2001
Problem - 2443
For any non-negative real numbers $x$ and $y$, the function $f(x+y^2)=f(x) + 2[f(y)]^2$ always holds, $f(x)\ge 0$, $f(1)\ne 0$. Find the value of $f(2+\sqrt{3})$.
\begin{align}
x=0, y=0 &\implies f(0) = 0\\
x=0, y=1 &\implies f(1) = \frac{1}{2}\\
x=1, y=1 &\implies f(2) = 1\\
x=2, y=1 &\implies f(\sqrt{3})=\frac{\sqrt{3}}{2}\\
x=0, y=\sqrt[4]{3} &\implies [f(\sqrt[4]{4})]^2=\frac{\sqrt{3}}{4}
\end{align}
Finally: $$f(2+\sqrt{3})=f(2)+2[f(\sqrt[4]{3})]^2 = \boxed{1 + \frac{\sqrt{3}}{2}}$$