# Can the solutions to this beautiful equation always be expressed in terms of algebraic numbers?

For $$n\geq1$$, the largest solution to this lovely equation is a local extremum on a function related to the Fibonacci sequence:

$$\sum_{k=1}^{n} k{(-1)^{k}} \cdot \frac{\sin(\frac{k\pi}{x} )}{3+2\cos(\frac{k\pi}{x} )} = 0$$

For $$n=1$$, the largest solution is $$1$$. For $$n=2$$, the largest solution is:

$$\frac{\pi}{2\arctan(\frac{\sqrt{4\sqrt{10}-5}}{3})}$$

Can it be proven that for any integer $$n\geq2$$, the solution to the above equation can be expressed as

$$\frac{\pi}{2\arctan(A)}$$

where $$A$$ is algebraic? If so, what are some minimal polynomials for $$A$$ when $$n\geq3$$?

• Beauty sure is in the eye of the beholder. Mar 12, 2021 at 22:52
Yes, you can write it in terms of an algebraic number as you desire. Both $$\sin\frac{k\pi}{x}$$ and $$\cos\frac{k\pi}{x}$$ can be written as polynomials in $$\cos\frac{\pi}{x},\sin\frac{\pi}{x}$$, which can in turn be written as rational functions in $$\tan\frac{\pi}{2x}$$. It follows that for any solution $$x$$ of your equation, $$A=\tan\frac{\pi}{2x}$$ will be algebraic. But since $$x=\frac{\pi}{2\arctan A}$$, this is your desired result.
From this method you can easily figure out a polynomial of which $$A$$ is a root of, but I doubt there is an easy way to find its minimal polynomial.