Standard Type II as the square-divisor completion of López Type A

Geometry

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Status: proved exact equivalence

Date: 2026-08-15

Depends on: ES-TWO-TARGET-DIVISOR-SQUARE.md, ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md

External background: Miguel Angel López, A Complete Congruence System for the Erdos-Straus Conjecture, arXiv:2404.01508, especially Theorem 7

Claim boundary: this identifies an exact structural relation between standard prime Type II and the López Type-A congruence family. It does not prove universal existence and no literature-priority claim is made without a separate prior-art review.

1. Exact Type-II divisor-square starting point

Let

p\equiv1\pmod4

be prime.

ES-TWO-TARGET-DIVISOR-SQUARE.md proves that a standard Type-II solution at an admissible shift

k\equiv3\pmod4, \qquad \gcd(k,p)=1

is equivalent to the existence of

C=\frac{p+k}{4}

and a divisor

\boxed{d\mid C^2}

such that

\boxed{d\equiv-C\pmod k.}

Write

\boxed{d+C=ak}

for a positive integer a.

2. Eliminate the shift

Using

4C=p+k,

multiply d+C=ak by four:

4d+p+k=4ak.

Hence

\boxed{p+4d=(4a-1)k.}

Therefore every Type-II certificate gives the López-shaped congruence

\boxed{4a-1\mid p+4d,}

or equivalently

\boxed{p\equiv-4d\pmod{4a-1}.}

This is exactly the residue shape occurring in López Type A.

3. The square-divisor condition collapses to d | a^2

Because

d\mid C^2

and

\gcd(C,k)=1,

we have

\gcd(d,k)=1.

Modulo d, the equation

d+C=ak

gives

C\equiv ak\pmod d.

Squaring,

C^2\equiv a^2k^2\pmod d.

Since d|C^2 and k is invertible modulo d,

\boxed{d\mid a^2.}

Thus every standard Type-II solution yields positive integers a,d satisfying

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

Also p does not divide d, because d|C^2 and p does not divide C.

4. Converse construction

Now suppose positive integers a,d satisfy

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

Define

\boxed{k=\frac{p+4d}{4a-1}.}

Because

p+4d\equiv1\pmod4

and

4a-1\equiv3\pmod4,

we obtain

\boxed{k\equiv3\pmod4.}

If p|k, then from

(4a-1)k=p+4d

we would get p|d, contradiction. Hence

\boxed{\gcd(k,p)=1.}

Put

\boxed{C=\frac{p+k}{4}.}

The defining equations give

\boxed{(4a-1)C=ap+d.}

Since any common divisor of d and 4a-1 divides a and then divides 1,

\boxed{\gcd(d,4a-1)=1.}

Modulo d,

(4a-1)C\equiv ap\pmod d.

Squaring and using d|a^2, together with p and 4a-1 invertible modulo d, gives

\boxed{d\mid C^2.}

Finally,

\begin{aligned} (4a-1)(d+C) &=(4a-1)d+ap+d\\ &=a(p+4d)\\ &=a(4a-1)k, \end{aligned}

\boxed{d+C=ak.}

Therefore

d\equiv-C\pmod k,

and the exact Type-II divisor-square criterion applies.

5. Exact theorem

Theorem — square-completed Type A equals standard prime Type II

For a prime

p\equiv1\pmod4,

the following are equivalent:

p has a standard Type-II Erdős--Straus solution;
there exist positive integers a,d such that

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

Equivalently,

\boxed{ \text{Type II} \iff \exists a,d>0: d\mid a^2, \quad p\equiv-4d\pmod{4a-1}, \quad p\nmid d.}

6. Relation to López Type A

López Theorem 7 states that a prime has a Type-A solution exactly when there exist positive d,n such that

\boxed{p\equiv-4d\pmod{4dn-1}.}

Put

\boxed{a=dn.}

Then López Type A is precisely the subfamily

\boxed{ d\mid a, \qquad p\equiv-4d\pmod{4a-1}.}

The standard Type-II theorem above enlarges only the divisor condition:

\boxed{ \begin{array}{ccl} \text{López Type A} &:& d\mid a,\\[1mm] \text{standard Type II} &:& d\mid a^2, \end{array}}

with the same modulus 4a-1 and the same target residue -4d.

Therefore:

\boxed{ \text{standard prime Type II is the square-divisor completion of López Type A}.}

Every López Type-A solution is automatically contained in the square-completed family, because

d\mid a\Longrightarrow d\mid a^2.

The converse need not hold.

7. Genuine square-only witness: p = 2521

López records 2521 among the exceptional primes lacking Type-A solutions in the finite analysis of that paper.

Take

\boxed{p=2521, \qquad a=12, \qquad d=16.}

Then

16\mid12^2=144,

but

\boxed{16\nmid12.}

Thus this is genuinely outside the ordinary López Type-A divisor condition.

Nevertheless

4a-1=47

and

p+4d=2521+64=2585=47\cdot55.

Hence

\boxed{47\mid p+4d.}

The corresponding exact signed-box shift is

\boxed{k=55,}

with

C=\frac{2521+55}{4}=644.

Indeed

16\mid644^2

and

16+644=660=12\cdot55.

So the square-completed congruence supplies a standard Type-II solution for 2521 even though this certificate is not a López Type-A certificate.

The complementary divisor is

\frac{644^2}{16}=25921=161^2,

so the normalized factor pair is especially transparent:

16=4^2, \qquad 25921=161^2, \qquad 55\mid4+161.

8. Strategic consequence

The old Type-A/B program and the complete Type-I/II program are not separate languages.

At least on the Type-A side there is an exact completion map:

\boxed{ d\mid a \quad\leadsto\quad d\mid a^2.}

The congruence itself does not change.

This suggests a new direct proof strategy:

retain all López Type-A congruence and shadow machinery;
replace the divisor lattice Div(a) by the square divisor lattice Div(a^2);
study whether the additional square-only residues close the zero-density composite-rescue core;
compare the resulting square-completed trap system with the exact Type-II signed-divisor box and its Kneser quotient defects.

The square completion may therefore provide the missing bridge by which the mature Type-A/B machinery can be reused inside the exact prime Erdős--Straus formulation rather than abandoned.