# Perfect Cuboid Radical-FDRS R10

## $2$-Adic Local Openness, Sharp Boundary Neighborhoods, and the Failure of Pure Formal-Group Isolation

**Series:** Perfect Cuboid Global Closure  
**Branch:** Radical-FDRS / Geometry-Coupling Branch R  
**Phase:** R10 — Local Boundary Analysis  
**Date:** 2026-10-04  
**Status:** Exact local theorem and strategy correction — not a proof of Perfect Cuboid existence or nonexistence

---

# 0. Executive summary

R9 forced every hypothetical rational point on the historical fixed fiber

$$
n=\frac7{32}
$$

into four thin $2$-adic Mordell-Weil tubes centered at the known geometric boundary points.

The proposed R10 target was to use the formal group over

$$
\mathbb Q_2
$$

to prove that the only points in those tubes satisfying both square gates are the boundary centers themselves.

R10 proves that this target is impossible by purely local squarehood arguments.

The reason is structural.

At every boundary center, both square gates take nonzero square values in

$$
\mathbb Q_2.
$$

The subgroup

$$
\mathbb Q_2^{\times2}
$$

is open in

$$
\mathbb Q_2^\times.
$$

Therefore every such boundary point has a nontrivial $2$-adic neighborhood in which both gates remain squares.

This is not merely a qualitative topology statement.

Using the one-dimensional fixed-fiber coordinate

$$
m,
$$

R10 computes explicit sharp sufficient neighborhoods.

Let

$$
\boxed{
Q_1(m)
=
(1024m^2+49)(49m^2+1024),
}
$$

and

$$
\boxed{
Q_2(m)
=
\left[
1073(m^2+1)-1950m
\right]
\left[
1073(m^2+1)+1950m
\right].
}
$$

The Perfect Cuboid fixed-fiber square conditions are exactly

$$
Q_1(m)\in\mathbb Q^2,
$$

and

$$
Q_2(m)\in\mathbb Q^2.
$$

Over

$$
\mathbb Q_2,
$$

the following hold.

### Boundary $m=0$

If

$$
\boxed{
v_2(m)\ge7,
}
$$

then both

$$
Q_1(m),
\qquad
Q_2(m)
$$

are $2$-adic squares.

The threshold is sharp for the first gate in the sense that

$$
v_2(m)=6
$$

forces

$$
Q_1(m)
$$

to be nonsquare.

### Boundary $m=\infty$

By reciprocal symmetry, if

$$
\boxed{
v_2(1/m)\ge7,
}
$$

then both gates are $2$-adic squares.

### Boundaries $m=\pm1$

If

$$
\boxed{
v_2(m-1)\ge8
}
$$

or

$$
\boxed{
v_2(m+1)\ge8,
}
$$

then both gates are $2$-adic squares.

The threshold is sharp for the second gate in the sense that

$$
v_2(m\mp1)=7
$$

forces

$$
Q_2(m)
$$

to be nonsquare.

Thus the four geometric boundary points are centers of genuine positive-radius $2$-adic solution tubes.

Consequently the R9 congruence

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

cannot be upgraded to

$$
c=\pm1
$$

by local square-unit analysis alone.

The correct next stage must use global arithmetic:

- multi-prime Mordell-Weil compatibility;
- elliptic Chabauty on one of the genus-$2$ covers;
- simultaneous covering descent;
- or a global height / logarithmic argument.

The main R10 conclusion is therefore a rigorous strategy correction:

$$
\boxed{
\text{Pure }\mathbb Q_2\text{ boundary isolation is impossible.}
}
$$

---

# 1. Fixed-fiber square gates

For

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

the first Twin-Psi quartic is, up to a rational square denominator,

$$
\boxed{
Q_1(m)
=
(1024m^2+49)(49m^2+1024).
}
$$

Expanding:

$$
\boxed{
Q_1(m)
=
50176m^4
+
1050977m^2
+
50176.
}
$$

The second square gate is

$$
\boxed{
Q_2(m)
=
\left[
1073(m^2+1)-1950m
\right]
\left[
1073(m^2+1)+1950m
\right].
}
$$

Expanding:

$$
\boxed{
Q_2(m)
=
1151329m^4
-
1499842m^2
+
1151329.
}
$$

The denominators removed in passing from the normalized Twin-Psi equations to

$$
Q_1,
\qquad
Q_2
$$

are rational squares.

Therefore squarehood is unchanged.

---

# 2. Boundary values

At

$$
m=0,
$$

we have

$$
\boxed{
Q_1(0)
=
50176
=
224^2,
}
$$

and

$$
\boxed{
Q_2(0)
=
1151329
=
1073^2.
}
$$

At

$$
m=1,
$$

we have

$$
\boxed{
Q_1(1)
=
1151329
=
1073^2,
}
$$

and

$$
\boxed{
Q_2(1)
=
802816
=
896^2.
}
$$

Because both quartics are even:

$$
Q_i(-1)=Q_i(1).
$$

Because both quartics are reciprocal:

$$
m^4Q_i(1/m)=Q_i(m),
$$

the boundary at

$$
m=\infty
$$

is equivalent to the boundary at

$$
m=0.
$$

Thus all four geometric boundary points have both gate values equal to nonzero rational squares.

