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

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    Thank you all. I saw this problem a long time ago, but do not know about the answers. The answer Apolo wrote is correct, and there are infinitely many solutions for this one, but I don't know how to find (all) the form of the solutions.2011-04-01

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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.2011-04-18