TITLE: Can a sum of roots of unity be an integer?
QUESTION [13 upvotes]: Let $n \geq 2$, $H \lneq (\mathbb{Z}/n\mathbb{Z})^*$, $\zeta_k$ a primitive $k$-th root of unity. Is it possible that $$\sum_{h \in H} \zeta_k^{h} \in \mathbb{Z}$$ for every $k$ dividing $n$ such that $n/k \leq |H|$?
I do not know the answer even if $k=n$. That is, can it be that $$\sum_{h \in H} \zeta_n^{h} \in \mathbb{Z}$$

REPLY [12 votes]: We suppose towards a contradiction that all the sums in the first question are integers. Set $$f(x) = \prod_{h \in H}(x - \zeta_n^h)$$
We claim that all symmetric polynomials in the roots of $f$ are integers, for which it is enough to show that for each $1 \leq d \leq |H|$ $$\sum_{h \in H} \zeta_n^{dh} \in \mathbb{Z}$$ but this is true by our assumption since $\zeta_n^d = \zeta_k$ for some $k$ dividing $n$ with $n/k \leq |H|$ from the inequality for $d$. This shows that $f \in \mathbb{Z}[x]$ is of degree smaller than $\phi(n)$ (since $H$ is proper) having $\zeta_n$ as a root, contradiction.