# Perfect Cuboid Radical-FDRS R1

## Squareclass Transition Algebra, Rational-Sphere Moves, Cayley Defect Ratios, and the Unification with C2

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R1 — Transition Algebra and Cayley-Root Reduction  
**Date:** 2026-10-03  
**Status:** Exact reformulation and structural analysis — not a proof of Perfect Cuboid existence or nonexistence

---

# 0. Executive statement

R0 showed that arithmetic squareclass defects can migrate across the flattened Boolean-cube state.

R1 separates two different state spaces that must not be confused:

1. the **ambient rational square-state simplex**, where

$$
A+B+C=1
$$

and the basic cells themselves need not be rational squares;

2. the **classical rational-edge shell**, where

$$
A=x^2,
\qquad
B=y^2,
\qquad
C=z^2,
$$

with

$$
x,y,z\in\mathbb Q_{>0},
$$

and

$$
x^2+y^2+z^2=1.
$$

Only the second space corresponds directly to normalized rational-edge cuboids.

On this shell, the six-channel arithmetic problem collapses from six squareclasses to three complementary face defects.

The central R1 result is that each complementary defect has a canonical Cayley ratio.

For:

$$
0<x<1,
$$

define:

$$
\boxed{
\omega(x)
=
\frac{1-x}{1+x}.
}
$$

Then:

$$
\boxed{
1-x^2\in\mathbb Q^2
\iff
\omega(x)\in\mathbb Q^2.
}
$$

Moreover:

$$
[1-x^2]
=
[\omega(x)]
$$

in:

$$
\mathbb Q^\times/\mathbb Q^{\times2}.
$$

Thus the three face-defect squareclasses are exactly the three Cayley-ratio squareclasses.

For the current seed:

$$
(x,y,z)
=
\left(
\frac{448}{1073},
\frac{495}{1073},
\frac{840}{1073}
\right),
$$

one obtains:

$$
\boxed{
\omega(x)
=
\frac{625}{1521}
=
\left(\frac{25}{39}\right)^2,
}
$$

$$
\boxed{
\omega(y)
=
\frac{289}{784}
=
\left(\frac{17}{28}\right)^2,
}
$$

while:

$$
\boxed{
\omega(z)
=
\frac{233}{1913}.
}
$$

Hence the seed is exactly:

$$
\boxed{
\text{two rational Cayley roots}
+
\text{one quadratic irrational Cayley root}.
}
$$

The old defect:

$$
233\cdot1913
$$

is therefore not merely an unexplained factorization.

It comes directly from:

$$
\boxed{
g-c=233,
\qquad
g+c=1913.
}
$$

This also produces a direct unification of Radical-FDRS with the previous C2 formulation.

---

# 1. Ambient Radical-FDRS state versus classical shell

R0 allowed:

$$
A,B,C\in\mathbb Q_{>0}
$$

subject to:

$$
A+B+C=1.
$$

Define:

$$
D=A+B=1-C,
$$

$$
E=A+C=1-B,
$$

$$
F=B+C=1-A.
$$

The ambient squareclass state is:

$$
K
=
(
[A],
[B],
[C],
[D],
[E],
[F]
).
$$

This space is useful for studying general squareclass migration.

However, a classical rational-edge cuboid requires:

$$
A=x^2,
$$

$$
B=y^2,
$$

$$
C=z^2
$$

with rational:

$$
x,y,z.
$$

Thus define the classical shell:

$$
\boxed{
\mathcal S_{\mathbb Q}
=
\left\{
(x,y,z)\in\mathbb Q_{>0}^3:
x^2+y^2+z^2=1
\right\}.
}
$$

On this shell:

$$
[A]=[B]=[C]=1.
$$

Therefore only the face channels can fail.

