# Phase 1F-B — Global Elimination of the $R(g)=4$ Prime-Fourth-Power Stratum

**Research-draft theorem:** no primitive Perfect Cuboid can have

$$
g=p^4,
\qquad
p\equiv1\pmod4
$$

with $p$ prime.

This removes one of the two possible primitive factorization types at
representation multiplicity

$$
R(g)=4.
$$

---

## 1. Why $g=p^4$ has exactly four Pythagorean representation states

Let

$$
p=r^2+s^2
$$

with

$$
r>s>0,
\qquad
\gcd(r,s)=1,
$$

and opposite parity.

Write

$$
r+is=\sqrt p\,e^{i\theta}.
$$

For

$$
g=p^4,
$$

we have

$$
g^2=p^8.
$$

Up to units, conjugation, and exchange of the two coordinates, the four
nontrivial positive representations of $g^2$ arise from Gaussian exponent
differences

$$
2,\ 4,\ 6,\ 8.
$$

Therefore the four even legs, normalized by $g$, have absolute values

$$
|\sin2\theta|,
\quad
|\sin4\theta|,
\quad
|\sin6\theta|,
\quad
|\sin8\theta|.
$$

Define

$$
\phi=2\theta
$$

and

$$
X=\sin^2\phi.
$$

Since

$$
\tan\theta=\frac sr\in\mathbb Q,
$$

we have

$$
\sin\phi=\sin2\theta=\frac{2rs}{p}\in\mathbb Q.
$$

Hence any admissible $X$ must satisfy

$$
\boxed{
X\in\mathbb Q^2.
}
$$

---

## 2. The four squared even-leg states

Let

$$
S_k=\sin^2(k\phi).
$$

Then

$$
S_1=X,
$$

$$
S_2=4X(1-X),
$$

$$
S_3=X(3-4X)^2,
$$

and

$$
S_4=16X(1-X)(1-2X)^2.
$$

By the Phase 1E equivalence, a Perfect Cuboid would require three of these
four even legs to form an internal Pythagorean triple.

Therefore one of the twelve ordered relations

$$
S_i+S_j=S_k
$$

with distinct $i,j,k\in\{1,2,3,4\}$ must hold.

---

## 3. Exhaustive factorization of all twelve cases

The twelve differences factor over $\mathbb Q$ as follows.

| Relation | Exact factorization |
|---|---|
| $S_1+S_2-S_3$ | $-4X(X-1)(4X-1)$ |
| $S_1+S_2-S_4$ | $X(64X^3-128X^2+76X-11)$ |
| $S_1+S_3-S_2$ | $2X(2X-1)(4X-3)$ |
| $S_1+S_3-S_4$ | $2X(4X-3)(8X^2-8X+1)$ |
| $S_1+S_4-S_2$ | $-X(64X^3-128X^2+76X-13)$ |
| $S_1+S_4-S_3$ | $-8X(X-1)(2X-1)(4X-1)$ |
| $S_2+S_3-S_1$ | $4X(X-1)(4X-3)$ |
| $S_2+S_3-S_4$ | $X(4X-3)(16X^2-16X+1)$ |
| $S_2+S_4-S_1$ | $-X(64X^3-128X^2+84X-19)$ |
| $S_2+S_4-S_3$ | $-X(64X^3-112X^2+60X-11)$ |
| $S_3+S_4-S_1$ | $-8X(X-1)(2X-1)(4X-3)$ |
| $S_3+S_4-S_2$ | $-X(4X-3)(16X^2-24X+7)$ |

Thus every possible internal Pythagorean ordering has been reduced to a
finite algebraic root condition for $X$.

---

## 4. Rational linear roots

The nondegenerate rational roots that occur in the factored linear pieces
are

$$
X=\frac14,\qquad
X=\frac12,\qquad
X=\frac34.
$$

The values

$$
\frac12
$$

and

$$
\frac34
$$

are not squares in $\mathbb Q$.

Thus they contradict

$$
X\in\mathbb Q^2.
$$

For

$$
X=\frac14,
$$

we have

$$
\sin\phi=\frac12.
$$

Since

$$
\phi=2\theta
$$

and

$$
t=\tan\theta=\frac sr\in\mathbb Q,
$$

the identity

$$
\sin2\theta=\frac{2t}{1+t^2}
$$

gives

$$
\frac{2t}{1+t^2}=\frac12.
$$

Hence

$$
t^2-4t+1=0
$$

and therefore

$$
t=2\pm\sqrt3,
$$

which is irrational.

Contradiction.

The roots

$$
X=0
$$

and

$$
X=1
$$

are degenerate and do not produce a valid primitive Pythagorean state.

---

## 5. Quadratic factors

The remaining quadratic factors are

$$
8X^2-8X+1,
$$

$$
16X^2-16X+1,
$$

and

$$
16X^2-24X+7.
$$

Their roots are respectively

$$
X=\frac{2\pm\sqrt2}{4},
$$

$$
X=\frac{2\pm\sqrt3}{4},
$$

and

$$
X=\frac{3\pm\sqrt2}{4}.
$$

All are irrational.

Therefore none can equal

$$
X=\sin^2(2\theta)\in\mathbb Q.
$$

---

## 6. Cubic factors

The four cubic factors are

$$
64X^3-128X^2+76X-11,
$$

$$
64X^3-128X^2+76X-13,
$$

$$
64X^3-128X^2+84X-19,
$$

and

$$
64X^3-112X^2+60X-11.
$$

By the rational-root theorem, a rational root in lowest terms must have
numerator dividing the constant term and denominator dividing $64$.

Every such candidate was checked exactly, and none is a root.

Hence all four cubics have no rational root.

Because an admissible $X$ must be rational, none can produce a
Perfect-Cuboid state.

---

## 7. Theorem

All twelve possible internal Pythagorean orderings have been exhausted.

Each leads only to:

- a degenerate root;
- a rational value that is not a rational square;
- a value forcing $\tan\theta$ to be irrational;
- an irrational quadratic root; or
- an irreducible cubic with no rational root.

Therefore

$$
\boxed{
g=p^4,\quad p\equiv1\pmod4
\Longrightarrow
\text{no primitive Perfect Cuboid}.
}
$$

---

## 8. Consequence for the $R(g)=4$ stratum

For a primitive admissible common hypotenuse,

$$
R(g)=4
$$

means

$$
\prod_{p\equiv1\pmod4}(2e_p+1)=9.
$$

There are only two factorization patterns:

$$
g=p^4
$$

or

$$
g=pq,
\qquad
p\ne q,
\qquad
p\equiv q\equiv1\pmod4.
$$

The first pattern has now been eliminated globally.

Thus the entire unresolved $R(g)=4$ stratum has been reduced to the mixed
two-prime family

$$
\boxed{
g=pq.
}
$$

This is the next structural target.