Let $A \subset N$ be an infinite set for which there exists some $ϵ > 0$ such that in any subset of $A$ of size $n$ there is a subset of size at least $ϵ n$ which contains no three-term arithmetic progression.Is it true that $A$ is the union of a finite number of sets which contain no three-term arithmetic progression?

Worked, still open.

additive combinatorics · solved · formalized (Lean) · 0 attempts

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vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

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unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

A counterexample is given by a recent construction of **Reiher, Rödl, and Sales**. Specialising their result to $k=3$, they construct [[nomath]](for any $\mu\in(0,2/3)$)[[/nomath]] a set (X\subset\mathbb N) such that:

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

formal

AMS 11 · solved (literature)

theorem erdos_847 : answer(False) ↔ ∀ (A : Set ℕ), Infinite A → HasFew3APs A →
    ∃ n, ∃ (S : Fin n → Set ℕ), (∀ i, ThreeAPFree (S i)) ∧ A = ⋃ i : Fin n, S i

formal-conjectures/847.lean ↗

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solved

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formal

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