# Phase 1E — Internal Pythagorean Closure of the Even-Leg Set

**Experiment:** `PCGC-P1E-EVEN-LEG-INTERNAL-TRIPLE`  
**Scope:** primitive common-hypotenuse states with $g\le5{,}000{,}000$  
**Arithmetic:** exact integer arithmetic

## Definition

For an admissible odd primitive common hypotenuse $g$, define

$$
L(g)=\left\{x>0:g^2-x^2\in\mathbb Z^2\right\}.
$$

Every representation of $g^2$ as two positive squares contributes one odd leg and one even leg. Define the even-leg subset

$$
E(g)=\left\{x\in L(g):2\mid x\right\}.
$$

For primitive admissible $g$, every member of $E(g)$ is in fact divisible by $4$.

## Internal-triple equivalence

Assume a primitive Perfect Cuboid exists with unique odd edge $a$ and even edges $b,c$.

The face diagonal opposite $a$ is

$$
f^2=b^2+c^2.
$$

Since

$$
a^2+f^2=g^2,
$$

the value $f$ is itself an even Pythagorean leg with common hypotenuse $g$.

Therefore

$$
b,c,f\in E(g)
$$

and

$$
b^2+c^2=f^2.
$$

Conversely, suppose there exist

$$
b,c,f\in E(g)
$$

with

$$
b^2+c^2=f^2.
$$

Because $f\in E(g)$, there is an odd integer $a$ such that

$$
a^2+f^2=g^2.
$$

Hence

$$
a^2+b^2+c^2=g^2.
$$

Because $b,c\in E(g)$, the quantities

$$
g^2-b^2=a^2+c^2
$$

and

$$
g^2-c^2=a^2+b^2
$$

are also integer squares. Together with

$$
b^2+c^2=f^2,
$$

all three face diagonals are integral.

Thus

$$
\boxed{
\text{Primitive Perfect Cuboid at }g
\iff
E(g)\text{ contains an internal Pythagorean triple.}
}
$$

This is an exact reformulation.

## Scaled-down version

Since every element of $E(g)$ is divisible by $4$, define

$$
E_4(g)=\left\{e/4:e\in E(g)\right\}.
$$

Then the target is equivalently an ordinary Pythagorean triple entirely inside $E_4(g)$.

The outer Perfect Cuboid problem therefore becomes a nested closure problem:

$$
\boxed{
\text{Pythagorean representations with hypotenuse }g
\supset
\text{an internal Pythagorean triple among their even legs}.
}
$$

## Computational validation through the existing bound

- Eligible fixed-$g$ states: **241,741**
- Old Phase 1D candidate-pair checks: **24,813,268**
- P1E internal even-leg pair checks: **5,734,358**
- Pair-check reduction factor: **4.327x**
- Internal Pythagorean hits: **0**
- Even-leg divisibility-by-4 validation: **PASS**
- One-odd / one-even representation validation: **PASS**

No internal Pythagorean triple was found in any $E(g)$ for the current certified range.

Therefore the previously certified finite exclusion remains unchanged, but the search architecture is now substantially simpler.

## Research consequence

The next proof-safe target is no longer the original three-edge subset problem.

It is:

> characterize finite sets $E_4(g)$ arising from common-hypotenuse representations, and determine whether their arithmetic structure can forbid an internal Pythagorean triple.

This creates a nested-Pythagorean formulation that may admit stronger modular, valuation, graph, or Gaussian-integer invariants.