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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. For q∈(0,1)q\in(0,1) let Gq={qn:n≥1}G_q=\{q^n:n\ge1\}. For every q∈(0,1)q\in(0,1) and every η∈(0,1)\eta\in(0,1) there is a compact set E⊆[0,1]E\subseteq[0,1] of Lebesgue measure greater than 1−η1-\eta such that for every x∈Rx\in\mathbb{R} and every s∈R∖{0}s\in\mathbb{R}\setminus\{0\} some n≥1n\ge1 has x+sqn∉Ex+sq^n\notin E; that is, EE contains no set aGq+baG_q+b with a≠0a\ne0, for either sign of aa. This is Theorem 1.1 of The geometric case of the Erdős similarity conjecture, a manuscript of the OpenAI mathematics release dated 5 October 2026, whose author line is "OpenAI" and whose release README says the manuscripts were produced by an internal OpenAI model and that the collection includes results at different stages of verification; the claimant is the organization. The statement is compiled on the result page theorem_1_1 and the digest on the card openai_2026_geometric_case_erdos_similarity_conjecture. The set depends on qq as well as on η\eta, and the manuscript says that the theorem makes no simultaneous assertion for different ratios. The proof reduces the theorem to Proposition 2.1, a periodic hitting statement: for a fixed qq and every p∈(0,1)p\in(0,1) an open 11-periodic set of density at most 6p6p meets x+tGqx+tG_q for every real xx and every t∈[1,2]t\in[1,2]. The proposition is proved by a random routing construction on nested dyadic grids whose resolution at index bb is comparable to q−b/(1−q)q^{-b}/(1-q), a finite ordered tree carrying index windows, independent local tests at a stable center, a finite set of scale representatives, and an open-neighborhood repair of the exceptional centers; the hypothesis q<1q<1 enters through the grid separation and the decay qn→0q^n\to0. At q=1/2q=1/2 the theorem is the dyadic case, which the release also states and formalizes separately on its own claim page; this manuscript does not cite that one. Since Problem 120 asks for an avoiding set for every infinite AA, the theorem is the case of a geometric progression with a fixed ratio and no more.

Covers. The question, answered yes for each set A={qn:n≥1}A=\{q^n:n\ge1\} with a fixed ratio q∈(0,1)q\in(0,1), one ratio at a time: for every η∈(0,1)\eta\in(0,1) a compact E⊆[0,1]E\subseteq[0,1] of Lebesgue measure greater than 1−η1-\eta, so positive, that contains no set aA+baA+b for any real bb and any nonzero aa of either sign. The theorem says nothing about an infinite set that is not a geometric progression, including exponentially decaying sequences that are not geometric progressions (the survey's example 2−n2^{-n} is the case q=1/2q=1/2 claimed here) and the Cantor sets the surveys name as open, and nothing about the general question. Sets containing a nonzero affine copy of some GqG_q follow by a one-line argument that is not in the manuscript.

Standing. The manuscript has no refereed publication and no arXiv version, and no reviewer independent of the claimant has endorsed it; the formalization described below covers the ratio q=1/2q=1/2 only. The site's commentary names the dyadic sequence as an open case of the conjecture and its proof-claims tab carried no claim for this problem. The claim therefore stays claimed. Read depth: the theorem statement, clause by clause, on the result page; no proof step was checked.

Formalization. Only the instance q=1/2q=1/2 is formally verified, while this page claims every q∈(0,1)q\in(0,1). The release's Lean catalog lists no formalization of this manuscript, and its family page describes a formalization of the dyadic manuscript only, so no formalization link is carried here. That formalization's declaration OAI.Problem310.dyadic_affine_avoidance, built by this corpus's verification at the pinned revision with the toolchain leanprover/lean4:v4.34.1, with axioms exactly propext, Classical.choice and Quot.sound and a fingerprint identical to its comparator challenge, is accepted on the dyadic claim page. It fixes the ratio through dyadicPoint n = (2:ℝ)⁻¹ ^ n and has no ratio parameter: for every η∈(0,1)\eta\in(0,1) it gives a compact E⊆[0,1]E\subseteq[0,1] of Lebesgue measure greater than 1−η1-\eta containing no copy x+s{2−n:n≥1}x+s\{2^{-n}:n\ge1\} with s≠0s\ne0, which is this page's Claim at q=1/2q=1/2 clause by clause, and it says nothing about any other ratio. It is therefore not formal evidence for this page's claim, and no formalized evidence is listed.

Depends on. Nothing beyond the cited manuscript.