# Phase 1F-D — Global Diagonal Restriction Re-derived in the Fixed-$g$ Framework

**Status:** independently re-derived; known in the literature (Korec, 1992)  
**Framework:** common-hypotenuse leg sets, internal Pythagorean closure, Gaussian-angle reduction

---

## 1. Prime-power diagonals are impossible

Assume a primitive Perfect Cuboid has

$$
g=p^e,
\qquad
p\equiv1\pmod4.
$$

Write

$$
p=r^2+s^2
$$

with

$$
r>s>0,
\qquad
\gcd(r,s)=1,
$$

and opposite parity.

Let

$$
r+is=\sqrt p\,e^{i\theta}.
$$

For a fixed hypotenuse

$$
g=p^e,
$$

the nontrivial Pythagorean representations of

$$
x^2+y^2=g^2
$$

are indexed by

$$
k=1,\dots,e
$$

and have normalized even-leg magnitudes

$$
|\sin(k\phi)|,
\qquad
\phi=2\theta.
$$

Because

$$
\tan\theta=\frac sr\in\mathbb Q,
$$

both

$$
\sin\phi
$$

and

$$
\cos\phi
$$

are rational.

By the internal-even-leg equivalence, a primitive Perfect Cuboid would require distinct positive integers

$$
m,n,k
$$

such that

$$
\sin^2(m\phi)+\sin^2(n\phi)=\sin^2(k\phi).
$$

Using

$$
\sin^2 t=\frac{1-\cos2t}{2},
$$

this becomes

$$
1-\cos(2m\phi)-\cos(2n\phi)+\cos(2k\phi)=0.
$$

Set

$$
y=2\cos(2\phi).
$$

Since

$$
\sin\phi,\cos\phi\in\mathbb Q,
$$

we have

$$
y\in\mathbb Q.
$$

Define the monic integer polynomials

$$
C_0(y)=2,
$$

$$
C_1(y)=y,
$$

$$
C_{j+1}(y)=yC_j(y)-C_{j-1}(y).
$$

Then

$$
C_j(2\cos\psi)=2\cos(j\psi).
$$

Therefore the internal Pythagorean condition becomes

$$
C_k(y)-C_m(y)-C_n(y)+2=0.
$$

Because $m,n,k$ are distinct, the largest degree among them occurs exactly once. Hence this polynomial is monic or anti-monic with integer coefficients.

By the rational-root theorem,

$$
y\in\mathbb Z.
$$

But

$$
|y|\le2,
$$

so

$$
y\in\{-2,-1,0,1,2\}.
$$

Equivalently,

$$
X=\sin^2\phi=\frac{2-y}{4}
$$

belongs to

$$
\left\{
0,\frac14,\frac12,\frac34,1
\right\}.
$$

The endpoints $0,1$ are degenerate.

For the remaining values:

- if $X=1/4$, then $1-X=3/4$ is not a square in $\mathbb Q$;
- if $X=1/2$, then $X$ is not a square in $\mathbb Q$;
- if $X=3/4$, then $X$ is not a square in $\mathbb Q$.

But both

$$
\sin\phi
$$

and

$$
\cos\phi
$$

must be rational.

Contradiction.

Thus

$$
\boxed{
g=p^e
\Longrightarrow
\text{no primitive Perfect Cuboid}
}
$$

for every positive integer $e$.

This single argument subsumes the earlier $R(g)=3$ and $g=p^4$ eliminations.

---

## 2. The two-prime branch $g=pq$ is impossible

Let

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

Write

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

with

$$
r>s>0,
\qquad
u>v>0,
$$

primitive opposite-parity representations.

Define

$$
R=r^2-s^2,
\qquad
U=u^2-v^2,
$$

and

$$
A=rs,
\qquad
B=uv.
$$

The four even-leg states reduce, after clearing a common positive denominator, to the four quantities

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

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

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

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

where

$$
a=\frac{2rs}{R},
\qquad
b=\frac{2uv}{U}.
$$

The twelve possible internal Pythagorean orderings reduce, after positivity and degeneracy exclusions and symmetry under $p\leftrightarrow q$, to two essential equations.

### Branch I

$$
a(1+b^2)=4b.
$$

Since

$$
1+b^2=\frac{q^2}{U^2},
$$

this is equivalent to

$$
Aq^2=4BRU.
$$

Because

$$
\gcd(q,BU)=1,
$$

$q$ cannot divide $A$ in a solution; otherwise $q$ would fail to divide $R$, contradicting the equation.

Hence

$$
q^2\mid R.
$$

Write

$$
R=q^2h.
$$

Then

$$
A=4BUh.
$$

Using

$$
p^2=R^2+4A^2,
$$

we obtain

$$
p^2
=
h^2
\left(
q^4+64B^2U^2
\right).
$$

Thus

$$
h\mid p.
$$

Since $p$ is prime and

$$
R<p,
$$

we cannot have $h=p$.

Therefore

$$
h=1
$$

and

$$
R=q^2.
$$

Now

$$
R=(r-s)(r+s).
$$

The factors $r-s$ and $r+s$ are positive odd coprime integers, so

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

Hence

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

But the equation $A=4BU$ gives

$$
q^4-1=16uv(u^2-v^2).
$$

Since

$$
uv\le\frac q2
$$

and

$$
u^2-v^2<q,
$$

the right side is strictly less than

$$
8q^2.
$$

For every admissible prime

$$
q\ge5,
$$

we have

$$
q^4-1>8q^2.
$$

Contradiction.

Therefore Branch I has no solution.

### Branch II

$$
a^2(1-b^2)+2b^2=0.
$$

After clearing denominators this becomes

$$
A^2(8B^2-q^2)=2B^2R^2.
$$

Since

$$
\gcd(B,q)=1,
$$

we have

$$
\gcd(B^2,8B^2-q^2)=1.
$$

Therefore

$$
B^2\mid A^2,
$$

hence

$$
B\mid A.
$$

Write

$$
A=hB.
$$

Since

$$
\gcd(A,R)=1,
$$

we also have

$$
\gcd(h,R)=1.
$$

The equation becomes

$$
h^2(8B^2-q^2)=2R^2.
$$

Thus

$$
h^2\mid2,
$$

so

$$
h=1.
$$

Therefore

$$
A=B.
$$

The equation reduces to

$$
8A^2-q^2=2R^2.
$$

Using

$$
p^2=R^2+4A^2,
$$

we obtain

$$
q^2+2p^2=16A^2.
$$

But $p,q$ are odd, so modulo $8$,

$$
q^2+2p^2\equiv1+2\equiv3\pmod8,
$$

while

$$
16A^2\equiv0\pmod8.
$$

Contradiction.

Therefore Branch II has no solution.

Hence

$$
\boxed{
g=pq
\Longrightarrow
\text{no primitive Perfect Cuboid}.
}
$$

---

## 3. Consolidated diagonal gate

Combining both results:

$$
\boxed{
\text{the primitive space diagonal is neither a prime power nor a product of two primes.}
}
$$

This restriction is known in the classical literature; the present derivation is an independent reconstruction inside the fixed-$g$/internal-leg framework.

The next surviving low-complexity factorization type is

$$
\boxed{
g=p^2q
}
$$

or symmetrically

$$
\boxed{
g=pq^2,
}
$$

with distinct primes

$$
p,q\equiv1\pmod4.
$$

Its representation multiplicity is

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

Thus the next structural frontier is the seven-state common-hypotenuse system.