Find all positive integer triples $(l,m,n)$ such that $\sin^2\frac{\pi}{n}+\sin^2\frac{\pi}{m}=\sin^2\frac{\pi}{l}$.
I have found the solutions $(m,m,1)$ for any $m\in\mathbb{Z}^+$, and also $(2,3,6),(2,6,3),(2,4,4),(4,6,6)$. It seems like these are the only ones (unless I missed a few other simple solutions), but I'm not sure how to prove it.
The article I found this problem was on cyclotomic polynomials, and the hint for this problem was to rewrite the equation in something that relates to cyclotomic polynomials, so that is what I tried to do:
Write the equation as $1-\cos^2\frac{\pi}{n}+1-\cos^2\frac{\pi}{m}=1-\cos^2\frac{\pi}{l}\rightarrow \cos^2\frac{\pi}{n}+\cos^2\frac{\pi}{m}=1+\cos^2\frac{\pi}{l}$. Now use $\cos^2x=\frac{1+\cos2x}{2}$ to rewrite the equation as $\cos\frac{2\pi}{l}+1=\cos\frac{2\pi}{m}+\cos\frac{2\pi}{n}$. Let $\omega_x=e^{i\frac{2\pi}{x}}$ for $x\in\mathbb{Z}^+$. Then $\frac{\omega_l+\overline{\omega_l}}{2}=\frac{\omega_m+\overline{\omega_m}}{2}+\frac{\omega_n+\overline{\omega_n}}{2}-1$. I do not know how to proceed from here.
Any ideas? A solution with cyclotomic polynomials would be nice!