Kneser defect theorem for the fixed-k FAB divisor box

Corridor

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Status: proved application of classical Kneser addition theory

Date: 2026-08-15

Depends on: FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md, FAB-FIXED-K-SIGNED-DIVISOR.md, SHIFTED-NONRESIDUE-TRANSFER.md

Claim boundary: this theorem classifies the structure forced by failure of one fixed-k signed divisor box. It does not prove that some k must succeed for every prime and therefore does not prove Erdős-Straus.

Imported theorem: Kneser's addition theorem for finite subsets of abelian groups. A convenient modern proof is Matt DeVos, A short proof of Kneser's addition theorem for abelian groups, arXiv:1303.3539. Kneser's theorem is classical; only its application to the present FAB divisor box is part of this note.

1. Fixed-k signed divisor box

Let p be an odd prime with

p\equiv1\pmod4,

and let

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

Put

C=\frac{p+k}{4} =\prod_{i=1}^m r_i^{e_i}.

The strong fixed-k divisor-ratio theorem gives the finite multiplicative box

\boxed{ \mathcal R_k(C) =\left\{ \prod_{i=1}^m r_i^{z_i}\pmod k: -e_i\le z_i\le e_i \right\} \subseteq G_k=(\mathbb Z/k\mathbb Z)^\times. }

In the strong sufficient lane the exact target is

\boxed{\tau=-p^{-1}\pmod k.}

Thus a fixed-k strong certificate exists exactly when

\boxed{\tau\in\mathcal R_k(C).}

For each prime-power factor of C, define the symmetric local set

A_i=\{r_i^{-e_i},\ldots,r_i^{-1},1,r_i,\ldots,r_i^{e_i}\}\subseteq G_k.

Then

\mathcal R_k(C)=A_1A_2\cdots A_m.

2. Kneser applied to the product box

Let

R=\mathcal R_k(C)

and let

H=\operatorname{Stab}(R) =\{g\in G_k:gR=R\}.

Kneser's theorem, written multiplicatively and iterated over the finite family A_i, gives

\boxed{

|R| \ge \sum_{i=1}^m |A_iH|-(m-1)|H|. }</div>

Let

\pi:G_k\to G_k/H

be the quotient map and put

\boxed{n=[G_k:H].}

For each factor define

d_i=\operatorname{ord}_{G_k/H}(r_iH).

Because the exponents in A_i form the consecutive interval [-e_i,e_i], its image in the cyclic subgroup generated by r_iH has exact size

\boxed{ s_i:=|\pi(A_i)|=\min(2e_i+1,d_i).}

Therefore

|A_iH|=s_i|H|

and Kneser becomes

\boxed{

|R| \ge

|H|\left(1+\sum_{i=1}^m(s_i-1)\right). }</div>

3. The defect-budget theorem

Assume the fixed-k target is missed:

\boxed{\tau\notin R.}

Then R is a proper H-periodic subset of G_k. Hence it occupies at most n-1 quotient cosets:

|R|\le(n-1)|H|.

Combining this with the Kneser lower bound gives the central inequality.

Theorem — FAB Kneser defect budget

If the fixed-k strong divisor-ratio target is missed, then for the stabilizer H of the signed divisor box,

\boxed{ \sum_{i=1}^m \left( \min(2e_i+1,\operatorname{ord}_{G_k/H}(r_iH))-1 \right) \le n-2, \qquad n=[G_k:H]. }

In words:

A failed exact divisor placement is possible only if all prime-power factors of (p+k)/4 have very little total expansion in one proper quotient of the unit group.

This turns an exact target miss into a quantitative quotient defect rather than an unstructured factorization accident.

4. Prime external-nonresidue specialization

Now let

\boxed{k=q}

be an odd prime satisfying

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

Then

G_q=(\mathbb Z/q\mathbb Z)^\times

is cyclic of order q-1.

Quadratic reciprocity gives

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

The fixed-k target

\tau=-p^{-1}\pmod q

is a quadratic residue because

\left(\frac{-p^{-1}}q\right) =\left(\frac{-1}q\right) \left(\frac pq\right)^{-1} =(-1)(-1)=+1.

Also

C=\frac{p+q}{4} \equiv\frac p4\pmod q

is a quadratic nonresidue modulo q.

This is exactly the external-nonresidue placement regime isolated elsewhere in the repository.

5. Index two cannot support a failure

The cyclic group G_q has a unique index-two subgroup, namely the quadratic residues

Q_q=G_q^2.

Suppose the stabilizer H of a failed box had index 2. Then

H=Q_q.

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

H\subseteq R.

The target tau is quadratic-residue-side, so

\tau\in H\subseteq R,

contradicting failure.

