Does $\sin^2{\frac{k\pi}{n}}=\sin{\frac{\pi}{n}}$ have infinitely many integer solutions? 
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.) 
 A: 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


*

*$n=d_1=d_2=1$ corresponding to the trivial solution $n=1$;

*$n=d_1=2$ and $d_2=1$, corresponding to $n=2$ and $k$ odd;

*$n=30$ and $d_1=d_2=3$, impossible since $\sin(\pi/30)\neq 3/4$.

