Even-index FAB defects expose every quadratic-nonresidue factor

Corridor

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Status: proved universal theorem in the external-nonresidue prime-shift lane

Date: 2026-08-15

Depends on: FAB-KNESER-FULL-STABILIZER-DEFECT.md, EXTERNAL-NR-FACTOR-CYCLE.md, SHIFTED-NONRESIDUE-TRANSFER.md

Claim boundary: gives a sharp lower bound on every even-index placement defect in terms of the quadratic-nonresidue valuation mass of the shifted factorization. It does not eliminate all odd-index defects and therefore does not prove Erdős-Straus.

1. Setup

Let p be a Mordell-hard prime and let

q\equiv3\pmod4

be a prime satisfying

\left(\frac qp\right)=-1.

Put

\boxed{C=\frac{p+q}{4}}

and let

R\subseteq G=(\mathbb Z/q\mathbb Z)^\times

be the fixed-q signed divisor box.

Assume the exact FAB target is missed. Let

\boxed{H=\operatorname{Stab}(R)}

be the full stabilizer and set

\boxed{n=[G:H].}

This note treats the case

\boxed{2\mid n.}

Because q≡3 mod4,

q-1=2m

with m odd. Hence every even divisor n of q-1 satisfies

\boxed{n\equiv2\pmod4.}

2. The stabilizer lies inside the quadratic residues

The group G is cyclic. Choose a generator g.

The unique subgroup of index n is

H=\langle g^n\rangle.

Since n is even,

g^n=(g^2)^{n/2}

is a square. Therefore every element of H is a square:

\boxed{H\subseteq G^2.}

So every quadratic nonresidue modulo q lies outside the full defect stabilizer.

Corollary — the factor-cycle edge is always visible

EXTERNAL-NR-FACTOR-CYCLE.md supplies, from this 3 mod4 source vertex q, a prime factor

r\mid C

with

\left(\frac rp\right)=-1.

The factorwise transfer theorem gives

\left(\frac rq\right)=-1.

Hence in every even-index failure,

\boxed{r\notin H.}

Thus the external-nonresidue descent edge cannot be hidden inside an even-index stabilizer.

3. Quadratic-nonresidue valuation mass

Write

C=\prod_s s^{e_s}.

Define the total valuation mass carried by quadratic nonresidue prime factors modulo q:

\boxed{ E_q(C) := \sum_{ s^{e_s}\parallel C, (s/q)=-1 } e_s. }

Every such prime factor lies outside H. Therefore

E_q(C) \le \sum_{s^{e_s}\parallel C,\ s\notin H} e_s.

The full-stabilizer defect theorem gives

2\sum_{s^{e_s}\parallel C,\ s\notin H} e_s \le n-2.

Hence

\boxed{2E_q(C)\le n-2.}

or equivalently

\boxed{n\ge2E_q(C)+2.}

4. Parity sharpens the bound by two more units

Because (q/p)=-1 and both p≡1 mod4, q≡3 mod4, quadratic reciprocity gives

\left(\frac pq\right)=-1.

Since 4 is a square modulo q,

\boxed{\left(\frac Cq\right)=-1.}

Thus the total nonresidue valuation mass has odd parity:

\boxed{E_q(C)\equiv1\pmod2.}

Consequently

2E_q(C)+2\equiv0\pmod4.

But every even defect index satisfies

n\equiv2\pmod4.

Therefore the first allowable even index at or above 2E_q(C)+2 is two units larger.

Theorem — even-defect edge bound

Every even-index failed fixed-q FAB box in the external-nonresidue lane satisfies

\boxed{ n\ge2E_q(C)+4.}

Equivalently,

\boxed{ E_q(C)\le\frac{n-4}{2}. }

This is sharper than the generic full-stabilizer bound because the external shift forces odd quadratic-nonresidue parity while q≡3 mod4 forces the quotient index to have exactly one factor of two.

5. Individual projected-order bound for every nonresidue factor

Let

s^e\parallel C

with

\left(\frac sq\right)=-1.

Then s notin H. FAB-KNESER-FULL-STABILIZER-DEFECT.md gives

\operatorname{ord}_{G/H}(sH)>2e+1.

Because H⊆G^2 and s is a nonresidue, the quotient class sH remains outside the quotient's square subgroup. Its order is therefore even.

Hence

\boxed{ \operatorname{ord}_{G/H}(sH) \ge2e+2. }

In particular the full quotient index obeys

\boxed{n\ge2e+2}

for every nonresidue prime-power factor separately.

Thus high valuation of even one quadratic-nonresidue factor forces a large defect quotient.

6. Small-index consequences

Index 6

The theorem gives

E_q(C)\le1.

Since E_q(C) is positive and odd,

\boxed{E_q(C)=1.}

So an index-six failure contains exactly one unit of quadratic-nonresidue valuation mass. This is consistent with, and is strengthened by, the exact unique order-six atom classification in FAB-KNESER-INDEX6-CLASSIFICATION.md.

Index 10

E_q(C)\le3.

Hence

\boxed{E_q(C)\in\{1,3\}.}

So every index-ten failure has either one or three total nonresidue valuation units, never two or four.

Index 14

E_q(C)\le5,

and therefore

E_q(C)\in\{1,3,5\}.

Index 18

E_q(C)\le7,

E_q(C)\in\{1,3,5,7\}.

The same parity ladder holds for every even defect index.

7. Contrapositive elimination rule

The theorem gives an exact test for excluding a candidate even defect index.

If the shifted integer C=(p+q)/4 has quadratic-nonresidue valuation mass E, then every even failed stabilizer quotient must satisfy

\boxed{n\ge2E+4.}

Therefore all even quotient indices

2,6,10,\ldots,2E+2

are automatically impossible.

As the factorization accumulates more nonresidue valuation mass, the entire low-index even defect spectrum is pushed upward.

This is the precise entropy side of the external factor-cycle program: every additional visible quadratic-nonresidue factor consumes stabilizer room that cannot be recovered by hiding it inside H.

8. Updated entropy-or-descent picture

At every 3 mod4 vertex of the external-nonresidue factor cycle:

if the FAB target is hit, the prime is solved;
if the target is missed with even defect index, the outgoing nonresidue edge is necessarily visible and the quotient index satisfies the sharp mass bound

n\ge2E_q(C)+4;

if the target is missed with odd defect index, the nonresidue edge may be hidden inside the stabilizer, and the prime-index/cubic defect theorems control the first cases.

Thus the remaining all-prime problem splits naturally into:

\boxed{ \text{odd high-power-residue defects} \quad\text{versus}\quad \text{even visible-edge defects with growing index}.}

A universal closure theorem can now attack these two branches separately rather than treating all fixed-q placement failures as one undifferentiated phenomenon.