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 $n$.  (Malaysia)  Answer. $f(n)=n$.
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.