Therefore:

Corollary — no quadratic defect

For an external nonresidue prime shift,

\boxed{ \tau\notin\mathcal R_q(C) \Longrightarrow [G_q:H]\ne2. }

Since index 1 would mean R=G_q, every failure has

\boxed{[G_q:H]\ge3.}

Thus the familiar quadratic-character obstruction has been removed completely. Any remaining failure lives in a strictly finer quotient.

6. Exact index-three classification

Assume

\boxed{[G_q:H]=3.}

The defect budget is

\sum_i(s_i-1)\le1.

In the cyclic quotient of order 3, every nonidentity element has order 3. Therefore if any prime factor r_i lies outside H, then e_i>=1 and

s_i=\min(2e_i+1,3)=3,

so that single factor contributes

s_i-1=2,

contradicting the budget.

Hence every prime factor of C lies in H.

Because H is the unique index-three subgroup of the cyclic unit group, this says:

\boxed{ q\equiv1\pmod3 \quad\text{and}\quad r_i\in G_q^3 \text{ for every prime }r_i\mid C. }

Moreover every local set lies in H, while R is H-periodic and contains 1; therefore

\boxed{R=H.}

So failure also requires

\boxed{\tau\notin G_q^3.}

Since C in H and p≡4C (mod q), we have

\tau=-p^{-1} \equiv-4^{-1}C^{-1}\pmod q.

The element -1 lies in every index-three subgroup of G_q, so the target lies outside H exactly when 4 does. In an order-three quotient, squaring is an automorphism, hence

4\in H\iff2\in H.

Thus:

Corollary — cubic-defect normal form

An index-three placement failure at an external nonresidue shift can occur only if

\boxed{ \begin{aligned} &3\mid q-1,\\ &\text{every prime factor of }C=(p+q)/4\text{ is a cubic residue mod }q,\\ &2\text{ is not a cubic residue mod }q. \end{aligned}}

In particular 2 cannot divide C. For a Mordell-hard prime p≡1 mod8, this forces

\boxed{q\equiv3\pmod8.}

So the entire index-three failure geometry collapses to a very rigid cubic-splitting condition on the shifted integer.

7. Index four is absent

For

q\equiv3\pmod4,

we have

v_2(q-1)=1.

Hence

4\nmid q-1.

The cyclic group G_q therefore has no subgroup of index 4.

Thus there is no index-four defect case to classify.

8. Exact index-five sparsity

Assume

\boxed{[G_q:H]=5.}

The defect budget is

\sum_i(s_i-1)\le3.

Every nonidentity element of the quotient has order 5.

If r_i notin H and e_i=1, then

s_i=3, \qquad s_i-1=2.

If instead e_i>=2, then

s_i=5, \qquad s_i-1=4,

which already exceeds the total budget.

Therefore:

Corollary — fifth-power defect sparsity

An index-five failure can have

\boxed{\text{at most one prime factor of }C\text{ outside }G_q^5,}

and if such an exceptional factor exists then

\boxed{v_r(C)=1.}

All other prime factors of C are fifth-power residues modulo q.

If no exceptional factor exists, then R=G_q^5, and failure requires tau outside that subgroup.

If one exceptional prime r exists, write bars for classes in G_q/H. Then

\boxed{ R/H=\{\bar r^{-1},1,\bar r\}, }

and the exact remaining miss condition is

\boxed{ \bar\tau\notin\{\bar r^{-1},1,\bar r\}. }

This is a three-coset defect, not an arbitrary factorization pattern.

9. General consequence: entropy or quotient defect

The theorem gives a dichotomy for every fixed-k product box:

Expansion

If no proper quotient G_k/H can satisfy the defect budget, then the box cannot miss the target and the fixed-k FAB certificate exists.

Structure

If the target is missed, there is a proper quotient of index n for which

\boxed{ \sum_i \left( \min(2e_i+1,\operatorname{ord}(r_iH))-1 \right) \le n-2. }

Thus the factorization of (p+k)/4 is forced into a low-entropy multiplicative configuration.

For adaptive external nonresidue primes, the first possible defect is already cubic, and the next prime-index defect is fifth-power-sparse.

This is the precise bridge needed for an entropy-or-descent attack:

enough independent factor residues force exact placement;
insufficient expansion forces a rigid higher-power-residue quotient;
that quotient structure can be transported along the external-nonresidue factor ladder / dual descent instead of restarting an unrelated search.

10. Next theorem target

The all-prime burden has now narrowed to proving that the quotient defects cannot persist for every adaptive external nonresidue shift.

A particularly concrete first target is:

\boxed{ \text{external nonresidue }q +\text{ cubic-defect conditions} \Longrightarrow \text{a new external shift where the defect index changes or placement succeeds}. }

The index-three corollary gives exact arithmetic data to exploit:

q≡3 mod8;
q≡1 mod3;
every prime factor of (p+q)/4 is a cube modulo q;
2 is not a cube modulo q.

A successful transfer/descent theorem from this configuration would close the first nontrivial placement defect rather than merely enlarge another finite census.