Aug 20, 2014 at 7:29

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