# Perfect Cuboid G2D.3 — Exact Rank-One Theorem for $E_-$

## Local $2$-Selmer closure and a nontrivial Cassels pairing

**Series:** Perfect Cuboid Global Closure  
**Phase:** G2D.3  
**Fixed slice:** $k=7/32$  
**Status:** exact rank theorem  
**Date:** 2026-09-30

---

# 1. Main theorem

For the final distinct elliptic factor in

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

we prove:

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

Together with the independently certified results

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

and

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

this yields:

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

Since

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

the fixed slice satisfies

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

Thus ordinary Chabauty–Coleman, which requires

$$
r<g,
$$

does not apply directly to this slice.

---

# 2. Corrigendum to G2D.2

An earlier G2D.2 draft used a root-number parity argument to conclude

$$
\operatorname{rank}E_0=1.
$$

That proof route is superseded here.

The rank-one result for $E_0$ has a purely descent-based proof.

Use the $2$-isogenous full-$2$-torsion model

$$
\boxed{
E_0':
\qquad
Y^2
=
X
(X-448^2)
(X+1073^2).
}
$$

Its full-$2$-descent squareclass search contains

$$
128^2
=
16384
$$

raw covering pairs.

Exact local gates reduce:

$$
16384
\rightarrow
2048
\rightarrow
512
\rightarrow
128
\rightarrow
32
\rightarrow
\boxed{8}.
$$

Hence

$$
\dim_{\mathbf F_2}
\operatorname{Sel}^{(2)}(E_0'/\mathbb Q)
\le3.
$$

Because $E_0'$ has full rational $2$-torsion,

$$
\dim_{\mathbf F_2}E_0'(\mathbb Q)[2]=2.
$$

Therefore:

$$
\operatorname{rank}E_0'\le1.
$$

The explicit point

$$
\boxed{
(480704,468686400)
}
$$

lies on $E_0'$ and is non-torsion.

Thus:

$$
\boxed{
\operatorname{rank}E_0'
=
\operatorname{rank}E_0
=
1.
}
$$

No parity conjecture or root-number inference is required.

---

# 3. Integral model for $E_-$

Use:

$$
E_-:
\qquad
Y^2
=
(X-e_1)(X-e_2)(X-e_3),
$$

where:

$$
e_1=-8195043371142,
$$

$$
e_2=8256895726659,
$$

$$
e_3=-61852355517.
$$

The root differences are exact squares:

$$
e_2-e_1
=
4056099^2,
$$

$$
e_3-e_1
=
2851875^2,
$$

$$
e_2-e_3
=
2884224^2.
$$

The exact torsion group was already certified as:

$$
\boxed{
E_-(\mathbb Q)_{\rm tors}
\cong
\mathbb Z/2\mathbb Z
\times
\mathbb Z/4\mathbb Z.
}
$$

---

# 4. Full-$2$-Selmer upper bound

G2D.1 enumerated the full squareclass support:

$$
\{
-1,2,3,5,7,13,29,37,73,18521
\}.
$$

The raw covering space contains:

$$
1024^2
=
1048576
$$

candidate pairs.

Exact local conditions reduce the space to at most:

$$
\boxed{32}
$$

classes.

Therefore:

$$
\boxed{
\dim_{\mathbf F_2}
\operatorname{Sel}^{(2)}(E_-/\mathbb Q)
\le5.
}
$$

---

# 5. Five independent Selmer directions

Choose the following two torsion Kummer directions:

$$
T_2=(1,-1,-1),
$$

and:

$$
T_4=
(
1352033,
20310239726,
15022
).
$$

Choose three additional directions:

$$
\boxed{
A=(1,406,406),
}
$$

$$
\boxed{
B=(1,74,74),
}
$$

$$
\boxed{
C=(65,1,65).
}
$$

Their squareclass vectors are linearly independent over:

$$
\mathbf F_2.
$$

Hence if $A,B,C$ are genuine Selmer classes, then:

$$
\dim_{\mathbf F_2}
\operatorname{Sel}^{(2)}
\ge5.
$$

Together with the previous upper bound this will force equality.

---

# 6. The class $A$ comes from a global rational point

The point:

$$
\boxed{
P_-=
\left(
\frac{5132247118878268203}{606841},
\frac{2520702383000286338446320000}{472729139}
\right)
}
$$

satisfies:

$$
Y^2
=
(X-e_1)(X-e_2)(X-e_3)
$$

exactly.

Its Kummer image is:

$$
\boxed{
\delta(P_-)
=
(1,406,406)=A.
}
$$

This squareclass direction is independent of the torsion image.

