# Perfect Cuboid G2A — Nested-X Double-Square Closure

## A Two-Parameter Exact Reduction of the Perfect Cuboid Problem

**Series:** Perfect Cuboid Global Closure  
**Main line:** Geometry Branch G  
**Verification line:** Verification Branch A  
**Phase:** G2A  
**Date:** 2026-09-28

---

# Abstract

The Center-Axis Rational Closure Theorem reduces Perfect Cuboid existence to a rational point problem on the three-parameter CARCE surface.

This note performs a second exact reduction.

Instead of working directly with three half-angle parameters, construct the cuboid geometry in two nested rational X / half-angle stages.

First choose a rational parameter

$$
p\in\mathbb Q\cap(0,1)
$$

to generate the center-axis right triangle

$$
(a,f,g).
$$

Then choose a second rational parameter

$$
h\in\mathbb Q\cap(0,1)
$$

to split the opposite face diagonal

$$
f
$$

into the rational right-triangle pair

$$
(b,c,f).
$$

This construction makes

$$
a,b,c,f,g
$$

rational automatically and satisfies the complete real cuboid geometry identically.

Only the two remaining face diagonals must still close arithmetically.

After a Möbius change of variable

$$
k=\frac{1-p}{1+p},
$$

define

$$
F_k(h)
=
\frac{h^2+k^2}{1+k^2h^2},
$$

and the edge-swap involution

$$
M(h)
=
\frac{1-h}{1+h}.
$$

Then Perfect Cuboid existence is exactly equivalent to the existence of

$$
k,h\in\mathbb Q\cap(0,1)
$$

such that

$$
\boxed{
F_k(h)\in\mathbb Q^2
}
$$

and

$$
\boxed{
F_k(M(h))\in\mathbb Q^2.
}
$$

Thus the original seven-integer problem becomes:

$$
\boxed{
\text{two rational parameters}
+
\text{two exact square gates}.
}
$$

For the current near-arithmetic seed

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

the two nested parameters are

$$
\boxed{
k=\frac7{32},
\qquad
h=\frac47.
}
$$

The first square gate passes exactly:

$$
F_{7/32}\left(\frac47\right)
=
\left(\frac{17}{28}\right)^2,
$$

while the second gives

$$
F_{7/32}\left(\frac3{11}\right)
=
\frac{233}{1913},
$$

which is not a rational square.

An exact finite scan with fixed

$$
k=\frac7{32}
$$

and every reduced rational

$$
0<h<1
$$

with denominator at most

$$
20{,}000
$$

tested

$$
121{,}590{,}395
$$

states.

Only four states passed the first elliptic square gate, four symmetric states passed the second, and none passed both.

The individual gate is a genus-$1$ curve.

The simultaneous fixed-$k$ double-square closure is the normalization of a degree-$4$ biquadratic cover of the projective line and has genus

$$
\boxed{5}.
$$

This identifies the first exact algebraic-geometry object produced directly from the new Geometry Branch.

---

# 1. Start from one rational center-axis channel

Choose

$$
p\in\mathbb Q\cap(0,1).
$$

Define

$$
x_a
=
\frac{1-p^2}{1+p^2},
$$

and

$$
y_a
=
\frac{2p}{1+p^2}.
$$

Then

$$
x_a^2+y_a^2=1.
$$

Interpret

$$
x_a=\frac ag,
$$

and

$$
y_a=\frac fg.
$$

Thus the first rational X / center-axis channel automatically gives

$$
a,
f,
g
$$

as rational ratios.

