Union of Two Arithmetic Progressions with the Same Common Difference Is Not Sum-dominant

Hung Viet Chu

Published 2019-06-10Version 1

Given a finite set $A\subseteq \mathbb{N}$, define the sum set $$A+A = \{a_i+a_j\mid a_i,a_j\in A\}$$ and the difference set $$A-A = \{a_i-a_j\mid a_i,a_j\in A\}.$$ The set $A$ is said to be sum-dominant if $|A+A|>|A-A|$. We prove the following results 1) The union of two arithmetic progressions (with the same common difference) is not sum-dominant. This result partially proves a conjecture proposed by the author in a previous paper; that is, the union of any two arbitrary arithmetic progressions is not sum-dominant. The author is motivated by the anonymous referee's comment that the conjecture was marvelous and tantalizing. 2) Hegarty proved that a sum-dominant set must have at least $8$ elements with computers' help. The author of the current paper provided a human-verifiable proof that a sum-dominant set must have at least $7$ elements. A natural question is about the largest cardinality of sum-dominant subsets of an arithmetic progression. Fix $n\ge 16$. Let $N$ be the cardinality of the largest sum-dominant subset(s) of $\{0,1,\ldots,n-1\}$ that contain(s) $0$ and $n-1$. Then $n-7\le N\le n-4$; that is, from an arithmetic progression of length $n\ge 16$, we need to discard at least $4$ and at most $7$ elements (in a clever way) to have the largest sum-dominant set(s). 3) Let $R\in \mathbb{N}$ have the property that for all $r\ge R$, $\{1,2,\ldots,r\}$ can be partitioned into $3$ sum-dominant subsets, while $\{1,2,\ldots,R-1\}$ cannot. Then $24\le R\le 145$. This result answers a question by the author et al. in another paper on whether we can find a stricter upper bound for $R$. The previous upper bound was at least $888$.