go_placidly wrote: Denote by $\mathbb{Q}^+$ the set of positive rational numbers. A function $f : \mathbb{Q}^+ \to \mathbb{Q}$ satisfies • $f(p) = 1$ for all primes $p$, and • $f(ab) = af(b) + bf(a)$ for all $ a,b \in \mathbb{Q}^+ $. For which positive integers $n$ does the equation $nf(c) = c$ have at least one solution $c$ in $\mathbb{Q}^+$? Easy to get $f(\prod p_i^{n_i})=\prod p_i^{n_i}\sum\frac{n_i}{p_i}$ where $p_i$ are prime positive integers and $n_i$ nonzero (positive or negative) integers (with $f(1)=0$) And so equation $nf(c)=c$ is $\frac 1n=\sum\frac{n_i}{p_i}$ which has solutions if and only if $n=1$ or $n$ is squarefree.

Here's my solution, which is essentially same as pco's but in more detail. Define a new function $g(x) = \frac{f(x)}{x}$. Then, the first equation becomes $f(p) = \frac{1}{p}$ and the second one becomes $ab g(ab) = abf(b) + abf(a) \implies g(ab) = g(a) + g(b)$. So, we get, by induction, that $g(n) = \frac{\alpha_1}{p_1} + \frac{\alpha_2}{p_2} + ... + \frac{\alpha_k}{p_k}$ where $n = p_1^{\alpha_1} p_2^{\alpha_2} ....p_k^{\alpha_k}$ Since $g(p) = g(q \frac{p}{q}) = g(q) + g(\frac{p}{q})$, it means $g(\frac{p}{q}) = g(p) - g(q)$, so we can get, by induction that $g(\frac{p}{q}) = \frac{\alpha_1}{p_1} + \frac{\alpha_2}{p_2} + ... + \frac{\alpha_k}{p_k} - \frac{\beta_1}{q_1} - \frac{\beta_2}{q_2} - ... - \frac{\beta_k}{q_k}$ The problem is asking us when $g(c) = \frac{1}{n}$ has solutions. Obviously, is $n$ is not squarefree, then this is not possible since the denominators cannot have a prime with exponent $\ge 2$, so $n$ must be squarefree If $n$ is squarefree, it is indeed possible to construct a number $c$ by choosing the $\alpha_i$'s and $\beta_i$'s such that $f(c) = \frac{1}{n}$ So, the answer is only when $n$ is squarefree