19-th Macedonian Mathematical Olympiad

Let $f(0) = a$, then $f(a) = f(f(0)) = 2$, $f(2) = f(f(a)) = a + 2$. Continuing this procedure we get that $f(2k) = a + 2k$ and $f(a + 2k) = 2k + 2$. We get $2k + 2 = f(a + 2k) < f(a) + f(2k) = 2 + a + 2k$, from where we get that $a > 0$. If we put $x = y = a$ we get $a + 2a = f(2a) < f(a) + f(a) = 4$ so $3a < 4$. i.e. $a = 1$. Hence using $f(2k) = a + 2k$ and $f(a + 2k) = 2k + 2$ we get $f(x) = x + 1$ for all natural numbers $x$.

For $x = y = \frac{1}{2}$ in the inequality we get $2 = f(1) = f(\frac{1}{2} + \frac{1}{2}) < 2f(\frac{1}{2})$, so $f(\frac{1}{2}) > 1$. On the other hand $1 + f(\frac{1}{2}) = f(f(\frac{1}{2})) = 2$, so $f(\frac{1}{2}) = 1$, which is a contradiction. It follows that there exists no function satisfying the required conditions.