Prime-index hierarchy for FAB Kneser defects — Free Computation Foundation

Corridor

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Status: proved corollary of FAB-KNESER-DIVISOR-DEFECT.md

Date: 2026-08-15

Claim boundary: sharply bounds the non-power-residue valuation mass in any prime-index fixed-k placement defect. It does not prove that such defects cannot occur for every auxiliary k and therefore does not prove Erdős-Straus.

1. Setup

Use the notation of FAB-KNESER-DIVISOR-DEFECT.md.

For a fixed admissible k, write

C=\frac{p+k}{4}=\prod_i r_i^{e_i}

and let R be the signed divisor box. Assume the exact target is missed and let

H=\operatorname{Stab}(R).

The Kneser defect budget is

\boxed{ \sum_i \left( \min(2e_i+1,\operatorname{ord}_{G_k/H}(r_iH))-1 \right) \le [G_k:H]-2. }

2. Prime quotient index

Assume the stabilizer quotient has odd prime order

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

Then every nonidentity element of G_k/H has exact order ell.

Hence for every prime factor r_i lying outside H,

\operatorname{ord}(r_iH)=\ell

and its contribution to the defect budget is

\boxed{ \min(2e_i+1,\ell)-1 =\min(2e_i,\ell-1). }

e_i\ge\frac{\ell-1}{2},

then this one factor contributes ell-1, already larger than the entire available budget ell-2. Therefore every exceptional factor must satisfy

\boxed{e_i\le\frac{\ell-3}{2}.}

In that allowed range its contribution is exactly 2e_i.

Summing the defect inequality gives

2\sum_{r_i\notin H}e_i \le\ell-2.

The left side is even, so in fact

\boxed{ \sum_{r_i\notin H}e_i \le\frac{\ell-3}{2}. }

3. Prime-index defect theorem

Theorem

If a fixed-k FAB signed divisor box misses its target and the stabilizer quotient has odd prime index ell, then the total valuation mass of prime factors of C=(p+k)/4 outside the index-ell subgroup is bounded by

\boxed{ \sum_{ r^e\parallel C, r\notin H }e \le\frac{\ell-3}{2}. }

When the ambient unit group is cyclic and H is the unique index-ell subgroup, this says:

\boxed{ \text{total exponent of prime factors of }C \text{ that are not }\ell\text{-th-power residues} \le\frac{\ell-3}{2}. }

This is a valuation statement, not merely a bound on the number of exceptional distinct primes.

4. First cases

ell = 3

\frac{\ell-3}{2}=0.

Therefore every prime factor of C lies in the cubic-residue subgroup.

This recovers the cubic-defect theorem.

ell = 5

\frac{\ell-3}{2}=1.

There is at most one non-fifth-power prime factor, and it must occur to exponent one.

This recovers the fifth-power sparsity theorem.

ell = 7

\frac{\ell-3}{2}=2.

The complete exceptional valuation mass is at most two. The only possibilities are therefore:

one exceptional prime to exponent one;
one exceptional prime to exponent two;
two distinct exceptional primes, both simple.

Everything else in C is a seventh-power residue in the quotient.

ell = 11

The total exceptional valuation mass is at most four.

Thus even at larger prime defect index, most of the shifted factorization is forced into one high-power residue subgroup.

5. Contrapositive expansion criterion

The theorem has an immediately useful contrapositive.

Fix an odd prime ell dividing the order of the relevant cyclic unit group. If

\boxed{ \sum_{ r^e\parallel C, r\notin G^\ell }e >\frac{\ell-3}{2}, }

then index ell cannot be the stabilizer defect of a failed FAB box.

Thus every independent non-ell-power prime factor consumes two units of Kneser room, and enough such factors eliminate that defect index completely.

This turns higher-power residue diversity in the shifted integer into an exact placement weapon.

6. Relation to external nonresidue descent

At an external nonresidue prime q=3 mod4, the shifted integer

C_q=\frac{p+q}{4}

contains a prime factor that is a quadratic nonresidue modulo both p and q.

If the FAB box at q fails with odd prime defect index ell, then all but at most (ell-3)/2 units of prime-factor valuation of C_q must nevertheless lie in the ell-th-power subgroup modulo q.

Therefore a persistent failure along the external-nonresidue factor cycle requires a sequence of shifted factorizations that are simultaneously:

quadratic-sign rich enough to carry the nonresidue descent;
higher-power-residue sparse enough to remain inside the Kneser budget.

That tension is the next universal obstruction to exploit.