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Let and be the smallest integer such that for all . Prove that
for some .
Source: erdosproblems.com/981
An accepted solution exists. The statement is true.
PROVED, the site's label (page last edited 27 December 2025), credited to Elliott's 1969 paper [El69]; the accepted claim page is Elliott 1969, refereed in Indag. Math. and credited as the proof by the site's curator and by the introduction of a 2025 preprint (Tang and Zhang, arXiv:2512.24631v2, p. 2: "This conjecture was proved by Elliott [4]", the conjecture being Erdős's display (80) restated as their Conjecture 1.1). The paper's theorem (printed p. 165) is the two-sided form: for each with there is a constant with , where is the least with for every (p. 164); with this is the displayed asymptotic for in place of , and since every . The paper restates (80) as its display (1) with Erdős's one-sided , then replaces by and says of the one-sided form only that "Simple changes in the present argument yield a proof of a similar result for the earlier definition of " (p. 164); no adaptation is printed. The standing therefore rests on the printed theorem for the two-sided threshold, on the author's remark for the one-sided threshold of the page's wording, and on the readings of the site and of Tang and Zhang, who report having read the paper. The literal wording is not shown false and is not degenerate: every threshold is finite, the instances are established (Formulation above), and for the attested proof stands undisputed. Reopening condition: a reading that shows the announced adaptation to the one-sided threshold fails, or a dispute of the theorem.