# Perfect Cuboid Radical-FDRS R2

## Twin-Psi Square System, Fixed-Fiber Genus-5 Reconstruction, and the Unification of the $7/32$ and $11/45$ Slices

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R2 — Twin-Psi Reduction  
**Date:** 2026-10-03  
**Status:** Exact structural reformulation — not a proof of Perfect Cuboid existence or nonexistence

---

# 0. Abstract

R1 rewrote the remaining Perfect Cuboid arithmetic obstruction in Cayley-root coordinates.

R2 performs a second reduction.

Instead of starting from three Cayley roots simultaneously, choose one rational face diagonal and the rational body diagonal as an already closed nested Pythagorean pair.

This allows the normalized cuboid to be parameterized by only two rational half-angle variables:

$$
m,n\in\mathbb Q.
$$

The first parameter:

$$
m
$$

controls the decomposition of a chosen rational face diagonal into its two edges.

The second parameter:

$$
n
$$

controls the decomposition of the rational body diagonal into that face diagonal and the remaining edge.

After these two closures are built in, only two face-diagonal square conditions remain.

Both conditions are governed by the same rational function:

$$
\boxed{
\Psi(x,n)
=
\frac{x^2+n^2}{1+x^2n^2}.
}
$$

The second input is not independent.

It is obtained from the first by the Möbius involution:

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

Therefore Perfect Cuboid existence is equivalent to finding nondegenerate rational:

$$
m,n
$$

such that:

$$
\boxed{
\Psi(m,n)\in\mathbb Q^2,
}
$$

and:

$$
\boxed{
\Psi(\mu(m),n)\in\mathbb Q^2.
}
$$

This is the **Twin-Psi Square System**.

The current G2 seed produces two rational charts.

The first is:

$$
\boxed{
(m,n)
=
\left(
\frac35,
\frac{11}{45}
\right),
}
$$

and the second is:

$$
\boxed{
(m,n)
=
\left(
\frac47,
\frac7{32}
\right).
}
$$

These are exactly the two one-axis parameters previously found independently.

For the second chart:

$$
\Psi\left(\frac47,\frac7{32}\right)
=
\left(\frac{17}{28}\right)^2,
$$

while:

$$
\Psi\left(\frac3{11},\frac7{32}\right)
=
\frac{233}{1913}.
$$

For the first chart:

$$
\Psi\left(\frac35,\frac{11}{45}\right)
=
\left(\frac{25}{39}\right)^2,
$$

while:

$$
\Psi\left(\frac14,\frac{11}{45}\right)
=
\frac{233}{1913}.
$$

Thus both historical fixed slices are two coordinate charts of the same two-dimensional arithmetic surface.

Moreover, for fixed:

$$
n,
$$

each Twin-Psi square condition becomes a quartic double cover of the:

$$
m
$$

line.

Generically the simultaneous fiber product has genus:

$$
\boxed{
5.
}
$$

The three nontrivial quadratic subcovers have genera:

$$
1,
\qquad
1,
\qquad
3.
$$

Therefore the previously encountered genus-$5$ and genus-$3$ structures arise naturally from the Twin-Psi system.

The former high-genus program is consequently reinterpreted as fixed-fiber arithmetic on a simpler two-dimensional rational parameter space.

---

# 1. Starting point

Let normalized positive cuboid edges be:

$$
a,b,c\in\mathbb Q_{>0}
$$

with body diagonal normalized to:

$$
g=1.
$$

Then:

$$
a^2+b^2+c^2=1.
$$

Suppose one face diagonal:

$$
d
$$

is already rational and satisfies:

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

Since:

$$
d^2+c^2=1,
$$

we have two nested rational right triangles:

$$
(a,b,d),
$$

and:

$$
(d,c,1).
$$

R2 takes these two nested right triangles as the primary FDRS reconstruction skeleton.

---

# 2. Outer rational Pythagorean parameter

Parameterize:

$$
d^2+c^2=1
$$

by:

$$
n\in\mathbb Q.
$$

Choose:

$$
\boxed{
d
=
\frac{1-n^2}{1+n^2},
}
$$

and:

$$
\boxed{
c
=
\frac{2n}{1+n^2}.
}
$$

For the positive nondegenerate branch one may take:

$$
0<n<1.
$$

Then:

$$
d,c\in\mathbb Q_{>0},
$$

and:

$$
d^2+c^2=1
$$

holds identically.

Thus body-diagonal closure is built into the coordinate system.

