# Cyclotomic polynomials evaluated at roots of unity

Dear MO_World,

I'm working on an ergodic theory question (about a generalization of eigenfunctions for measure-preserving transformations) and have run into a number theory question concerning cyclotomic polynomials that I'm unable to tackle.

The question is this:

Let $p$ be a prime and let $p|n$. When is it the case that $\Phi_n(e^{2\pi i/p})=\pm e^{2\pi ij/p}$ for some $j$?

Here $\Phi_n$ denotes the $n$th cyclotomic polynomial.

I've experimented with Mathematica and have found there are non-trivial cases in which the condition holds, whereas for most cases it does not seem to hold.

Letting $c(n,p)=\Phi_n(e^{2\pi i/p})$, we have $c(105,3)=1$, but $c(105,5)$ and $c(105,7)$ are not on the unit circle. None of $c(15,3)$, $c(21,3)$, $c(15,5)$, $c(35,5)$, $c(21,7)$, $c(35,7)$ are on the unit circle; $c(40,2)=1$, but $c(50,2)=5$...

Not surprisingly it seems to be easiest for the condition to hold for small $p$.

Also, using the relations $\Phi_{p^2n}(x)=\Phi_{pn}(x^p)$; and $\Phi_n(1)=q$ if $n=q^k$ for some prime $q$ and an integer $k$, but $\Phi_n(1)=1$ otherwise, it's not hard to see that the condition holds whenever $p^2|n$, but $n$ is not a power of $p$.

Thanks for any more systematic suggestions...

Set $\zeta=e^{2\pi i/p}$. By what Anthony wrote, we may assume $n=pm$ where $p$ does not divide $m$. Note that if $\Phi_n(\zeta)=\pm\zeta^j$, then $\Phi_n(\zeta^k)=\pm\zeta^{jk}$ for each $k$ prime to $p$, because the absolute Galois group of $\mathbb Q$ acts transitively on the $\zeta^k$s (by irreducibility of $\Phi_p$).

Let $r$ be the radical of $m$ (the product of the distinct prime divisors of $m$). From $\Phi_n(\zeta)=\Phi_{pr}(\zeta^{m/r})$, we see that we furthermore may assume that $m$ is squarefree.

We cannot have the case $\Phi_n(\zeta)=-\zeta^j$ unless $p=2$: Suppose that happens. Then $\zeta$ is a root of $\Phi_n(X)+X^j$. On the other hand, $\Phi_p$ is irreducible, so $\Phi_p$ divides $\Phi_n(X)+X^j$. But then $p=\Phi_p(1)$ divides $\Phi_n(1)+1=2$, so $p=2$.

Using the formula $\Phi_n(X)=\prod_{d\mid n}(X^d-1)^{\mu(n/d)}$, we get $\Phi_{pm}(\zeta)=\Phi_m(1)/\Phi_m(\zeta)$. Clearly, if $m$ is a prime, then $\lvert\Phi_m(1)/\Phi_m(\zeta)\rvert>1$. Thus $m$ is not a prime, so $\Phi_m(1)=1$, and $\Phi_m(\zeta)$ is $\pm$ a power of $\zeta$.

Now $\Phi_m(\zeta)$ depends only on the residue class of each prime divisor of $m$ modulo $p$. So given $p$, and the number $t$ of prime factors of $m$, one only has to check which of the $(p-1)^t$ possibilities work. For instance, if each prime divisor of $m$ is $\equiv1\pmod{p}$, then $\Phi_m(\zeta)=1$. So by Dirichlet, for each prime $p$, there are infinitely many cases.

I don't believe that a more explicit description of the possibilities for $m$ is possible.

Added: Under the assumptions from above ($m$ is squarefree, prime to $p$, and not a prime) $\Phi_m(\zeta)$ (and then also $\Phi_{pm}(\zeta)$) is a power of $\zeta$ whenever $m$ has a prime divisor which is $\equiv1$ or $\equiv-1\pmod{p}$. This follows by induction from the identity $\Phi_m(X)=\Phi_k(X^q)/\Phi_k(X)$ for $m=kq$ and $q$ a prime.

Probably, this sufficient condition is necessary too. I only checked it if $m=uv$ with $u,v$ distinct primes: $\Phi_m(\zeta)=\zeta^j$ gives that $X^p-1$ divides $(X^{uv}-1)(X-1)-X^j(X^u-1)(X^v-1)$, and this easily yields $u\equiv\pm1\pmod{p}$ or $v\equiv\pm1\pmod{p}$.

• Thanks a lot @Peter. I'm going to take a little time to digest this. – Anthony Quas Sep 1 '12 at 19:23