# Perfect Cuboid G2E.6 — Local Exact Verification Handoff

## Full-$2$ Descent, Global-Compatible Covering Isolation, and Final Simultaneous Quartic Target

**Series:** Perfect Cuboid Global Closure  
**Main line:** Geometry Branch G  
**Verification line:** Local Exact Verification  
**Phase:** G2E.6  
**Status:** Local-compute handoff / provisional-to-certified transition  
**Date:** 2026-09-30

---

# 0. Purpose

This document is the canonical handoff for continuing the current fixed-slice Perfect Cuboid research locally.

The web-side task is now restricted to:

- deriving exact equations;
- proving symbolic reductions;
- defining finite verification targets;
- separating certified results from provisional ones.

The local-side task is:

- recompute all heavy finite searches independently;
- perform Sage / PARI / mwrank / Magma checks when available;
- verify descent spaces;
- verify local solubility;
- certify surviving global covering classes;
- independently re-check every candidate.

The current fixed slice is:

$$
\boxed{
k=\frac7{32}.
}
$$

---

# 1. Current proof architecture

The fixed-slice problem has been reduced through the chain:

$$
\text{Perfect Cuboid fixed slice}
$$

$$
\Downarrow
$$

$$
\text{genus-5 simultaneous double-square curve}
$$

$$
\Downarrow
$$

$$
\text{genus-3 intermediate curve}
$$

$$
\Downarrow
$$

$$
\text{genus-2 bielliptic square-value curve}
$$

$$
\Downarrow
$$

$$
\text{full-}2\text{ descent coverings}
$$

$$
\Downarrow
$$

$$
\text{one or very few global-compatible coverings}
$$

$$
\Downarrow
$$

$$
\text{simultaneous quartic-square system}.
$$

The immediate local goal is to convert the remaining provisional reductions into independent exact certificates.

---

# 2. Genus-$2$ square-value curve

Let:

$$
w\in\mathbb Q.
$$

The genus-$2$ curve can be written as:

$$
\boxed{
Y^2=
(2-w^2)
\left(
w^2+\frac{2146}{975}w+\frac{1352033}{950625}
\right)
\left(
w^2-\frac{2146}{975}w+\frac{1352033}{950625}
\right).
}
$$

Equivalently, after setting:

$$
w=\frac uv,
\qquad
\gcd(u,v)=1,
$$

define:

$$
F_1(u,v)
=
2v^2-u^2,
$$

$$
F_2(u,v)
=
950625u^2
+
2092350uv
+
1352033v^2,
$$

$$
F_3(u,v)
=
950625u^2
-
2092350uv
+
1352033v^2.
$$

Then a rational point requires:

$$
\boxed{
F_1F_2F_3
\in
\mathbb Z^2.
}
$$

---

# 3. Full-$2$ descent formulation

Write:

$$
F_1=d_1X_1^2,
$$

$$
F_2=d_2X_2^2,
$$

$$
F_3=d_3X_3^2,
$$

where:

$$
d_1,d_2,d_3
$$

are squarefree squareclass representatives.

Because:

$$
F_1F_2F_3
$$

is a square, one must have:

$$
d_1d_2d_3
\in
\mathbb Q^{\times2}.
$$

The squareclass support is expected to lie in:

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

This support comes from discriminants and pairwise resultants of the three quadratic factors.

---

# 4. First local-side verification target

Recompute the full-$2$ descent independently.

Do **not** assume the previous web-side provisional count.

The local program should:

1. enumerate all squareclass pairs supported on $S$;
2. impose
   $$
   d_3=d_1d_2
   $$
   modulo squares;
3. test local solubility at:
   $$
   p=\infty
   $$
   and every prime in $S$;
4. use increasing powers:
   $$
   p,\ p^2,\ p^3,\ldots
   $$
   as needed;
5. record exact surviving squareclass triples.

The provisional web-side expectation was:

$$
1048576
\longrightarrow
8
$$

positive locally admissible classes.

This number must be independently recomputed.

---

# 5. Important provisional structural observation

In the timed-out web-side run, all surviving positive classes appeared to satisfy:

$$
\boxed{
d_1=1.
}
$$

If independently confirmed, then every rational point must satisfy:

$$
\boxed{
2-w^2\in\mathbb Q^2.
}
$$

This is especially important because:

$$
x=2-w^2
$$

is the elliptic quotient coordinate used in G2E.5.

Thus the genus-$2$ full-$2$ descent would automatically force the additional genus-$3$ lift gate.

This statement is **not yet promoted to theorem** until local recomputation confirms the full descent.

---

# 6. Known boundary descent class

The known rational boundary points:

$$
w=\pm1
$$

give:

$$
F_1=1,
$$

and the two norm factors have squareclass:

$$
1073.
$$

Therefore the boundary class is:

$$
\boxed{
(d_1,d_2,d_3)
=
(1,1073,1073).
}
$$

This class is globally soluble.

