TITLE: Can the sum of two roots of unity be a root of unity?
QUESTION [6 upvotes]: Let $p$ be prime, and $z_0, z_1, ..., z_{p-1}$ be all the $p$-th roots of unity, i.e. solutions of the equation $z^p = 1.$
Is it true or false that a combination of two (or more, in general) of the roots can give us another root of the same order? 
In mathematical terms, does there exist indices $i_1,i_2,...,i_s, j,$
such that
$z_j = \sum_{k=1}^s z_{i_k}?$
It seems to be that this is not possible, but I also don't have proof of that.
Thank you!

REPLY [18 votes]: If $p$ is intended to be prime, you should say so explicitly. I'll assume $p$ is prime. Then the subset sums are distinct except that the sum of all $p$th roots of unity is $0$, the sum over the empty set. Any coincidence of subset sums $\sum_{i \in I} \zeta_p^i = \sum_{j\in J} \zeta_p^j $produces a polynomial of degree at most $p-1$ with coefficients in $\lbrace-1,0,1\rbrace$ so that $\zeta_p$ is a root. This polynomial must be a multiple of the minimum polynomial for $\zeta_p$, the $p$th cyclotomic polynomial $\Phi_p(x) =1 + x + ... + x^{p-1}$. The only possibilities are scalar multiples corresponding to $1 + \zeta_p + ... + \zeta_p^{p-1} = 0$.

REPLY [10 votes]: It is known that for every positive integer $n$, the primitive $n$-th roots of unity are linearly independent over $\mathbb{Q}$ if and only if $n$ is square free.