First exact shifted-factor filters for Mordell-hard primes

Corridor

Let p be a Mordell-hard prime:

Status: proved sufficient families and necessary restrictions on any prime counterexample

Date: 2026-08-15

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

Claim boundary: these filters remove infinite families and sharply constrain a hypothetical counterexample. They do not prove that the remaining intersection is empty.

1. Hard-prime facts

Let p be a Mordell-hard prime:

p\bmod840\in\{1,121,169,289,361,529\}.

Then

\boxed{p\equiv1\pmod{24}.}

In particular p≡1 mod8 and p≡1 mod3.

2. The p+1 / mod-4 filter

Take a=b=1 in the coprime divisor criterion. A certificate exists whenever

\exists k\mid p+1,\qquad k\equiv-p\equiv3\pmod4.

Thus a counterexample must satisfy:

\boxed{ \text{every odd prime factor of }(p+1)/2\text{ is }1\pmod4. }

This is the familiar simplest Type-B spine, included here to align the hierarchy.

3. Exact k=3 filter

Set

N_3^-:=\frac{p+3}{4}.

Because p≡1 mod24, this is an odd integer congruent to 1 mod3.

Theorem

If N_3^- has a divisor

a\equiv2\pmod3,

then fab(p,a,1)>0 with admissible divisor

\boxed{k=3.}

Proof

Since a|N_3^-,

4a\mid p+3.

Also p≡1 mod3 and a≡2 mod3, hence

3\mid p+a.

Thus k=3 divides a+p and

3\equiv-p\pmod{4a}

because 4a|p+3. The coprime divisor criterion with b=1 applies. QED.

Since N_3^-≡1 mod3, it has a divisor 2 mod3 if and only if it has a prime factor 2 mod3.

Therefore any prime counterexample must satisfy

\boxed{ \text{every prime factor of }\frac{p+3}{4}\text{ is }1\pmod3. }

An explicit decomposition is obtained by putting

t=\frac{p+3}{4a}, \qquad q=\frac{p+a}{3},

which gives

\frac4p = \frac1{at} + \frac1{aqt} + \frac1{pqt}.

4. Dual d=3 filter

Set

N_3^+:=\frac{3p+1}{4}.

Again N_3^+≡1 mod3.

Theorem

If N_3^+ has a divisor

c\equiv2\pmod3,

then p has a coprime fab certificate.

Construction

Put

q=\frac{N_3^+}{c}.

Because N_3^+≡1 mod3 and c≡2 mod3,

q\equiv2\pmod3.

Define

a=\frac{q+1}{3}, \qquad k=\frac{4c+1}{3}.

Both are positive integers, and since

3k=4c+1\equiv1\pmod4,

we have

\boxed{k\equiv3\pmod4.}

Now

3(a+p)=q+1+3p=q+(4cq-1)=q(4c+1)=3kq,

k\mid a+p.

Also

3(p+k)=3p+4c+1=4c(q+1)=12ac,

hence

4a\mid p+k.

Thus k satisfies the coprime divisor criterion for (a,b)=(a,1). QED.

Consequently any prime counterexample must also satisfy

\boxed{ \text{every prime factor of }\frac{3p+1}{4}\text{ is }1\pmod3. }

5. The two Eisenstein-split neighbours

A hypothetical Mordell-hard prime counterexample must therefore have

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

with every prime factor of both A and B equal to 1 mod3.

These numbers satisfy the exact relation

\boxed{B=3A-2.}

Hence the all-prime remainder is contained in the simultaneous splitting problem

\boxed{ A\text{ and }3A-2 \text{ are composed entirely of primes }1\pmod3. }

Equivalently, both lie in the multiplicative semigroup of rational primes that split in the Eisenstein quadratic field, with no ramified factor 3 and no inert factor 2 mod3.

This is a structural restriction, not a contradiction: examples exist. Its value is that every subsequent rescue theorem may assume this simultaneous split structure.

6. The p+2 / mod-8 filter

Take (a,b)=(2,1). Since hard p≡1 mod8, the target divisor class is

-p\equiv7\pmod8.

Thus

\boxed{ \exists k\mid p+2,\quad k\equiv7\pmod8 \Longrightarrow p\text{ is solved.} }

The absence of such a divisor has an exact factorization description.

Because

p+2\equiv3\pmod8,

a divisor 7 mod8 exists unless all prime factors of p+2 are in the classes

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

Indeed a prime factor 7 mod8 is itself a forbidden divisor, while a factor 3 mod8 times a factor 5 mod8 gives 7 mod8. If no 7 divisor exists, classes 3 and 5 cannot both occur; the total residue 3 mod8 then forces the nontrivial class to be 3, with odd total valuation parity.

So a counterexample must satisfy

\boxed{ \text{every prime factor of }p+2\text{ is }1\text{ or }3\pmod8. }

7. Counterexample sieve produced by theorem, not by range

Every prime counterexample in the six Mordell-hard classes must simultaneously satisfy:

(p+1)/2 has only odd prime factors 1 mod4;
(p+3)/4 has only prime factors 1 mod3;
(3p+1)/4 has only prime factors 1 mod3;
p+2 has only prime factors 1 or 3 mod8.

These are exact infinite restrictions. They should be used as assumptions in the next descent rather than merely as computational filters.