Exact index-6 FAB Kneser defect classification

Corridor

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

Date: 2026-08-15

Depends on: FAB-KNESER-DIVISOR-DEFECT.md, FAB-KNESER-PRIME-INDEX-DEFECT.md, SHIFTED-NONRESIDUE-TRANSFER.md

Claim boundary: classifies every failed fixed-q FAB box whose exact stabilizer quotient has order six. It does not prove that all possible stabilizer indices are impossible and therefore does not prove Erdős-Straus.

1. Setup

Let p be a Mordell-hard prime and let

q\equiv3\pmod4

be a prime with

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

Put

C=\frac{p+q}{4}=\prod_i r_i^{e_i}

and let

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

be the signed divisor box in

G=(\mathbb Z/q\mathbb Z)^\times.

Assume the exact FAB target

\tau=-p^{-1}\pmod q

is missed.

Let

H=\operatorname{Stab}(R)

and assume

\boxed{[G:H]=6.}

Because G is cyclic, the quotient

\bar G=G/H

is cyclic of order six.

By definition of H as the full stabilizer of R, the image

\bar R=R/H

has trivial stabilizer in \bar G.

2. Order-two projections are impossible

Suppose a prime factor r_i|C has image

\bar r_i\in\bar G

of order two.

For every exponent e_i>=1, its local signed set in the quotient is

\{\bar r_i^{-e_i},\ldots,1,\ldots,\bar r_i^{e_i}\} =\{1,\bar r_i\},

which is exactly the order-two subgroup of \bar G.

Multiplying any set by this local set makes the resulting product set invariant under that order-two subgroup.

Hence \bar R would have a nontrivial stabilizer, contradicting the fact that H was the full stabilizer of R.

Therefore:

\boxed{\operatorname{ord}(\bar r_i)\ne2\quad\text{for every }r_i|C.}

3. Order-three projections are impossible

The same argument applies to an image of order three.

\operatorname{ord}(\bar r_i)=3,

then already for e_i>=1 the exponent interval [-e_i,e_i] covers all three powers, so the local set is the complete order-three subgroup

\{1,\bar r_i,\bar r_i^2\}.

Its presence would force \bar R to be invariant under that subgroup, again contradicting trivial stabilizer.

Thus:

\boxed{\operatorname{ord}(\bar r_i)\ne3\quad\text{for every }r_i|C.}

Consequently every prime factor of C either lies in H or projects to an element of exact order six.

4. Kneser room for order-six factors

For an order-six image, the local Kneser contribution is

\min(2e_i+1,6)-1.

The index-six defect budget from FAB-KNESER-DIVISOR-DEFECT.md is

\sum_i(s_i-1)\le4.

Hence:

e_i=1 contributes 2;
e_i=2 contributes 4;
e_i>=3 contributes 5, impossible.

Therefore the only Kneser-allowed nontrivial patterns are:

one order-six factor of exponent one;
one order-six factor of exponent two;
two order-six factors, both of exponent one.

All remaining prime factors must lie in H.

5. Quadratic character kills the even patterns

Because

q\equiv3\pmod4

and (q/p)=-1, quadratic reciprocity gives

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

Since 4 is a square,

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

Thus C is a quadratic nonresidue modulo q.

Now H has index six in a cyclic group. Every element of H is a square because

H=G^6\subseteq G^2.

An order-six quotient element is represented by an odd exponent in the quotient and therefore carries the nontrivial quadratic parity.

One exponent-two exception

If the only nontrivial projected factor has exponent two, its quotient contribution to C is a square. Every other factor lies in H and is also a square.

Therefore C would be a quadratic residue, contradiction.

Two simple order-six exceptions

Let their quotient classes be generators x and y of C_6.

Every generator is either g or g^{-1} for a fixed quotient generator g. Thus

xy\in\{g^2,1,g^{-2}\},

which always has even quotient exponent and is quadratic-residue-side.

Again all remaining factors lie in H, so C would be a quadratic residue, contradiction.

Hence both even patterns are impossible.

6. Exact classification theorem

Theorem

Under the setup above, if the fixed-q FAB divisor box fails and its full stabilizer has quotient index six, then there is exactly one prime factor

\boxed{r\parallel C}

such that

\boxed{\operatorname{ord}_{G/H}(rH)=6.}

Every other prime factor of C lies in H:

\boxed{s|C,\ s\ne r\Longrightarrow s\in H.}

Equivalently, in the sixth-power quotient, the entire shifted factorization has exactly one simple primitive order-six defect.

Since G is cyclic and H=G^6, this can be stated as:

\boxed{ C=r\cdot U, \qquad v_r(C)=1, \qquad U\text{ is supported entirely on sixth-power residues mod }q, }

where r mod q has exact order six in G/G^6.

7. Quotient shape of the failed box

Let x=rH, a generator of the order-six quotient.

All other local sets collapse to the identity coset, while the exceptional simple factor contributes

\{x^{-1},1,x\}.

Therefore

\boxed{ R/H=\{x^{-1},1,x\}. }

The FAB target is quadratic-residue-side. In C_6, the quadratic-residue subgroup is

\{1,x^2,x^4\}.

Because the target is missed, it cannot be the identity coset already present in R/H. Hence its quotient class is exactly one of

\boxed{x^2\text{ or }x^4.}

Thus the index-six failure is a rigid three-coset picture:

\boxed{ \text{available}=\{x^{-1},1,x\}, \qquad \text{target}\in\{x^2,x^4\}. }

There is no other index-six geometry.

8. Why this matters for descent

At an external nonresidue vertex q, EXTERNAL-NR-FACTOR-CYCLE.md guarantees a prime factor of C that is also a quadratic nonresidue modulo p.

In an index-six failure, the theorem above says the quotient has exactly one simple factor carrying the nontrivial sixth-power class. All remaining factor mass is invisible in the quotient.

Therefore the descent has a canonical candidate:

if the unique order-six factor is the external-nonresidue factor, the factor edge is forced through the unique quotient defect;
if it is not, the external-nonresidue descent factor lies inside H=G^6, so it is simultaneously a sixth-power residue modulo q and a quadratic nonresidue modulo p.

Either way, index six no longer represents an arbitrary mixed character failure. It reduces to one distinguished simple factor and a sixth-power-residue background.

The next useful classification targets are indices 10 and 18, which are the other recurrent mixed defects seen in proof-mining and should admit analogous exact-stabilizer reductions.