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$\mathbb Z/p\mathbb Z=A\cup(A-A)$?
$\newcommand{\Z}{\mathbb Z/p\mathbb Z}$
Can one partition a group of prime order as $A\cup(A-A)$ where $A$ is a subset of the group, $A-A$ is the set of all differences $a'-a''$ with $a',a''\in A$, ...