Find all polynomials $p$ such that $p(x^2)=p(x)p(x+1)$

Question

Find all polynomials $p$ with complex coefficients such that
$$ p(x^2)=p(x)p(x+1).$$

The goal is to find a general formula for polynomials that satisfy the above equation. I believe the answer is $\left(x^2-x\right)^k$ with $k\in\mathbb{N}$, but I have no idea how to get this solution.

Answer 1

Suppose $r$ is a root of the polynomial. Then $p(r^2) = p(r) p(r+1) = 0$ and $p((r-1)^2) = p(r-1) p(r) = 0$, so $r^2$ and $(r-1)^2$ are roots as well.
In order to avoid generating an infinite sequence of distinct roots, each the square of the previous one, we need $|r| = 0$ or $|r| = 1$. Similarly we need $|r-1| = 0$ or $1$, and $|r^2-1| = 0$ or $1$. It's not hard to show that the only possible roots are $0$ and $1$.

EDIT: So let $p(x) = a x^m (x-1)^n$. By considering the leading coefficient of $p(x^2) - p(x)p(x+1)$, we find that $a = 1$. The zero of $p(x^2)$ at $0$ has order $2m$, while the zero of $p(x) p(x+1)$ there has order $m+n$, so $m = n$.
And finally, we find that $p(x) = x^m (x-1)^m$ does satisfy the equation.

Answer 2

If $p(x) = 0$ then $p(x^2) = 0$. Hence if there exists a root $r$ that is not $0$ or $1$ then we get roots $\{r,r^2,r^4,\dots \}$. But this would contradict the fundamental theorem of algebra ($p$ must, of course, have finite degree). Also note as $p(0) = p(0)p(1)$, $p(0) = 0$ or $p(1) = 1$. You can show a similar claim for $x=1$. Do you think you can conclude from here?