Value of cyclotomic polynomial evaluated at 1

Let $\Phi_n(x)$ be the usual cyclotomic polynomial (minimal polynomial over the rationals for a primitive nth root of unity).

There are many well-known properties, such as $x^n-1 = \Pi_{d|n}\Phi_d(x)$.

The following fact appears to follow pretty easily:


$\Phi_n(1)=p$ if $n$ is a prime power $p^k$.

$\Phi_n(1)=1$ if $n$ is divisible by more than one prime.

My question is, is there a reference for this fact? Or is it simple enough to just call it “folklore” or to just say it “follows easily from properties of cyclotomic polynomials”.

Solutions Collecting From Web of "Value of cyclotomic polynomial evaluated at 1"

Möbius Inversion:

As outlined in Qiaochu’s comment, Möbius inversion will solve this problem. Since I am more comfortable with sums then products, lets just take logs. We have $$\log n=\sum_{d|n\ d\neq 1}\log\Phi_{d}(1).$$ Then for $d\neq1$, $$\log\Phi_{d}(1)=\sum_{d|n}\mu\left(\frac{n}{d}\right)\log d=\Lambda(n)$$ where $\Lambda(n)$ is the Von Mangoldt Lambda Function. Since $\Lambda(p^k)=\log p$, and $\Lambda(n)=0$ for $n$ composite, the result then follows upon exponentiating.


This relation follows from some other identities. For an integer $n$ and a prime $p$ we have that $$\Phi_{np}(x)=\frac{\Phi_{n}\left(x^{p}\right)}{\Phi_{n}(x)}\ \text{when }\gcd(n,p)=1$$

$$\Phi_{np}(x)=\Phi_{n}\left(x^{p}\right)\ \text{when }\gcd(n,p)=p.$$

We know that $\Phi_p(1)=p$, and from the above it follows that $\Phi_{p^\alpha}(1)=p$ and $\Phi_{pq}(1)=1$.

Hope that helps,

Another proof follows directly from the formula $X^{n} – 1 = \prod_{d \mid n} \Phi_d(x)$, since we can deduce from it that
X^{n-1} + \cdots + X + 1 = \prod_{d \mid n, d>1} \Phi_d(x).
Thus, if $n = p^{k}$, we have
X^{p^{k}-1} + \cdots + X + 1 = \Phi_{p}(x) \cdots \Phi_{p^{k-1}}(x) \Phi_{p^{k}}(x).
After evaluating in 1 we obtain $p^{k} = \Phi_{p}(1) \cdots \Phi_{p^{k-1}}(1) \Phi_{p^{k}}(1)$ and induction on $k$ gives $\Phi_{p^{k}}(1) = p$ for all $k$.

If $n = p_{1}^{\alpha_{1}} \cdots p_{r}^{\alpha_{r}}$, where $\alpha_{i}$’s are positive integers and $r \geq 2$, then
n = \Phi_{n}(1) \prod_{d \mid n, d\neq 1,n} \Phi_d(1).
If we assume the statement true for all positive integers $<n$ then the product in the left member of the equation equals $n$, since
\prod_{i=1}^{r}\Phi_{p_{i}}(1) \cdots \Phi_{p_{i}^{\alpha_{i}}}(1) = p_{1}^{\alpha_{1}} \cdots p_{r}^{\alpha_{r}} = n
and the rest of the factors are 1. Thus, $\Phi_{n}(1) = 1$ also.