CN-Shared Theorem — Lift-Room, Tight Clusters, and Admissible Escape

Status: lift-room and totient-ratio lemmas proved; unrestricted shared-factor C2 is false; admissible program C2-shared certified on the remaining tight cluster

Date: 2026-08-15

Depends on: C1-THEOREM.md, C2-THEOREM.md, CN-THEOREM.md

Claim boundary: Does not prove Erdős-Straus, López-all-primes, or universal DSC-P. Closes the analytic reduction of shared-factor CN to a 3-adic complementary-cover problem, and certifies that problem on every admissible hard Type A/B candidate through k ≤ 1500.

Read with:

• cn_shared_analyzer.py
• verify_cn_shared.py

1. Setup

As in C1/C2, a directly novel candidate at target depth k has progression

with r the CRT combination of a Mordell-hard class h ∈ H and a trap t ∈ T_k. For each earlier layer j,

and R_j is the affine pullback of T_j to s \bmod q_j. Write S_j = U_j \setminus R_j with U_j = (\mathbb Z/q_j\mathbb Z)^\times.

A simultaneous reduced escape is an s with gcd(s,Q)=1 and s \bmod q_j \in S_j for every active j, where Q=\operatorname{lcm}(q_j).

2. Lemma (odd totient ratio)

Let d \mid q with 1 \le d < q and q odd. Then

Proof

Write q=dm with m\ge 3 odd. Euler's product gives

If every prime of m already divides d, the product is empty and the ratio equals m \ge 3. If m contributes a new prime, the product is at least φ(m_{\mathrm{new}})/m_{\mathrm{new}} times an integer ≥ 3 factor, and for odd m_{\mathrm{new}}≥3 one has φ(m_{\mathrm{new}})≥2. In all cases the ratio is an integer at least 2.

Checked for every odd q ≤ 5000 and every proper divisor: minimum ratio 2, zero failures.

3. Theorem (lift-room)

Let d=\gcd(q_1,q_2)>1. Write P_i for the set of residues a \bmod d that lift to at least one element of S_i. If

for some i, then P_i contains every unit modulo d. Combined with C1 on the other layer, P_1 ∩ P_2 \ne \emptyset, and CRT supplies a reduced simultaneous escape.

Proof

The reduction U_i \to (\mathbb Z/d\mathbb Z)^\times is surjective with fibres of size φ(q_i)/φ(d). A residue a \bmod d fails to lie in P_i only if that entire fibre sits in R_i, which requires at least φ(q_i)/φ(d) forbidden units. This is impossible if the displayed inequality holds. C1 supplies some a' ∈ S_{3-i}; its projection is a unit modulo d and therefore lies in the full P_i. QED.

Uniform form (independent of r)

Since |R_i| \le |T_{j_i}| \le 2\tau(j_i),

implies lift-room for every r.

4. Theorem (C2-thin reduction)

Assume |R_1| \le 1 and |R_2| \le 1. Then a shared-factor C2 failure can occur only if

and the two singleton forbidden sets are complementary in {1,2}.

Proof

If d < q_i, the totient-ratio lemma gives φ(q_i)/φ(d) ≥ 2 > |R_i|, so that layer has lift-room. Thus both layers fail lift-room only if q_1=d=q_2. For equal moduli, |R_1 ∪ R_2| ≤ 2, so the units are covered only if φ(q) ≤ 2. The only odd q>1 with φ(q)=2 is q=3. QED.

Census through k ≤ 1500 found |R| ∈ \{0,1\} on every single-active row. The thin reduction is therefore the generic remaining C2 hole.

5. Unrestricted C2-shared is false

The previous C2 write-up claimed zero shared-factor failures for “standard L and multiple r”. That scan did not hit the complementary-producing residue classes.

An exact search — every r satisfying

with {σ_1,σ_2}={1,2} — finds complementary q=3 covers. First examples:

So C2-shared is not a theorem for arbitrary (L,r,j_1,j_2).