---

# 7. Important competing provisional class

A second squareclass observed during the timed-out analysis was:

$$
\boxed{
d
=
1073\cdot1352033.
}
$$

Numerically:

$$
\boxed{
d=1450731409.
}
$$

The previously known clean primes:

$$
11,\ 59,\ 181,\ 197,\ 1669
$$

do not automatically distinguish:

$$
1073
$$

from:

$$
1073\cdot1352033
$$

via the simple quadratic-character test used in the provisional analysis.

Therefore the local computation should **not** assume the boundary class is already unique.

---

# 8. Global-compatible covering isolation

Once the locally soluble descent classes are recomputed, apply the corrected dual-quotient Mordell–Weil compatibility sieve.

The corrected Sylow-$2$ projection rule is:

if:

$$
\#E(\mathbb F_p)
=
2^s m,
\qquad
m\text{ odd},
$$

then both the candidate point and the known Mordell–Weil subgroup must be projected by the same map:

$$
[m].
$$

The correct membership test is:

$$
\boxed{
[m]P_{\rm candidate}
\in
\left\langle
[m]G,\,
[m]E(\mathbb Q)_{\rm tors}
\right\rangle.
}
$$

Do not project only the subgroup.

---

# 9. Previously revalidated clean primes

After correcting the projection implementation, the following primes remained valid clean primes:

$$
\boxed{
11,\ 59,\ 181,\ 197,\ 1669.
}
$$

At each of these primes, simultaneous compatibility of the two elliptic quotients forces the genus-$3$ parameter into boundary residue classes.

These primes may be reused as independent local filters, but all covering-class conclusions must be recomputed from the actual local squareclass image.

---

# 10. Recommended clean-prime search

For every surviving nonboundary descent class:

$$
(1,d,d),
$$

search for a corrected clean prime $p$ satisfying:

1. both elliptic quotient maps are defined in the relevant residue class;
2. the dual-quotient Mordell–Weil compatibility leaves only boundary residues;
3. the relative squareclass is detected by:
   $$
   \boxed{
   \left(
   \frac{d/1073}{p}
   \right)
   =
   -1.
   }
   $$

If such a prime exists, that covering class cannot contain a global rational point compatible with the boundary Mordell–Weil images.

Repeat until every nonboundary covering is eliminated or explicitly survives.

---

# 11. Desired theorem target for the descent stage

The strongest desired result is:

$$
\boxed{
\text{The only globally compatible descent class is }
(1,1073,1073).
}
$$

This must **not** be claimed until all other locally soluble classes have been independently eliminated.

If a second class survives globally, continue with that class separately.

---

# 12. Boundary-class parameterization

Assume the class:

$$
\boxed{
(1,1073,1073)
}
$$

is under study.

Then:

$$
F_2
=
1073A^2,
$$

$$
F_3
=
1073B^2,
$$

and:

$$
F_1
=
C^2.
$$

Start with:

$$
F_2
=
1073A^2.
$$

The conic has the rational point:

$$
\boxed{
(w,A)=(1,64).
}
$$

Use the line:

$$
A
=
64
+
m(w-1).
$$

The second intersection gives the exact rational parameterization:

$$
\boxed{
w(m)
=
\frac{
1073m^2
-
137344m
+
3042975
}{
1073m^2
-
950625
}.
}
$$

This formula should be independently symbolically verified.

---

# 13. Remaining $F_3$ square gate

Substituting:

$$
w=w(m)
$$

into:

$$
F_3
$$

gives:

$$
\frac{F_3}{1073}
=
\frac{
4A_4(m)
}{
(1073m^2-950625)^2
}.
$$

Therefore:

$$
F_3=1073B^2
$$

is equivalent to:

$$
\boxed{
A_4(m)\in\mathbb Q^2,
}
$$

where:

$$
\boxed{
A_4(m)
=
56415121m^4
+
6561609600m^3
+
3887248218750m^2
}
$$

$$
\boxed{
\qquad
-
248783886000000m
+
3745786306640625.
}
$$

---

# 14. Remaining $F_1$ square gate

The first factor becomes:

$$
2-w(m)^2
=
\frac{
B_4(m)
}{
(1073m^2-950625)^2
}.
$$

Therefore:

$$
F_1=C^2
$$

is equivalent to:

$$
\boxed{
B_4(m)\in\mathbb Q^2,
}
$$

where:

$$
\boxed{
B_4(m)
=
1151329m^4
+
294740224m^3
-
29473681186m^2
}
$$

$$
\boxed{
\qquad
+
835868716800m
-
7452321069375.
}
$$

---

# 15. Final simultaneous quartic system

Under the boundary covering class, the fixed-slice problem becomes:

$$
\boxed{
m\in\mathbb Q
}
$$

such that:

$$
\boxed{
Y^2=A_4(m),
}
$$

and:

$$
\boxed{
Z^2=B_4(m).
}
$$

Explicitly:

$$
\boxed{
Y^2
=
56415121m^4
+
6561609600m^3
+
3887248218750m^2
-
248783886000000m
+
3745786306640625,
}
$$

and:

$$
\boxed{
Z^2
=
1151329m^4
+
294740224m^3
-
29473681186m^2
+
835868716800m
-
7452321069375.
}
$$

This is the canonical local-compute system for the next stage.

