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We let $A,B\subseteq \mathbb{Z}$ such that $|A|=|B|=n$. I am trying to show that

$|A+B|\ge 2n-4$ for large $n$

where we define $A+B=\{a+b:a\in A, b\in B\}$.

If this is not true, I'd like to see a counter example. I am aware of the result that

$|A\widehat{+}B|\ge \min\{p,|A|+|B|-3\}$

where $\widehat{+}$ denotes a restricted sumset given subsets $A$ and $B$ of a finite abelian group $\mathbb{Z}_p$. I am quite sure this bound can be improved when $|A|=|B|$. I was wondering if I could perhaps take advantage of this result and assume that $p$ is large and assume that $A,B\subseteq \mathbb{Z}_p$ for large $p$. However, I am not sure how to proceed from this point forwards.

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    $\begingroup$ You can show that $ |A+B| \geq 2n-1$ directly. by finding $ 2n-1$ distinct values. $\endgroup$
    – Calvin Lin
    Commented Jan 31, 2022 at 3:18
  • $\begingroup$ As Calvin Lin says, a stronger bound is easy to prove directly for sets of integers, no need to go via a mod $p$ result. But in case you weren't aware, there is a stronger bound available for the full, unrestricted sumset, known as the Cauchy-Davenport inequality: $\lvert A+B\rvert \geq \min(p,\lvert A\rvert+\lvert B\rvert-1)$. (This does imply $\lvert A+B\rvert \geq \lvert A\rvert+\lvert B\rvert-1$ by taking $p$ large enough, but again this is proving the latter the long way round, no need to go via mod $p$ at all!) $\endgroup$
    – Thomas Bloom
    Commented Feb 4, 2022 at 15:44

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