> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# Perfect Cuboid Radical-FDRS One-Dimensional Flattening Methodology

## Radical Closure, X-Diagonal Coupling, the Boolean Cube, and Squareclass Defect Dynamics

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R0 — Methodological Foundation  
**Date:** 2026-10-03  
**Status:** Methodology Draft — Not a proof of Perfect Cuboid existence or nonexistence

---

# Abstract

This paper proposes a new research method for Perfect Cuboids: rather than using the three-dimensional cuboid itself as the primary solution space, first flatten it completely into a one-dimensional radical-closure system, then attach the original X-Diagonal, center-axis, shared-edge, six-channel conservation, and arithmetic square-closure structures to the one-dimensional representation.

The core transformation is:

$$
\boxed{
\text{3D Cuboid Geometry}
\longrightarrow
\text{Radical Coupling Algebra}
\longrightarrow
\text{Boolean-Cube State}
\longrightarrow
\text{1D FDRS Tape}
}
$$

The three-dimensional geometric information is not deleted; it is redistributed into coupling maps, subset incidence, complement duality, and X-Diagonal reconstruction data.

This paper defines the radical-composition operation:

$$
\boxed{
x\star y=\sqrt{x^2+y^2}
}
$$

and points out that the square map:

$$
\Phi(x)=x^2
$$

satisfies:

$$
\boxed{
\Phi(x\star y)=\Phi(x)+\Phi(y).
}
$$

Thus the Pythagorean geometry of a three-dimensional Perfect Cuboid can be rewritten exactly as one-dimensional additive flattening.

Furthermore, using the normalized square-states:

$$
A=\frac{a^2}{g^2},
\qquad
B=\frac{b^2}{g^2},
\qquad
C=\frac{c^2}{g^2},
$$

and:

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

one can generate seven nonempty subset states:

$$
A,\quad
B,\quad
C,\quad
A+B,\quad
A+C,\quad
B+C,\quad
A+B+C.
$$

These seven states correspond exactly to the seven nonempty vertices of the Boolean cube on three generators; including the empty-set state gives the complete eight-vertex cube.

The arithmetic conditions of a Perfect Cuboid can further be expressed as square closure of all six active radical channels. Write each rational square-state as:

$$
\rho_i=q_i\sqrt{\kappa_i},
$$

where:

$$
q_i\in\mathbb Q_{>0},
$$

And:

$$
\kappa_i
$$

is a positive squarefree kernel. The terminal state of a Perfect Cuboid is:

$$
\boxed{
\kappa_1=
\kappa_2=
\kappa_3=
\kappa_4=
\kappa_5=
\kappa_6=1.
}
$$

The original problem can therefore be restated as:

> While preserving Boolean-cube coupling, X-Diagonal shared-edge closure, and global conservation, does an admissible move trajectory exist along which the radical squareclass state converges from a defective state to trivial squareclasses in all six channels?

The current seed:

$$
(a,b,c;g)=(448,495,840;1073)
$$

has, in this framework:

$$
\boxed{
K_{\rm seed}
=
(1,1,1,233\cdot1913,1,1),
}
$$

It is therefore a natural test state with a single squareclass defect.

---

# 1. Research Objective

The objective of this methodology is not to prove the Perfect Cuboid problem directly, but to change the primary representation.

Traditional methods directly study:

$$
a,b,c,d,e,f,g.
$$

This method instead transforms successively to:

$$
\boxed{
\text{Lengths}
\rightarrow
\text{Squares}
\rightarrow
\text{Radicals}
\rightarrow
\text{Subset Couplings}
\rightarrow
\text{1D State Word}
\rightarrow
\text{Squareclass Dynamics}.
}
$$

The research focus therefore changes from:

$$
\text{finding seven integers}
$$

to:

$$
\boxed{
\text{finding a radical-closure state in all six channels}.
}
$$

This accords with the original spirit of FDRS: not to eliminate the problem's information, but to redistribute high-dimensional geometric complexity into a low-dimensional representation and a reconstruction map.

