FormalConjectures/ErdosProblems/429.lean

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Explanation legend Lean4 Domain Explanation Language (L4DEL)

L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.

L4DEL
Lean4 Domain Explanation Language

Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.

Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

1 declarations, 1 contain sorry, 0 declared as axioms

Erdős Problem 429

References:

Imports (1)
  1. theorem Erdos429.erdos_429 Documented declaration: erdos_429 sorry line 49 source

    Source documentation

    Is it true that, if $A\subseteq \mathbb{N}$ is sparse enough and does not cover all residue classes modulo $p$ for any prime $p$, then there exists some $n\in \mathbb{Z}$ such that $n+a$ is prime for all $a\in A$?

    Weisenberg [We24] has shown the answer is no: $A$ can be arbitrarily sparse and missing at least one residue class modulo every prime $p$, and yet $A+n$ is not contained in the primes for any $n\in \mathbb{Z}$. (Weisenberg gives several constructions of such an $A$.)

    depends on sorry what this means

    Statement uses: Iff, False, Exists, Nat, And, Filter.Tendsto, Filter.atTop, Set, Set.Infinite, LE.le, Set.ncard, Inter.inter, Set.Icc, OfNat.ofNat, Nat.Prime, ZMod, Membership.mem, Ne, Int, Int.toNat and 1 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    28 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_429 : answer(False) ↔
        ∃ f : ℕ → ℕ, Tendsto f atTop atTop ∧
          ∀ A : Set ℕ, A.Infinite → (∀ N, (A ∩ Set.Icc 1 N).ncard ≤ f N) →
            (∀ p : ℕ, p.Prime → ∃ b : ZMod p, ∀ a ∈ A, (a : ZMod p) ≠ b) →
            ∃ n : ℤ, ∀ a ∈ A, (n + a).toNat.Prime := by
    …