Binary defects on the full external-nonresidue factor cycle

Status: proved local reductions for both parity types of external prime vertices

Date: 2026-08-15

Depends on: EXTERNAL-NR-FACTOR-CYCLE.md, BINARY-R-DIVISOR-COLLISION.md, BINARY-R-KNESER-DEFECT.md

Claim boundary: puts every vertex of the external-nonresidue factor cycle into a binary-rescue defect framework and gives a necessary classification of any full-stabilizer index-two defect at a 1 mod 4 vertex. The broader Eisenstein support condition is an exact sufficient condition for binary failure, but it does not imply that the full stabilizer has index two. The file does not prove that a full cycle cannot consist entirely of failed vertices and does not prove Erdős-Straus.

1. One binary numerator for every external prime vertex

Let p be Mordell-hard and let

Define

Then R_q=3 mod 4. Put

Since q<p, one has A_q<p, and gcd(R_q,N_q)=1. Therefore the exact binary divisor-collision theorem applies at every external-prime vertex. The vertex is binary-rescued exactly when the signed divisor box of N_q modulo R_q contains -1.

2. Odd full-stabilizer quotient is impossible for every binary modulus

Let

let B be the signed divisor box, and let

If B misses -1, the quotient G/H cannot have odd order. Indeed, the image of -1 has order dividing two. In an odd-order quotient its image is trivial, so -1 in H. Since 1 in B and B is H-periodic, H subseteq B, forcing -1 in B, contradiction.

Hence

This holds for both R_q=q and R_q=3q.

3. The q = 3 mod 4 branch

If

then R_q=q is prime and BINARY-R-KNESER-DEFECT.md applies directly. Every failure satisfies

Every quadratic-nonresidue prime factor of

is visible outside H, and the sharp Kneser valuation-mass bound holds. Thus a 3 mod 4 cycle vertex has no index-two defect.

4. The q = 1 mod 4 branch

Now assume

Then R_q=3q, and CRT gives

There are three nontrivial quadratic characters

For hard p,

For the target -1,

5. Necessary classification of a full-stabilizer index-two defect

Assume a binary failure has full stabilizer of index two.

An index-two subgroup is the kernel of one of the three quadratic characters above.

H = ker(lambda) is impossible

Since lambda(-1)=+1, the target lies in H, hence in B, contradiction.

H = ker(chi) is impossible

Since chi(p)=-1, the exponent-one local set {p^{-1},1,p} already fills the two-element quotient. Thus the target quotient class is hit, contradiction.

Only H = ker(epsilon) can survive

Therefore any full-stabilizer index-two failure must have

If a prime factor s|N_q satisfies s=2 mod 3, then its local signed set fills the two-element quotient. Since hard p=1 mod3, index-two failure therefore requires

This implication is universal:

6. Eisenstein support is an exact coarse failure condition

There is a converse statement about binary failure, but not about the exact full-stabilizer index.

Assume every prime factor of

is 1 mod3. Since hard p is also 1 mod3, every divisor of N_q=pA_q is 1 mod3. Hence every signed divisor ratio lies in ker epsilon, while -1=2 mod3. Therefore

So Eisenstein support proves binary failure.

However, the full stabilizer of the failed box may be a proper subgroup of ker epsilon, giving a much larger quotient index. Thus the correct logic is

and neither reverse implication is asserted for the full stabilizer index.

Regression example

Take

Then

and 307=1 mod3. Hence Eisenstein support proves binary failure. Direct reconstruction of the signed box gives full-stabilizer quotient index

not 2.

This example is a permanent guard against conflating a coarse character obstruction with the exact stabilizer.

7. Parity refinement of the Eisenstein obstruction

If every prime factor of A_q is 1 mod3, then A_q is odd, since 2=2 mod3.

For hard p=1 mod8 and q=1 mod4,

is odd exactly when

Thus every Eisenstein-support failure satisfies q=1 mod8. If q=5 mod8, then A_q is even, so the literal factor 2=2 mod3 is present and the coarse mod-3 obstruction cannot occur.

8. Consequence for the outgoing factor-cycle edge under Eisenstein support

At a 1 mod4 source vertex, EXTERNAL-NR-FACTOR-CYCLE.md chooses a prime factor

with (s/p)=-1. The edge-character theorem gives

Under Eisenstein support, s=1 mod3, so

Thus even when the mod-3 support obstruction already proves binary failure, the outgoing edge remains a quadratic nonresidue modulo both p and the source prime q.

9. Correct local alphabet for a failed factor cycle

Every failed external-nonresidue cycle vertex now has one of the following forms.

Type I: q = 3 mod 4

An even Kneser defect of index at least six, with every q-nonresidue factor visible.

Type II: q = 1 mod 4

The full stabilizer quotient is even. If it is exactly two, Eisenstein support is necessary. More generally, Eisenstein support itself is a sufficient binary-failure obstruction and can coexist with a much larger full-stabilizer quotient.

There are no odd-index full-stabilizer defects anywhere on the binary cycle.

10. Remaining cycle target

The remaining cycle theorem must not identify Eisenstein support with full index two.

A valid closure target is:

Prove that no directed external-nonresidue factor cycle can carry binary failure at every vertex when every vertex has an even full-stabilizer defect, the 3 mod4 vertices obey the visible-nonresidue Kneser bounds, and any Eisenstein-support 1 mod4 vertex forces all factors of (p+3q)/4 into 1 mod3 while its outgoing edge remains a quadratic nonresidue to the source.

This cycle program remains supplementary to the stronger exact two-target signed-box reformulation of the prime problem. The latter is now the preferred global frontier because either the Type-I target -p^{-1} or the Type-II target -1 suffices.