74th Romanian Mathematical Olympiad

For $y=0$, the given relation reduces to $f(x)=f(f(x))$, which means that the given relation becomes: $$f(x+y^{2n})=f(x)+y^{2n-1}f(y),\forall x,y\in \mathbb{R}.(1)$$ If we consider $x=0$ and $y=1$ in (1), then $f(0)=0$. Since the equation $f(x)=0$ has a unique solution, we have the implication: $$f(x)=0\Rightarrow x=0.(2)$$ Putting $x=0$ in (1), we obtain $f(y^{2n})=y^{2n-1}f(y)$, for all $y\in \mathbb{R}$, which means that relation (1) rewrites as: $$f(x+y^{2n})=f(x)+f(y^{2n}),\forall x,y\in \mathbb{R}.(3)$$ Moreover, $y^{2n-1}f(y)=f(y^{2n})=f((-y{)}^{2n})=-y^{2n-1}f(-y)$, for all $y\in \mathbb{R}$, so $f$ is an odd function. Considering $t=\sqrt[2n]{y}\ge 0$, relation (3) becomes: $$f(x+t)=f(x)+f(t),\forall x\in \mathbb{R},t\ge 0.(4)$$

Using the fact that $f$ is odd, for $t<0$, we obtain: $$f(x+t)=-f(-x-t)=-(f(-x)+f(-t))=f(x)+f(t),$$ which, together with relation (4), implies: $$f(x+t)=f(x)+f(t),\forall x,t\in \mathbb{R}.(5)$$ If $f(x_1)=f(x_2)$, according to relation (5), we have $f(x_1-x_2)=0$, which, according to relation (2), implies $x_1=x_2$, meaning $f$ is one-to-one. But since $f(f(x))=f(x)$, we have $f(x)=x$, which is a solution to the given equation.