Composite successor `3r`: index-four defect elimination

Theorem

---

Status: proved universal theorem

Date: 2026-08-15

Depends on: ES-COMPOSITE-SUCCESSOR-3R.md, ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md, FAB-KNESER-FULL-STABILIZER-DEFECT.md

Claim boundary: eliminates full-stabilizer index four for a failed 3r successor. It does not eliminate the Eisenstein-split branch or all higher even indices and therefore does not prove Erdős--Straus.

1. Setup

Let p be Mordell-hard and let

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

Use the natural admissible composite successor

\boxed{k=3r}

and put

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

Let

G=(\mathbb Z/3r\mathbb Z)^\times, \qquad R=\mathcal R_{3r}(C), \qquad H=\operatorname{Stab}(R).

Assume both exact solution targets miss:

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

By inversion symmetry,

-p\notin R.

We prove that

\boxed{[G:H]\ne4.}

2. General full-stabilizer order gap

Pass to the quotient

\bar G=G/H, \qquad \bar R=R/H.

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

For every prime-power factor

s^e\parallel C

whose image sH is nontrivial, the full-stabilizer theorem gives

\boxed{ \operatorname{ord}_{\bar G}(sH)>2e+1. }

If |bar G|=4, the only possible nontrivial projected order exceeding three is order four, and it can occur only for

\boxed{e=1.}

Thus every factor outside H, if one exists, is a simple order-four atom.

3. Two order-four atoms already fill the quotient

A quotient group of order four containing an element of order four is cyclic:

\bar G\cong C_4.

For a simple order-four atom x, its signed local set is

\{0,x,-x\}

in additive notation.

Two such local sets sum to all of C_4:

\{0,\pm1\}+\{0,\pm1\}=C_4.

That would make bar R=bar G and hit both targets.

Therefore a failed index-four box has at most one nontrivial quotient atom.

So there are only two abstract cases:

no prime factor of C lies outside H, hence bar R={0};
exactly one simple order-four factor lies outside H, hence

\bar R=\{0,1,3\},

after orienting the generator.

We eliminate both.

4. The one-atom case is impossible

Assume

\bar R=\{0,1,3\}\subset C_4.

Its unique missing class is 2.

Since all three natural excluded residues miss the box, they must all lie in that one class:

\boxed{ [-1]=[-p^{-1}]=[-p]=2. }

Dividing the first two classes gives

[p^{-1}]=0,

\boxed{p\in H.}

Now use the exact identity

4C=p+3r.

Modulo 3r,

\boxed{4C\equiv p.}

Passing to C_4, and writing b=[2], gives

2b+[C]=[p]=0.

But the product C contains exactly one nontrivial quotient atom, so

[C]=1\text{ or }3,

an element of order four.

Thus

2b=-[C]

would make a doubled element of C_4 equal to an element of order four.

That is impossible: every double in C_4 is class 0 or 2 and has order at most two.

Therefore the one-atom index-four defect cannot occur.

5. The subgroup-box case forces p into H

Now assume every prime factor of C lies in H.

Then

R\subseteq H.

But 1 in R and R is H-periodic, so

H\subseteq R.

Hence

\boxed{R=H.}

The identity 4C≡p and C in H show that

[p]=[4]=2[2]

is a square in the order-four quotient.

The image of -1 is nontrivial of order two, since -1 is missed. Therefore in any order-four quotient its class is an order-two element.

If [p] were the nontrivial order-two class, then

[-p^{-1}]=[-1]-[p]=0,

so the Type-I target would lie in H=R, contradiction.

Thus target failure forces

\boxed{[p]=0,\quad p\in H.}

We now show that no index-four subgroup of G can contain p.

6. A quadratic nonresidue p cannot lie in an index-four subgroup

CRT gives

G \cong C_2\times C_{r-1}.

Choose a generator g of the cyclic r-component and write

p\equiv g^a\pmod r.

Because r≡1 mod4 and (r/p)=-1, quadratic reciprocity gives

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

Therefore

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

The mod-3 coordinate of hard p is +1, so the order of p in G is exactly its order in the cyclic r-component:

\operatorname{ord}_G(p) =\frac{r-1}{\gcd(a,r-1)}.

Since a is odd,

\gcd(a,r-1)

is odd. Hence ord_G(p) contains the entire 2-primary part of r-1.

Because r≡1 mod4, write

2^v\parallel r-1, \qquad v\ge2.

Then

2^v\mid\operatorname{ord}_G(p).

If H had index four, its order would be

|H| =\frac{|G|}{4} =\frac{2(r-1)}4 =\frac{r-1}{2},

whose 2-primary part is only

2^{v-1}.

By Lagrange's theorem, an element of order divisible by 2^v cannot lie in a subgroup whose order has only 2^{v-1}.

Therefore

\boxed{p\notin H.}

This contradicts the subgroup-box failure condition from Section 5.

7. Theorem

Both possible index-four quotient geometries are impossible.

Hence:

Composite-successor index-four elimination

For every Mordell-hard prime p and every external nonresidue prime

r\equiv1\pmod4, \qquad (r/p)=-1,

a failed exact two-target signed box at the natural successor shift

k=3r

cannot have full stabilizer index four:

\boxed{ \{-p^{-1},-1\}\cap\mathcal R_{3r}((p+3r)/4)=\varnothing \Longrightarrow [(\mathbb Z/3r\mathbb Z)^\times:H]\ne4. }

Combined with ES-COMPOSITE-SUCCESSOR-3R.md:

odd index is impossible;
index two is possible only in the pure Eisenstein-split obstruction;
index four is impossible universally.

Therefore in the non-Eisenstein-split branch, the first possible full-stabilizer defect is index at least six.

8. Next target

Finite proof-mining shows genuine index-six 3r defects do occur, so the next theorem must classify rather than simply exclude them.

The natural target is a composite analogue of FAB-INDEX6-COMBINED-DEFECT.md: classify the order-six quotient under the mod-3 parity constraint and determine whether its unique primitive atom is again the forced external-nonresidue successor.