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Are there solutions to $$\sin(x+y)\sin(x-y)=n\ \sin(x)\sin(y)$$ for $n\ge 3$ where $x$ and $y$ are rational multiples of $\pi$? (excluding the trivial solutions when both sides are $0$).

Known solutions are: $$n=1:\quad x=\frac{2}{5}\pi\quad y = \frac{1}{5}\pi$$ $$n=2:\quad x=\frac{3}{8}\pi\quad y = \frac{1}{8}\pi$$

I've reworked the equation a few times using trig identities with no luck. I used identities like $$3\sin(x)=\sin(3x)-4\sin^3(x)$$ $$2\sin(x)=\sin(x-\pi/4)+\sin(x+\pi/4)$$ $$2^k\sin(x)=\sin(2^kx)\prod_{i=0}^{k-1}\cos(2^ix)$$ One can work out that $$x=\sin^{-1}\Big[\frac{1}{2}\sin(y)(n\pm\sqrt{n^2+4})\Big]$$ But this was of no help either.

My last attempt was to use the identity $$a\sin(x)+b\sin(x)=\sqrt{a^2+b^2}\sin(x+a\tan 2(a,b))$$ to show that there is no solution when $n$ cannot be expressed as the sum of 2 squares. But I was unable to find an workable expressions for $\text{atan2}(a,b)$.

The problem is originally about ratios of diagonals of polygons. Such ratios may be written with the roots of unity as $$r=\Big|\frac{1-\zeta_c^a}{1-\zeta_d^b}\Big|$$ We wish to find integers $a,b,c,d$ such that $r$ satisfies $r-r^{-1}=n$ for some positive integer $n$. In the case $n=1$, we are looking for the golden ratio. The relationship to the former trig equations is that $$r=\frac{\sin(\pi\frac{a}{c})}{\sin(\pi\frac{b}{d})}$$

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