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.

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 $X^{n+1} - X^{n} - 1 = (X^2-X+1)X^{n-1}-(X^{n-1}+1)$.
