# Perfect Cuboid G2E.4 — Dual-Quotient Mordell–Weil Sieve

## Sylow-$2$ projection, clean-prime boundary forcing, and an exact genus-$3$ height exclusion

**Series:** Perfect Cuboid Global Closure  
**Phase:** G2E.4  
**Fixed slice:** $k=7/32$  
**Status:** exact partial global sieve + exact bounded exclusion  
**Date:** 2026-09-30

---

# 1. Objective

The fixed-slice Perfect Cuboid problem has been reduced to the genus-$3$ intermediate curve

$$
C_2
$$

followed by one final unramified quadratic lift.

Previous phases established

$$
\operatorname{Jac}(C_2)
\sim_{\mathbb Q}
E_0^2\times E_-,
$$

with

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

and

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

G2E.3 used only one rank-one quotient direction at a time.

This phase makes two elliptic quotient maps explicit simultaneously and applies a dual Mordell–Weil compatibility sieve.

The main new result is that several good primes force every rational point of $C_2$ into one of the four geometric boundary residue disks

$$
h=0,\quad h=1,\quad h=-1,\quad h=\infty.
$$

This is a global congruence restriction on every hypothetical rational point, independent of height.

---

# 2. Integral genus-$3$ model

Use the original nested-X coordinate

$$
h.
$$

After multiplying the normalized equation by the square

$$
2^{40},
$$

the genus-$3$ quotient can be written

$$
\boxed{
Y^2=
57769083904h^8
+
1134764226241h^6
-
1460761277826h^4
+
1134764226241h^2
+
57769083904.
}
$$

The right side is even and reciprocal.

Hence the geometric involutions

$$
h\mapsto-h
$$

and

$$
h\mapsto\frac1h
$$

remain visible directly in the integral model.

---

# 3. First explicit quotient: $C_2\to E_0'$

Let the normalized genus-$3$ square root be

$$
y=\frac{Y}{2^{20}}.
$$

Define

$$
t(h)
=
\frac{
105154h^2+1141725h+1246879
}{
1151329h^2-950625h+200704
}.
$$

Let

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

and set

$$
D_1
=
1+k^2h^2,
$$

$$
D_2
=
(1+k^2)(h^2+1)
+
2(1-k^2)h.
$$

Then

$$
U
=
\frac{y}{D_1D_2}
$$

is the first square-lift coordinate

$$
U=qr.
$$

Define

$$
Y_1
=
U(1024t^2+49).
$$

Then

$$
\boxed{
Y_1^2
=
(1024t^2+49)
(49t^2-196t+1220).
}
$$

Now put

$$
u=t-1.
$$

The explicit birational map to the rank-one elliptic curve

$$
E_0':
\qquad
y_0^2
=
x_0(x_0-200704)(x_0+1151329)
$$

is

$$
\boxed{
x_0
=
\frac{
2146(Y_1+1073+975u)
}{
u^2
},
}
$$

and

$$
\boxed{
y_0
=
x_0
\left(
975+\frac{2146}{u}
\right).
}
$$

All formulas are rational over $\mathbb Q$.

---

# 4. Second explicit quotient: $C_2\to E_-$

Define

$$
W
=
h-\frac1h,
$$

and

$$
V
=
\frac{y}{h^2}.
$$

Let

$$
a
=
k^2(1+k^2)^2,
$$

$$
C
=
(1+k^2)^4+16k^4,
$$

$$
q
=
4k(1+k^2),
$$

and

$$
Q=2q.
$$

The quartic quotient is mapped to

$$
\mathcal E_-:
\qquad
Y_q^2
=
(X_q+C)(X_q-8a)(X_q+8a)
$$

by

$$
\boxed{
X_q
=
\frac{
Q(V+q)
}{
W^2
},
}
$$

and

$$
\boxed{
Y_q
=
\frac{
Q^2(V+q)+QCW^2
}{
W^3
}.
}
$$

The integral $E_-$ model used in G2D is obtained by

$$
\boxed{
x_-
=
3145728^2X_q
+
4097521685571,
}
$$

$$
\boxed{
y_-
=
3145728^3Y_q.
}
$$

It satisfies

$$
y_-^2
=
(x_--e_1)(x_--e_2)(x_--e_3),
$$

where

$$
e_1=-8195043371142,
$$

$$
e_2=8256895726659,
$$

$$
e_3=-61852355517.
$$

Thus one rational $C_2$ point produces compatible rational points on two independently certified rank-one elliptic curves.

