Team Selection Test for IMO 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 $$\begin{array}{rl}f\left(\frac{m}{n}\right)& =f\left(\frac{m/(n-m)}{m/(n-m)+1}\right)=\frac{1}{m/(n-m)+1}f\left(\frac{m}{n-m}\right)\\ & =\frac{n-m}{n}\cdot \frac{m^2}{n-m}f(1)=\frac{m^2}{n}f(1)\end{array}$$ 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.