Primitive sextic defect chains and the neighbor-square recurrence

Theorem

---

Status: proved local transition theorem with exact finite witness

Date: 2026-08-15

Depends on: ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md, FAB-TWO-TARGET-KNESER.md, FAB-INDEX6-COMBINED-DEFECT.md, EXTERNAL-NR-FACTOR-CYCLE.md

Claim boundary: this gives an exact recurrence for consecutive primitive index-six failures at 3 mod 4 external prime shifts. Consecutive primitive failures do occur, so this note does not prove Erdős--Straus.

1. Primitive index-six edge

Let p be Mordell-hard and let

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

Assume the exact two-target signed box at the prime shift k=q misses both Type I and Type II and has full stabilizer index six.

Then FAB-INDEX6-COMBINED-DEFECT.md gives

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

where

r is a simple prime factor;
r is the unique quadratic-nonresidue factor relative to p;
modulo q, the class of r generates

(\mathbb Z/q\mathbb Z)^\times/((\mathbb Z/q\mathbb Z)^\times)^6\cong C_6;

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

Thus the external factor edge is forced:

\boxed{q\longrightarrow r.}

If the source itself has index six, then necessarily

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

2. What the previous vertex becomes at the successor

Assume now that the forced successor also satisfies

r\equiv3\pmod4

and that the exact two-target box at k=r also has a primitive index-six failure.

Because r divides (p+q)/4,

p+q\equiv0\pmod r.

Therefore

\boxed{q\equiv-p\pmod r.}

But -p is the inverse orientation of the exact Type-I target at the successor modulus r.

For an index-six combined failure, the quotient box occupies the classes

0,\ \pm1

while the inverse Type-I targets occupy

\boxed{\pm2}

and Type II occupies class 3.

Hence the previous vertex has exact quotient order three:

\boxed{ q\,((\mathbb Z/r\mathbb Z)^\times)^6 \text{ has order }3.}

Equivalently,

\boxed{ q\text{ is a quadratic residue but a cubic nonresidue modulo }r.}

By contrast, the forward exceptional factor at the previous vertex has quotient order six:

\boxed{ r\text{ is both a quadratic and cubic nonresidue modulo }q.}

Thus a consecutive primitive edge has the asymmetric sextic signature

\boxed{ \begin{array}{c|cc} & \text{quadratic} & \text{cubic}\\ \hline r\pmod q & - & -\\ q\pmod r & + & - \end{array}}

which is consistent with ordinary quadratic reciprocity for two 3 mod 4 primes.

3. Three consecutive primitive vertices

Suppose one step farther that

q_-\longrightarrow q\longrightarrow q_+

are three consecutive external primes, all congruent to 3 mod 4, and the exact two-target signed box has primitive index-six failure at the middle vertex q.

The forward exceptional factor q_+ occupies one of the quotient classes

\pm1.

The previous vertex satisfies

q_-\equiv-p\pmod q

and therefore occupies one of the inverse Type-I classes

\pm2.

In the cyclic quotient C_6, every class ±2 is the square or inverse square of a class ±1.

Therefore:

Theorem — neighbor-square recurrence

There exists a sign

\varepsilon\in\{+1,-1\}

and a unit u mod q such that

\boxed{ q_- \equiv q_+^{\,2\varepsilon}u^6 \pmod q.}

Equivalently,

\boxed{ q_-\,q_+^{-2\varepsilon} \in ((\mathbb Z/q\mathbb Z)^\times)^6.}

Thus every internal primitive sextic vertex obeys an exact second-neighbor recurrence modulo sixth powers.

This is stronger than the pair of quadratic/cubic character statements: it identifies the complete quotient relation in C_6.

4. Concrete consecutive-defect witness

Consecutive combined primitive defects genuinely occur.

Take

\boxed{p=808369.}

This is Mordell-hard because

808369\equiv289\pmod{840}.

Vertex q = 43

\frac{p+43}{4} =202103 =11\cdot19\cdot967.

The exact two-target signed box misses both targets and has full stabilizer index

\boxed{6.}

The unique external nonresidue factor is

\boxed{19,}

so the edge is

43\to19.

With primitive root 3 mod 43, quotient exponents modulo six may be chosen so that

19\mapsto1,

while

-p^{-1}\mapsto2, \qquad -1\mapsto3, \qquad -p\mapsto4.

Vertex q = 19

\frac{p+19}{4} =202097 =7\cdot28871.

Again the exact two-target signed box misses both targets and has full stabilizer index

\boxed{6.}

The unique external factor is

\boxed{28871,}

19\to28871.

With primitive root 2 mod 19, the quotient classes modulo six satisfy

28871\mapsto5,

and the previous vertex

43\equiv-p\pmod{19}

has class

\boxed{4=2\cdot5\pmod6.}

Thus the neighbor-square recurrence holds explicitly.

Next vertex

q=28871,

the exact two-target box still misses, but the full stabilizer index jumps to

\boxed{28870,}

rather than remaining six.

So the finite chain begins

\boxed{ 43\xrightarrow{\,6\,}19\xrightarrow{\,6\,}28871\xrightarrow{\,28870\,}\cdots }

where the labels indicate the defect index at the source vertex.

5. Consequence for the proof search

The conjecture

\text{“primitive index six cannot occur twice consecutively”}

is false.

The correct local object is the recurrence

q_{i-1}\equiv q_{i+1}^{\pm2}\cdot(\text{sixth power})\pmod{q_i}.

A cycle-level proof must therefore exploit one of:

higher reciprocity for the complete sextic quotient data;
incompatibility of these recurrences around a closed cycle;
forced growth or inflation of the full stabilizer index;
a transition to the composite 3r parity-fibre regime when a successor is 1 mod 4.

The exact witness above shows why a one-edge contradiction is too strong, but also shows the phenomenon to explain: primitive sextic behavior can repeat locally and then collapse into a much larger defect quotient.