$\newcommand{\Z}{\mathbb Z/p\mathbb Z}$
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 $a',a''\in A$, and the union is disjoint?


As stated, the answer is "yes", at least if the order of the group is $p\equiv 2\pmod 3$, in which case one can take $A$ to be an appropriately located interval of an appropriate length: namely, $A=[n,2n-1]$ where $n=(p+1)/3$. One also can dilate $A$ replacing intervals with arithmetic progressions.


Are there any other examples where $A$ and $A-A$ partition the whole group?



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> Suppose that $A\cup(A-A)$ is a partition of a group of prime order; does it follow that $A$ is an arithmetic progression?
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**Added two days later.** A set found by Peter Mueller in the comments can be generalized to produce infinitely many counterexamples, essentially answering the original question. Specifically, for a prime $p\equiv 5\pmod 8$ let $m:=(p+3)/8$ and define $A:=[-(2m-1),-m]\cup[m,2m-1]$. It is easily verified that then $A-A=[-(m-1),m-1]\cup[2m,p-2m]$, so that $A-A$ is disjoint from $A$, and $A\cup(A-A)$ is the whole group, while $A$ is not an arithmetic progression.



$\mathbb Z/p\mathbb Z=A\cup(A-A)$?
*This is a previous comment which was moved to [chat](https://chat.stackexchange.com/rooms/141236/discussion-on-question-by-seva-mathbb-z-p-mathbb-za-cupa-a) by Ben Webster:* In fact for every prime $p$ if $A=[-(2m-1),-m] \cup [m, 2m-1]$ for $(p+3)/8\le m<(p+3)/6$, then $\mathbb Z/p\mathbb Z$ is a disjoint union of $A$ and $A-A$, and in addition this set $A$ is not an arithmetic progression if and only if $m\ne(p+1)/6$. There are many more examples though. One peculiar sporadic one happens for $p=41$ with $A$ the multiplicative subgroup of order $10$ of $(\mathbb Z/p\mathbb Z)^\star$.


