# Perfect Cuboid Radical-FDRS R8

## Exact $E_0'$ Synchronization, the Biquadratic Cover Lattice, and the Double Square-Value Reduction

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R8 — Shared-$E_0'$ Synchronization  
**Date:** 2026-10-04  
**Status:** Exact structural reduction for the historical $n=7/32$ fiber — not a proof of Perfect Cuboid existence or nonexistence

---

# 0. Executive summary

R7 produced a new genus-$2$ quotient of the historical fixed fiber

$$
n=\frac7{32}
$$

and mapped it explicitly to the full-$2$-torsion elliptic curve

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

The older G2E.4 route had independently mapped the historical genus-$3$ intermediate curve to the same elliptic curve.

R8 compares the two maps on the same parent point.

The result is exact:

$$
\boxed{
X_{\rm new}
=
X_{\rm old},
}
$$

and

$$
\boxed{
Y_{\rm new}
=
-
Y_{\rm old}.
}
$$

Therefore the synchronization correspondence on $E_0'$ is simply elliptic negation:

$$
\boxed{
P_{\rm new}
=
[-1]P_{\rm old}.
}
$$

There is no hidden translation, no unknown torsion offset, and no higher-degree correspondence.

This synchronization has a group-theoretic explanation.

Let

$$
C
$$

be the historical genus-$5$ fixed fiber.

Let

$$
\delta
$$

be the central deck involution changing both square-root signs.

Let

$$
\alpha
$$

be the R6 gate-swap involution.

They commute.

The quotient lattice is:

$$
\boxed{
C/\langle\delta\rangle
=
H,
}
$$

where

$$
g(H)=3,
$$

and:

$$
\boxed{
C/\langle\alpha\rangle
=
G,
}
$$

where

$$
g(G)=2.
$$

The common quotient is:

$$
\boxed{
C/\langle\alpha,\delta\rangle
=
E_0',
}
$$

with genus:

$$
1.
$$

Consequently:

$$
\boxed{
C
\simeq
\operatorname{Norm}
\left(
H\times_{E_0'}G
\right).
}
$$

Thus the original genus-$5$ curve is the normalization of a genus-$3$/genus-$2$ fiber product over one elliptic curve.

R8 then derives the two corresponding quadratic lift functions directly on $E_0'$.

Define:

$$
A=448^2,
$$

$$
B=1073^2,
$$

$$
M=B-A=975^2,
$$

and:

$$
L=A+B=1352033.
$$

Then:

$$
E_0':
\qquad
Y^2=X(X-A)(X+B).
$$

The new genus-$2$ lift is controlled by:

$$
\boxed{
f_G(X)
=
\frac{2X+M}{L}.
}
$$

The complementary quadratic extension may be written using:

$$
\boxed{
f_K(X)
=
L\,X(MX-2AB).
}
$$

The complete historical genus-$5$ rational-point problem therefore reduces to:

$$
\boxed{
P=(X,Y)\in E_0'(\mathbb Q),
}
$$

such that:

$$
\boxed{
f_G(P)\in\mathbb Q^2,
}
$$

and:

$$
\boxed{
f_K(P)\in\mathbb Q^2.
}
$$

Since:

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

the historical fixed-slice problem is now a **double square-value problem on one rank-one elliptic Mordell-Weil group**.

This is the principal R8 reduction.

---

# 1. Historical full-$2$-torsion elliptic core

Use:

$$
\boxed{
A=448^2=200704,
}
$$

$$
\boxed{
B=1073^2=1151329.
}
$$

Define:

$$
\boxed{
M=B-A=950625=975^2,
}
$$

and:

$$
\boxed{
L=A+B=1352033.
}
$$

The shared elliptic curve is:

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

The historical rank theorem established:

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

Its rational torsion subgroup is:

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

---

# 2. Historical genus-$3$ product model

At:

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

the R4 product curve is:

$$
H:
\qquad
q^2
=
F_1(m)F_2(m).
$$

Multiplying by the square:

$$
2^{40}
$$

and defining:

$$
\boxed{
Q=2^{20}q,
}
$$

gives exactly the historical integral genus-$3$ model:

$$
\boxed{
Q^2
=
57769083904m^8
+
1134764226241m^6
}
$$

$$
\boxed{
\qquad
-
1460761277826m^4
+
1134764226241m^2
+
57769083904.
}
$$

Therefore the old nested-X coordinate:

$$
h
$$

and the Radical-FDRS coordinate:

$$
m
$$

are literally identical on this model:

$$
\boxed{
h=m.
}
$$

No further birational change is required at the genus-$3$ level.