---

# 2. Split the opposite face by a second rational X parameter

Choose

$$
h\in\mathbb Q\cap(0,1).
$$

Define

$$
u_h
=
\frac{1-h^2}{1+h^2},
$$

and

$$
v_h
=
\frac{2h}{1+h^2}.
$$

Then

$$
u_h^2+v_h^2=1.
$$

Use these to split

$$
f
$$

into

$$
b
$$

and

$$
c:
$$

$$
x_b
=
y_a u_h
=
\frac{
2p(1-h^2)
}{
(1+p^2)(1+h^2)
},
$$

$$
x_c
=
y_a v_h
=
\frac{
4ph
}{
(1+p^2)(1+h^2)
}.
$$

Interpret

$$
x_b=\frac bg,
$$

$$
x_c=\frac cg.
$$

Because

$$
u_h^2+v_h^2=1,
$$

we have identically

$$
x_b^2+x_c^2=y_a^2.
$$

Together with

$$
x_a^2+y_a^2=1,
$$

this gives

$$
\boxed{
x_a^2+x_b^2+x_c^2=1.
}
$$

Therefore the body-diagonal geometry is already closed.

---

# 3. What is automatic after the nested construction

For rational

$$
p,h,
$$

all of the following are rational:

$$
\frac ag,
\qquad
\frac bg,
\qquad
\frac cg,
\qquad
\frac fg.
$$

And they satisfy:

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

as well as

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

So after one common scaling,

$$
a,b,c,f,g
$$

can all be made integers.

Only the two remaining face diagonals:

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

and

$$
e=\sqrt{a^2+c^2}
$$

remain to be forced rational.

This is the whole residual arithmetic problem.

---

# 4. Remaining half-angle square gates

The remaining center-axis half-angle parameters satisfy

$$
t_b^2
=
\frac{1-x_b}{1+x_b},
$$

and

$$
t_c^2
=
\frac{1-x_c}{1+x_c}.
$$

A Perfect Cuboid requires:

$$
t_b\in\mathbb Q,
$$

and

$$
t_c\in\mathbb Q.
$$

Equivalently:

$$
\boxed{
\frac{1-x_b}{1+x_b}
\in
\mathbb Q^2
}
$$

and

$$
\boxed{
\frac{1-x_c}{1+x_c}
\in
\mathbb Q^2.
}
$$

---

# 5. Möbius compression

Define

$$
k
=
\frac{1-p}{1+p}.
$$

Since

$$
p\in\mathbb Q\cap(0,1),
$$

we also have

$$
k\in\mathbb Q\cap(0,1).
$$

Conversely,

$$
p
=
\frac{1-k}{1+k}.
$$

In the new coordinate,

$$
x_a
=
\frac{2k}{1+k^2},
$$

and

$$
y_a
=
\frac{1-k^2}{1+k^2}.
$$

The nested geometry becomes

$$
x_b
=
\frac{
(1-k^2)(1-h^2)
}{
(1+k^2)(1+h^2)
},
$$

$$
x_c
=
\frac{
2h(1-k^2)
}{
(1+k^2)(1+h^2)
}.
$$

---

# 6. First square gate

Define

$$
F_k(h)
=
\frac{h^2+k^2}{1+k^2h^2}.
$$

Direct algebra gives

$$
\frac{1-x_b}{1+x_b}
=
\boxed{
F_k(h).
}
$$

Hence:

$$
\boxed{
t_b^2=F_k(h).
}
$$

Therefore the first remaining face diagonal is rational if and only if:

$$
\boxed{
F_k(h)\in\mathbb Q^2.
}
$$

---

# 7. Edge-swap involution

Define:

$$
M(h)
=
\frac{1-h}{1+h}.
$$

Then:

$$
M(M(h))=h.
$$

Geometrically, this is exactly the interchange:

$$
b\leftrightarrow c.
$$

Indeed, if

$$
h=\tan\frac{\delta}{2},
$$

then swapping the two legs of the right triangle sends

$$
\delta
\mapsto
\frac{\pi}{2}-\delta,
$$

and therefore:

$$
\tan
\left(
\frac{\pi/2-\delta}{2}
\right)
=
\frac{1-h}{1+h}.
$$

---

# 8. Second square gate

Direct substitution gives:

$$
\frac{1-x_c}{1+x_c}
=
F_k(M(h)).
$$

Thus:

$$
\boxed{
t_c^2
=
F_k(M(h)).
}
$$

The second remaining face diagonal is rational if and only if:

$$
\boxed{
F_k(M(h))
\in
\mathbb Q^2.
}
$$

---

# 9. Main two-parameter theorem

## Theorem — Nested-X Double-Square Closure

A Perfect Cuboid exists if and only if there exist

$$
k,h\in\mathbb Q\cap(0,1)
$$

such that:

$$
\boxed{
F_k(h)\in\mathbb Q^2
}
$$

and

$$
\boxed{
F_k(M(h))\in\mathbb Q^2,
}
$$

where:

$$
F_k(h)
=
\frac{h^2+k^2}{1+k^2h^2},
$$

and:

$$
M(h)
=
\frac{1-h}{1+h}.
$$

This is exact.

