25
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ Something to begin with $$\prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$ $\endgroup$
    – rtybase
    May 12, 2018 at 14:35

1 Answer 1

12
$\begingroup$

Consider a product: $$ \prod_{m}^{}2\sin\frac{m}{n}\pi, $$ where $m$ are integer numbers: $$ 0<m< n.$$ 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 $n>210$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .