Can the sum of two roots of unity be a root of unity? 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!
 A: True. For three terms  $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…
Edit: Now that you edited the question and $p$ becomes prime, there is a more general answer. As Douglas pointed out the key word here is cyclotomic polynomials $P_p(z)=\sum_{n=0}^{p-1}z^n$, whose roots are precisely the $p^{th}$-root of unity except $1$. In that case
$1+P(z_j)=1$
gives you such a relation as long as $z_j\neq 1$. 
A: 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 $1 + \zeta_p + ... + \zeta_p^{p-1} = 0$.
A: 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.  
