q=3 Fiber Injectivity and Three-Layer Minimum — Free Computation Foundation

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Status: proved universal theorem

Date: 2026-08-15

Depends on: REDUCED-PARAMETER-DOMAIN.md, Q3-ABSORPTION.md, Q3-WEAK-REDUNDANCY.md, Q3-POINTWISE-ABSORPTION.md

Claim boundary: proves that every q=3 Type A/B pullback is a singleton or empty and therefore that a corrected-domain q=3 cover needs at least three distinct layers. It does not rule out such a three-layer cover universally and does not prove DSC-P, López-all-primes, or Erdős-Straus.

1. Setup

Let

m=4j-1

and assume

3\mid m.

Put

\boxed{n=m/3.}

The Type A/B trap set is

T_j=\{-e,-4e\pmod m:e\mid j\}.

It is convenient to negate it:

S_j=-T_j=\{e,4e\pmod m:e\mid j\}.

Since

4j=m+1=3n+1,

we have

\boxed{j=(3n+1)/4<n}

for every nontrivial case.

2. Theorem — trap reduction modulo n is injective

Theorem

The natural reduction map

\boxed{T_j\longrightarrow\mathbb Z/n\mathbb Z}

is injective.

Equivalently, if two Type A/B traps agree modulo n=m/3, then they were already the same trap modulo m.

Proof

Work with the normalized set S_j=-T_j.

For every divisor e|j:

the plain value e satisfies 1<=e<=j<n;
if e<j, then e<=j/2, hence

4e\le2j=(3n+1)/2<2n;

if e=j, then

4e=4j=m+1\equiv1\pmod m,

which is already the same normalized trap as the divisor 1.

Thus every distinct element of S_j has a representative in [1,2n).

Suppose two distinct representatives are congruent modulo n. Their difference must therefore be exactly n or -n.

There are only three structural cases.

Case A: plain/plain

e-f=\pm n,

then |e-f|<n because e,f<n, impossible.

Case B: fourfold/fourfold

4e-4f=\pm n,

then the left side is divisible by 4, while n is odd. Impossible.

Case C: fourfold/plain

The only potentially nontrivial equation is

4e-f=n.

(The opposite sign would force the plain divisor to exceed n>j; the other mixed orientation is symmetric.)

Since f>0, this gives

4e>n>\frac{4j-1}{3},

e>j/3-1/12.

As e is a proper divisor of j in a genuinely distinct mixed pair, the integer quotient j/e is at least 2. The displayed lower bound forces

\boxed{j/e=2},

so e=j/2.

Substituting into 4e-f=n gives

f=2j-n =2j-\frac{4j-1}{3} =\frac{2j+1}{3}.

For j>1, this satisfies

\frac j2<f<j.

But a positive proper divisor of j cannot lie strictly between j/2 and j. Contradiction.

The tiny endpoint is checked by the same formulas and produces no distinct trap collision.

Therefore no two distinct normalized traps can agree modulo n. Negation is a bijection, so the same is true for T_j. QED.

3. Corollary — every q=3 pullback is empty or singleton

Fix a candidate progression

x(s)=r+Ls

with

q_j=\frac{m_j}{\gcd(L,m_j)}=3.

Then

\gcd(L,m_j)=n=m_j/3.

The three parameter classes modulo 3 map bijectively to the three lifts modulo m_j above the single residue r mod n.

By the theorem, at most one of those three lifts can belong to T_j.

Hence

\boxed{|R_j|\le1.}

This is universal. No hard-class assumption, finite search, or primitivity assumption is needed.

4. Corollary — corrected q=3 coverage needs at least three layers

For the Mordell-hard Type A/B program,

3\mid840\mid L.

The exact Dirichlet condition therefore imposes no forbidden parameter class modulo 3 merely for reducedness. The local reduced domain is the full ring

\boxed{\mathbb Z/3\mathbb Z=\{0,1,2\}.}

Since every q=3 layer forbids at most one class, any q=3 cover of the corrected domain requires at least three distinct nonempty layers:

\boxed{ \bigcup_jR_j=\mathbb Z/3\mathbb Z \Longrightarrow \#\{j:R_j\ne\varnothing\}\ge3. }

In particular, the old two-singleton R={1} / R={2} complementary-pair obstruction was an artifact of restricting the parameter to units modulo 3. In the exact Dirichlet domain it leaves the third class.

5. Combination with the absorption hierarchy

On a directly novel candidate:

strong q=3 descendants cannot be nonempty (Q3-ABSORPTION.md);
weak q=3 descendants add no new residue (Q3-WEAK-REDUNDANCY.md);
every actually used trap must be pointwise primitive (Q3-POINTWISE-ABSORPTION.md);
by this theorem, every surviving layer contributes at most one class.

Therefore a genuine directly novel q=3 obstruction has the exact form

\boxed{ \text{three or more pointwise-primitive base singleton rows} \text{ whose classes cover }\{0,1,2\}. }

That is now the residual q=3 theorem target.

6. Proof-mining observation

A direct reconstruction of the corrected primitive system through target depth k<=100000 found no admissible candidate with three primitive q=3 classes simultaneously present. This observation is not promoted here as a certificate because the present file is a universal theorem note; it is recorded only as motivation for the next independently replayed finite attack.

The theorem burden is now much narrower than the former shared-factor problem: explain why three pointwise-primitive base singleton rows cannot align on one directly novel admissible target, or find the first such alignment and analyze it.