---

# 3. Historical G2E.4 auxiliary coordinate

The old elliptic quotient used:

$$
\boxed{
t_{\rm old}(m)
=
\frac{
105154m^2
+
1141725m
+
1246879
}{
1151329m^2
-
950625m
+
200704
}.
}
$$

Define the R6 gate-swap invariant:

$$
\boxed{
\xi
=
m+\mu(m),
}
$$

where:

$$
\mu(m)
=
\frac{1-m}{1+m}.
$$

Then:

$$
\boxed{
\xi
=
\frac{m^2+1}{m+1}.
}
$$

Using:

$$
m^2
=
\xi(m+1)-1,
$$

the historical coordinate collapses to:

$$
\boxed{
t_{\rm old}
=
\frac{
105154\xi+1141725
}{
1151329\xi-950625
}.
}
$$

Thus the old auxiliary variable is already a rational function of the R6 gate-swap quotient coordinate.

---

# 4. Exact inverse relation

Solving for:

$$
\xi
$$

gives:

$$
\boxed{
\xi
=
\frac{
950625t_{\rm old}
+
1141725
}{
1151329t_{\rm old}
-
105154
}.
}
$$

The historical inverse elliptic map was:

$$
\boxed{
t_{\rm old}
=
1
+
\frac{
2146X
}{
Y-975X
}.
}
$$

Substituting this into the previous relation gives the simple identity:

$$
\boxed{
\xi
=
\frac{
1950Y
}{
1352033X+975Y
}.
}
$$

Therefore the gate-swap invariant has an explicit rational expression on:

$$
E_0'.
$$

---

# 5. The old genus-$3$ lift condition

The two preimages:

$$
m,
\qquad
\mu(m)
$$

are the roots of:

$$
\boxed{
T^2-\xi T+(1-\xi)=0.
}
$$

Hence an elliptic point lifts from:

$$
E_0'
$$

to the genus-$3$ product layer if and only if the discriminant:

$$
\boxed{
\Delta_H
=
\xi^2+4\xi-4
}
$$

is a rational square.

Substitute:

$$
\xi
=
\frac{
1950Y
}{
LX+975Y
}.
$$

Then:

$$
\boxed{
\Delta_H
=
\frac{
4X(2X+M)(MX-2AB)
}{
(LX+975Y)^2
}.
}
$$

Therefore the squareclass of the historical genus-$3$ lift is:

$$
\boxed{
[\Delta_H]
=
\left[
X(2X+M)(MX-2AB)
\right].
}
$$

---

# 6. Historical genus-$3$ square function

Define:

$$
\boxed{
f_H(P)
=
X(2X+M)(MX-2AB),
}
$$

for:

$$
P=(X,Y)\in E_0'(\mathbb Q).
$$

Away from the finite exceptional denominator locus:

$$
\boxed{
P
\text{ lifts to }H(\mathbb Q)
\iff
f_H(P)\in\mathbb Q^2.
}
$$

The exceptional points are handled by projective normalization.

Thus the complicated historical genus-$3$ map is controlled by one cubic square-value function on:

$$
E_0'.
$$

---

# 7. The new R7 genus-$2$ lift condition

The R7 hyperelliptic quotient specialized to:

$$
n=\frac7{32}
$$

was:

$$
\boxed{
Y_G^2
=
2(T^4-1)
(1352033T^2-950625).
}
$$

Its elliptic quotient map was:

$$
\boxed{
X
=
\frac{
1352033T^2-950625
}{2}.
}
$$

Therefore:

$$
\boxed{
T^2
=
\frac{
2X+950625
}{
1352033
}.
}
$$

Hence define:

$$
\boxed{
f_G(P)
=
\frac{
2X+M
}{
L
}.
}
$$

Then:

$$
\boxed{
P
\text{ lifts to }G(\mathbb Q)
\iff
f_G(P)\in\mathbb Q^2.
}
$$

This is the new genus-$2$ square gate.