---

# 3. Inner rational Pythagorean parameter

Now parameterize:

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

using:

$$
m\in\mathbb Q.
$$

Choose:

$$
\boxed{
a
=
d
\frac{1-m^2}{1+m^2},
}
$$

and:

$$
\boxed{
b
=
d
\frac{2m}{1+m^2}.
}
$$

Again for the positive branch take:

$$
0<m<1.
$$

Then:

$$
a,b\in\mathbb Q_{>0},
$$

and:

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

holds identically.

Therefore:

$$
a^2+b^2+c^2=1
$$

also holds identically.

At this stage:

- all three edges are rational;
- one face diagonal is rational;
- the body diagonal is rational.

Only the other two face diagonals remain to be tested.

---

# 4. Remaining square gates

The remaining face diagonals satisfy:

$$
e^2=a^2+c^2=1-b^2,
$$

and:

$$
f^2=b^2+c^2=1-a^2.
$$

Therefore the remaining closure conditions are:

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

and:

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

Using the R1 Cayley criterion:

$$
1-x^2\in\mathbb Q^2
\iff
\frac{1-x}{1+x}\in\mathbb Q^2,
$$

it is sufficient and necessary to study:

$$
\omega(a)
=
\frac{1-a}{1+a},
$$

and:

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

---

# 5. First exact collapse

Substitute:

$$
a
=
\frac{1-n^2}{1+n^2}
\frac{1-m^2}{1+m^2}.
$$

Then:

$$
1-a
=
\frac{
(1+n^2)(1+m^2)
-
(1-n^2)(1-m^2)
}{
(1+n^2)(1+m^2)
}.
$$

The numerator simplifies to:

$$
2(m^2+n^2).
$$

Similarly:

$$
1+a
=
\frac{
2(1+m^2n^2)
}{
(1+n^2)(1+m^2)
}.
$$

Therefore:

$$
\boxed{
\omega(a)
=
\frac{m^2+n^2}{1+m^2n^2}.
}
$$

Define:

$$
\boxed{
\Psi(x,n)
=
\frac{x^2+n^2}{1+x^2n^2}.
}
$$

Then:

$$
\boxed{
\omega(a)
=
\Psi(m,n).
}
$$

Thus:

$$
\boxed{
f\in\mathbb Q
\iff
\Psi(m,n)\in\mathbb Q^2.
}
$$

---

# 6. Second exact collapse

Now:

$$
b
=
\frac{1-n^2}{1+n^2}
\frac{2m}{1+m^2}.
$$

Then:

$$
1-b
=
\frac{
(1+n^2)(1+m^2)
-
2m(1-n^2)
}{
(1+n^2)(1+m^2)
},
$$

and:

$$
1+b
=
\frac{
(1+n^2)(1+m^2)
+
2m(1-n^2)
}{
(1+n^2)(1+m^2)
}.
$$

The two numerators can be written as:

$$
(1-m)^2+n^2(1+m)^2,
$$

and:

$$
(1+m)^2+n^2(1-m)^2.
$$

Hence:

$$
\omega(b)
=
\frac{
(1-m)^2+n^2(1+m)^2
}{
(1+m)^2+n^2(1-m)^2
}.
$$

Define the Möbius involution:

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

Since:

$$
\mu(m)^2
=
\frac{(1-m)^2}{(1+m)^2},
$$

we obtain:

$$
\boxed{
\omega(b)
=
\Psi(\mu(m),n).
}
$$

Therefore:

$$
\boxed{
e\in\mathbb Q
\iff
\Psi(\mu(m),n)\in\mathbb Q^2.
}
$$

---

# 7. Twin-Psi theorem

We therefore obtain the central R2 theorem.

## Theorem — Twin-Psi formulation

A nondegenerate rational Perfect Cuboid exists if and only if there exist:

$$
m,n\in\mathbb Q,
$$

with an admissible positive branch such as:

$$
0<m,n<1,
$$

such that:

$$
\boxed{
\Psi(m,n)\in\mathbb Q^2,
}
$$

and:

$$
\boxed{
\Psi(\mu(m),n)\in\mathbb Q^2,
}
$$

where:

$$
\Psi(x,n)
=
\frac{x^2+n^2}{1+x^2n^2},
$$

and:

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

The corresponding normalized cuboid is reconstructed by:

$$
d
=
\frac{1-n^2}{1+n^2},
$$

$$
c
=
\frac{2n}{1+n^2},
$$

$$
a
=
d
\frac{1-m^2}{1+m^2},
$$

$$
b
=
d
\frac{2m}{1+m^2}.
$$

The remaining two rational face diagonals follow from the two Twin-Psi square roots.

