# Phase 1F-C — Structural Reduction of the Remaining $R(g)=4$ Branch $g=pq$

**Status:** reduction complete; global elimination not yet proved.

Let

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

Choose Gaussian arguments $\alpha,\beta$ for the two primes and set

$$
A=2\alpha,
\qquad
B=2\beta.
$$

Because the prime representations are rational,

$$
a=\tan A\in\mathbb Q,
\qquad
b=\tan B\in\mathbb Q,
$$

and, crucially,

$$
1+a^2\in\mathbb Q^2,
\qquad
1+b^2\in\mathbb Q^2.
$$

The four normalized squared even-leg states can be multiplied by the common positive denominator

$$
(1+a^2)(1+b^2)
$$

without changing any Pythagorean equality. The resulting four quantities are

$$
T_1=a^2(1+b^2),
$$

$$
T_2=b^2(1+a^2),
$$

$$
T_3=(a+b)^2,
$$

$$
T_4=(a-b)^2.
$$

Testing all twelve ordered relations

$$
T_i+T_j=T_k
$$

gives exact factorizations.

After removing:

- sign-impossible cases for $a,b>0$;
- $a=b$, which makes $T_4=0$ degenerate;
- $ab=1$, which makes the $A+B$ representation degenerate;

all remaining nondegenerate possibilities reduce, up to swapping $a$ and $b$, to only two essential equations:

## Family PQ-I

$$
\boxed{
a(1+b^2)=4b.
}
$$

## Family PQ-II

$$
\boxed{
a^2(1-b^2)+2b^2=0.
}
$$

Equivalently,

$$
\boxed{
a^2=\frac{2b^2}{b^2-1},
\qquad
b>1.
}
$$

Their swapped versions give no new structural class.

Therefore the entire unresolved $R(g)=4$ problem has been reduced to:

> Do either PQ-I or PQ-II admit positive rational solutions $a,b$ such that

$$
1+a^2
$$

and

$$
1+b^2
$$

are both rational squares, with the additional requirement that the corresponding half-angle Gaussian norms are distinct primes $p,q\equiv1\pmod4$?

A bounded exact rational scan found no solution to either essential equation for half-angle rational parameters through denominator $500$ for PQ-I and denominator $1000$ for PQ-II, but this is only exploratory evidence and is not a proof.

The next task is therefore no longer the original Perfect Cuboid problem. It is the arithmetic study of two explicit rational curves.