# Phase 1G — The First Surviving Diagonal Type: $g=p^2q$

**Status:** exact bounded scan + complete seven-state symbolic map  
**Prime bound:** $p,q\le 20,000$, $p\ne q$, $p\equiv q\equiv1\pmod4$

## Why this is the next frontier

The fixed-$g$ framework independently re-derived the classical restrictions that a primitive Perfect Cuboid body diagonal is neither a prime power nor a product of two primes.

The first surviving factorization type by representation multiplicity is therefore

$$
g=p^2q
$$

or, by exchanging prime labels,

$$
g=pq^2.
$$

For

$$
g=p^2q,
$$

the number of nontrivial Pythagorean representations is

$$
R(g)
=
\frac12
\left(
(2\cdot2+1)(2\cdot1+1)-1
\right)
=
7.
$$

## Seven exact Gaussian states

Write

$$
p=r^2+s^2,
\qquad
q=u^2+v^2,
$$

and define

$$
\phi=2\arg(r+is),
\qquad
\psi=2\arg(u+iv).
$$

The seven even-leg states are represented by the angle lattice

$$
(1,0),
(2,0),
(0,1),
(1,1),
(1,-1),
(2,1),
(2,-1),
$$

corresponding to

$$
\phi,\ 2\phi,\ \psi,\ \phi+\psi,\ \phi-\psi,\ 2\phi+\psi,\ 2\phi-\psi.
$$

The implementation generates these seven states directly with Gaussian integer multiplication, with no search over integers below $g$.

Direct comparison against exhaustive fixed-$g$ leg enumeration for every small test case with

$$
g\le120{,}000
$$

gave:

**Gaussian seven-state formula validation: PASS**

## Bounded exact scan

- Primes $1\pmod4$ up to the bound: **1,125**
- Ordered distinct $(p,q)$ states: **1,264,500**
- Exact internal Pythagorean triples found inside the seven even legs: **0**
- Exact near-hits where two seven-state legs form a Pythagorean pair but the resulting hypotenuse is not another state: **21**
- Runtime for the bounded state scan: **9.777 s**

No $p^2q$ seven-state internal closure was found in this bounded prime range.

This is exploratory evidence only; it is **not** a global exclusion.

## Symbolic reduction

Let

$$
a=\tan\phi,
\qquad
b=\tan\psi.
$$

Because both base prime representations are rational,

$$
a,b\in\mathbb Q
$$

and

$$
1+a^2,\quad1+b^2
$$

are rational squares.

After multiplying all normalized squared even-leg values by the common positive denominator

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

the seven states become

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

$$
T_2=4a^2(1+b^2),
$$

$$
T_3=b^2(1+a^2)^2,
$$

$$
T_4=(a+b)^2(1+a^2),
$$

$$
T_5=(a-b)^2(1+a^2),
$$

$$
T_6=(2a+b(1-a^2))^2,
$$

$$
T_7=(2a-b(1-a^2))^2.
$$

Every possible internal Pythagorean closure is one of the exact equations

$$
T_i+T_j=T_k
$$

for distinct indices.

There are

$$
105
$$

ordered relations.

All 105 have been expanded and factorized exactly. Their total degrees are:

{'4': 39, '6': 66}

The complete factor map is stored in `phase1g_p2q_factor_map.csv`.

## Next structural task

The problem has now been reduced from the original seven cuboid lengths to a finite family of bivariate rational equations subject to the two conic conditions

$$
1+a^2\in\mathbb Q^2,
$$

$$
1+b^2\in\mathbb Q^2.
$$

The next target is to quotient the 105 relations by the angle symmetries

$$
b\mapsto-b
$$

and representation degeneracies, then identify which remaining algebraic curves can be eliminated by divisibility, congruence, or rational-point arguments.