TITLE: How to solve the functional equation $f\left(x^2+f(y)\right)=y+f(x)^2$
QUESTION [10 upvotes]: How to solve the following functional equation:
Find all $f:\mathbb{R}\to\mathbb{R}
$ such that:
$$
f\left(x^2+f(y)\right)=y+f(x)^2
$$
Holds for every $x,y\in\mathbb{R}$.
A friend gave it to me, probably its an olympiad question.
I started with the equation $f(f(y))=y+f(0)^2$ which seems to be quite helpful, but I couldn't do it effectively. How to solve it properly?

REPLY [5 votes]: Here is an elementary way to show that $ f=1_\mathbb R$
Claim 1: $f$ is an odd function.
If we substitute $x=y=0$ in the given equation, then we get $$f(f(0))=(f(0))^{2}.$$
We then substitute $x=0$ in the given equation and find out that $\forall y \in \Bbb R$, $$f(f(y))=y+(f(0))^{2}.$$ Now, we observe that, for all $x,y \in \Bbb R$, $$y+(f(x))^{2}=f(x^{2}+f(y))=f((-x)^{2}+f(y))=y+(f(-x))^{2}.$$
Hence $\forall x \in \Bbb R,$ $$f(-x)=f(x) \space \space or \space \space f(-x)=-f(x).$$
 But, if for some $x \in \Bbb R$, $f(-x)=f(x)$, then we would have, $$x+(f(y))^{2}=f(y^{2}+f(x))=f(y^{2}+f(-x))=-x+(f(y))^{2}$$ for any $y \in \Bbb R$ implying $x=0$. So, for any $x \neq 0$, $f(-x)=-f(x).$.Hence $f$ is an odd function.
Claim 2: $f(x)>0$ if $x>0$ and $f$ is increasing.
Note that $f(x)\neq 0$ if $x\neq 0$. Indeed, if $f(x)=0$, then 
$$0=f(0)=f(f(x))=x.$$ Furthermore, if $x>0$, then $f(x)>0$. Indeed, $f(x)=f(\sqrt{x}^2)=f(\sqrt{x})^2>0$.
Then in fact, $f$ is increasing. Since $f$ is odd, it suffices to show it is increasing on $(0,\infty)$. Well, if $x>y>0$, $$f(x)-f(y)=f(\sqrt{x}^2)+f(-y)=f(\sqrt x)^2+f(-y)=f(x+f(f(-y)))=f(x-y)>0.$$
But then, if $f(x) >x$, $x>f(f(x))>x$. Likewise, if $f(x)<x$, $x=f(f(x))<f(x)<x$.
As these are impossible, we must have $f(x)=x$ for all $x$