---

# 3. General $2$-adic openness lemma

Let

$$
f
$$

be a rational function defined at

$$
x_0\in\mathbb Q_2,
$$

and suppose

$$
f(x_0)\in\mathbb Q_2^{\times2}.
$$

Because

$$
f
$$

is continuous and

$$
\mathbb Q_2^{\times2}
$$

is an open subgroup of

$$
\mathbb Q_2^\times,
$$

there exists an open neighborhood

$$
U
$$

of

$$
x_0
$$

such that

$$
\boxed{
f(U)\subseteq\mathbb Q_2^{\times2}.
}
$$

For finitely many functions

$$
f_1,\ldots,f_r
$$

with nonzero square values at

$$
x_0,
$$

the intersection of the corresponding neighborhoods is again open.

Therefore simultaneous nonzero squarehood is locally stable.

This already implies that a boundary point at which both R8 gates are nonzero squares cannot be isolated by local squarehood alone.

The remaining sections make the neighborhoods explicit.

---

# 4. Square criterion for odd $2$-adic units

For

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

the standard criterion is

$$
\boxed{
u\in\mathbb Q_2^2
\iff
u\equiv1\pmod8.
}
$$

More generally, a nonzero

$$
q\in\mathbb Q_2
$$

is a square if and only if

1. its valuation is even;

2. its odd unit part is congruent to

$$
1\pmod8.
$$

Therefore if

$$
q_0
$$

is already a nonzero square and

$$
\frac q{q_0}
\in
1+8\mathbb Z_2,
$$

then

$$
q
$$

is a square.

This is the criterion used below.

---

# 5. Neighborhood of $m=0$: first gate

Write

$$
m=\delta.
$$

Normalize by the boundary square:

$$
Q_1(0)=224^2.
$$

Then

$$
\boxed{
\frac{Q_1(\delta)}{224^2}
=
1
+
\frac{1050977}{50176}\delta^2
+
\delta^4.
}
$$

Now

$$
\boxed{
v_2(50176)=10,
}
$$

and

$$
1050977
$$

is odd.

Hence

$$
\boxed{
v_2
\left(
\frac{1050977}{50176}
\right)
=
-10.
}
$$

If

$$
v_2(\delta)=k,
$$

the first correction has valuation

$$
2k-10.
$$

The quartic correction has valuation

$$
4k.
$$

Therefore if

$$
k\ge7,
$$

then

$$
2k-10\ge4.
$$

Thus

$$
\boxed{
\frac{Q_1(\delta)}{224^2}
\in
1+16\mathbb Z_2
\subset
1+8\mathbb Z_2.
}
$$

Hence:

$$
\boxed{
v_2(m)\ge7
\Longrightarrow
Q_1(m)\in\mathbb Q_2^2.
}
$$

---

# 6. Sharpness at $m=0$

Suppose

$$
v_2(m)=6.
$$

Then

$$
2v_2(m)-10=2.
$$

The quartic correction has valuation

$$
24.
$$

Therefore the first nonzero correction to

$$
Q_1(m)/224^2
$$

has exact valuation

$$
2.
$$

Its coefficient is odd.

Consequently

$$
\boxed{
\frac{Q_1(m)}{224^2}
\equiv
5
\pmod8.
}
$$

Therefore:

$$
\boxed{
v_2(m)=6
\Longrightarrow
Q_1(m)\notin\mathbb Q_2^2.
}
$$

So the radius

$$
2^7\mathbb Z_2
$$

is sharp at the first transition shell for the first gate.

---

# 7. Neighborhood of $m=0$: second gate

Normalize by:

$$
Q_2(0)=1073^2.
$$

Then

$$
\boxed{
\frac{Q_2(\delta)}{1073^2}
=
1
-
\frac{1499842}{1151329}\delta^2
+
\delta^4.
}
$$

Now

$$
1151329
$$

is odd and

$$
\boxed{
v_2(1499842)=1.
}
$$

Therefore:

$$
\boxed{
v_2
\left(
\frac{1499842}{1151329}
\right)
=
1.
}
$$

If

$$
v_2(\delta)\ge1,
$$

the quadratic correction already has valuation at least

$$
3.
$$

Thus the second gate is locally much less restrictive near

$$
m=0
$$

than the first gate.

In particular:

$$
\boxed{
v_2(m)\ge7
\Longrightarrow
Q_2(m)\in\mathbb Q_2^2.
}
$$

Combining with the previous section:

$$
\boxed{
v_2(m)\ge7
\Longrightarrow
Q_1(m),Q_2(m)\in\mathbb Q_2^2.
}
$$

---

# 8. Neighborhood of $m=\infty$

Because:

$$
\boxed{
m^4Q_i(1/m)=Q_i(m),
}
$$

and:

$$
m^4
$$

is a square, squarehood at

$$
m
$$

is equivalent to squarehood at

$$
1/m.
$$

Therefore the result at zero immediately gives:

$$
\boxed{
v_2(1/m)\ge7
\Longrightarrow
Q_1(m),Q_2(m)\in\mathbb Q_2^2.
}
$$

The first transition shell is again sharp through the first gate.

