For which integers $n \geqslant 1$ is the polynomial $1+x^{2}+x^{4}+\cdots+x^{2 n-2}$ divisible by the polynomial $1+x+x^{2}+\cdots+x^{n-1}$?

D'Alembert-Gauss Theorem We admit the following theorem (also known as the fundamental theorem of algebra):

Theorem 3.48. Every non-constant polynomial in $\mathbb{C}[X]$ has at least one root. We say that $\mathbb{C}$ is algebraically closed.

By induction on the degree, we deduce:

Corollary 3.49. Let $\mathrm{P} \in \mathbb{C}[X]$. Then $\mathrm{P}$ can be written in the form:

$$
P(x)=c\left(x-\alpha_{1}\right)^{m_{1}} \cdots\left(x-\alpha_{k}\right)^{m_{k}}
$$

where $c, \alpha_{1}, \ldots, \alpha_{k}$ are complex numbers and $m_{1}, \ldots, m_{k}$ are strictly positive integers.

We finally define complex conjugation, which will be useful when we want to determine the irreducible polynomials of $\mathbb{R}[X]$.

Definition 3.50. Let $z=a+b i \in \mathbb{C}$. We define its conjugate $\bar{z}$ by $\bar{z}=a-b i$.

Proposition 3.51. For all $w, z \in \mathbb{C}$, we have $\overline{w z}=\bar{w} \cdot \bar{z}$.

Proof. Exercise.

## - Arithmetic of $\mathbb{K}[X]$ -

Just as in the case of integers, Euclidean division between polynomials allows us to prove Bézout's theorem, and by consequence to define the notion of GCD and to have access to Gauss's lemma. The proofs being similar to the case of integers, we do not reproduce them. In what follows, $\mathbb{K}=\mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$.

## Bézout's Theorem in $\mathbb{K}[X]$

Definition 3.52. Let $P, Q \in \mathbb{K}[X]$. When $P$ is non-zero, we recall that $P$ divides $Q$ if there exists $R \in \mathbb{K}[X]$ such that $Q=P R$. We say that $P$ and $Q$ are coprime if they have no common divisors (in $\mathbb{K}[X]$) other than non-zero constants. We also use these definitions in the case of $\mathbb{Z}[X]$.

Remark 3.53. The previous definition suggests that the notion of coprimality between two polynomials depends on the set chosen for their coefficients: thus, a priori, nothing prevents two polynomials with integer coefficients from being coprime when viewed as elements of $\mathbb{Q}[X]$, but not being so when viewed as elements of $\mathbb{C}[X]$.

Theorem 3.54 (Bézout). Let $P, Q \in \mathbb{K}[X]$. Then $P$ and $Q$ are coprime if, and only if, there exist $\mathrm{U}, \mathrm{V} \in \mathbb{K}[X]$ such that $\mathrm{PU}+\mathrm{QV}=1$.
Let $\mathrm{P}(\mathrm{x})=1+x+\cdots+x^{n-1}=\left(1-x^{n}\right) /(1-x)$. Thus, $P(x)$ divides $\mathrm{P}\left(\mathrm{x}^{2}\right)$ if and only if there exists a polynomial $\mathrm{Q}(\mathrm{x})$ such that

$$
\frac{1-x^{2 n}}{1-x^{2}}=Q(x) \frac{1-x^{n}}{1-x}
$$

or equivalently,

$$
\left(1+x^{\mathfrak{n}}\right)=Q(x)(1+x)
$$

Thus, $P(X)$ divides $P\left(X^{2}\right)$ if and only if -1 is a root of $1+X^{n}$, in other words, if and only if $n$ is odd.