The classical shell state is:

$$
\boxed{
K_{\rm shell}
=
(
[D],
[E],
[F]
).
}
$$

where:

$$
D=1-z^2,
$$

$$
E=1-y^2,
$$

$$
F=1-x^2.
$$

Perfect Cuboid closure becomes:

$$
\boxed{
[D]=[E]=[F]=1.
}
$$

---

# 2. Why the R0 linear move is not sufficient for the classical problem

The ambient conservation move:

$$
(A,B,C)
\mapsto
(A+\delta_A,B+\delta_B,C+\delta_C)
$$

with:

$$
\delta_A+\delta_B+\delta_C=0
$$

preserves:

$$
A+B+C=1.
$$

But it does not generally preserve:

$$
A,B,C\in\mathbb Q^2.
$$

Therefore it can move through valid Radical-FDRS states while leaving the classical rational-edge shell.

This distinction is essential.

R0 defect migration remains valid as an ambient arithmetic phenomenon.

For actual Perfect Cuboid repair, R1 replaces linear simplex moves by shell-preserving rational rotations.

---

# 3. Shell-preserving rational rotation

For rational:

$$
t,
$$

define the rational plane rotation:

$$
\boxed{
R(t)
=
\frac{1}{1+t^2}
\begin{pmatrix}
1-t^2 & -2t\\
2t & 1-t^2
\end{pmatrix}.
}
$$

Its entries are rational whenever:

$$
t\in\mathbb Q.
$$

Also:

$$
R(t)^\top R(t)=I.
$$

Therefore it preserves:

$$
u^2+v^2.
$$

Embedding this rotation in coordinate planes gives elementary moves:

$$
R_{xy}(t),
$$

$$
R_{xz}(t),
$$

$$
R_{yz}(t).
$$

Each move preserves:

$$
x^2+y^2+z^2=1.
$$

Hence:

$$
\boxed{
R_{ij}(t):
\mathcal S_{\mathbb Q}
\rightarrow
\mathcal S_{\mathbb Q}.
}
$$

These are legitimate rational-edge-shell moves.

---

# 4. Same-axis move composition

For one plane:

$$
R(t)R(s)
=
R(t\oplus s),
$$

where:

$$
\boxed{
t\oplus s
=
\frac{t+s}{1-ts}
}
$$

whenever:

$$
1-ts\ne0.
$$

This is the tangent-half-angle group law.

The inverse is:

$$
\boxed{
R(t)^{-1}=R(-t).
}
$$

Projectively the parameter belongs naturally to:

$$
\mathbb P^1(\mathbb Q).
$$

Rotations in different coordinate planes generally do not commute.

Thus the shell-preserving move algebra is qualitatively closer to the original Magic-Cube intuition than the ambient additive translation algebra.

---

# 5. Anchored-pair move

Consider a move that fixes:

$$
y=\frac bg.
$$

Then:

$$
E=1-y^2
$$

is fixed.

Suppose:

$$
E
=
\left(\frac eg\right)^2.
$$

Therefore:

$$
b^2+e^2=g^2.
$$

The remaining coordinates satisfy:

$$
x^2+z^2
=
\left(\frac eg\right)^2.
$$

Using unnormalized rational lengths:

$$
a^2+c^2=e^2.
$$

Parameterize the entire rational circle by:

$$
\boxed{
a(t)
=
e
\frac{1-t^2}{1+t^2},
}
$$

$$
\boxed{
c(t)
=
e
\frac{2t}{1+t^2}.
}
$$

This preserves:

$$
a(t)^2+c(t)^2=e^2
$$

exactly.

Thus:

$$
b,
\quad
e,
\quad
g
$$

remain fixed.

The only two face channels that must be retested are:

$$
d(t)^2=a(t)^2+b^2,
$$

and:

$$
f(t)^2=b^2+c(t)^2.
$$

Hence a one-defect repair has become a one-parameter two-gate problem.

---

# 6. First anchored square gate

Define:

$$
N_D(t)
=
e^2(1-t^2)^2
+
b^2(1+t^2)^2.
$$

Since:

$$
b^2+e^2=g^2,
$$

exact factorization gives:

$$
\boxed{
N_D(t)
=
Q_-(t)Q_+(t),
}
$$

where:

$$
\boxed{
Q_-(t)
=
g(1+t^2)-2et,
}
$$

$$
\boxed{
Q_+(t)
=
g(1+t^2)+2et.
}
$$

Because:

$$
(1+t^2)^2
$$

is already a rational square denominator,

$$
d(t)
$$

is rational if and only if:

$$
N_D(t)
$$

is a rational square.