No floating-point approximation is part of the theorem.

---

# 10. Why the theorem is sufficient

Given rational

$$
k,h,
$$

recover:

$$
p
=
\frac{1-k}{1+k}.
$$

The nested construction gives rational:

$$
x_a,x_b,x_c,y_a.
$$

If:

$$
F_k(h)=q^2,
$$

and:

$$
F_k(M(h))=r^2
$$

for:

$$
q,r\in\mathbb Q,
$$

then the missing opposite-face half-angle channels are rational.

Therefore every normalized edge and diagonal ratio is rational.

A common denominator then scales the configuration to an integer Perfect Cuboid.

---

# 11. Integer-only square gates

Write:

$$
k=\frac mn,
$$

$$
h=\frac uv,
$$

in lowest terms.

Then:

$$
F_k(h)
=
\frac{
n^2u^2+m^2v^2
}{
n^2v^2+m^2u^2
}.
$$

Define:

$$
N_1
=
n^2u^2+m^2v^2,
$$

$$
D_1
=
n^2v^2+m^2u^2.
$$

Reduce by:

$$
g_1=\gcd(N_1,D_1).
$$

Then the first square gate passes exactly if:

$$
\boxed{
\frac{N_1}{g_1}
\text{ and }
\frac{D_1}{g_1}
\text{ are both perfect squares}.
}
$$

No factorization is required.

---

# 12. Integer-only second square gate

Since:

$$
M(h)
=
\frac{v-u}{v+u},
$$

we obtain:

$$
N_2
=
n^2(v-u)^2
+
m^2(v+u)^2,
$$

$$
D_2
=
n^2(v+u)^2
+
m^2(v-u)^2.
$$

Let:

$$
g_2=\gcd(N_2,D_2).
$$

Then the second square gate passes exactly if:

$$
\boxed{
\frac{N_2}{g_2}
\text{ and }
\frac{D_2}{g_2}
\text{ are both perfect squares}.
}
$$

Thus the search kernel needs only:

- integer addition;
- multiplication;
- gcd;
- exact integer-square tests.

---

# 13. Current seed in the new coordinates

The current exact-real seed is:

$$
(a,b,c;f,g)
=
(448,495,840;975,1073).
$$

The first nested center-axis parameter is:

$$
p
=
\frac{f}{g+a}
=
\frac{975}{1521}
=
\frac{25}{39}.
$$

Therefore:

$$
k
=
\frac{1-p}{1+p}
=
\boxed{
\frac7{32}.
}
$$

The second nested parameter comes from:

$$
\frac bf
=
\frac{495}{975}
=
\frac{33}{65},
$$

$$
\frac cf
=
\frac{840}{975}
=
\frac{56}{65}.
$$

Thus:

$$
h
=
\frac{56}{65+33}
=
\boxed{
\frac47.
}
$$

Hence the current seed is:

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

---

# 14. First gate at the seed

Compute:

$$
F_{7/32}\left(\frac47\right)
=
\frac{
(4/7)^2+(7/32)^2
}{
1+(7/32)^2(4/7)^2
}.
$$

After exact reduction:

$$
\boxed{
F_{7/32}\left(\frac47\right)
=
\frac{289}{784}
=
\left(
\frac{17}{28}
\right)^2.
}
$$

So the first remaining face-diagonal channel passes exactly.

---

# 15. Second gate at the seed

The involution gives:

$$
M\left(\frac47\right)
=
\frac{1-4/7}{1+4/7}
=
\boxed{
\frac3{11}.
}
$$

Then:

$$
F_{7/32}\left(\frac3{11}\right)
=
\boxed{
\frac{233}{1913}.
}
$$

This is exactly the old shared-core defect.

