(IMO 1992)(M) Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f\left(x^{2}+f(y)\right)=y+f(x)^{2}$
We have the bijectivity of $f$. We set $x=0$ and get $f(f(y))=y+f(0)^{2}$, then we set $x=f(0)$ and get $f\left(f(0)^{2}+f(y)\right)=f^{3}(y)=y+f(f(0))^{2}$, and with $x=y=0$ we have $f(f(0))=f(0)^{2}$, so $f^{3}(y)=y+f(0)^{4}$. With $y=0$ we have $f\left(x^{2}+f(0)\right)=f(x)^{2}$ and thus $f(x)=0$ implies $f(-x)=0$. But by bijectivity, we must then have $f(0)=0$. From there, we have $f^{3}=f^{2}$ and by injectivity $f=I d$.