---

# 8. Completeness of the parameterization

This is not merely a special family.

Suppose a nondegenerate rational Perfect Cuboid exists.

Choose any rational face diagonal:

$$
d=\sqrt{a^2+b^2}.
$$

Because:

$$
a,b,d
$$

form a rational Pythagorean triple, there exists rational:

$$
m
$$

giving:

$$
a
=
d\frac{1-m^2}{1+m^2},
$$

$$
b
=
d\frac{2m}{1+m^2}
$$

up to standard sign and permutation choices.

Likewise:

$$
d,c,g
$$

form another rational Pythagorean triple.

After setting:

$$
g=1,
$$

there exists rational:

$$
n
$$

with:

$$
d
=
\frac{1-n^2}{1+n^2},
$$

$$
c
=
\frac{2n}{1+n^2}.
$$

Therefore every nondegenerate Perfect Cuboid admits a Twin-Psi chart.

Hence the formulation is globally complete up to the usual Pythagorean chart boundaries and permutations.

---

# 9. FDRS interpretation

The original three-dimensional object is reconstructed from two one-dimensional half-angle coordinates:

$$
(m,n).
$$

The nested geometry is:

$$
\boxed{
m
\longrightarrow
(a,b,d)
}
$$

and:

$$
\boxed{
n
\longrightarrow
(d,c,g).
}
$$

All remaining arithmetic complexity is transferred into two square gates:

$$
\boxed{
\Psi(m,n),
\qquad
\Psi(\mu(m),n).
}
$$

Thus:

$$
\boxed{
\text{3D Perfect Cuboid}
}
$$

is FDRS-reduced to:

$$
\boxed{
\text{2D rational parameter plane}
+
\text{two coupled one-dimensional square tests}.
}
$$

---

# 10. Tanh-addition algebra

Define on rational variables:

$$
X,Y
$$

the operation:

$$
\boxed{
X\boxplus Y
=
\frac{X+Y}{1+XY}.
}
$$

Then:

$$
\Psi(m,n)
=
m^2\boxplus n^2.
$$

Therefore the Twin-Psi problem is:

$$
\boxed{
m^2\boxplus n^2\in\mathbb Q^2,
}
$$

and:

$$
\boxed{
\mu(m)^2\boxplus n^2\in\mathbb Q^2.
}
$$

This is formally the addition law of hyperbolic tangents:

$$
\tanh(\alpha+\beta)
=
\frac{
\tanh\alpha+\tanh\beta
}{
1+\tanh\alpha\tanh\beta
}.
$$

No transcendental interpretation is required for the arithmetic program.

The relevant point is that the nonlinear rational map possesses an exact multiplicative Cayley linearization.

---

# 11. Multiplicative Cayley linearization

Define:

$$
\boxed{
\mathcal C(X)
=
\frac{1+X}{1-X}.
}
$$

Then direct algebra gives:

$$
\boxed{
\mathcal C(X\boxplus Y)
=
\mathcal C(X)\mathcal C(Y).
}
$$

Therefore if:

$$
\Psi(m,n)=u^2,
$$

then:

$$
\boxed{
\mathcal C(u^2)
=
\mathcal C(m^2)
\mathcal C(n^2).
}
$$

Likewise, if:

$$
\Psi(\mu(m),n)=v^2,
$$

then:

$$
\boxed{
\mathcal C(v^2)
=
\mathcal C(\mu(m)^2)
\mathcal C(n^2).
}
$$

Thus the same multiplier:

$$
\mathcal C(n^2)
$$

must simultaneously send two base points into the rational-square Cayley locus.

The Perfect Cuboid problem therefore admits a second interpretation:

$$
\boxed{
\text{simultaneous multiplicative closure problem}.
}
$$

---

# 12. Elimination of the outer parameter

Let:

$$
\Psi(m,n)=u^2,
$$

and:

$$
\Psi(\mu(m),n)=v^2.
$$

Then:

$$
n^2
=
\frac{u^2-m^2}{1-u^2m^2}.
$$

Also:

$$
n^2
=
\frac{
v^2-\mu(m)^2
}{
1-v^2\mu(m)^2
}.
$$

Equating them and simplifying gives:

$$
\boxed{
(1-m^2)
\frac{1-v^2}{1+v^2}
=
2m
\frac{1-u^2}{1+u^2}.
}
$$

Equivalently:

$$
\boxed{
\frac{
\mathcal C(u^2)
}{
\mathcal C(v^2)
}
=
\frac{2m}{1-m^2}.
}
$$

This relation is independent of:

$$
n.
$$

Hence the two output square roots are not independent.

