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The Erdos similarity conjecture for geometric progressions

A set A⊆RA\subseteq\mathbb R is measure universal if every measurable set of positive Lebesgue measure contains an affine copy x+sAx+sA with s≠0s\ne0. Every finite set is universal. Erdos (1974, Problem 4.33.7*) conjectured that no infinite set is. Falconer and Eigen proved it for sequences an→0a_n\to0 with an+1/an→1a_{n+1}/a_n\to1, and later criteria (Kolountzakis, Humke-Laczkovich, Chlebik) do not apply to geometric progressions, so even the dyadic sequence {2−n}\{2^{-n}\} remained open. Is it true that for every q∈(0,1)q\in(0,1) the progression {qn:n≥1}\{q^n:n\ge1\} is not measure universal, that is, some set of positive measure contains no affine copy of it?

Result
Proved(see note)
Status
Partial result
AI contribution
AI-discovered
Method
Construction
Field
Real analysis; measure theory and affine copies
Posed by
Paul Erdos (Problem 4.33.7*, Mathematica Balkanica 4, 1974)
Year posed
1974
Years open
52y
Solved
2026-10-05
Model
Unreleased internal OpenAI model
Vendor
OpenAI
Collaborators
—
Verification
Unreviewed
Publication
Announced
Collection
OpenAI math release (October 2026), version adc7f12
Significance
34 / 100
Disclosed cost
—
Wikipedia
No dedicated article

What was actually shown

Theorem 1.1: for every q∈(0,1)q\in(0,1) and η∈(0,1)\eta\in(0,1) there is a compact Eq,η⊆[0,1]E_{q,\eta}\subseteq[0,1] with m(Eq,η)>1−ηm(E_{q,\eta})>1-\eta such that for all x∈Rx\in\mathbb R, s≠0s\ne0, some x+sqn∉Eq,ηx+sq^n\notin E_{q,\eta}. So no geometric progression is measure universal; the dyadic companion first proved q=1/2q=1/2. The construction is probabilistic (random routing tables on a finite tree). It does not prove the conjecture for other infinite sets, and makes no claim of one set avoiding all ratios at once.

What the AI did

The release README says every result in openai/math was produced by an unreleased internal OpenAI model with a fixed procedure, on average about three hours of ChatGPT Pro thinking compute per result. This result is not one of the README's two exceptions (the Hodge conjecture for CM abelian varieties and the Re(s) > 11/12 zero-free region). The manuscript is authored 'OpenAI' and names no human author. The family's earlier manuscript, 'The dyadic case of the Erdos similarity conjecture' (September 25, 2026), proves the case q=1/2q=1/2 and carries the release's Lean formalization.

Verification

No independent mathematician has checked this yet. Checked here: Theorem 1.1 of the geometric-case manuscript was read against the special case of Erdos's conjecture. For each q∈(0,1)q\in(0,1) and η∈(0,1)\eta\in(0,1) it gives a compact E⊆[0,1]E\subseteq[0,1] of measure >1−η>1-\eta with no copy x+sGqx+sG_q, either sign of ss; the set may depend on qq. Lean: lean/ComparatorChallenges/DyadicAvoidance.json (solution_module OAI.MeasureTheory.DyadicAvoidance.Main, file present at the pinned commit; not in formalization.yaml) was read here. It states only the dyadic case q=1/2q=1/2 (from the companion manuscript), a special case of this entry's claim, so the entry stays Unreviewed under the tier rule. Not rebuilt here.

Sources

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