3. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that

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

for all $m, n \in \mathbb{N}$. (We denote the set of natural numbers by $\mathbb{N}$.)

(Albania)
3. From the condition of the problem, it follows that $n+f(m) \mid f(n)+n f(m)-n(n+f(m))=f(n)-n^{2}$. For $m=n=1$, we get $f(1)+1 \mid f(1)-1$, from which $f(1)=1$.

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

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