Romanian Mathematical Olympiad

For $y=0$ we have $f(f(x))=x+(f(0){)}^n$, for any $x\in \mathbb{R}$, implying that $f$ is bijective. Let $a$ such that $f(a)=0$. For $y=a$ we deduce $f(f(x))=x$, for all real $x$, so $f(0)=0$.

For $x=0$ we get $f(yf(y))=f(y{)}^n$, for real $y$. Substituting $y$ with $f(y)$, we get $f(y{)}^n=y^n$.

i) For odd $n$ we get $f(y)=y$, which gives a contradiction in the given relation.

ii) For even $n$ we obtain $f(y)\in \{-y,y\}$, for any $y\in \mathbb{R}$. Thus $y^n\in \{-y^2,y^2\}$ for all real $y$, which simply imply $n=2$. Thus $f(f(x)+yf(y))=x+y^2$, for $x,y\in \mathbb{R}$.

It is easy to check that the identity function and minus the identity function, verify the relation. We shall show that they are the only ones. If not, if there are $a,b\ne 0$ such that $f(a)=a$ and $f(b)=-b$, than for $x=a,y=b$ we should have $f(a-b^2)=a+b^2\in \{a-b^2,-a+b^2\}$, that is $a=0$ or $b=0$, a contradiction.