2015 Ninth STARS OF MATHEMATICS Competition

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 ${\ell }^4=(m^4-2n^4{)}^2=(m^4+2n^4{)}^2-8m^4{n}^4$, and ${\ell }^4-8m^4{n}^4-(m^4+2n^4{)}^2=-16m^4{n}^4=-(2mn{)}^4$. Multiply the latter by ${\ell }^4-8m^4{n}^4+(m^4+2n^4{)}^2=2{\ell }^4$ to get $({\ell }^4-8m^4{n}^4+(m^4+2n^4{)}^2)({\ell }^4-8m^4{n}^4-(m^4+2n^4{)}^2)=-2\cdot (2\ell mn{)}^4$;

that is, $({\ell }^4-8m^4{n}^4{)}^2-(m^4+2n^4{)}^4=-2\cdot (2\ell mn{)}^4$. Letting $m^{\prime }=m^4+2n^4$ and $n^{\prime }=2\ell mn$, clearly $m^{\prime }>m$, $m^{\prime }$ is odd, $n^{\prime }>n$, the difference $m^{\prime 4}-2n^{\prime 4}$ is a perfect square, and it is readily checked that $m^{\prime }$ and $n^{\prime }$ are relatively prime. The conclusion follows.