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Value of cyclotomic polynomial evaluated at 1

I have to show $\Phi_n(1)=1$ for $n\neq p^k$ with $p$ is prime.
(I already proved to easy part $\Phi_n(1)=p$ for $n=p^k$)

For the proof I would start with: For an arbitrary natural number $n$ we have the unique factorization of primes $n=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$ ($gcd(p_i,p_j)=1$ for $i\neq j$)

I know that there are $\beta_i$ divisors of $n$, which are the powers of each prime $p_i$ which divide $n$. But how can I say that $\Phi_d(1)=1$ for any $d$ with more than one prime divisor?

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marked as duplicate by bgins, Brandon Carter, lhf, Pete L. Clark, Zev Chonoles May 6 '12 at 3:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

start with what you know, like perhaps $$x^n-1=\prod_{0<d|n}\Phi_d(x)$$ and $$\Phi_d(x)=\prod_{0<d|n}\left(x^d-1\right)^{\mu(n/d)}=\prod_{(d,n)=1}\left(x-e^‌​{2\pi id/n}\right)$$ – bgins May 5 '12 at 16:36
Given your factorization of $n$, there are $\tau(n)=\prod(\beta_i+1)$ divisors of $n$. – bgins May 5 '12 at 16:52
up vote 2 down vote accepted

Hint: For $n>1$, $$\Phi_n(x)=\frac{\sum_{k=0}^{n-1}x^k}{\prod_{1 < d < n,~d|n}\Phi_d(x)}$$ is the ratio of $\frac{x^n-1}{x-1}$ to the product of $\Phi_d(x)$ over all proper divisors $d$ of $n$. The numerator is just $n$ when $x=1$ (which also follows from L'Hopital's rule for the limit of the ratio as $x\to1$), and the denominator gives you an inductive statement about $$a_n=\left\{\matrix{1&n=1~\\\Phi_n(1)&n>1,}\right.$$ namely $$ a_1=1,\quad a_n=\frac{n}{\prod_{1<d<n,~d|n}a_d} $$ where, again, the product is over all proper divisors $d$ of $n$. For example the result that $a_n=p$ for $n=p^k$ a positive power of a prime follows by an easy induction.

For the general case, write $a(n)=\Phi_n(1)+\epsilon(n)$, as a function, where $\epsilon(n)=\delta_{1n}$ is $1$ for $n=1$ and $0$ for $n\ne1$. Then $a$ is an arithmetic function, in fact a unit, and $\epsilon$ is the multiplicative identity, in the Dirichlet ring of such functions, and we have the identity $n=\prod_{0<d|n}a(n)$ (i.e., $\log(n)=\log(a(n))*1$ using Dirichlet convolution and natural logarithms). This has solution given by the Möbius inversion formula, which states that $f=g*1\iff g=f*\mu$ for arithmetic functions $f,g$: $$ \eqalign{ \log(a(n))& =\sum_{0<d|n}\mu\left(\frac{n}{d}\right)\log d =\href{}{\Lambda(n)}\\ a(n)& =\prod_{0<d|n}\mu\left(\frac{n}{d}\right)\log d =e^{\Lambda(n)}\\ &=n^{\epsilon(\href{}{\omega(n)})/\href{}{\Omega(n)}} =\left\{\matrix{p&\text{if }n=p^k\text{ is a power of a single prime} \\1&\text{ otherwise, i.e. if }n=1\text{ or }\omega(n)>1}\right. }$$ so that $$ \Phi_n(1)=\left\{\matrix{ 0&\text{if }n=1~~\\ p&\text{if }n=p^k\\ 1&\omega(n)>1}\right. $$


Further proofs and discussion can be found in Naslund's and Bontea's solutions in this thread of the identical question!

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thank you very much. Now it is clear. – wieschoo May 5 '12 at 16:56

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