# Primes $p_i$ such that $\sin(\frac{\pi}{n})\sin(\frac{p_1\pi}{n})\cdots\sin(\frac{p_k\pi}{n})$ is rational

The following amazing identity can be checked directly by hand. $$\sin\left(\frac{\pi}{30}\right)\sin\left(\frac{7\pi}{30}\right)\sin\left(\frac{11\pi}{30}\right)\sin\left(\frac{13\pi}{30}\right)\sin\left(\frac{17\pi}{30}\right)\sin\left(\frac{19\pi}{30}\right)\sin\left(\frac{23\pi}{30}\right)\sin\left(\frac{29\pi}{30}\right) = \frac{1}{2^8}$$ This makes me wonder:

Question. Can we find arbitrarily long sequences of prime numbers $p_1<p_2<\cdots<p_k$ such that the product $$\sin\left(\frac{\pi}{n}\right)\sin\left(\frac{p_1\pi}{n}\right)\sin\left(\frac{p_2\pi}{n}\right)\cdots\sin\left(\frac{p_k\pi}{n}\right)$$ is rational for some integer $n>p_k$.

Other examples are: \begin{align} \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{3\pi}{4}\right) &= \frac{1}{2} \\ \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{3\pi}{6}\right)\sin\left(\frac{5\pi}{6}\right) &= \frac{1}{2^2} \\ \sin\left(\frac{\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)\sin\left(\frac{7\pi}{8}\right) &= \frac{1}{2^3} \\ \sin\left(\frac{\pi}{12}\right)\sin\left(\frac{2\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)\sin\left(\frac{7\pi}{12}\right)\sin\left(\frac{11\pi}{12}\right) &= \frac{1}{2^5} \\ \sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right)\sin\left(\frac{11\pi}{18}\right)\sin\left(\frac{13\pi}{18}\right)\sin\left(\frac{17\pi}{18}\right) &= \frac{1}{2^6} \\ \sin\left(\frac{\pi}{18}\right)\sin\left(\frac{3\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right)\sin\left(\frac{11\pi}{18}\right)\sin\left(\frac{13\pi}{18}\right)\sin\left(\frac{17\pi}{18}\right) &= \frac{1}{2^7} \\ \sin\left(\frac{\pi}{30}\right)\sin\left(\frac{7\pi}{30}\right)\sin\left(\frac{11\pi}{30}\right)\sin\left(\frac{13\pi}{30}\right)\sin\left(\frac{17\pi}{30}\right)\sin\left(\frac{19\pi}{30}\right)\sin\left(\frac{23\pi}{30}\right)\sin\left(\frac{29\pi}{30}\right) &= \frac{1}{2^8} \\ \sin\left(\frac{\pi}{30}\right)\sin\left(\frac{5\pi}{30}\right)\sin\left(\frac{7\pi}{30}\right)\sin\left(\frac{11\pi}{30}\right)\sin\left(\frac{13\pi}{30}\right)\sin\left(\frac{17\pi}{30}\right)\sin\left(\frac{19\pi}{30}\right)\sin\left(\frac{23\pi}{30}\right)\sin\left(\frac{29\pi}{30}\right) &= \frac{1}{2^9} \end{align} The longest I found is $$\sin\left(\frac{\pi}{38}\right)\sin\left(\frac{3\pi}{38}\right)\sin\left(\frac{5\pi}{38}\right)\sin\left(\frac{7\pi}{38}\right)\sin\left(\frac{11\pi}{38}\right)\sin\left(\frac{13\pi}{38}\right)\sin\left(\frac{17\pi}{38}\right)\sin\left(\frac{19\pi}{38}\right)\sin\left(\frac{23\pi}{38}\right)\sin\left(\frac{29\pi}{38}\right) = \frac{1}{2^9}$$ Note that it breaks the pattern of $n=p_k+1$. Note also that not all examples have consecutive primes.

• Something to begin with $$\prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$ May 12, 2018 at 14:35

Consider a product: $$\prod_{m}^{}2\sin\frac{m}{n}\pi,$$ where $$m$$ are integer numbers: $$0 As all multipliers are algebraic integers the product is rational if and only if it is integer.