Therefore the seed status is:

$$
\boxed{
\text{Gate 1 PASS}
}
$$

and:

$$
\boxed{
\text{Gate 2 FAIL at }233/1913.
}
$$

The old and new formulations are therefore identical at the exact arithmetic level.

---

# 16. Seed integer gate values

For:

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

and:

$$
h=\frac47,
$$

the first integer pair is:

$$
N_1=18785,
$$

$$
D_1=50960.
$$

Their gcd is:

$$
g_1=65.
$$

Therefore:

$$
\frac{N_1}{g_1}=289=17^2,
$$

$$
\frac{D_1}{g_1}=784=28^2.
$$

Gate 1 passes.

The second pair is:

$$
N_2=15145,
$$

$$
D_2=124345.
$$

Again:

$$
g_2=65.
$$

Reduction gives:

$$
\frac{N_2}{g_2}=233,
$$

$$
\frac{D_2}{g_2}=1913.
$$

Neither is a square.

Gate 2 fails.

---

# 17. The individual square gate is genus 1

The first gate:

$$
q^2
=
F_k(h)
=
\frac{h^2+k^2}{1+k^2h^2}
$$

can be rewritten as:

$$
Y^2
=
(h^2+k^2)(1+k^2h^2),
$$

where:

$$
Y=q(1+k^2h^2).
$$

For:

$$
0<k<1,
$$

the quartic has four distinct roots:

$$
\pm ik,
$$

$$
\pm\frac{i}{k}.
$$

Therefore its smooth projective model has genus:

$$
\boxed{1}.
$$

Since:

$$
h=1
$$

gives a rational point,

the curve is an elliptic curve over:

$$
\mathbb Q.
$$

---

# 18. Explicit elliptic model

Divide the quartic by:

$$
k^2
$$

and set:

$$
y_0=\frac{Y}{k}.
$$

Then:

$$
y_0^2
=
h^4
+
A h^2
+
1,
$$

where:

$$
A
=
k^2+k^{-2}.
$$

Define:

$$
X
=
\frac{y_0+1}{h^2}.
$$

Then:

$$
h^2
=
\frac{2X+A}{X^2-1}.
$$

Define:

$$
x=2X,
$$

and:

$$
y=2h(X^2-1).
$$

Then the quartic is birational to:

$$
\boxed{
E_k:
\quad
y^2
=
(x^2-4)(x+A).
}
$$

That is:

$$
\boxed{
y^2
=
(x-2)(x+2)(x+A).
}
$$

The inverse recovers:

$$
\boxed{
h
=
\frac{2y}{x^2-4}.
}
$$

Thus every single-gate rational closure belongs to an explicit elliptic curve.

---

# 19. Fixed current first channel

For:

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

we have:

$$
A
=
\left(\frac7{32}\right)^2
+
\left(\frac{32}{7}\right)^2
=
\frac{1050977}{50176}.
$$

The seed:

$$
h=\frac47,
$$

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

maps to the elliptic point:

$$
x
=
\frac{1497}{64},
$$

$$
y
=
\frac{2224625}{14336}.
$$

This point is not one of the visible rational $2$-torsion points.

So the fixed-$k$ first-gate curve already has nontrivial rational structure.

---

# 20. Simultaneous double-square curve

For fixed rational:

$$
k\in(0,1),
$$

Perfect Cuboid closure requires:

$$
q^2=F_k(h),
$$

and:

$$
r^2=F_k(M(h)).
$$

This is the fiber product of two double covers of:

$$
\mathbb P^1_h.
$$

The first cover branches at:

$$
\left\{
\pm ik,
\pm i/k
\right\}.
$$

The second branches at the Möbius preimages:

$$
M^{-1}
\left(
\left\{
\pm ik,
\pm i/k
\right\}
\right).
$$

For rational:

$$
0<k<1,
$$

these two four-point branch sets are disjoint.

Therefore the combined cover has eight distinct branch locations.

