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Let n, k be integers, $n>1$ and $k \perp n$ denote that k, n are coprime and let $S_n = \{1 \le k \le \lfloor n / 2 \rfloor : k \perp n \}.$ Then $$ n \left( \prod_{k \in S_{n}} \sin \left( k \frac {\pi}{n} \right) \right)^{-2} \in \mathbb{Z}. $$

I think this is surprising but I have no proof.

Jonas Alomo
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1 Answers1

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There are a number of results for products that are closely connected to your product. There are some inessential differences (inverse is not taken), and generally products are over all $k$ from $1$ to $n-1$ relatively prime to $n$, but that would be taken care of by your squaring. Here is a link to a fully available paper by Steven Galovich.

André Nicolas
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  • Thank you very much for this nice and highly relevant paper. Yes, I think the above formula can be justified by Galovich's theorems. – Jonas Alomo Jun 24 '11 at 21:48
  • @Andre Nicolas I tried clicking on the link you provided, but the page is not found. Would you kindly provide an updated link to this paper by Steven Galovich? Thank you, – DDS Jun 24 '19 at 19:56
  • @Andre Nicolas Also, would you know of any other formulas similar to the one posted of (almost) integer values? If you could provide them or point me in their direction I would be grateful. Thanks again. – DDS Jun 24 '19 at 19:58
  • @mlchristians You can try here: https://www.jstor.org/stable/2690306?seq=1#page_scan_tab_contents – user Jun 24 '19 at 23:15
  • @user Many thanks. – DDS Jun 24 '19 at 23:43