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Erdős problem 26

Let A⊂NA\subset\mathbb{N} be infinite. Must there exist some k≥1k\geq 1 such that almost all integers have a divisor of the form a+ka+k for some a∈Aa\in A? The question as posed follows negatively from Davenport–Erdős (1951). The AI result settles Tenenbaum's harder variant, also negatively: there is an infinite AA such that for every k≥1k\geq 1 the set of multiples of A+kA+k has upper density below 0.340.34.

Sources

The source's own material, retained exactly.

11 retained statements36c6d0fef71a

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FormalConjectures/ErdosProblems/

26.lean

Retained formal statement3 of 10

Tenenbaum asked the weaker variant where for every ϵ>0\epsilon>0 there is some k=k(ϵ)k=k(\epsilon) such that at least 1−ϵ1-\epsilon density of all integers have a divisor of the form a+ka+k for some a∈Aa\in A.

The DeepMind prover agent has found a formal disproof of this statement.

FormalConjectures/ErdosProblems/26.leanErdos26.erdos_26.variants.tenenbaum1 lineExact file
False ↔ ∀ (A : ℕ → ℕ), StrictMono A → Erdos26.IsThick A → ∀ ε > 0, ∃ k, Erdos26.IsWeaklyBehrend (fun x => A x + k) ε
SolvedProof has a holeformal conjecturesexternal proof

The proof uses `sorry`: part of the argument is written but not proved. Lean accepts the file; it does not accept the theorem.