# $N^2=2M^4-2p^2e^4$ has no integer solution

If $\gcd(M,e)=\gcd(N,e)=1$ and $p$ is prime and $p‎\equiv 5 \mod(16)$ then how I can show that $N^2=2M^4-2p^2e^4$ has no integer solution.

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You're clearly meant to consider the equation modulo $16$. Looking at it modulo $2$ is often a good way to start doing so. –  Hurkyl Jan 10 '13 at 9:55

Write $N = 2n$ and divide through by $2$. Write the right hand side as a product, and show that one of the factors is a square, the other one twice a square. Reduce the equation involving twice the square mod $p$.

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Franz's answer is the best one, and works for p that are 5 mod 8. One could also argue as follows. (N^2/2)=(M^4)-(p^2)(e^4) and 2(M^2)(p)(e^2) are the sides of a Pythagorean right triangle whose area is 2p(NMe/2)^2. So 2p would be a "congruent number", and as Fermat showed by an infinite descent this is impossible.(But the Fermat argument essentially uses Franz's calculation, which is the determination of a 2-Selmer group, and the infinite descent isn't at all needed to answer the proposer's question)

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