Let $a, b, c \in \mathbf C$ such that $|a| = |b| = |c| = 1$, and $a + b + c = 0$. Prove that if $k \in \mathbf N$ is not divisible by $3$, then
$$
a^k + b^k + c^k = 0.
$$
Consider the triangle with vertices at $a$, $b$ and $c$. Because $(a+b+c)/3 = 0$, its barycenter lies at the origin; likewise, since all its vertices
belong the unit circle, its circumcenter is also located at $0$. This is enough to claim that the triangle is equilateral, but for completeness
we prove it:
Recall that the barycenter is the intersection of the medians, whereas the circumcenter is the intersection of the perpendicular bisectors;
since they share a point by construction – i.e., the midpoints of the sides –, in order for them to pass through the same other, they must
be equal. We thus have a triangle where the cevians joining the vertices to the midpoints from the opposite sides form right angles, so by
definition the heights coincide with the medians and the perpendicular bisectors. The triangle then has all its sides equal by applying Pythagoras' theorem:
If $m=n$, then $a = \sqrt{m^2 + h^2} = \sqrt{n^2 + h^2} = b$.
Finally, this translates into the complex plane as the existence of a $z \in \mathbf C$ with $|z|=1$ such that $a=z$, $b=z\xi_3$, and $c=z\xi_3^2$,
where $\xi_3 = e^{i2\pi / 3}$; hence
$$
a^k + b^k + c^k = z^k (1 + \xi_3^k + \xi_3^{2k}) = z^k \frac{1 - \xi_3^{3k}}{1 - \xi_3^k} = 0.
$$