For positive rational:

$$
U,V,
$$

we have:

$$
UV\in\mathbb Q^2
\iff
\frac UV\in\mathbb Q^2.
$$

Therefore:

$$
\boxed{
d(t)\in\mathbb Q
\iff
\frac{Q_-(t)}{Q_+(t)}
\in
\mathbb Q^2.
}
$$

The square gate has become a conjugate-factor ratio gate.

---

# 7. Second anchored square gate

Similarly define:

$$
N_F(t)
=
4e^2t^2
+
b^2(1+t^2)^2.
$$

Exact factorization gives:

$$
\boxed{
N_F(t)
=
S_-(t)S_+(t),
}
$$

with:

$$
\boxed{
S_-(t)
=
(g+e)t^2+(g-e),
}
$$

$$
\boxed{
S_+(t)
=
(g-e)t^2+(g+e).
}
$$

Hence:

$$
\boxed{
f(t)\in\mathbb Q
\iff
\frac{S_-(t)}{S_+(t)}
\in
\mathbb Q^2.
}
$$

Thus the anchored shell-repair problem is exactly:

$$
\boxed{
\frac{Q_-(t)}{Q_+(t)}
\in\mathbb Q^2,
\qquad
\frac{S_-(t)}{S_+(t)}
\in\mathbb Q^2.
}
$$

---

# 8. Current seed on the first anchored fiber

Use:

$$
b=495,
$$

$$
e=952,
$$

$$
g=1073.
$$

The variable pair is initially:

$$
(a,c)=(448,840).
$$

Since:

$$
448^2+840^2=952^2,
$$

the rational-circle parameter is:

$$
t_0
=
\frac{c}{e+a}
=
\frac{840}{952+448}
=
\boxed{
\frac35.
}
$$

The failed gate factors are:

$$
Q_-(t)
=
1073t^2-1904t+1073,
$$

$$
Q_+(t)
=
1073t^2+1904t+1073.
$$

At:

$$
t_0=\frac35,
$$

we obtain:

$$
Q_-(t_0)
=
\frac{7922}{25}
=
\boxed{
\frac{2\cdot17\cdot233}{5^2},
}
$$

and:

$$
Q_+(t_0)
=
\frac{65042}{25}
=
\boxed{
\frac{2\cdot17\cdot1913}{5^2}.
}
$$

Therefore:

$$
\boxed{
\frac{Q_-(t_0)}{Q_+(t_0)}
=
\frac{233}{1913}.
}
$$

The two defect primes are no longer hidden inside one squarefree kernel.

They are separated into the two conjugate factors.

---

# 9. Successful gate on the same fiber

The successful gate factors are:

$$
S_-(t)
=
2025t^2+121,
$$

$$
S_+(t)
=
121t^2+2025.
$$

At:

$$
t_0=\frac35,
$$

their ratio is:

$$
\boxed{
\frac{S_-(t_0)}{S_+(t_0)}
=
\frac{625}{1521}
=
\left(\frac{25}{39}\right)^2.
}
$$

Thus the closed channel automatically recovers the previous C2 parameter:

$$
\boxed{
r_0=\frac{25}{39}.
}
$$

This is not an additional numerical coincidence.

It is the same half-angle coordinate expressed in the anchored FDRS factorization.

---

# 10. Second independent anchored fiber

Now instead fix:

$$
a=448
$$

and:

$$
f=975.
$$

Since:

$$
448^2+975^2=1073^2,
$$

this complementary pair stays closed.

The variable pair is:

$$
(b,c)=(495,840),
$$

and:

$$
495^2+840^2=975^2.
$$

The corresponding rational-circle parameter is:

$$
s_0
=
\frac{c}{f+b}
=
\frac{840}{975+495}
=
\boxed{
\frac47.
}
$$

The failed gate factors become:

$$
Q_-^{(2)}(s)
=
1073s^2-1950s+1073,
$$

$$
Q_+^{(2)}(s)
=
1073s^2+1950s+1073.
$$

At:

$$
s_0=\frac47,
$$

we obtain:

$$
Q_-^{(2)}(s_0)
=
\frac{15145}{49}
=
\boxed{
\frac{5\cdot13\cdot233}{7^2},
}
$$

and:

$$
Q_+^{(2)}(s_0)
=
\frac{124345}{49}
=
\boxed{
\frac{5\cdot13\cdot1913}{7^2}.
}
$$

Therefore again:

$$
\boxed{
\frac{Q_-^{(2)}(s_0)}
{Q_+^{(2)}(s_0)}
=
\frac{233}{1913}.
}
$$

The same canonical defect ratio appears on a different rational-shell flattening.