The Type A/B program does not use arbitrary r. A candidate is admissible when

for some hard h and some t ∈ T_k, and L=\operatorname{lcm}(840,4k-1).

6. Theorem (205-ancestor absorption)

Let j=205. Then

and the divisor-child theorem gives T_{205} \bmod 39 \subseteq T_{10}.

Theorem

If q_{205}=3 and R_{205} \ne \emptyset, the candidate is directly shadowed by layer 10.

Proof

m_{205}=3^2\cdot 7\cdot 13. The condition q_{205}=3 means

so 13 \mid L. Already 3 \mid 840 \mid L, hence 39 \mid L. Therefore q_{10}=1 and the entire progression is frozen modulo 39:

Nonempty R_{205} supplies some parameter whose point lies in T_{205}. Reducing modulo 39 and applying the divisor-child inclusion puts that point in T_{10}. But every point of the progression has the same residue modulo 39, so the whole progression lies in T_{10}. QED.

Corollary

Layer 205 cannot participate in a complementary q=3 cover on any directly novel candidate.

7. Admissible complementary covers through k ≤ 1500

Replayable tight scan on all 73,814 hard-compatible admissible candidates with k ≤ 1500 (this is the same candidate population as the frozen DSC-P k≤1500 bundle):

tight q<=9 shared pairs checked:     1,365,222
pair escapes:                        1,365,201
pair failures:                              21
C1 failures:                                 0
tight q<=9 triples checked:          3,994,891
triple failures:                             0

The 21 pair failures are all complementary q=3 covers, and all contain layer 205:

Each of the 21 is directly shadowed by layer 10 (and usually by one more q=1 ancestor such as 6, 8, or 36), exactly as the absorption theorem predicts.

Therefore:

A residue-class search that only reduces complementary r modulo lcm(m_{j_1},m_{j_2}) can miss these 21 lifts; the admissible-first scan is the complete check.

8. Theorem CN-shared (program form)

For every directly novel admissible Type A/B candidate with target depth k:

1. Coprime active moduli escape by CN-coprime (already proved).
2. Any layer with lift-room over gcd(q, lcm(q_{\mathrm{rest}})) extends a solution of the complementary cluster (lift-room theorem).
3. Uniformly roomy pairs (2τ(j) < φ(q)/φ(d), or same-q with 2τ(j_1)+2τ(j_2) < φ(q)) escape independently of r.
4. The only remaining C2-thin obstruction is a complementary q=3 pair. Every admissible example through k ≤ 1500 uses layer 205 and is directly shadowed by layer 10. Directly novel candidates have zero such covers in the range.
5. Tight q ≤ 9 triples on admissible candidates: 3,994,891 checks, 0 failures.

Thus every directly novel admissible shared-factor pair in the completed range has a reduced simultaneous escape.

This upgrades the C2/CN shared-factor story from “sampled r, zero fails” to:

• a proved reduction to one local obstruction;
• an explicit counterexample to the unrestricted statement;
• a proved absorption theorem that kills the 205 family on novel candidates;
• an exact certificate that no other complementary family appears through k ≤ 1500.

9. Finite certificates (replayable)

Independent control flow: verify_cn_shared.py rebuilds blocked projections from Euler's product and rechecks the totient-ratio lemma, the 4j-1 \mid qL classification of exact-q layers, and a program-L sample.

10. Correction to earlier C2/C3/C4 claims

C2-THEOREM.md and CN-THEOREM.md reported zero shared-factor failures for standard L and sampled r. That is true of those samples and is not a universal C2-shared theorem. Complementary-aligned r produce covers. The correct claim is the program-admissible statement in §7.

11. Scoreboard

12. Next

1. Classify every q=3 layer as an ancestry child of a q=1 anchor, generalising the 205 → 10 absorption.
2. Peel every roomy layer from an arbitrary finite active core; only a 3/5/7-adic tight cluster remains.
3. Bound |N^{act} ∩ tight| on Class-C residuals, then assemble DSC-P.