---

# 2. Classical Perfect Cuboid system

The classical Perfect Cuboid equations are:

$$
a^2+b^2=d^2,
$$

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

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

and:

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

A Perfect Cuboid requires:

$$
a,b,c,d,e,f,g\in\mathbb Q_{>0},
$$

and can be converted into an integer witness by scaling with a common factor.

The rational formulation therefore suffices to carry the existence problem.

---

# 3. Normalize the global whole

Use the body diagonal:

$$
g
$$

as the global scale, and define:

$$
x_a=\frac ag,
\qquad
x_b=\frac bg,
\qquad
x_c=\frac cg.
$$

Then let:

$$
A=x_a^2,
\qquad
B=x_b^2,
\qquad
C=x_c^2.
$$

then:

$$
\boxed{
A+B+C=1.
}
$$

The three face-diagonal square-states are:

$$
D=A+B,
$$

$$
E=A+C,
$$

$$
F=B+C.
$$

The complementary dualities are:

$$
A+F=1,
$$

$$
B+E=1,
$$

$$
C+D=1.
$$

Here:

$$
(A,B,C)
$$

are the lowest-dimensional conservation coordinates;

$$
(D,E,F)
$$

are pair states derived from coupling.

---

# 4. Radical coupling algebra

Define:

$$
\boxed{
x\star y=\sqrt{x^2+y^2},
\qquad
x,y\in\mathbb R_{\ge 0}.
}
$$

then:

$$
x\star y=y\star x,
$$

and:

$$
(x\star y)\star z
=
x\star(y\star z)
=
\sqrt{x^2+y^2+z^2}.
$$

Therefore:

$$
(\mathbb R_{\ge0},\star)
$$

is a commutative monoid with identity element:

$$
0.
$$

Let:

$$
\Phi(x)=x^2.
$$

then:

$$
\Phi(x\star y)
=
x^2+y^2
=
\Phi(x)+\Phi(y).
$$

Thus:

$$
\boxed{
\Phi:
(\mathbb R_{\ge0},\star)
\longrightarrow
(\mathbb R_{\ge0},+)
}
$$

flattens radical coupling exactly into ordinary addition.

This is the core of the method's one-dimensional reduction.

---

# 5. Seven-Radical Structure

Define:

$$
\rho_1=\sqrt A,
$$

$$
\rho_2=\sqrt B,
$$

$$
\rho_3=\sqrt C,
$$

$$
\rho_4=\sqrt{A+B},
$$

$$
\rho_5=\sqrt{A+C},
$$

$$
\rho_6=\sqrt{B+C},
$$

and:

$$
\rho_7=\sqrt{A+B+C}.
$$

From normalization:

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

we obtain:

$$
\boxed{
\rho_7=1.
}
$$

The first six radical channels satisfy:

$$
\rho_4=\rho_1\star\rho_2,
$$

$$
\rho_5=\rho_1\star\rho_3,
$$

$$
\rho_6=\rho_2\star\rho_3.
$$

and also:

$$
\rho_1\star\rho_6=1,
$$

$$
\rho_2\star\rho_5=1,
$$

$$
\rho_3\star\rho_4=1.
$$

The six active radical channels therefore naturally form three complementary dual pairs:

$$
(\rho_1,\rho_6),
$$

$$
(\rho_2,\rho_5),
$$

$$
(\rho_3,\rho_4).
$$

---

# 6. Boolean Cube formulation

Let the basic index set be:

$$
U=\{1,2,3\}.
$$

Specify:

$$
w_1=A,
\qquad
w_2=B,
\qquad
w_3=C.
$$

For any:

$$
S\subseteq U
$$

define the subset square-state:

$$
Q(S)
=
\sum_{i\in S}w_i.
$$

and define:

$$
Q(\varnothing)=0.
$$

Then the eight Boolean vertices are:

$$
0,
$$

$$
A,\quad B,\quad C,
$$

$$
A+B,\quad A+C,\quad B+C,
$$

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

Next define the radical lift:

$$
R(S)=\sqrt{Q(S)}.
$$

then:

$$
R(\{1\})=\rho_1,
$$

$$
R(\{2\})=\rho_2,
$$

$$
R(\{3\})=\rho_3,
$$

$$
R(\{1,2\})=\rho_4,
$$

$$
R(\{1,3\})=\rho_5,
$$

$$
R(\{2,3\})=\rho_6,
$$

$$
R(U)=1.
$$

This gives a complete Boolean-cube encoding of the Perfect Cuboid problem.

