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