# Perfect Cuboid G2E.1 — Square-Lift Tower and Quadratic-Chabauty Admissibility

## From the fixed-$k$ genus-$5$ curve to a rational conic and a depth-$2$ Chabauty target

**Series:** Perfect Cuboid Global Closure  
**Phase:** G2E.1  
**Fixed slice:** $k=7/32$  
**Status:** exact structural reduction  
**Date:** 2026-09-30

---

# 1. Starting point

G2D completed the Mordell–Weil rank certification for the three distinct elliptic factors:

$$
\operatorname{rank}E_f(\mathbb Q)=1,
$$

$$
\operatorname{rank}E_0(\mathbb Q)=1,
$$

$$
\operatorname{rank}E_-(\mathbb Q)=1.
$$

The fixed-slice genus-$5$ curve satisfies

$$
\operatorname{Jac}(C_k)
\sim_{\mathbb Q}
E_f^2\times E_0^2\times E_-.
$$

Hence

$$
\boxed{
\operatorname{rank}\operatorname{Jac}(C_{7/32})(\mathbb Q)=5.
}
$$

Since

$$
g(C_{7/32})=5,
$$

ordinary Chabauty–Coleman does not apply because its strict inequality

$$
r<g
$$

fails by equality.

This phase shows that the same splitting that causes the rank equality also makes the curve strongly suitable for quadratic Chabauty.

Before entering $p$-adic heights, we first quotient the square symmetries as far as possible.

---

# 2. Fixed first channel

Let

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

be fixed and write

$$
K=k^2.
$$

Let the two remaining half-angle variables be

$$
q,r.
$$

Define their squares

$$
Q=q^2,
$$

$$
R=r^2.
$$

The corresponding normalized edge coordinates are

$$
x_Q=
\frac{1-Q}{1+Q},
$$

and

$$
x_R=
\frac{1-R}{1+R}.
$$

For the first fixed channel,

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

so the remaining radius is

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

Therefore the fixed-slice CARCE equation is

$$
\boxed{
\left(
\frac{1-Q}{1+Q}
\right)^2
+
\left(
\frac{1-R}{1+R}
\right)^2
=
\left(
\frac{1-K}{1+K}
\right)^2.
}
$$

The genus-$5$ problem is produced by demanding that both

$$
Q
$$

and

$$
R
$$

are rational squares.

---

# 3. Symmetric quotient

Define the elementary symmetric coordinates

$$
S=Q+R,
$$

and

$$
P=QR.
$$

Exact elimination gives the conic

$$
\boxed{
A(S^2+P^2+1)
-
B\,S(P+1)
-
C\,P
=
0,
}
$$

where

$$
A=K^2+6K+1,
$$

$$
B=2(K-1)^2,
$$

and

$$
C=14K^2+20K+14.
$$

Thus after quotienting:

- the signs of $q$ and $r$;
- the exchange $q\leftrightarrow r$;

the underlying fixed-slice geometry becomes genus $0$.

This is the lowest-level rational quotient of the simultaneous double-square problem.

---

# 4. Projection discriminant

Treat the conic as a quadratic equation in $S$.

Its discriminant is

$$
\Delta_S
=
64(K+1)^2
\left[
-KP^2
+
(K^2+4K+1)P
-
K
\right].
$$

Therefore introduce

$$
Z
$$

by

$$
\boxed{
Z^2
=
-KP^2
+
(K^2+4K+1)P
-
K.
}
$$

Then

$$
S
=
\frac{
B(P+1)
\pm
8(K+1)Z
}{
2A
}.
$$

The conic has the obvious rational boundary point

$$
\boxed{
(P,Z)=(K,2K).
}
$$

---

# 5. Rational parameterization of the quotient conic

Draw a rational line through the boundary point:

$$
Z
=
2K+t(P-K).
$$

The second intersection with the conic is

$$
\boxed{
P(t)
=
\frac{
K(t-2)^2+1
}{
K+t^2
}.
}
$$

Then

$$
Z(t)
=
2K+t(P(t)-K).
$$

Thus the entire symmetric quotient is rationally parameterized by one rational variable

$$
t.
$$

The remaining difficulty is entirely in lifting this genus-$0$ quotient back through the square maps.

