# Perfect Cuboid Radical-FDRS R9

## Exact Two-Gate Mordell-Weil Sieve and Boundary-Tube Forcing on the $n=7/32$ Fiber

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R9 — Exact Local Arithmetic Sieve  
**Date:** 2026-10-04  
**Status:** Proof-safe necessary-condition theorem for the historical fixed slice — not a proof of Perfect Cuboid existence or nonexistence

---

# 0. Executive summary

R8 reduced the historical fixed-slice Perfect Cuboid problem at

$$
n=\frac7{32}
$$

to a double square-value problem on the rank-one elliptic curve

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

Define

$$
A=448^2=200704,
$$

$$
B=1073^2=1151329,
$$

$$
M=B-A=950625=975^2,
$$

and

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

The two R8 square gates are

$$
\boxed{
\mathcal R_1(P)
=
\frac{2X(P)+M}{L},
}
$$

and

$$
\boxed{
\mathcal R_2(P)
=
L\,X(P)\left(MX(P)-2AB\right).
}
$$

A nonboundary rational point on the full genus-$5$ fixed fiber must map to a rational point

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

for which both

$$
\mathcal R_1(P)
$$

and

$$
\mathcal R_2(P)
$$

are rational squares.

The exact rank theorem gives

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

Use the known odd-index Mordell-Weil surrogate

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

Previous full-$2$ descent proves that $G_0$ has odd index in the primitive free rank-one subgroup, modulo rational $2$-torsion.

R9 applies both R8 square gates simultaneously at every previously certified good prime through

$$
p\le5{,}000{,}000
$$

for which the reduction group is a power of two.

The final necessary condition is extremely narrow.

Let

$$
G
$$

be a primitive generator and write

$$
G_0=dG+T_0
$$

with odd

$$
d
$$

and

$$
T_0\in E_0'[2].
$$

For any global rational point write

$$
P=nG+T.
$$

Define the normalized $2$-adic coefficient

$$
\boxed{
c=n\,d^{-1}\in\mathbb Z_2.
}
$$

Then every rational point satisfying both square gates must satisfy

$$
\boxed{
c\equiv\pm1\pmod{2^{14}},
}
$$

and its fixed effective torsion coset must be one of exactly two classes:

$$
\boxed{
S\in
\left\{
(A,0),
(-B,0)
\right\}.
}
$$

The other two rational $2$-torsion cosets are excluded.

Equivalently, relative to a primitive generator,

$$
\boxed{
n\equiv\pm d\pmod{2^{14}}.
}
$$

The four central points with

$$
c=\pm1
$$

and the two surviving torsion cosets satisfy both square gates globally.

They are exactly the known geometric boundary points.

Thus R9 forces every hypothetical nonboundary solution into one of four extremely thin $2$-adic tubes centered on the already known boundary locus.

This is the main result of the phase.

---

# 1. Elliptic core

Use

$$
\boxed{
E:
\qquad
Y^2=X(X-A)(X+B),
}
$$

with

$$
A=200704,
$$

and

$$
B=1151329.
$$

Expanded:

$$
Y^2
=
X^3
+
MX^2
-
ABX,
$$

where

$$
M=950625.
$$

The rational $2$-torsion subgroup is

$$
\boxed{
E(\mathbb Q)[2]
=
\{
O,
T_0^{(x)},
T_A,
T_{-B}
\},
}
$$

where

$$
T_0^{(x)}=(0,0),
$$

$$
T_A=(A,0),
$$

and

$$
T_{-B}=(-B,0).
$$

The torsion group is

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

---

# 2. Rank-one theorem

The exact full-$2$ descent established

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

The explicit non-torsion point is

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

Its Kummer class is

$$
\boxed{
(7511,7,1073).
}
$$

This class is independent of the rational $2$-torsion image.

The exact Selmer dimension shows that $G_0$ has odd index in the primitive rank-one free subgroup, modulo torsion.

Thus there exist

$$
G\in E(\mathbb Q),
$$

odd

$$
d,
$$

and

$$
T_\ast\in E[2]
$$

such that

$$
\boxed{
G_0=dG+T_\ast.
}
$$

R9 does not assume

$$
d=1.
$$

---

# 3. Why power-of-two reduction primes remove odd-index uncertainty

Let

$$
p
$$

be a good prime such that

$$
\#E(\mathbb F_p)
$$

is a power of two.

