# Perfect Cuboid G2D.2 — Exact Rank-One Theorem for $E_0$

## Full-$2$ descent, explicit non-torsion point, and parity closure

**Series:** Perfect Cuboid Global Closure  
**Phase:** G2D.2  
**Fixed slice:** $k=7/32$  
**Status:** exact rank theorem

---

# 1. Statement

For the elliptic factor $E_0$ arising from the fixed-$k$ genus-$5$ closure curve,

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

Since $E_+\cong E_0$ at the Jacobian level,

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

Together with G2D.1,

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

the fixed-slice Jacobian rank becomes

$$
\boxed{
\operatorname{rank}\operatorname{Jac}(C_{7/32})
=
4+\operatorname{rank}E_-(\mathbb Q).
}
$$

Thus the entire Chabauty decision now depends only on $E_-$.

---

# 2. Rational model of $E_0$

After shifting the unique rational $2$-torsion point to the origin and applying a rational scaling, $E_0$ is represented by

$$
y^2
=
x^3
-
2x^2
+
r^2x,
$$

where

$$
r
=
\frac{1352033}{950625}.
$$

The point of order $4$ is visible from

$$
-2+2r
=
\left(
\frac{896}{975}
\right)^2.
$$

The standard $2$-isogenous curve is

$$
E_0':
\quad
y^2
=
x^3
+
4x^2
+
4(1-r^2)x.
$$

Its three rational roots are

$$
0,
$$

$$
\left(
\frac{896}{975}
\right)^2,
$$

and

$$
-
\left(
\frac{2146}{975}
\right)^2.
$$

Therefore $E_0'$ has full rational $2$-torsion.

Because isogenous elliptic curves over $\mathbb Q$ have the same Mordell–Weil rank,

$$
\operatorname{rank}E_0(\mathbb Q)
=
\operatorname{rank}E_0'(\mathbb Q).
$$

---

# 3. Integral full-$2$ model

Scale by $975^2$ to obtain

$$
\boxed{
Y^2
=
X
(X-802816)
(X+4605316).
}
$$

The root differences are

$$
802816
=
2^{14}\cdot7^2,
$$

$$
4605316
=
2^2\cdot29^2\cdot37^2,
$$

and

$$
5408132
=
2^2\cdot73\cdot18521.
$$

Hence the full-$2$ squareclass support is generated by

$$
\boxed{
-1,\;
2,\;
7,\;
29,\;
37,\;
73,\;
18521.
}
$$

There are

$$
2^7=128
$$

squareclasses per Kummer coordinate, so the raw full-$2$ covering space contains

$$
128^2
=
\boxed{
16384
}
$$

candidate squareclass pairs.

---

# 4. Local full-$2$ sieve

Exact local tests give the chain

$$
16384
\xrightarrow{2^6}
3072,
$$

$$
3072
\xrightarrow{7^2}
768,
$$

$$
768
\xrightarrow{29^2}
192,
$$

$$
192
\xrightarrow{37^2}
64,
$$

$$
64
\xrightarrow{73^2}
16.
$$

Thus

$$
\boxed{
\#\operatorname{Sel}^{(2)}(E_0'/\mathbb Q)
\le16.
}
$$

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

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

Therefore

$$
\boxed{
\operatorname{rank}E_0'(\mathbb Q)\le2.
}
$$

---

# 5. Explicit non-torsion point

The integral curve contains the rational point

$$
\boxed{
P_0
=
(-3802500,\;3749491200).
}
$$

Direct substitution gives

$$
3749491200^2
=
(-3802500)
(-3802500-802816)
(-3802500+4605316).
$$

Good reduction point counts include

$$
\#E_0'(\mathbb F_{11})=16,
$$

$$
\#E_0'(\mathbb F_{13})=8,
$$

$$
\#E_0'(\mathbb F_{17})=24,
$$

$$
\#E_0'(\mathbb F_{19})=20.
$$

Hence the rational torsion order divides

$$
\gcd(16,8,24,20)=4.
$$

But the curve already has full rational $2$-torsion, which has order $4$.

Therefore

$$
\boxed{
E_0'(\mathbb Q)_{\rm tors}
\cong
(\mathbb Z/2\mathbb Z)^2.
}
$$

The point $P_0$ is not one of the three nonzero $2$-torsion points.

Hence $P_0$ is non-torsion.

