Composite successor `3r`: parity-constrained signed-box theorem

Theorem

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Status: proved exact local reduction and index-two classification

Date: 2026-08-15

Depends on: ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md, FAB-TWO-TARGET-KNESER.md, FAB-INDEX6-COMBINED-DEFECT.md, FAB-HARD-FIRST-FILTERS.md

Claim boundary: this treats the natural composite successor when the forced external-nonresidue prime has residue 1 mod 4. It does not prove that the successor shift must hit a target and therefore does not prove Erdős--Straus.

1. Why `3r` is the natural successor

Let p be a Mordell-hard prime. Then

\boxed{p\equiv1\pmod{12}.}

Let r<p be an external quadratic-nonresidue prime for p with

\boxed{r\equiv1\pmod4, \qquad \left(\frac rp\right)=-1.}

The prime shift k=r is not admissible in the two-target formulation because it is 1 mod 4.

Multiply by the hard-residue prime 3 and put

\boxed{k=3r.}

Then

k\equiv3\pmod4, \qquad \gcd(k,p)=1.

Set

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

Because p≡1 mod3,

\boxed{C\equiv1\pmod3,}

and because r\ne p,

\boxed{\gcd(C,3r)=1.}

Thus the exact two-target signed box is defined at the composite shift 3r.

2. CRT split of the signed box

Factor

C=\prod_i s_i^{e_i}.

CRT gives

\boxed{ (\mathbb Z/3r\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/r\mathbb Z)^\times.}

The first factor has order two. For every prime divisor s_i of C, define

\boxed{ \epsilon_i= \begin{cases} +1,&s_i\equiv1\pmod3,\\ -1,&s_i\equiv2\pmod3. \end{cases}}

For a signed exponent vector

z=(z_i), \qquad -e_i\le z_i\le e_i,

the mod-3 coordinate of the corresponding signed divisor is

\boxed{\epsilon(z)=\prod_i\epsilon_i^{z_i}\in\{+1,-1\}.}

Define the two parity fibres projected to the prime modulus r:

\boxed{ \mathcal R_{r}^{\pm}(C) = \left\{ \prod_i s_i^{z_i}\pmod r: -e_i\le z_i\le e_i, \quad \epsilon(z)=\pm1 \right\}.}

The full signed box modulo 3r is exactly the CRT union of these two fibres.

3. All three natural targets lie in the negative parity fibre

The exact two-target theorem uses

\tau_I=-p^{-1}, \qquad \tau_{II}=-1,

and inversion symmetry adds

\tau_I^{-1}=-p.

Since p≡1 mod3, all three satisfy

\boxed{ -p^{-1}\equiv-p\equiv-1\equiv-1\pmod3.}

Therefore target membership modulo 3r reduces exactly to the negative parity fibre modulo r:

\boxed{ -p^{-1}\in\mathcal R_{3r}(C) \iff -p^{-1}\pmod r\in\mathcal R_r^{-}(C),}

\boxed{ -1\in\mathcal R_{3r}(C) \iff -1\pmod r\in\mathcal R_r^{-}(C).}

By inversion symmetry the same holds for -p.

Theorem — exact composite-successor fibre criterion

The shift k=3r produces an Erdős--Straus certificate if and only if

\boxed{ \mathcal R_r^{-}(C) \cap \{-p^{-1},-1\} \ne\varnothing, \qquad C=\frac{p+3r}{4}.}

Thus the composite modulus introduces only one binary parity constraint beyond a prime-modulus signed divisor problem.

4. The empty-fibre obstruction is exactly Eisenstein splitting

The negative fibre is empty precisely when no signed exponent vector has odd mod-3 parity.

If every prime factor of C is 1 mod 3, then every epsilon_i=+1, so

\boxed{\mathcal R_r^{-}(C)=\varnothing.}

Conversely, if some prime factor

s\mid C

satisfies

s\equiv2\pmod3,

then choosing exponent z_s=1 and all other exponents zero gives a negative-parity element. Hence

\boxed{ \mathcal R_r^{-}(C)=\varnothing \iff \text{every prime factor of }C\text{ is }1\pmod3.}

So the first obstruction at the composite successor is exactly another simultaneous-splitting condition in the Eisenstein direction already visible in FAB-HARD-FIRST-FILTERS.md.