Therefore:

$$
P_-
$$

is non-torsion and:

$$
\boxed{
\operatorname{rank}E_-(\mathbb Q)\ge1.
}
$$

---

# 7. The two remaining Selmer classes

The directions $B$ and $C$ reduce to the genus-$1$ quartics:

$$
\mathcal Q_B:
\qquad
Y^2
=
(31117T^2+401408Z^2)
(100352T^2+42599173Z^2),
$$

and:

$$
\mathcal Q_C:
\qquad
Y^2
=
(78125T^2-30074733Z^2)
(177957T^2-1953125Z^2).
$$

Their discriminants are supported exactly on:

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

For every prime outside this set, the projective quartics have smooth genus-$1$ reduction.

The Hasse bound gives a residue-field point, and Hensel lifting gives a local point.

The real loci are nonempty because the values at:

$$
(T,Z)=(0,1)
$$

are positive.

---

# 8. Explicit local certificates for $B$ and $C$

For $p=2$ one may use:

$$
\mathcal Q_B:
\quad
(T,Z)=(-100,1),
$$

and:

$$
\mathcal Q_C:
\quad
(T,Z)=(-100,1).
$$

After removing the even $2$-adic valuation, both values have odd unit:

$$
1\pmod8.
$$

Hence both are squares in:

$$
\mathbb Q_2.
$$

For $p=3$ use:

$$
\mathcal Q_B:
\quad
(T,Z)=(-95,1),
$$

and:

$$
\mathcal Q_C:
\quad
(T,Z)=(-99,1).
$$

Both quartic values have even $3$-adic valuation and unit:

$$
1\pmod3.
$$

At the remaining odd bad primes one may use the following smooth residue points:

| $p$ | $\mathcal Q_B$ | $\mathcal Q_C$ |
|---:|---:|---:|
| $5$ | $(0,2)$ | $(1,2)$ |
| $7$ | $(1,3)$ | $(0,2)$ |
| $13$ | $(0,4)$ | $(1,5)$ |
| $29$ | $(1,5)$ | $(0,1)$ |
| $37$ | $(1,16)$ | $(0,8)$ |
| $73$ | $(0,1)$ | $(0,18)$ |
| $18521$ | $(0,8669)$ | $(0,4658)$ |

Every listed $Y$ is nonzero modulo $p$, so Hensel lifting applies.

Therefore:

$$
\boxed{
B,C
\in
\operatorname{Sel}^{(2)}(E_-/\mathbb Q).
}
$$

Consequently:

$$
\boxed{
\dim_{\mathbf F_2}
\operatorname{Sel}^{(2)}(E_-/\mathbb Q)
=
5.
}
$$

---

# 9. Cassels pairing

Modulo the two torsion directions, the pure $2$-Selmer space is:

$$
\langle A,B,C\rangle.
$$

Because $A$ is represented by a global rational point, it lies in the radical of the Cassels pairing.

It remains to evaluate:

$$
\langle B,C\rangle.
$$

For the $B$-cover, use the dependent quadrics:

$$
H_1:
\quad
c^2w^2+74u_2^2-74u_3^2=0,
$$

$$
H_2:
\quad
b^2w^2-u_1^2+74u_3^2=0,
$$

$$
H_3:
\quad
-a^2w^2+u_1^2-74u_2^2=0,
$$

with:

$$
a=4056099,
\qquad
b=2851875,
\qquad
c=2884224.
$$

Choose tangent forms:

$$
L_1=u_2-u_3,
$$

$$
L_2=bw-u_1,
$$

$$
L_3=u_1-aw.
$$

For:

$$
C=(65,1,65),
$$

Cassels' split-$2$-torsion formula reduces the local pairing to:

$$
\boxed{
(L_1L_3,65)_v.
}
$$

Only a finite set of places can contribute.

---

# 10. Local pairing at $p=3$

Parametrize $H_1$ by:

$$
\tau=-87.
$$

Then the induced value of:

$$
u_1^2
$$

has:

$$
v_3(u_1^2)=2
$$

and square unit:

$$
1\pmod3.
$$

Thus:

$$
v_3(u_1)=1.
$$

For the tangent factors:

$$
v_3(L_1)=-2,
$$

and:

$$
v_3(aw)=-2.
$$

Hence:

$$
v_3(L_3)=-2
$$

and:

$$
v_3(L_1L_3)=-4.
$$

The second Hilbert-symbol argument $65$ is a $3$-adic nonsquare unit, but the first argument has even valuation.

