Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$ 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

share|improve this question
1  
What have you tried? What is the problem (specific)? –  Mixxiphoid Aug 6 '13 at 8:16
    
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
show 1 more comment

1 Answer

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

$$|D(\Phi_n(x))|=\frac{n^{\phi(n)}}{\prod_{p|n}p^{\frac{\phi(n)}{p-1}}}$$

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.