Let $u$ be a integer which has an odd prime $m$ as a divisor. So we let $u=u_1m$ where $u_1=\frac{u}{m}$.
Consider the set of $u$th roots of unity, i.e., $A=\{e^{j\frac{2\pi}{u}n} | 0 \leq n \leq u-1\}=\{e^{j\frac{2\pi}{u_1m}n} | 0 \leq n \leq u_1m-1\}$.
For convenience, let $e^{j\frac{2\pi}{u_1m}n}=x_n$.
And I am interested in the following summation:
\begin{equation} \sum_{i=0}^{m-1}x_{f(i)}=0 (1) \end{equation} where $f(i)$ is an integer-valued function. That is, I want to choose $m$ elements in $A$ such that the summation becomes zero (either allowing repetition or not).
I guess, to do that, I have to choose $m$ 'distinct' elements in $A$ that are at a distance $u_1$ from each other. In other words,
\begin{equation} \{x_0, x_{u_1-1}, x_{2u_1-1}, \cdots, x_{(m-1)u_1-1}\} (2) \end{equation} which is the same as the summation of all $m$th roots of unity that is trivially equal to zero.
Or \begin{equation} \{x_1, x_{u_1}, x_{2u_1}, \cdots, x_{(m-1)u_1}\} (3) \end{equation} also makes the summation zero because (3) is only the phase-rotating version of (2).
My question: I think there is no other patterns that make the summation (1) zero. Am I right?
Thank you in advance.
Unfortunately your intuition is incorrect, for kind of an annoying reason: you can find smaller sets of elements, of cardinalities $k$ and $m-k$, which individually sum to $0$.
For example, take $u=30$ and $m=5$. Then $x_0+x_{15}=0$ and $x_1+x_{11}+x_{21}=0$, and so the set $\{x_0,x_1,x_{11},x_{15},x_{21}\}$ has the right sum but not the nice form you hoped for.