Bulgarian Mathematical Olympiad

Solution:

a) It follows from $f(x+y)-f(x-y)>0$ that $f$ is an increasing function. Therefore the function $f(x)$ has a limit $l \geq 0$ when $x \rightarrow 0$, $x>0$ (prove!). Thus letting $x, y \rightarrow 0$, $x>y>0$, we get $l-l=4 \sqrt{l^{2}}$, i.e. $l=0$. Fixing $x$ and letting $y \rightarrow 0$, $y>0$ we conclude that $f(x+y)-f(x-y) \rightarrow 0$. Since the function $f$ is increasing we conclude that it is continuous at $x$. Finally, letting $y \rightarrow x$, $y<x$, we get $f(2 x)=4 f(x)$.

b) Setting $x=n y>0$, where $n \geq 2$ is an integer, we obtain from the given identity that
$$f((n+1) y)=f((n-1) y)+4 \sqrt{f(n y) f(y)}$$
Using $f(2 y)=4 f(y)$, it follows by induction that $f(n y)=n^{2} f(y)$. Set $f(1)=c>0$. Then $f(n)=n^{2} c$. Now, for any positive integers $p$ and $q$ we have $c p^{2}=f(q \cdot p / q)=q^{2} f(p / q)$, i.e. $f(p / q)=c(p / q)^{2}$. Since $f$ is a continuous function, we conclude that $f(x)=c x^{2}$ for any $x>0$. Conversely, any function of the form $f(x)=c x^{2}$ satisfies the condition.

Remark. It is possible to show that any function satisfying the condition is differentiable. Thus, letting $y \rightarrow 0$, $y>0$, in the identity
$$\frac{f(x+y)-f(x-y)}{2 y}=2 \frac{\sqrt{f(y)}}{y} \sqrt{f(x)}$$
we get $f'(x)=2 c \sqrt{f(x)}$. Therefore $(\sqrt{f(x)})'=c$, i.e. $f(x)=c^{2} x^{2}$.