BxMO Team Selection Test

Substituting $m=n$ gives $f(n)\mid f(n)-n$, so for all natural numbers $n$ we have $f(n)\mid n$. Applying this to the original condition, it follows that $f(n)\mid f(m)$ if and only if $n\mid m$.

We show that $f(n)=n$ by induction on the number of prime factors of $n$. The base of the induction is the case $n=1$. In this case, we have $f(1)\mid 1$, therefore $f(1)=1$.

Suppose that $f(k)=k$ for all natural numbers $k$ with fewer prime factors than $n$, and suppose for a contradiction that $f(n)\mid n$ is a strict divisor of $n$. Then there exists a prime number $p$ such that $f(n)\mid \frac{n}{p}=f(\frac{n}{p})$, by the induction hypothesis. But $n$ does not divide $\frac{n}{p}$, which contracts our assumption that $f(n)\ne n$. Therefore $f(n)=n$, and this concludes the induction.

Note that $f(n)=n$ is indeed a solution, since $n\mid m$ holds if and only if $n\mid m-n$. $\square$