---

# 11. Successful gate on the second fiber

For the already closed:

$$
E
$$

channel, the corresponding factor ratio is:

$$
\boxed{
\frac{289}{784}
=
\left(\frac{17}{28}\right)^2.
}
$$

Hence the second C2 parameter is recovered:

$$
\boxed{
s_{\rm C2}
=
\frac{17}{28}.
}
$$

Thus the two previously discovered C2 rational parameters emerge directly from the FDRS conjugate-factor ratios.

---

# 12. Why the same defect ratio must appear

The repeated ratio:

$$
\frac{233}{1913}
$$

has a simple exact origin.

For the first anchored fiber:

$$
t_0
=
\frac{c}{e+a}.
$$

Using:

$$
a^2+c^2=e^2,
$$

we obtain:

$$
1+t_0^2
=
\frac{2e}{e+a}.
$$

Therefore:

$$
Q_-(t_0)
=
\frac{2e(g-c)}{e+a},
$$

and:

$$
Q_+(t_0)
=
\frac{2e(g+c)}{e+a}.
$$

Thus:

$$
\boxed{
\frac{Q_-(t_0)}{Q_+(t_0)}
=
\frac{g-c}{g+c}.
}
$$

The anchor:

$$
b,e
$$

has disappeared completely from the final ratio.

Therefore any valid anchored flattening that targets the complementary pair associated with:

$$
c
$$

must recover the same canonical ratio:

$$
\boxed{
\omega_c
=
\frac{g-c}{g+c}.
}
$$

This explains why both independent fibers returned exactly:

$$
\frac{233}{1913}.
$$

---

# 13. Cayley defect-ratio theorem

Let:

$$
x\in\mathbb Q,
\qquad
0<x<1.
$$

Define:

$$
\boxed{
\omega(x)
=
\frac{1-x}{1+x}.
}
$$

Then:

$$
1-x^2
=
(1-x)(1+x).
$$

Moreover:

$$
\frac{1-x^2}{\omega(x)}
=
(1+x)^2.
$$

Therefore:

$$
\boxed{
[1-x^2]
=
[\omega(x)]
}
$$

in:

$$
\mathbb Q^\times/\mathbb Q^{\times2}.
$$

Consequently:

$$
\boxed{
1-x^2\in\mathbb Q^2
\iff
\omega(x)\in\mathbb Q^2.
}
$$

This gives a canonical one-dimensional defect coordinate for every complementary pair.

---

# 14. Cayley root

Define:

$$
\boxed{
r_x
=
\sqrt{
\frac{1-x}{1+x}
}.
}
$$

If:

$$
r_x\in\mathbb Q,
$$

then inversion gives:

$$
\boxed{
x
=
\frac{1-r_x^2}{1+r_x^2}.
}
$$

Also:

$$
\boxed{
\sqrt{1-x^2}
=
\frac{2r_x}{1+r_x^2}.
}
$$

Thus:

$$
r_x
$$

is exactly the rational unit-circle half-angle parameter.

The Radical-FDRS root language and the classical rational-circle parameterization are therefore the same arithmetic object in different representations.

