If $p \geq 5$ is a prime, are there any integers $x, y, z > p$ such that $(x, y) = 1$ and $$x^{p}  4y^{p} = z^{2}$$
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 BennettSkinner 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 ShimuraTaniyama) for $p=11$ or $p\ge 17$. 

