46th Austrian Mathematical Olympiad National Competition (Final Round, part 2)

We claim that $$f(q_1\cdots q_s)=q_1\cdots q_s\left(\frac{1}{q_1}+\cdots +\frac{1}{q_s}\right)(1)$$ holds for (not necessarily distinct) prime numbers $q_1,\dots ,q_s$. We prove the claim by induction on $s$. For $s=0$, the claim reduces to $f(1)=0$, which is true by assumption. If (1) holds for some $s$, then $$\begin{array}{rl}f(q_1\cdots q_s{q}_{s+1})& =f((q_1\cdots q_s)q_{s+1})=q_{s+1}f(q_1\cdots q_s)+q_1\cdots q_{s}f(q_{s+1})\\ & =q_1\cdots q_{s+1}\left(\frac{1}{q_1}+\cdots +\frac{1}{q_s}\right)+q_1\cdots q_s\\ & =q_1\cdots q_{s+1}\left(\frac{1}{q_1}+\cdots +\frac{1}{q_s}+\frac{1}{q_{s+1}}\right).\end{array}$$

2. It is easily verified that the function given by (1) fulfills the given functional equation.

3. Let $p_1,\dots ,p_r$ be distinct primes and ${\alpha }_1,\dots ,{\alpha }_r$ be positive integers. Then collecting equal primes in (1) leads to $$f(p_1^{{\alpha }_1}\cdots p_r^{{\alpha }_r})=p_1^{{\alpha }_1}\cdots p_r^{{\alpha }_r}\sum _{j=1}^r\frac{{\alpha }_j}{p_j}.$$

4. We now determine all $n\ge 2015$ with $f(n)=n$. We write $n=p_1^{{\alpha }_1}\cdots p_r^{{\alpha }_r}$. Then $$\frac{{\alpha }_1}{p_1}+\cdots +\frac{{\alpha }_r}{p_r}=1.(2)$$ We write $$\frac{{\alpha }_1}{p_1}+\cdots +\frac{{\alpha }_{r-1}}{p_{r-1}}=\frac{a}{p_1\cdots p_{r-1}}$$ for some non-negative integer $a$. Then $$\frac{a}{p_1\cdots p_{r-1}}+\frac{{\alpha }_r}{p_r}=1⟺a{p}_r+{\alpha }_r{p}_1\cdots p_{r-1}=p_1\cdots p_r.$$ As $p_r$ is coprime to $p_1\cdots p_{r-1}$, we conclude that $p_r\mid {\alpha }_r$. As (2) implies ${\alpha }_r\le p_r$, we conclude that $r=1$ and ${\alpha }_r=p_r$. Thus $f(n)=n$ holds if and only if $n=p^r$ for some prime number $p$. We have $$2^2=4<3^3=27<2015<5^5=3125,$$ so the smallest such $n$ is $3125$.