Functional equation of divisibility: $m^2+f(n)\mid mf(m)+n$

Question

I am struggling with the following issue:

Find all functions $f:\mathbb{N}_{+}\to\mathbb{N}_{+} $, such that for all positive integers $m$ and $n$, there is the divisibility $$m^2+f(n)\mid mf(m)+n\text.$$ $\mathbb{N}_{+}$ stands for the set of positive integers.

I've tried various substitutions but I dont know how to solve functional equations of this form therefore I couldn't manage to find any $f$. I think this is a interesting problem so I'd like to know the answer.

Answer 1

Thanks @lulu for showing that $f(1)=1$. Now I can finish the problem by proving that the identity is the only possibility for $f$.

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Suppose that there exists $m$ such that $f(m)<m$. Then taking $n=1$ we have $$m^2+1=m^2+f(1)\le mf(m)+1<m^2+1\ ,$$ a contradiction. Now suppose there exists $n$ such that $f(n)>n$. Taking $m=1$, we have $$1+n<1+f(n)\le f(1)+n=1+n\ ,$$ another contradiction. Thus $f(n)=n$ for all $n$.

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Addendum. Proof that $f(1)=1$. This proof was provided in a comment by Lulu, I'm copying it verbatim here in case the comment disappears in the future.

Setting $n=1$ we see that $m^2+f(1)$ divides $mf(m)+1$ for all $m$. But if there were a prime $p\mid f(1)$ then taking $m=p$ gives a contradiction, hence there is no such $p$ which implies that $f(1)=1$.