# Phase 1F-A — Global Elimination of the Minimal Representation Stratum

**Result:** the entire primitive stratum

$$
R(g)=3
$$

is impossible for a Perfect Cuboid.

This is a global structural exclusion, not a finite-height observation.

## Step 1 — classify primitive $g$ when $R(g)=3$

From the primitive common-hypotenuse gate,

$$
g=\prod_{p\equiv1\pmod4}p^{e_p}.
$$

The number of unordered positive Pythagorean representations with
hypotenuse $g$ is

$$
R(g)=
\frac12
\left(
\prod_p(2e_p+1)-1
\right).
$$

If

$$
R(g)=3,
$$

then

$$
\prod_p(2e_p+1)=7.
$$

Because every factor $2e_p+1$ is an odd integer at least $3$, the only
possibility is one prime with exponent $3$:

$$
\boxed{
g=p^3,
\qquad
p\equiv1\pmod4.
}
$$

## Step 2 — write the three even-leg states

Write the prime as

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

with

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

and opposite parity.

Let

$$
\theta=\arg(r+is).
$$

Then

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

For

$$
g=p^3,
$$

the three nontrivial Pythagorean representations of $g^2$ arise from the
Gaussian states with arguments

$$
2\theta,\qquad4\theta,\qquad6\theta.
$$

Hence the three even legs, divided by $g$, have absolute values

$$
|\sin2\theta|,
\qquad
|\sin4\theta|,
\qquad
|\sin6\theta|.
$$

By the Phase 1E equivalence, a Perfect Cuboid in this stratum would require
these three even legs themselves to form a Pythagorean triple.

Set

$$
X=\sin^2(2\theta).
$$

Then

$$
S_1=X,
$$

$$
S_2=\sin^2(4\theta)=4X(1-X),
$$

and by the triple-angle identity,

$$
S_3=\sin^2(6\theta)=X(3-4X)^2.
$$

## Step 3 — exhaust the three possible Pythagorean orderings

There are only three possibilities.

### Case A

$$
S_1+S_2=S_3.
$$

Direct factorization gives

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

Thus

$$
X\in
\left\{
0,\frac14,1
\right\}.
$$

The values $X=0$ and $X=1$ are degenerate.

If

$$
X=\frac14,
$$

then, since $0<2\theta<\pi$,

$$
\sin(2\theta)=\frac12.
$$

Writing

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

we have

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

Hence

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

so

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

which is irrational. Contradiction.

### Case B

$$
S_1+S_3=S_2.
$$

Then

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

Thus

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

But

$$
\sin(2\theta)=\frac{2rs}{p}\in\mathbb Q.
$$

Therefore $X$ must be the square of a rational number.

Neither

$$
\frac12
$$

nor

$$
\frac34
$$

is a square in $\mathbb Q$.

Contradiction.

### Case C

$$
S_2+S_3=S_1.
$$

Then

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

Thus

$$
X\in
\left\{
0,\frac34,1
\right\}.
$$

Again $0$ and $1$ are degenerate, while $3/4$ is not a rational square.

Contradiction.

## Theorem

All possible internal Pythagorean orderings fail.

Therefore

$$
\boxed{
R(g)=3
\Longrightarrow
\text{no primitive Perfect Cuboid at }g.
}
$$

Combining this with the previous multiplicity gate yields the stronger
necessary condition

$$
\boxed{
R(g)\ge4.
}
$$

Equivalently,

$$
\boxed{
\prod_{p\equiv1\pmod4}(2e_p+1)\ge9.
}
$$

## Consequence

The minimal representation stratum is removed globally.

The next unresolved representation stratum is

$$
R(g)=4,
$$

which has two primitive factorization types:

$$
g=p^4
$$

or

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

These are the next natural structural targets.