Shifted-factor descent for the post-DSC Erdős–Straus wall

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Status: working theorem checkpoint; exact reductions proved below, universal existence step open

Date: 2026-08-15

Project: Free Computation Foundation / CENTL

Depends on: FAB-COPRIME-DIVISOR-CRITERION.md, FAB-HARD-NONRESIDUE-BRIDGE.md, FAB-HARD-FIRST-FILTERS.md

Claim boundary: this document does not prove Erdős–Straus or López-all-primes. It records exact algebraic reductions that survive the post-DSC pivot and isolates a narrower divisor-placement theorem.

1. Starting criterion

For a prime

p\equiv1\pmod4,

and coprime positive integers a,b<p, the current fab reduction says that a positive divisor k|a+bp gives a certificate exactly when

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

Write

q=\frac{a+bp}{k}, \qquad p+k=4abc.

Then the Egyptian-fraction identity is

\frac4p = \frac1{abc} + \frac1{aqc} + \frac1{bpqc}.

The post-DSC goal is existence of at least one such certificate for every hard prime, not exact-depth realizability.

2. The b=1 four-cycle

Set b=1. Then

kq=a+p, \qquad p+k=4ac.

From

kq=a+p=a+4ac-k

we obtain

k(q+1)=a(1+4c).

The coprime criterion gives gcd(a,k)=1, hence

\boxed{k\mid1+4c.}

Write

\boxed{1+4c=kd.}

Since k≡3 mod4, necessarily

\boxed{d\equiv3\pmod4.}

Then

\boxed{q+1=ad.}

Thus every b=1 certificate sits in the exact four-link cycle

\boxed{ p+k=4ac, \quad 4c+1=kd, \quad p+a=kq, \quad q+1=ad. }

Eliminating c,q gives the single bilinear surface

\boxed{p=akd-a-k.}

Conversely, if positive integers a,k,d satisfy

k\equiv d\equiv3\pmod4, \qquad p=akd-a-k>0,

then

k\mid p+a, \qquad 4a\mid p+k,

so k is a valid b=1 divisor certificate.

Therefore:

Theorem — b=1 surface

For prime p≡1 mod4, the b=1 lane is equivalent to finding

\boxed{ a,k,d>0,\quad k\equiv d\equiv3\pmod4,\quad p=akd-a-k. }

This is an exact equivalence, not a heuristic.

3. Shifted-factor identity

Multiplying the surface relation by d gives

dp+1 =akd^2-ad-kd+1 =(ad-1)(kd-1).

Hence every b=1 certificate satisfies

\boxed{dp+1=(ad-1)(kd-1),\qquad d\equiv3\pmod4.}

This packages the remaining existence problem as a self-referential shifted-factor problem.

The already-proved d=3 filter is the first special case. Failure at d=3 forces

\frac{3p+1}{4}

to have only prime factors 1 mod3.

The mirror k=3 filter similarly forces

\frac{p+3}{4}

to have only prime factors 1 mod3 on a hypothetical counterexample.

Thus a hard counterexample must make the two neighbours

A=\frac{p+3}{4}, \qquad B=\frac{3p+1}{4}=3A-2

simultaneously Eisenstein-split.

4. Fixed-k divisor-square reduction

Return to general coprime a,b and fix a prospective divisor k≡3 mod4. Put

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

A certificate with this k has

C=abc.

Using

p=4abc-k,

we get

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

Because gcd(a,k)=1, the condition k|a+bp is equivalent to

\boxed{4b^2c\equiv-1\pmod k.}

Define

u=b^2c.

Then u|C^2.

The converse divisor realization is exact:

Lemma — every divisor of C^2 is a b^2 c realization

For every positive divisor

u\mid C^2,

there exist positive integers a,b,c such that

abc=C, \qquad \gcd(a,b)=1, \qquad b^2c=u.

Proof

Work prime by prime. If

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

then choose exponent triples (alpha,beta,gamma) for (a,b,c) as follows.

If U<=E, take

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

If U>=E, take

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

In both cases

\alpha+\beta+\gamma=E, \qquad 2\beta+\gamma=U,

and never both alpha,beta are positive, so gcd(a,b)=1. Combining the local choices proves the lemma. QED.