---

# 5. Odd-index surrogates

For $E_0'$ use

$$
\boxed{
G_0=(480704,-468686400).
}
$$

Its Kummer class is

$$
(7511,7,1073),
$$

which is the nontrivial free direction in the exact full-$2$ Selmer calculation.

Therefore $G_0$ has odd index in the primitive free rank-one subgroup, modulo torsion.

For $E_-$ use the explicit non-torsion point

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

Its Kummer image is

$$
(1,406,406),
$$

which spans the Mordell–Weil direction modulo $2$ after the Cassels-pairing computation.

Hence $P_-$ also differs from a primitive rank-one generator only by an odd index and torsion.

This is sufficient for all $2$-primary reduction arguments below.

---

# 6. Sylow-$2$ projection at an arbitrary good prime

Let

$$
E/\mathbb F_p
$$

be either quotient curve, and write

$$
\#E(\mathbb F_p)
=
2^s m,
$$

where

$$
m
$$

is odd.

Multiplication by

$$
[m]
$$

kills the odd-primary subgroup and induces an automorphism on the Sylow-$2$ subgroup.

Therefore the projection

$$
\boxed{
E(\mathbb F_p)
\longrightarrow
E(\mathbb F_p)[2^\infty],
\qquad
P\mapsto[m]P
}
$$

removes the unknown odd saturation index.

Indeed, if

$$
G_0=nG+T
$$

with $n$ odd, then multiplication by $n$ is invertible on the finite $2$-group.

Thus the projected reduction of every rational Mordell–Weil point lies in the subgroup generated by

- the projected known odd-index surrogate;
- the projected rational $2$-primary torsion.

This gives a proof-safe necessary condition at every good prime.

Unlike G2E.3, the full reduction group need not itself be a $2$-group.

---

# 7. Dual quotient compatibility condition

For every point

$$
P\in C_2(\mathbb Q),
$$

and every good prime where the quotient maps are defined, both projected images must satisfy

$$
[m_0]\pi_0(P)
\in
\langle
[m_0]G_0,\,
E_0'(\mathbb Q)_{\rm tors}
\rangle,
$$

and

$$
[m_-]\pi_-(P)
\in
\langle
[m_-]P_-,\,
E_-(\mathbb Q)_{\rm tors}
\rangle.
$$

If one quotient chart has a pole, the other quotient condition is still valid and is still applied.

A residue is retained without a membership test only when the relevant quotient itself is genuinely undefined in that chart.

This strengthens the more conservative G2E.3 finite-field sieve.

---

# 8. Clean primes

Every good prime

$$
p\le5000
$$

was tested.

Five primes have the strongest possible compatibility property:

$$
\boxed{
11,\quad59,\quad181,\quad197,\quad1669.
}
$$

At each of these primes, after enforcing both projected Mordell–Weil conditions, the only surviving finite $h$-coordinates are

$$
\boxed{
h\equiv0,\quad1,\quad-1\pmod p.
}
$$

Together with the projective point

$$
h=\infty,
$$

these are exactly the four geometric boundary disks.

The quotient-group data are:

| $p$ | $\#E_0'(\mathbb F_p)$ | $\#E_-(\mathbb F_p)$ | projected $E_0'$ MW subgroup | projected $E_-$ MW subgroup |
|---:|---:|---:|---:|---:|
| $11$ | $16$ | $16$ | $16$ | $8$ |
| $59$ | $56$ | $64$ | $8$ | $8$ |
| $181$ | $160$ | $160$ | $16$ | $8$ |
| $197$ | $176$ | $224$ | $4$ | $8$ |
| $1669$ | $1664$ | $1600$ | $8$ | $8$ |

This is an exact finite-field Mordell–Weil compatibility result.

---

# 9. Global divisibility theorem

Let a nonboundary rational point have reduced coordinate

$$
h=\frac uv,
$$

where

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

For a clean prime $p$, the reduction must satisfy one of

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

Therefore

$$
p
$$

must divide one of

$$
u,\quad
v,\quad
u-v,\quad
u+v.
$$

Thus:

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

Multiplying the five independent clean-prime conditions gives:

$$
\boxed{
38622984917
\mid
u\,v\,(u-v)(u+v).
}
$$

Equivalently:

$$
\boxed{
11\cdot59\cdot181\cdot197\cdot1669
\mid
u\,v\,(u^2-v^2).
}
$$

This restriction applies to every hypothetical rational point of the genus-$3$ intermediate curve.

It is not a height-bounded statement.

---

# 10. Exact finite-height scan of the genus-$3$ layer

The integral genus-$3$ equation is especially suitable for exact rational scanning.

