6. (IND 2) ${ }^{\mathrm{IMO}(F 2}$ Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f\left(x^{2}+f(y)\right)=y+f(x)^{2} \quad$ for all $x, y$ in $\mathbb{R}$.
6. It is easy to see that $f$ is injective and surjective. From $f\left(x^{2}+f(y)\right)=$ $f\left((-x)^{2}+f(y)\right)$ it follows that $f(x)^{2}=(f(-x))^{2}$, which implies $f(-x)=$ $-f(x)$ because $f$ is injective. Furthermore, there exists $z \in \mathbb{R}$ such that $f(z)=0$. From $f(-z)=-f(z)=0$ we deduce that $z=0$. Now we have $f\left(x^{2}\right)=f\left(x^{2}+f(0)\right)=0+(f(x))^{2}=f(x)^{2}$, and consequently $f(x)=f(\sqrt{x})^{2}>0$ for all $x>0$. It also follows that $f(x)0$ and $y=-f(x)$ in the given functional equation we obtain
$$
f(x-f(x))=f\left(\sqrt{x}^{2}+f(-x)\right)=-x+f(\sqrt{x})^{2}=-(x-f(x)) .
$$

But since $f$ preserves sign, this implies that $f(x)=x$ for $x>0$. Moreover, since $f(-x)=-f(x)$, it follows that $f(x)=x$ for all $x$. It is easily verified that this is indeed a solution.