IMO 2016 Shortlisted Problems

It is given that $$f(m)+f(n)-mn\mid mf(m)+nf(n).\text{\hspace{1em}(1)}$$ Taking $m=n=1$ in (1), we have $2f(1)-1\mid 2f(1)$. Then $2f(1)-1\mid 2f(1)-(2f(1)-1)=1$ and hence $f(1)=1$. Let $p⩾7$ be a prime. Taking $m=p$ and $n=1$ in (1), we have $f(p)-p+1\mid pf(p)+1$ and hence $$f(p)-p+1\mid pf(p)+1-p(f(p)-p+1)=p^2-p+1$$ If $f(p)-p+1=p^2-p+1$, then $f(p)=p^2$. If $f(p)-p+1\ne p^2-p+1$, as $p^2-p+1$ is an odd positive integer, we have $p^2-p+1⩾3(f(p)-p+1)$, that is, $$\begin{array}{c}f(p)⩽\frac{1}{3}\left(p^2+2p-2\right)\end{array}\text{\hspace{1em}(2)}$$ Taking $m=n=p$ in (1), we have $2f(p)-p^2\mid 2pf(p)$. This implies $$2f(p)-p^2\mid 2pf(p)-p\left(2f(p)-p^2\right)=p^3.$$ By (2) and $f(p)⩾1$, we get $$-p^2<2f(p)-p^2⩽\frac{2}{3}\left(p^2+2p-2\right)-p^2<-p$$ since $p⩾7$. This contradicts the fact that $2f(p)-p^2$ is a factor of $p^3$. Thus we have proved that $f(p)=p^2$ for all primes $p⩾7$. Let $n$ be a fixed positive integer. Choose a sufficiently large prime $p$. Consider $m=p$ in (1). We obtain $$f(p)+f(n)-pn\mid pf(p)+nf(n)-n(f(p)+f(n)-pn)=pf(p)-nf(p)+p{n}^2.$$ As $f(p)=p^2$, this implies $p^2-pn+f(n)\mid p\left(p^2-pn+n^2\right)$. As $p$ is sufficiently large and $n$ is fixed, $p$ cannot divide $f(n)$, and so $\left(p,p^2-pn+f(n)\right)=1$. It follows that $p^2-pn+f(n)\mid p^2-pn+n^2$ and hence $$p^2-pn+f(n)\mid \left(p^2-pn+n^2\right)-\left(p^2-pn+f(n)\right)=n^2-f(n).$$ Note that $n^2-f(n)$ is fixed while $p^2-pn+f(n)$ is chosen to be sufficiently large. Therefore, we must have $n^2-f(n)=0$ so that $f(n)=n^2$ for any positive integer $n$. Finally, we check that when $f(n)=n^2$ for any positive integer $n$, then $$f(m)+f(n)-mn=m^2+n^2-mn$$ and $$mf(m)+nf(n)=m^3+n^3=(m+n)\left(m^2+n^2-mn\right).$$ The latter expression is divisible by the former for any positive integers $m,n$. This shows $f(n)=n^2$ is the only solution.