Find all non-constant polynomials with coefficients in $\mathbb{R}$ such that, for all $z \in \mathbb{C}$,

$$
\mathrm{P}\left(z^{2}\right)=\mathrm{P}(z) \mathrm{P}(z-1)
$$
Suppose by contradiction that P is not constant. If $z_{0}$ is a root, then $z_{0}^{2}, z_{0}^{4}, \ldots$ are also roots. If we have roots $\left|z_{0}\right|<1$ or $\left|z_{0}\right|>1$, then P would have an infinite number of roots. Therefore, all roots must have a modulus of 1.

The equation can also be written as $P\left((x+1)^{2}\right)=P(x+1) P(x)$. Therefore, if $z_{0}$ is a root, then $(z+1)^{2}$ is also a root. Since all roots have a modulus of 1, we have $\left|z_{0}+1\right|=1$. To find the complex numbers that satisfy $|z|=|z+1|=1$, we set $z=a+ib$. We find $a^{2}+b^{2}=1=a^{2}+b^{2}+2a+1$. Then $a=\frac{-1}{2}$ and $b= \pm \frac{\sqrt{3}}{2}$. Therefore, $P(x)=(x-\epsilon)^{a}\left(x-\frac{1}{\epsilon}\right)^{b}$ for $\epsilon=\frac{-1}{2}+i \frac{\sqrt{3}}{2}$. Since $P \in \mathbb{R}[x]$, $a=b$, so $P=\left(x^{2}+x+1\right)^{a}$. Conversely, any polynomial of this form works.