---

# 21. Genus of the fixed-$k$ simultaneous closure

The full simultaneous curve is a degree-$4$ cover of:

$$
\mathbb P^1.
$$

At each of the eight branch locations:

- two points lie above the base point;
- each has ramification index $2$.

Thus each branch location contributes:

$$
2
$$

to the Riemann-Hurwitz ramification sum.

Total ramification:

$$
8\times2=16.
$$

Riemann-Hurwitz gives:

$$
2g-2
=
4(-2)+16.
$$

Hence:

$$
2g-2=8,
$$

so:

$$
\boxed{
g=5.
}
$$

Therefore for each fixed rational:

$$
0<k<1,
$$

the simultaneous two-square Perfect Cuboid closure lives on a genus-$5$ curve.

This is the first direct algebraic-geometric complexity classification of the new two-parameter Geometry Branch.

---

# 22. Consequence of genus 5

For any fixed rational:

$$
k,
$$

the simultaneous closure curve has genus greater than:

$$
1.
$$

Therefore its rational points are finite by Faltings' theorem.

This does not provide an effective list.

But it explains why a fixed first channel can have extremely sparse exact double-square closures.

The global Perfect Cuboid problem remains open because:

$$
k
$$

itself is free to vary over:

$$
\mathbb Q\cap(0,1).
$$

---

# 23. Exact fixed-$k$ scan

The first computational experiment fixes:

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

the value induced by the current seed.

Every reduced rational:

$$
h=\frac uv
$$

with:

$$
0<u<v,
$$

$$
\gcd(u,v)=1,
$$

and:

$$
v\le20{,}000
$$

was tested.

Number of tested states:

$$
\boxed{
121{,}590{,}395.
}
$$

The search used only the integer square gates from Sections 11 and 12.

No floating-point acceptance criterion was used.

---

# 24. First-gate rational points found in the scan

Exactly four reduced rational states in the tested range passed the first square gate:

$$
h=\frac47,
\qquad
q=\frac{17}{28},
$$

$$
h=\frac4{33},
\qquad
q=\frac14,
$$

$$
h=\frac{33}{119},
\qquad
q=\frac{269}{763},
$$

$$
h=\frac{256}{817},
\qquad
q=\frac{61}{160}.
$$

No additional first-gate state with denominator:

$$
v\le20{,}000
$$

was found.

---

# 25. Second-gate rational points found in the scan

Exactly four states passed the second square gate:

$$
h=\frac3{11},
\qquad
r=\frac{17}{28},
$$

$$
h=\frac{29}{37},
\qquad
r=\frac14,
$$

$$
h=\frac{43}{76},
\qquad
r=\frac{269}{763},
$$

$$
h=\frac{561}{1073},
\qquad
r=\frac{61}{160}.
$$

These are exactly the involution images of the four first-gate states.

For example:

$$
M\left(\frac47\right)
=
\frac3{11},
$$

and:

$$
M\left(\frac4{33}\right)
=
\frac{29}{37}.
$$

---

# 26. No simultaneous closure in the finite scan

Within:

$$
v\le20{,}000,
$$

the number of states satisfying both square gates was:

$$
\boxed{0}.
$$

Therefore:

$$
\boxed{
\text{No Perfect Cuboid occurs on the fixed }
k=\frac7{32}
\text{ slice with }
\operatorname{den}(h)\le20{,}000.
}
$$

This is only a bounded exact computational exclusion.

It is not a proof that the entire fixed-$k$ genus-$5$ curve has no positive rational point.

---

# 27. Independent small-range cross-check

A second implementation using exact Python integer arithmetic independently scanned all reduced:

$$
h=\frac uv
$$

with:

$$
v\le1000.
$$

It tested:

$$
304{,}191
$$

states.

It reproduced exactly the same first-gate points visible in that range:

$$
\frac47,
\quad
\frac4{33},
\quad
\frac{33}{119},
\quad
\frac{256}{817},
$$

and the second-gate points with denominator at most $1000$:

$$
\frac3{11},
\quad
\frac{29}{37},
\quad
\frac{43}{76}.
$$

No simultaneous closure was found.