---

# 15. Current seed in Cayley-root coordinates

Normalize:

$$
x=\frac{448}{1073},
$$

$$
y=\frac{495}{1073},
$$

$$
z=\frac{840}{1073}.
$$

Then:

$$
\omega(x)
=
\frac{1073-448}{1073+448}
=
\frac{625}{1521}
=
\left(\frac{25}{39}\right)^2.
$$

Therefore:

$$
\boxed{
r_x=\frac{25}{39}.
}
$$

Likewise:

$$
\omega(y)
=
\frac{1073-495}{1073+495}
=
\frac{578}{1568}
=
\frac{289}{784}
=
\left(\frac{17}{28}\right)^2.
$$

Thus:

$$
\boxed{
r_y=\frac{17}{28}.
}
$$

For:

$$
z,
$$

we have:

$$
\omega(z)
=
\frac{1073-840}{1073+840}
=
\boxed{
\frac{233}{1913}.
}
$$

Therefore:

$$
\boxed{
r_z
=
\sqrt{\frac{233}{1913}}
\notin\mathbb Q.
}
$$

The seed can therefore be written as:

$$
\boxed{
(r_x,r_y,r_z)
=
\left(
\frac{25}{39},
\frac{17}{28},
\sqrt{\frac{233}{1913}}
\right).
}
$$

This is a considerably sharper representation of the old one-defect state.

---

# 16. Direct explanation of the old squarefree kernel

The missing face diagonal satisfies:

$$
d^2
=
g^2-c^2.
$$

Therefore:

$$
d^2
=
(g-c)(g+c).
$$

For the seed:

$$
g-c
=
1073-840
=
\boxed{
233,
}
$$

and:

$$
g+c
=
1073+840
=
\boxed{
1913.
}
$$

Hence:

$$
\boxed{
d^2
=
233\cdot1913.
}
$$

So the squarefree defect:

$$
233\cdot1913
$$

is exactly the product of the two Cayley conjugate walls:

$$
g-c
$$

and:

$$
g+c.
$$

The corresponding ratio is:

$$
\boxed{
\frac{g-c}{g+c}
=
\frac{233}{1913}.
}
$$

Thus the product representation and the ratio representation carry the same squareclass obstruction.

---

# 17. Product defect versus ratio defect

For positive rational:

$$
u,v,
$$

the squareclasses satisfy:

$$
[uv]
=
\left[\frac uv\right],
$$

because:

$$
\frac{uv}{u/v}
=
v^2.
$$

Therefore:

$$
\boxed{
[(g-c)(g+c)]
=
\left[
\frac{g-c}{g+c}
\right].
}
$$

Hence either of the following can encode the same defect:

$$
\boxed{
\kappa_D
=
\operatorname{sf}\big((g-c)(g+c)\big),
}
$$

or:

$$
\boxed{
\omega_c
=
\frac{g-c}{g+c}.
}
$$

The ratio form is often superior because square closure becomes simply:

$$
\boxed{
\omega_c\in\mathbb Q^2.
}
$$

---

# 18. Triple Cayley-root formulation of Perfect Cuboid

Let:

$$
r,s,t\in\mathbb Q_{>0}.
$$

Define:

$$
x(r)
=
\frac{1-r^2}{1+r^2},
$$

$$
y(s)
=
\frac{1-s^2}{1+s^2},
$$

$$
z(t)
=
\frac{1-t^2}{1+t^2}.
$$

Each coordinate automatically has a rational complementary coordinate:

$$
\sqrt{1-x(r)^2}
=
\frac{2r}{1+r^2},
$$

and similarly for:

$$
s,t.
$$

Therefore a nondegenerate Perfect Cuboid exists if and only if there exist positive rational:

$$
r,s,t
$$

such that:

$$
\boxed{
\left(
\frac{1-r^2}{1+r^2}
\right)^2
+
\left(
\frac{1-s^2}{1+s^2}
\right)^2
+
\left(
\frac{1-t^2}{1+t^2}
\right)^2
=
1.
}
$$

This is the symmetric Triple Cayley-Root formulation.

The classical seven-length problem has therefore been reduced to one symmetric rational surface in three half-angle variables.

---

# 19. Seed as a one-coordinate rationality defect on the Cayley surface

The current seed satisfies the Triple Cayley-Root equation over:

$$
\mathbb Q
\left(
\sqrt{\frac{233}{1913}}
\right).
$$

Explicitly:

$$
r=\frac{25}{39},
$$

$$
s=\frac{17}{28},
$$

$$
t=
\sqrt{\frac{233}{1913}}.
$$

The first two coordinates are rational.

