Integer solutions to $2m^4 = n^4 + 1$?

Question

The problem looks deceptively simple. I know of solutions to variations of that such as $m^4 = 2n^4 + 1$. I have tried various approaches but didnt get far and so not worth reproducing it here.

To be clear, I am looking for integer solutions to $2m^4 = n^4 + 1$.

Thanks

Answer 1

Ljunggren has proved that the Diophantine equation $2m^4=n^2+1$ has only two (positive) integral solutions, namely $(1,1)$ and $(13,239)$. Since $(13,239)$ is not a solution to the equation of the question, the only positive solution for $2m^4=n^4+1$ is $(1,1)$; so the only integral solutions are $(\pm 1,\pm 1)$.

References: L. J. Mordell, The Diophantine equation $y^2 = Dx^4 +1$, J. London Math. Soc. 39 (1964), 161-164.

Hua: A new solution of the Diophantine equation $x^2+1=2y^4$.

Answer 2

It has been shown (although the general proof is difficult, being related to that of Fermat's Last Theorem) that there are no non-trivial solutions of $x^n + y^n = 2z^n$ for any $n \geq 3$ (Darmon & Merel 2001 pp 2 & 24). Hence, putting $n=4$ and $y=1$ (and re-lettering), there are no solutions of $2m^4=n^4+1$ other than $(\pm1,\pm1)$.