51st Ukrainian National Mathematical Olympiad, 3rd Round

Take $x=y$, we get: $$f(2x)=f(0)f(2x)+f(0)f(-2x).(1)$$ Take in (1) $x=0$ and we get $f(0)=2f^2(0)$. So $f(0)=0$ or $f(0)=\frac{1}{2}$. If $f(0)=0$, (1) implies that $f(2x)=0$, hence $f=0$.

If $f(0)=\frac{1}{2}$, (1) implies that $f(2x)=\frac{1}{2}f(2x)+\frac{1}{2}f(-2x)$ and $f(2x)=f(-2x)$. Taking $x=0$ and using the fact that our function is even, we arrive at $$\frac{1}{2}=f(0)=(f(y){)}^2+(f(-y){)}^2=2(f(y){)}^2.$$ This gives us that $f=\frac{1}{2}$.

It is easy to check that both functions satisfy all the requirements.