FormalConjectures.ErdosProblems.«871»

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import FormalConjecturesUtil

Erdős Problem 871

References:

erdosproblems.com/871

[ErNa88] Erdős, Paul and Nathanson, Melvyn B., Partitions of bases into disjoint unions of bases. J. Number Theory (1988), 1-9.

[ErNa89] Erdős, Paul and Nathanson, Melvyn B., Additive bases with many representations. Acta Arith. (1989), 399-406.


namespace Erdos871

Let $A$ be an additive basis of order $2$, and suppose $1_A\ast 1_A(n)\to \infty$ as $n\to \infty$. Can $A$ be partitioned into two disjoint additive bases of order $2$?

A question of Erdős and Nathanson [ErNa88], who proved this is true if $1_A\ast 1_A(n) > c\log n$ (for all large $n$) for some constant $c>(\log\frac{4}{3})^{-1}$. Erdős and Nathanson [ErNa89] also proved that for every $t$ there exists a basis $A$ of order $2$ such that $1_A\ast 1_A(n)\geq t$ for all large $n$ and yet $A$ cannot be partitioned into two disjoint additive bases. This has been disproved by Larsen using Claude Opus 4.5 - in fact only a small modification of the argument of [ErNa89] is required.

@[category research solved, AMS 11, formal_proof using lean4 at "https://github.com/plby/lean-proofs/blob/main/src/v4.29.1/ErdosProblems/Erdos871.lean"]
theorem declaration uses 'sorry'erdos_871 :
    answer(False) ↔
      ∀ (A : Set ℕ),
        (∀ᶠ n in Filter.atTop, ∃ a ∈ A, ∃ b ∈ A, a + b = n) ∧
        (∀ t, ∀ᶠ n in Filter.atTop, ∃ pairs : Finset (ℕ × ℕ),
          pairs.card ≥ t ∧
            ∀ p ∈ pairs, p.1 ∈ A ∧ p.2 ∈ A ∧ p.1 + p.2 = n ∧ p.1 ≤ p.2) →
        ∃ (B C : Set ℕ),
          (∀ x, x ∈ A ↔ x ∈ B ∨ x ∈ C) ∧
          Disjoint B C ∧
          (∀ᶠ n in Filter.atTop, ∃ a ∈ B, ∃ b ∈ B, a + b = n) ∧
          (∀ᶠ n in Filter.atTop, ∃ a ∈ C, ∃ b ∈ C, a + b = n) := by⊢ False ↔
  ∀ (A : Set ℕ),
    ((∀ᶠ (n : ℕ) in Filter.atTop, ∃ a ∈ A, ∃ b ∈ A, a + b = n) ∧
        ∀ (t : ℕ),
          ∀ᶠ (n : ℕ) in Filter.atTop,
            ∃ pairs, pairs.card ≥ t ∧ ∀ p ∈ pairs, p.1 ∈ A ∧ p.2 ∈ A ∧ p.1 + p.2 = n ∧ p.1 ≤ p.2) →
      ∃ B C,
        (∀ (x : ℕ), x ∈ A ↔ x ∈ B ∨ x ∈ C) ∧
          Disjoint B C ∧
            (∀ᶠ (n : ℕ) in Filter.atTop, ∃ a ∈ B, ∃ b ∈ B, a + b = n) ∧
              ∀ᶠ (n : ℕ) in Filter.atTop, ∃ a ∈ C, ∃ b ∈ C, a + b = n
  sorryAll goals completed! 🐙


end Erdos871