Let $q=p^r$ be an odd prime power. Let $\epsilon \in \Bbb C$ be a primitive $(q-1)$-th root of unity and let $d\in \Bbb Z_{\ge 1}$.
I want to prove that $\sum\limits_{k=1}^{\frac{q-3}{2}} (-1)^k (\epsilon^{kd}+\epsilon^{-kd})$ is an integer.
I tried to prove it for $d=1$:
$\sum\limits_{k=1}^{\frac{q-3}{2}} (-1)^k (\epsilon^{k}+\epsilon^{-k}) = \sum\limits_{k=1, k\ even}^{q}\epsilon^{k}-\sum\limits_{k=1, k\ odd}^{q}\epsilon^{-k} = 0$ which follows from the fact that $\sum\limits_{k=0}^{q-2} \epsilon^k =0$.
But how to prove that it is an integer for every $d\in \Bbb Z_{\ge 1}$.
Many thanks for your help.