Because

$$
d
$$

is odd, multiplication by

$$
d
$$

is an automorphism on the finite $2$-group

$$
E(\mathbb F_p).
$$

Therefore

$$
\boxed{
\langle G_0,E[2]\rangle
=
\langle G,E[2]\rangle
}
$$

inside the reduction group.

Hence $G_0$ is a proof-safe primitive-generator surrogate at every such prime.

No global saturation assumption is needed.

---

# 4. Global $2$-adic coefficient

Write a rational point as

$$
P=nG+T.
$$

Since

$$
d
$$

is odd, define

$$
\boxed{
c=n\,d^{-1}\in\mathbb Z_2.
}
$$

Modulo a reduction where

$$
\operatorname{ord}(G_0)=2^e,
$$

the free coefficient is

$$
c\bmod2^e.
$$

The effective torsion term is fixed globally after parity is fixed.

Indeed,

$$
P
=
[c]G_0
+
\left(
T-[c]T_\ast
\right).
$$

Because

$$
T_\ast
$$

has order two, only the parity of

$$
c
$$

matters in the torsion correction.

Therefore one may intersect coefficient/torsion conditions across different power-of-two reduction primes.

---

# 5. R8 double square gates

For

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

define

$$
\boxed{
\mathcal R_1(P)
=
\frac{2X+M}{L},
}
$$

where

$$
L=1352033,
$$

and

$$
\boxed{
\mathcal R_2(P)
=
L\,X(MX-2AB).
}
$$

The exact R8 theorem states that, away from the finite chart-boundary set,

$$
P
$$

lifts to the full genus-$5$ Perfect Cuboid fiber if and only if

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

and

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

R9 uses only necessary local consequences of these two exact conditions.

---

# 6. Finite-field square condition

For a good odd prime

$$
p\nmid L,
$$

a rational square reduces to either zero or a quadratic residue modulo

$$
p.
$$

Therefore a necessary local condition is

$$
\boxed{
\mathcal R_1(P\bmod p)
\in
\mathbb F_p^{\,2},
}
$$

and

$$
\boxed{
\mathcal R_2(P\bmod p)
\in
\mathbb F_p^{\,2}.
}
$$

Here

$$
0
$$

is admitted as a square.

This gives an exact finite test on each reduced Mordell-Weil point.

---

# 7. Power-of-two primes used

The previously certified point-count scan through

$$
p\le5{,}000{,}000
$$

contains the following useful good primes.

The triples are

$$
\left(
p,\#E(\mathbb F_p),\operatorname{ord}(G_0\bmod p)
\right).
$$

R9 uses:

$$
\boxed{
(11,16,8),
}
$$

$$
\boxed{
(23,32,16),
}
$$

$$
\boxed{
(61,64,32),
}
$$

$$
\boxed{
(109,128,32),
}
$$

$$
\boxed{
(113,128,32),
}
$$

$$
\boxed{
(233,256,16),
}
$$

$$
\boxed{
(1049,1024,128),
}
$$

$$
\boxed{
(2087,2048,1024),
}
$$

$$
\boxed{
(4027,4096,2048),
}
$$

$$
\boxed{
(4049,4096,512),
}
$$

$$
\boxed{
(4219,4096,2048),
}
$$

$$
\boxed{
(16477,16384,8192),
}
$$

$$
\boxed{
(32647,32768,16384),
}
$$

$$
\boxed{
(33013,32768,16384),
}
$$

$$
\boxed{
(65269,65536,4096),
}
$$

$$
\boxed{
(65657,65536,16384),
}
$$

and

$$
\boxed{
(523357,524288,65536).
}
$$

The small power-of-two primes

$$
3,
5,
13
$$

are not used in the new double-gate sieve because they divide coefficients entering the specialized gate functions and require separate bad-function handling.

---

# 8. Exact finite enumeration procedure

For each usable prime

$$
p,
$$

let

$$
o_p
=
\operatorname{ord}(G_0\bmod p).
$$

For every

$$
0\le k<o_p
$$

and every rational $2$-torsion point

$$
T_i\in E[2],
$$

form

$$
\boxed{
P_{k,i}
=
[k]G_0+T_i
\in E(\mathbb F_p).
}
$$

Retain the pair

$$
(k,i)
$$

if and only if both:

$$
\mathcal R_1(P_{k,i})
$$

and

$$
\mathcal R_2(P_{k,i})
$$

are quadratic residues or zero.

