$z^n=(z+1)^n=1$, show that $n$ is divisible by $6$. we are given $z^n=(z+1)^n=1$, $z$ complex number. we want to prove that $n$ is divisible by $6$. I showed that $|z|=|z+1|=1$. Hence $z$ is on the intersection of two unit circles, one centered at $(0,0)$ and the other one centered at $(-1,0)$. Then $z$ can take two values $z=\operatorname{cis}(2 \pi /3)$ or $z=\operatorname{cis}(4 \pi/3)$. Then from $z^n=1=\operatorname{cis}(2 \pi k)$, $k$ integer I showed that $n$ should be divisible by $3$. I am left to show that $n$ should be divisible by $2$. I think if I figure out a way to show that $\operatorname{arg}(z+1)=\pi/3$ or $-\pi/3$ then I would be done, by using $(z+1)^n=\operatorname{cis}(2\pi k)$. but how can I do this? Any ideas are welcome! Thanks in advance!

If $z$ is one of the two nontrivial roots of unity, then $z+1$ is...?
Geometrically: Consider the hexagon inscribed inside the unit circle.
Algebraically: if $z^3=1$ then $z^2+z+1=0$ so $z+1=-z^2$ is a $6$th root.
