Find all polynomials with real coefficients satisfying:

$$
P(x) P(x-1)=P\left(x^{2}\right) \quad \forall x \in \mathbb{C}
$$
If $P$ is constant, then $P(x)=0$ or $P(x)=1$. We now assume that $P$ is non-constant, so $P$ has a complex root denoted by $a$. By substituting $x$ with $a$ in the polynomial equation, we find:

$$
P\left(a^{2}\right)=P(a) P(a-1)=0
$$

thus $a^{2}$ is also a root of $P$. It follows that $a^{2^{n}}$ is a root of $P$ for any natural number $n$. Since $P$ has a finite number of roots, it follows that $|a|=1$. However, by substituting $x$ with $a+1$, we see that $(a+1)^{2}$ is also a root of $P$, which means that $(a+1)^{2^{n}}$ is also a root of $P$ for any natural number $n$, so $|a|=\left|(a+1)^{2}\right|=1$ or $|a|=|a+1|$. We quickly find that $a=j$ or $a=j^{2}$, so the only roots of $P$ are $j$ and $j^{2}$, hence $P$ is of the form $(x-j)^{n}\left(x-j^{2}\right)^{n}=\left(x^{2}+x+1\right)^{n}$. Conversely, polynomials of this form are all solutions to the polynomial equation.

## Roots of Unity

What are the complex numbers satisfying $z^{n}=1$?

First, the modulus of such a number is 1. Let's now set $z=e^{i \theta}$ and assume that $z^{n}=1$. In this case, $n \theta=0 \bmod 2 \pi$ so there exists an integer $k$ such that $n \theta=2 k \pi$ or $\theta=\frac{2 k \pi}{n}$. Since $\frac{2 n \pi}{n}=0 \bmod 2 \pi$, we see that the $n$ numbers $z$ satisfying $z^{n}=1$ are the $n$ numbers of the form $e^{i \frac{2 k \pi}{n}}$ with $0 \leq k \leq n-1$ or the first $n$ powers of the number $e^{\frac{2 \pi}{n}}$. These numbers are called the $n$-th roots of unity. An amusing observation is that $e^{i \frac{2(k+1) \pi}{n}}$ is obtained by multiplying $e^{i \frac{2 k \pi}{n}}$ by $e^{i \frac{2 \pi}{n}}$, so the point with affix $e^{i \frac{2(k+1) \pi}{n}}$ is the image of $e^{i \frac{2 k \pi}{n}}$ under the rotation centered at the origin of the coordinate system and with angle $\frac{2 \pi}{n}$. Therefore, the $n$ $n$-th roots of unity form a regular $n$-gon in the complex plane.