Their Cayley-hyperbolic ratio is controlled entirely by the inner face parameter:

$$
m.
$$

---

# 13. First rational chart of the G2 seed

Use the already closed face:

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

Relabel this nested triangle as:

$$
(a,b,d)
=
(448,840,952).
$$

The remaining edge is:

$$
c=495,
$$

and:

$$
g=1073.
$$

The inner parameter is:

$$
m
=
\frac{b}{d+a}
=
\frac{840}{952+448}
=
\boxed{
\frac35.
}
$$

The outer parameter is:

$$
n
=
\frac{c}{g+d}
=
\frac{495}{1073+952}
=
\boxed{
\frac{11}{45}.
}
$$

Thus the first G2 chart is:

$$
\boxed{
(m,n)
=
\left(
\frac35,
\frac{11}{45}
\right).
}
$$

---

# 14. First-chart Twin-Psi values

Compute:

$$
\Psi\left(
\frac35,
\frac{11}{45}
\right)
=
\boxed{
\frac{625}{1521}
}
=
\boxed{
\left(
\frac{25}{39}
\right)^2.
}
$$

Also:

$$
\mu\left(\frac35\right)
=
\frac{1-\frac35}{1+\frac35}
=
\boxed{
\frac14.
}
$$

Then:

$$
\Psi\left(
\frac14,
\frac{11}{45}
\right)
=
\boxed{
\frac{233}{1913}.
}
$$

Thus this chart is exactly:

$$
\boxed{
\text{one closed Twin-Psi gate}
+
\text{one failed gate}.
}
$$

The failed gate carries precisely the old defect ratio.

---

# 15. Second rational chart of the G2 seed

Use the second already closed face:

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

Relabel:

$$
(a,b,d)
=
(495,840,975).
$$

The remaining edge is:

$$
c=448,
$$

with:

$$
g=1073.
$$

Then:

$$
m
=
\frac{840}{975+495}
=
\boxed{
\frac47.
}
$$

The outer parameter is:

$$
n
=
\frac{448}{1073+975}
=
\boxed{
\frac7{32}.
}
$$

Thus the second chart is:

$$
\boxed{
(m,n)
=
\left(
\frac47,
\frac7{32}
\right).
}
$$

---

# 16. Second-chart Twin-Psi values

Compute:

$$
\Psi\left(
\frac47,
\frac7{32}
\right)
=
\boxed{
\frac{289}{784}
}
=
\boxed{
\left(
\frac{17}{28}
\right)^2.
}
$$

Also:

$$
\mu\left(\frac47\right)
=
\frac{1-\frac47}{1+\frac47}
=
\boxed{
\frac3{11}.
}
$$

Then:

$$
\Psi\left(
\frac3{11},
\frac7{32}
\right)
=
\boxed{
\frac{233}{1913}.
}
$$

Again:

$$
\boxed{
\text{one closed gate}
+
\text{the same failed gate}.
}
$$

---

# 17. Meaning of the old $11/45$ and $7/32$ slices

The previously discovered parameters:

$$
\boxed{
\frac{11}{45}
}
$$

and:

$$
\boxed{
\frac7{32}
}
$$

are therefore not arbitrary fixed constants.

They are the two outer half-angle coordinates:

$$
n
=
\frac{\text{remaining edge}}{\text{body diagonal}+\text{chosen closed face diagonal}}
$$

obtained from the two already closed rational faces of G2.

The G2 seed has exactly two rational closed face diagonals.

Therefore it produces exactly two rational nested-Pythagorean charts.

This explains structurally why C1 found exactly two useful one-axis repair orbits.

---

# 18. Why there is no third rational G2 chart

The missing face diagonal is:

$$
\sqrt{445729}.
$$

It is irrational.

Therefore the nested pair built from that face cannot be parameterized by a rational outer half-angle:

$$
n.
$$

Hence the one-defect seed naturally possesses:

- two rational nested-face charts;
- one irrational nested-face chart.

The two rational charts are precisely:

$$
n=\frac{11}{45},
$$

and:

$$
n=\frac7{32}.
$$

Thus the old orbit count is a direct arithmetic consequence of the face-closure pattern.

