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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".

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Seems pretty straightforward to me. It should follow from $n = \prod_{d | n, d > 1} \Phi_d(1)$ by Mobius inversion. – Qiaochu Yuan Jul 6 '11 at 21:06
If you need a reference, see the proof of the corollary to Theorem 1 in Section 1 of Chapter IV of Lang's Algebraic Number Theory (page 74 of the 2nd edition). But "it follows easily..." is good enough too. – KCd Apr 29 '12 at 20:44
up vote 7 down vote accepted

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,

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Thank you. There probably isn't much more to say about this question, so I'll accept your answer. – idmercer Jul 7 '11 at 20:33

Another proof follows directly from the formula $X^{n} - 1 = \prod_{d \mid n} \Phi_d(x)$, since we can deduce from it that \begin{equation} X^{n-1} + \cdots + X + 1 = \prod_{d \mid n, d>1} \Phi_d(x). \end{equation} 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.

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