2001
Problem - 3648
If for any non-negative real numbers $x$ and $y$, function $f(x)$ satisfies the properties that $f(x)\ge 0$, $f(1)\ne 0$, and $f(x+y^2)=f(x)+2f^2(y)$ , compute the value of $f(2+\sqrt{3})$.
Because
\begin{equation}
f(2+\sqrt{3})=f(2+(\sqrt[4]{3})^2)=f(2)+2[f(\sqrt[4]{3})]^2,
\end{equation}
we will need to compute $f(2)$ and $2[f(\sqrt[4]{3})]^2$.
Setting $x=y=0$, then $$f(0+0^2)=f(0)+2[f(0)]^2\implies f(0)=0$$
Setting $x=1, y=0$, then $$f(1+0^2)=f(1)+2[f(0)]^2\implies f(1)=\frac{1}{2}$$
Setting $x=y=1$, the
\begin{equation}
f(1+1^2)=f(1)+2[f(1)]^2 \implies f(2) = 1
\end{equation}
Setting $x=2, y=1$, then $$f(2+1^2)=f(2)+2[f(1)]^2 \implies f(3)= \frac{3}{2}$$
Setting $x=0, y=\sqrt{3}$, then $$f(0+(\sqrt{3})^2=f(0)+2[f(\sqrt{3})]^2\implies f(\sqrt{3})=\frac{\sqrt{3}}{2}$$
Setting $x=0, y=\sqrt[4]{3}$, then
\begin{equation}
f(0+(\sqrt[4]{3}))=f(0) + 2[f(\sqrt[4]{3})]^2\implies 2[f(\sqrt[4]{3})]^2=\frac{\sqrt{3}}{2}
\end{equation}
$$\therefore\quad f(2+\sqrt{3})=\boxed{1+\frac{\sqrt{3}}{2}}$$