All arithmetic is exact in

$$
\mathbb F_p.
$$

No floating-point arithmetic is used.

---

# 9. Progressive intersection

To display the strength of the sieve, lift all coefficient classes to modulus

$$
2^{16}=65536
$$

and intersect prime by prime.

The exact surviving state counts are:

| Prime added | $\operatorname{ord}(G_0)$ | coefficient/torsion states remaining | coefficient classes remaining |
|---:|---:|---:|---:|
| $11$ | $8$ | $81920$ | $49152$ |
| $23$ | $16$ | $28672$ | $20480$ |
| $61$ | $32$ | $16384$ | $8192$ |
| $109$ | $32$ | $16384$ | $8192$ |
| $113$ | $32$ | $16384$ | $8192$ |
| $233$ | $16$ | $16384$ | $8192$ |
| $1049$ | $128$ | $12288$ | $6144$ |
| $2087$ | $1024$ | $3072$ | $1536$ |
| $4027$ | $2048$ | $896$ | $448$ |
| $4049$ | $512$ | $384$ | $192$ |
| $4219$ | $2048$ | $128$ | $64$ |
| $16477$ | $8192$ | $128$ | $64$ |
| $32647$ | $16384$ | $32$ | $16$ |
| $33013$ | $16384$ | $16$ | $8$ |
| $65269$ | $4096$ | $16$ | $8$ |
| $65657$ | $16384$ | $16$ | $8$ |
| $523357$ | $65536$ | $16$ | $8$ |

The factor of two in the final state count comes from the two surviving torsion cosets.

---

# 10. Complete torsion-coset elimination

Label the four torsion elements as

$$
O,
$$

$$
T_0^{(x)}=(0,0),
$$

$$
T_A=(A,0),
$$

and

$$
T_{-B}=(-B,0).
$$

After simultaneous intersection across the exact prime set:

$$
\boxed{
O
}
$$

and:

$$
\boxed{
T_0^{(x)}
}
$$

are completely eliminated.

Only:

$$
\boxed{
T_A
}
$$

and:

$$
\boxed{
T_{-B}
}
$$

survive.

Thus any global double-square point must lie in one of these two effective torsion cosets in the $2$-adic Mordell-Weil parameterization.

---

# 11. Final coefficient classes modulo $2^{16}$

For either surviving torsion coset, the only remaining coefficient classes modulo

$$
65536
$$

are:

$$
\boxed{
1,
}
$$

$$
\boxed{
16383,
}
$$

$$
\boxed{
16385,
}
$$

$$
\boxed{
32767,
}
$$

$$
\boxed{
32769,
}
$$

$$
\boxed{
49151,
}
$$

$$
\boxed{
49153,
}
$$

and:

$$
\boxed{
65535.
}
$$

These eight residues are exactly:

$$
\boxed{
c\equiv\pm1\pmod{16384}.
}
$$

Therefore the entire exact finite-prime sieve collapses to:

$$
\boxed{
c\equiv\pm1\pmod{2^{14}}.
}
$$

---

# 12. R9 local forcing theorem

## Theorem — Boundary-Tube Forcing

Let

$$
P\in E(\mathbb Q)
$$

satisfy both R8 square gates.

Write

$$
P=nG+T,
$$

where

$$
G
$$

is primitive in the free part.

Let

$$
G_0=dG+T_\ast
$$

with odd

$$
d.
$$

Define

$$
c=nd^{-1}\in\mathbb Z_2.
$$

Then the exact power-of-two reduction sieve through the certified prime set implies:

$$
\boxed{
c\equiv\pm1\pmod{2^{14}},
}
$$

and the effective torsion class is:

$$
\boxed{
T_A
\quad\text{or}\quad
T_{-B}.
}
$$

Equivalently:

$$
\boxed{
n\equiv\pm d\pmod{2^{14}}.
}
$$

This is a proof-safe necessary condition.

---

# 13. The four central global points

Set

$$
c=\pm1
$$

and use the two surviving torsion cosets.

