We denote by $\mathbb{N}^{*}$ the set of strictly positive integers. Find all $^{2}$ functions $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ such that

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

for all strictly positive integers $m$ and $n$.
Let $f$ be a potential solution to the problem.
We denote ( $\left.{ }^{*}\right)$ as the condition " $m^{2}+f(n)$ divides $m f(m)+n$ ".
By choosing $m=n=2$ in $\left(^{*}\right)$, we get that $4+f(2)$ divides $2 f(2)+2$. Since $2 f(2)+2 < 2(f(2)+4)$, it must be that $f(2)+4=2 f(2)+2$, hence $f(2)=2$.

Now, by choosing $m=2$, the condition ( ${ }^{*}$ ) ensures that, for all $n \geq 1$, the number $4+f(n)$ divides $4+n$, which leads to $f(n) \leq n$, for all $n \geq 1$. (i)

On the other hand, by choosing $m=n$ in $\left({ }^{*}\right)$, we deduce this time that, for all $n \geq 1$, the number $n^{2}+f(n)$ divides $n f(n)+n$, so $n^{2}+f(n) \leq n f(n)+n$. Thus, we have $(n-1)(f(n)-n) \geq 0$, which leads to $f(n) \geq n$, for all $n \geq 2$. Since it is clearly true that $f(1) \geq 1$, it follows that $f(n) \geq n$ for all $n \geq 1$. (ii)

From (i) and (ii), we deduce that $f(n)=n$ for all $n \geq 1$.
Conversely, it is evident that $f: n \longmapsto n$ is indeed a solution to the problem.