---

# 8. The third quadratic character

Because:

$$
f_H
=
X(2X+M)(MX-2AB),
$$

and:

$$
f_G
=
\frac{2X+M}{L},
$$

their quotient is:

$$
\boxed{
\frac{f_H}{f_G}
=
L\,X(MX-2AB).
}
$$

Define:

$$
\boxed{
f_K(P)
=
L\,X(MX-2AB).
}
$$

Then:

$$
\boxed{
[f_H]
=
[f_Gf_K].
}
$$

Thus the three nontrivial quadratic characters of the biquadratic extension are represented by:

$$
\boxed{
f_G,
\qquad
f_K,
\qquad
f_H=f_Gf_K
}
$$

in:

$$
\mathbb Q(E_0')^\times/
\mathbb Q(E_0')^{\times2}.
$$

---

# 9. The V4 cover lattice

Let:

$$
C
$$

be the genus-$5$ fixed fiber.

Let:

$$
\delta
$$

be the central simultaneous square-root sign involution.

Let:

$$
\alpha
$$

be the R6 gate-swap involution.

They satisfy:

$$
\boxed{
\delta^2
=
\alpha^2
=
1,
}
$$

and:

$$
\boxed{
\delta\alpha
=
\alpha\delta.
}
$$

Therefore:

$$
\boxed{
\langle\delta,\alpha\rangle
\cong
(\mathbb Z/2\mathbb Z)^2.
}
$$

The quotient lattice is:

$$
\boxed{
H
=
C/\langle\delta\rangle,
}
$$

$$
\boxed{
G
=
C/\langle\alpha\rangle,
}
$$

$$
\boxed{
K
=
C/\langle\alpha\delta\rangle,
}
$$

and:

$$
\boxed{
E
=
C/\langle\delta,\alpha\rangle.
}
$$

Here:

$$
\boxed{
E\cong E_0'.
}
$$

---

# 10. Genera of the quotient lattice

The known genera are:

$$
\boxed{
g(C)=5,
}
$$

$$
\boxed{
g(H)=3,
}
$$

$$
\boxed{
g(G)=2,
}
$$

and:

$$
\boxed{
g(E)=1.
}
$$

The V4 dimension relation gives:

$$
g(C)+2g(E)
=
g(H)+g(G)+g(K).
$$

Therefore:

$$
5+2
=
3+2+g(K).
$$

Hence:

$$
\boxed{
g(K)=2.
}
$$

Thus the three nontrivial intermediate quotients have genera:

$$
\boxed{
3,
\quad
2,
\quad
2.
}
$$

---

# 11. Function-field realization

Let:

$$
\mathbb Q(E)
$$

be the function field of:

$$
E_0'.
$$

Then the three intermediate quadratic extensions are:

$$
\boxed{
\mathbb Q(G)
=
\mathbb Q(E)
\left(
\sqrt{f_G}
\right),
}
$$

$$
\boxed{
\mathbb Q(K)
=
\mathbb Q(E)
\left(
\sqrt{f_K}
\right),
}
$$

and:

$$
\boxed{
\mathbb Q(H)
=
\mathbb Q(E)
\left(
\sqrt{f_Gf_K}
\right).
}
$$

The full genus-$5$ function field is:

$$
\boxed{
\mathbb Q(C)
=
\mathbb Q(E)
\left(
\sqrt{f_G},
\sqrt{f_K}
\right).
}
$$

This is the canonical biquadratic description.

---

# 12. Fiber-product theorem

Because:

$$
\mathbb Q(C)
$$

is the compositum of:

$$
\mathbb Q(H)
$$

and:

$$
\mathbb Q(G)
$$

over:

$$
\mathbb Q(E),
$$

we obtain:

## Theorem — Shared-$E_0'$ fiber product

$$
\boxed{
C
\simeq
\operatorname{Norm}
\left(
H
\times_{E_0'}
G
\right).
}
$$

Equivalently, the genus-$5$ fixed fiber is the normalization of the genus-$3$/genus-$2$ fiber product over:

$$
E_0'.
$$

This is an exact quotient-lattice statement.

