FormalConjectures/ErdosProblems/429.lean
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1 declarations, 1 contain sorry, 0 declared as axioms
Erdős Problem 429
References:
- erdosproblems.com/429
- [ErGr80] Erdős, P. and Graham, R., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathematique (1980).
- [Er80] Erdős, Paul, A survey of problems in combinatorial number theory. Ann. Discrete Math. (1980), 89-115.
- [We24] D. Weisenberg, Sparse Admissible Sets and a Problem of Erdős and Graham. Integers (2024).
Imports (1)
- theorem
Erdos429.erdos_429Documented declaration: erdos_429 sorry line 49 sourceSource 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.toNatand 1 moreProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous28 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 …