Coprime fab certificates reduce to one divisor congruence

Corridor

Let p be a prime with

Status: proved exact reformulation

Date: 2026-08-15

External framework: Bello-Hernández, Benito, Fernández, arXiv:2606.10922v1

Claim boundary: this is a reformulation of the admissibility conditions in the recent divisor-parametrization framework. It does not prove that such a divisor always exists and therefore does not prove Erdős-Straus.

1. Setup

Let p be a prime with

p\equiv1\pmod4.

Let a,b be positive integers satisfying

\gcd(a,b)=1, \qquad a<p, \qquad b<p.

For a positive divisor k of a+bp, put

q=\frac{a+bp}{k}.

The fab admissibility conditions are

k\equiv3\pmod4,

4b\mid q(p+k),

and

4a\mid p q(p+k).

2. Coprimality collapse

Because

kq=a+bp,

any common divisor of b and kq divides a; since gcd(a,b)=1,

\boxed{\gcd(b,kq)=1.}

Likewise any common divisor of a and kq divides bp. Since a<p, p is coprime to a, and gcd(a,b)=1, so

\boxed{\gcd(a,kq)=1.}

In particular

\gcd(a,pq)=1, \qquad \gcd(b,q)=1.

If k≡3 mod4, then with p≡1 mod4,

p+k\equiv0\pmod4.

Write

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

The two remaining fab conditions become

b\mid qc, \qquad a\mid pqc.

By the coprimalities above these are equivalent to

\boxed{b\mid c,\qquad a\mid c.}

Since gcd(a,b)=1, this is equivalent to

\boxed{ab\mid c.}

Thus

4ab\mid p+k,

or equivalently

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

Because p≡1 mod4, this congruence already forces k≡3 mod4.

3. The theorem

Coprime divisor criterion

For prime p≡1 mod4 and coprime a,b<p, a positive divisor k of a+bp is fab-admissible if and only if

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

Equivalently,

\boxed{ \operatorname{fab}(p,a,b)>0 \iff \exists k\mid(a+bp): k\equiv-p\pmod{4ab}. }

The right side is now a pure divisor-in-residue-class condition.

4. Explicit decomposition

Write

p+k=4abt

and

q=\frac{a+bp}{k}.

The divisor identity becomes

\boxed{ \frac4p = \frac1{abt} + \frac1{aqt} + \frac1{bpqt}. }

Indeed,

4abqt=a+p(b+q)

follows from kq=a+bp and k=4abt-p, and clearing denominators verifies the identity.

5. Edge cases a=1 or b=1

b=1

A certificate is equivalent to finding a divisor

k\mid p+a

with

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

a=1

A certificate is equivalent to finding a divisor

k\mid1+bp

with

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

For a=b=1, this says exactly that p+1 has a divisor 3 mod4, recovering the familiar simplest Type-B spine.

6. New all-prime target

The 2026 divisor-parametrization paper reports that every tested prime

5\le p\equiv1\pmod4, \qquad p<10^{14},

has a certificate with 1<=a,b<=11.

For coprime pairs, the theorem above translates that phenomenon into the concrete statement:

For each tested prime, at least one small linear form a+bp has a divisor in the single target class -p mod 4ab.

This suggests a cleaner proof target than universal exact-depth realizability:

\boxed{ \forall p\equiv1\pmod4\text{ prime}, \quad \exists\ (a,b)=1 \text{ with a controlled size and } \exists k\mid(a+bp), \ k\equiv-p\pmod{4ab}. }

No bounded universal theorem is claimed here. The point is that the ES wall has been reduced to a precise divisor-distribution statement that can be attacked with reciprocity, shifted-factor structure, or a descent argument.