Only the third is not.

Therefore the old G2 seed can be reinterpreted as:

$$
\boxed{
\text{a point on the Cayley-root surface with exactly one irrational coordinate}.
}
$$

The repair problem becomes:

> Can the point be moved along the rational-shell surface so that all three Cayley-root coordinates become rational simultaneously?

---

# 20. Möbius complement operator

Define:

$$
\boxed{
\mu(t)
=
\frac{1-t}{1+t}.
}
$$

Then:

$$
\mu(\mu(t))=t.
$$

Thus:

$$
\mu
$$

is an involution.

Define:

$$
\chi(t)
=
\left(
\frac{1-t^2}{1+t^2}
\right)^2.
$$

Direct substitution gives:

$$
\boxed{
\chi(\mu(t))
=
1-\chi(t).
}
$$

Hence Boolean complement duality in square-state space corresponds to a Möbius involution in Cayley-root parameter space.

This is the exact algebraic bridge between the Radical-FDRS root formulation and the previous C2 dual-square formulation.

---

# 21. Recovery of C2

The Triple Cayley-Root equation is:

$$
\chi(r)+\chi(s)+\chi(t)=1.
$$

Therefore:

$$
\chi(r)+\chi(s)
=
1-\chi(t).
$$

Using the complement identity:

$$
1-\chi(t)
=
\chi(\mu(t)),
$$

we obtain:

$$
\boxed{
\chi(r)+\chi(s)
=
\chi(\mu(t)).
}
$$

This is exactly the structural form used in C2:

$$
\boxed{
\chi(r)+\chi(s)=\chi(u).
}
$$

with:

$$
\boxed{
u=\mu(t).
}
$$

Therefore:

$$
\boxed{
\text{C2 dual-square additive closure}
}
$$

and:

$$
\boxed{
\text{R1 Triple Cayley-root closure}
}
$$

are the same global surface written in complementary coordinates.

---

# 22. Interpretation of the C2 parameters

For the seed:

$$
r_0=\frac{25}{39}
$$

is not an arbitrary parameter.

It is:

$$
\boxed{
r_0
=
\sqrt{
\frac{g-a}{g+a}
}.
}
$$

Likewise:

$$
s_0=\frac{17}{28}
$$

is:

$$
\boxed{
s_0
=
\sqrt{
\frac{g-b}{g+b}
}.
}
$$

The missing third rational parameter would be:

$$
\boxed{
t_0
=
\sqrt{
\frac{g-c}{g+c}
}
=
\sqrt{
\frac{233}{1913}
}.
}
$$

Thus the previous C2 state can now be read immediately as:

$$
\boxed{
(r_0,s_0,t_0)
=
(\mathbb Q,\mathbb Q,\mathbb Q(\sqrt{\Delta})\setminus\mathbb Q).
}
$$

---

# 23. Defect splitting across conjugate walls

The first anchored fiber gave:

$$
Q_-(t_0)
=
\frac{2\cdot17\cdot233}{5^2},
$$

$$
Q_+(t_0)
=
\frac{2\cdot17\cdot1913}{5^2}.
$$

The second anchored fiber gave:

$$
Q_-^{(2)}(s_0)
=
\frac{5\cdot13\cdot233}{7^2},
$$

$$
Q_+^{(2)}(s_0)
=
\frac{5\cdot13\cdot1913}{7^2}.
$$

Thus in both flattenings:

- one conjugate wall carries the factor:
  $$
  233;
  $$
- the other carries:
  $$
  1913;
  $$
- slice-dependent common squareclasses cancel in the ratio.

This gives an explicit realization of the R0 idea of defect-factor separation.

The defect:

$$
233\cdot1913
$$

does not merely behave as one atomic label.

Under the conjugate-wall representation it naturally splits into two wall factors.

