Unique-active valuation excess is a first prime-power lift

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Status: proved universal theorem

Date: 2026-08-14

Project: Free Computation Foundation / CENTL

Coordinator: Operator-01 / primary research lead

Partner input: Operator-02 active fixed-negative core and valuation criterion

Claim boundary: this theorem classifies the shape of the excess quotient of a unique active fixed-negative row. It does not prove the observed hard-class restriction q in {3,5,9}, universal Direct-Shadow Completeness, López Type A/B coverage, or the Erdős-Straus conjecture.

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1. Setup

Fix a target progression modulus L and a fixed-negative earlier layer j with

m_j=4j-1.

Because the layer is fixed at squareclass level, write

\boxed{m_j=d s^2}

with d squarefree and every prime dividing d already dividing L.

The Jacobi sign of the target residue on this entire square-lift tower depends only on d:

\left(\frac r{d s^2}\right)=\left(\frac r d\right).

Thus replacing s by a divisor of s preserves the fixed-negative character sign whenever the resulting layer is earlier and active.

Define the active excess quotient

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

Assume j is the unique member of the active fixed-negative core:

\boxed{|\mathcal N^{\rm act}_{k,r}|=1.}

2. The theorem

Theorem

For the unique active fixed-negative layer,

\boxed{q_j=p^a}

for one prime p, with

\boxed{a\in\{1,2\}.}

Equivalently,

\boxed{q_j=p\quad\text{or}\quad p^2.}

Proof

For a prime p, write

v_p(m_j)=v_p(d)+2v_p(s).

Let

b_p=v_p(L).

Then

v_p(q_j)=\max(v_p(m_j)-b_p,0).

Because j is active, at least one exponent is positive.

Step 1: only one prime can divide q_j

Suppose two distinct primes p and ell divide q_j.

Then both satisfy

v_p(m_j)>b_p, \qquad v_\ell(m_j)>b_\ell.

Since the squarefree part d is already supported on primes dividing L, every valuation excess comes from the square factor s^2 beyond the amount absorbed by L.

Choose one excess prime, say p, and replace

s\mapsto s/p.

The resulting modulus

m'=d(s/p)^2

is smaller than m_j, hence its corresponding depth is earlier than j and therefore earlier than the target depth.

Its squarefree part is still d, so it has the same fixed-negative Jacobi sign.

Removing one factor of p from s reduces the p-valuation of the modulus by exactly two but leaves the ell-valuation unchanged. Since ell was already in valuation excess,

v_\ell(m')=v_\ell(m_j)>b_\ell.

Therefore the earlier layer m' remains active.

This produces a second active fixed-negative layer, contradicting uniqueness.

Hence q_j has support on only one prime:

q_j=p^a.

Step 2: the excess exponent is at most two

Suppose

a=v_p(q_j)\ge3.

Again replace s by s/p. The new layer has the same squarefree part d and therefore the same fixed-negative sign.

Its p-valuation is lower by exactly two, so its remaining excess above L is

a-2\ge1.

Thus the earlier layer is still active, again contradicting uniqueness.

Therefore

a\le2.

Since the layer is active, a>=1, giving

\boxed{a\in\{1,2\}.}

QED.

3. Class-B corollary

Recall Operator-02 Class B means the excess prime is absent from L and appears to even exponent in the fixed-squareclass layer.

Corollary

If the unique active fixed-negative layer is Class B, then necessarily

\boxed{q_j=p^2}

for a prime p not dividing L.

Proof

If p does not divide L, then

v_p(q_j)=v_p(m_j).

Fixed-squareclass support requires a prime absent from L to occur to even exponent. The theorem restricts the positive exponent to 1 or 2, so it must equal 2. QED.

Thus the unique-active universe splits cleanly into:

\boxed{ \begin{array}{ll} \text{Class A:}&q=p\text{ or }p^2,\quad p\mid L,\\ \text{Class B:}&q=p^2,\quad p\nmid L. \end{array}}

4. First-excess interpretation

The theorem says a unique active fixed-negative row is literally the first active prime-power lift of its negative squarefree ancestor along one prime direction.

If more than one prime direction were already excessive, a smaller active row would exist by removing one direction.

If one direction had more than two units of valuation excess, a smaller active row would exist by removing one square factor.

So uniqueness forces the active layer onto the first shell of the valuation lattice.

5. Relation to the k <= 1500 census

The independently verified census found

2,770 single-active candidates
q=3: 1,322
q=5:    34
q=9: 1,414
Class A only: 2,770
Class B/mixed: 0

The theorem explains why only prime or prime-square quotients can occur.

It does not yet explain why the only observed prime directions are 3 and 5, why 7 is absent, why no prime >=11 occurs, or why 25 is absent.

Those are now isolated as a sharper theorem candidate rather than being mixed together with the already solved prime-power question.

6. Hard-class small-prime collapse conjecture

The finite data suggests:

For Mordell-hard target classes modulo 840, if a fixed-negative active core has exactly one layer, then its first-excess quotient belongs to <div class="math" role="math">> \boxed{\{3,5,9\}}.

></div>

In particular the unique active row is Class A, never Class B.

This statement is not proved here.

The next attack should exploit the additional hard-class facts at primes 3,5,7, the square-lift ancestor structure, and target-modulus divisibility. A counterexample would be a hard-compatible target with |N^act|=1 and quotient 7, 25, p>=11, or p^2 for a free prime.

7. Why this matters

The Class-C problem began with arbitrary active valuation excess.

The theorem reduces the single-active branch to one prime and at most one square-lift step:

\boxed{

|\mathcal N^{act}|=1 \Longrightarrow \text{one prime direction} \Longrightarrow q=p\text{ or }p^2. }</div>

That is a genuine universal compression of C1 and gives the 3,5,9 observation a precise remaining burden of proof.