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Prove that $\prod\limits_{1 \le k \le n-1,gcd(n,k)=1} \sin \frac{k \pi}{n}=\frac{1}{2^{\phi (n)}}$ where $n$ is not a power of a prime number.

My attempt:

$S=\prod\limits_{1 \le k \le n-1,gcd(n,k)=1} \sin \frac{k \pi}{n}=(2i)^{-\phi (n)}*\prod\limits_{1 \le k \le n-1,gcd(n,k)=1}(e^{\frac{ik \pi}{n}}-e^{\frac{-ik \pi}{n}})=(2i)^{-\phi (n)}*e^{-\sum\limits_{1 \le k \le n-1,gcd(n,k)=1} \frac{k \pi}{n}}*\prod\limits_{1 \le k \le n-1,gcd(n,k)=1} (e^{\frac{2ik \pi}{n}}-1)$

No we use thislemma and we have:

$S=(2i)^{- \phi (n)}*(-i)^{\phi(n)}*\prod\limits_{1 \le k \le n-1,gcd(n,k)=1} (e^{\frac{2ik \pi}{n}}-1)=(-2)^{- \phi (n)}*\prod\limits_{1 \le k \le n-1,gcd(n,k)=1} (e^{\frac{2ik \pi}{n}}-1)=2^{- \phi (n)}*\prod\limits_{1 \le k \le n-1,gcd(n,k)=1}(1-e^{\frac{2ik \pi}{n}})$

So it only remains to prove $\prod\limits_{1 \le k \le n-1,gcd(n,k)=1}(1-e^{\frac{2ik \pi}{n}})=1$.Which I can't do.My attempt was based on thisproblem but the rest can't be continued with that method.And also I don't know how to use $n$ not being power of a prime.

Mikasa
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Taha Akbari
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  • Plz change the title if it looks inappropriate. – Mikasa Sep 21 '17 at 10:25
  • Only a note: A question about $\prod\limits_{1 \le k \le n-1,gcd(n,k)=1}(x-e^{\frac{2ik \pi}{n}})$ may fit better to https://mathoverflow.net/ where some experts are answering . – user90369 Sep 22 '17 at 12:14
  • @user90369 Really not I am a high school student who is studying for IMO and I got this from one of the books I am studying so indeed it is not appropriate for math overflow. – Taha Akbari Sep 22 '17 at 16:22
  • Then something of that you have studied should give an idea how to solve it. I hope you'll find the answer in you books, because it's (at least for me) an interesting question. :-) – user90369 Sep 22 '17 at 16:59
  • @user90369 It is not solved in that book so I will add a bounty to get a result. – Taha Akbari Sep 22 '17 at 18:57
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    Once you know the name, results are easy to find: You are trying to evaluate a cyclotomic polynomial at 1. The question has been answered here: https://math.stackexchange.com/questions/49956/value-of-cyclotomic-polynomial-evaluated-at-1 – Long Sep 22 '17 at 20:48
  • @Long How should we prove $x^n-1 = \Pi_{d|n}\Phi_d(x)$? – Taha Akbari Sep 23 '17 at 12:15

1 Answers1

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This question has been answered here: Value of cyclotomic polynomial evaluated at 1

I'll provide a proof for $x^n-1 = \prod_{d|n} \Phi_d(x)$:

$x^n -1$ has $n$ distinct roots $1,a=e^{2\pi i/n},a^2,...,a^{n-1}$, so we can write $x^n-1 = \prod_{k=0}^{n-1} (x-a^k)$.

Next we can partition the numbers $0,..,n-1$ by the greatest common divisor with $n$:

$$\prod_{d|n} \prod_{0\leq k \leq n-1 \\ \gcd(k,n)=d} (x-a^k) =\prod_{d|n} \prod_{0\leq k \leq n-1 \\ \gcd(k,n)=d} (x-(a^d)^{k/d}) =\prod_{d|n} \prod_{0\leq l \leq n/d-1 \\ \gcd(l,n)=1} (x-(a^d)^l) $$

Note that $a^d$ generates the $n/d$-th roots of unity, so $$\prod_{d|n} \prod_{0\leq l \leq n/d-1 \\ \gcd(l,n)=1} (x-(a^d)^l) = \prod_{d|n} \Phi_{n/d}(x) = \prod_{d|n} \Phi_d(x)$$

Long
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