---

# 13. Branch structure of the new genus-$2$ cover

The new cover:

$$
G\to E_0'
$$

is generated by:

$$
\sqrt{f_G}.
$$

Its only non-even zero divisor comes from:

$$
2X+M=0.
$$

Thus the branch $X$-value is:

$$
\boxed{
X=-\frac M2.
}
$$

There are two geometric points of:

$$
E_0'
$$

above this $X$-value.

Hence the ramification count is:

$$
\boxed{
r_G=2.
}
$$

For a degree-$2$ cover of a genus-$1$ curve:

$$
g(G)
=
1+\frac{r_G}{2}
=
2.
$$

This recovers the R7 genus.

---

# 14. Branch structure of the third genus-$2$ cover

The third character is:

$$
f_K
=
L\,X(MX-2AB).
$$

The divisor of:

$$
X
$$

is even because:

$$
X=0
$$

is a rational $2$-torsion point of:

$$
E_0'.
$$

Therefore this factor produces no odd ramification.

The nontrivial branch value is:

$$
\boxed{
X
=
\frac{
2AB
}{
M
}.
}
$$

Again there are two geometric points above this $X$-value.

Hence:

$$
\boxed{
r_K=2,
}
$$

and:

$$
\boxed{
g(K)=2.
}
$$

---

# 15. Branch structure of the genus-$3$ product layer

The historical product character is:

$$
f_H
=
f_Gf_K.
$$

Its odd branch divisor contains both branch pairs:

$$
X=-\frac M2,
$$

and:

$$
X=\frac{2AB}{M}.
$$

Thus:

$$
\boxed{
r_H=4.
}
$$

Therefore:

$$
g(H)
=
1+\frac{4}{2}
=
\boxed{
3.
}
$$

The three quotient genera are now derived directly from the two elliptic square-value functions.

---

# 16. Why the genus-$5$ compositum has genus $5$

The two quadratic covers do not have disjoint branch architecture in the naive projective-line sense.

They are covers of the elliptic base:

$$
E_0'.
$$

The V4 cover is determined by the two independent characters:

$$
[f_G],
$$

and:

$$
[f_K].
$$

The quotient genus relation gives:

$$
\boxed{
5+2
=
2+2+3.
}
$$

Therefore:

$$
\boxed{
g(C)=5.
}
$$

This rederives the fixed-fiber genus without starting from the simultaneous quartic model.

---

# 17. Exact synchronization of old and new elliptic maps

Let:

$$
P_C
=
(m,Y_1,Y_2)
$$

be a point on the parent genus-$5$ curve.

Define:

$$
q=Y_1Y_2.
$$

Let:

$$
Q=2^{20}q.
$$

Then:

$$
Q
$$

is the historical integral genus-$3$ square-root coordinate.

Define:

$$
\boxed{
D(m)
=
m^2-2m-1.
}
$$

Define:

$$
\boxed{
S(m,Q)
=
AB(m^2+1)^2
+
M^2m(m^2-1)
+
LQ.
}
$$

Set:

$$
\boxed{
X_\ast
=
\frac{
2S(m,Q)
}{
M D(m)^2
}.
}
$$

Also define:

$$
\boxed{
Y_\ast
=
\frac{
L(m^2+1)
}{
975D(m)
}
X_\ast.
}
$$

Then direct substitution gives:

$$
\boxed{
Y_\ast^2
=
X_\ast
(X_\ast-A)
(X_\ast+B).
}
$$

Thus:

$$
(X_\ast,Y_\ast)
\in
E_0'.
$$

---

# 18. New R7 route

The R7 genus-$2$ quotient followed by its elliptic quotient sends the parent point to:

$$
\boxed{
P_{\rm new}
=
(X_\ast,Y_\ast).
}
$$

All dependence on the individual signs:

$$
Y_1,
\qquad
Y_2
$$

enters only through the product:

$$
Q.
$$

---

# 19. Historical G2E.4 route

The old genus-$3$ elliptic quotient formulas send the same parent point to:

$$
\boxed{
P_{\rm old}
=
(X_\ast,-Y_\ast).
}
$$

Therefore:

$$
\boxed{
X_{\rm new}=X_{\rm old},
}
$$

and:

$$
\boxed{
Y_{\rm new}=-Y_{\rm old}.
}
$$

Hence:

$$
\boxed{
P_{\rm new}
=
[-1]P_{\rm old}.
}
$$

This identity is exact in the common function field.