This particularly happens if $$m$$ is a set of numbers coprime to $$n$$: $$\gcd(m,n)=1$$. It holds: $$P_n:=\prod_{m}^{\gcd(m,n)=1}2\sin\frac{m}{n}\pi= \begin{cases} 1,& n\ne p^k\\ p,& n=p^k \end{cases},\tag{1}$$ where $$p$$ is a prime number, and $$k$$ is a positive integer. The expression (1) can be proved similarly to the result cited in a comment above, using cyclotomic polynomials $$\Phi_n(z)$$ instead of $$\frac{z^n-1}{z-1}$$.

If all $$m$$ coprime to $$n$$ happen to be prime numbers we find an example of the set in question. This should however fail for large $$n$$. The next idea may be: let us try the product $$P_kP_\ell$$ for distinct $$k,\ell$$. This however does not work. Indeed, in accord with question the arguments of sines have to be reduced to common denominator $$n=\frac{k\cdot\ell}{(k,\ell)}$$. Since $$k$$ and $$\ell$$ are distinct at least one of them is not a divisor of the other. Without lost of generality we may assume it is $$\ell$$. Consider now the numerators $$m_k$$ which after reducing to common denominator $$n$$ are $$m_k\equiv m_k^{(n)}=m_k^{(k)}\frac{\ell}{(k,\ell)}$$ (here $$m^{(n)}$$ means the value of the numerator for denominator $$n$$). Thus $$m_k$$ can be a prime number for $$\frac{\ell}{(k,\ell)}\ne1$$ only if $$m_k^{(k)}=1$$ and $$\frac{\ell}{(k,\ell)}$$ is a prime number. This is clearly impossible for all $$m_k$$ since $$m_k^{(k)}$$ (except for a single one) are distinct from $$1$$. The only exception is $$P_2$$ which consists of single factor $$2\sin\frac12\pi=2$$. Except for this case we cannot therefore use a product $$P_k P_\ell$$. By the same argument this is true for the products of a larger number of $$P$$'s as well.

After closer inspection of $$(1)$$ one can note that if $$n$$ is not a prime power one can use only half of sines picking up one from the pair ($$\sin\frac{m}{n}\pi,\sin\frac{n-m}{n}\pi$$) to obtain still integer result (equal to $$1$$). We denote such products as $$Q_n$$. Similarly to the case of $$P_n$$ there is an exceptional product $$Q_6$$ consisting of the single factor $$2\sin\frac16\pi=1$$.

Thus if $$m$$ are required to be prime or 1 the simplest possible combinations are $$P_n, Q_n,\tag{2}$$ where the latter form can be used only if $$n$$ is not a prime power. If $$n=2p$$ or $$n=6p$$ with $$p$$ being a prime number the expressions $$(2)$$ can be multiplied additionally by $$P_2$$ or $$Q_6$$, respectively.

And indeed all examples given in question are of the form $$(2)$$. In the same order as they appear in question, they can be symbolically represented as: $$P_4, P_6P_2, P_8, P_{12}Q_6, P_{18}, P_{18}Q_6, P_{30}Q_6, Q_{38} P_2.$$

The reason for the equalities in question can be seen now in the already mentioned fact that the sets of $$m$$ coprime to given $$n$$ consist in all cases except for the last one exclusively of prime numbers.

Considering in the case $$n=38$$ the table of coprimes to $$38$$: $$\begin{matrix} 1&3&5&7&9&11&13&15&17\\ 37&35&33&31&29&27&25&23&21 \end{matrix},$$ one easily observes that every pair contains at least one prime (observe that $$38=2\cdot19$$ is not a prime power).

The point of the above consideration is the following. The construction of long rational product of prime $$m$$ on these lines is hardly possible, as the list of numbers coprime to large $$n$$ cannot consist exclusively of prime numbers (in fact $$n=30$$ is the largest such number), and even the trick similar to the case $$n=38$$ will not work on a long run. Thus, the construction of very long sequences on the base of $$(2)$$ is impossible.

There exist however much longer sequences than those listed in question. The longest ones by numerical tests are: $$Q_{105}, Q_{140}, Q_{180}, Q_{210},$$ all consisting of 24 multipliers. It can be claimed that there are no sequences of required structure for $$n>210$$.