Small sumsets in a prime order group (Q1432026)

``Let \(A, B \subset {\mathbb Z}/p{\mathbb Z}\) such that \(3\leq | A| \), \(4\leq | B| \) and \(| A+B| =| A| +| B| \leq p-4\). Then \(A\) (resp. \(B\)) can be obtained from an arithmetic progression by deleting one element.'' This result is due to \textit{Ø. Rødseth} and the reviewer [cf. ``An inverse theorem mod \(p\)'', Acta Arith. 92, No. 3, 251--262 (2000; Zbl 0945.11003)]. The author gives an incorrect classification for subsets \(A,B\subset {\mathbb Z}/ p{\mathbb Z}\) with \(| A+B| =| A| +| B| +i\), where \(0\leq i \leq 1\). One may construct infinite families counter-examples to this classification. The author's theorem 4.1 implies that the above result of Rødseth and the reviewer holds under the weaker hypothesis \(| A| +| B| \leq p-2\). This claim contradicts the following example: Put \(p=11\), \(A=\{0,1,4\}\) and \(B=\{-4,0,1,3,4\}\). We have \(A+B=\{-4,-3,0,1,2,3,4,5\}\). We have \(| A+B| =8=| A| +| B| =p-3\). On the other side \(A\) is not a progression with one hole. To my opinion the author did not use enough powerful tools. In any case, the proofs obtained by the author are based on the following false claim formulated in Page 68, lines 27--31. I quote from the paper: ``Observe that lemmas (3.7), (3.16) and (3.18) when the set \(B\) has every gap except one of small length compared with the interval lengths of \(A\). The same method used in the proofs before can be applied to show that if the length of the gaps of \(B\) increases, then \(| A+B| \geq | A| +| B| +2.\)'' This claim is logically equivalent to the author's theorems and contradicts the above example.