Erdős Problem a Daythe open tracker

Erdős Problem #1185

SOLVEDdatabase updated 2026-04-04

Resolved in some way other than a straight proof or disproof — usual for open-ended questions, or for multi-part problems whose parts went different ways.

What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).

`HasAPWithStepInDiff k A B` says that `A` contains a non-trivial \(k\)-term arithmetic progression \(a, a+d, \ldots, a+(k-1)d\) whose common difference \(d>0\) lies in \(B-B\), that is, \(d=b_1-b_2\) with \(b_1,b_2\in B\). Subtraction on `ℕ` is truncated, but \(d>0\) forces \(b_2<b_1\).

Let \(\delta>0\) and \(k\geq 3\). Is it true that there exists \(m\geq 1\) (depending only on \(\delta\) and \(k\)) such that, for all large \(N\), if \(A,B\subseteq \{1,\ldots,N\}\) with \(\lvert A\rvert \geq \delta N\) and \(\lvert B\rvert \geq m\) then there is a non-trivial \(k\)-term arithmetic progression in \(A\) whose common difference is in \(B-B\)? A problem of Erdős and Mauldin [Er80, p. 92], who ask the case \(k=3\) and say that the same probably holds for \(k\) terms. The answer is no, already for \(k=3\): to disprove it suffices to give an infinite set \(B\) such that \(B-B\) is not \(2\)-intersective (a set \(S\) is \(k\)-intersective if every subset of \(\mathbb{N}\) with positive density contains an arithmetic progression of length \(k+1\) whose common difference is in \(S\)), which was provided by Furstenberg [Fu81]. The statement below is the negation of the claim for all \(\delta>0\) and \(k\geq 3\), with \(m\) chosen before the threshold \(N_0\) for "large \(N\)".
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 1 further variant / known partial result in the Lean file. The authoritative wording is on erdosproblems.com/1185.
Prizenone
Topicsadditive combinatorics, arithmetic progressions
Statement in LeanFormal Conjectures
Solution in Leanno
Original pageerdosproblems.com/1185
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