---

# 6. The square-lift tower

A genuine Perfect Cuboid point requires

$$
Q=q^2,
\qquad
R=r^2.
$$

The symmetric quotient gives only

$$
S=Q+R,
\qquad
P=QR.
$$

The full rational lift can be reconstructed through three exact square conditions.

## Lift 1 — product square

Because

$$
P=QR=(qr)^2,
$$

we require

$$
\boxed{
P=U^2
}
$$

for some

$$
U\in\mathbb Q.
$$

After substituting into the projection conic:

$$
\boxed{
Z^2
=
-KU^4
+
(K^2+4K+1)U^2
-
K.
}
$$

This is a genus-$1$ quartic for nondegenerate $K$.

## Lift 2 — split the two squared coordinates

Because $Q,R$ are roots of

$$
X^2-SX+P=0,
$$

we require

$$
\boxed{
S^2-4P=V^2
}
$$

with

$$
V\in\mathbb Q.
$$

Then

$$
Q=\frac{S+V}{2},
$$

$$
R=\frac{S-V}{2}.
$$

## Lift 3 — one coordinate is itself a square

Finally require

$$
\boxed{
Q=W^2.
}
$$

Since

$$
QR=P=U^2,
$$

we automatically obtain

$$
R
=
\left(
\frac{U}{W}
\right)^2.
$$

Thus the three lifts are sufficient:

$$
\boxed{
P=U^2,
}
$$

$$
\boxed{
S^2-4P=V^2,
}
$$

$$
\boxed{
\frac{S+V}{2}=W^2.
}
$$

The final lifted curve is precisely the nondegenerate genus-$5$ simultaneous closure problem.

---

# 7. Specialized conic for $k=7/32$

For

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

we have

$$
K=\frac{49}{1024}.
$$

Multiplying the symmetric conic by

$$
2^{20}=1048576
$$

gives the exact integer equation

$$
\boxed{
1352033(S^2+P^2+1)
-
1901250S(P+1)
-
15717198P
=
0.
}
$$

The projection discriminant becomes

$$
\boxed{
W^2
=
-50176P^2
+
1251681P
-
50176,
}
$$

where

$$
W=1024Z.
$$

The rational boundary point is

$$
\boxed{
\left(
P,W
\right)
=
\left(
\frac{49}{1024},
98
\right).
}
$$

Equivalently, the rational parameterization can be written

$$
\boxed{
P(t)
=
\frac{
49(t-2)^2+1024
}{
49+1024t^2
}.
}
$$

---

# 8. Exact projective model of the fixed slice

For

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

the two square gates can be written without rational denominators as

$$
\boxed{
Y_1^2
=
(1024h^2+49)(49h^2+1024),
}
$$

and

$$
\boxed{
Y_2^2
=
\left[
1073(h^2+1)+1950h
\right]
\left[
1073(h^2+1)-1950h
\right].
}
$$

The second equation is exactly the first gate after the geometric involution

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

with square denominators cleared.

The normalization of this biquadratic fiber product is the genus-$5$ curve

$$
C_{7/32}.
$$

---

# 9. Rational boundary divisor

There are four degenerate $h$-locations:

$$
h=0,
$$

$$
h=1,
$$

$$
h=-1,
$$

and

$$
h=\infty.
$$

At

$$
h=0,
$$

the two square values are

$$
Y_1^2=224^2,
$$

$$
Y_2^2=1073^2.
$$

At

$$
h=\pm1,
$$

they are

$$
Y_1^2=1073^2,
$$

$$
Y_2^2=896^2.
$$

At infinity the leading coefficients again give

$$
224^2
$$

and

$$
1073^2.
$$

Each $h$-location therefore contributes four rational sign choices.

