BMO 2017

We will consider 2 cases, whether the range of the functions is infinite or finite or in other words the function takes infinite or finite values.

Case 1. The function has an infinite range. Let's fix a random natural number $n$ and let $m$ be any natural number. Then using (1) we have $$n+f(m)\mid f(n)+nf(m)=f(n)-n^2+n(f(m)+n)\Rightarrow n+f(m)\mid f(n)-n^2$$ Since $n$ is a fixed natural number, then $f(n)-n^2$ is as well a fixed natural number, and since the above result is true for any $m$ and the function $f$ has an infinite range, we can choose $m$ such that $n+f(m)>|f(n)-n^2|$. This implies that $f(n)=n^2$ for any natural number $n$. We now check that it is a solution. Since $$n+f(m)=n+m^2$$ and $$f(n)+nf(m)=n^2+n{m}^2=n(n+m^2)$$ it is straightforward that $n+f(m)\mid f(n)+nf(m)$.

Case 2. The function has a finite range. Since the function takes finite values, then there exists a natural number $k$ such that $1\le f(n)\le k$ for any natural number $n$. It is clear that there exists at least one natural number $s$ (where $1\le s\le k$) such that $f(n)=s$ for infinitely many natural numbers $n$. Let $m,n$ be any natural numbers such that $f(m)=f(n)=s$. Using (1) we have $$n+s\mid s+ns=s-s^2+s(n+s)\Rightarrow n+s\mid s^2-s.$$ Since this is true for any natural number $n$ such that $f(n)=s$ and since there exist infinitely many natural numbers $n$ such that $f(n)=s$, we can choose the natural number $n$ such that $n+s>s^2-s$, which implies that $s^2=s\Rightarrow s=1$, or in other words $f(n)=1$ for infinitely many natural numbers $n$. Let's fix a random natural number $m$ and let $n$ be any natural number with $f(n)=1$. Then using (1) we have $$n+f(m)\mid 1+nf(m)=1-(f(m){)}^2+f(m)(n+f(m))\Rightarrow n+f(m)\mid (f(m){)}^2-1$$ Since $m$ is a fixed random natural number, then $(f(m){)}^2-1$ is a fixed non-negative integer and since $n$ is any natural number such that $f(n)=1$ and since there exist infinitely many numbers $n$ such that $f(n)=1$, we can choose the natural number $n$ such that $n+f(m)>(f(m){)}^2-1$. This implies $f(m)=1$ for any natural number $m$. We now check that it is a solution. Since $$n+f(m)=n+1$$ and $$f(n)+nf(m)=1+n$$ it is straightforward that $n+f(m)\mid f(n)+nf(m)$.

So, all the functions that satisfy the given condition are $f(n)=n^2$ for any $n\in \mathbb{N}$ or $f(n)=1$ for any $n\in \mathbb{N}$.