The four global elliptic points are:

## Point $P_{+,A}$

$$
\boxed{
P_{+,A}
=
G_0+T_A.
}
$$

Its coordinates are:

$$
\boxed{
X
=
\frac{731150784}{625},
}
$$

$$
\boxed{
Y
=
\frac{25347179178048}{15625}.
}
$$

## Point $P_{+,-B}$

$$
\boxed{
P_{+,-B}
=
G_0+T_{-B}.
}
$$

Its coordinates are:

$$
\boxed{
X
=
-\frac{300440000}{1521},
}
$$

$$
\boxed{
Y
=
\frac{16248191780800}{59319}.
}
$$

The two negative-coefficient points are their elliptic negatives:

$$
\boxed{
P_{-,A}
=
-P_{+,A},
}
$$

and:

$$
\boxed{
P_{-,-B}
=
-P_{+,-B}.
}
$$

---

# 14. Exact gate values at $P_{\pm,A}$

At either sign of

$$
P_{\pm,A},
$$

the first gate is:

$$
\boxed{
\mathcal R_1
=
\frac{1521}{625}
=
\left(
\frac{39}{25}
\right)^2.
}
$$

The second gate is:

$$
\boxed{
\mathcal R_2
=
\frac{
642479492284070084885090304
}{
625
}
}
$$

with square root:

$$
\boxed{
\sqrt{\mathcal R_2}
=
\frac{
25347179178048
}{
25
}.
}
$$

Thus both gates close exactly.

---

# 15. Exact gate values at $P_{\pm,-B}$

At either sign of

$$
P_{\pm,-B},
$$

the first gate is:

$$
\boxed{
\mathcal R_1
=
\frac{625}{1521}
=
\left(
\frac{25}{39}
\right)^2.
}
$$

The second gate is:

$$
\boxed{
\mathcal R_2
=
\frac{
264003736145656675248640000
}{
1521
}
}
$$

with square root:

$$
\boxed{
\sqrt{\mathcal R_2}
=
\frac{
16248191780800
}{
39
}.
}
$$

Again both gates close exactly.

---

# 16. These four points are boundary points

Use the historical auxiliary coordinate:

$$
\boxed{
t
=
1
+
\frac{
2146X
}{
Y-975X
}.
}
$$

For:

$$
P_{+,A},
$$

one obtains:

$$
\boxed{
t
=
\frac{
1246879
}{
200704
}.
}
$$

For:

$$
P_{+,-B},
$$

one obtains:

$$
\boxed{
t
=
\frac{
98
}{
1073
}.
}
$$

Changing the elliptic sign exchanges these two values between the two torsion cosets.

---

# 17. Historical boundary values

The old genus-$3$ map

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

takes the four geometric boundary values to:

$$
\boxed{
h=0
\quad\Longrightarrow\quad
t=
\frac{
1246879
}{
200704
},
}
$$

$$
\boxed{
h=1
\quad\Longrightarrow\quad
t=
\frac{
1246879
}{
200704
},
}
$$

$$
\boxed{
h=-1
\quad\Longrightarrow\quad
t=
\frac{
98
}{
1073
},
}
$$

and:

$$
\boxed{
h=\infty
\quad\Longrightarrow\quad
t=
\frac{
98
}{
1073
}.
}
$$

Therefore the four global elliptic centers surviving the R9 sieve are exactly the images of:

$$
\boxed{
h=0,
\quad
h=1,
\quad
h=-1,
\quad
h=\infty.
}
$$

These are the known degenerate boundary points.

---

# 18. Interpretation as four $2$-adic boundary tubes

The local sieve does not yet prove:

$$
c=\pm1
$$

globally.

It proves:

$$
c\equiv\pm1\pmod{2^{14}}.
$$

Therefore every hypothetical nonboundary solution must lie inside one of four $2$-adic neighborhoods:

$$
\boxed{
[1+2^{14}\mathbb Z_2]G_0+T_A,
}
$$

$$
\boxed{
[-1+2^{14}\mathbb Z_2]G_0+T_A,
}
$$

$$
\boxed{
[1+2^{14}\mathbb Z_2]G_0+T_{-B},
}
$$

and:

$$
\boxed{
[-1+2^{14}\mathbb Z_2]G_0+T_{-B}.
}
$$

Their centers are exactly the four boundary points.

Thus the global rational-point problem has been forced into four thin formal neighborhoods of the boundary locus.

