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Let $\zeta_k$ be the first complex primitive $k$-th root of unity, i.e. $\zeta_k = e^{2\pi i/k}$. Then the last complex primitive $k$-th root of unity is given by $\zeta_k^{k-1} = \zeta_k^{-1} = e^{-2\pi i/k}$. Summing these two elements yields a real number, for the imaginary parts cancel. For which $k$ is is true that $\zeta_k + \zeta_k^{-1} \in \mathbb{Q}$? I suspect that the answer is, "only for $k=1,2,3,4$" but I can't seem to show that, or even to see why it's not true for larger $k$.

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Hint: The sum is an algebraic integer which, by virtue of being $2\cos(\frac{2\pi}{k})$, is in $[-2,2]$. How many rational, algebraic integers do you know in that range?

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I didn't know what an algebraic integer was but I just looked it up and I see that a) there are precisely five and b) this immediately illustrates why it's not true for higher $k$. Thank you. –  Julien Clancy Mar 28 '13 at 5:24
    
@JulienClancy No problem! If you're happy with my answer (i.e. you don't want a different take, etc.) you can upvote/accept it. –  Alex Youcis Mar 28 '13 at 5:27
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Hint:

$e^{i \theta}= \cos \theta+i \sin \theta$

$e^{-i \theta}= \cos \theta-i \sin \theta$

Note: When you add $e^{-i \theta}$ and $e^{i \theta}$, imaginary terms gets cancelled and only real terms remain.

$e^{-i \theta}+e^{i \theta}= 2\cos \theta$ , Now $\theta= \dfrac{2 \pi}{k}$

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