No use of numerical approximation is involved.

---

# 20. Symbolic verification form

If:

$$
Q^2
=
57769083904m^8
+
1134764226241m^6
-
1460761277826m^4
+
1134764226241m^2
+
57769083904,
$$

then the residual:

$$
Y_\ast^2
-
X_\ast(X_\ast-A)(X_\ast+B)
$$

vanishes identically.

Likewise, substituting the full old G2E.4 map and the full new R7 map gives:

$$
\boxed{
X_{\rm R7}
-
X_{\rm G2E.4}
=
0,
}
$$

and:

$$
\boxed{
Y_{\rm R7}
+
Y_{\rm G2E.4}
=
0.
}
$$

These are rational-function identities.

---

# 21. Elliptic-group interpretation

The negation map on:

$$
E_0'
$$

is:

$$
\boxed{
[-1](X,Y)
=
(X,-Y).
}
$$

Therefore the complete old/new synchronization is the group automorphism:

$$
\boxed{
\Theta=[-1].
}
$$

The R7 question from the previous phase:

> Is the correspondence sign, torsion translation, multiplication, or a higher-degree relation?

is now answered:

$$
\boxed{
\text{It is exactly sign.}
}
$$

---

# 22. Compatibility after the $2$-isogeny

Let:

$$
\phi:
E_0'
\longrightarrow
E_0
$$

be the historical degree-$2$ isogeny.

Because:

$$
\phi
$$

is a group homomorphism:

$$
\phi(-P)
=
-\phi(P).
$$

Therefore on the normalized historical:

$$
E_0
$$

curve the two genus-$2$/genus-$3$ routes also differ only by negation.

Thus no hidden synchronization ambiguity appears after the isogeny.

---

# 23. Canonical double square-value formulation

The full rational point problem on the historical fixed fiber is equivalent, away from the finite boundary loci, to:

$$
\boxed{
P=(X,Y)
\in
E_0'(\mathbb Q)
}
$$

such that:

$$
\boxed{
\frac{2X+M}{L}
\in
\mathbb Q^2,
}
$$

and:

$$
\boxed{
L\,X(MX-2AB)
\in
\mathbb Q^2.
}
$$

Explicitly:

$$
\boxed{
\frac{
2X+950625
}{
1352033
}
\in
\mathbb Q^2,
}
$$

and:

$$
\boxed{
1352033
\,X
\left(
950625X
-
462152671232
\right)
\in
\mathbb Q^2.
}
$$

This is the R8 canonical fixed-slice target.

---

# 24. Why these two conditions are sufficient

Suppose:

$$
P\in E_0'(\mathbb Q)
$$

satisfies:

$$
f_G(P)\in\mathbb Q^2,
$$

and:

$$
f_K(P)\in\mathbb Q^2.
$$

Then:

$$
P
$$

has a rational lift to:

$$
G,
$$

and a rational lift to:

$$
K.
$$

Their product character satisfies:

$$
f_Gf_K=f_H
$$

in squareclass.

Hence the same:

$$
P
$$

also lifts to:

$$
H.
$$

The compositum of the two independent quadratic extensions is:

$$
\mathbb Q(C).
$$

Therefore the compatible rational square roots define a rational point on the normalized fiber product:

$$
C.
$$

Thus the two square-value conditions recover the full genus-$5$ lift.

---

# 25. Why these two conditions are necessary

Conversely, let:

$$
P_C\in C(\mathbb Q).
$$

Its image:

$$
P\in E_0'(\mathbb Q)
$$

has rational coordinates.

