# Phase 1F-C — Reduction of the Mixed $R(g)=4$ Stratum

The only unresolved representation-multiplicity-four family after P1F-B is

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

with $p,q$ prime.

Write

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

using primitive representations with

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

and opposite parity in each pair. Define

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

Let

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

The four normalized even-leg states are, up to absolute value,

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

A Perfect Cuboid in this stratum requires three of these four values to form an internal Pythagorean triple.

## Six globally eliminated orderings

The following six orderings factor into elementary conditions and are globally impossible in the primitive mixed-prime domain:

1. $A^2+B^2=C^2$.
   The only interior factor is

   $$
   st+s+t=1.
   $$

   Homogenizing gives

   $$
   (m+n)(u+v)=2nv.
   $$

   Because primitive sum-of-two-squares pairs have opposite parity, both $m+n$ and $u+v$ are odd. Thus the left side is odd and the right side even: contradiction.

2. $A^2+B^2=D^2$.
   Both nontrivial factors are strictly positive for $0<t,s<1$.

3. $A^2+C^2=B^2$.
   All roots lie on forbidden boundaries or sign-incompatible loci.

4. $A^2+D^2=B^2$.
   The only interior solution is $s=t$, which for reduced primitive slopes forces $p=q$.

5. $B^2+C^2=A^2$.
   All roots lie on forbidden boundaries or sign-incompatible loci.

6. $B^2+D^2=A^2$.
   The only interior solution is again $s=t$, forcing $p=q$.

Hence

$$
\boxed{
6\text{ of the }12\text{ internal Pythagorean orderings are globally eliminated.}
}
$$

## Six surviving core curves

The unresolved problem is exactly the following six bivariate rational-point equations:

$$
H_1=
s^4t+4s^3t^2-4s^3+2s^2t-4st^2+4s+t=0,
$$

$$
H_2=
s^4t-4s^3t^2+4s^3+2s^2t+4st^2-4s+t=0,
$$

$$
H_3=
4s^2t^3-4s^2t+st^4+2st^2+s-4t^3+4t=0,
$$

$$
H_4=
4s^2t^3-4s^2t-st^4-2st^2-s-4t^3+4t=0,
$$

$$
H_5=
s^4t^2+2s^2t^4-10s^2t^2+2s^2+t^2=0,
$$

$$
H_6=
2s^4t^2+s^2t^4-10s^2t^2+s^2+2t^2=0.
$$

## Exact bounded cross-check on the actual prime-slope domain

The six equations were homogenized in the integer variables $(m,n,u,v)$ and evaluated with integer arithmetic only.

Bound:

$$
p,q\le 30,000.
$$

Results:

- primitive Gaussian-prime slopes: **1,611**
- distinct prime pairs tested: **1,296,855**
- exact zeros among $H_1,\ldots,H_6$: **0**
- runtime: **6.875 seconds**

No zero was found in this finite prime-slope domain.

This bounded check is **not** a global proof for $H_1$--$H_6$.

## Current frontier

The entire $R(g)=4$ problem has now been reduced to six explicit bivariate curves subject to:

$$
0<t,s<1,
$$

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

with

$$
\gcd(m,n)=\gcd(u,v)=1,
$$

opposite parity in each pair, and

$$
m^2+n^2,
\quad
u^2+v^2
$$

distinct primes congruent to $1\pmod4$.

The next proof target is to determine whether these six curves have any rational points satisfying those arithmetic side conditions.

The natural next attacks are local-solvability sieves, valuation constraints, and birational classification of each curve.