# Phase 1J — $2$-Adic Valuation Quotient of the $g=p^2q$ Seven-State System

## Result

Starting from the Phase 1I quotient:

$$
105
\text{ raw relations}
\rightarrow
57
\text{ conjugation orbits},
$$

the exact $2$-adic valuation structure of the seven integer even-leg states eliminates another

$$
\boxed{20}
$$

previously surviving orbits.

Combined with the 15 proof-safe P1I eliminations:

$$
\boxed{
57
\rightarrow
22
}
$$

quotient orbits remain.

In terms of the original 105 relations:

$$
\boxed{
105
\rightarrow
44
}
$$

raw relations remain.

No curve theory is used in this reduction.

---

## 1. Integer even-leg states

Let

$$
p=m^2+n^2,
\qquad
q=u^2+v^2,
$$

where each pair is primitive and of opposite parity.

Write

$$
\Delta_p=n^2-m^2,
\qquad
\Delta_q=v^2-u^2.
$$

Both are odd.

For

$$
g=p^2q,
$$

the seven actual integer even-leg states can be written as

$$
A_g=2mnpq,
$$

$$
B_g=4mn\Delta_p q,
$$

$$
C_g=2uvp^2,
$$

$$
D_g=2p\left(mn\Delta_q+uv\Delta_p\right),
$$

$$
E_g=2p\left(mn\Delta_q-uv\Delta_p\right),
$$

and

$$
F_g=
2\left(
2mn\Delta_p\Delta_q
+
uvC_p
\right),
$$

$$
G_g=
2\left(
2mn\Delta_p\Delta_q
-
uvC_p
\right),
$$

where

$$
C_p=n^4-6m^2n^2+m^4
$$

is odd.

Define

$$
a=v_2(mn)\ge1,
$$

$$
b=v_2(uv)\ge1.
$$

Then

$$
v_2(A_g)=a+1,
$$

$$
v_2(B_g)=a+2,
$$

$$
v_2(C_g)=b+1.
$$

---

## 2. The $D/E$ cancellation pair

Inside $D_g,E_g$, the two summands have valuations $a$ and $b$.

If

$$
a<b,
$$

then

$$
v_2(D_g)=v_2(E_g)=a+1.
$$

If

$$
a>b,
$$

then

$$
v_2(D_g)=v_2(E_g)=b+1.
$$

If

$$
a=b,
$$

write the two reduced summands as odd integers $X,Y$.

Then one of

$$
X+Y,
\qquad
X-Y
$$

has $2$-adic valuation exactly $1$, while the other has valuation at least $2$.

Therefore the unordered pair is

$$
\boxed{
\left\{
v_2(D_g),
v_2(E_g)
\right\}
=
\left\{
a+2,\ \ge a+3
\right\}.
}
$$

---

## 3. The $F/G$ cancellation pair

Inside $F_g,G_g$, the two summands before the outer factor $2$ have valuations

$$
a+1
$$

and

$$
b.
$$

Therefore:

if

$$
a+1<b,
$$

then

$$
v_2(F_g)=v_2(G_g)=a+2;
$$

if

$$
a+1>b,
$$

then

$$
v_2(F_g)=v_2(G_g)=b+1.
$$

When

$$
b=a+1,
$$

the same odd-sum/odd-difference lemma gives

$$
\boxed{
\left\{
v_2(F_g),
v_2(G_g)
\right\}
=
\left\{
a+3,\ \ge a+4
\right\}.
}
$$

Thus the entire seven-state valuation pattern depends only on the relative class of

$$
b-a
$$

plus the orientation of the two cancellation pairs.

---

## 4. Necessary valuation pattern for an internal Pythagorean triple

Suppose

$$
x^2+y^2=z^2.
$$

Let

$$
r=v_2(x),
\qquad
s=v_2(y),
$$

and assume

$$
r\le s.
$$

If

$$
r=s,
$$

then after dividing by the common power of $2$, both legs are odd and

$$
x_0^2+y_0^2\equiv2\pmod4,
$$

which cannot be a square.

Hence

$$
r<s.
$$

The hypotenuse must satisfy

$$
v_2(z)=r.
$$

If

$$
s=r+1,
$$

then modulo $8$ the reduced equation has the form

$$
1+4\equiv5\pmod8,
$$

again impossible for a square.

Therefore every integer Pythagorean triple has the necessary valuation shape

$$
\boxed{
\{h,h,h+d\},
\qquad
d\ge2,
}
$$

where one minimal-valuation entry is the hypotenuse and the other is one leg.

---

## 5. Finite valuation quotient

Only six relative classes of $b-a$ matter to this necessary test:

$$
b-a\le-2,
$$

$$
b-a=-1,
$$

$$
b-a=0,
$$

$$
b-a=1,
$$

$$
b-a=2,
$$

$$
b-a\ge3.
$$

The two equal-valuation cancellation cases are split only by whether the
higher branch is one additional power of $2$ above the lower cancellation
branch or at least two.

This produces a finite exact list of valuation states.

Every one of the 57 conjugation-orbit relations was tested against all of
these abstract valuation states.

If no state can realize the necessary Pythagorean valuation pattern, the
entire orbit is impossible.

---

## 6. New eliminations

The valuation gate eliminates 22 quotient orbits in total.

Two of those had already been eliminated in P1I by stronger independent
arguments.

Therefore the new contribution is

$$
\boxed{
20
}
$$

additional quotient-orbit eliminations, representing

$$
\boxed{
34
}
$$

additional raw relations.

Combined state:

$$
\boxed{
35
\text{ of }57
\text{ quotient orbits eliminated}
}
$$

and

$$
\boxed{
61
\text{ of }105
\text{ raw relations eliminated}.
}
$$

---

## 7. Remaining exact frontier

The surviving quotient representatives are:

$$
A^2+C^2=F^2,
$$

$$
A^2+D^2=G^2,
$$

$$
A^2+F^2=C^2,
$$

$$
A^2+F^2=E^2,
$$

$$
A^2+F^2=G^2,
$$

$$
B^2+C^2=D^2,
$$

$$
B^2+D^2=C^2,
$$

$$
B^2+D^2=F^2,
$$

$$
B^2+D^2=G^2,
$$

$$
B^2+F^2=D^2,
$$

$$
B^2+F^2=E^2,
$$

$$
B^2+F^2=G^2,
$$

$$
C^2+D^2=F^2,
$$

$$
C^2+D^2=G^2,
$$

$$
C^2+F^2=G^2,
$$

$$
D^2+F^2=C^2,
$$

$$
D^2+F^2=E^2,
$$

$$
D^2+F^2=G^2,
$$

$$
D^2+G^2=A^2,
$$

$$
D^2+G^2=C^2,
$$

$$
D^2+G^2=E^2,
$$

$$
D^2+G^2=F^2.
$$

These 22 orbits represent 44 raw relations after undoing conjugation.

---

## 8. Next step

The next reduction should classify these 22 orbit polynomials by:

- factorization after removing impossible boundary factors;
- total degree and symmetry;
- local solvability modulo odd primes and prime powers;
- Gaussian-prime norm divisibility;
- birational equivalence between orbit curves.

Only the residual classes that survive those tests should be promoted to
elliptic or higher-genus analysis.