---

# 24. Canonical versus slice-dependent data

The following are slice-dependent:

$$
t_0=\frac35,
$$

$$
s_0=\frac47,
$$

and the common factors:

$$
2\cdot17,
$$

$$
5\cdot13.
$$

The following are slice-independent:

$$
g-c=233,
$$

$$
g+c=1913,
$$

$$
\omega_c=\frac{233}{1913},
$$

and the squareclass:

$$
233\cdot1913.
$$

Therefore R1 separates:

$$
\boxed{
\text{representation artifacts}
}
$$

from:

$$
\boxed{
\text{canonical arithmetic defect data}.
}
$$

This is an important FDRS requirement: the flattening may change representation, but the reconstructed obstruction must remain invariant.

---

# 25. Three canonical defect ratios

For any rational-shell state:

$$
(x,y,z)\in\mathcal S_{\mathbb Q},
$$

define:

$$
\boxed{
\Omega
=
(\omega_x,\omega_y,\omega_z),
}
$$

where:

$$
\omega_x
=
\frac{1-x}{1+x},
$$

$$
\omega_y
=
\frac{1-y}{1+y},
$$

$$
\omega_z
=
\frac{1-z}{1+z}.
$$

Then:

$$
[F]=[\omega_x],
$$

$$
[E]=[\omega_y],
$$

$$
[D]=[\omega_z].
$$

Perfect Cuboid closure is exactly:

$$
\boxed{
\omega_x,\omega_y,\omega_z
\in
\mathbb Q^2.
}
$$

Thus the three face square gates are simultaneously represented by three one-dimensional ratio gates.

---

# 26. New canonical state descriptor

For the classical rational-edge shell, the six-component squareclass vector contains redundancy.

R1 therefore proposes the reduced state:

$$
\boxed{
\mathfrak C
=
(x,y,z;\omega_x,\omega_y,\omega_z).
}
$$

Because:

$$
x^2+y^2+z^2=1,
$$

the geometry lives on the rational sphere.

The arithmetic closure lives in the squarehood of:

$$
\omega_x,\omega_y,\omega_z.
$$

Alternatively use the root coordinates:

$$
\boxed{
\mathfrak R
=
(r_x,r_y,r_z),
}
$$

where:

$$
r_i^2=\omega_i.
$$

A Perfect Cuboid is exactly a rational point:

$$
\boxed{
(r_x,r_y,r_z)\in\mathbb Q^3
}
$$

on the Triple Cayley-Root surface.

---

# 27. Revised transition problem

The R0 transition question was:

$$
K\longrightarrow K'.
$$

R1 sharpens it.

On the classical shell, the actual move is:

$$
(x,y,z)
\overset{R_{ij}(t)}{\longrightarrow}
(x',y',z'),
$$

which induces:

$$
(\omega_x,\omega_y,\omega_z)
\longrightarrow
(\omega_x',\omega_y',\omega_z').
$$

The exact target is:

$$
\boxed{
(\omega_x',\omega_y',\omega_z')
\in
(\mathbb Q^2)^3.
}
$$

Thus squareclass dynamics should be studied as the image of rational orthogonal moves under the nonlinear Cayley map.

---

# 28. Why this is closer to a Magic-Cube move system

The ambient linear moves commute.

The shell-preserving rational rotations do not generally commute across different coordinate planes.

Thus a sequence:

$$
R_{xy}(u)
R_{xz}(v)
$$

need not equal:

$$
R_{xz}(v)
R_{xy}(u).
$$

The arithmetic state after intermediate moves can therefore depend strongly on move ordering.

This produces a genuine move-word problem:

$$
\boxed{
W
=
R_{i_1j_1}(t_1)
R_{i_2j_2}(t_2)
\cdots
R_{i_nj_n}(t_n).
}
$$

The geometry remains exactly on:

$$
\mathcal S_{\mathbb Q},
$$

while squareclass closure can change.

This is the proper shell-level realization of the earlier Magic-Cube intuition.

---

# 29. R1 main structural result

R1 gives the following chain:

$$
\boxed{
\text{Perfect Cuboid}
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
\text{rational point on }S^2
+
\text{three rational complementary roots}
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
\omega_x,\omega_y,\omega_z
\text{ are all rational squares}
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
r_x,r_y,r_z\in\mathbb Q
}
$$

subject to:

$$
\boxed{
\left(
\frac{1-r_x^2}{1+r_x^2}
\right)^2
+
\left(
\frac{1-r_y^2}{1+r_y^2}
\right)^2
+
\left(
\frac{1-r_z^2}{1+r_z^2}
\right)^2
=
1.
}
$$

This formulation preserves the original geometry but relocates all arithmetic obstruction into the rationality of three one-dimensional Cayley roots.

