Status
On this page
Status
Topics
Status
On this page
Status
Topics
If has then must contain arbitrarily long arithmetic progressions?
Source: erdosproblems.com/3
An accepted solution exists. The statement is true.
OPEN, the site's label (page last edited 4 April 2026, as accessed 2026-09-04). The OpenAI mathematics release of 23 September 2026 answers the question yes. Its manuscript states for every fixed (Theorem 1.1) and sums that bound over dyadic intervals (Corollary 1.2). The release's Lean proves the yes answer from a weaker bound, . This corpus built that declaration, checked its axioms and found it identical to the release's comparator challenge, so the result is accepted on its claim page (OpenAI, 2026) and the problem stands solved and proved here. Theorem 1.1's bound is not formalized; it is a claimed partial result on Problem 142's claim page. The case , proved by Bloom and Sisask, is a claimed partial result on its own claim page (Bloom and Sisask, 2020), and the case the set of primes, proved by Green and Tao, is an accepted partial result on its own claim page (Green and Tao, 2004).