Kneser defect theorem for external binary-r rescue

Theorem

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Status: proved universal theorem in the external-nonresidue binary-rescue lane

Date: 2026-08-15

Depends on: BINARY-R-DIVISOR-COLLISION.md, FAB-KNESER-FULL-STABILIZER-DEFECT.md, SHIFTED-NONRESIDUE-TRANSFER.md

Claim boundary: proves that every failed external binary-r collision has an even stabilizer quotient and sharply bounds its visible nonresidue valuation mass. It does not prove that some r must succeed for every hard prime and therefore does not prove Erdős-Straus.

1. Setup

Let p be a Mordell-hard prime and let

r\equiv3\pmod4

be a prime satisfying

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

Put

A=\frac{p+r}{4}, \qquad N=pA.

BINARY-R-DIVISOR-COLLISION.md proves that the exact binary rescue exists if and only if the signed divisor exponent box

\boxed{ R= \left\{ \prod_{s^e\parallel N}s^z\pmod r: -e\le z\le e \right\} \subseteq G=(\mathbb Z/r\mathbb Z)^\times }

contains

\boxed{-1.}

Assume the binary rescue fails:

\boxed{-1\notin R.}

Let

H=\operatorname{Stab}(R), \qquad n=[G:H].

Because r is prime, G is cyclic of order r-1.

2. Odd defect index is impossible

Suppose

2\nmid n.

Write G=<g>. The unique subgroup of index n is

H=\langle g^n\rangle.

Since n|(r-1) and n is odd,

n\mid\frac{r-1}{2}.

Therefore

-1 =g^{(r-1)/2} \in H.

But 1 in R, and R is H-periodic. Hence

H\subseteq R.

Thus

-1\in R,

contradicting binary-rescue failure.

Theorem — no odd binary defect

Every failed external binary-r divisor collision has

\boxed{2\mid n.}

Since

r\equiv3\pmod4,

we have

v_2(r-1)=1.

Therefore every even defect index satisfies

\boxed{n\equiv2\pmod4.}

So the complete defect spectrum is

\boxed{n=2,6,10,14,18,\ldots}

before the additional Kneser constraints are imposed.

3. Index two is also impossible

n=2,

then H is the quadratic-residue subgroup of G.

Quadratic reciprocity gives

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

because p≡1 mod4.

Thus the distinguished prime factor p||N lies outside H.

Its local signed set is

\{p^{-1},1,p\}.

Modulo H, the element pH has order two, so this local set already fills the whole quotient

G/H.

That would make the full signed box project onto the whole quotient, including the class of -1, contradicting failure.

Equivalently, this is the projected-order gap from the full-stabilizer theorem: an exponent-one factor outside H would require quotient order strictly greater than 3, not 2.

Hence

\boxed{n\ne2.}

Combining with the previous section:

\boxed{n\ge6,\qquad n\equiv2\pmod4.}

4. Both p and A are quadratic nonresidues modulo r

We already have

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

Also

A=\frac{p+r}{4} \equiv\frac p4\pmod r,

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

Therefore the total quadratic-nonresidue valuation mass inside the factorization of A is odd.

Define

\boxed{ E_r(A) = \sum_{ s^e\parallel A, (s/r)=-1 }e. }

Then

\boxed{E_r(A)\equiv1\pmod2.}

Because p does not divide A, the corresponding nonresidue valuation mass in

N=pA

is exactly

\boxed{E_r(N)=1+E_r(A).}

Thus E_r(N) is positive and even.

5. Every quadratic-nonresidue factor is visible

Every failed defect index is even. As in FAB-KNESER-EVEN-DEFECT-EDGE.md, this implies

H\subseteq G^2.

Therefore every prime factor of N that is a quadratic nonresidue modulo r lies outside H.

The full-stabilizer Kneser theorem gives

2\sum_{s^e\parallel N,\ s\notin H}e \le n-2.

Hence

2E_r(N)\le n-2.