---

# 16. Known boundary solutions

The following rational values are known to satisfy both quartic-square equations and correspond to degenerate boundary geometry:

$$
\boxed{
m=25,
}
$$

$$
\boxed{
m=39,
}
$$

and:

$$
\boxed{
m=\frac{31200}{1073}.
}
$$

These should be verified independently.

For:

$$
m=25,
$$

the previous symbolic reduction gave:

$$
A_4(25)
=
8960000^2,
$$

and:

$$
B_4(25)
=
280000^2.
$$

For:

$$
m=39,
$$

the previous symbolic reduction gave:

$$
A_4(39)
=
21805056^2,
$$

and:

$$
B_4(39)
=
681408^2.
$$

For:

$$
m=\frac{31200}{1073},
$$

the previous symbolic reduction gave:

$$
A_4(m)
=
\left(
\frac{326064375}{1073}
\right)^2,
$$

and:

$$
B_4(m)
=
\left(
\frac{46580625}{1073}
\right)^2.
$$

All of these must be independently rechecked locally before publication.

---

# 17. Local-side recommended workflow

## Stage A — Descent recomputation

Recompute the full-$2$ descent on:

$$
F_1F_2F_3.
$$

Output:

- exact squareclass support;
- exact local covering equations;
- locally soluble classes;
- local witnesses;
- local obstructions.

## Stage B — Global covering elimination

For every surviving class:

- apply corrected dual-quotient Mordell–Weil compatibility;
- search for clean primes;
- record Legendre-symbol separation;
- certify whether the class is globally impossible or survives.

## Stage C — Unique covering reduction

If only:

$$
(1,1073,1073)
$$

survives, independently verify the rational parameterization:

$$
w(m).
$$

## Stage D — Simultaneous quartic analysis

Study:

$$
Y^2=A_4(m),
$$

$$
Z^2=B_4(m).
$$

Recommended tools:

- Sage `HyperellipticCurve`;
- Magma `RankBounds`, `TwoCoverDescent`, `Chabauty`;
- PARI/GP;
- explicit elliptic Jacobian conversion;
- resultants;
- local solubility;
- Mordell–Weil sieve;
- elliptic Chabauty;
- quadratic Chabauty if needed.

## Stage E — Certificate generation

If all rational simultaneous solutions are boundary:

$$
\boxed{
\text{the fixed slice }
k=\frac7{32}
\text{ contains no nondegenerate Perfect Cuboid point}.
}
$$

If a nonboundary rational solution is found, reconstruct:

$$
w,
$$

then the genus-$2$ point, genus-$3$ point, full double-square state, and finally the Perfect Cuboid rational/integer witness.

---

# 18. What is certified vs provisional

## Already certified before this handoff

- CARCE equivalence;
- nested-X double-square reduction;
- fixed-$k$ genus-$5$ structure;
- elliptic rank:
  $$
  r_f=r_0=r_-=1;
  $$
- corrected dual-quotient clean primes:
  $$
  11,59,181,197,1669;
  $$
- exact finite-height genus-$3$ exclusion through denominator:
  $$
  20000.
  $$

## Provisional and must be reverified locally

- full-$2$ genus-$2$ descent survivor count:
  $$
  8;
  $$
- all surviving positive classes having:
  $$
  d_1=1;
  $$
- global uniqueness of:
  $$
  (1,1073,1073);
  $$
- the explicit quartic coefficients $A_4,B_4$;
- the listed rational boundary $m$-solutions.

No provisional item should be promoted without an independent exact rerun.

---

# 19. Canonical local theorem target

The preferred final local theorem is:

> For the fixed slice $k=7/32$, every rational point on the genus-$2$ square-value curve belongs to the unique descent class $(1,1073,1073)$, and every rational solution of the resulting simultaneous quartic system is degenerate.

Symbolically:

$$
\boxed{
Y^2=A_4(m),
\qquad
Z^2=B_4(m)
}
$$

has only the known boundary rational solutions.

If proved, then:

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

is globally excluded as a nondegenerate Perfect Cuboid first-channel slice.

---

# 20. One-line handoff

$$
\boxed{
\text{Recompute full }2\text{-descent}
\rightarrow
\text{isolate global covering}
\rightarrow
\text{solve }
Y^2=A_4(m),\;
Z^2=B_4(m)
\text{ over }\mathbb Q.
}
$$

---

**End of G2E.6 Local Verification Handoff**