Therefore, whenever k<3p (so C<p and the reconstructed a,b automatically satisfy a,b<p), we have:

Theorem — fixed-k divisor box

\boxed{ \text{a coprime fab certificate with divisor }k \iff \exists u\mid C^2: 4u\equiv-1\pmod k, \quad C=\frac{p+k}{4}. }

For larger k, the same arithmetic equivalence holds provided the reconstructed a,b satisfy the framework's size hypotheses.

This removes the apparent three-parameter search. At fixed k, the wall is one divisor-box hit in Div(C^2).

5. Character alignment for an external nonresidue prime

Let k now be an odd prime with

k\equiv3\pmod4

and suppose

\left(\frac{k}{p}\right)=-1.

Because p≡1 mod4, quadratic reciprocity gives

\left(\frac{p}{k}\right)=-1.

Hence

\boxed{ \left(\frac{-p}{k}\right)=+1. }

Also

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

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

The target residue

-4^{-1}\pmod k

is also a quadratic nonresidue because (-1/k)=-1 and 4 is a square.

Thus the fixed-k divisor-box problem has no quadratic-character obstruction: both C and the target class lie on the same nonresidue side.

What remains is exact divisor placement inside that coset.

This is a useful narrowing, not an existence proof.

6. Prescribing the leftover nonresidue c = ell

Let ell be an odd prime with

\left(\frac{\ell}{p}\right)=-1.

Suppose k is an odd prime satisfying

k\equiv3\pmod4, \qquad k\equiv-p\pmod\ell.

Then

\boxed{ \left(\frac{-\ell}{k}\right)=+1. }

Proof

Since p≡1 mod4, reciprocity gives (p/ell)=(ell/p)=-1. From k≡-p mod ell,

\left(\frac{k}{\ell}\right) = \left(\frac{-p}{\ell}\right).

Applying reciprocity between k and ell and using k≡3 mod4 yields (ell/k)=-1; multiplying by (-1/k)=-1 gives (-ell/k)=+1. QED.

Therefore the quadratic congruence

\boxed{4\ell b^2\equiv-1\pmod k}

is automatically solvable.

Again, the remaining issue is not character theory. It is choosing a square-root representative b compatible with the exact factorization

\frac{p+k}{4}=ab\ell.

This isolates the missing theorem as a divisor-placement problem.

7. Bezout and norm form when c = ell

Assume a certificate has c=ell. From

1+4b^2\ell=kd, \qquad q+b=ad,

put

t=q=ad-b.

Then

\boxed{4\ell bt-pd=1.}

Indeed,

p=4ab\ell-k

and multiplication by d gives

pd=4ab\ell d-(1+4b^2\ell) =4b\ell(ad-b)-1.

There is also the exact factorization

\boxed{ (4a\ell b-p)(4a\ell t-p) =p^2+4a^2\ell. }

The first factor is exactly

4a\ell b-p=k.

Thus any certificate with prescribed external nonresidue ell gives a factorization of the norm-like integer

\boxed{p^2+4a^2\ell.}

This is the cleanest present bridge to the repository's quadratic-field / norm machinery.

8. Immediate universal target

The current reductions point to one narrow statement.

For a hypothetical Mordell-hard counterexample p:

the first exact shifted-factor filters force simultaneous splitting restrictions on (p+1)/2, (p+3)/4, (3p+1)/4, p+2, and related small shifts;
any coprime certificate must import an external quadratic nonresidue prime ell>=11;
after fixing a prospective divisor k, the entire fab search becomes

\exists u\mid ((p+k)/4)^2:\quad4u\equiv-1\pmod k;

when k is a 3 mod4 quadratic nonresidue of p, the character obstruction to this congruence vanishes automatically;
prescribing the external nonresidue c=ell converts the problem into the Bezout/norm equations above.

The next theorem should therefore not be another finite k census. It should prove a divisor-placement result of one of the following forms:

\boxed{ \text{character alignment} +\text{hard-prime split restrictions} \Longrightarrow \text{target divisor in }\operatorname{Div}(C^2), }

or, equivalently, a norm-factor selection theorem for

p^2+4a^2\ell.

That is the present shortest route toward the all-prime wall.