Unbounded strong sufficient certificates and the fixed-k divisor-ratio box

Certificate

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Status: proved exact algebraic reduction for a strong sufficient subclass

Date: 2026-08-15

Project: Free Computation Foundation / CENTL

Depends on: FAB-COPRIME-DIVISOR-CRITERION.md, FAB-SHIFTED-FACTOR-DESCENT.md

Claim boundary: Bello-Hernández–Benito–Fernández define fab(n,a,b) for arbitrary positive a,b; they do not impose a,b<n. The bounds a,b<p entered the FCF coprime simplification in FAB-COPRIME-DIVISOR-CRITERION.md. This note removes that auxiliary size restriction only for the stronger sufficient congruence/divisibility subclass below. It does not claim to characterize every fab certificate and does not prove Erdős–Straus.

1. Strong sufficient identity

Let n,a,b,k be positive integers and assume

\boxed{k\mid a+bn}

and

\boxed{4ab\mid n+k.}

Put

q=\frac{a+bn}{k}, \qquad t=\frac{n+k}{4ab}.

Then

kq=a+bn, \qquad k=4abt-n.

Substitution gives

(4abt-n)q=a+bn,

\boxed{4abqt=a+n(b+q).}

Therefore

\boxed{ \frac4n = \frac1{abt} + \frac1{aqt} + \frac1{bnqt}. }

This identity is exact and requires no size bound on a,b.

Provenance correction

The original 2026 fab definition already allows arbitrary positive a,b. The restriction a,b<p was used only in the FCF coprime divisor criterion to collapse the full fab admissibility conditions to the single congruence

k\equiv-p\pmod{4ab}.

The two conditions in this section are stronger than general fab admissibility, but they are sufficient for an Erdős–Straus decomposition and remain valid for arbitrarily large auxiliaries.

We call such data a strong sufficient certificate in this note.

2. Fixed-k setup

Let p be an odd prime and k a positive integer with

\gcd(p,k)=1, \qquad p+k\equiv0\pmod4.

Put

\boxed{C=\frac{p+k}{4}.}

Then

\gcd(C,k)=1.

A strong sufficient certificate using this k is equivalent to positive a,b,c satisfying

abc=C

and

k\mid a+bp,

because p+k=4abc then automatically gives 4ab|p+k.

3. Divisor-square equivalence for the strong subclass

For

C=abc

we have

p=4abc-k.

Hence

a+bp =a+4ab^2c-bk =a(1+4b^2c)-bk.

Since a|C and gcd(C,k)=1,

\gcd(a,k)=1.

Therefore

\boxed{ k\mid a+bp \iff k\mid1+4b^2c.}

Put

u=b^2c.

Then u|C^2.

Conversely, every divisor u|C^2 can be realized as b^2c inside a factorization abc=C with gcd(a,b)=1. Prime by prime, if

v_r(C)=E, \qquad v_r(u)=U, \qquad 0\le U\le2E,

choose exponent triples (alpha,beta,gamma) for (a,b,c) by

(\alpha,\beta,\gamma)=(E-U,0,U)

when U<=E, and

(\alpha,\beta,\gamma)=(0,U-E,2E-U)

when U>=E.

Thus:

Theorem — fixed-k strong divisor-square certificate

For odd prime p and positive k with

\gcd(p,k)=1, \qquad p+k\equiv0\pmod4,

there exists a strong sufficient certificate using this k if and only if

\boxed{ \exists u\mid C^2: \quad 4u\equiv-1\pmod k, \qquad C=\frac{p+k}{4}. }

This is not asserted to characterize every possible fab certificate with that k; it characterizes the stronger 4ab|p+k subclass.

4. Divisor-ratio box

Because

\frac{u}{C} = \frac{b^2c}{abc} = \frac ba,

and

4C\equiv p\pmod k,

the target congruence becomes

\boxed{ \frac ba\equiv-p^{-1}\pmod k. }

C=\prod_r r^{E_r},

the possible ratios b/a are exactly the signed exponent box

\boxed{ \mathcal R_k(C) = \left\{ \prod_r r^{z_r}\bmod k: -E_r\le z_r\le E_r \right\}. }

Hence:

Theorem — fixed-k strong divisor-ratio box

\boxed{ \text{strong sufficient certificate with divisor }k \iff -p^{-1}\in\mathcal R_k(C), \qquad C=\frac{p+k}{4}. }

This gives a precise multiplicative-box subproblem inside the larger all-prime fab wall.

5. Three canonical points of the strong box

Center: b/a = 1

Here u=C. The target condition gives

4C\equiv-1\pmod k.

Since 4C=p+k,

\boxed{k\mid p+1.}

This is the familiar simplest Type-B / p+1 spine.

Upper endpoint: b/a = C

Here u=C^2. The target becomes

k\mid4C^2+1,

which modulo k is equivalent to

\boxed{k\mid p^2+4.}

after multiplication by 4.

For a hard prime p≡1 mod4, any k with p+k≡0 mod4 satisfies k≡3 mod4. Such a k>1 has a prime divisor q≡3 mod4. But a 3 mod4 prime cannot divide the coprime sum of two squares

p^2+2^2.

Therefore the upper endpoint cannot solve a hard prime.

Lower endpoint: b/a = C^{-1}

Here u=1, so k|5. Positive divisors of 5 are 1 mod4, while the hard-prime shift k is 3 mod4. Thus the lower endpoint also cannot solve a hard prime.

6. Structural consequence

Within this strong sufficient subclass, a hard prime escaping the p+1 spine can be rescued only by a genuinely asymmetric interior signed divisor of

C=\frac{p+k}{4}.

The three canonical box points are

\boxed{ C^{-1}:\text{ impossible}, \qquad 1:\text{ the }p+1\text{ spine}, \qquad C:\text{ impossible by sums of two squares}. }

This is a useful local theorem target for the repository's multiplicative quotient / defect / zero-sum machinery, but the all-prime proof must remember that general fab admissibility is broader than this strong box.