Find all the functions $f:\mathbb R \to \mathbb R$ which satisfy $$f(x^2+f(y))=y+(f(x))^2$$ for all $x, y$ in $\mathbb R$.
I have the following proof from my math book and want to see if I can get another one.
Proof: Let $x=0$. Then we have $$f(f(y))=y + f^2(0),$$ so $f$ is bijective. Let's suppose $f(x)$ is not equal to $x$. If $f(x)>x$ then we have $y\in \mathbb R$ that $f(x)= x+ (f(y))^2$. From our condition we can say that $f(x)=f(y^2 + f(x))$ and because the function is bijcetive we have $x=y^2 + f(x)$ so $x>f(x)$, which is impossible. Hence $f(x)=x$.
