Consider the $n$th roots of unity for $\frac{n}{2}$ odd. By an odd root we mean an odd power of $e^{\frac{2\pi i}{n}}$, and by an even root we mean any other root of unity. Denote the set of odd roots $O$ and even roots $E$. Since $n$ is even, as we go through all the roots as powers of $e^{\frac{2\pi i}{n}}$ starting with $1$, we alternate even, odd, even, odd... Is it possible for any such $n$ for the set $O+E:=\{o+e\ |\ o\in O,e\in E\}$ to additively generate a ring. That is, can the set of sums of equal numbers of odd and even roots of unity be a ring? It can be shown that this is equivalent to $-1$ being such a sum. It is not possible for all $n$, for example $n=6$. If it is possible, for which $n$?
Motivation (not necessary for the question): I am interested in what happens if you tile regular $n$-gons by joining them at the edges, for arbitrary $n\geq 3$, so that for $n\neq 3,4,6$ they intersect in their interiors. If we have an edge from $(0,0)$ to $(1,0)$, then we can add an $n$-gon at that edge, and then another at the edge containing $(0,0)$ just created by adding the last $n$-gon, and so on, until we add the $n$-gon on the other side of the original end, and loop around. When we consider the plane as $\mathbb{C}$ he edges passing through $(0,0)$ will clearly be $m$th roots of unity for some $m$ since we are just applying the same rotation again and again. For odd $n$, $m=2n$, for $n$ of the form $4k$, $m=n$ and the relevant case is for $n$ of the form $4k+2$, where we have odd $m=\frac{n}{2}$. If we look at the other vertex attached to one of these edges, we may add $n$-gons in the same manner, and obtain edges the same angles apart. In this way, we are simply adding the negative $m$th roots of unity to the vertex. For one of the new vertices, we add the $m$th roots, then their negatives, and so on. All vertices can be obtained through a sequence of such edges. For $n$ not of the form $4k+2$, the negatives are just the $m$th roots of unity since $m$ is even. Therefore the possible edges that can through a vertex are precisely the $m$th roots of unity added to that vertex. However, for $n$ of the form $4k+2$, we have $m$ odd, so that the negatives are not $m$th of unity, but rather the odd $2m$th roots of unity. So we can have edges consisting of $m$th roots of unity (even $2m$th roots of unity), odd $2m$th roots, or all the $2m$th roots of unity around each vertex. If the last case holds for one vertex, it holds for all of them. It will only hold if we can reach $(0,0)$ at an odd and an even (done) stage of our process of choosing edges. If we reach zero at an odd stage, we may obtain vertex $(-1,0)$ at an even stage. The set of vertices found at even stages, sums of equal numbers of even and odd roots of unity, are closed under multiplication, so this is equivalent to the question.
Long time since I thought about this, but it is impossible if and only if $\frac{n}{2}$ is prime power. If not it has two coprime factors $f,g$ and we can add the $f$ and $g$th roots of unity repeated until eventually all numbers of even roots of unity can sum to $0$, including some number and that number plus one. Rearranging we get a member of $O+E$ equal to an odd root, and multiply the latter to get $-1$, and doing the same to the former, we have what we want. We can see that having a sum of a number of even roots equal to a sum of one more even roots is equivalent to what we want, and in the prime power case, this paper can be used to show we can’t have that. Lam and Leung