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$ n\geq2 $, prove that the product $$\prod_{1 \leq j<k\leq n \atop \gcd(j,n)=\gcd(k,n)=1}4 \sin^2\frac{(k-j)\pi}{n}=\dfrac{n^{\varphi(n)}}{\prod\limits_{p\mid n, p\; prime}{p^{\frac{\varphi(n)}{p-1}}}}$$ where $\varphi(n)$ is the Euler's totient function

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Using the imperative mood might not endear you to potential answerers. – user17794 Aug 6 '13 at 8:19
Zero personal input after 25+ questions asked on the site is rather curious, no? – Did Aug 6 '13 at 9:03
Personal input $\ne$ adding wanted formula. – Did Aug 6 '13 at 10:09
I don't think it's a homework question – ziang chen Aug 6 '13 at 22:18
It is a difficult problem – ziang chen Aug 6 '13 at 22:25
up vote 3 down vote accepted

Let $\Phi_n(x)$ be the $n$th cyclotomic polynomial. So we have:

$$\Phi_n(x)=\prod_{\substack{1\le k\le n\\\gcd(k,n)=1}}(x-e^{\frac{2k\pi i}{n}})$$

The roots of $\Phi_n(x)$ are exactly the primitive $n$th roots of unity, so its discriminant is:

$$D(\Phi_n(x))=\prod_{\substack{1\le j<k\le n\\\gcd(j,n)=\gcd(k,n)=1}}(e^{\frac{2k\pi i}{n}}-e^{\frac{2j\pi i}{n}})^2$$

Now see if you can show the identity:

$$|e^{\frac{2k\pi i}{n}}-e^{\frac{2j\pi i}{n}}|^2=4\sin^2\left(\frac{(k-j)\pi}{n}\right)$$

It follows that your desired product is the absolute value of the discriminant of the $n$th cyclotomic polynomial.

Edit: To address the edit to your question, after an internet search, I've learned it is actually true that


so that this answer is consistent with what you want to prove. However, the calculation of the discriminant of the cyclotomic polynomial seems fairly complicated.

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