Because:

$$
P_C
$$

maps rationally to both:

$$
G
$$

and:

$$
K,
$$

the corresponding quadratic characters must be trivial:

$$
[f_G(P)]=1,
$$

and:

$$
[f_K(P)]=1.
$$

Therefore:

$$
\boxed{
f_G(P),f_K(P)\in\mathbb Q^2.
}
$$

Hence the reduction is bidirectional.

---

# 26. Rank-one Mordell-Weil reduction

The exact rank theorem gives:

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

Therefore all rational elliptic points have the form:

$$
\boxed{
P
=
T
+
rG,
}
$$

where:

$$
T\in E_0'(\mathbb Q)_{\rm tors},
$$

and:

$$
r\in\mathbb Z,
$$

after choosing a saturated generator:

$$
G
$$

of the free part.

Thus the fixed-slice Perfect Cuboid problem becomes:

> Determine whether there is a nonboundary Mordell-Weil point $T+rG$ for which two explicit rational functions are simultaneously squares.

This is now a one-dimensional arithmetic orbit problem.

---

# 27. The two square-value gates

Define:

$$
\boxed{
\mathcal R_1(P)
=
\frac{
2X(P)+950625
}{
1352033
}.
}
$$

Define:

$$
\boxed{
\mathcal R_2(P)
=
1352033
\,X(P)
\left(
950625X(P)
-
462152671232
\right).
}
$$

The exact target is:

$$
\boxed{
\mathcal R_1(P)\in\mathbb Q^2,
}
$$

and:

$$
\boxed{
\mathcal R_2(P)\in\mathbb Q^2.
}
$$

This is the **R8 Double Square Gate**.

---

# 28. Relation to the historical genus-$3$ gate

The old genus-$3$ lift was controlled by:

$$
f_H
=
X(2X+M)(MX-2AB).
$$

Because:

$$
f_H
=
f_Gf_K,
$$

the old single product-square condition is weaker than requiring both:

$$
f_G
$$

and:

$$
f_K
$$

to be squares individually.

This is exactly analogous to the earlier distinction:

$$
[F_1]=[F_2]
$$

versus:

$$
[F_1]=[F_2]=1.
$$

R8 therefore gives a clean elliptic-base interpretation of the genus-$3$ intermediate layer.

---

# 29. Shared branch factor and defect coupling

The old product function contains:

$$
2X+M.
$$

This is exactly the new genus-$2$ gate factor.

Thus the genus-$3$ and genus-$2$ covers are not arithmetically independent.

Their squareclasses share one branch factor.

Removing the shared factor gives the third quadratic character:

$$
f_K.
$$

This is why the V4 fiber product has genus:

$$
5
$$

rather than behaving like a generic pair of unrelated quadratic covers.

---

# 30. The two nontrivial branch values

The entire biquadratic cover is controlled by only two nontrivial $X$-values:

$$
\boxed{
X_1
=
-\frac{
950625
}{2},
}
$$

and:

$$
\boxed{
X_2
=
\frac{
2\cdot448^2\cdot1073^2
}{
975^2
}.
}
$$

The first controls:

$$
f_G.
$$

The second controls:

$$
f_K.
$$

The genus-$3$ product character sees both.

Thus the complicated high-genus branch geometry collapses to two elliptic $X$-fibers.

---

# 31. Squareclass form after the first gate

Suppose:

$$
\mathcal R_1(P)
=
r^2.
$$

Then:

$$
2X+M
=
Lr^2.
$$

Substitute into the historical product squareclass:

$$
f_H
=
X(2X+M)(MX-2AB).
$$

Modulo rational squares:

$$
[f_H]
=
\left[
L\,X(MX-2AB)
\right].
$$

Therefore after the first gate closes, the second remaining obstruction is exactly:

$$
\boxed{
\mathcal R_2(P).
}
$$

This gives the sequential interpretation:

$$
\boxed{
\mathcal R_1
\longrightarrow
\mathcal R_2.
}
$$

---

# 32. Elimination of $X$ after the first gate

Let:

$$
\mathcal R_1(P)=r^2.
$$

Then:

$$
\boxed{
X
=
\frac{
Lr^2-M
}{2}.
}
$$

Also:

$$
MX-2AB
=
\frac{
L(Mr^2-L)
}{2},
$$

using the exact identity:

$$
\boxed{
M^2+4AB=L^2.
}
$$

Therefore:

$$
\mathcal R_2(P)
=
\frac{
L^2
}{
4
}
\left(
Lr^2-M
\right)
\left(
Mr^2-L
\right).
$$

