
3. Find all pairs $(m, n)$ of positive integers such that $(m+n)^{2}$ divides $4(m n+1)$. (Tomáš Jurík)


Solution. The problem is symmetric in $(m, n)$ and we can wlog assume $m \geqslant n$.

If positive integer $A=(m+n)^{2}$ divides positive integer $B=4(m n+1)$, we have

$$
(m+n)^{2} \leqslant 4(m n+1), \quad \text { or } \quad(m-n)^{2} \leqslant 4
$$

Thus $0 \leqslant m-n \leqslant 2$ and we are left with one of the following:
$\triangleright m=n$, then $A=4 n^{2}, B=4 n^{2}+4$, and $A$ divides $B$ if and only if $4 n^{2}$ divides 4 , that is $n=1$, and $(m, n)=(1,1)$.

$\triangleright m=n+1$, then $A=4 n^{2}+4 n+1, B=4 n^{2}+4 n+4=A+3$, and $A$ divides $B$ if and only if $4 n^{2}+4 n+1$ divides 3 . But for positive integers $n$ there is $4 n^{2}+4 n+1 \geqslant 4+4+1=9$, and thus we get no solution in this case.

$\triangleright m=n+2$, then $A=4 n^{2}+8 n+4=B$, thus any pair $(n+2, n)$ of positive integers is a solution.

Conclusion. The solutions are pair $(1,1)$ and any pair of the form $(n+2, n)$ or $(n, n+2)$, where $n$ is a positive integer.
