Additional filters on the parity-cubic `3r` index-six defect

Theorem

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Status: proved corollaries of the Branch-B normal form

Date: 2026-08-15

Depends on: ES-COMPOSITE-SUCCESSOR-INDEX6.md

Claim boundary: these restrictions sharpen the remaining parity-cubic index-six branch but do not eliminate it and therefore do not prove Erdős--Straus.

1. Branch-B setup

Let p be Mordell-hard and let

r<p, \qquad r\equiv1\pmod4, \qquad \left(\frac rp\right)=-1.

At the natural composite shift

k=3r, \qquad C=\frac{p+3r}{4},

assume a combined full-stabilizer index-six failure in Branch B of ES-COMPOSITE-SUCCESSOR-INDEX6.md.

Then

\boxed{ H=\ker\varepsilon\cap\ker\kappa,}

where

\varepsilon(x)=\left(\frac x3\right)

is the mod-3 parity character and kappa is the cubic-residue quotient modulo r.

The quotient is

\bar G=G/H\cong C_6.

The Branch-B theorem gives:

p∈H, so p is a cubic residue modulo r;
every exceptional prime has order six in bar G;
every exceptional prime is 2 mod 3 and a cubic nonresidue modulo r;
the total exceptional valuation mass is exactly two;
every hidden factor is 1 mod 3 and cubic-residue-side;
the quotient box is

\bar R=C_6\setminus\{3\}.

We now exploit the exact identity

\boxed{4C\equiv p\pmod{3r}.}

Since p∈H, this becomes in the quotient

\boxed{2[2]+[C]=0.}

2. The shifted integer C must be odd

Suppose

2\mid C.

Because

\varepsilon(2)=-1,

the prime 2 cannot lie in H. Hence it is one of the exceptional factors.

Every exceptional factor has quotient order six. Therefore

\boxed{[2]\text{ has order }6.}

Let

e=v_2(C).

The total exceptional valuation mass is exactly two, so

e\in\{1,2\}.

Case e = 2

Then 2 uses the entire exceptional mass. All other factors lie in H, so

[C]=2[2].

The identity 2[2]+[C]=0 gives

4[2]=0.

But an element of order six is not annihilated by four. Contradiction.

Case e = 1

Then there is exactly one further simple exceptional prime s, whose quotient class also has order six.

Thus

[C]=[2]+[s].

The same identity gives

3[2]+[s]=0.

For an order-six element, 3[2] is the unique element of order two, namely class 3 in C_6. Hence

[s]=3,

which has order two rather than six. Contradiction.

Therefore:

Theorem — Branch B is odd

\boxed{2\nmid C.}

So every parity-cubic Branch-B defect satisfies

\boxed{C\text{ is odd}.}

3. The source prime satisfies r = 1 mod 24

Mordell-hard primes satisfy

p\equiv1\pmod8.

Since

C=\frac{p+3r}{4}

is odd, we must have

p+3r\equiv4\pmod8.

With p≡1 mod8, this gives

3r\equiv3\pmod8,

\boxed{r\equiv1\pmod8.}

The index-six classification already gives

r\equiv1\pmod3.

Therefore:

Corollary — exact source congruence

\boxed{r\equiv1\pmod{24}.}

Thus the parity-cubic defect occupies only one of the six odd residue classes modulo 24 available to a general external prime.

4. The class of 2 records the exceptional orientation

Although 2 does not divide C, its quotient class still enters through

2[2]+[C]=0.

Because

\varepsilon(2)=-1,

the class [2] is one of the odd classes

1,3,5

in C_6.

The exceptional valuation mass two has two structural possibilities.

Configuration I: opposite simple atoms

Suppose there are two distinct simple exceptional primes with quotient classes

1\quad\text{and}\quad5.

Then their total contribution to C is zero:

[C]=0.

Hence

2[2]=0.

Among the odd classes of C_6, only class 3 is killed by multiplication by two. Thus

\boxed{[2]=3.}

So 2 has trivial cubic coordinate:

\boxed{2\text{ is a cubic residue modulo }r.}

Configuration II: aligned mass

If instead the exceptional contribution is

[C]=2 \quad\text{or}\quad [C]=4,

which happens either when the two simple atoms have the same orientation or when one exceptional prime occurs to exponent two, then

2[2]=-[C]

forces

[2]=1 \quad\text{or}\quad [2]=5.

Therefore

\boxed{2\text{ is a cubic nonresidue modulo }r.}

We obtain:

Theorem — cubic character of 2 distinguishes the two Branch-B shapes

In a parity-cubic Branch-B defect:

\boxed{ 2\text{ cubic residue mod }r \iff \text{the two simple exceptional classes are opposite}.}

If 2 is a cubic nonresidue modulo r, then the exceptional valuation is aligned in one of the two classes ±1: either one squared exceptional prime or two simple exceptional primes with the same orientation.

5. Strengthened Branch-B normal form

Every Branch-B index-six defect at k=3r therefore satisfies all of the following:

\boxed{ r\equiv1\pmod{24}, \qquad C=\frac{p+3r}{4}\text{ odd}.}

Moreover:

p is a cubic residue modulo r;
exactly two units of factor valuation are carried by primes 2 mod 3 that are cubic nonresidues modulo r;
every other factor is 1 mod 3 and cubic-residue-side;
the cubic character of 2 modulo r determines whether the two visible order-six atoms cancel or align in the quotient.

The remaining branch is now a simultaneous cubic splitting problem involving

p,\quad 2,\quad\text{and the complete factorization of }\frac{p+3r}{4}.

6. Next target

The most concrete closure targets are now:

use cubic reciprocity to compare the condition

p\in((\mathbb Z/r\mathbb Z)^\times)^3

with the prescribed cubic characters of the two inert-mod-3 exceptional primes;

use an explicit criterion for the cubic character of 2 mod r together with

r\equiv1\pmod{24}

to separate or eliminate the aligned and cancelling configurations;

show that either configuration forces a new exact Type-I or Type-II divisor-square hit at a related admissible shift.

This is substantially narrower than the original composite-successor problem: the first non-primitive index-six obstruction now lives on one congruence class of source primes and only two units of visible factor valuation.