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 $mn\ne 0$, 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 $$d^3{a}^5-d^3{b}^5=16ab(1)$$ 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 $$ab{r}^5-ab{r}^3=16ab\Rightarrow r(a^5-b^5)=16$$ 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 $$\begin{array}{rl}|a^5-b^5|& =|(x+1{)}^5-b^5|\ge |(x+1{)}^5-x^5|=\\ & =|5x^4+10x^3+10x^2+5x+1|\ge 31\end{array}$$ which is impossible. Thus, the only solutions are $(m,n)=(0,0)$ or $(-2,2)$.