---

# 19. Fixed-$n$ first quartic cover

Fix:

$$
n\in\mathbb Q.
$$

The first gate:

$$
\Psi(m,n)=u^2
$$

is:

$$
\frac{m^2+n^2}{1+m^2n^2}
=
u^2.
$$

Multiply by:

$$
(1+m^2n^2)^2.
$$

Define:

$$
Y_1
=
u(1+m^2n^2).
$$

Then:

$$
\boxed{
Y_1^2
=
F_1(m;n),
}
$$

where:

$$
\boxed{
F_1(m;n)
=
(m^2+n^2)(1+m^2n^2).
}
$$

Expanded:

$$
\boxed{
F_1(m;n)
=
n^2m^4
+
(1+n^4)m^2
+
n^2.
}
$$

For generic nondegenerate:

$$
n,
$$

this is a separable quartic double cover of the:

$$
m
$$

line and hence a genus-$1$ curve after smooth projective completion.

---

# 20. Fixed-$n$ second quartic cover

The second gate is:

$$
\Psi(\mu(m),n)=v^2.
$$

Define:

$$
A(m;n)
=
(1-m)^2+n^2(1+m)^2,
$$

and:

$$
B(m;n)
=
(1+m)^2+n^2(1-m)^2.
$$

Then:

$$
\Psi(\mu(m),n)
=
\frac{A(m;n)}{B(m;n)}.
$$

Define:

$$
Y_2
=
vB(m;n).
$$

Then:

$$
\boxed{
Y_2^2
=
F_2(m;n),
}
$$

where:

$$
\boxed{
F_2(m;n)
=
A(m;n)B(m;n).
}
$$

This simplifies to:

$$
\boxed{
F_2(m;n)
=
(1+n^2)^2(1+m^2)^2
-
4m^2(1-n^2)^2.
}
$$

Equivalently:

$$
\boxed{
F_2(m;n)
=
(1+n^2)^2(m^4+1)
-
2(1-6n^2+n^4)m^2.
}
$$

Again, generically this is a separable quartic and defines a genus-$1$ double cover.

---

# 21. Symmetries of the quartics

Both:

$$
F_1
$$

and:

$$
F_2
$$

are even in:

$$
m.
$$

Thus:

$$
m\mapsto-m
$$

is an involution.

Both also possess reciprocal-type structure.

For:

$$
F_2,
$$

the leading and constant coefficients are equal.

For:

$$
F_1,
$$

the leading and constant coefficients are both:

$$
n^2.
$$

Thus fixed fibers have built-in dihedral-like involutive structure.

This helps explain why lower-genus quotients appear naturally in the historical fixed-slice program.

---

# 22. Simultaneous fiber product

A fixed-$n$ Perfect Cuboid candidate requires:

$$
Y_1^2=F_1(m;n),
$$

and:

$$
Y_2^2=F_2(m;n).
$$

Therefore the complete fixed fiber is the simultaneous quadratic cover:

$$
\boxed{
C_n:
\quad
Y_1^2=F_1(m;n),
\qquad
Y_2^2=F_2(m;n).
}
$$

This is a:

$$
(\mathbb Z/2\mathbb Z)^2
$$

cover of the projective:

$$
m
$$

line.

The three nontrivial quadratic subcovers are:

$$
Y_1^2=F_1,
$$

$$
Y_2^2=F_2,
$$

and:

$$
\boxed{
Y_3^2=F_1F_2.
}
$$

---

# 23. Generic genus calculation

Assume:

- both quartics are separable;
- their branch loci are disjoint;
- no degeneration occurs at infinity.

Then:

$$
Y_1^2=F_1
$$

has genus:

$$
1.
$$

Likewise:

$$
Y_2^2=F_2
$$

has genus:

$$
1.
$$

The product cover:

$$
Y_3^2=F_1F_2
$$

has degree:

$$
8
$$

and therefore genus:

$$
3.
$$

For a smooth:

$$
(\mathbb Z/2\mathbb Z)^2
$$

fiber product, the genus is:

$$
\boxed{
g(C_n)
=
1+1+3
=
5.
}
$$

Thus the generic fixed-$n$ Twin-Psi fiber is genus:

$$
\boxed{
5.
}
$$

This is exactly the high-genus complexity previously encountered in fixed-slice research.

---

# 24. Interpretation of the old genus-$5$ curve

The old fixed:

$$
k=\frac7{32}
$$

program can now be reinterpreted as:

$$
\boxed{
n=\frac7{32}
}
$$

inside the Twin-Psi surface.

Fixing:

$$
n
$$

removes one degree of freedom.

The remaining two square conditions produce the genus-$5$ simultaneous quartic fiber:

$$
C_n.
$$

Therefore the genus-$5$ object was not an accidental complication caused by a poor derivation.

It is the natural arithmetic fiber of a simpler two-dimensional surface.

