Exact `2p+1` factor filter — Free Computation Foundation

Corridor

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Status: proved elementary sufficient family / necessary counterexample restriction

Date: 2026-08-15

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

Claim boundary: this removes an infinite family and adds one exact linear-form restriction on any prime counterexample. It does not prove Erdős--Straus.

1. The pair `(a,b)=(1,2)`

Let p be a Mordell-hard prime, so

p\equiv1\pmod8.

The coprime pair

(a,b)=(1,2)

has linear form

a+bp=1+2p=2p+1

and modulus

4ab=8.

The coprime divisor criterion says that this pair yields an Erdős--Straus certificate if and only if 2p+1 has a positive divisor

k\equiv-p\pmod8.

Because p≡1\pmod8,

\boxed{-p\equiv7\pmod8.}

2. Theorem

If 2p+1 has a divisor

k\equiv7\pmod8,

then p satisfies Erdős--Straus.

Proof

The displayed congruence is exactly the coprime divisor criterion for (1,2). Write

p+k=8t, \qquad q=\frac{2p+1}{k}.

The general reconstruction gives the explicit decomposition

\boxed{ \frac4p = \frac1{2t} + \frac1{qt} + \frac1{2pqt}. }

QED.

In particular any prime factor of 2p+1 that is itself 7\bmod8 solves p.

3. Exact miss restriction

For hard p,

2p+1\equiv3\pmod8.

A divisor congruent to 7\bmod8 exists unless every prime factor of 2p+1 lies in the classes

\boxed{1\text{ or }3\pmod8.}

Indeed a prime factor 7\bmod8 is itself a forbidden divisor, while a factor 3\bmod8 times a factor 5\bmod8 produces a divisor 7\bmod8. If no 7\bmod8 divisor exists, the classes 3 and 5 cannot both occur. The total residue 3\bmod8 then forces the nontrivial class to be 3.

Thus a hard-prime counterexample must satisfy

\boxed{ q\mid(2p+1),\ q\text{ prime} \Longrightarrow q\equiv1\text{ or }3\pmod8. }

This is the same pair of residue classes forced on p+2 by the existing (2,1) filter, now imposed on the dual linear form 2p+1.

4. Forced factor `3`

Hard primes satisfy p≡1\pmod{24}, hence

\boxed{3\mid(2p+1).}

The prime 3 is itself 3\bmod8, so it is allowed by the miss restriction. A miss therefore means that the cofactor

\frac{2p+1}{3^v}

is composed entirely of primes 1\bmod8, except possibly further primes 3\bmod8 whose total 3\bmod8 valuation keeps every partial product out of the class 7.

5. Place in the linear-form sieve

A hypothetical hard-prime counterexample must now simultaneously place all of the following forms in restricted quadratic-residue semigroups:

(p+1)/2 — primes 1\bmod4;
(p+3)/4 — primes 1\bmod3;
(3p+1)/4 — primes 1\bmod3;
p+2 — primes 1 or 3\bmod8;
4p+1 — primes 1\bmod4;
p+4 — primes 1\bmod4;
2p+1 — primes 1 or 3\bmod8.

These are exact infinite restrictions, not range-limited computations.

6. Finite regression signal

Among Mordell-hard primes through 500{,}000, the first four filters leave 202 survivors, 4p+1 and p+4 then leave 78, and the present theorem removes 37 of those 78. The remaining 41 are all solved by some two-target shift

k\in\{7,11,15,19,23,31\}.

That last count is finite evidence only.

1. The pair (a,b)=(1,2)