Hence the curve has an explicit rational boundary set

$$
\boxed{
\mathcal B
\subset C_{7/32}(\mathbb Q)
}
$$

with

$$
\boxed{
\#\mathcal B=16.
}
$$

All these points correspond to degenerate cuboid geometry and are excluded from the physical interval

$$
0<h<1.
$$

---

# 10. Good-reduction support

Let

$$
F(h)
=
(1024h^2+49)(49h^2+1024),
$$

and

$$
G(h)
=
\left[
1073(h^2+1)+1950h
\right]
\left[
1073(h^2+1)-1950h
\right].
$$

Exact discriminant and resultant factorization shows that every prime of potentially bad reduction lies in

$$
\boxed{
\{
2,3,5,7,13,29,37,73,18521
\}.
}
$$

Therefore

$$
11,\quad19,\quad31
$$

are good reduction primes for the fixed-slice curve.

---

# 11. Boundary-only reduction at $p=11,19,31$

Exact enumeration gives:

$$
\boxed{
\#C_{7/32}(\mathbb F_{11})=16,
}
$$

$$
\boxed{
\#C_{7/32}(\mathbb F_{19})=16,
}
$$

and

$$
\boxed{
\#C_{7/32}(\mathbb F_{31})=16.
}
$$

At each of the three primes, the only finite $h$-coordinates are

$$
\boxed{
h\equiv0,\;1,\;-1.
}
$$

Together with the four points above

$$
h=\infty,
$$

these are exactly the reductions of the sixteen known rational boundary points.

Thus:

$$
\boxed{
C_{7/32}(\mathbb F_p)
=
\operatorname{red}_p(\mathcal B)
\qquad
\text{for }
p=11,19,31.
}
$$

There are no nonboundary residue classes at any of these three good primes.

---

# 12. Immediate rational consequence

Let a nondegenerate rational candidate be written

$$
h=\frac uv
$$

with

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

For each

$$
p\in\{11,19,31\},
$$

its reduction must be one of

$$
0,\quad1,\quad-1,\quad\infty.
$$

Therefore:

$$
\boxed{
p
\mid
u\,v\,(u-v)(u+v).
}
$$

Hence every hypothetical nondegenerate rational point must satisfy

$$
\boxed{
11\cdot19\cdot31
\mid
u\,v\,(u-v)(u+v).
}
$$

This is only a necessary condition, not a contradiction.

Its importance is that every rational point is forced into one of the known boundary residue disks at all three primes.

---

# 13. Néron–Severi rank lower bound

G2C and G2D give the $\mathbb Q$-isogeny

$$
\operatorname{Jac}(C_{7/32})
\sim_{\mathbb Q}
E_f^2
\times
E_0^2
\times
E_-.
$$

On a product

$$
E^2,
$$

the divisor classes

- $E\times\{0\}$;
- $\{0\}\times E$;
- the diagonal;

give at least three independent Néron–Severi classes defined over $\mathbb Q$.

Therefore the two repeated elliptic blocks contribute at least

$$
3+3
$$

classes.

The final elliptic factor contributes at least one more polarization class.

Hence

$$
\boxed{
\rho
\left(
\operatorname{Jac}(C_{7/32})/\mathbb Q
\right)
\ge7.
}
$$

The Néron–Severi rank is invariant after tensoring with $\mathbb Q$ under $\mathbb Q$-isogeny, so the lower bound transfers from the split product to the Jacobian.

---

# 14. Quadratic Chabauty inequality

We have

$$
g=5,
$$

$$
r=5,
$$

and

$$
\rho\ge7.
$$

Therefore

$$
g+\rho-1
\ge
5+7-1
=
11.
$$

Hence

$$
\boxed{
r=5<11\le g+\rho-1.
}
$$

The Balakrishnan–Dogra quadratic-Chabauty criterion is therefore satisfied with a large margin.

Thus, although classical Chabauty fails at

$$
r=g,
$$

the fixed slice lies inside the rank range where the depth-$2$ quadratic-Chabauty locus is finite.

This is a structural theorem about the fixed slice.

It does not yet compute that finite locus.

