This is a problem I played around with several years ago. I proved a few things, but the proofs were long and messy and not worth reproducing here.
Given a positive integer $n$ and a set $S$ such that $S \subset Z_n$, define a focal point of $n$ and $S$ to be any $a \in Z_n$ such that for each $x \in S$, there exists $y \in S$ such that $x+y \equiv a \mod n$. It is possible to have $x = y$.
Example 1. $n=15$, $S_1=\{1,2,6,7,10,11,12,13\}$. Then $n$ and $S_1$ have the focal point $8$ because $1+7 = 2+6 = 10+13 = 11+12 = 8$. By exhaustion one can show that there are no others.
Example 2: $n=15$, $S_2=\{1,2,6,7,11,12,13\}$. These have no focal points, even though $S_2$ is simply $S_1$ with $10$ removed. In fact, removing any element from $S_1$ results in a set that has no focal points with $n$.
Example 3: $n = 15$, $S_3=\{1,2,6,7,11,12\}$. $S_3$ is obtained by removing both $10$ and $13$ from $S_1$. $S_3$ and $n$ have 3 focal points: $3, 8$, and $13$.
Let $F(n,S)$ be the set of focal points of $n$ and $S$. My questions are these:
- What can we say about the size of $F(n,S)$? I was able to prove that it must be $0$ or a divisor of $n$, but surely one can say more.
- Even better, is there a nice way to characterize $F(n,S)$?
My hunch is that the right algebraic structure will make what's going on here crystal clear. Unfortunately, my algebra is quite rusty.
Added: One thing to observe is that in the case of Example 3, $S_3$ splits nicely into residue classes $\bmod 5$: $\{1, 6, 11\}$ and $\{2, 7, 12\}$. If $n_4 = 5$ and $S_4 = \{1,2\}$, then clearly $n_4$ and $S_4$ have one focal point: $3$. Surely it's not a coincidence, then, that $n_3 = 15$ and $S_3=\{1,2,6,7,11,12\}$ have 3 focal points. So perhaps there's a characterization that has to do with when $S$ can be divided into residue classes modulo a divisor of $n$?