

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Let φ(n) denote Euler's totient function and k $\perp$ n denote that k, n are integers and relatively prime. Let N = φ(n) + 1. If n is not a prime power $$ \prod_{\substack{0 < k < n, \\ k \perp n }} \Gamma \left(\frac{k}{n}\right) = \sqrt{N}\prod_{ 0 < k < N}\Gamma \left(\frac{k}{N}\right) \quad (n \neq p^a) . $$ This identity was proved here as a corollary to some other identities, but the authors asked: ``is there some natural direct proof of this formula?´´

share|cite|improve this question
up vote 7 down vote accepted

Denote $$f(n)=\prod_{k=1}^{n-1}\Gamma \left(\frac{k}{n}\right)$$ and $$ F(n)=\prod_{1\le k\le n-1, k\perp n}\Gamma \left(\frac{k}{n}\right)$$ We have $f(n)=\prod_{d|n}F(n)$ and therefore by Mobius inversion $F(n)=\prod_{d|n}f(d)^{\mu(n/d)}$

By the multiplication theorem we have $f(n)=\frac{1}{\sqrt{n}}(2\pi)^{\frac{n-1}{2}}$, so if $n$ is not a prime power $$F(n)=\prod_{d|n}\left(\frac{1}{\sqrt{d}}(2\pi)^{\frac{d-1}{2}}\right)^{\mu(n/d)}=(2\pi)^{\frac{1}{2}\varphi (n)}$$ The formula $F(n)=\sqrt{\varphi(n)+1}f(\varphi(n)+1)$ follows.

share|cite|improve this answer

Your Answer