Since:

$$
\frac{L^2}{4}
$$

is a rational square, the second gate reduces to:

$$
\boxed{
\left(
Lr^2-M
\right)
\left(
Mr^2-L
\right)
\in
\mathbb Q^2.
}
$$

This is a symmetric quartic square condition in the first-gate root:

$$
r.
$$

The elliptic condition:

$$
P\in E_0'
$$

must still be imposed.

---

# 33. The constants satisfy a Pythagorean discriminant identity

The constants obey:

$$
\boxed{
M^2+4AB=L^2.
}
$$

Explicitly:

$$
950625^2
+
4
(448^2)
(1073^2)
=
1352033^2.
$$

This identity is the algebraic reason the second square gate simplifies after the first one is imposed.

The same constant:

$$
1352033
$$

appeared repeatedly in the earlier:

- $E_0$ model;
- genus-$2$ bielliptic transformation;
- R7 hyperelliptic quotient;
- R8 synchronization functions.

It is therefore a canonical fixed-slice invariant, not an elimination artifact.

---

# 34. Exact V4 genus check

Let:

$$
E=E_0'
$$

with:

$$
g(E)=1.
$$

The three quadratic subcovers have genera:

$$
g(G)=2,
$$

$$
g(K)=2,
$$

and:

$$
g(H)=3.
$$

For a V4-cover:

$$
C\to E,
$$

the Kani-Rosen dimension relation is:

$$
\boxed{
g(C)+2g(E)
=
g(G)+g(K)+g(H).
}
$$

Therefore:

$$
g(C)+2
=
2+2+3.
$$

Hence:

$$
\boxed{
g(C)=5.
}
$$

The entire genus-$5$ count is recovered from the elliptic square-value lattice.

---

# 35. Relation to R0 defect migration

R0 described arithmetic defect migration on a six-channel squareclass state.

R8 gives a much smaller fixed-slice representation.

On the selected fiber, all remaining arithmetic data are encoded by only two squareclasses:

$$
\boxed{
[\mathcal R_1(P)],
\qquad
[\mathcal R_2(P)].
}
$$

The four possible states are:

$$
(1,1),
$$

$$
(1,\kappa_2),
$$

$$
(\kappa_1,1),
$$

and:

$$
(\kappa_1,\kappa_2).
$$

The Perfect Cuboid target on this fiber is:

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

Thus the original Magic-Cube state has collapsed to a two-bit squareclass gate state on one elliptic orbit.

---

# 36. Relation to the seed defect $233\cdot1913$

The original G2 seed itself is not a rational point on the complete genus-$5$ fiber because one Twin-Psi gate remains nonsquare.

Its defect was:

$$
\boxed{
\frac{233}{1913}.
}
$$

R8 does not replace that diagnostic.

Instead it describes the exact closure curve whose rational points would repair it.

The two square functions:

$$
\mathcal R_1,
\qquad
\mathcal R_2
$$

are the fixed-slice arithmetic gates that any successful repair point must satisfy simultaneously.

Thus:

$$
233\cdot1913
$$

remains the seed marker, while:

$$
\mathcal R_1,
\mathcal R_2
$$

define the repair locus.

---

# 37. Proof-safe boundary handling

The formulas for:

$$
\xi,
$$

the historical quotient maps, and the birational charts contain finite denominator loci.

These include points where:

$$
m=-1,
$$

$$
m^2-2m-1=0,
$$

or where a chosen affine elliptic denominator vanishes.

Such points must be checked separately on the projective normalizations.

R8 does not discard them silently.

The double square-value theorem is stated on the generic affine open, with the finite exceptional set retained as an explicit boundary-verification obligation.

