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Let be minimal such that for any with and any set of consecutive integers there exists some with such that
Is it true that
Or perhaps even ?
Let be minimal such that for any with and any set of consecutive integers there exists some with such that
Is it true that
Or perhaps even ?
Source: erdosproblems.com/708
No claim settles this problem.
Open, the site's label; both questions of the corrected Statement are open. A partial proof claim of 5 September 2026 gives explicit linear upper bounds in the at-most convention, for every with a Lean development its author reports kernel-checked, and the conjectured for sets with ; it is pending on its claim page (Chen, 2026), unreviewed outside its author's pipeline and not built here, and it leaves both displayed questions, and , unresolved. The site's exact-size wording admits no for large , an observation that answers only the printed wording (Notes).
Read as printed, the condition asks for a subset of exactly elements in every instance, and then no exists once is large: for and the interval has only elements, which caps at , while the 1959 lower bound gives instances that need more than selected elements, so neither displayed bound holds as worded (Progress). The change replaces by ; nothing else changes. The evidence is the convention the sources assume: Erdős and Surányi (1959), section 1, p.39, fixes the interval at the largest given integer and asks how many of its integers must be selected, and section 9, p.44, and the summaries, pp.47-48, bound that selected count from below by ; Erdős (1992), p.34, uses the same loose exact-number wording for the least number of integers one can find, and its equation (1) is the same lower bound on that count. Lech Mazur posted the exact-size failure in comment 6301, 6 May 2026, attributing the observation to GPT-5.5 Pro. The page's standing judges the corrected Statement.