# Phase 1H — Global Elimination of the Prime-Power Strata $R(g)=5$, $R(g)=6$, and the $p^7$ Branch of $R(g)=7$

## Result

The following primitive common-hypotenuse families are impossible:

$$
\boxed{
g=p^5,
\qquad
g=p^6,
\qquad
g=p^7,
\qquad
p\equiv1\pmod4.
}
$$

Consequently:

- the entire $R(g)=5$ stratum is eliminated;
- the entire $R(g)=6$ stratum is eliminated;
- within $R(g)=7$, the pure prime-power branch $g=p^7$ is eliminated.

The only remaining factorization type at $R(g)=7$ is therefore

$$
\boxed{
g=p^2q
}
$$

up to interchange of the distinct primes $p,q\equiv1\pmod4$.

---

## 1. One-angle structure of a prime power

Let

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

be the primitive sum-of-two-squares representation of the prime, and write

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

For

$$
g=p^e,
$$

the $e$ positive Pythagorean representations of $g^2$ generate normalized even-leg states

$$
|\sin\phi|,
|\sin2\phi|,
\ldots,
|\sin e\phi|,
$$

where

$$
\phi=2\theta.
$$

Because

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

we have

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

Define

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

Then every admissible $X$ must be a square in $\mathbb Q$.

---

## 2. Exact polynomial encoding

For each $k$,

$$
\sin(k\phi)
=
\sin\phi\,
U_{k-1}(\cos\phi),
$$

where $U_{k-1}$ is the Chebyshev polynomial of the second kind.

Therefore

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

is an integer polynomial in

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

An internal Pythagorean triple among the even-leg states requires

$$
S_i(X)+S_j(X)=S_k(X)
$$

for distinct indices $i,j,k$.

For $e=5,6,7$, every such ordered relation was constructed exactly over $\mathbb Z[X]$, factored exactly, and all rational roots were extracted exactly.

The number of ordered internal-Pythagorean relations is:

$$
30
$$

for $e=5$,

$$
60
$$

for $e=6$,

and

$$
105
$$

for $e=7$.

---

## 3. Exact rational-root result

For each of

$$
e=5,6,7,
$$

the union of every rational root appearing in every internal-Pythagorean relation is exactly

$$
\boxed{
\left\{
0,\frac14,\frac12,\frac34,1
\right\}.
}
$$

No other rational value of $X$ occurs.

Now:

$$
X=0
$$

and

$$
X=1
$$

are degenerate boundary states.

The values

$$
X=\frac12
$$

and

$$
X=\frac34
$$

are not squares in $\mathbb Q$, contradicting the necessary condition

$$
X=\sin^2\phi\in\mathbb Q^2.
$$

The only nondegenerate rational-square candidate is

$$
X=\frac14.
$$

Because

$$
0<\phi<\frac{\pi}{2},
$$

this means

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

Since

$$
\phi=2\theta
$$

and

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

we have

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

Thus

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

so

$$
t=2\pm\sqrt3.
$$

The admissible root in $(0,1)$ is

$$
2-\sqrt3,
$$

which is irrational.

Contradiction.

Therefore none of the three prime-power families can contain a primitive Perfect Cuboid.

---

## 4. Consequences by representation multiplicity

For primitive $g$,

$$
R(g)
=
\frac12
\left(
\prod_{p\equiv1\pmod4}(2e_p+1)-1
\right).
$$

### $R(g)=5$

Then

$$
\prod_p(2e_p+1)=11,
$$

so necessarily

$$
g=p^5.
$$

This family is eliminated.

Hence

$$
\boxed{
R(g)\ne5.
}
$$

### $R(g)=6$

Then

$$
\prod_p(2e_p+1)=13,
$$

so necessarily

$$
g=p^6.
$$

This family is eliminated.

Hence

$$
\boxed{
R(g)\ne6.
}
$$

### $R(g)=7$

Then

$$
\prod_p(2e_p+1)=15.
$$

There are two exponent patterns:

$$
15
$$

itself, giving

$$
g=p^7,
$$

or

$$
15=3\cdot5,
$$

giving

$$
g=pq^2
$$

up to renaming the distinct primes.

The $p^7$ branch is eliminated by the exact polynomial sweep.

Therefore the only unresolved $R(g)=7$ family is

$$
\boxed{
g=pq^2,
\qquad
p\ne q,
\qquad
p,q\equiv1\pmod4.
}
$$

---

## 5. Current global frontier

The eliminated representation multiplicities are now:

$$
R(g)=3,
$$

$$
R(g)=4,
$$

$$
R(g)=5,
$$

$$
R(g)=6.
$$

At

$$
R(g)=7,
$$

the prime-power branch is also eliminated.

Thus any primitive Perfect Cuboid must lie either in the mixed family

$$
g=pq^2
$$

at $R(g)=7$, or at a still higher representation multiplicity.

The next structural target is therefore not a generic integer search.

It is the two-angle mixed-exponent family

$$
\boxed{
g=pq^2.
}
$$

That family should be attacked by the same sequence used successfully at $R=4$:

1. enumerate its exact common-hypotenuse angle states;
2. identify the even-leg subset;
3. reduce internal Pythagorean closure to a finite set of two-angle equations;
4. apply parity, valuation, sign, and prime-square discriminant obstructions before invoking heavier curve machinery.