Condition for a polynomial to have root of modulus 1 
Prove that the polynomial$$P(X) = X^{n+1} - X^{n} - 1,\text{ }P \in \mathbb {C}[X]$$has a root $z$ with $\left|z\right|=1$ if and only if $6\,|\,(n+2)$.


One implication, from left to right, is quite simple because$$z=\cos \alpha + i\sin \alpha$$and$$z^n=\cos n\alpha + i\sin n\alpha$$allows me to find out the $\alpha$ value. It is the other implication that I wasn't able to prove.
 A: We have $z^n(z-1) = 1$, so $\left|z-1\right| = 1$, which is only possible if $z = w$, $w^5$, where $w$ is the first primitive $6$th root of unity. Note that $w - 1 = w^2$ and $w^5 - 1 = w^4$, so $w^n = w^4$ and $(w^5)^n = w^2$, both of which give $n \equiv 4 \text{ (mod }6\text{)}$.
A: Note that $X^{n+1} - X^{n} - 1 = (X^2-X+1)X^{n-1}-(X^{n-1}+1)$.
Let $\omega$ be a root of $X^2-X+1$. Then $\omega$ is a primitive $6$-root of unity and so $\omega^3=-1$.
If $6 \mid (n+2)$, then $n=6k-2$ and $n-1=6k-3$.
Therefore, $\omega^{n-1}=\omega^{6k-3}=\omega^{-3}=-1$, and so $\omega$ is a root of $X^{n-1}+1$.
Thus, $\omega$ is a root of $X^{n+1} - X^{n} - 1 = (X^2-X+1)X^{n-1}-(X^{n-1}+1)$.
A: We can calculate at least two unitary roots as follows:
Let $z = \frac{1 + \sqrt{3}i}{2}$, then $z^4 = -z$, $z^5 = \bar{z}$,  and $z^6 = 1$. If $6 \mid n + 2$, there is a constant $k$ such that $n = 6k + 4$. Then,
$\begin{align*}P(z) &= z^{6k + 5} - z^{6k + 4} - 1\\
&= z^5 - z^4 - 1\\
&= 0.\end{align*}$
$z$ and $\bar{z}$ are roots of $P(X)$ with unitary modulus.
