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[P1 exact] Prove the #659 five-cell onsets from geometry, without enumerating L=6 #673

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@LightChainr

Parent: #650. Follow-up to PR #669 / #659. No new census. No L=6.

Why

#669 extracted exact integer identities from the committed 4×4 tables and left most of them empirical. Upgrade the cheap ones to combinatorial proofs, and isolate the two-point diamond conjectures.

The reader scripts/wrapping_five_cell_reader.py on PR #669 re-verifies every identity as integers. Use that as the check, not as the proof.

Prove or kill (in this order)

  1. I1 (low-k binomial). Axis: n×b(k) = C(N,k) for k < L. Diamond: for k < 2L. The blocking-line reading is already in the #659 [P2 exact] Integer structure of the five wrapping-type cells #669 note; write it as a proof that does not mention the tables except as a check.
  2. I2 (high-k binomial). b×n(k) = C(N,k) for k ≥ N−L+1 on both geometries.
  3. Axis I3. d0(L) = L (the L full rows) and b×n(2L−1) = L² (one full row plus one full column).
  4. Axis I4. Pre-both-wrap deficit C(N,k) − n×b = d0 + d1 for L ≤ k < 2L−1.
  5. Diamond I3. d0(2L) = 4·C(2L,4), b×b(2L) = 2L, b×n onset 3L−1 with value 4L². These rest on two sizes. Give a geometric reading or explicitly leave them as two-point conjectures. Do not enumerate diamond L=5 (2^50 is not a ticket).

Also record: M_L irreducible over ℚ at the five committed sizes is a factorization fact, not a wrapping theorem, unless you find a reason it must be. Do not spend the ticket on Galois theory.

Tripwire

A "proof" that uses an unstated fact from one L is not a proof. If you cannot prove an item, write OPEN and point at the #669 falsification kit. Do not enumerate the next size.

No Huawei. No STATUS. Do not close #659 / #640. Draft PR against main.

Wait for ASSIGNED_MACHINE.

Related: #659, #640, #651, PRs #669, #653, #657.

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