Substituting E_r(N)=1+E_r(A) yields:

Theorem — external binary defect mass bound

Every failed external binary-r collision satisfies

\boxed{ n\ge2E_r(A)+4.}

Equivalently,

\boxed{ E_r(A)\le\frac{n-4}{2}. }

Because E_r(A) is odd and n≡2 mod4, this parity is exact, not merely an inequality artifact.

Every additional nonresidue valuation unit in the shifted integer (p+r)/4 eliminates another low even defect index.

6. Projected-order bound

Let

s^e\parallel N

with

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

Then s notin H. The full-stabilizer theorem gives

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

Because sH remains quadratic-nonresidue-side in the even quotient, its order is even. Therefore

\boxed{ \operatorname{ord}_{G/H}(sH) \ge2e+2. }

In particular, the distinguished factor p has exponent one, so

\boxed{ \operatorname{ord}_{G/H}(pH)\ge4. }

But the quotient order is 2 mod4, so the first possible order is actually

\boxed{6.}

This is another direct proof that no index-two defect can survive.

7. Exact index-six classification

Assume

\boxed{n=6.}

The mass bound gives

E_r(A)\le1.

Since E_r(A) is positive and odd,

\boxed{E_r(A)=1.}

Thus A contains exactly one unit of quadratic-nonresidue valuation mass: there is a unique prime

s\parallel A

with (s/r)=-1, and every other prime factor of A is a quadratic residue modulo r.

The two nonresidue prime factors of N=pA are therefore precisely the simple factors

p \quad\text{and}\quad s.

They already consume the full outside-stabilizer valuation budget:

2(1+1)=4=n-2.

Hence every other prime factor of N lies in H.

The projected-order gap excludes orders 2 and 3; in the cyclic quotient of order six, both pH and sH therefore have exact order six.

Let x generate G/H. Every generator is x or x^{-1}, so each of the two simple nontrivial local sets is

\{x^{-1},1,x\}.

Their product is

\boxed{ R/H = \{x^{-2},x^{-1},1,x,x^2\}, }

which is every quotient class except the unique order-two class

x^3.

Because r≡3 mod4 and H⊂G^2, the image of -1 in G/H is exactly that order-two class.

Therefore:

Theorem — exact binary index-six defect

A binary-r failure has stabilizer index six if and only if, in the quotient geometry forced above, the shifted factorization has exactly two simple primitive order-six atoms p and s, all other factor mass lies in H, and

\boxed{ R/H=(G/H)\setminus\{-1H\}. }

So index six is the extremal one-hole failure: the signed divisor box covers five of the six quotient classes and misses only the required ES target.

8. Why the binary route is cleaner than the FAB target route

For the strong fixed-q FAB target, odd-index defects can survive and must be treated separately.

For the binary-r target, the distinguished class is exactly -1. Since -1 belongs to every odd-index subgroup of a cyclic group of order 2 mod4, all odd-index defects vanish immediately.

Thus every external binary-r failure has the same architecture:

\boxed{ \text{even quotient} \Longrightarrow H\subseteq\text{quadratic residues} \Longrightarrow \text{every nonresidue factor is visible} \Longrightarrow \text{Kneser mass forces the quotient upward}. }

This removes the odd high-power-residue branch from the entropy-or-descent program entirely.

9. New direct ES target

For every Mordell-hard prime p, choose an external nonresidue prime

r\equiv3\pmod4, \qquad (r/p)=-1.

If the binary collision at r fails, its full stabilizer index obeys

\boxed{ n\ge2E_r((p+r)/4)+4, \qquad n\mid r-1, \qquad n\equiv2\pmod4. }

Therefore a universal proof can aim to force

2E_r((p+r)/4)+4>r-1

for some external r, or more generally show that the finite external-nonresidue factor cycle cannot support the required sequence of increasingly structured even defects.

Unlike the FAB route, there is no odd-defect escape hatch.