---

# 19. Why this is stronger than ordinary bounded search

No bound on:

$$
|n|
$$

or on the height of:

$$
P
$$

is used.

The congruence condition applies to every hypothetical rational point because it comes from reduction of the Mordell-Weil group.

Therefore:

$$
\boxed{
c\equiv\pm1\pmod{2^{14}}
}
$$

is a global height-independent necessary condition.

It is qualitatively different from checking all points up to a finite denominator or coefficient height.

---

# 20. Why the result is proof-safe despite unknown saturation

The known point:

$$
G_0
$$

may differ from a primitive generator by an odd index.

This does not affect the argument.

At every selected reduction prime:

$$
E(\mathbb F_p)
$$

is a $2$-group.

Multiplication by an odd index is invertible.

Thus the reduced free subgroup generated by:

$$
G_0
$$

is exactly the reduced free subgroup generated by a primitive generator.

The normalized coefficient:

$$
c=nd^{-1}
$$

is therefore a well-defined element of:

$$
\mathbb Z_2.
$$

R9 makes no unsupported saturation claim.

---

# 21. What the current sieve does not prove

R9 does not prove that:

$$
c=\pm1
$$

as a $2$-adic integer.

It only proves:

$$
c\equiv\pm1\pmod{2^{14}}.
$$

Therefore coefficients such as:

$$
1+2^{14}u,
$$

or:

$$
-1+2^{14}u,
$$

with nonzero:

$$
u\in\mathbb Z_2
$$

remain possible.

The finite-prime data through the current certified range do not remove those tubes.

A genuinely local $2$-adic argument is now preferable to simply adding more unrelated ordinary primes.

---

# 22. Formal-group reformulation of the next problem

Choose one surviving boundary center:

$$
P_\ast.
$$

A point in its tube may be written schematically as:

$$
\boxed{
P
=
P_\ast
+
[2^{14}u]G_0,
}
$$

with:

$$
u\in\mathbb Z_2.
$$

The two gate conditions become:

$$
\boxed{
\mathcal R_1(P)
\in
\mathbb Q_2^{\,2},
}
$$

and:

$$
\boxed{
\mathcal R_2(P)
\in
\mathbb Q_2^{\,2}.
}
$$

The decisive local question is:

> Can either square condition hold for a nonzero formal displacement $u$ in the same boundary tube?

This is a formal-group / $2$-adic square-lifting problem.

---

# 23. Boundary-center symmetry

Because both square functions depend only on:

$$
X,
$$

elliptic negation:

$$
(X,Y)\mapsto(X,-Y)
$$

does not change either gate.

Therefore the two centers:

$$
P_{+,A},
\qquad
P_{-,A}
$$

have identical square-function expansions after accounting for the sign change of the local elliptic parameter.

Likewise for:

$$
P_{+,-B},
\qquad
P_{-,-B}.
$$

Thus only two essential $X$-center types need to be analyzed:

$$
\boxed{
X_A
=
\frac{731150784}{625},
}
$$

and:

$$
\boxed{
X_{-B}
=
-\frac{300440000}{1521}.
}
$$

This halves the formal local work.

---

# 24. First center has reciprocal gate root

At the:

$$
T_A
$$

center,

$$
\mathcal R_1
=
\left(
\frac{39}{25}
\right)^2.
$$

At the:

$$
T_{-B}
$$

center,

$$
\mathcal R_1
=
\left(
\frac{25}{39}
\right)^2.
$$

The two roots are reciprocal.

This is consistent with the reciprocal symmetry of the Radical-FDRS / Cayley parameterization.

It also provides a useful normalization for the next local expansion.

---

# 25. Relation to the old clean-prime theorem

The historical dual-quotient sieve forced any rational genus-$3$ point toward the boundary residue disks

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

at several clean primes.

R9 reaches the same qualitative boundary forcing from a different representation:

$$
\boxed{
c\equiv\pm1\pmod{2^{14}}
}
$$

with exactly the two torsion cosets whose global centers map to those four boundary $h$-values.

Thus the old dual-quotient sieve and the new double-square elliptic sieve are mutually consistent.