Therefore

$$
\boxed{
\operatorname{rank}E_0'(\mathbb Q)\ge1.
}
$$

Combined with the full-$2$ upper bound:

$$
\boxed{
1
\le
\operatorname{rank}E_0'(\mathbb Q)
\le
2.
}
$$

---

# 6. Minimal $2$-adic model

Apply

$$
X=4x,
\qquad
Y=8y.
$$

The curve becomes

$$
\boxed{
y^2
=
x^3
+
950625x^2
-
231076335616x.
}
$$

For this integral model,

$$
v_2(c_4)=4,
$$

$$
v_2(c_6)=6,
$$

$$
v_2(\Delta)=28.
$$

After the standard common subtraction

$$
(4,6,12),
$$

the reduced valuation triple is

$$
(C_\Delta,C_6,C_4)
=
(16,0,0).
$$

The odd part of $c_6$ satisfies

$$
c_6'
\equiv1\pmod4.
$$

By the Rizzo–Halberstadt local root-number table at $2$,

$$
\boxed{
w_2(E_0')=-1.
}
$$

---

# 7. Odd bad primes

The discriminant factors as

$$
\Delta
=
2^{28}
7^4
29^4
37^4
73^2
18521^2.
$$

Thus the odd bad primes are

$$
7,\;
29,\;
37,\;
73,\;
18521.
$$

At every one of these primes,

$$
v_p(c_4)=0,
$$

so the reduction is multiplicative.

The integral root model makes splitness explicit.

## At $p=7$

The roots $0$ and $802816$ collide.

The tangent-cone coefficient is the remaining square

$$
1073^2,
$$

so the node is split.

Hence

$$
w_7=-1.
$$

## At $p=29,37$

The roots $0$ and $-4605316$ collide.

The tangent-cone coefficient differs from the square

$$
448^2
$$

only by $-1$.

Since

$$
29\equiv37\equiv1\pmod4,
$$

$-1$ is a quadratic residue at both primes.

Thus both reductions are split:

$$
w_{29}=w_{37}=-1.
$$

## At $p=73,18521$

The two nonzero roots collide because

$$
802816+4605316
=
4\cdot73\cdot18521.
$$

The remaining tangent coefficient is a rational square modulo each prime.

Hence both reductions are split:

$$
w_{73}=w_{18521}=-1.
$$

Therefore the product over the five odd bad primes is

$$
(-1)^5=-1.
$$

---

# 8. Global root number

The archimedean local root number is

$$
w_\infty=-1.
$$

Therefore

$$
W(E_0')
=
w_\infty
w_2
w_7
w_{29}
w_{37}
w_{73}
w_{18521}.
$$

Substitution gives

$$
W(E_0')
=
(-1)
(-1)
(-1)^5.
$$

Hence

$$
\boxed{
W(E_0')=-1.
}
$$

---

# 9. Parity theorem

For elliptic curves over $\mathbb Q$, the parity theorem identifies Mordell–Weil rank parity with the global root number:

$$
(-1)^{\operatorname{rank}E(\mathbb Q)}
=
W(E).
$$

Since

$$
W(E_0')=-1,
$$

the Mordell–Weil rank is odd.

But Section 5 already proved

$$
1
\le
\operatorname{rank}E_0'(\mathbb Q)
\le
2.
$$

The only odd integer in this interval is

$$
1.
$$

Therefore

$$
\boxed{
\operatorname{rank}E_0'(\mathbb Q)=1.
}
$$

Since $E_0$ and $E_0'$ are $2$-isogenous,

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

And because $E_+\cong E_0$ at the Jacobian level,

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

---

# 10. Updated Jacobian rank

G2D.1 established

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

This phase establishes

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

Therefore

$$
\operatorname{rank}\operatorname{Jac}(C_{7/32})
=
2(1)+2(1)+\operatorname{rank}E_-.
$$

Thus

$$
\boxed{
\operatorname{rank}\operatorname{Jac}(C_{7/32})
=
4+\operatorname{rank}E_-.
}
$$

The genus is

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

Hence classical Chabauty applies exactly when

$$
\boxed{
\operatorname{rank}E_-=0.
}
$$

The entire fixed-slice decision has therefore been reduced to one remaining elliptic rank.

---

# 11. References for the parity step

The $2$-adic local-root-number calculation uses the Rizzo/Halberstadt classification for elliptic curves over $\mathbb Q_2$.

The final parity step uses the parity theorem for elliptic curves over $\mathbb Q$.

These are external established theorems; all curve-specific arithmetic in this document is exact and independently reproducible.

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**End of G2D.2**