This cross-check validates the integer gate implementation on the overlapping bounded domain.

---

# 28. Why the involution pattern matters

The relation:

$$
F_k(M(h))
$$

is not a separate unrelated equation.

It is the same elliptic square gate evaluated after the geometric edge swap:

$$
b\leftrightarrow c.
$$

Therefore Perfect Cuboid closure at fixed:

$$
k
$$

can be stated as:

> find a rational $h$ such that both $h$ and $M(h)$ belong to the rational-square projection set of the same elliptic gate.

This is the exact reason the two finite hit sets appear as involution pairs.

---

# 29. New canonical search object

After G2A, the recommended primary exact search state is no longer:

$$
(t_a,t_b,t_c).
$$

For constructive search, use:

$$
\boxed{
(k,h).
}
$$

The geometry is generated identically.

The verifier only needs to evaluate:

$$
F_k(h)
$$

and:

$$
F_k(M(h)).
$$

The formal target is:

$$
\boxed{
\exists
k,h\in\mathbb Q\cap(0,1)
:
F_k(h),F_k(M(h))
\in
\mathbb Q^2.
}
$$

---

# 30. Geometry Branch interpretation

The two parameters have direct geometric meaning.

The first parameter:

$$
k
$$

controls the first center-axis decomposition:

$$
(a,f,g).
$$

The second:

$$
h
$$

controls the internal face decomposition:

$$
(b,c,f).
$$

Therefore software can display:

1. the body-diagonal X / center-axis triangle;
2. the opposite-face X split;
3. the two remaining arithmetic face channels;
4. the two exact square-gate indicators.

The visual target becomes:

$$
\boxed{
\text{Gate 1}
\quad
\checkmark
}
$$

and:

$$
\boxed{
\text{Gate 2}
\quad
\checkmark.
}
$$

When both flip to PASS at the same rational state, the Perfect Cuboid certificate exists.

---

# 31. Immediate next mathematical directions

## A. Fixed-$k$ genus-$5$ arithmetic

For:

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

determine the rational points on the simultaneous genus-$5$ curve.

Possible tools:

- Jacobian decomposition;
- Chabauty-Coleman;
- Mordell-Weil sieve;
- covering collections;
- elliptic quotient analysis;
- local obstruction.

## B. Elliptic gate generation

Use:

$$
E_k:
y^2=(x-2)(x+2)(x+A)
$$

to generate rational first-gate states without denominator brute force.

Then test their involution images against the second square gate.

## C. Vary $k$

Search over rational:

$$
k
$$

using low-height exact arithmetic, but use elliptic-gate structure instead of raw $(k,h)$ grid enumeration.

## D. Jacobian symmetry

The genus-$5$ curve has multiple involutions:

- sign changes in the two square roots;
- the edge swap $h\mapsto M(h)$;
- elliptic double-cover symmetries.

These may split the Jacobian into lower-dimensional factors.

That decomposition is a high-priority proof target.

---

# 32. Current interpretation of progress

The original classical problem:

$$
7
\text{ integer lengths}
$$

has now been compressed through the following exact chain:

$$
\text{Perfect Cuboid}
$$

$$
\Longleftrightarrow
$$

$$
\text{CARCE rational point}
$$

$$
\Longleftrightarrow
$$

$$
\boxed{
(k,h)
\text{ rational}
+
2
\text{ simultaneous square gates}.
}
$$

For the current seed:

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

one gate already passes.

The unresolved defect is exactly:

$$
\boxed{
\frac{233}{1913}
\notin
\mathbb Q^2.
}
$$

So the research problem has become substantially more localized.

---

# 33. Canonical one-line target

$$
\boxed{
\exists\,
k,h,q,r
\in
\mathbb Q\cap(0,1)
:
q^2
=
\frac{h^2+k^2}{1+k^2h^2},
\qquad
r^2
=
\frac{M(h)^2+k^2}{1+k^2M(h)^2},
\quad
M(h)=\frac{1-h}{1+h}.
}
$$

Any constructive solution of this system yields an exact integer Perfect Cuboid.

---

**End of G2A**