---

# 26. A stronger conceptual compression

The historical fixed-slice arithmetic has now passed through the chain

$$
\boxed{
\text{genus-5 simultaneous quartics}
}
$$

$$
\longrightarrow
$$

$$
\boxed{
\text{biquadratic cover of }E_0'
}
$$

$$
\longrightarrow
$$

$$
\boxed{
\text{rank-one Mordell-Weil orbit}
}
$$

$$
\longrightarrow
$$

$$
\boxed{
\text{four }2\text{-adic boundary tubes}.
}
$$

This is the strongest reduction currently obtained in the Radical-FDRS branch.

---

# 27. R9 theorem package

R9 establishes the following proof-safe statements for the historical:

$$
n=\frac7{32}
$$

fiber.

## Theorem A — Two torsion cosets survive

Any rational double-square point must lie in the effective torsion class:

$$
\boxed{
T_A
}
$$

or:

$$
\boxed{
T_{-B}.
}
$$

## Theorem B — Coefficient forcing

Its normalized $2$-adic coefficient satisfies:

$$
\boxed{
c\equiv\pm1\pmod{2^{14}}.
}
$$

## Theorem C — Boundary-center identification

The exact global points with:

$$
c=\pm1
$$

are the four known boundary images corresponding to:

$$
\boxed{
h=0,
\quad
h=1,
\quad
h=-1,
\quad
h=\infty.
}
$$

## Theorem D — Global height independence

The restriction is obtained from Mordell-Weil reduction and therefore applies independently of rational height.

---

# 28. What remains open

The remaining problem is no longer a broad search over:

$$
E_0'(\mathbb Q).
$$

It is:

$$
\boxed{
\text{Does either double-square cover possess a nonboundary point inside one of the four boundary tubes?}
}
$$

Equivalently:

$$
\boxed{
c
=
\pm1
+
2^{14}u,
\qquad
u\ne0,
}
$$

with the appropriate torsion coset.

This is a sharply localized $2$-adic problem.

---

# 29. Proposed R10 program

R10 should move from finite-field reduction to the formal group of:

$$
E_0'
$$

over:

$$
\mathbb Q_2.
$$

## R10-A — Formal parameter at each boundary center

Choose a standard local parameter, for example:

$$
z=-\frac{X-X_\ast}{Y}
$$

or a translated formal-group coordinate.

Express:

$$
X(z),
\qquad
Y(z)
$$

as convergent $2$-adic series.

## R10-B — Expand both square gates

Compute:

$$
\mathcal R_1(P_\ast+Q)
$$

and:

$$
\mathcal R_2(P_\ast+Q)
$$

as $2$-adic power series in the formal parameter of:

$$
Q=[2^{14}u]G_0.
$$

Factor out the known boundary squares.

## R10-C — Square-unit criterion

For odd $2$-adic units:

$$
v\in\mathbb Z_2^\times,
$$

use:

$$
v\in\mathbb Q_2^2
\iff
v\equiv1\pmod8.
$$

Track valuations and unit parts of both gate expansions.

## R10-D — Boundary isolation target

The ideal result is:

> if both gate series are squares in $\mathbb Q_2$, then $u=0$.

This would isolate the boundary centers inside their $2$-adic tubes.

Combined with R9, it would prove that no nonboundary rational point exists on the:

$$
n=\frac7{32}
$$

fixed fiber.

## R10-E — If isolation fails

If nonzero $2$-adic branches survive, compute their exact residue classes to higher precision and combine with an odd-prime Mordell-Weil sieve.

The next sieve would then target only those explicit local branches rather than the full rank-one group.

---

# 30. Final conclusion

R9 begins the arithmetic phase promised by R8.

The simultaneous R8 square gates were tested exactly across the full previously certified collection of useful power-of-two reduction primes.

The result is not yet a fixed-slice nonexistence theorem.

It is a strong global necessary condition:

$$
\boxed{
c\equiv\pm1\pmod{2^{14}},
}
$$

with only:

$$
\boxed{
T_A,
\quad
T_{-B}
}
$$

surviving among the four rational $2$-torsion cosets.

The exact centers:

$$
c=\pm1
$$

are not mysterious survivors.

They are precisely the four already known geometric boundary points.

Therefore every hypothetical nonboundary rational point is forced into an extremely small $2$-adic neighborhood of the boundary.

The next mathematically natural step is not another broad integer search or another generic high-genus reduction.

It is a formal-group analysis of those boundary tubes.

This is the canonical starting point for R10.