Brazilian Math Olympiad

Let $f(0) = k$. Plugging $a = \sqrt{2}$ and $b = -\sqrt{2}$ one obtain $f(\sqrt{2} + (-\sqrt{2})) = f(\sqrt{2}(-\sqrt{2})) \iff f(-2) = k$.
Let $\alpha \in \mathbb{R} \setminus \mathbb{Q}$. Since the quadratic equation $x^2-\alpha x-2=0$ has discriminant $\Delta = \alpha^2+8>0$, its roots have sum $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ and product $-2 \in \mathbb{Q}$, at least one of its roots $m$ is irrational; the other root, $n = -2/m$, is also irrational. Thus we can plug $m$ and $n$, obtaining $f(m+n) = f(mn) \iff f(\alpha) = f(-2) = k$.
Now let $q \in \mathbb{Q}$. The quadratic equation $x^2 - qx - \sqrt{2} = 0$ has discriminant $\Delta = q^2 + 4\sqrt{2} > 0$, sum of the roots $q \in \mathbb{Q}$ and product of the roots $-\sqrt{2} \in \mathbb{R} \setminus \mathbb{Q}$, so one of its roots $r$ is irrational; the other root, $s = q-r$, is also irrational. We can plug $r$ and $s$, obtaining $f(r+s) = f(rs) \iff f(q) = f(-\sqrt{2}) = k$.
So all the functions are the constant functions $f(x) = k, k \in \mathbb{R}$.