How many polynomial functions exist such that $f(x^2) = (f(x))^2 = f(f(x))$ How many polynomial functions $f$ of degree $\geq1$ satisfy
$f(x^2) = (f(x))^2 = f(f(x))$ for all real $x$?
Hint: if $f(x)$ has degree $n \geq 1$ then $f(f(x))$ has degree $n^2$, but $f(x^2)$ and
$f(x)^2$ have degree $2n$. For what $n \in \mathbb{N}$ is $2n=n^2$?

Once you find out just solve the equations for the coefficients.
May $x_1$ be a complex root of the polynomial $f$.

By the property $f(x^2) = (f(x))^2$, then, $f(x_1^2) = f(x_1)^2 = 0$.

Side note: By the last relation, we get $f(0) = 0$, but it's not needed.

Therefore, $\forall k \in \mathbb{R} ^+ , x_1^k$ is a root of $f$. So $f$ has infinitely many roots. So $f$ is equal to zero on its whole interval, because it is a polynomial with infinitely many roots.

It satisfies the last relation as well.

Edit: Side note: If you need a proof of the property for infinitely many roots polynomial, there it is: Polynomial with infinite roots

Edit: 
my solution doesn't apply whenever $0$ is the only root. I can apply it for a polynomial whose only root is 0: $x^m$. Using the first equality, we get $x^{2m} = x^{2m}$ which is useless. The second equation gives $x^{2m} = x^{m^2}$ which leads to $m=0$ and $m=2$.