Prime-index spectrum of the square-completed Type-II layers

Geometry

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Status: proved exact classification

Date: 2026-08-15

Depends on: ES-SQUARE-COMPLETION-TRAP-GEOMETRY.md, ES-SQUARE-COMPLETION-BACKBONE.md, THEORY.md

Claim boundary: classifies square-completed minimal-depth realizability when the layer index itself is prime. It does not classify composite layer indices and does not prove universal Erdős--Straus coverage.

1. Setup

For every positive layer index a, put

m_a=4a-1

and define the square-completed Type-II trap

\boxed{ S_a=\{-4D\pmod{m_a}:D\mid a^2\}.}

Call a an exact completed depth if there exists a prime p whose first hit among the layers S_1,S_2,... occurs exactly at a.

The prime-modulus backbone already proves that every a with m_a prime greater than 7 is realized infinitely often, including by Mordell-hard primes.

Here we prove the converse when a itself is prime.

2. A prime-index completed layer has only three parameters

Let

\boxed{a\text{ be prime}.}

Then

\operatorname{Div}(a^2)=\{1,a,a^2\}.

Therefore

S_a = \{-4,-4a,-4a^2\} \pmod{m_a}.

Since

4a\equiv1\pmod{m_a},

we have

-4a\equiv-1

and

-4a^2 =-a(4a) \equiv-a.

Hence:

Lemma — three-point prime layer

For prime a,

\boxed{ S_a=\{-4,-1,-a\}\pmod{4a-1}.}

3. Reduction along every modulus-ancestry edge

Suppose

1\le j<a

and

\boxed{m_j=4j-1\mid m_a=4a-1.}

Since both

4a\equiv1\pmod{m_j}

and

4j\equiv1\pmod{m_j},

subtracting gives

4(a-j)\equiv0\pmod{m_j}.

The modulus m_j is odd, so 4 is invertible modulo it. Thus

\boxed{a\equiv j\pmod{m_j}.}

Reducing the three-point prime layer gives

S_a\bmod m_j = \{-4,-1,-j\}.

But all three residues lie in the earlier completed layer S_j:

D=1 gives -4;
D=j gives

-4j\equiv-1;

D=j^2 gives

-4j^2\equiv-j.

Therefore:

Theorem — prime-index ancestry absorption

If a is prime and

4j-1\mid4a-1,

then

\boxed{ S_a\bmod(4j-1) \subseteq S_j.}

In fact the reduction is exactly the three canonical residues

\boxed{\{-4,-1,-j\}.}

Thus every modulus-ancestry edge into a prime-index completed layer is a complete direct shadow.

4. Every composite modulus 4a-1 has an earlier 3 mod 4 prime divisor

Assume now that

a\text{ is prime}

and

m_a=4a-1

is composite.

Because

m_a\equiv3\pmod4,

its prime factorization contains at least one prime factor

q\equiv3\pmod4

with odd total multiplicity.

Since m_a is composite, such a prime factor may be chosen with

q<m_a.

Write

\boxed{q=4j-1}

with

j=\frac{q+1}{4}.

Then

1\le j<a

and

4j-1=q\mid4a-1.

By the prime-index ancestry theorem,

S_a\bmod q\subseteq S_j.

Therefore every candidate captured by layer a is already captured by the earlier layer j.

Hence:

Corollary — composite-modulus prime layers are structural gaps

\boxed{a\text{ is prime}, \qquad 4a-1\text{ is composite},}

then a cannot be a minimal square-completed depth for any integer, and in particular for any prime.

The entire layer is directly redundant.

5. Converse: prime modulus gives infinitely many exact first hits

If instead

\boxed{4a-1>7\text{ is prime},}

ES-SQUARE-COMPLETION-BACKBONE.md applies the CRT, the neutral-residue theorem

1\notin S_j,

and Dirichlet's theorem to produce infinitely many primes whose first completed hit is exactly a.

Those primes may all be chosen in the Mordell-hard class

\boxed{p\equiv1\pmod{840}.}

6. Exact classification

Combining the two directions gives the main result.

Theorem — prime-index completed spectrum

Let a be prime. Then

\boxed{ a\text{ is an exact square-completed first-hit depth} \iff 4a-1\text{ is prime}.}

Moreover, in the positive case the depth is realized by infinitely many Mordell-hard primes.

In the negative case the whole completed layer is directly shadowed by an earlier layer associated with any 3 mod 4 prime divisor of 4a-1.

7. Relation to the prime-modulus backbone

The prime-modulus backbone is therefore not merely a sufficient infinite family inside the completed spectrum.

Among prime-valued layer indices, it is complete.

The remaining completed-depth spectrum problem is entirely concentrated on composite layer indices:

\boxed{ \text{new completed spectrum beyond the backbone} \subseteq \{a:\ a\text{ composite}\}.}

This is a substantial sharpening of the spectrum geometry.

8. Why the completed layer makes the proof short

The theorem depends crucially on the square completion.

At a prime index a, the full Type-II square layer contains the three natural divisor parameters

1,\ a,\ a^2,

which reduce along ancestry to

1,\ j,\ j^2.

Those are automatically available in every earlier completed layer S_j.

In centered signed-box language, the prime layer is the one-dimensional box

\{-1,0,1\}

and ancestry sends its three points directly into the canonical three-point subset of the earlier box.

This is the first exact theorem obtained by combining the new square-completed internal geometry with the old modulus-ancestry shadow framework.

9. Next theorem target

The natural next question is the composite-index analogue:

For which composite a does every completed signed exponent vector reduce into an earlier completed layer along some ancestry divisor of 4a-1?

The prime-index theorem suggests separating the composite spectrum by the factorization of both

and

4a-1.

The most promising first families are:

semiprime a=uv, where the mixed region has low dimension;
prime powers, where S_a=T_a and no mixed completion occurs;
indices whose modulus 4a-1 has a small 3 mod4 prime divisor, giving a strong ancestry coordinate.