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How does one find all integer solutions to the equation $3a^2-4b^3=7^c$?

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Could you reformulate this as a real question so people will answer. I'd like to see what the answer is. I mean like show what you think about it, what you've tried, what's giving you trouble and stuff like that. There's two answers that are very obvious. Could you at least find it and put that in there? –  knucklebumpler Mar 31 '11 at 7:18
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This question was crossposted to MO here, and was closed. Amir, you should know that in general, it is not considered polite to post a question in multiple forums simultaneously - you should post in one, and wait a while for answers before trying somewhere else. –  Zev Chonoles Mar 31 '11 at 7:38
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Also, knucklebumpler's comment is correct - you should explain where this question is coming from, what you have tried, etc. It is not considered polite to simply post an imperative sentence as your question - it isn't conducive to people wanting to help you. –  Zev Chonoles Mar 31 '11 at 7:40
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If no attempt is made by the OP, I think this should be closed here too. –  picakhu Mar 31 '11 at 13:25
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@picakhu, no don't close it. Maybe nothing they tried worked. –  quanta Mar 31 '11 at 16:16

1 Answer 1

up vote 3 down vote accepted

I have a feeling that the solutions I gave are the only ones, but I don't have a proof of that. (Those solutions are the families $(\pm6^{3m},-7^{2m},1+6m)$ and $(\pm13\cdot6^{3m},-5\cdot7^{2m},1+6m)$ for $m\in\mathbb{N}$.)

Maybe somebody who knows more about elliptic curves can pick this up. For any value of $c$, you can look at the elliptic curve $y^2=x^3+2^4\,3^3\,7^c$. For fixed values of $c$ it seems you can show that the torsion part is trivial and that the curves are of rank 0 unless $c\equiv1\pmod{3}$. For $c\equiv1\pmod{6}$ you'll get the two solutions given above coming from powers of the generator with the rest of the powers giving non-integral rational solutions. (For $c\equiv4\pmod{6}$ points on the curve do not correspond to integral solutions, since one can easily check the requirements $a\equiv1\pmod2$, $b\equiv2\pmod3$ and $c\equiv1\pmod2$ for integer solutions to the original equation.)

I don't have a general argument for these facts (just checked numerous special cases) so this is just hand-waving for now, but I'm not a number-theorist, so I'll let the experts fill in the details...

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BTW Twh article "Computing integral points on Mordell’s elliptic curves" by Gebel, Petho, Zimmer shows that the torsion groups are trivial (Prop.3.1) and the article may give the tools to complete the proof of the result. –  Apollo Apr 18 '11 at 14:27

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