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Let $n$ be a positive integer, $k=0,\cdots,n-1$, $\omega_k=e^{\tfrac{2\pi i}{n}k}$ be the roots of unity, $c_k \in \mathcal{Z}$ be integer coefficients, trivial and non-trivial be two subcategories for $c_k$ defined as follows: Trivial is when $c_k$ contains equal integers that are evenly spaced under modulo addition of a composite divisor of $n$, and non-trivial $c_k$ is when evenly spaced integers under modulo addition of a composite divisor of $n$ are not equal. An example of a trivial case for $n=6$ is $$ (2)\omega_0+(2)\omega_1+(1)\omega_2+(2)\omega_3+(2)\omega_4+(1)\omega_5=0 $$ because $(2)\omega_0+(2)\omega_3=0$, and $(2)\omega_1+(2)\omega_4=0$, and $(1)\omega_2+(1)\omega_5=0$. Then regarding the linear independence of $\sum_k c_k \omega_k = 0$ consider the following: Does their exist a non-trivial case?

Note: Evenly spaced under modulo addition of a composite divisor of $n$ means the following: Let $m_p$ be a composite divisors of $n$ such that $n=m_0 m_1 \cdots$, then the trivial case of $c_k$ satisfies $c_k=c_j$ for $(k-j) \mod m_p =0$.

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    $\begingroup$ This is somewhat poorly worded. I think that what you want to ask is whether given integers $a_0,\ldots,a_{n-1}$ such that $\sum a_i\omega_i = 0$, must is follow that the nonzero $a_i$ are all equal. Correct? $\endgroup$ – Arturo Magidin Apr 16 at 18:56
  • $\begingroup$ @ArturoMagidin The case when all $a_i$ are equal is not the most general case in the trivial case mentioned in the OP. This is because $n$ can be a composite number. $\endgroup$ – linuxfreebird Apr 16 at 20:13
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    $\begingroup$ I'm trying to understand what you wrote; "that area evenly spaced" is a typo, but even taking that into account I'm not sure what you mean. I don't know what 'evenly spaced under modulo addition' means.If my attempt at understanding it is incorrect, fine, but just telling me I'm wrong doesn't clarify what you are trying to say. $\endgroup$ – Arturo Magidin Apr 16 at 20:34
  • $\begingroup$ @ArturoMagidin. Thank you for pointing out my poor wording and the typo. I have corrected the typo and provide clarification as a note in the OP. $\endgroup$ – linuxfreebird Apr 16 at 22:00
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I am not sure I understood your definition correctly but I believe the fact that the $n$th cyclotomic field extension of $Q$ has degree $\phi(n)$ says that there are no non trivial combinations.

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  • $\begingroup$ Thanks for the answer and the reference. May I ask: how can I improve the wording of my question to address your confusion? $\endgroup$ – linuxfreebird Apr 16 at 18:52
  • $\begingroup$ @linuxfreebird sure, use more maths in your definitions and less words. This is not a general rule but it would help in this case. Could you make sense of my answer? Do you think it answers your original question? $\endgroup$ – Bananach Apr 17 at 5:49

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