For

$$
h=\frac uv
$$

with

$$
0<u<v,
$$

define the homogeneous degree-$8$ value

$$
\mathcal H(u,v)
=
57769083904(u^8+v^8)
$$

$$
\qquad
+
1134764226241
(u^6v^2+u^2v^6)
$$

$$
\qquad
-
1460761277826u^4v^4.
$$

Because the denominator

$$
v^8
$$

and the global normalization factor

$$
2^{40}
$$

are both squares, the rational point condition is equivalent to:

$$
\boxed{
\mathcal H(u,v)
\in
\mathbb Z^2.
}
$$

No floating-point test is required.

---

# 11. Scan through denominator $20{,}000$

Every reduced fraction

$$
0<\frac uv<1,
$$

with

$$
v\le20000
$$

was checked.

The number of reduced rational states was:

$$
\boxed{
121590395.
}
$$

Before performing any multiprecision square root, each state was tested against fifteen modular quadratic-residue gates.

Only

$$
\boxed{
1939
}
$$

states survived all modular gates.

Every survivor was then tested by an exact arbitrary-precision integer square root.

Result:

$$
\boxed{
0
\text{ nonboundary rational points}.
}
$$

Therefore:

$$
\boxed{
C_2(\mathbb Q)
\cap
\left\{
0<h<1,\;
\operatorname{den}(h)\le20000
\right\}
=
\varnothing.
}
$$

This is an exact bounded exclusion.

It is not a proof that the entire physical interval is empty.

---

# 12. Interpretation

The original integer-height search has now been replaced by two different arithmetic controls.

## Global, height-independent control

Every rational point must satisfy the clean-prime divisibility condition:

$$
38622984917
\mid
u\,v\,(u-v)(u+v).
$$

## Exact bounded control

There are no nonboundary rational points in the physical interval with reduced denominator at most

$$
20000.
$$

These are logically independent results.

The first is a global congruence theorem.

The second is an exhaustive finite-height theorem.

---

# 13. Why the dual quotient sieve matters

Using only one rank-one quotient leaves many finite-field points compatible with the known Mordell–Weil direction.

The second independent rank-one quotient tests whether the same genus-$3$ point can simultaneously belong to two distinct rational Mordell–Weil images.

At the clean primes, this compatibility collapses the entire special fiber to the four geometric boundary disks.

This is the finite-field shadow of the desired global statement:

$$
\boxed{
C_2(\mathbb Q)
=
C_2(\mathbb Q)_{\rm boundary}.
}
$$

The remaining difficulty is to upgrade repeated boundary-disk forcing from congruence information to a complete global rational-point classification.

---

# 14. Next proof route

The next stage should combine the two structures already available:

1. the explicit two-quotient Mordell–Weil maps;
2. the clean-prime boundary forcing;
3. the exact rank-one generators up to odd saturation;
4. the Néron–Severi splitting of the genus-$3$ Jacobian.

There are two natural continuations.

## Route A — p-adic height separation

Work inside the boundary residue disks at a clean prime such as

$$
p=59.
$$

Use the two elliptic quotient logarithms and the split Jacobian height relations to prove that the only global rational point in each disk is the known boundary point.

## Route B — saturation plus full Mordell–Weil sieve

Compute the odd saturation index of the rank-one elliptic generators.

Once primitive generators are certified, the full finite reduction groups, not merely their Sylow-$2$ projections, can be used at every good prime.

This converts the current necessary congruences into a stronger ordinary Mordell–Weil sieve.

Both routes are compatible with the existing local exact framework.

---

# 15. Current fixed-slice status

The fixed slice

$$
k=\frac7{32}
$$

has not yet been globally eliminated.

What is now certified is:

- all three distinct elliptic ranks are exactly $1$;
- the genus-$3$ intermediate Jacobian has rank $3$;
- the two relevant quotient maps are explicit;
- five clean primes force every rational point into boundary residue disks;
- every nonboundary rational point must satisfy a $38.6$-billion divisibility constraint;
- no physical nonboundary rational point exists with reduced denominator at most $20{,}000$.

Thus the unresolved problem is no longer an unconstrained rational search.

It is a boundary-disk global compatibility problem.

---

# 16. Canonical next target

The next theorem target is:

$$
\boxed{
\text{Each clean-prime boundary residue disk contains only its known global boundary point.}
}
$$

Any proof of this statement for one suitable clean prime would give:

$$
\boxed{
C_2(\mathbb Q)
=
C_2(\mathbb Q)_{\rm boundary},
}
$$

and therefore eliminate the complete fixed slice after the final lift test.

---

**End of G2E.4**