Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$ Let $\mathbb R$ denote the set of all real numbers. Find all function $f: R\to \ R$ such that   $$f\left(x^2+f(y)\right)=y+(f(x))^2$$   It is the problem. I tried to it by putting many at the place of $x$ and $y$ but I can't proceed. Please somebody help me. 

Here is a short approach without Cauchy equation. Starting from $f(0)=0$ by @Leo163, we obtain $f(f(x))=x$ by plugging $y=0$. This shows $f$ is a bijection. Now plug $y=f^{-1}(\zeta)$, we see
$$f(x^2+\zeta)=f^{-1}(\zeta)+(f(x))^2\geq f^{-1}(\zeta)=f(\zeta)$$
Hence $f$ is monotone increasing. Now suppose for some $x_0$ we have $f(x_0)> x_0$, then $x_0=f(f(x_0))\geq f(x_0)> x_0$, a contradiction. Thus we have $f(x)\leq x$. Similarly we have $f(x)\geq x$, so $f(x)=x$.
