# Phase 1G — Complete Global Elimination of the $R(g)=4$ Stratum

## Main result

$$
\boxed{
R(g)=4
\Longrightarrow
\text{no primitive Perfect Cuboid}.
}
$$

Together with the earlier elimination of $R(g)=3$, this yields the necessary condition

$$
\boxed{
R(g)\ge5.
}
$$

The proof combines the previous elimination of the $g=p^4$ family with a direct elimination of the only remaining $R=4$ family

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

---

## 1. The last mixed-prime ordering

For the mixed family write

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

with primitive opposite-parity pairs and

$$
0<m<n,
\qquad
0<u<v.
$$

Set

$$
t=\frac mn,
\qquad
s=\frac uv,
$$

and

$$
\alpha=2\arctan t,
\qquad
\beta=2\arctan s.
$$

The four normalized even-leg states are

$$
A=\sin\alpha,
\quad
B=\sin\beta,
\quad
C=\sin(\alpha+\beta),
\quad
D=\sin(\alpha-\beta)
$$

up to absolute value.

After the elementary sign, boundary, parity, symmetry, and $2$-adic eliminations from P1F-C, the only remaining ordering up to interchange of $p$ and $q$ is

$$
A^2+D^2=C^2.
$$

Using

$$
C^2-D^2=\sin(2\alpha)\sin(2\beta),
$$

this gives

$$
\sin^2\alpha
=
\sin(2\alpha)\sin(2\beta),
$$

hence

$$
\tan\alpha=2\sin(2\beta).
$$

Since

$$
\tan\alpha=\frac{2t}{1-t^2}
$$

and

$$
\sin(2\beta)=\frac{4uv(v^2-u^2)}{q^2},
$$

we obtain

$$
\frac{t}{1-t^2}
=
\frac{4uv(v^2-u^2)}{q^2}.
$$

Let

$$
K=4uv(v^2-u^2).
$$

Then

$$
Kt^2+q^2t-K=0.
$$

For rational $t$, the discriminant must be a rational square:

$$
q^4+4K^2=Z^2
$$

for some integer $Z$. Setting

$$
M=2K=8uv(v^2-u^2)
$$

gives

$$
(q^2)^2+M^2=Z^2.
$$

Because $q$ is an odd prime sum of two coprime squares,

$$
\gcd(q^2,M)=1.
$$

Thus $(q^2,M,Z)$ is a primitive Pythagorean triple with odd leg $q^2$.

Write

$$
q^2=r^2-s^2=(r-s)(r+s).
$$

The positive odd factors $r-s$ and $r+s$ are coprime, and their product is the prime square $q^2$. Therefore

$$
r-s=1,
\qquad
r+s=q^2.
$$

Hence the even leg is forced to be

$$
M=2rs=\frac{q^4-1}{2}.
$$

But

$$
\frac{4uv(v^2-u^2)}{q^2}
=
\sin(2\beta)\le1,
$$

so

$$
M=8uv(v^2-u^2)\le2q^2.
$$

For every prime

$$
q\equiv1\pmod4,
$$

we have $q\ge5$, and therefore

$$
\frac{q^4-1}{2}>2q^2.
$$

Contradiction.

The symmetric remaining ordering is eliminated by interchanging $p$ and $q$.

Thus the mixed $pq$ family is impossible, completing the global elimination of $R(g)=4$.