The conceptual hierarchy is now:

$$
\boxed{
\text{Twin-Psi surface}
}
$$

$$
\Downarrow
$$

$$
\boxed{
n=\text{constant}
}
$$

$$
\Downarrow
$$

$$
\boxed{
\text{genus-5 fiber}.
}
$$

---

# 25. Natural genus-$3$ intermediate

The third quadratic subcover is:

$$
\boxed{
H_n:
\quad
Y_3^2
=
F_1(m;n)F_2(m;n).
}
$$

Generically:

$$
\deg(F_1F_2)=8.
$$

Hence:

$$
\boxed{
g(H_n)=3.
}
$$

Therefore the previously encountered genus-$3$ intermediate layer is also predicted directly by the Twin-Psi decomposition.

The chain:

$$
\text{genus-5 simultaneous double-square curve}
$$

$$
\Downarrow
$$

$$
\text{genus-3 product-square curve}
$$

is not ad hoc.

It is the standard quadratic-subcover structure of:

$$
C_n.
$$

---

# 26. Specialization to $n=7/32$

Let:

$$
n=\frac7{32}.
$$

Then:

$$
n^2=\frac{49}{1024}.
$$

The first quartic factors as:

$$
F_1
=
\left(
m^2+\frac{49}{1024}
\right)
\left(
1+\frac{49}{1024}m^2
\right).
$$

After clearing squares:

$$
\boxed{
F_1
\sim
(1024m^2+49)
(49m^2+1024).
}
$$

The second gate factors into:

$$
A
=
(1-m)^2+\frac{49}{1024}(1+m)^2,
$$

$$
B
=
(1+m)^2+\frac{49}{1024}(1-m)^2.
$$

After multiplying by:

$$
1024,
$$

these become:

$$
\boxed{
1024A
=
1073m^2-1950m+1073,
}
$$

and:

$$
\boxed{
1024B
=
1073m^2+1950m+1073.
}
$$

Thus the characteristic constants:

$$
49,
\quad
1024,
\quad
1073,
\quad
1950
$$

emerge directly from the Twin-Psi specialization.

---

# 27. Seed point on the $7/32$ fiber

At:

$$
m=\frac47,
$$

the first gate is:

$$
\boxed{
\Psi\left(
\frac47,
\frac7{32}
\right)
=
\left(
\frac{17}{28}
\right)^2.
}
$$

The Möbius partner is:

$$
\mu\left(\frac47\right)
=
\frac3{11}.
$$

The second gate is:

$$
\boxed{
\Psi\left(
\frac3{11},
\frac7{32}
\right)
=
\frac{233}{1913}.
}
$$

Therefore the known seed is a rational point on one elliptic gate cover whose partner gate remains in the nontrivial squareclass:

$$
233\cdot1913.
$$

---

# 28. Specialization to $n=11/45$

Let:

$$
n=\frac{11}{45}.
$$

Then:

$$
n^2=\frac{121}{2025}.
$$

The first quartic satisfies:

$$
\boxed{
F_1
\sim
(2025m^2+121)
(121m^2+2025).
}
$$

The second gate factors into:

$$
\boxed{
2025A
=
2146m^2-3808m+2146,
}
$$

and:

$$
\boxed{
2025B
=
2146m^2+3808m+2146.
}
$$

Dividing by the common factor:

$$
2
$$

gives:

$$
\boxed{
1073m^2-1904m+1073,
}
$$

and:

$$
\boxed{
1073m^2+1904m+1073.
}
$$

These are precisely the conjugate factors found independently in R1.

---

# 29. Seed point on the $11/45$ fiber

At:

$$
m=\frac35,
$$

the first gate is:

$$
\boxed{
\Psi\left(
\frac35,
\frac{11}{45}
\right)
=
\left(
\frac{25}{39}
\right)^2.
}
$$

The Möbius partner is:

$$
\mu\left(\frac35\right)
=
\frac14.
$$

The second gate is:

$$
\boxed{
\Psi\left(
\frac14,
\frac{11}{45}
\right)
=
\frac{233}{1913}.
}
$$

Thus the same defect appears on both rational charts.

---

# 30. Why both charts expose the same defect

The two charts differ in which already closed face diagonal is chosen as the nested face:

- one uses:
  $$
  952;
  $$
- the other uses:
  $$
  975.
  $$

But both leave the same missing face:

$$
\sqrt{448^2+495^2}.
$$

Therefore both must ultimately test the same complementary channel:

$$
1-\left(\frac{840}{1073}\right)^2.
$$

Its canonical Cayley defect ratio is:

$$
\boxed{
\frac{1073-840}{1073+840}
=
\frac{233}{1913}.
}
$$

The Twin-Psi formulation makes this invariance transparent.

