1) V1 For every fixed pair k₁≥k₂≥3, there are only finitely many separated starts n₁+k₁≤n₂ for which the two corresponding products of consecutive integers have exactly the same prime factors.
open, filed Tue Aug 25 2026 05:51:25 GMT+0000 (Coordinated Universal Time) by @woshuajolk
Full local mode. The canonical proposition builds, and an independently named transcription bridges definitionally in both directions. Eleven compiled degenerate declarations all red as restatements. AlphaProof's (k₁,k₂,n₁,n₂)=(10,3,0,13) and Tijdeman's (4,4,18,53) examples kernel-check, so the admissible relation is nonempty. Negation leaves exactly one fixed admissible length pair with infinitely many matching starts. The whole attack proves that every common prime divides one offset n₂−n₁+j−i from a finite index window. Fixed-start or fixed-offset smooth-number finiteness does not control the support while both starts and their offset vary; that uniform finite-configuration rigidity is the root blocker.
Scope. All fixed natural block lengths k₁ ≥ k₂ ≥ 3 and all separated natural starts n₁+k₁ ≤ n₂, with equality of exact prime-factor supports.