, 2011

Let $n$ be a positive integer and $x = 1/n$. Then $f(\frac{1}{n}) = \frac{n+1}{n} f(\frac{1}{n+1})$. It follows by induction that $f(\frac{1}{n}) = \frac{f(1)}{n}$ for all positive integers $n$.

We claim that $f(\frac{m}{n}) = \frac{m^2}{n} f(1)$ for all relatively prime positive integers $m$ and $n$. This time we use induction on $\max(m, n)$. If $m < n$, then
$$\begin{aligned} f\left(\frac{m}{n}\right) &= f\left(\frac{m/(n-m)}{m/(n-m)+1}\right) = \frac{1}{m/(n-m)+1} f\left(\frac{m}{n-m}\right) \\ &= \frac{n-m}{n} \cdot \frac{m^2}{n-m} f(1) = \frac{m^2}{n} f(1) \end{aligned}$$
where we used the induction hypothesis. On the other hand, if $m > n$, then first
$$f\left(\frac{m}{n}\right) = \frac{m^3}{n^3} f\left(\frac{n}{m}\right), \text{ and then } \frac{m^3}{n^3} f\left(\frac{n}{m}\right) = \frac{m^3}{n^3} \cdot \frac{n^2}{m} f(1) = \frac{m^2}{n} f(1).$$

Conversely, it can be easily verified that the function defined by $f(\frac{m}{n}) = \frac{m^2}{n}$ for all relatively prime positive integers $m$ and $n$, where $r$ a positive rational number, satisfies the conditions of the question.