Consider $\mathbb{Z}_m^n$, an $n$-dimensional vector space over $\mathbb{Z}_m$. For two sets of vectors $P = \left\{ p^i \right\}$ and $Q = \left\{ q^j \right\}$ and a skew-symmetric matrix $S_{ij}=\mathrm{sgn}(i-j)$ (i.e., +1 above diagonal, -1 below, and 0 on diagonal) define a set $$ M = \left\{ p^i \pm q^j | \forall i,j \quad p^i \cdot S \cdot q^j \neq 0 \right\} $$ i.e., set of sums and differences of all pairs of vectors not orthogonal under the product induced by $S$.
I'm looking for a lower bound on size of $M$ as a function of size of $P$. Specifically, assume that for every $p^i$ there exist at least one $q^j$ such that $p^i \cdot S \cdot q^j \neq 0$ (so that we cannot increase the size of $P$ without affecting $M$). My conjecture is that for odd $m>2$ $$ \left| M \right| \geqslant \left| P \right|. $$ In my specific case, there is an additional condition on $P$ and $Q$: $$ \sum_k p^i_k=1\;\text{ and }\;\sum_k q^j_k = 0\quad\text{ for all }i, j. $$ I'm not sure if it is relevant to the problem. The conjecture is clearly wrong for $m=2$, due to the fact that sum and difference are the same and we have twice less elements to begin with.
For $M$ to be small, vectors in $P$ should be orthogonal to the most of vectors in $q$; ideally every $p$ is not orthogonal only to one vector in $q$; this by itself will give $2\left|P\right|$ elements. This means that $\left|Q\right|$ can't be too big: there won't be a vector orthogonal to almost all $q^j$ in this case.
Moreover, there should be "a lot" of pairs $(i_1, j_1)$ and $(i_2, j_2)$ such that $p^{i_1}+q^{j+1}=p^{i_2}+q^{j+2}$.
However it is not clear how to prove that this is impossible.
I've tried to run numerical experiments to find such sets, and didn't find any; my intuition says that the conjecture is true but I'm not sure about it. Finding counterexample also will be helpful.
