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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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    I'm not sure what you're trying to say, but given the equation $\omega^n - 1 = 0$, we can factor it as $(\omega-1)(1+\omega + \omega^2 + \ldots + \omega^{n-1}) = 0$. If $\omega \neq 1$, then the latter factor must be zero.2012-12-07
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    Look up [*cyclotomic polynomials*](http://en.wikipedia.org/wiki/Cyclotomic_polynomial). In characteristic zero (presumably your main case of interest) that is all there is to is. In positive characteristic the cyclotomic polynomials are no longer irreducible, and there is more.2012-12-07
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    @Ben Here for example for $3$rd root of unity, $1+\omega_{3} +\omega_{3}^{2}$ determine all linear relation means if $a_0 + a_1\omega_{3} + a_2\omega_{3}^{2} =0$ means $a_0 = a_1 = a_2$2012-12-07
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    Do you insist on integer coefficients in your linear relations?2012-12-08
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    may be I should add for integer coefficients and non-negative integer coefficients to the question.2012-12-08
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    For integer coefficients, see the comment by @Jyrki.2012-12-08
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    @JyrkiLahtonen, please post as an answer for visibility, there's no sense for everyone else to do it.2013-02-16
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    @AndreasCaranti: Your vote of confidence is appreciated. Yet, the OP edited the question 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, but such a description doesn't feel very useful 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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    I suspect OP wants integer coefficients, although I grant this is not stated in the question.2012-12-08
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    when I faced the question I thought about integer coefficient only, if you have answer for that will you like to share it. Also I like to know for the non negative integral coefficients. Currently, I do not know whether that follows from integral coefficients or not.2012-12-08
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    OK, sorry. I'll leave this answer up anyway. For some reason the word linear made me think we're working over a field.2012-12-08
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    Oh, unless you'd like me to delete it because it's more likely to attract new answers that way?2012-12-08
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    @Ben Let the answer be there, it's useful. No need to be sorry, I am thankful for the answer. Let's see how interesting is the answer for integral and non-negative integral cases.2012-12-08