Therefore:

$$
\boxed{
(L_1L_3,65)_3=+1.
}
$$

---

# 11. Local pairing at $p=5$

Use the local parameter:

$$
\tau=2.
$$

Modulo $5$:

$$
u_1^2\equiv4,
$$

and choose:

$$
u_1\equiv2.
$$

The tangent residues are:

$$
L_1\equiv4,
$$

$$
aw\equiv3,
$$

so:

$$
L_3\equiv4.
$$

Therefore:

$$
L_1L_3\equiv1\pmod5.
$$

Hence:

$$
\boxed{
(L_1L_3,65)_5=+1.
}
$$

---

# 12. Local pairing at $p=13$

Use:

$$
\tau=1.
$$

Modulo $13$:

$$
u_1^2\equiv9.
$$

Choose:

$$
u_1\equiv3.
$$

Then:

$$
L_1\equiv1,
$$

$$
aw\equiv8,
$$

and:

$$
L_3
\equiv
3-8
\equiv8
\pmod{13}.
$$

Thus:

$$
L_1L_3
\equiv8
\pmod{13}.
$$

But:

$$
\left(
\frac8{13}
\right)
=
-1.
$$

Since:

$$
v_{13}(65)=1,
$$

we obtain:

$$
\boxed{
(L_1L_3,65)_{13}=-1.
}
$$

---

# 13. All other local contributions

At $p=2$:

$$
65\equiv1\pmod8,
$$

so $65$ is a square in:

$$
\mathbb Q_2.
$$

At:

$$
p\in
\{
7,29,37,73,18521
\},
$$

one has:

$$
\left(
\frac{65}{p}
\right)=1.
$$

Thus $65$ is already a local square and the Hilbert symbol is trivial.

At the real place:

$$
65>0,
$$

so the contribution is trivial.

Outside the finite Cassels support set, the pairing is unramified and the local contribution is trivial.

Therefore the complete local sign product is:

$$
(+1)_\infty
(+1)_2
(+1)_3
(+1)_5
(+1)_7
(-1)_{13}
(+1)_{29}
(+1)_{37}
(+1)_{73}
(+1)_{18521}.
$$

Hence:

$$
\boxed{
\langle B,C\rangle
\neq0.
}
$$

---

# 14. Pairing matrix and Mordell–Weil rank

On the pure Selmer basis:

$$
(A,B,C),
$$

the alternating Cassels pairing has matrix:

$$
\boxed{
\begin{pmatrix}
0&0&0\\
0&0&1\\
0&1&0
\end{pmatrix}.
}
$$

Its radical is exactly:

$$
\boxed{
\langle A\rangle.
}
$$

The image of:

$$
E_-(\mathbb Q)/2E_-(\mathbb Q)
$$

inside the pure $2$-Selmer group lies in the radical.

Therefore:

$$
\operatorname{rank}E_-(\mathbb Q)\le1.
$$

But Section 6 already supplied the non-torsion point $P_-$.

Thus:

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

---

# 15. Complete fixed-slice rank closure

We now have exact ranks:

$$
\boxed{
r_f=1,
\qquad
r_0=1,
\qquad
r_-=1.
}
$$

Therefore:

$$
\operatorname{rank}
\operatorname{Jac}(C_{7/32})(\mathbb Q)
=
2r_f+2r_0+r_-.
$$

Hence:

$$
\boxed{
\operatorname{rank}
\operatorname{Jac}(C_{7/32})(\mathbb Q)
=
2+2+1
=
5.
}
$$

Since:

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

we obtain:

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

The entire G2D rank-certification stage for the current slice is therefore complete.

---

# 16. Consequence

The classical Chabauty inequality:

$$
r<g
$$

fails by equality.

So the next proof stage should not perform another rank descent.

The fixed-slice structure is unusually favorable in another sense:

$$
\operatorname{Jac}(C_{7/32})
\sim
E_f^2\times E_0^2\times E_-,
$$

and every distinct elliptic factor has rank exactly one.

The natural next target is therefore:

$$
\boxed{
\text{G2E — Elliptic Quotient Compatibility / Mordell–Weil Sieve}.
}
$$

Instead of asking whether each quotient separately has infinitely many rational points, G2E must ask whether rational points on the five rank-one elliptic factors can arise from one common point on the original genus-$5$ curve.

That compatibility condition is the next place where the Perfect Cuboid square gates may fail globally.

---

# References

1. J. W. S. Cassels, *Second descents for elliptic curves*, Journal für die reine und angewandte Mathematik 494 (1998), 101–127.
2. Tom Fisher, *On binary quartics and the Cassels–Tate pairing*, Research in Number Theory 8, 74 (2022).
3. Himanshu Shukla and Michael Stoll, *The Cassels–Tate pairing on 2-Selmer groups of elliptic curves*, 2023.

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**End of G2D.3**