Fixed-k fab certificates are signed divisor products

Status: proved exact theorem

Date: 2026-08-15

Depends on: FAB-COPRIME-DIVISOR-CRITERION.md

Claim boundary: this is an exact reformulation of the coprime fab problem for a fixed admissible divisor k. It does not prove that the signed product target is always attained and therefore does not prove Erdős-Straus.

1. Setup

Let p be a prime with

and fix an odd positive integer

with

Put

Then N_k is a positive integer and

Indeed, multiplication by 4 is invertible modulo odd k, and

Factor

All prime factors r are therefore units modulo k.

We consider coprime positive integers a,b satisfying

When N_k<p (for example whenever k<3p), the size hypotheses a,b<p in the coprime fab criterion are automatic.

2. From a,b to signed exponents

Because gcd(a,b)=1 and ab|N_k, each prime power r^{e_r} of N_k is distributed in exactly one of three ways:

1. some exponent z_r with 1<=z_r<=e_r goes into a;
2. some exponent -z_r with 1<=-z_r<=e_r goes into b;
3. none of that prime enters a or b.

Equivalently, choose

and define

This gives every coprime pair a,b with ab|N_k, and gives it uniquely.

Because every r is invertible modulo k, the ratio satisfies

3. Exact fixed-k criterion

By FAB-COPRIME-DIVISOR-CRITERION.md, the fixed integer k is a fab-admissible divisor for (p,a,b) exactly when

and

The second condition is precisely

Since b is invertible modulo k, the first condition is equivalent to

Combining with the signed-exponent representation gives the theorem.

Theorem — fixed-k signed-divisor criterion

Assume N_k<p (or separately require the resulting a,b<p). Then there is a coprime fab certificate using the fixed admissible divisor k if and only if

In words: the target class -p mod k must be representable as a signed product of prime factors of (p+k)/4, using each prime no more often than its actual valuation.

The zero exponent means that prime power is left in the unused factor

4. Explicit reconstruction

Given signed exponents attaining -p mod k, set

Then

and in fact

The congruence

is equivalent to

Put

The resulting Erdős-Straus decomposition is

Thus a signed-product witness is immediately constructive.

5. Group-theoretic interpretation

Let

For each prime power r^{e_r}||N_k, define the symmetric local set

Then fixed-k solvability is exactly

This is a finite restricted product-set problem in an abelian group.

Consequently, failure at a fixed k is not vague factorization bad luck. It says that the symmetric product set generated by the actual prime factors of N_k misses one explicitly prescribed group element.

6. Shift form

For the especially useful family

write

Then

So the fixed-divisor sequence

corresponds to the consecutive shifted integers

For prime or otherwise convenient k≡3 mod4, each shift imposes a finite signed-product obstruction in G_k.

This converts the all-prime problem into a simultaneous factor-residue problem across short shifts of one integer A.

7. Research target

A prime counterexample must make the target product element fail for every admissible k.

A promising finite-window theorem would therefore have the form:

for a fixed finite set K, or else prove a descent showing that simultaneous failure over an expanding set of shifts is impossible.

Computational success for a bounded K is only evidence; a proof must use the group/product-set structure above.