---

# 30. R2 target

The next phase should not return to blind integer-height search.

R2 should study the **Triple Cayley-Root Surface**:

$$
\boxed{
\mathcal C:
\chi(r)+\chi(s)+\chi(t)=1,
}
$$

where:

$$
\chi(u)
=
\left(
\frac{1-u^2}{1+u^2}
\right)^2.
$$

The principal questions are:

1. Can the surface be reduced birationally to a simpler family?
2. What are its natural fibrations?
3. What happens if two coordinates are rationally parameterized first?
4. Does fixing:
   $$
   r=\frac{25}{39}
   $$
   or:
   $$
   s=\frac{17}{28}
   $$
   produce a tractable curve?
5. Can the irrational coordinate:
   $$
   \sqrt{\frac{233}{1913}}
   $$
   be moved to a rational coordinate through a rational-shell rotation?
6. Is there a rational invariant obstructing simultaneous squarehood of all three Cayley ratios?
7. Can the Möbius complement involution:
   $$
   \mu(t)=\frac{1-t}{1+t}
   $$
   expose a lower-genus representation of the C2 surface?
8. Do the conjugate wall ratios admit a useful descent in:
   $$
   \mathbb Q^\times/\mathbb Q^{\times2}?
   $$

---

# 31. Recommended R2 decomposition

R2 should be divided into three subprograms.

## R2-A — Cayley surface algebra

Study:

$$
\chi(r)+\chi(s)+\chi(t)=1
$$

as an algebraic surface.

Goal:

$$
\boxed{
\text{find rational fibrations / birational reductions}.
}
$$

## R2-B — shell move dynamics

Generate short words in:

$$
R_{xy},
\quad
R_{xz},
\quad
R_{yz}
$$

with rational half-angle parameters.

Track:

$$
\Omega
=
(\omega_x,\omega_y,\omega_z).
$$

Goal:

$$
\boxed{
\text{find exact squareclass-reducing trajectories}.
}
$$

## R2-C — defect-ratio descent

Start from:

$$
\omega_z
=
\frac{233}{1913}.
$$

Study whether conjugate-wall factorizations permit a descent toward:

$$
\omega_z'=q^2.
$$

Goal:

$$
\boxed{
\text{replace product-kernel search by ratio-square descent}.
}
$$

---

# 32. Final conclusion

The main R1 advance is not a new large numerical near-solution.

It is a coordinate collapse.

The old seed:

$$
(448,495,840;1073)
$$

can now be represented as:

$$
\boxed{
\left(
\frac{25}{39},
\frac{17}{28},
\sqrt{\frac{233}{1913}}
\right).
}
$$

The first two coordinates are exactly the previously observed C2 rational parameters.

The third coordinate is exactly the remaining arithmetic defect.

Furthermore:

$$
233=g-c,
$$

$$
1913=g+c.
$$

The squarefree kernel:

$$
233\cdot1913
$$

and the Cayley ratio:

$$
\frac{233}{1913}
$$

are two representations of the same squareclass obstruction.

Two independent FDRS anchored fibers recover the same ratio, while the already closed channels recover:

$$
\left(\frac{25}{39}\right)^2
$$

and:

$$
\left(\frac{17}{28}\right)^2.
$$

Therefore the Radical-FDRS, X-Diagonal coupling, six-channel conservation framework, and C2 dual-square formulation are no longer separate approaches.

They are now linked by the common Cayley-root coordinate system:

$$
\boxed{
x
\longleftrightarrow
\omega(x)
=
\frac{1-x}{1+x}
\longleftrightarrow
r_x
=
\sqrt{\omega(x)}.
}
$$

The next research object is consequently:

$$
\boxed{
\mathcal C:
\chi(r)+\chi(s)+\chi(t)=1
}
$$

together with rational orthogonal move dynamics on its geometric reconstruction.

This becomes the canonical starting point for R2.