---

# 15. Why $p=11$ is an unusually favorable quadratic-Chabauty prime

At the good prime

$$
p=11,
$$

the complete special fiber has only

$$
16
$$

rational points.

Every one is the reduction of a known rational boundary point.

Therefore a quadratic-Chabauty computation at $p=11$ needs to analyze only sixteen residue disks, each already anchored by a known rational point.

There is no unexplained nonboundary residue disk.

The same boundary-only phenomenon independently occurs at

$$
p=19
$$

and

$$
p=31.
$$

This gives three independent good primes for later Mordell–Weil and $p$-adic-height sieving.

---

# 16. Relation to the five rank-one elliptic quotients

The Jacobian splitting is not merely useful for rank accounting.

Each rational point on

$$
C_{7/32}
$$

has compatible images on five rank-one elliptic factors.

Individually, each factor has infinitely many rational points.

A point on the genus-$5$ curve exists only when the five elliptic images arise from one common value of the geometric parameter

$$
h.
$$

Thus G2E changes the question from:

> Does each quotient have rational points?

to:

> Can five rank-one quotient points satisfy the common pullback compatibility conditions?

This is the correct Mordell–Weil sieve / quadratic-Chabauty problem.

---

# 17. Recommended depth-$2$ computation

The next computational stage should choose

$$
p=11
$$

and perform:

1. a basis of the five-dimensional Mordell–Weil space from the three rank-one factors with multiplicities;
2. the Néron–Severi correspondences induced by the split Jacobian;
3. Coleman integrals of the five holomorphic differentials;
4. $p$-adic height pairings associated with independent Néron–Severi classes;
5. quadratic-Chabauty functions on each of the sixteen boundary residue disks;
6. a zero test inside every disk;
7. independent confirmation at $p=19$ or $p=31$.

The desired outcome is:

$$
\boxed{
C_{7/32}(\mathbb Q)
=
\mathcal B.
}
$$

If obtained, it would prove that the complete fixed slice

$$
k=\frac7{32}
$$

contains no nondegenerate Perfect Cuboid point.

---

# 18. What has and has not been proved

## Proved in this phase

- the fixed squared-coordinate quotient is genus $0$;
- its exact conic equation;
- its exact rational parameterization;
- the three-stage square-lift tower;
- the sixteen rational boundary points;
- the complete bad-prime support of the biquadratic model;
- boundary-only good reduction at $11,19,31$;
- the lower bound
  $$
  \rho\ge7;
  $$
- the quadratic-Chabauty rank inequality.

## Not yet proved

- that the sixteen boundary points are all rational points over $\mathbb Q$;
- that the physical interval
  $$
  0<h<1
  $$
  contains no rational simultaneous double-square point;
- that the full Perfect Cuboid problem has no solution.

The remaining task for this fixed slice is now finite $p$-adic compatibility computation, not another height-enumeration search.

---

# 19. Canonical G2E target

The next theorem target is:

$$
\boxed{
C_{7/32}(\mathbb Q)
=
\mathcal B.
}
$$

Equivalently:

$$
\boxed{
\text{the fixed slice }
k=\frac7{32}
\text{ has no nondegenerate rational double-square closure}.
}
$$

The preferred route is now:

$$
\boxed{
\text{quadratic Chabauty}
+
\text{elliptic quotient compatibility}
+
\text{Mordell–Weil sieve}.
}
$$

---

# 20. References

- Jennifer S. Balakrishnan and Netan Dogra, *Quadratic Chabauty and rational points II: Generalised height functions on Selmer varieties*.
- Jennifer S. Balakrishnan and Netan Dogra, *An effective Chabauty–Kim theorem*.
- Jennifer Balakrishnan, Netan Dogra, Jan Steffen Müller, Jan Tuitman, and Jan Vonk, *Quadratic Chabauty for modular curves: Algorithms and examples*.
- Juanita Duque-Rosero, Sachi Hashimoto, and Pim Spelier, *Geometric quadratic Chabauty and $p$-adic heights*.

---

**End of G2E.1**