TITLE: Linear relations satisfied by nth root of unity
QUESTION [5 upvotes]: How do you characterize all the linear relations satisfied by $n$th roots of unity with real, integral and non-negative integral coefficients?
Here are two examples for 3rd and 4th root:
Let $\omega_{i}$ be the primitive $i$th root of unity, then 
For $2$nd root, $1 + \omega_{2} = 0$;
For $3$rd root, $1+ \omega_{3} + \omega_{3}^{2}=0$

REPLY [2 votes]: From the comments, it looks like we're solving the equation
$$\sum_{i=0}^{n-1} a_i \omega_n^i= 0$$
where $\omega_n$ is a primitive $n$th root of unity in $\mathbb{C}$ and the $a_i$s are real (if they can be complex, the problem seems pretty trivial; I'm not sure how to do it if they can only be rational). Take $\omega_n = e^{2\pi i/n}$. This splits into two equations:
$$\sum_{i=0}^{n-1} a_i \cos(2\pi i/n) = 0$$
$$\sum_{i=0}^{n-1} a_i \sin(2\pi i/n) = 0.$$
$\sin 0 = 0$, so the second sum has no $a_0$ term. This means that $a_2, a_3, \ldots, a_{n-1}$ determine $a_1$:
$$a_1 = -\frac{\sum_{i=2}^{n-1} a_i \sin(2\pi i/n)}{\sin(2\pi/n)}.$$
Now, in the same way, $a_1, a_2, \ldots, a_{n-1}$ determine $a_0$ from the first equation:
$$a_0 = -\sum_{i=1}^{n-1} a_i \cos(2\pi i/n) = \frac{\sum_{i=2}^{n-1} a_i \sin(2\pi i/n)}{\sin(2\pi/n)} - \sum_{i=2}^{n-1} a_i \cos(2\pi i/n).$$
This means that $a_2, a_3, \ldots, a_{n-1}$ can be picked arbitrarily, and they determine $a_0$ and $a_1$ uniquely.
(Originally I had a lot of nonsense in this post, it has since been edited)