jred wrote: Let $n$ be a natural number. Prove that a necessary and sufficient condition for the equation $z^{n+1}-z^n-1=0$ to have a complex root whose modulus is equal to $1$ is that $n+2$ is divisible by $6$. Problem is equivalent to $e^{i(n+1)t}-e^{int}-1=0$ $\iff$ $\sin(n+1)t=\sin nt$ and $\cos(n+1)t=1+\cos nt$ $\iff$ $t=\frac{2k+1}{2n+1}\pi$ and $\cos nt=-\frac 12$ $\iff$ $\exists p,k$ such that $(2k+1)\frac{n}{2n+1}=2p\pm \frac 43$ $\iff$ $n=3m+1$ and $\exists p,k$ such that $(2k+1)\frac{3m+1}{2m+1}=6p\pm 4$ $\iff$ $n=3m+1$ $m=2u+1$ and $\exists p,k$ such that $(2k+1)\frac{3u+2}{4u+3}=3p\pm 2$ $\iff$ $n=6u+4$ $\iff$ $6|n+2$ Q.E.D.

If $|z| = 1$ is root then $z^n (z - 1) = 1$, take modulus get $|z - 1| = 1$, so plot on unit circle points $z, z-1$ to show they form equilateral triangle with one side parallel to real axis, so $z = \omega$ or $\omega^5$ where $\omega = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}.$ If $z = \omega$ then $\omega^n \cdot \omega^2 = 1$ because $\omega^2 = \omega - 1$; otherwise if $z = \omega^5$ then $\omega^{5n+4} = 1$. In either case $6 | n + 2$. The converse is clearly true.