Junior Balkan Mathematical Olympiad

If one of $m$, $n$ is $0$, the other has to be $0$ too, and $(m,n) = (0,0)$ is one solution.

If $mn \neq 0$, let $d = \gcd(m,n)$ and we write $m = da$, $n = db$, $a, b \in \mathbb{Z}$ with $(a,b) = 1$. Then, the given equation is transformed into
$$d^3 a^5 - d^3 b^5 = 16ab \quad (1)$$
So, by the above equation, we conclude that $a | d^3 b^5$ and thus $a | d^3$. Similarly $b | d^3$. Since $(a,b) = 1$, we get that $ab | d^3$, so we can write $d^3 = abr$ with $r \in \mathbb{Z}$. Then, equation (1) becomes
$$abr^5 - abr^3 = 16ab \Rightarrow r(a^5 - b^5) = 16$$
Therefore, the difference $a^5 - b^5$ must divide $16$. This means that
$$a^5 - b^5 = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16.$$
The smaller values of $|a^5 - b^5|$ are $1$ or $2$. Indeed, if $|a^5 - b^5| = 1$ then $a = \pm 1$ and $b = 0$ or $a = 0$ and $b = \pm 1$, a contradiction. If $|a^5 - b^5| = 2$, then $a = 1$ and $b = -1$ or $a = -1$ and $b = 1$. Then $r = -8$, and $d^3 = -8$ or $d = -2$. Therefore, $(m,n) = (-2,2)$.

If $|a^5 - b^5| > 2$ then, without loss of generality, let $a > b$ and $a \ge 2$. Putting $a = x+1$ with $x \ge 1$, we have
$$\begin{aligned} |a^5 - b^5| &= |(x+1)^5 - b^5| \ge |(x+1)^5 - x^5| = \\ &= |5x^4 + 10x^3 + 10x^2 + 5x + 1| \ge 31 \end{aligned}$$
which is impossible. Thus, the only solutions are $(m,n) = (0,0)$ or $(-2,2)$.