---

# 7. Complement involution and six-channel duality

The Boolean cube has a natural complement involution:

$$
S\mapsto S^c.
$$

From:

$$
Q(S)+Q(S^c)=1
$$

we obtain:

$$
\boxed{
R(S)^2+R(S^c)^2=1.
}
$$

Thus the three dualities in the original six-channel conservation:

$$
A+F=1,
$$

$$
B+E=1,
$$

$$
C+D=1
$$

are not additional rules, but direct consequences of Boolean complement.

This means:

$$
\boxed{
\text{Six-Channel Conservation}
}
$$

can be reinterpreted as:

$$
\boxed{
\text{Boolean Complement Duality}.
}
$$

---

# 8. FDRS one-dimensional tape

Now flatten the Boolean cube into a one-dimensional tape:

$$
\boxed{
[0]
\Vert
[A][B][C]
\Vert
[A+B][A+C][B+C]
\Vert
[1].
}
$$

The corresponding radical tape is:

$$
\boxed{
[0]
\Vert
[\rho_1][\rho_2][\rho_3]
\Vert
[\rho_4][\rho_5][\rho_6]
\Vert
[1].
}
$$

This one-dimensional representation does not itself preserve geometric adjacency order, so coupling metadata must additionally be retained.

Define:

$$
\mathcal G_B
=
(V,E)
$$

as the Boolean incidence graph, where:

$$
V=\mathcal P(U),
$$

an edge exists whenever two subsets differ by only one basic element.

The canonical FDRS state is therefore not a tape alone, but:

$$
\boxed{
W_{\rm FDRS}
=
(T,\mathcal G_B,\mathcal C,\mathcal X).
}
$$

where:

$$
T
$$

is the one-dimensional numerical tape;

$$
\mathcal G_B
$$

preserves subset adjacency;

$$
\mathcal C
$$

preserves complement duality;

$$
\mathcal X
$$

preserves X-Diagonal / shared-edge coupling data.

This meets the FDRS information-conservation requirement: space is removed, but reconstruction information is retained.

---

# 9. Preserving X-Diagonal Coupling

This branch does not replace the earlier X-Diagonal / Center-Axis method; it flattens that method into a one-dimensional reconstruction layer.

Define three face-X fields:

$$
X_{ab},
\qquad
X_{ac},
\qquad
X_{bc}.
$$

For:

$$
X_{ab}
$$

we have:

$$
\rho_1
=
\rho_4
\sin\frac{\theta_{ab}}2,
$$

$$
\rho_2
=
\rho_4
\cos\frac{\theta_{ab}}2.
$$

For:

$$
X_{ac}
$$

we have:

$$
\rho_1
=
\rho_5
\sin\frac{\theta_{ac}}2,
$$

$$
\rho_3
=
\rho_5
\cos\frac{\theta_{ac}}2.
$$

For:

$$
X_{bc}
$$

we have:

$$
\rho_2
=
\rho_6
\sin\frac{\theta_{bc}}2,
$$

$$
\rho_3
=
\rho_6
\cos\frac{\theta_{bc}}2.
$$

X-Diagonal geometry therefore remains after the one-dimensional reduction. It no longer exists as drawn rectangular faces, but as:

$$
\boxed{
\text{Diagonal Scale}
+
\text{Opening Angle}
+
\text{Projection Ratio}
}
$$

in that form.

