Source labels openChecked July 26, 2026

Erdős Problem 203

Is there an integer $m$ with $(m, 6) = 1$ such that none of $2^k \cdot 3^\ell \cdot m + 1$ are prime, for any $k, \ell \ge 0$ ?

Combinatorics AMS 5

Mathematical statement

Is there an integer $m$ with $(m, 6) = 1$ such that none of $2^k \cdot 3^\ell \cdot m + 1$ are prime, for any $k, \ell \ge 0$ ?

Statement source: Erdős Problems statement material

Statement terms: Source-specific

Source-specific terms. Therefore does not assert reuse rights beyond attributed display.

Statement artifacts, not proofs

These records expose exact Lean propositions and statement-only wrappers. Defining a proposition does not supply a proof of it. A placeholder-bearing target also contains no proof. Elaboration checks syntax and types; it does not certify that a formalization perfectly captures every nuance of the informal problem.

Pinned Lean formulation 1

erdos_203

Canonical source

Complete statement target, proof intentionally absentLean 4

theorem erdos_203 : answer(sorry) ↔ ∃ m, m.Coprime 6 ∧ ∀ k l, ¬ (2^k * 3^l * m + 1).Prime := by  sorry

Statement source: Formal Conjectures
Lean version: v4.27.0
Placeholder: Present; no proof artifact
Source evidence: Pinned source index
Fidelity review: Community formulation

References

Pinned Formal Conjectures sourceformalization erdosproblems.com/203background