Let $A, B \subseteq \mathbb{Z}$ be finite subsets of the integers. Then there exists an element in $A+B$ with a unique representation as a sum of an element in $A$ and an element in $B$, namely $\max(A+B)$ (or $\min(A+B)$).


If $A, B \subseteq \mathbb{Z}\_m$ with $|A+B| < |A|+|B|-1$ or $|A|+|B| \geq m+2$ then $r\_{A+B}(x) \geq 2$ for all $x \in A+B$.


So, suppose $A, B \subseteq \mathbb{Z}\_m$, $|A+B| \geq |A| + |B| -1$ and $|A|+|B| < m+2$, under which conditions can we expect the existance of an element having unique representation in $A+B$?


Clearly, if $A$ and $B$ are union of cosets of the same subgroup $H$ of $\mathbb{Z}\_m$, such an element cannot exist, so I am interested in sets $A$ and $B$ with low density in every coset, say $|A \cap x+H| < |H|/2$ for all $x \in \mathbb{Z}\_m$ and subgroup $H$ of $\mathbb{Z}\_m$.



Element with unique representation in A+B
First, a technical remark: you do not need to assume $|A|+|B|<m+2$ as this is implied by your first assumption $|A+B|\ge|A|+|B|-1$, and indeed you do not need to assume $|A+B|\ge |A|+|B|-1$ either since otherwise, as you mention, there are no elements with a unique representation in $A+B$.


To the essence of your question. This problem has been studied in a number of papers. The latest I am aware of is *Unique Sums and Differences in Finite Abelian Groups* by K. H. Leung and B. Schmidt, available [here](http://www.ntu.edu.sg/home/Bernhard/Publications/pub/UniqueSums.pdf). It contains a number of further references. In brief, the results known are of the following sort: if $A$ and $B$ are "small" in some sense, then a uniquely representable element exists.




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### Added April 03, 2015


It may be worth adding that there is some hidden philosophy behind this question. It is a well-known phenomenon that combinatorial problems in abelian groups are much easier to handle in the case where the underlying group is torsion-free. Interestingly, it seems that the crucial property of torsion-free groups responsible for this phenomenon is the unique representation property: for any finite, non-empty subsets $A$ and $B$ of a torsion-free abelian group, there exists a group element $g$ uniquely representable as $g=a-b$ with $a\in A$ and $b\in B$. Here are some examples in support of this claim.


(1) If there is an element with exactly one representation as $a+b$ with $a\in A$ and $b\in B$, then $|A-B|\ge|A|+|B|-1$. This is immediate by observing that $|(A-b)\cap(a-B)|=1$ and that both $A-b$ and $a-B$ are contained in $A-B$. Of course, the inequality $|A-B|\ge|A|+|B|-1$ need not hold in general, and when it holds under some assumptions, this may be not easy to prove.


(2) As the OP mention in his question, if there is an element with exactly one representtion as $a+b$ with $a\in A$ and $b\in B$, then $|A+B|\ge|A|+|B|-1$. This is a basic (but non-trivial) result by Kemperman and Scherk. 


(3) It is known that for any finite integer set $A$ and any integer $h\ge 2$, one has the following inequality relating the sizes of the $(h-1)$-fold and the $h$-fold sumsets of $A$:
 $$ \frac{|hA|-1}h\ge\frac{|(h-1)A|-1}{h-1}. \tag{$\ast$} $$
It is not clear whether and to which extent this inequality stays true in the groups other than the group of integers; see [the paper](http://www.cs.elte.hu/~gykati/publ/2A3A.ps) *Double and triple sums modulo a prime* by Gyarmati, Konyagin, and Ruzsa for some partial results in this direction. Interestingly, it can be shown that ($\ast$) holds true under the assumption that the sumset $(h-1)A$ has the unique representation property.


