Type-II root geometry: López A/B are the comparable-root cases

Theorem

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Status: proved exact parametrization and structural corollary

Date: 2026-08-15

Depends on: ES-TYPEII-SQUARE-COMPLETION-LOPEZ-A.md, ES-SQUARE-COMPLETION-TRAP-GEOMETRY.md

Claim boundary: this is an exact reparametrization of the square-completed standard Type-II theorem. It does not prove universal Type-II existence.

1. Squarefree-root decomposition

Let p≡1 mod4 be prime and suppose a square-completed Type-II certificate is given by positive integers a,d with

\boxed{ d\mid a^2, \qquad p\nmid d, \qquad 4a-1\mid p+4d.}

Because

d\cdot\frac{a^2}{d}=a^2

is a square, the two integers

d \qquad\text{and}\qquad \frac{a^2}{d}

have the same squarefree kernel.

Therefore there are unique positive integers s,b,c with s squarefree such that

\boxed{ d=sb^2, \qquad \frac{a^2}{d}=sc^2.}

Multiplying gives

a^2=s^2b^2c^2,

\boxed{a=sbc.}

Thus every square-completed Type-II certificate has canonical root data

\boxed{(s,b,c),\qquad s\text{ squarefree}.}

2. The exact congruence in root coordinates

Let

q=rac{p+4d}{4a-1}.

Substituting

d=sb^2, \qquad a=sbc

gives

\boxed{ p+4sb^2=(4sbc-1)q.}

Define

\boxed{t=cq-b.}

Then

\begin{aligned} p+q &=4sbcq-4sb^2\\ &=4sb(cq-b)\\ &=4sbt, \end{aligned}

\boxed{p+q=4sbt.}

Since the left side is positive,

\boxed{t>0.}

Also by definition

\boxed{b+t=cq.}

Thus the square-completed congruence is equivalent to the four positive parameters

\boxed{ p+q=4sbt, \qquad b+t=cq.}

3. Explicit Type-II decomposition

The two equations immediately give

\begin{aligned} \frac1{sctp} +\frac1{sbt} +\frac1{sbcp} &= \frac{b+p c+t}{sbc tp}\\ &= \frac{cq+pc}{sbc tp}\\ &= \frac{c(p+q)}{sbc tp}\\ &= \frac4p. \end{aligned}

Therefore:

Theorem — explicit root decomposition

Every square-completed certificate

p+4sb^2=(4sbc-1)q

with

t=cq-b>0

gives the exact standard Type-II identity

\boxed{ \frac4p = \frac1{sctp} + \frac1{sbt} + \frac1{sbcp}.}

Two of the three denominators carry the factor p, as expected for Type II.

4. López Type A is b | c

Recall

d=sb^2, \qquad a=sbc.

The ordinary López Type-A boundary condition is

d\mid a.

This is equivalent to

sb^2\mid sbc,

hence

\boxed{b\mid c.}

Therefore the lower López orthant is exactly the region where the first root divides the second.

5. López Type B is c | b

The upper square-divisor boundary corresponding to López Type B is

a\mid d.

In root variables this is

sbc\mid sb^2,

\boxed{c\mid b.}

Thus the upper López orthant is exactly the opposite divisibility order.

6. Exact comparability theorem

Combining the two cases:

Theorem — López A/B are the comparable-root Type-II certificates

A square-completed standard Type-II certificate belongs to one of the two López boundary families at the same layer if and only if

\boxed{b\mid c\quad\text{or}\quad c\mid b.}

The genuinely square-only Type-II certificates are exactly those with

\boxed{b\nmid c \qquad\text{and}\qquad c\nmid b.}

So the cross-orthant geometry has an elementary interpretation:

the two square roots are incomparable in the divisibility poset.

7. Complement simply swaps the roots

The divisor complement is

d^*=\frac{a^2}{d}=sc^2.

Thus

\boxed{d\longleftrightarrow d^*}

is exactly

\boxed{b\longleftrightarrow c.}

This makes all previous symmetry transparent:

Type A (b|c) is sent to Type B (c|b);
mixed certificates (b,c incomparable) remain mixed;
inverse residue pairing is just root exchange.

The entire square-divisor complement theory therefore becomes a two-root symmetry.

8. Example p = 2521

The mixed certificate

a=12, \qquad d=16

has

d=1\cdot4^2, \qquad \frac{a^2}{d}=9=1\cdot3^2.

Thus

\boxed{s=1, \qquad b=4, \qquad c=3.}

The roots are incomparable:

4\nmid3, \qquad 3\nmid4.

The congruence quotient is

q=55

and

t=cq-b=3\cdot55-4=161.

Hence

p+q=2521+55=2576=4\cdot1\cdot4\cdot161

and

b+t=4+161=165=3\cdot55.

The explicit decomposition is

\boxed{ \frac4{2521} = \frac1{3\cdot161\cdot2521} + \frac1{4\cdot161} + \frac1{4\cdot3\cdot2521}.}

This is a standard Type-II solution represented by incomparable roots.

9. Strategic consequence

The López all-prime conjecture can now be interpreted inside standard Type II as a divisibility-comparability conjecture:

for every prime, find a Type-II certificate whose canonical roots can be chosen comparable by divisibility.

The complete Type-II problem drops that comparability requirement.

This suggests two distinct proof directions:

completion route: use all incomparable-root certificates directly, which is enough for standard Type II;
descent route: try to transform any incomparable-root certificate into another certificate with a smaller incomparability measure until one root divides the other.

A natural descent statistic is obtained after writing

g=\gcd(b,c), \qquad b=gB, \qquad c=gC, \qquad \gcd(B,C)=1.

The mixed case is precisely

B>1, \qquad C>1.

Whether the exact equations

p+q=4sbt, \qquad b+t=cq

admit a transformation that reduces BC while preserving p is now a concrete theorem target.