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

We claim that
$$f(q_1 \cdots q_s) = q_1 \cdots q_s \left( \frac{1}{q_1} + \cdots + \frac{1}{q_s} \right) \quad (1)$$
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
$$\begin{aligned} f(q_1 \cdots q_s q_{s+1}) &= f((q_1 \cdots q_s)q_{s+1}) = q_{s+1} f(q_1 \cdots q_s) + q_1 \cdots q_s f(q_{s+1}) \\ &= q_1 \cdots q_{s+1} \left( \frac{1}{q_1} + \cdots + \frac{1}{q_s} \right) + q_1 \cdots q_s \\ &= q_1 \cdots q_{s+1} \left( \frac{1}{q_1} + \cdots + \frac{1}{q_s} + \frac{1}{q_{s+1}} \right). \end{aligned}$$

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
$$\frac{\alpha_1}{p_1} + \cdots + \frac{\alpha_r}{p_r} = 1. \quad (2)$$
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
$$\frac{a}{p_1 \cdots p_{r-1}} + \frac{\alpha_r}{p_r} = 1 \iff a p_r + \alpha_r p_1 \cdots p_{r-1} = p_1 \cdots p_r.$$
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 $3125$.

1. We claim that
$$f(q_1 \cdots q_s) = q_1 \cdots q_s \left( \frac{1}{q_1} + \cdots + \frac{1}{q_s} \right)$$
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 (4) holds for some $s$, then
$$\begin{aligned} f(q_1 \cdots q_s q_{s+1}) &= f((q_1 \cdots q_s)q_{s+1}) = q_{s+1} f(q_1 \cdots q_s) + q_1 \cdots q_s f(q_{s+1}) \\ &= q_1 \cdots q_{s+1} \left( \frac{1}{q_1} + \cdots + \frac{1}{q_s} \right) + q_1 \cdots q_s \\ &= q_1 \cdots q_{s+1} \left( \frac{1}{q_1} + \cdots + \frac{1}{q_s} + \frac{1}{q_{s+1}} \right). \end{aligned}$$

2. It is easily verified that the function given by (4) 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 (4) 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
$$\frac{\alpha_1}{p_1} + \cdots + \frac{\alpha_r}{p_r} = 1.$$
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
$$\frac{a}{p_1 \cdots p_{r-1}} + \frac{\alpha_r}{p_r} = 1 \iff a p_r + \alpha_r p_1 \cdots p_{r-1} = p_1 \cdots p_r.$$
As $p_r$ is coprime to $p_1 \cdots p_{r-1}$, we conclude that $p_r \mid \alpha_r$. As (5) 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 $n = p^\alpha$ for some prime number $p$. We have
$$2^2 = 4 < 3^3 = 27 < 2015 < 5^5 = 3125,$$
so the smallest such $n$ is $3125$.