Two-divisibility constructor for the fab master equation

Theorem

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Status: proved exact sufficient construction

Date: 2026-08-15

Depends on: FAB-DUAL-DESCENT-SYSTEM.md, FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md

Claim boundary: this converts one four-variable certificate search into two elementary divisibilities and generates explicit congruence families. It does not prove that one such family covers every Mordell-hard prime and therefore does not prove Erdős-Straus.

1. Constructor

Let p be a positive odd integer. Choose positive integers

B,Q,s

such that

\boxed{s\mid pQ+B}

and

\boxed{4BQ\mid p+s.}

Define

\boxed{A=\frac{pQ+B}{s}}

and

\boxed{c=\frac{p+s}{4BQ}}.

Then A,c are positive integers.

The two defining divisibilities imply the exact master identity

\boxed{4ABcQ=B+p(A+Q).}

Proof

Because sA=pQ+B,

A(p+s) =pA+sA =pA+pQ+B =B+p(A+Q).

But p+s=4BQc, so

A(p+s)=4ABQc.

Therefore

4ABcQ=B+p(A+Q).

QED.

2. Immediate Erdős-Straus certificate

Put

\boxed{k=4ABc-p.}

From the master identity,

Qk =Q(4ABc-p) =B+pA.

Hence

\boxed{k\mid B+pA}

with quotient exactly Q.

Also

\boxed{p+k=4ABc.}

Thus the unbounded sufficient fab identity applies with

a=B, \qquad b=A, \qquad q=Q, \qquad t=c.

The resulting positive decomposition is

\boxed{ \frac4p = \frac1{ABc} + \frac1{BQc} + \frac1{ApQc}. }

Therefore:

Two-divisibility rescue theorem

If there exist positive B,Q,s satisfying

\boxed{ s\mid pQ+B, \qquad 4BQ\mid p+s, }

then p satisfies the Erdős-Straus equation.

No primality condition on s, Q, or the resulting k is required.

3. Congruence-family form

For fixed B,Q,s, the two conditions are simply

\boxed{p\equiv-s\pmod{4BQ}}

and

\boxed{Qp\equiv-B\pmod s.}

Thus every fixed triple (B,Q,s) defines either an empty congruence system or an explicit arithmetic progression of integers solved by one closed formula.

g=\gcd(Q,s),

the second congruence is soluble exactly when

\boxed{g\mid B.}

After division by g, it becomes one residue class modulo s/g; compatibility with the first congruence is then an ordinary CRT test.

This gives a systematic generator of exact ES congruence families directly from the master equation.

4. Relation to the dual system

The variables are exactly the dual variables already present in FAB-DUAL-DESCENT-SYSTEM.md.

Under the identification

a=B, \qquad b=A, \qquad q=Q,

the new variable s is the swapped dual certificate divisor k', because the dual identity is

\boxed{k' b=a+pq.}

Indeed here

\boxed{sA=B+pQ.}

The second constructor condition is exactly the corresponding dual divisibility

\boxed{p+s=4BcQ,}

which is the swapped form of

p+k'=4acq.

The hidden 3 mod 4 cofactor from the dual-descent system is a different variable, say d, and is recovered only after the certificate exists through

\boxed{A+Q=Bd.}

So the constructor should be read as choosing the dual certificate divisor s=k' first, not as choosing the hidden cofactor.

The point is operational: instead of searching four positive variables subject to

4ABcQ=B+p(A+Q),

one may choose the three simpler variables B,Q,s and check only two divisibilities.

5. Finite-cover falsification checkpoint

As a theorem-mining test, small triples were generated and converted to their exact CRT progressions. Restricting to families whose combined moduli divide

840\cdot11\cdot13\cdot17\cdot19,

a search over B,Q<=30 and s<=100 produced 3,126 distinct compatible progression families in that modulus envelope.

Their union does not cover the six Mordell-hard progressions. Large exact residue cores remain in every hard class.

This finite negative result is not a theorem that no finite covering exists. It only rules out the tempting small-parameter cover tested here and prevents confusing the constructor itself with a completed proof.

6. Research use

The theorem creates a clean fork for the remaining attack:

covering route: find a genuinely complete finite or structured infinite family of triples (B,Q,s);
descent route: assume the two divisibilities fail throughout a controlled family and translate that failure into factor/character restrictions;
external-nonresidue route: choose the dual divisor s or the original certificate divisor k from the synchronized external-nonresidue packet and use the hard shield to force the complementary divisibility.

The useful object is now the pair

\boxed{ s\mid pQ+B, \qquad p\equiv-s\pmod{4BQ}, }

rather than the original four-variable master equation.