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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. The answer to Problem 1185 is no, already for k=3k=3: there are δ>0\delta>0 and, for every mm, arbitrarily large NN with sets A,B⊆{1,…,N}A,B\subseteq\{1,\ldots,N\}, ∣A∣≥δN\lvert A\rvert\ge\delta N and ∣B∣≥m\lvert B\rvert\ge m, such that no nontrivial 33-term arithmetic progression in AA has its common difference in B−BB-B. The claimed input is the example in H. Furstenberg, Recurrence in ergodic theory and combinatorial number theory (M. B. Porter Lectures, Princeton University Press, 1981), pp. 177--178: an infinite set S⊆NS\subseteq\mathbb{N} whose difference set S−SS-S is a set of recurrence but not a set of 22-recurrence. Call T⊆NT\subseteq\mathbb{N} kk-intersective if every set of positive upper density contains a (k+1)(k+1)-term arithmetic progression with difference in TT; by Furstenberg's correspondence principle a set of kk-recurrence is the same as a kk-intersective set, so there is a set A′⊆NA'\subseteq\mathbb{N} of positive upper density 2δ2\delta with no 33-term progression whose difference lies in S−SS-S. Given mm, let BB be the mm smallest elements of SS; for infinitely many NN the set A=A′∩{1,…,N}A=A'\cap\{1,\ldots,N\} has at least δN\delta N elements, $B\subseteq {1,\ldots,N}$, and B−B⊆S−SB-B\subseteq S-S meets no difference of a 33-term progression in AA. This is the deduction the site's commentary records and the page restates in its own words; since a kk-term progression contains a 33-term progression with the same difference, the same AA and BB refute the question for every k≥3k\ge3. Erdős and Mauldin asked it as a question, motivated by a problem in measure theory ([Er80], p. 92, on the card erdos_1980_survey_problems_combinatorial_number_theory); its answer is no, so the claim is a disproof. Frantzikinakis, Lesigne and Wierdl (Ann. Inst. Fourier 56 (2006), 839--849, arXiv:math/0503367) locate Furstenberg's example on those pages and extend it to a set of kk-recurrence that is not a set of (k+1)(k+1)-recurrence for every kk; their explicit sets Sk={n:{nkα}∈[1/4,3/4]}S_k=\{n:\{n^k\alpha\}\in[1/4,3/4]\} (k≥2k\ge2, α\alpha irrational; their Theorem A) are not presented as difference sets, and Furstenberg's example alone already answers the question no for every k≥3k\ge3, so they are context here. The DOI linked above is the publisher's record of the book.

Formalization. Boris Alexeev's lean-proofs repository holds, since 2026-08-17, a Lean 4 development that declares itself a formalization of a solution, with Furstenberg as its informal author and Codex and GPT-5.6 Sol as its formal authors (src/latest/ErdosProblems/Erdos1185.lean, linked above at the pinned commit). Its theorem not_erdos_1185 shows that the universal statement fails at δ=1/200\delta=1/200 and k=3k=3: for every proposed mm and every cutoff there are NN beyond the cutoff and sets A,B⊆{1,…,N}A,B\subseteq\{1,\ldots,N\} with ∣A∣≥N/200\lvert A\rvert\ge N/200 and ∣B∣≥m\lvert B\rvert\ge m and no nontrivial 33-term progression in AA with difference in B−BB-B, built from a finite periodic form of Furstenberg's quadratic skew-shift example. This corpus has not built it, so no formalized evidence is listed.

Depends on. Nothing in this wiki.

Acceptance. Reviewed: the site's curator, Thomas Bloom, labels the problem solved, states that it is false already at k=3k=3, and credits Furstenberg's example [Fu81] for the infinite set whose difference set is not 22-intersective (page last edited 5 April 2026; empty discussion thread and proof-claims tab); he is independent of Furstenberg, and the deduction from the example to the finite statement is his own commentary. Not refereed: the source is a published monograph, not a journal article, and the deduction from it to the finite statement appears only in the site's commentary; the 2006 Annales paper that cites the example is refereed but is not the source of the claim. The example's property is stated as the commentary and the 2006 paper give it.