3. Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:

$$
n+f(m) \text { divides } f(n)+n f(m)
$$

for all $m, n \in \mathbb{N}$.

(Albania)
3. The problem condition gives us $n+f(m) \mid f(n)+n f(m)-n(n+f(m))=f(n)-n^{2}$. Setting $m=n=1$ yields $f(1)+1 \mid f(1)-1$ and hence $f(1)=1$.

Function $f(x) \equiv x^{2}$ satisfies the conditions. Suppose that $f\left(n_{0}\right) \neq n_{0}^{2}$ for some $n_{0}$. Then from $n_{0}+f(m)|| f\left(n_{0}\right)-n_{0}^{2} \mid$ it follows that $f(m) \leqslant A=\left|f\left(n_{0}\right)-n_{0}^{2}\right|-n_{0}$. On the other hand, the problem condition for $m=1$ yields $n+1 \mid f(n)+n$, i.e. $f(n) \equiv 1(\bmod n)$, but $f(n)A$ it holds that $n+f(m) \mid f(m)(n+f(m))-(f(n)+n f(m))=$ $f(m)^{2}-1$, so $f(m)=1$ for all $m \in \mathbb{N}$, which is also a solution.

Therefore, the only solutions are functions $f(x) \equiv x^{2}$ and $f(x) \equiv 1$.