---

# 31. New interpretation of fixed-$k$ research

The old fixed-$k$ program can now be classified as:

$$
\boxed{
\text{outer half-angle fiber selection}.
}
$$

Specifically:

$$
k=n.
$$

Fixing:

$$
n
$$

means fixing one rational Pythagorean decomposition:

$$
d^2+c^2=1.
$$

The remaining variable:

$$
m
$$

moves along the inner face circle:

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

Thus the old one-axis searches were literally one-dimensional motions inside the nested FDRS geometry.

This gives a direct geometric interpretation to the former arithmetic slice parameter.

---

# 32. Why global search should return to two dimensions

A fixed:

$$
n
$$

fiber is generically genus:

$$
5.
$$

That explains why proving a single slice can become expensive.

But the complete Twin-Psi problem lives before the slicing step:

$$
\boxed{
(m,n)\in\mathbb Q^2.
}
$$

Therefore R2 suggests that the global problem should not be attacked only by proving many isolated genus-$5$ fibers.

Instead one should study the two-dimensional arithmetic surface:

$$
\boxed{
\mathfrak T:
\quad
\Psi(m,n)=u^2,
\qquad
\Psi(\mu(m),n)=v^2.
}
$$

The high genus appears only after:

$$
n=\text{constant}.
$$

This distinction is central.

---

# 33. Twin-Psi surface as simultaneous square-preserving addition

Set:

$$
M=m^2,
$$

$$
N=n^2.
$$

Then:

$$
\Psi(m,n)
=
\frac{M+N}{1+MN}.
$$

Define:

$$
M\boxplus N
=
\frac{M+N}{1+MN}.
$$

Then:

$$
\boxed{
M\boxplus N=u^2.
}
$$

The second gate is:

$$
\boxed{
\mu(m)^2\boxplus N=v^2.
}
$$

The arithmetic problem is therefore:

> Find a square input $N$ that simultaneously maps two Möbius-related square inputs into squares under the same $\boxplus$ operation.

This is a much smaller algebraic statement than the original seven-length Diophantine formulation.

---

# 34. Hyperbolic square locus

Define:

$$
\mathcal H_{\rm sq}
=
\left\{
\mathcal C(q^2):
q\in\mathbb Q
\right\},
$$

where:

$$
\mathcal C(X)
=
\frac{1+X}{1-X}.
$$

Each element:

$$
h=\mathcal C(q^2)
$$

satisfies:

$$
h^2-1
=
\left(
\frac{2q}{1-q^2}
\right)^2.
$$

Therefore:

$$
\mathcal H_{\rm sq}
$$

lies on the rational unit hyperbola.

Twin-Psi closure becomes:

$$
\boxed{
\mathcal C(m^2)\mathcal C(n^2)
\in
\mathcal H_{\rm sq},
}
$$

and:

$$
\boxed{
\mathcal C(\mu(m)^2)\mathcal C(n^2)
\in
\mathcal H_{\rm sq}.
}
$$

Thus one common rational multiplier must simultaneously return two hyperbolic points to the square-parameterized hyperbolic locus.

---

# 35. A possible R3 descent direction

Because:

$$
\mathcal C(X\boxplus Y)
=
\mathcal C(X)\mathcal C(Y),
$$

one may attempt to replace quartic squarehood by multiplicative squareclass data.

Define:

$$
H_m=\mathcal C(m^2),
$$

$$
H_\mu=\mathcal C(\mu(m)^2),
$$

$$
H_n=\mathcal C(n^2).
$$

Then seek:

$$
H_u=H_mH_n,
$$

$$
H_v=H_\mu H_n,
$$

with:

$$
H_u,H_v\in\mathcal H_{\rm sq}.
$$

The ratio:

$$
\frac{H_u}{H_v}
$$

does not depend on:

$$
n.
$$

Indeed:

$$
\boxed{
\frac{H_u}{H_v}
=
\frac{2m}{1-m^2}.
}
$$

This suggests a possible elimination/descent strategy in which:

1. choose rational output hyperbolic points;
2. reconstruct:
   $$
   m;
   $$
3. recover the common multiplier:
   $$
   H_n;
   $$
4. test whether:
   $$
   H_n
   $$
   itself comes from a rational square parameter.

---

# 36. R2 canonical theorem statement

The central result of R2 is:

> Perfect Cuboid existence is equivalent to the existence of rational nondegenerate $m,n$ for which two Möbius-related inputs are simultaneously square-preserved by the rational addition law $\Psi$.