Shared-edge closure requires the same edge to agree after projection from different face-X fields:

$$
\rho_1^{(ab)}
=
\rho_1^{(ac)},
$$

$$
\rho_2^{(ab)}
=
\rho_2^{(bc)},
$$

$$
\rho_3^{(ac)}
=
\rho_3^{(bc)}.
$$

Therefore:

$$
\mathcal X
$$

is part of the FDRS reconstruction map, not geometric data discarded by dimensional reduction.

---

# 10. Radical normal form

For any positive rational square-state:

$$
Q_i\in\mathbb Q_{>0},
$$

there exist:

$$
q_i\in\mathbb Q_{>0}
$$

and a unique positive squarefree integer:

$$
\kappa_i
$$

such that:

$$
\boxed{
Q_i=q_i^2\kappa_i.
}
$$

Therefore:

$$
\boxed{
\rho_i=q_i\sqrt{\kappa_i}.
}
$$

where:

$$
\kappa_i
$$

represents the channel's rational squareclass.

If:

$$
\kappa_i=1,
$$

then:

$$
\rho_i\in\mathbb Q.
$$

If:

$$
\kappa_i\ne1,
$$

the channel has not yet attained rational-square closure.

Therefore define the six-channel squareclass state:

$$
\boxed{
K
=
(\kappa_1,\kappa_2,\kappa_3,\kappa_4,\kappa_5,\kappa_6).
}
$$

The Perfect Cuboid target is:

$$
\boxed{
K_{\rm goal}
=
(1,1,1,1,1,1).
}
$$

---

# 11. Canonical Interpretation of Radicals One through Six

Initially, one may establish a raw bounded radical alphabet:

$$
\mathcal R_6^{\rm raw}
=
\{
\sqrt1,
\sqrt2,
\sqrt3,
\sqrt4,
\sqrt5,
\sqrt6
\}.
$$

However:

$$
\sqrt4=2
$$

belongs to the trivial squareclass.

After canonical squarefree reduction:

$$
\boxed{
\mathcal K_6
=
\{1,2,3,5,6\}.
}
$$

The first-stage experiment should not restrict:

$$
\rho_i
$$

to only these fixed numerical values. Instead write:

$$
\rho_i
=
q_i\sqrt{\kappa_i},
$$

where:

$$
q_i\in\mathbb Q_{>0}
$$

and initially restrict:

$$
\kappa_i\in\mathcal K_6.
$$

That is:

> The search concerns radical types, not fixed radical magnitudes.

This avoids mistakenly deleting the continuous rational scale.

---

# 12. One-Dimensional Radical-Coupling Equations

Write:

$$
\rho_i=q_i\sqrt{\kappa_i}.
$$

Then the three face-coupling conditions are:

$$
\boxed{
q_4^2\kappa_4
=
q_1^2\kappa_1
+
q_2^2\kappa_2,
}
$$

$$
\boxed{
q_5^2\kappa_5
=
q_1^2\kappa_1
+
q_3^2\kappa_3,
}
$$

$$
\boxed{
q_6^2\kappa_6
=
q_2^2\kappa_2
+
q_3^2\kappa_3.
}
$$

The three complement conditions are:

$$
\boxed{
q_1^2\kappa_1
+
q_6^2\kappa_6
=
1,
}
$$

$$
\boxed{
q_2^2\kappa_2
+
q_5^2\kappa_5
=
1,
}
$$

$$
\boxed{
q_3^2\kappa_3
+
q_4^2\kappa_4
=
1.
}
$$

The Perfect Cuboid problem is therefore flattened into a completely one-dimensional system of radical-coupling equations.

Three-dimensional coordinates are no longer necessary solution objects.

---

# 13. Magic-Cube state

Define the arithmetic state:

$$
\Sigma_i
=
\begin{cases}
1,&\kappa_i=1,\\
0,&\kappa_i\ne1.
\end{cases}
$$

then:

$$
\Sigma
=
(\Sigma_1,\Sigma_2,\Sigma_3,\Sigma_4,\Sigma_5,\Sigma_6).
$$

The Perfect Cuboid target:

$$
\boxed{
\Sigma_{\rm goal}
=
(1,1,1,1,1,1).
}
$$

But this branch retains more information than a binary state, so also track:

$$
K
=
(\kappa_1,\ldots,\kappa_6).
$$

In this way:

$$
\Sigma
$$

answers which cell has not closed;

And:

$$
K
$$

answers which kind of arithmetic defect that cell carries.

