Evaluate sum roots of unity $\zeta_{17}+\zeta_{17}^{-1}+\zeta_{17}^{2}+\zeta_{17}^{-2}+\zeta_{17}^{4}+\zeta_{17}^{-4}+\zeta_{17}^{8}+\zeta_{17}^{-8}$

Question

For $\zeta_{17}$ a primitive $17^{\text{th}}$ root of unity, I need to reduce $\zeta_{17}+\zeta_{17}^{-1}+\zeta_{17}^{2}+\zeta_{17}^{-2}+\zeta_{17}^{4}+\zeta_{17}^{-4}+\zeta_{17}^{8}+\zeta_{17}^{-8}$ into a simpler form. I believe it should be an integer actually. I just can't work it out, even using complex exponentials. There must be a trick I don't know about.

Answer 1

Let $a=\frac\pi{17}$ and

$$\begin{align} & S = \cos2a+\cos4a+\cos8a+\cos16a \\ & T = \cos6a+\cos10a+\cos12a+\cos14a \\ \end{align} $$

Then, we have $$\zeta_{17}+\zeta_{17}^{-1}+\zeta_{17}^{2}+\zeta_{17}^{-2}+\zeta_{17}^{4}+\zeta_{17}^{-4}+\zeta_{17}^{8}+\zeta_{17}^{-8}=2S$$

Evaluate $$\begin{align} & 2\sin a( S+T )\\ & = 2\sin a( \cos2a+\cos4a+\cos6a+\cos8a+\cos10a+\cos12a+\cos14a+\cos16a ) \\ & = \sin17a-\sin a = 0-\sin a = -\sin a \\ \end{align}$$

which leads to $$S+T= -\frac12$$

On the other hand, it can also be evaluated, albeit more involved, that

$$ST = -1$$

Therefore, $S$ satisfies the quadratic equation

$$S^2+\frac12S-1=0$$

which yields $S=\frac{\sqrt{17}-1}4$. Thus,

$$\zeta_{17}+\zeta_{17}^{-1}+\zeta_{17}^{2}+\zeta_{17}^{-2}+\zeta_{17}^{4}+\zeta_{17}^{-4}+\zeta_{17}^{8}+\zeta_{17}^{-8}=\frac{\sqrt{17}-1}2$$

Answer 2

Let $\zeta_{17}+\zeta_{17}^{-1}+\zeta_{17}^{2}+\zeta_{17}^{-2}+\zeta_{17}^{4}+\zeta_{17}^{-4}+\zeta_{17}^{8}+\zeta_{17}^{-8}=a$ and

$\zeta_{17}^3+\zeta_{17}^{-3}+\zeta_{17}^{5}+\zeta_{17}^{-5}+\zeta_{17}^{6}+\zeta_{17}^{-6}+\zeta_{17}^{7}+\zeta_{17}^{-7}=b.$

Thus, $$a+b+1=0$$ and calculate $ab$.

I got $$ab=-4.$$

Now, easy to check that $a>0$, which since $$a^2+a-4=0,$$ gives $$a=\frac{-1+\sqrt{17}}{2}.$$