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A lower bound concerning subsets sums which do not cover all the residues modulo \(p\). (English) Zbl 1222.11037

Summary: Let \(p\) be a prime number, \(c>\sqrt{2}\) a positive constant, and \({\mathcal A}\) a subset of \(\mathbb{Z}/p \mathbb{Z}\), with cardinality larger than \(c\sqrt{p}\), such that its subset sums do not cover \(\mathbb{Z}/p\mathbb{Z}\). Then there is an isomorphic shift of this subset, which is concentrated near the beginning or the end of the residue classes mod \(p\). More precisely, it was proved in [J.-M. Deshouillers and G. A. Freiman, J. Number Theory 104, No. 2, 255–262 (2004; Zbl 1048.11077)] that there exists \(s\) prime to \(p\)p such that \[ \sum_{a\in{\mathcal A}}\left\|\frac{as}{p}\right\|<1+O(p^{-1/4}\ln p), \]
the constant in \(O\) depending at most on \(c\). Deshouillers and Freiman conjectured that the upper bound for the error term may be replaced by \(O(p^{-1/2})\). In the paper under review, the author proves that for given \(c\), \(\sqrt{2}<c<2\), there exists a positive constant \(K\), such that for a sufficiently large prime number \(p\), there exists a subset \({\mathcal A}\) of cardinality larger that \(c\sqrt{p}\), such that its subset sums do not cover all residue classes, and for every \(s\) prime to \(p\), one has
\[ \sum_{a\in{\mathcal A}}\left\|\frac{as}{p}\right\|>1+Kp^{-1/2}. \]
The proof is elementary, and such a subset is constructed explicitly.

MSC:

11B75 Other combinatorial number theory
11B13 Additive bases, including sumsets

Citations:

Zbl 1048.11077
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