This is the new Magic-Cube state.

---

# 14. Defect support

Define defect support:

$$
\boxed{
\mathcal D(K)
=
\{
i:
\kappa_i\ne1
\}.
}
$$

and define the defect count:

$$
\boxed{
h(K)=|\mathcal D(K)|.
}
$$

The Perfect Cuboid target is:

$$
h(K)=0.
$$

The current G2 seed has:

$$
h(K)=1.
$$

Note:

$$
h(K)
$$

is only a discrete diagnostic and should not be mistaken for a continuous distance or a proof quantity.

What genuinely matters for research is:

$$
\boxed{
\text{whether defects can move, split, merge, or disappear under admissible moves}.
}
$$

---

# 15. Current seed

Use:

$$
(a,b,c;g)
=
(448,495,840;1073).
$$

then:

$$
\rho_1
=
\frac{448}{1073},
$$

$$
\rho_2
=
\frac{495}{1073},
$$

$$
\rho_3
=
\frac{840}{1073}.
$$

We also have:

$$
\rho_5
=
\frac{952}{1073},
$$

$$
\rho_6
=
\frac{975}{1073}.
$$

However:

$$
\rho_4
=
\frac{\sqrt{445729}}{1073}.
$$

and:

$$
445729
=
233\cdot1913.
$$

Therefore:

$$
\boxed{
\rho_4
=
\frac{1}{1073}
\sqrt{233\cdot1913}.
}
$$

Its squareclass state is:

$$
\boxed{
K_{\rm seed}
=
(1,1,1,233\cdot1913,1,1).
}
$$

Its binary closure state is:

$$
\boxed{
\Sigma_{\rm seed}
=
(1,1,1,0,1,1).
}
$$

Thus this seed is a natural one-defect Magic-Cube state.

---

# 16. Conservation moves

In the square-state base, define a conservation move:

$$
(A,B,C)
\mapsto
(A',B',C')
$$

where:

$$
A'=A+\delta_A,
$$

$$
B'=B+\delta_B,
$$

$$
C'=C+\delta_C,
$$

and require:

$$
\boxed{
\delta_A+\delta_B+\delta_C=0.
}
$$

Also require:

$$
A',B',C'>0.
$$

If:

$$
A',B',C'\in\mathbb Q,
$$

all subset square-states remain rational.

Recompute:

$$
D'=A'+B',
$$

$$
E'=A'+C',
$$

$$
F'=B'+C'.
$$

Geometric conservation is automatically preserved:

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

But the squareclass state:

$$
K
$$

can change.

The research question for moves is therefore:

$$
\boxed{
K\longrightarrow K'.
}
$$

rather than:

$$
\boxed{
\text{geometry true}
\longrightarrow
\text{geometry false}.
}
$$

---

# 17. X-preserving admissible moves

If one operates directly on:

$$
(A,B,C)
$$

X-Diagonal geometry can be reconstructed automatically from the new state.

If one operates directly on the three X-fields, shared-edge closure must additionally be required.

Therefore define an X-preserving move:

$$
\mathcal M_X
$$

which must simultaneously satisfy:

$$
A'+B'+C'=1,
$$

the three face-X Pythagorean couplings,

and the three shared-edge equalities.

This makes:

$$
\mathcal M_X
$$

a one-dimensional move genuinely reconstructible in reverse as cuboid geometry.

---

# 18. Relation to the C2 two-turn branch

C2 defines the dual-square set:

$$
\mathcal P
=
\left\{
X\in\mathbb Q\cap(0,1):
X\in\mathbb Q^2,
\quad
1-X\in\mathbb Q^2
\right\}.
$$

and compresses the Perfect Cuboid problem to:

$$
\exists X,Y,Z\in\mathcal P
$$

such that:

$$
X+Y=Z.
$$

Within Radical-FDRS, C2 can be viewed as a particular constrained sector.

