Exact fixed-k=7 filter for Mordell-hard primes

Status: proved exact factor-residue classification

Date: 2026-08-15

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

Claim boundary: this completely classifies when the fixed fab divisor k=7 succeeds for a Mordell-hard prime. It is one filter, not an all-prime proof.

1. Setup

Let p be a Mordell-hard prime. Then

so p is always a quadratic residue modulo 7.

Set

Because 4^{-1}=2 mod7,

Use 3 as a primitive root modulo 7:

The target signed-product element is

Its discrete-log exponent is:

p mod 7	N_7 mod 7	-p mod 7	target exponent mod 6
1	2	6	3
2	4	5	5
4	1	3	1

Thus the target is always an odd exponent, as expected from quadratic character.

By FAB-FIXED-K-SIGNED-DIVISOR.md, k=7 succeeds exactly when that target exponent is a signed sum of the exponents of prime factors of N_7, with each prime usable up to its valuation.

Below, “occurrence” counts prime factors with multiplicity.

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2. Cases p = 2 or 4 mod 7

Here the target exponent is 5 or 1, i.e. ±1 mod6.

Immediate successes

A prime factor of N_7 congruent to

has exponent ±1, so by using that factor in a or b we hit the target immediately.

If no such factor occurs, the only nonzero exponent types are:

• residues 2,4: exponent ±2;
• residue 6: exponent 3.

A 6 mod7 factor together with any 2 or 4 mod7 factor gives

so that also solves the target.

Conversely:

• using only residues 1,2,4 produces only even exponents;
• using only residues 1,6 produces only exponents 0 or 3.

Neither can produce ±1.

Theorem A

If

then fixed k=7 succeeds if and only if at least one of the following holds:

1. N_7 has a prime factor 3 or 5 mod7;
2. N_7 has both a prime factor 6 mod7 and a prime factor 2 or 4 mod7.

Equivalently, failure occurs exactly when the prime support of N_7 lies wholly in one of

or

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3. Case p = 1 mod 7

Now the target is

which has exponent 3 mod6.

A 6 mod7 factor

Any prime factor

already has exponent 3, so it solves the target by itself.

Hence assume no 6 mod7 factor occurs.

Mixing a ±1 factor with a ±2 factor

A factor 3 or 5 mod7 contributes ±1. A factor 2 or 4 mod7 contributes ±2.

Choosing signs appropriately gives

So the presence of both types solves the target.

Three ±1 occurrences

Any three occurrences from residue classes 3 or 5 mod7 can each be signed to contribute +1, yielding

This remains true when several occurrences come from the valuation of the same prime, because the signed exponent may use any integer between -e and e.

Thus multiplicity at least 3 among the 3/5 classes solves the target.

Necessity

If none of the three mechanisms above occurs, then:

• there is no exponent-3 factor;
• either there is no odd-exponent (±1) factor at all, in which case every signed sum is even;
• or there are at most two ±1 occurrences and no nonzero even (±2) factor. Their possible signed sums are among 0,±1,±2, never 3 mod6.

Hence no other solution is possible.

Theorem B

If

then fixed k=7 succeeds if and only if at least one of:

1. N_7 has a prime factor 6 mod7;
2. N_7 has a prime factor 3 or 5 mod7 and also one 2 or 4 mod7;
3. the total valuation multiplicity of prime factors 3 or 5 mod7 is at least 3.

This is an exact classification.

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4. Counterexample consequences

A Mordell-hard prime surviving the k=7 filter must have a highly restricted factorization of

Combined with FAB-HARD-FIRST-FILTERS.md, any genuine prime counterexample must simultaneously satisfy exact restrictions on

The first, second, third, and fifth restrictions come from mod 4, mod 8, and Eisenstein-splitting filters; the present theorem completely specifies the mod-7 signed-divisor obstruction on the fourth shifted form.

5. Why this is useful

The theorem demonstrates that the fixed-k signed-product formulation is not merely computational packaging. For k=7 it yields a complete human-scale obstruction classification.

The natural next targets are k=11, 19, and other small 3 mod4 primes, followed by a simultaneous theorem showing that the resulting shifted-factor obstruction classes cannot all hold for one Mordell-hard prime.