Formally:

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

if and only if:

$$
\boxed{
\exists m,n,u,v\in\mathbb Q_{>0}
}
$$

such that:

$$
\boxed{
u^2
=
\frac{m^2+n^2}{1+m^2n^2},
}
$$

and:

$$
\boxed{
v^2
=
\frac{
\left(\frac{1-m}{1+m}\right)^2+n^2
}{
1+
\left(\frac{1-m}{1+m}\right)^2n^2
}.
}
$$

The original cuboid is then reconstructed rationally by the nested Pythagorean formulas.

---

# 37. Main conceptual unification

The following objects are now identified as different views of the same structure:

$$
\boxed{
\text{FDRS nested geometry}
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
\text{X-diagonal / shared-edge coupling}
}
$$

$$
\Longleftrightarrow
$$

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

$$
\Longleftrightarrow
$$

$$
\boxed{
\text{Twin-Psi square system}
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
\text{fixed-}n\text{ genus-5 fibers}
}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
\text{previous }k=\frac7{32}
\text{ and }
k=\frac{11}{45}
\text{ programs}.
}
$$

This is the principal R2 advance.

---

# 38. Revised position of the old high-genus work

The previous genus-$5$, genus-$3$, genus-$2$, elliptic-quotient, and descent computations remain useful.

They are not discarded.

Their new role is:

$$
\boxed{
\text{fiber-level arithmetic certification}.
}
$$

The Twin-Psi system becomes the global two-dimensional parent object.

Thus future work can alternate between:

$$
\boxed{
\text{global surface structure}
}
$$

and:

$$
\boxed{
\text{certified fixed-fiber analysis}.
}
$$

This separation may prevent repeatedly rediscovering the same high-genus obstruction in different coordinates.

---

# 39. R3 proposed program

R3 should proceed in four directions.

## R3-A — Twin-Psi birational geometry

Study:

$$
u^2
=
\Psi(m,n),
$$

$$
v^2
=
\Psi(\mu(m),n)
$$

as a two-dimensional algebraic surface.

Determine:

- singularities;
- rational fibrations;
- involutions;
- quotient surfaces;
- possible conic-bundle structures;
- possible elliptic-surface structures.

## R3-B — Multiplicative Cayley descent

Use:

$$
\mathcal C(X\boxplus Y)
=
\mathcal C(X)\mathcal C(Y)
$$

to move the problem into multiplicative arithmetic.

Study the simultaneous multiplier problem on:

$$
\mathcal H_{\rm sq}.
$$

## R3-C — Fiber comparison

Compare the two seed fibers:

$$
n=\frac7{32},
$$

and:

$$
n=\frac{11}{45}.
$$

Determine whether their fixed-fiber curves are related by an explicit birational correspondence induced by axis permutation.

## R3-D — Dynamic FDRS transitions

Allow both:

$$
m
$$

and:

$$
n
$$

to change under shell-preserving moves.

Track whether the failed gate:

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

can move toward:

$$
q^2
$$

without fixing either historical slice.

---

# 40. Final conclusion

R2 reduces the Perfect Cuboid problem to a two-variable Twin-Psi square system:

$$
\boxed{
\Psi(m,n)\in\mathbb Q^2,
\qquad
\Psi(\mu(m),n)\in\mathbb Q^2.
}
$$

The G2 seed has two rational nested-face charts:

$$
\boxed{
\left(
\frac35,
\frac{11}{45}
\right),
}
$$

and:

$$
\boxed{
\left(
\frac47,
\frac7{32}
\right).
}
$$

These are exactly the two historical one-axis branches.

Each chart has:

- one square Twin-Psi gate;
- one failed Twin-Psi gate;
- the same canonical defect:
  $$
  \frac{233}{1913}.
  $$

Fixing:

$$
n
$$

turns the two gates into quartic double covers.

Their simultaneous fiber product is generically genus:

$$
5,
$$

with quadratic subcover genera:

$$
1,
\qquad
1,
\qquad
3.
$$

Therefore the old genus-$5$ and genus-$3$ structures are reconstructed directly from the new global two-dimensional parent system.

The previous fixed-$k$ theory is now interpreted as fiber arithmetic.

The next target is no longer another blind fixed slice.

It is the global arithmetic geometry of:

$$
\boxed{
\mathfrak T:
\quad
u^2=\Psi(m,n),
\qquad
v^2=\Psi(\mu(m),n).
}
$$

This is the canonical starting point for R3.