A problem about $e^{2\pi i \alpha_1}+e^{2\pi i \alpha_2}+\cdots+e^{2\pi i \alpha_N}=0$ Let $\alpha_i\in [0,1),\; i\in \{1,\cdots,N\}$ for some positive integer $N$, such that
$$e^{2\pi i \alpha_1}+e^{2\pi i \alpha_2}+\cdots+e^{2\pi i \alpha_N}=0$$
and if for any non-empty proper subset $E\subset \{1,\cdots,N\}$ satisfy $\sum_{k\in E}e^{2\pi i\alpha_{k}}\neq 0$, then $N$ be a prime number, and $\{\alpha_i: i\in \{1,\cdots,N\}\}=\rho+\{\frac{i}{N}:i\in\{0,\cdots,N-1\}\}$  for some $\rho \in [0,1)$.
 A: No, this is false. For $N=5$, there are many other solutions. Heuristic reasoning: There are $5$ angles, and the condition they impose is a codimension two condition. Thus, we expect the solutions to the problem to form a three-dimensional set. However, for each way of dividing the $5$ angles into a pair $D$ and a triple $T$, there is only a $2$ dimensional space of ways to have $\sum_{i \in D} \alpha_i = \sum_{i \in T} \alpha_i = 0$. Also, there is only a one-dimensional space of ways to have a rotation of a pentagon. We expect there to be lots of points in a three-dimensional set which are not in finitely many one- and two-dimensional sets.
To give an explicit example, in the picture below, choose $\pi/3 < x < \pi/2$ and $x \neq 2 \pi/5$ and take $y = \cos^{-1}(1/2+\cos x)$, to get an equilateral pentagon which is not regular and no subset of whose sides add up to zero.

A: Your conjecture is false, see mercio's answer here for a 6-term counterexample with all $\alpha_i$ rational:
$$\alpha_1 = \frac{3}{30},\quad \alpha_2 = \frac9{30},\quad \alpha_3 = \frac{10}{30},\quad \alpha_4 = \frac{20}{30},\quad \alpha_5 = \frac{21}{30},\quad \alpha_6 = \frac{27}{30}.$$
I believe there's no non-trivial proper subset of these which sums to zero (if there were, then it would decompose into regular polygons).  However, it is composed of sums and differences of regular polygons, so it does not contradict Mann's result cited in the comments.
