Team Selection Test for IMO 2024

Answer: $f(x)=0$ and $f(x)=x$.

$f(x+y{)}^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$.

By putting $(x,y)=(0,0)$ to the original equation we get $f(0)=0$.

By putting $(x,y)=(x,0)$ to the original equation we get $f(x{)}^3=xf(x^2)$.

By putting $(x,y)=(0,x)$ to the original equation we get $f(x{)}^3=f(f(x))x^2$.

Replace this in the original equation to get $$f(x+y{)}^3=(x^2+2xy)f(f(x))+(x^2+3xy+y^2)f(f(y)).$$ Reversing the $x,y$ order, we get $$f(x+y{)}^3=(x^2+3xy+y^2)f(f(x))+(2xy+y^2)f(f(y)).$$ Since the RHS in both expressions are equal we get $$f(f(x))(x+y)y=f(f(y))(x+y)x$$ for all real numbers $x,y$.

Setting $y=1$ and letting $f(f(1))=k$, we have $f(f(x))=kx$ for every $x\in \mathbb{R}\setminus \{-1\}$, setting $y=2$ and $x=-1$ gives that this is true for all real numbers.

Putting this into $f(x{)}^3=f(f(x))x^2$, we get $f(x)=cx$ for a constant $c=\sqrt[3]{k}$, and putting this into $xf(x^2)=x^{2}f(f(x))$ we get $c=0$ or $c=1$.

Therefore, the solutions are $f(x)=0$ and $f(x)=x$ and they satisfy the original equation.