Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
4  
$78^3-4\times29^3=614^2$. $93^3-4\times53^3=457^2$. –  Gerry Myerson Aug 20 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. –  Kurt Aug 20 at 2:00
1  
Note that my earlier comment referred to an earlier version of the question. –  Gerry Myerson Aug 20 at 2:02
1  
If there were infinitely many, it would contradict the abc conjecture. –  Felipe Voloch Aug 20 at 3:14
1  
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 at 3:34

1 Answer 1

up vote 9 down vote accepted

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 $p\ge 17$.

share|improve this answer
2  
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 at 3:48
1  
@GHfromMO: Thanks for pointing that out! I'll also add a link to Darmon's paper. –  Lucia Aug 20 at 3:51
1  
And very similar result were proved by Ivorra, see Acta Arith. 108 (2003), 327–338. –  GH from MO Aug 20 at 3:52
1  
And also by Siksek, see J. Théor. Nombres Bordeaux 15 (2003), 839–846. –  GH from MO Aug 20 at 3:55
    
Thanks so much. I wish I can vote you up more than once. –  Kurt Aug 20 at 7:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.