Japan Mathematical Olympiad

We prove that the function $f(n)=n$ is the unique function satisfying the condition in the problem. It is easy to see that this $f$ satisfies the condition in the problem.

Suppose that $f$ is the function that satisfies the condition in the problem. First, for any positive integer $k$, we prove that $f(k)$ is a multiple of $k$. Let $r$ be the remainder when dividing $f(k)$ by $k$. Substituting $m=k$ and $n=k-r+1$ into the equation in the problem, we obtain $$\mathrm{lcm}(k,f(k+f(k-r+1)))=\mathrm{lcm}(f(k),f(k)+k-r+1).$$ Therefore, $\mathrm{lcm}(f(k),f(k)+k-r+1)$ is a multiple of $k$. Here, $f(k)+k-r+1$ leaves a remainder of $1$ when divided by $k$, and hence, $f(k)+k-r+1$ is coprime to $k$. Therefore, we conclude that $f(k)$ is a multiple of $k$.

Let $m$ be a positive integer. We prove that $f(m)=m$. Substituting $n=f(m)$ into the equation in the problem, we get $$\mathrm{lcm}(m,f(m+f(f(m))))=\mathrm{lcm}(f(m),2f(m))=2f(m).$$ Therefore, $2f(m)$ is a multiple of $f(m+f(f(m)))$. Since $f(m+f(f(m)))$ is a multiple of $m+f(f(m))$, $2f(m)$ is a multiple of $m+f(f(m))$. Note that we have $$2f(m)\le 2f(f(m))<2(m+f(f(m))).$$ Here, the first inequality follows from the fact that $f(f(m))$ is a multiple of $f(m)$. Therefore, we conclude that $2f(m)=m+f(f(m))$. Since $$f(f(m))=2f(m)-m<2f(m)$$ and $f(f(m))$ is a multiple of $f(m)$, we conclude that $f(f(m))=f(m)$. Hence, we have $2f(m)=m+f(m)$, and hence $f(m)=m$. We complete the proof.