C283 — All K splits in the retained state cube

24 September2026. Finite comparison following C282's positive2K/15K route.

Exactly one of the eight state configurations gives both legs as integer multiples of529. It repeats at all six retained positions. Four configurations retain a17K total, so a17K total alone does not select the all-K subdivision.

Let s and f select cumulative Cainan OFF0/ON1 for Shem and Flood. Let r select regular SP OFF/Terah0 at0 or MT ON/Terah60 at1, both full430. Keep C279's tagged indices j and j−2. The two leg widths and total are:

598+460(s−f), 8395+460(f−r), 8993+460(s−r).

All these widths remain divisible by23. In23 units the three expressions are26+20(s−f),365+20(f−r),391+20(s−r). Divisibility by529 asks for an additional factor23. In this binary domain, the first leg acquires it only when s−f=1; the second only when f−r=−1. Together they select s=1,f=0,r=1.

The calculation retains all48 rows: eight state choices at each of six positions. It introduces no new source state, event, Gear or independent translation. The uniqueness is bounded by this declared state cube and the chosen source indices; it is not a probability statement or global chronology theorem.

The source operations stay distinct. The regular profile selector combines a tradition change with130 and60; it is not a460 regular Cainan toggle. The140/600/1060 same-index separations are not substituted for the138/598/1058 cross-index widths used here.

Ninety-eight mathematical/domain checks and eight authentication checks pass; exact replay agrees. Author review only; canonical controls unchanged.

Best next bounded step — C284: remove the held j−2 pairing and test all independent annual indices in the same frozen fields and six Flood endpoints. Determine whether the six2K/15K paths exhaust that larger bounded domain. High is sufficient.