Problem 1143

By taking $m=n=1$, we find $f(1)^{2}+f(1) \mid 4$, so $f(1)=1$. By taking $m=1$, we then obtain

$$1+f(n) \mid (n+1)^{2}$$

To minimize the number of divisors as much as possible, let's choose $n=p-1$ with $p$ being a prime. Then $1+f(p-1) \mid p^{2}$, so $f(p-1)$ can only be $p-1$ or $p^{2}-1$. In the latter case, by taking $n=1$ and $m=p-1$ in the initial equation, we get

$$1+(p^{2}-1)^{2} \mid ((p-1)^{2}+1)^{2}$$

Expanding this, it becomes $p^{4}-2 p^{2}+2 \leqslant p^{4}-4 p^{3}+6 p^{2}-8 p+4$, which is impossible for $p$ large enough. Therefore, we get $f(p-1)=p-1$ for $p$ a large enough prime.

Finally, let $n \in \mathbb{N}^{*}$. For a large enough prime $p$, we have

$$(p-1)^{2}+f(n) \mid ((p-1)^{2}+n)^{2}$$

We aim to put this in a more convenient form: we have $(p-1)^{2} \equiv -f(n) \pmod{(p-1)^{2}+f(n)}$, so the last equation becomes

$$(p-1)^{2}+f(n) \mid ((-f(n)+n)^{2}$$

and this for any large enough prime $p$. The number $(f(n)-n)^{2}$ thus has an infinite number of divisors, so $f(n)=n$.