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If is a set of integers such that
for all then must be subcomplete? That is, must
contain an infinite arithmetic progression?
Source: erdosproblems.com/344
An accepted solution exists. The statement is true.
Proved, the site's label (PROVED; page last edited 28 December 2025, accessed 2026-10-07). Szemerédi and Vu [SzVu06] proved that there is an absolute constant such that every increasing sequence with for all is subcomplete, which answers the question yes in the reading the Formulation records; their first proof is in the Annals of Mathematics (2006), and the cited paper gives a second, shorter proof. The refereed papers and the site's acceptance are recorded on the claim page (Szemerédi and Vu, 2005). Y.-G. Chen proved the same theorem in Acta Arithmetica (2003) by a different method; it is recorded on his claim page (Chen, 2003) and is not credited by the site. Folkman [Fo66] had proved the statement under the stronger hypothesis , a refereed partial result recorded on his claim page (Folkman, 1966).