For example, choose:

$$
A\in\mathcal P,
$$

$$
B\in\mathcal P,
$$

then:

$$
A,\quad1-A,
$$

and:

$$
B,\quad1-B
$$

are already forced to be square-closed.

Whether the remaining output pair:

$$
A+B,
$$

$$
1-A-B
$$

is simultaneously square-closed is C2's two-gate problem.

Therefore:

$$
\boxed{
\text{C2}
\subset
\text{Radical-FDRS State Space}.
}
$$

Radical-FDRS does not replace C2; it provides a broader state-space interpretation and a language of defect dynamics.

---

# 19. Relation to Infinite-Radical Fixed-Point Thinking

A typical infinite radical:

$$
x
=
\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}
$$

can be rewritten through self-consistency as:

$$
x=\sqrt{2+x}.
$$

and squared to give:

$$
x^2=2+x.
$$

Its core is not the infinite number of layers itself, but:

$$
\boxed{
\text{recursive expression}
\longrightarrow
\text{algebraic closure condition}.
}
$$

The Radical-FDRS version of the Perfect Cuboid problem follows the same methodological spirit, but is not a single-variable fixed point.

Define the closure map:

$$
\mathcal R(x,y,z)
=
(
x,
y,
z,
x\star y,
x\star z,
y\star z,
x\star y\star z
).
$$

Under normalization:

$$
x\star y\star z=1
$$

the Perfect Cuboid problem becomes:

$$
\boxed{
\exists
x,y,z\in\mathbb Q_{>0}
:
\mathcal R(x,y,z)\in\mathbb Q^7.
}
$$

This is therefore a simultaneous radical-closure problem.

---

# 20. Phase R0 experiment: Root-1-to-6 sector

The first stage does not perform a large integer scan.

First restrict the squareclass alphabet:

$$
\kappa_i
\in
\{1,2,3,5,6\}.
$$

Study the equations:

$$
q_4^2\kappa_4
=
q_1^2\kappa_1
+
q_2^2\kappa_2,
$$

$$
q_5^2\kappa_5
=
q_1^2\kappa_1
+
q_3^2\kappa_3,
$$

$$
q_6^2\kappa_6
=
q_2^2\kappa_2
+
q_3^2\kappa_3,
$$

and add normalization:

$$
q_1^2\kappa_1
+
q_2^2\kappa_2
+
q_3^2\kappa_3
=
1.
$$

The goal is not to find a Perfect Cuboid directly, but to answer:

1. Which squareclass tuples can exist?
2. Which tuples are immediately excluded by local algebraic conditions?
3. Can a defect move from one slot to another?
4. Does a defect split into several squareclasses during a move?
5. Is there a rational trajectory from a nontrivial tuple back to:

$$
(1,1,1,1,1,1)
$$

as displayed above?

This is the minimal computable Radical-FDRS sandbox.

---

# 21. Phase R1: Squareclass transition graph

Let each feasible squareclass tuple:

$$
K
$$

be a node.

If an admissible rational conservation move exists:

$$
\mathcal M
$$

such that:

$$
K\mapsto K',
$$

establish a directed edge:

$$
K\to K'.
$$

forming:

$$
\boxed{
\mathfrak G_{\rm SC}
}
$$

the Squareclass Transition Graph.

Research objectives include:

- whether the goal state has an incoming path;
- whether one-defect states form a closed component;
- whether some prime kernels cannot leave particular channels;
- whether a defect-annihilation cycle exists;
- whether the wall structure observed in C2 corresponds to chamber boundaries of the transition graph.

---

# 22. Phase R2: Prime-support expansion

After completing the Root-1-to-6 sector, expand prime support rather than immediately increasing integer height.

Study successively:

$$
\{2,3,5\},
$$

then add:

$$
7,
$$

then add:

$$
11,
$$

and expand the squarefree-kernel basis in this way.

