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Let $n$ be an integer greater $2$. What is the value of the following product for $a=3, \dots , n-1$:

$$\prod_{k = 1}^{n-1} \left( 1 - \sum_{j = 1}^{a-1} \zeta^{jk} \right)$$

where $\zeta$ is some complex $n$th root of unity. I can prove that:

$$\prod_{k = 1}^{n-1} \left( 1 + \sum_{j = 1}^{a-1} \zeta^{jk} \right) = 1$$

and for $a = 2$:

$$\prod_{k = 1}^{n-1} \left( 1 - \sum_{j = 1}^{a-1} \zeta^{jk} \right) = \prod_{k = 1}^{n-1} (1-\zeta^k) = n$$

However, I have problem with the general case.

Marcin
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  • It is maybe more advisable not to use $i$ as an index of summation when dealing with complex numbers. – Jean Marie Dec 16 '16 at 22:42
  • I find it is $\displaystyle\frac{P(\zeta^{a-1})}{P(\zeta^{-1})}$ where $P(x) = \prod_{k=1}^{n-1} (1-x^k)$ – reuns Dec 17 '16 at 18:36
  • I believe that your claim would imply that $\prod_{k = 1}^{p-1} \left(1 - \sum_{j = 1}^{p-1} \zeta^{jk} \right) = 1$ for an odd prime $p$. And this contradicts every example I've computed. Here is the sketch of the proof. Let $a \in {1, 2, \dots, p-1}$. Then $x \mapsto ax$ is a permutation of ${1, 2, \dots, p-1}$. Therefore: $P(\zeta^{a}) = \prod_{k = 1}^{p-1} (1 - (\zeta^{a})^k) = \prod_{k = 1}^{p-1} (1-\zeta^{ak}) = \prod_{k = 1}^{p-1} (1 - \zeta^k) = p$. Therefore $P(\zeta^{a-1}) = P(\zeta^{-1}) = p$. – Marcin Dec 17 '16 at 20:11
  • @user1952009 Thank you for your comment. Could you give a sketch of the argument? This results seem to contradict all the examples I've computed, as I explained above. – Marcin Dec 18 '16 at 20:31

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