Exact index-six normal form for the combined FAB targets

Corridor

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Status: proved theorem

Date: 2026-08-15

Depends on: FAB-TWO-TARGET-KNESER.md, SHIFTED-NONRESIDUE-TRANSFER.md, EXTERNAL-NR-FACTOR-CYCLE.md

Claim boundary: classifies the first possible combined fixed-shift Kneser defect. It does not prove that index-six defects cannot occur, and therefore does not prove Erdős--Straus.

1. Setup

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

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,

and let

R=\mathcal R_q(C)

be the signed divisor box.

Assume that both exact solution targets are missed:

\boxed{-p^{-1}\notin R,\qquad -1\notin R.}

By inversion symmetry this also forces

\boxed{-p\notin R.}

Let

H=\operatorname{Stab}(R)

and assume the first possible combined index:

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

By FAB-TWO-TARGET-KNESER.md, the symmetric combined defect budget is

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

where

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

Because G is cyclic,

\boxed{G/H\cong C_6.}

Since H has index six, it is the unique sixth-power subgroup

\boxed{H=G^6.}

2. Local contributions in C6

Write the quotient additively as Z/6Z.

A prime factor whose image is trivial contributes 0 to the Kneser budget.

For a nontrivial image there are only three possible orders.

Order two

The local signed interval projects to

\{0,3\},

for every positive exponent, so

\boxed{s_i-1=1.}

Order three

The local signed interval fills

\{0,2,4\},

and

\boxed{s_i-1=2.}

Order six

If e_i=1, the local image is

\{0,g,-g\}

for a generator g of C_6, giving

\boxed{s_i-1=2.}

If e_i>=2, then

s_i\ge5,

\boxed{s_i-1\ge4,}

which is impossible under the budget 2.

Thus every order-six factor must occur to exponent exactly one.

3. Aperiodicity forces an order-six factor

Pass to

\bar R=R/H\subseteq C_6.

Because H is the full stabilizer of R, the quotient set bar R has trivial stabilizer.

Suppose no prime factor has order six in the quotient.

Then every nontrivial local set is one of the two proper subgroups

\{0,3\} \quad\text{or}\quad \{0,2,4\}.

If only order-two factors occur, their sum remains the order-two subgroup and has nontrivial stabilizer.

If only order-three factors occur, their sum remains the order-three subgroup and has nontrivial stabilizer.

If both types occur, their sum is all of C_6, contradicting the target misses.

Therefore

\boxed{\text{at least one prime factor has order six modulo }H.}

4. The order-six factor consumes the full budget

An order-six factor already contributes exactly 2, the entire available budget.

Hence every other prime factor must have trivial image modulo H.

There can be no second order-six factor, no order-three factor, and not even an order-two factor.

Therefore:

Theorem — single primitive sextic defect

A combined index-six failure has exactly one prime factor

\boxed{r\mid C}

outside H=G^6. It satisfies

\boxed{v_r(C)=1}

and

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

Every other prime factor of C lies in G^6.

Equivalently,

\boxed{ \bar R=\{H,rH,r^{-1}H\}. }

After orienting the generator rH as class 1,

\boxed{\bar R=\{0,1,5\}\subset C_6.}

5. Residue interpretation

Because rH has order six, r is neither a square nor a cube modulo q:

\boxed{ \left(\frac rq\right)=-1, \qquad r\notin G^3. }

Every other prime factor s|C lies in G^6, hence is simultaneously a square and a cube modulo q.

Thus

\boxed{ C=rS, \qquad v_r(C)=1, }

where every prime factor of S is a sixth-power residue modulo q.

Since an index-six subgroup exists only when

6\mid q-1,

and q≡3 mod4, necessarily

\boxed{q\equiv7\pmod{12}.}

6. The exceptional prime is the unique external nonresidue factor

For every prime factor s of

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

the shifted-nonresidue transfer theorem gives

\boxed{ \left(\frac sp\right) = \left(\frac sq\right).}

Every prime factor of S is a sixth power modulo q, hence a quadratic residue modulo q and therefore modulo p.

The exceptional factor r is a quadratic nonresidue modulo q, hence

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

Therefore r is the unique external-nonresidue prime factor of C.

The external factor graph has the forced edge

\boxed{q\longrightarrow r,}

with

\boxed{r\ne q,\qquad r<p.}

At the first surviving combined defect, the formerly nondeterministic factor descent becomes deterministic.

7. The three missing quotient classes are exact

The quotient box occupies

\boxed{\{0,1,5\}.}

The Type-II target

-1H

is the unique order-two class, namely

\boxed{3.}

The Type-I target -p^{-1} is a quadratic residue, so its quotient class is even. It is missed and nontrivial, therefore it is class 2 or 4.

Its inverse -p occupies the other one of those two classes.

Consequently the six quotient classes split exactly as

\boxed{ \begin{array}{c|c} \text{hit by }R & 0,1,5\\ \hline -p^{-1}H & 2\text{ or }4\\ -1H & 3\\ -pH & 4\text{ or }2 \end{array}}

Thus the three excluded natural targets are precisely the three missing quotient classes.

8. A small-prime consequence

The unique exceptional factor r is a quadratic nonresidue modulo p. Mordell-hard primes satisfy

\left(\frac2p\right) = \left(\frac3p\right) = \left(\frac5p\right) = \left(\frac7p\right)=+1.

Therefore none of 2,3,5,7 can be the exceptional factor.

If any of these primes divides

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

it must lie in the sixth-power subgroup modulo q.

For example, if

q\equiv7\pmod8,

then C is even, so a combined index-six defect forces

\boxed{2\in G^6.}

Since (2/q)=+1 in this congruence class, this adds the nontrivial requirement that 2 also be a cubic residue modulo q.

This supplies an immediate local filter on index-six candidates.

9. Strategic consequence

The first possible combined Kneser obstruction is a one-prime defect:

\boxed{ \frac{p+q}{4}=rS, }

where

r occurs exactly once;
r is simultaneously a quadratic and cubic nonresidue modulo q;
every prime factor of S is a sixth power modulo q;
r is the unique external quadratic-nonresidue factor relative to p;
the external factor cycle is forced to take q -> r;
the quotient box occupies exactly the three classes 0,±1, while the three natural solution targets occupy exactly the complementary classes.

The next descent problem is therefore sharply defined: transport this primitive sextic defect through the forced successor r, including the r≡1 mod4 case where the natural next admissible shift is composite (3r, or another 3 mod4 hard-residue multiplier times r).