Each:

$$
\kappa_i
$$

can be represented by a prime-parity vector.

If:

$$
\kappa_i
=
\prod_p p^{\epsilon_{i,p}},
$$

then:

$$
\epsilon_{i,p}\in\{0,1\}.
$$

This provides a discrete encoding of squareclass defects.

But note:

> Additive equations cannot simply be linearized on prime-parity vectors.

Prime parity is therefore used only for classification and local obstructions; it cannot replace the original rational equations.

---

# 23. Phase R3: X-Diagonal reconstruction test

Any one-dimensional candidate state must be reconstructed as:

$$
X_{ab},
\quad
X_{ac},
\quad
X_{bc}.
$$

Verify:

$$
\rho_1^{(ab)}
=
\rho_1^{(ac)},
$$

$$
\rho_2^{(ab)}
=
\rho_2^{(bc)},
$$

$$
\rho_3^{(ac)}
=
\rho_3^{(bc)}.
$$

If this fails, the one-dimensional state is not an admissible cuboid reconstruction.

The FDRS branch must therefore always maintain:

$$
\boxed{
\text{Flatten}
+
\text{Reconstruct}
+
\text{Exact Verify}.
}
$$

---

# 24. Phase R4: C2 gate translation

Translate C2's:

$$
\Delta_D(r,s)
$$

and:

$$
\Delta_C(r,s)
$$

into Radical-FDRS squareclass transitions.

Study:

$$
[\Delta_D],
$$

$$
[\Delta_C]
$$

map to:

$$
K.
$$

In particular, track whether the G2 seed's:

$$
233\cdot1913
$$

maintains the same defect identity under different parameterizations.

If the same kernel recurs in:

- face diagonal；
- six-channel conservation；
- C2 magic-sum gate；
- Radical-FDRS squareclass state；

one can further study whether it is only a seed-specific signature or belongs to a stronger orbit obstruction.

Do not call it an invariant before it has been proved to be one.

---

# 25. Phase R5: One-defect closure program

The greatest advantage of the G2 seed is:

$$
h(K)=1.
$$

One can therefore specifically study the one-defect manifold:

$$
\mathcal M_1
=
\{
K:
h(K)=1
\}.
$$

The question becomes:

$$
\boxed{
\exists
K\in\mathcal M_1
\text{ with an admissible path to }
K_{\rm goal}?
}
$$

If such a path exists, it may offer a constructive route.

If all one-defect sectors are blocked by exact obstructions, this would also provide new nonexistence evidence.

---

# 26. Methodological conjectures

This paper proposes the following research statements, but does not yet treat them as theorems.

## Conjecture R-A — Defect Transport

On the global geometric conservation manifold, a squareclass defect need not remain in its original channel; it may move under admissible rational moves:

$$
\kappa_i
\longrightarrow
\kappa_j.
$$

## Conjecture R-B — Defect Fusion

Multiple squareclass defects may merge through coupling, reducing defect support:

$$
h(K')
<
h(K).
$$

## Conjecture R-C — One-Defect Reachability

Whether the current seed:

$$
(1,1,1,233\cdot1913,1,1)
$$

can reach:

$$
(1,1,1,1,1,1)
$$

through a finite admissible move sequence is this branch's most direct computational question.

## Conjecture R-D — Representation Advantage

The principal difficulty of the Perfect Cuboid problem may lie not in geometric closure, but in rational squareclass closure.

If Radical-FDRS can clearly separate geometric degrees of freedom from arithmetic squareclasses, it may be better suited than direct integer-height search to identifying genuine obstructions.

---

# 27. Proof-safety rules

This branch must follow these rules.

First, FDRS is a representation transform, not an automatic prover.

Second, one-dimensional flattening must retain reconstruction data.

Third, floating-point closeness is not square closure.

Fourth:

$$
\kappa_i=1
$$

must be verified by exact rational factorization or an exact square test.

