Fixed-k fab criterion collapses to one divisor of N²

Corridor

Let

Status: proved exact theorem

Date: 2026-08-15

Depends on: FAB-FIXED-K-SIGNED-DIVISOR.md

Claim boundary: this is an exact fixed-k reformulation. It does not prove that some k always succeeds and therefore does not prove Erdős-Straus.

1. Setup

Let

p\equiv1\pmod4

be prime, and let

k\equiv3\pmod4, \qquad \gcd(k,p)=1.

Put

\boxed{N=N_k=\frac{p+k}{4}.}

Then

\gcd(N,k)=1.

Assume N<p, for example k<3p, so every divisor of N automatically satisfies the size hypotheses in the coprime fab criterion.

2. The theorem

Fixed-k square-divisor criterion

There exists a coprime fab certificate using the fixed admissible divisor k if and only if

\boxed{ \exists D\mid N^2 \quad\text{such that}\quad 4D\equiv-1\pmod k. }

Equivalently,

\boxed{ D\equiv-4^{-1}\pmod k. }

For

k=4s+3,

one has

4^{-1}\equiv s+1\pmod k,

so the fixed target class is

\boxed{D\equiv3s+2\pmod{4s+3}.}

Crucially, this target residue depends only on k, not on p.

3. Proof from signed exponents

Factor

N=\prod r^{e_r}.

By FAB-FIXED-K-SIGNED-DIVISOR.md, fixed-k solvability is equivalent to choosing integers

-e_r\le z_r\le e_r

such that

\prod r^{z_r}\equiv-p\pmod k.

But

4N=p+k\equiv p\pmod k,

hence

-p\equiv-4N\pmod k.

Write

d_r=e_r-z_r.

Then

0\le d_r\le2e_r.

Therefore

D:=\prod r^{d_r}

runs through exactly all positive divisors of N^2.

Also

\prod r^{z_r} =N D^{-1}\pmod k.

Thus the target equation becomes

N D^{-1}\equiv-4N\pmod k.

Since N is invertible modulo k, cancel it:

D^{-1}\equiv-4\pmod k,

or equivalently

\boxed{4D\equiv-1\pmod k.}

This proves the equivalence. QED.

4. Direct factor reconstruction

The divisor D|N^2 contains the complete fab data.

For each prime r^e||N, let

d=v_r(D), \qquad0\le d\le2e.

Define the exponents

v_r(a)=\max(e-d,0),

v_r(b)=\max(d-e,0),

v_r(t)=e-|e-d|.

Then

\boxed{N=abt,}

\boxed{D=b^2t,}

and

\gcd(a,b)=1.

The congruence

4D\equiv-1\pmod k

4b^2t\equiv-1\pmod k.

Since

p+k=4N=4abt,

the fixed-k fab equations follow exactly.

Let

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

Then the decomposition is

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

5. Immediate examples

k=3

Here

-4^{-1}\equiv2\pmod3.

Thus fixed k=3 succeeds iff

\boxed{ N_3^2 \text{ has a divisor }2\pmod3. }

That occurs exactly when N_3=(p+3)/4 has a prime factor 2 mod3, recovering the first Eisenstein filter.

k=7

Here

-4^{-1}\equiv5\pmod7.

Thus fixed k=7 succeeds iff

\boxed{ N_7^2 \text{ has a divisor }5\pmod7. }

This is exactly the residue-product problem classified in FAB-K7-EXACT-FILTER.md.

k=11

Here

-4^{-1}\equiv8\pmod{11}.

So the entire fixed-11 problem is simply:

\boxed{ \exists D\mid((p+11)/4)^2, \quad D\equiv8\pmod{11}. }

No separate search over a,b is needed.

6. Shift formulation

Let

A=\frac{p+3}{4}.

For k=4s+3,

N_k=A+s.

Therefore the all-prime problem contains the following exact subproblem:

Given A=(p+3)/4, prove that for some s>=0, the square of the shifted integer A+s has a divisor congruent to <div class="math" role="math">> 3s+2\pmod{4s+3}.

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A finite set of s values would give a finite universal fab menu if proved sufficient. An expanding-set descent would also suffice.

This is a substantially simpler arithmetic target than the original two-parameter fab conditions.