# Phase 1G-B — Proof-Safe Quotient of the $g=p^2q$ Seven-State System

**Status:** exact reduction; four state classes remain open  
**Diagonal type:** $g=p^2q$, $p\ne q$, $p\equiv q\equiv1\pmod4$

---

## 1. The seven outer even-leg states

For

$$
p=r^2+s^2,
\qquad
q=u^2+v^2,
$$

the seven nontrivial common-hypotenuse representations are indexed by

$$
(1,0),(2,0),(0,1),(1,1),(1,-1),(2,1),(2,-1).
$$

Write the corresponding even legs as

$$
E_1,\ldots,E_7.
$$

Their angle labels are

$$
\phi,\ 2\phi,\ \psi,\ \phi+\psi,\ \phi-\psi,\ 2\phi+\psi,\ 2\phi-\psi.
$$

The rational common factors forced by Gaussian exponent splitting give the
following valuation strata.

For the prime $p$:

$$
v_p(E_2)=v_p(E_6)=v_p(E_7)=0,
$$

$$
v_p(E_1)=v_p(E_4)=v_p(E_5)=1,
$$

and

$$
v_p(E_3)\ge2.
$$

For the prime $q$:

$$
v_q(E_3)=v_q(E_4)=v_q(E_5)=v_q(E_6)=v_q(E_7)=0,
$$

while

$$
v_q(E_1)\ge1,
\qquad
v_q(E_2)\ge1.
$$

The inequalities are intentional: cross-prime accidental divisibility is
allowed. No unsupported exact valuation is assumed.

---

## 2. Minimum-valuation lemma

Let

$$
X^2+Y^2=Z^2
$$

and let $\ell$ be an odd prime.

Among

$$
v_\ell(X),\quad v_\ell(Y),\quad v_\ell(Z),
$$

the minimum value must occur at least twice.

Indeed, if exactly one term had strictly smaller valuation, divide the equation
by the corresponding smallest even power of $\ell$. Modulo $\ell$, exactly one
of the three squared terms would remain nonzero, which is impossible.

This lemma is independent of whether $\ell\equiv1$ or $3\pmod4$.

---

## 3. First finite quotient

There are initially

$$
\binom73=35
$$

unordered ways to choose three states.

Applying the minimum-valuation lemma simultaneously at $p$ and $q$ leaves
exactly

$$
\boxed{15}
$$

state triples.

Conjugating the $q$-Gaussian prime exchanges

$$
4\leftrightarrow5,
\qquad
6\leftrightarrow7.
$$

Quotienting by this exact symmetry leaves only

$$
\boxed{10}
$$

essential unordered state classes.

---

## 4. The ten classes

Using state numbers, the ten classes are represented by

$$
\{1,3,4\},
\{1,4,5\},
\{1,6,7\},
\{2,3,6\},
\{2,4,6\},
$$

$$
\{2,4,7\},
\{2,6,7\},
\{3,4,5\},
\{3,6,7\},
\{4,6,7\}.
$$

The omitted conjugate partners are obtained by

$$
4\leftrightarrow5,
\qquad
6\leftrightarrow7.
$$

---

## 5. Six classes eliminated

### Class $\{1,3,4\}$

The three possible internal Pythagorean orderings reduce to positivity or to

$$
\tan\phi\,\tan\psi=1.
$$

The latter makes

$$
\phi+\psi=\frac\pi2
$$

and therefore makes one outer Pythagorean representation degenerate.

The conjugate class $\{1,3,5\}$ similarly reduces to positivity or
$\phi=\psi$, incompatible with distinct prime Gaussian states.

Hence the class is eliminated.

### Class $\{1,4,5\}$

Put

$$
A=rs,\quad R=r^2-s^2,
$$

$$
B=uv,\quad U=u^2-v^2.
$$

After dividing the common factor, the two nontrivial orderings reduce to

$$
Aq^2=4RBU
$$

or

$$
A^2(8B^2-q^2)=2B^2R^2.
$$

The first forces $q^2\mid R$, then a prime-square factorization and size
contradiction.

The second forces $B\mid A$, then $A=B$, and finally

$$
q^2+2p^2=16A^2,
$$

which is impossible modulo $8$.

Hence this class is eliminated.

### Class $\{2,3,6\}$

Its two nontrivial algebraic conditions are exactly the vanishing of the
numerator or denominator of

$$
\tan(2\phi+\psi).
$$

Thus the state $2\phi+\psi$ has either zero even leg or zero odd complement.

Both are degenerate.

The conjugate class with state $7$ is eliminated identically.

### Class $\{2,6,7\}$

Let

$$
R_2=R^2-4A^2,
\qquad
A_2=2AR.
$$

Then

$$
p^4=R_2^2+4A_2^2,
\qquad
\gcd(R_2,A_2)=1.
$$

This class has exactly the same plus/minus structure as the previous
two-prime calculation, but with the primitive $p^2$ representation in place
of $p$.

One branch again yields a modulo-$8$ contradiction.

The other forces a common divisor that is either $1$, $p$, or $p^2$.
The $p$ and $p^2$ allocations contradict primitiveness or size; the remaining
allocation reduces to the same prime-square size contradiction.

Hence the class is eliminated.

### Class $\{3,4,5\}$

This is the prime-swapped counterpart of class $\{1,4,5\}$.

The same two Branch-I / Branch-II arguments eliminate it.

### Class $\{3,6,7\}$

Again use the primitive $p^2$ representation $(R_2,2A_2)$.

One branch gives the same modulo-$8$ contradiction.

The remaining branch forces

$$
U=p^4
$$

and

$$
B=4A_2R_2.
$$

Writing

$$
x=R_2,
\qquad
y=2A_2,
$$

we have

$$
p^4=x^2+y^2
$$

and the $q$-norm condition becomes

$$
q^2=x^4+18x^2y^2+y^4.
$$

Pocklington's classical quartic obstruction states that

$$
x^4+18x^2y^2+y^4=z^2
$$

has no nonzero integer solution.

Hence this class is eliminated.

---

## 6. Only four state classes remain

After all proof-safe reductions above, the entire $p^2q$ seven-state problem
has been reduced to four conjugacy classes:

$$
\boxed{
\{1,6,7\},
\quad
\{2,4,6\},
\quad
\{2,4,7\},
\quad
\{4,6,7\}.
}
$$

Their conjugates add no new structural class.

Thus the original

$$
105
$$

oriented equations have been compressed first to $35$ unordered triples,
then to $15$ valuation-compatible triples, then to $10$ conjugacy classes,
and finally to only four unresolved classes.

---

## 7. Research status

This does **not** yet prove the global exclusion of $g=p^2q$.

A current ProofAtlas research record independently reports a $p^2q$ exclusion
inside a retained low-prime-support route, but explicitly marks that exclusion
as source-reported and recommends an independent audit of its coprimality and
square-divisibility steps.

The present reduction is therefore best treated as an independent audit route.

The next task is to resolve the four boxed classes without assuming the
source-reported exclusion.