Question 117
Let $f: \mathbb{N} \to \mathbb{N}_0$ be a non-zero function from the set of positive integers to the set of non-negative integers such that for all $a, b \in \mathbb{N}$ it holds that
$$
2f(ab) = (b+1)f(a) + (a+1)f(b).
$$
Prove that for any prime $p$ there are a prime $q$, positive integers $x_1, \dots, x_n$ and a non-negative integer $m$ so that
$$
\frac{f(q^p)}{f(q)} = (px_1 + 1) \cdots (px_n + 1) \cdot p^m,
$$
where all of the numbers $px_1 + 1, \dots, px_n + 1$ are primes.
$$
2f(ab) = (b+1)f(a) + (a+1)f(b).
$$
Prove that for any prime $p$ there are a prime $q$, positive integers $x_1, \dots, x_n$ and a non-negative integer $m$ so that
$$
\frac{f(q^p)}{f(q)} = (px_1 + 1) \cdots (px_n + 1) \cdot p^m,
$$
where all of the numbers $px_1 + 1, \dots, px_n + 1$ are primes.