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Let $k\geq 3$ and $g_k(N)$ be minimal such that if $A\subseteq \{1,\ldots,2N\}$ has $\lvert A\rvert \geq N+g_k(N)$ then there exist integers $b_1,\ldots,b_k$ such that all $\binom{k}{2}$ pairwise sums are in $A$ (but the $b_i$ themselves need not be in $A$).

Estimate $g_k(N)$.
Disclaimer: The open status of this problem reflects the current belief of the owner of this website. There may be literature on this problem that I am unaware of, which may partially or completely solve the stated problem. Please do your own literature search before expending significant effort on solving this problem. If you find any relevant literature not mentioned here, please add this in a comment.
A problem of Choi, Erdős, and Szemerédi. It is clear that, for the set of odd numbers in $\{1,\ldots,2N\}$, no such $b_i$ exist, whence $g_k(N)\geq 0$ always. Choi, Erdős, and Szemerédi proved that $g_3(N)=2$ and $g_4(N) \ll 1$. van Doorn has shown that $g_4(N)\leq 2338$.

Choi, Erdős, and Szemerédi also proved that\[g_5(N)\asymp \log N\]and\[g_6(N)\asymp N^{1/2}.\]In general they proved that\[g_k(N) \ll_k N^{1-2^{-k}}\]and for every $\epsilon>0$ if $k$ is sufficiently large then\[g_k(N) > N^{1-\epsilon}.\]As an example, taking $A$ to be the set of all odd integers and the powers of $2$ shows that $g_5(N)\gg \log N$.

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Formalised statement? No (Create a formalisation here)
Related OEIS sequences: Possible

Additional thanks to: Wouter van Doorn

When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:

T. F. Bloom, Erdős Problem #866, https://www.erdosproblems.com/866, accessed 2025-11-15
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  • I got interested in trying to find an explicit upper bound for $g_4(N)$. It should still be considered a work in progress, but what I have written so far can be found here, and gives an upper bound of $g_4(N) \le 2032$. If I find any improvements (or mistakes that change my currently claimed bound), or if I decide to put it on the arXiv, I'll be sure to update this comment.

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