(JBMO 2018, 1) Find all pairs of integers $(m, n)$ such that $m^{5}-n^{5}=16 m n$
First, let's divide both sides by what we can. The idea is to make $m$ and $n$ coprime. Let $d$ be the gcd of $m$ and $n$, we write $m=d x$ and $n=d y$. The equation then becomes $d^{3}\left(x^{5}-y^{5}\right)=16 x y$. We notice that $x y$ is coprime with $x^{5}-y^{5}$. Indeed, if a prime $p$ divides $x y$, then $p$ divides $x$ or $p$ divides $y$. Without loss of generality, we can assume that $p$ divides $x$. In this case, if $p$ divides $x^{5}-y^{5}$, then $p$ divides $y^{5}$, so $p$ divides $y$, which is absurd. We conclude that $x^{5}-y^{5}$ divides 16, hence $\left|x^{5}-y^{5}\right| \leqslant 16$. If $x \neq y$, then the gap between two fifth powers can be 1 or 2 if $|x| \leqslant 1$ and $|y| \leqslant 1$. Otherwise, it is greater than the gap between two consecutive powers, which is minimal between 1 and 32 (or -1 and -32). If $x$ or $y$ is 0, we have the solution $(0,0)$. Otherwise, $x \neq y$ (since $16 x y \neq 0$), so $x^{5}-y^{5}=2$, so $d^{3}=8$, which gives $d=2$, yielding the solutions $(2,-2)$ and $(-2,2)$.