Team Selection Test for IMO 2009

Substituting $x=2$ in the second equation gives $3f(1)=2f(2)$. In particular, $f(1)$ is even. It follows by induction that, for $n\in {\mathbb{Z}}^{+}$, $f(n)=(n+1)f(1)/2$.

Now we show by induction on $p+q$ that for all $p,q\in {\mathbb{Z}}^{+}$ and $(p,q)=1$, $f(p/q)=(p+q)f(1)/2$. If $p>q$, $p/q\cdot f(p/q)=(p+q)/q\cdot f((p-q)/q)=(p+q)/q\cdot (p-q+q)f(1)/2$; and if $p<q$, then $f(p/q)=f(q/p)$ and we are in the first case.

To summarize, $f$ satisfies the conditions of the problem if and only if $m$ is a positive integer and $f(p/q)=(p+q)m$ for all $p,q\in {\mathbb{Z}}^{+}$ with $(p,q)=1$.