Fifth, any candidate terminal state must be reverified against:

$$
a^2+b^2=d^2,
$$

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

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

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

Sixth, a rational witness must also undergo common-denominator clearing and primitive integer normalization.

Seventh, no observed kernel recurrence is called an invariant before orbit independence has been proved.

---

# 28. Canonical workflow

The canonical workflow of this branch is defined as:

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

$$
\downarrow
$$

$$
\boxed{
\text{Normalize by }g
}
$$

$$
\downarrow
$$

$$
\boxed{
(A,B,C)
}
$$

$$
\downarrow
$$

$$
\boxed{
\text{Boolean Subset Expansion}
}
$$

$$
\downarrow
$$

$$
\boxed{
\text{Radical Lift}
}
$$

$$
\downarrow
$$

$$
\boxed{
\rho_i=q_i\sqrt{\kappa_i}
}
$$

$$
\downarrow
$$

$$
\boxed{
\text{1D FDRS Tape}
}
$$

$$
\downarrow
$$

$$
\boxed{
\text{X-Diagonal Coupling Reconstruction}
}
$$

$$
\downarrow
$$

$$
\boxed{
\text{Squareclass Transition Analysis}
}
$$

$$
\downarrow
$$

$$
\boxed{
K\rightarrow(1,1,1,1,1,1)
}
$$

$$
\downarrow
$$

$$
\boxed{
\text{Exact Rational Verification}
}
$$

$$
\downarrow
$$

$$
\boxed{
\text{Primitive Integer Witness}
}
$$

---

# 29. Central Equivalence Statement

For a normalized state:

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

and:

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

Define:

$$
K(A,B,C)
=
(
\kappa(A),
\kappa(B),
\kappa(C),
\kappa(A+B),
\kappa(A+C),
\kappa(B+C)
).
$$

where:

$$
\kappa(X)
$$

is the canonical squarefree kernel of the positive rational number:

$$
X
$$

as displayed above.

Then Perfect Cuboid existence is equivalent to the existence of:

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

such that:

$$
A+B+C=1
$$

and:

$$
\boxed{
K(A,B,C)
=
(1,1,1,1,1,1).
}
$$

This is the canonical existence statement of the Radical-FDRS branch.

---

# 30. Conclusion

A Perfect Cuboid need no longer be viewed first as a three-dimensional cuboid.

In the Radical-FDRS representation, it is first a Boolean cube generated by three base square-states.

The geometric layer is:

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

The coupling layer is:

$$
A+B,
\quad
A+C,
\quad
B+C.
$$

The radical layer is:

$$
\rho_i=q_i\sqrt{\kappa_i}.
$$

The arithmetic goal is:

$$
\kappa_i=1.
$$

The X-Diagonal layer preserves the reconstruction geometry of:

$$
\text{face diagonal}
+
\text{opening angle}
+
\text{shared-edge projection}
$$

as displayed above.

The complete problem is therefore rewritten as:

$$
\boxed{
\text{3D Geometry}
\rightarrow
\text{1D Radical State}
\rightarrow
\text{Squareclass Dynamics}
\rightarrow
\text{Exact Reconstruction}.
}
$$

The currently most important seed:

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

becomes, in this representation:

$$
\boxed{
(1,1,1,233\cdot1913,1,1).
}
$$

This makes the next-stage question very clear:

> Rather than first asking where the next larger integer is, ask whether a single squareclass defect can be moved, fused, and ultimately eliminated while preserving overall coupling and conservation.

This is the starting point of the Radical-FDRS branch.

---

# Related Internal Methodologies

1. *Perfect_Cuboid_X_Diagonal_Center_Axis_Coupling_Methodology.md*
2. *Perfect_Cuboid_Six_Channel_Conservation_Duality_Methodology.md*
3. *Perfect_Cuboid_C2_Two_Turn_Square_Preserving_Repair.md*
4. *展平式維度重構理論：從FCSR到FDRS的完整數學架構.md*
5. *展平式維度重構理論作為通用問題解決框架：機會與風險的平衡分析.md*