n-th roots of unity summing to $0$

Let $\zeta = e^{2\pi i/n}$ be an $n$-th root of unity, and let $S = \{\zeta^m|m=0,1,\ldots,n-1\}$ be the corresponding sets of all $n$-th roots of unity.

Let $k \leq z$. Let $C \subseteq S$ such that $k=|C|$.

I made following conjecture, but so far I'm unable to prove it:

Then $\sum_{c\in C} c = 0$ implies that $k= |C|$ is a $\mathbb Z$-linear combination of strict divisors (divisors strictly greather than 1) of $n$.

This seems to be plausible, and I checked it up to $n=15$. For $n=15$ we have the interesting case that the converse does not hold for $k=11 = 1\cdot 5 + 2 \cdot 3$. Another observation we can use is that for $C \subset S$ we have the equivalence $$\sum_{c \in C} c = 0 \iff \sum_{d \in S \setminus C} d= 0$$ which is quite obvious when you consider that $\sum_{s\in S} s = 0$.

So can anyone prove or disprove this conjecture?

• Where did you fall on this? – Did Jul 6 '16 at 16:34
• I was thinking about a problem on when it is possible to balance a given number of objecs on the vertices of a regular polyhedron. See this challenge on PPCG.SE or where it originated on reddit. – flawr Jul 6 '16 at 16:42
• @Did Why are you asking? =) – flawr Jul 6 '16 at 16:42
• Because I find troubling the coincidence with the paper @quid mentions. – Did Jul 6 '16 at 16:47
• I just tried to reduce my problem and my conjecture to a way that can easily be expressed and understood. You can see it is coming from an entierly different background so I hope this is not causing any trouble! – flawr Jul 6 '16 at 16:52

This is true, in fact slightly more is true, namely that it is a combination with positive coefficients (which is likely intended). If one allows repetitions of the roots, then the converse is true too.

This is a consequence of the main result of the following paper.

T. Y. Lam and K. H. Leung, MR 1736695 On vanishing sums of roots of unity, J. Algebra 224 (2000), no. 1, 91--109.

Below is its abstract:

An unsolved problem in number theory asked the following: For a given natural number $m$, what are the possible integers $n$ for which there exist $m$-th roots of unity $\alpha_1, \dots, \alpha_n \in \mathbb{C}$ such that $\alpha_1 + \dots + \alpha_n=0$? We show in this paper that the set of all possible $n$'s is exactly the collection of $\mathbb{N}$-combinations of the prime divisors of $m$, where $\mathbb{N}$ denotes the set of all non-negative integers. The proof is long and involves a subtle analysis of minimal vanishing sums of mth roots of unity, couched in the setting of integral group rings of finite cyclic groups. Our techniques also recovered with ease some of the classical results on vanishing sums of roots of unity, such as those of Rédei, de Bruijn, and Schoenberg.

• One more remark: that the combination is with positive coeeficients is likely what you meant to begin with. Note that if $n$ is divisible by two primes, then every number can be written as a Z-linear combination. . – quid Jul 6 '16 at 16:55