Ninth STARS OF MATHEMATICS Competition, 2015

Let $m$ and $n$ be relatively prime positive integers such that $m$ is odd and $m^4 - 2n^4$ is a perfect square, e.g., $m = 3$ and $n = 2$. Write $\ell^2 = m^4 - 2n^4$, so $\ell^4 = (m^4 - 2n^4)^2 = (m^4 + 2n^4)^2 - 8m^4n^4$, and $\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2 = -16m^4n^4 = -(2mn)^4$. Multiply the latter by $\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2 = 2\ell^4$ to get $(\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2)(\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2) = -2 \cdot (2\ell mn)^4$; that is, $(\ell^4 - 8m^4n^4)^2 - (m^4 + 2n^4)^4 = -2 \cdot (2\ell mn)^4$. Letting $m' = m^4 + 2n^4$ and $n' = 2\ell mn$, clearly $m' > m$, $m'$ is odd, $n' > n$, the difference $m'^4 - 2n'^4$ is a perfect square, and it is readily checked that $m'$ and $n'$ are relatively prime. The conclusion follows.