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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, $1+ \omega_{3} + \omega_{3}^{2}=0$

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    @AndreasCaranti: Yo$u$r vote of confidence is appreciated. Yet, the OP edited the q$u$estion an added *non-negativity* of the coefficients. Yes, we can still describe the answer as the set of polynomials with non-neg coefficients in the ideal generated by the relevant cyclotomic polynomial, b$u$t such a description doesn't feel ver$y$ $u$seful at this point :-)2013-02-17

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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)

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    @Ben Let the answer be there, it's useful. No need to be sorr$y$, I am thankful for the answer. Let's see how interesting is the answer for integral and non-negative integral cases.2012-12-08