Let $ \zeta_n = e^{2 \pi i / n} $ be the n-th root of unity.


$$ P(z) = \sum_{k = 0}^{n-1} s_k z^k $$

be a monic polynomial over $ z \in \mathbb{C} $, specified by integer coefficients $ s_k \in \mathbb{Z} $.

I am looking for a way to show if (or if not) a given polynomial $P$ has a zero $ P(\zeta) = 0 $.

In particular I am interested in the special case $ s_k \in \{0,1\} $. So given the coefficients $s_k$, does $P$ has the zero $\zeta$?


A polynomial $P\in\Bbb{Z}[X]$ has the primitive $n$-th root of unity $\zeta_n$ as its root if and only if it is divisible by the $n$-th cyclotomic polynomial $\Phi_n$. So polynomial long division of $\sum_{k=0}^{n-1}s_kX^k$ by $\Phi_k$ has remainder $0$ if and only if $P$ has $\zeta$ as a root.

  • $\begingroup$ Thanks! So in general $P(z)$ has root $\zeta_n$ iff $P(z)$ is divisible by $\Phi_n(z)$. Any ideas regarding the special case $s_k \in \{0,1\}$? Note that this is related to the problem of balancing less than n glasses on a round tray with n equidistant holders for glasses. $\endgroup$ – TomS Feb 12 at 22:16
  • $\begingroup$ Yes, if and ony if $P$ is divisible by $\Phi_n$. $\endgroup$ – Servaes Feb 12 at 22:21

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