Two-target Kneser collapse for external nonresidue shifts

Corridor

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Status: proved theorem

Date: 2026-08-15

Depends on: FAB-KNESER-DIVISOR-DEFECT.md, FAB-TYPE-II-SIGNED-DIVISOR.md, FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md, EXTERNAL-NR-FACTOR-CYCLE.md

Claim boundary: this removes every odd-index Kneser stabilizer defect from the combined Type-I/Type-II Erdős--Straus problem at an external nonresidue prime shift and strengthens the remaining Kneser budget by exploiting inversion symmetry. It does not eliminate the remaining even-index defects and therefore does not prove Erdős--Straus.

1. Same box, two solution targets

Let p be a Mordell-hard prime and let q<p be an external quadratic-nonresidue prime satisfying

q\equiv3\pmod4, \qquad \left(\frac qp\right)=-1.

Put

C=\frac{p+q}{4} =\prod_i r_i^{e_i}, \qquad G=(\mathbb Z/q\mathbb Z)^\times.

The fixed-q signed divisor box is

\boxed{ R= \left\{ \prod_i r_i^{z_i}\pmod q: -e_i\le z_i\le e_i \right\}. }

It is inversion-symmetric:

\boxed{R^{-1}=R.}

Two exact sufficient targets live in the same box:

\boxed{\tau_I=-p^{-1}\pmod q}

from the strong fixed-q FAB / normalized Type-I lane, and

\boxed{\tau_{II}=-1\pmod q}

from the normalized Type-II lane.

Hence

\boxed{ \tau_I\in R \quad\text{or}\quad \tau_{II}\in R \Longrightarrow p\text{ satisfies Erdős--Straus}.}

Because R=R^{-1}, a Type-I miss automatically also misses

\boxed{\tau_I^{-1}=-p.}

Thus a genuine combined failure excludes three natural residues:

\boxed{-p^{-1},\quad -p,\quad -1.}

2. Quadratic character positions

Because p≡1 mod4, quadratic reciprocity gives

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

Also q≡3 mod4, so

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

Therefore

\boxed{ \left(\frac{-p^{-1}}q\right)=+1, \qquad \left(\frac{-p}q\right)=+1, \qquad \left(\frac{-1}q\right)=-1. }

The two inverse Type-I orientations lie on the quadratic-residue side, while the Type-II target lies on the quadratic-nonresidue side.

3. Stabilizer and Kneser setup

Let

H=\operatorname{Stab}(R)

and put

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

Because 1∈R and R is H-periodic,

\boxed{H\subseteq R.}

For each prime factor r_i of C, define

d_i=\operatorname{ord}_{G/H}(r_iH), \qquad s_i=\min(2e_i+1,d_i).

Kneser's theorem gives

\boxed{

|R| \ge

|H|\left(1+\sum_i(s_i-1)\right). }</div>

4. Odd stabilizer index is impossible

Assume both solution targets are missed.

If n were odd, then because

q-1=2m

with m odd, the subgroup H would have even order. The cyclic group G has a unique element of order two, namely -1, so every even-order subgroup contains -1.

Hence

-1\in H\subseteq R,

contradicting the Type-II miss.

Therefore

\boxed{n\text{ is even}.}

This eliminates every odd-index Type-I Kneser defect from the combined Erdős--Straus obstruction.

5. Index two is impossible

If n=2, then H is the quadratic-residue subgroup. The Type-I target -p^{-1} is a quadratic residue, so

-p^{-1}\in H\subseteq R,

contradicting failure.

Thus

\boxed{ \text{combined failure} \Longrightarrow n\ge6\text{ and }n\text{ is even}.}

Since v_2(q-1)=1, every possible combined defect index has the form

\boxed{n=2m,\qquad m\ge3\text{ odd}.}

The first possible index is 6, not 3.

6. Three distinct missed H-cosets

For an even combined defect index, H has odd order. Hence

H\subseteq G^2,

the quadratic-residue subgroup.

The Type-II target -1 is a quadratic nonresidue, while -p^{-1} and -p are quadratic residues. Therefore

(-1)H

is distinct from both Type-I target cosets.

It remains to compare the inverse Type-I cosets.

Because H has odd order, the quotient

G^2/H

also has odd order. The class of -p^{-1} lies in this quotient. If

(-p^{-1})H=(-p)H,

then the class of -p^{-1} would equal its inverse and therefore have order at most two. An odd-order group has no nontrivial element of order two, so this would force

-p^{-1}\in H\subseteq R,

contradicting the assumed Type-I miss.

Hence

\boxed{ (-p^{-1})H, \quad (-p)H, \quad (-1)H \text{ are three distinct }H\text{-cosets}.}

7. Symmetric three-coset Kneser budget

A combined failure therefore misses at least three distinct H-cosets. Since R is H-periodic,

\boxed{|R|\le(n-3)|H|.}

Combining with the Kneser lower bound gives

1+\sum_i(s_i-1) \le n-3.

Thus:

Theorem — symmetric combined defect budget

If both Type I and Type II miss at an external nonresidue prime shift, then

\boxed{ \sum_i \left( \min(2e_i+1,\operatorname{ord}_{G/H}(r_iH))-1 \right) \le n-4, \qquad n=[G:H]. }

This improves the original one-target budget n-2 by two full units of quotient room.

8. Odd-index defects are automatically Type-II rescued

The one-target analysis previously identified cubic index 3 as the first possible Type-I defect, followed by prime indices 5,7,11,....

For the full equation, none of those odd-index defects can survive. At any odd index, -1∈H⊆R, so the Type-II target is already hit at the same fixed shift.

In particular,

\boxed{ \text{Type-I cubic defect} \Longrightarrow \text{Type-II rescue at the same }q.}

The same implication holds for every odd stabilizer index.

9. First combined case: index six

n=6,

the symmetric combined budget becomes

\boxed{ \sum_i(s_i-1)\le2.}

Moreover the quotient has six classes. Because a combined failure misses exactly the three distinguished target-side classes at minimum,

(-p^{-1})H, \quad (-p)H, \quad (-1)H,

only three quotient classes can remain occupied by the signed box.

FAB-INDEX6-COMBINED-DEFECT.md proves that this forces a single primitive sextic factor.

10. Strategic consequence

The direct ES search should no longer classify odd-prime Kneser defects one by one.

The residual wall is now

\boxed{ \text{external nonresidue }q +\text{three target cosets missed} +\text{even stabilizer index }n\ge6 +\text{symmetric budget }\le n-4.}

The first case is index six with budget only 2. If index six is eliminated, the search jumps directly to the next even quotient 2m under the same stronger symmetric budget.