q=3 Pullbacks Are Singleton — Free Computation Foundation

Theorem

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Status: proved universal theorem

Date: 2026-08-15

Depends on: Type A/B trap definition

Claim boundary: proves a single q=3 row forbids at most one parameter class in the full parameter ring. It does not prove that three distinct rows cannot cover all three classes, universal DSC-P, López-all-primes, or Erdős-Straus.

Theorem

Let

m=4j-1, \qquad g=\gcd(L,m), \qquad q=m/g=3.

For

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

define the pullback

R_j=\{s\bmod3:r+Ls\bmod m\in T_j\}.

Then

\boxed{|R_j|\le1.}

Equivalently, no two distinct points in the attained three-point fibre differ by a nonzero element of that fibre and both lie in T_j.

Step 1 — what two hits would imply

Since q=3,

m=3g.

The map s -> r+Ls mod m is injective on Z/3Z; its three attained values form one coset modulo g. Any two distinct attained values differ modulo m by

\boxed{g\quad\text{or}\quad2g=-g\pmod m.}

Thus two pullback classes would produce two distinct trap residues whose difference is congruent to ±g modulo m.

We prove that this cannot occur.

Because m=4j-1=3g, for j>1

j<g<2j, \qquad g\equiv1\pmod4.

The case j=1 has only one distinct trap and is immediate.

Step 2 — same-box differences

Take divisors e,f|j.

A/A

A difference of two -e traps has representative

e-f.

But

|e-f|<j<g,

so it cannot represent ±g mod 3g.

B/B

A difference of two -4e traps has representative

4(e-f).

A residue congruent to ±g mod 3g can be represented in the range of such a difference only by one of ±g,±2g.

But

g\equiv1\pmod4, \qquad 2g\equiv2\pmod4,

whereas 4(e-f) is divisible by 4. Hence none of those values is possible.

Step 3 — mixed differences

It remains to consider

D=4e-f

(the other mixed orientation is its negative after swapping the two traps).

Since e,f<=j,

4-j\le D\le4j-1=3g.

Therefore D≡±g (mod 3g) can only occur through

D\in\{g,-g,2g,-2g\}.

The ±2g cases are impossible by size

If e=j, then

D=4j-f.

The equation D=2g would require

f=4j-2g=g+1>j,

impossible. If e<j, then e<=j/2, so

D<2j<2g.

Thus D=2g is impossible.

Also

D\ge4-j>-2g

for j>1, so D=-2g is impossible.

The +g case

Suppose

4e-f=g=\frac{4j-1}{3}.

Then

12e-3f=4j-1.

Reducing modulo e gives

3f\equiv1\pmod e.

Hence

\gcd(e,f)=1.

Since e|j and f|j, coprimality implies

ef\mid j.

Write

j=efh, \qquad h\ge1.

Substitution gives

4efh-12e+3f=1,

\boxed{e(4fh-12)+3f=1.}

If fh>=3, the left side is at least 3, impossible. If fh=1, then f=h=1 and the equation becomes -8e+3=1, impossible. If fh=2, the possibilities (f,h)=(1,2) or (2,1) give respectively -4e+3=1 or -4e+6=1, again impossible.

Therefore D=g cannot occur.

The -g case

Suppose

4e-f=-g.

Then

12e-3f=-4j+1,

4j+12e-3f=1.

Reducing modulo e again forces gcd(e,f)=1, hence j=efh. Therefore

4efh+12e-3f=1,

i.e.

\boxed{f(4eh-3)+12e=1.}

But e,h>=1 gives 4eh-3>=1, so the left side is at least f+12e>=13, contradiction.

Thus D=-g is impossible.

All mixed cases are excluded.

Conclusion

No two distinct Type A/B traps differ by the nonzero displacement of a q=3 fibre. Hence the fibre contains at most one trap:

\boxed{|R_j|\le1.}

QED.

Corrected-domain consequence

For the Mordell-hard program,

3\mid840\mid L,

so the exact Dirichlet parameter domain on a q=3 coordinate is the full ring

\mathbb Z/3\mathbb Z.

Therefore:

one q=3 row forbids at most one of the three classes;
two q=3 rows can never cover the exact local domain;
any genuine q=3 local obstruction requires at least three distinct rows, with singleton pullbacks covering exactly

\boxed{\{0,1,2\}.}

after Q3-ABSORPTION.md, Q3-WEAK-REDUNDANCY.md, and Q3-POINTWISE-ABSORPTION.md, a directly novel obstruction would require three pointwise-primitive base contributions carrying three distinct classes.

This is now the precise q=3 theorem target.