---

# 38. R8 canonical theorem

## Theorem — Double Square-Value Reduction

Away from the finite chart-boundary set, the rational points of the historical fixed Perfect Cuboid genus-$5$ fiber:

$$
C_{7/32}
$$

are in bijective correspondence with compatible choices of square roots above rational points:

$$
P=(X,Y)\in E_0'(\mathbb Q)
$$

satisfying:

$$
\boxed{
\frac{
2X+950625
}{
1352033
}
\in
\mathbb Q^2,
}
$$

and:

$$
\boxed{
1352033
\,X
\left(
950625X
-
462152671232
\right)
\in
\mathbb Q^2.
}
$$

The two old/new elliptic quotient maps synchronize by:

$$
\boxed{
[-1]:
(X,Y)\mapsto(X,-Y).
}
$$

This is the canonical R8 form of the fixed-slice problem.

---

# 39. Computational consequence

The old direct target was a genus-$5$ rational-point problem.

R8 replaces it by:

$$
\boxed{
\text{rank-one elliptic group}
+
\text{two square-value tests}.
}
$$

This suggests the following exact computational methods:

- Mordell-Weil coefficient sieving;
- local squareclass tests;
- $2$-adic logarithms;
- elliptic divisibility sequences;
- finite-field compatibility;
- elliptic Chabauty on either square-value cover;
- simultaneous covering descent.

The high-genus curve remains the canonical total space, but no longer needs to be the primary search space.

---

# 40. Proposed R9 program

R9 should stop expanding the structural hierarchy and begin the exact arithmetic sieve on the two R8 gates.

## R9-A — Generator and saturation checkpoint

Use the certified rank-one theorem for:

$$
E_0'.
$$

Verify a saturated generator:

$$
G
$$

of the free part, or record the exact finite index of the known non-torsion point.

Keep the four rational $2$-torsion cosets explicit.

## R9-B — Local squareclass sieve for $\mathcal R_1$

For good primes:

$$
p,
$$

compute the subset of:

$$
E_0'(\mathbb F_p)
$$

on which:

$$
\mathcal R_1
$$

is a quadratic residue.

Pull the condition back to Mordell-Weil coefficient classes.

## R9-C — Local squareclass sieve for $\mathcal R_2$

Repeat for:

$$
\mathcal R_2.
$$

Intersect the two allowed coefficient sets.

## R9-D — $2$-adic refinement

Because both covers are quadratic and the curve has full rational $2$-torsion, perform a synchronized:

$$
2
$$

-primary analysis.

Track:

- torsion cosets;
- coefficient parity;
- higher $2$-adic residue disks;
- valuations of the two square functions.

## R9-E — Boundary certification

Check all finite exceptional chart points independently.

No generic denominator argument may be used to discard a boundary point.

## R9-F — Exact fixed-slice target

The desired theorem is:

> Every rational point of $E_0'$ satisfying both R8 square gates is boundary.

If proved, then:

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

contains no nondegenerate Perfect Cuboid point.

If a nonboundary point survives, reconstruct the full genus-$5$ point and then the rational/integer cuboid candidate.

---

# 41. Final conclusion

R8 resolves the synchronization problem left by R7.

The new genus-$2$ quotient and the old genus-$3$ quotient do not merely share an isogenous elliptic factor.

They share the exact same elliptic quotient:

$$
E_0'.
$$

On a common parent point their maps differ only by:

$$
\boxed{
[-1].
}
$$

The complete genus-$5$ fixed fiber is therefore a biquadratic cover of one rank-one elliptic curve.

Its function field is:

$$
\boxed{
\mathbb Q(C)
=
\mathbb Q(E_0')
\left(
\sqrt{\mathcal R_1},
\sqrt{\mathcal R_2}
\right),
}
$$

with:

$$
\boxed{
\mathcal R_1
=
\frac{
2X+950625
}{
1352033
},
}
$$

and:

$$
\boxed{
\mathcal R_2
=
1352033
\,X
\left(
950625X
-
462152671232
\right).
}
$$

Thus the historical fixed-slice Perfect Cuboid problem has been reduced from a genus-$5$ curve to a simultaneous double-square problem on a rank-one elliptic Mordell-Weil orbit.

This is the canonical starting point for R9.