Problem (Number Theory):
Let $\mathbb{Z}_{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that

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

for all positive integers $m$ and $n$.

Solution:
Setting $m=n=2$ tells us that $4+f(2) \mid 2 f(2)+2$. Since $2 f(2)+2<2(4+f(2))$, we must have $2 f(2)+2=4+f(2)$, so $f(2)=2$. Plugging in $m=2$ then tells us that $4+f(n) \mid 4+n$, which implies that $f(n) \leqslant n$ for all $n$.

Setting $m=n$ gives $n^{2}+f(n) \mid n f(n)+n$, so $n f(n)+n \geqslant n^{2}+f(n)$ which we rewrite as $(n-1)(f(n)-n) \geqslant 0$. Therefore $f(n) \geqslant n$ for all $n \geqslant 2$. This is trivially true for $n=1$ also.

It follows that $f(n)=n$ for all $n$. This function obviously satisfies the desired property.

Setting $m=f(n)$ we get $f(n)(f(n)+1) \mid f(n) f(f(n))+n$. This implies that $f(n) \mid n$ for all $n$.

Now let $m$ be any positive integer, and let $p>2 m^{2}$ be a prime number. Note that $p>m f(m)$ also. Plugging in $n=p-m f(m)$ we learn that $m^{2}+f(n)$ divides $p$. Since $m^{2}+f(n)$ cannot equal 1, it must equal $p$. Therefore $p-m^{2}=f(n) \mid n=p-m f(m)$. But $p-m f(m)<p<2\left(p-m^{2}\right)$, so we must have $p-m f(m)=p-m^{2}$, i.e., $f(m)=m$.

Plugging $m=1$ we obtain $1+f(n) \leqslant f(1)+n$, so $f(n) \leqslant n+c$ for the constant $c=$ $f(1)-1$. Assume that $f(n) \neq n$ for some fixed $n$. When $m$ is large enough (e.g. $m \geqslant \max (n, c+1)$ ) we have

$$
m f(m)+n \leqslant m(m+c)+n \leqslant 2 m^{2}<2\left(m^{2}+f(n)\right)
$$

so we must have $m f(m)+n=m^{2}+f(n)$. This implies that

$$
0 \neq f(n)-n=m(f(m)-m)
$$

which is impossible for $m>|f(n)-n|$. It follows that $f$ is the identity function.

Final answer: $f(n)=n$