5. Target character positions modulo r

Because both p and r are 1 mod 4, quadratic reciprocity gives

\left(\frac pr\right) = \left(\frac rp\right) =-1.

Also

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

Therefore, modulo r,

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

So the two Type-I orientations and Type II still occupy opposite quadratic sides, but the roles are reversed from the q≡3 mod4 prime-shift case.

6. Odd stabilizer index still cannot support failure

Let

\widetilde G=(\mathbb Z/3r\mathbb Z)^\times

and let

H=\operatorname{Stab}(\mathcal R_{3r}(C)).

[\widetilde G:H]

is odd, then the quotient \widetilde G/H has odd order. Every element of order two in \widetilde G must therefore map to the identity.

In particular the element -1 mod 3r lies in H.

Since the signed box contains 1 and is H-periodic,

H\subseteq\mathcal R_{3r}(C).

Hence

-1\in\mathcal R_{3r}(C),

which is a Type-II hit.

Therefore:

\boxed{ \text{combined failure at }3r \Longrightarrow [\widetilde G:H]\text{ is even}.}

The odd-index collapse is not a prime-modulus accident.

7. Exact index-two classification

Assume now

\boxed{[\widetilde G:H]=2}

and that both exact targets are missed.

Because

\widetilde G \cong C_2\times(\mathbb Z/r\mathbb Z)^\times

with the second factor cyclic of even order, there are exactly three nontrivial quadratic characters on \widetilde G:

the mod-3 parity character epsilon;
the Legendre character eta=(\cdot/r);
their product epsilon eta.

Thus the three index-two subgroups are the kernels of these characters.

The target signatures are

\boxed{ \begin{array}{c|ccc} & \epsilon & \eta & \epsilon\eta\\ \hline -p^{-1} & -1 & -1 & +1\\ -p & -1 & -1 & +1\\ -1 & -1 & +1 & -1 \end{array}}

because p≡1 mod3, (p/r)=-1, and (-1/r)=+1.

Case 1: H = ker eta

Then -1 has eta=+1, so

-1\in H\subseteq\mathcal R_{3r}(C),

a Type-II hit. Impossible.

Case 2: H = ker(epsilon eta)

Then -p^{-1} has epsilon eta=+1, so

-p^{-1}\in H\subseteq\mathcal R_{3r}(C),

a Type-I hit. Impossible.

Case 3: H = ker epsilon

This is the only index-two subgroup that contains neither solution target.

But if any prime factor s|C satisfies

s\equiv2\pmod3,

then the signed box contains the element represented by exponent z_s=1, which lies outside H=ker epsilon.

Since the box is H-periodic, contains H, and H has index two, one element outside H forces

\mathcal R_{3r}(C)=\widetilde G,

contradicting target failure.

Therefore every prime factor of C must be 1 mod3.

Theorem — index-two successor defect is pure Eisenstein splitting

If the composite successor shift k=3r misses both exact targets and its signed box has stabilizer index two, then necessarily

\boxed{ H=\ker\epsilon }

and

\boxed{ \text{every prime factor of } \frac{p+3r}{4} \text{ is }1\pmod3.}

Equivalently, the entire negative parity fibre is empty.

Thus every non-Eisenstein-split composite successor automatically eliminates the index-two defect.

8. New successor dichotomy

For a forced external-nonresidue successor r≡1 mod4, the natural admissible shift 3r therefore has a clean first dichotomy:

Split obstruction

\frac{p+3r}{4}

is composed entirely of primes 1 mod3, then the negative target fibre is empty and the shift cannot solve p.

Non-split regime

If the shifted integer has even one prime factor 2 mod3, then the negative fibre is nonempty and stabilizer index two is impossible. Any combined failure must move to a finer even quotient.

This is the composite analogue of the earlier prime-shift Kneser collapse.

9. Next theorem target

The forced-successor problem has now split into two precise cases:

eliminate or descend through the Eisenstein-split condition

\operatorname{supp}\left(\frac{p+3r}{4}\right)\subseteq\{\ell:\ell\equiv1\pmod3\};

in the non-split case, classify the first possible even stabilizer defect above index two inside

(\mathbb Z/3r\mathbb Z)^\times,

using the same three-target symmetry and Kneser expansion.

The important reduction is that a 1 mod4 successor no longer requires a general composite-modulus search. It is a prime-r divisor-placement problem with one parity bit.

1. Why 3r is the natural successor