# Phase 1F-B — Elimination of the $R(g)=4$ Prime-Power Branch $g=p^4$

**Status:** research-draft theorem with exact symbolic cross-check  
**Branch:** $R(g)=4$, $g=p^4$

## 1. Representation structure

Let

$$
p=r^2+s^2,
\qquad
p\equiv1\pmod4,
$$

and write

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

For

$$
g=p^4,
$$

the four nontrivial representations of

$$
x^2+y^2=g^2
$$

are obtained from Gaussian states whose arguments, up to signs and conjugation, are

$$
2\theta,\quad4\theta,\quad6\theta,\quad8\theta.
$$

Because $g$ is odd, each representation has one odd leg and one even leg. The even legs have normalized absolute values

$$
|\sin\phi|,
\quad
|\sin2\phi|,
\quad
|\sin3\phi|,
\quad
|\sin4\phi|,
$$

where

$$
\phi=2\theta.
$$

Since

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

both

$$
\sin\phi
$$

and

$$
\cos\phi
$$

are rational.

Set

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

Then

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

and

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

The four normalized squared even legs are

$$
S_1=X,
$$

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

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

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

By the Phase 1E equivalence, a Perfect Cuboid in this branch would require some three distinct members among

$$
S_1,S_2,S_3,S_4
$$

to satisfy a Pythagorean relation

$$
S_i+S_j=S_k.
$$

There are exactly twelve ordered possibilities.

## 2. Complete factor table

The twelve exact differences factor over $\mathbb Q[X]$ as follows:

$$
S_1+S_2-S_3=-4X(X-1)(4X-1),
$$

$$
S_1+S_2-S_4
=
X(64X^3-128X^2+76X-11),
$$

$$
S_1+S_3-S_2
=
2X(2X-1)(4X-3),
$$

$$
S_1+S_3-S_4
=
2X(4X-3)(8X^2-8X+1),
$$

$$
S_1+S_4-S_2
=
-X(64X^3-128X^2+76X-13),
$$

$$
S_1+S_4-S_3
=
-8X(X-1)(2X-1)(4X-1),
$$

$$
S_2+S_3-S_1
=
4X(X-1)(4X-3),
$$

$$
S_2+S_3-S_4
=
X(4X-3)(16X^2-16X+1),
$$

$$
S_2+S_4-S_1
=
-X(64X^3-128X^2+84X-19),
$$

$$
S_2+S_4-S_3
=
-X(64X^3-112X^2+60X-11),
$$

$$
S_3+S_4-S_1
=
-8X(X-1)(2X-1)(4X-3),
$$

$$
S_3+S_4-S_2
=
-X(4X-3)(16X^2-24X+7).
$$

The non-linear factors that remain after removing the displayed linear factors have no rational roots:

- the three quadratic discriminants are respectively
  $$
  32,\quad192,\quad128,
  $$
  none of which is a rational square;
- the four cubic factors have no rational root by the rational-root theorem.

Therefore the only nondegenerate rational candidates in

$$
0<X<1
$$

are

$$
X\in
\left\{
\frac14,\frac12,\frac34
\right\}.
$$

## 3. Rational-angle gate

But $X$ and $1-X$ must both be rational squares.

For

$$
X=\frac14,
$$

we have

$$
1-X=\frac34,
$$

which is not a square in $\mathbb Q$.

For

$$
X=\frac12,
$$

$X$ itself is not a square in $\mathbb Q$.

For

$$
X=\frac34,
$$

$X$ itself is not a square in $\mathbb Q$.

Hence all twelve internal Pythagorean possibilities fail.

## 4. Conclusion

Therefore

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

This eliminates the prime-power half of the $R(g)=4$ stratum.

The only remaining $R(g)=4$ primitive factorization type is

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