Pham 2024 sharp bound erdos straus non averaging

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theorem_1: The sharp upper bound for non-averaging sets, which also bounds non-dividing sets from above.

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Huy Tuan Pham, Dmitrii Zakharov, Sharp bound for the Erdős-Straus non-averaging set problem. Geom. Funct. Anal. 35 (2025), no. 6, 1712--1738, DOI 10.1007/s00039-025-00728-8 (published online 3 December 2025; Crossref record read). arXiv:2410.14624. The arXiv record names arXiv's non-exclusive distribution license (arXiv:2410.14624), every other right reserved; that is the term of the arXiv v2 read here. The journal's version of record is under the Creative Commons Attribution 4.0 license: its Crossref record (https://api.crossref.org/works/10.1007/s00039-025-00728-8, read 2026-10-07) deposits https://creativecommons.org/licenses/by/4.0 for the version of record from 3 December 2025.

Pham and Zakharov prove that a non-averaging subset of {1,...,n} (no element is the average of a nonempty subset of the other elements) has at most n^{1/4+o(1)} elements; with Bosznay's construction this fixes the maximum at n^{1/4+o(1)} and settles the Erdos-Straus non-averaging set problem. For each fixed d they also find how large a non-averaging set inside a d-dimensional box can be. The proof combines the subset-sums structure theorem of Conlon, Fox and Pham with a result on point sets in nearly convex position. The copy read for this card is arXiv:2410.14624v2 (10 September 2025, 20 pp.), whose pagination is used here; the journal text was not compared. Read status: claims checked for the definition of a non-averaging set and Theorem 1 (p. 2, |A| <= n^{1/4+o(1)} for every non-averaging A in [n], so h(n) = n^{1/4+o(1)}), read in the text layer; the proof was not read. Result page: theorem_1. For problem 789 it is context only: it resolves the sibling non-averaging problem (Problem 186; arXiv 2024, Geom. Funct. Anal. 2025) and states no bound for the different quantity in 789.

Source: https://arxiv.org/abs/2410.14624.

Bears on. #789, #131 (a set in which no element divides the sum of any distinct other elements is non-averaging, so Theorem 1 gives $F(N)\le N^{1/4+o(1)}$ for the problem's $F(N)$, answering its displayed question $F(N)>N^{1/2-o(1)}$ in the negative; the order of $F(N)$ stays open), #186: Theorem 1 (p. 2 of arXiv v2, text layer), $|A|\le n^{1/4+o(1)}$ for every non-averaging $A\subseteq[n]$, with Bosznay's $h(n)=\Omega(n^{1/4})$ recalled on p. 1, gives $h(n)=n^{1/4+o(1)}$; the paper's non-averaging condition is the problem's (a one-element subset averages to itself, so the problem's "at least two" and the paper's "average of a nonempty subset of $A$ not containing $a$" (p. 1) agree), and $h(n)$ is the problem's $F(N)$, whose order of growth is thereby determined up to the $o(1)$ in the exponent (theorem_1).

No file of this source is held: the arXiv v2 read here carries no license that permits its redistribution, and the journal's version of record, under CC BY 4.0, was not read; page numbers are those of the arXiv v2.