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Determine if $\sin^2{\frac{k\pi}{n}}=\sin{\frac{\pi}{n}}$ given that $\\n,k\in\mathbb{N},\\ n \geq 3,\\n\geq k-1 .$ has a finite number of different pairs $(n,k)$, that satisfy the equation

I got this query trying to solve this problem

I don't know which branch of mathematics deals with this kind of problems. (I'm sorry if the tags are irrelevant.)

metamorphy
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John Kall
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  • Are you meaning an infinite number of zero's ? If you speak about roots, do you want to solve for $k$ or for $n$ ? – Claude Leibovici Apr 20 '20 at 05:00
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    Hi @ClaudeLeibovici, solve for pairs $(n,k)$ with $n\ge3,k-1$. Or when is $\frac n\pi\arcsin\sqrt{\sin\frac\pi n}$ an integer ? –  Apr 20 '20 at 06:28
  • I have just changed the wording. With the word roots I meant pairs of x and y, which satisfy the equation. Definitely not considering one as variable and the other as constant. – John Kall Apr 20 '20 at 06:40
  • Using @Yves Daoust's suggestion, I did not find any $n$ such that $\frac n\pi\arcsin\sqrt{\sin\frac\pi n}$ is an integer (for sure, I did not explore many but up to $10,000$) – Claude Leibovici Apr 20 '20 at 06:56
  • @ClaudeLeibovici: sorry, I discarded this comment, was useless. –  Apr 20 '20 at 07:04
  • @ClaudeLeibovici: $\sin kx$ can be expressed by a Chebyshev polynomial of the second kind. For even $k$, the equation $\sin^2kx=\sin x$ is polynomial in $\sin x$, of degree $2k-1$, with integer coefficients. At the same time, $\sin\frac\pi n$ are also algebraic numbers. I don't know how to exploit this. –  Apr 20 '20 at 07:21

1 Answers1

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There is a working approach I've used here and here.

Squaring both sides of $\sin^2(k\pi/n)=\sin(\pi/n)$ and introducing $\zeta:=\exp(2\pi\mathrm{i}/n)$, we get $$\zeta^{2k}+\zeta^{-2k}-4(\zeta^k+\zeta^{-k})+4(\zeta+\zeta^{-1})=2.$$ As a polynomial equation with rational coefficients, it holds with $\zeta$ replaced by $\zeta^m$ with $\gcd(m,n)=1$ (repeating my answers, write it as $P(\zeta)=0$ and observe that the polynomial $P$ has a common root $\zeta$ with the $n$-th cyclotomic polynomial, which is irreducible over the rationals, hence divides $P$).

Now we sum it over $1\leqslant m\leqslant n$ with $\gcd(m,n)=1$, using the formula $$\frac{1}{\varphi(n)}\sum_{\substack{1\leqslant m\leqslant n\\\gcd(m,n)=1}}\zeta^{km}=\rho(d):=\frac{\mu(d)}{\varphi(d)},\qquad d=\frac{n}{\gcd(n,k)};$$ denoting $d_1=n/\gcd(n,k)$ and $d_2=n/\gcd(n,2k)$, we get $\color{blue}{4\rho(n)-4\rho(d_1)+\rho(d_2)=1}$.

Observe that $|\rho(d)|\leqslant 1/10$ if $d>30$, and we can't have $\rho(d_2)=0$ (since otherwise $\rho(d_1)=\rho(n)=0$), so we have not too many values of $d_2$ to try. Enumeration shows that the last equality holds only if either

  1. $n=d_1=d_2=1$ corresponding to the trivial solution $n=1$;
  2. $n=d_1=2$ and $d_2=1$, corresponding to $n=2$ and $k$ odd;
  3. $n=30$ and $d_1=d_2=3$, impossible since $\sin(\pi/30)\neq 3/4$.
metamorphy
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