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.

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1The exponents are all the quadratic residues modulo $17$. So, by Galois theory, the sum is, indeed, an element of a quadratic field. Let $S$ be your sum. Compute $S^2$. You can write it in terms of $S$ and an integer using the minimal polynomial of $\zeta_{17}$. Then you have a quadratic equation and can solve for $S$. – Jyrki Lahtonen Apr 09 '20 at 05:26
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Lol, I'm actually building that minimal polynomial. This would be the coefficient of degree 1 – Jos van Nieuwman Apr 09 '20 at 05:27
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3I used the (standard) idea in this old answer. Then the root of unity was of order seven. You have seventeen, so you get a little bit more tems, but the principle is the same. – Jyrki Lahtonen Apr 09 '20 at 05:27
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1The minimal polynomial of $\zeta_{17}$ reads $$\sum_{j=-8}^8\zeta_{17}^j=0.$$ When you calculate $S^2$, the missing exponents also appear. – Jyrki Lahtonen Apr 09 '20 at 05:29
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Sorry, but what do you take for $S$? – Jos van Nieuwman Apr 09 '20 at 05:35
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1You take the above thing you want to reduce as $S$. Don't worry when you calculate $S^2$, because you can see a pattern in this. – Sarvesh Ravichandran Iyer Apr 09 '20 at 05:42
2 Answers
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,
$$\zeta_{17}+\zeta_{17}^{-1}+\zeta_{17}^{2}+\zeta_{17}^{-2}+\zeta_{17}^{4}+\zeta_{17}^{-4}+\zeta_{17}^{8}+\zeta_{17}^{-8}=\frac{\sqrt{17}-1}2$$

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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 $$a=\frac{-1+\sqrt{17}}{2}.$$

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