Thus the boundary at infinity also has a genuine open square-solution neighborhood.

---

# 9. Neighborhood of $m=1$: first gate

Write

$$
m=1+\delta.
$$

Exact factorization gives:

$$
\boxed{
\frac{Q_1(1+\delta)}{1073^2}
=
1
+
\frac{
\delta(\delta+2)
\left(
50176\delta^2
+
100352\delta
+
1151329
\right)
}{
1151329
}.
}
$$

If

$$
v_2(\delta)=k\ge2,
$$

then:

$$
v_2(\delta+2)=1,
$$

and the final parenthesis is odd.

Therefore the correction has valuation

$$
k+1.
$$

Hence for

$$
k\ge2,
$$

the correction belongs to

$$
8\mathbb Z_2.
$$

Thus:

$$
\boxed{
v_2(m-1)\ge2
\Longrightarrow
Q_1(m)\in\mathbb Q_2^2.
}
$$

Again the first gate is not the restrictive one near this boundary.

---

# 10. Neighborhood of $m=1$: second gate

Normalize by

$$
Q_2(1)=896^2.
$$

Exact factorization gives:

$$
\boxed{
\frac{Q_2(1+\delta)}{896^2}
=
1
+
\frac{
\delta(\delta+2)
\left(
1151329\delta^2
+
2302658\delta
+
802816
\right)
}{
802816
}.
}
$$

The relevant valuations are:

$$
\boxed{
v_2(802816)=14,
}
$$

$$
\boxed{
v_2(2302658)=1,
}
$$

and:

$$
1151329
$$

is odd.

Let:

$$
v_2(\delta)=k.
$$

For:

$$
k\ge8,
$$

we have:

$$
v_2(\delta+2)=1.
$$

Inside the final parenthesis the three terms have valuations:

$$
2k,
$$

$$
k+1,
$$

and:

$$
14.
$$

At:

$$
k=8,
$$

the minimum is:

$$
9.
$$

Therefore the total numerator correction has valuation at least:

$$
8+1+9=18.
$$

After dividing by the boundary value of valuation:

$$
14,
$$

the relative correction has valuation at least:

$$
4.
$$

Hence:

$$
\boxed{
v_2(m-1)\ge8
\Longrightarrow
Q_2(m)\in\mathbb Q_2^2.
}
$$

Combining both gates:

$$
\boxed{
v_2(m-1)\ge8
\Longrightarrow
Q_1(m),Q_2(m)\in\mathbb Q_2^2.
}
$$

---

# 11. Sharpness at $m=1$

Suppose:

$$
v_2(m-1)=7.
$$

Then the quadratic term in the expansion of:

$$
Q_2(1+\delta)-Q_2(1)
$$

has exact valuation:

$$
16,
$$

while:

$$
v_2(Q_2(1))=14.
$$

All competing terms have strictly larger valuation.

Therefore:

$$
\boxed{
\frac{Q_2(1+\delta)}{Q_2(1)}
\equiv
5
\pmod8.
}
$$

Hence:

$$
\boxed{
v_2(m-1)=7
\Longrightarrow
Q_2(m)\notin\mathbb Q_2^2.
}
$$

Thus the radius:

$$
m\in1+2^8\mathbb Z_2
$$

is sharp at the first transition shell for the second gate.

---

# 12. Neighborhood of $m=-1$

Because both quartics are even:

$$
Q_i(-m)=Q_i(m),
$$

the analysis at:

$$
m=-1
$$

is identical.

Therefore:

$$
\boxed{
v_2(m+1)\ge8
\Longrightarrow
Q_1(m),Q_2(m)\in\mathbb Q_2^2.
}
$$

And:

$$
\boxed{
v_2(m+1)=7
\Longrightarrow
Q_2(m)\notin\mathbb Q_2^2.
}
$$

---

# 13. Explicit local solution tubes

The historical fixed-fiber double-square equations therefore have the following entire open $2$-adic neighborhoods of solutions:

## Tube around $m=0$

$$
\boxed{
m\in2^7\mathbb Z_2.
}
$$

## Tube around $m=\infty$

$$
\boxed{
m^{-1}\in2^7\mathbb Z_2.
}
$$

## Tube around $m=1$

$$
\boxed{
m\in1+2^8\mathbb Z_2.
}
$$

## Tube around $m=-1$

$$
\boxed{
m\in-1+2^8\mathbb Z_2.
}
$$

Every point in each displayed local neighborhood satisfies both fixed-fiber square equations over:

$$
\mathbb Q_2.
$$

These are genuine analytic families, not isolated points.

---

# 14. Consequence for the R9 boundary tubes

R9 proved a global necessary Mordell-Weil coefficient condition:

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

with only two torsion cosets.

The central points correspond to:

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

R10 shows that the local double-square cover already contains open neighborhoods around these four boundary points.

Therefore no argument of the following form can close the fixed slice:

> Increase the $2$-adic precision around the boundary and wait until only the center survives.

That cannot occur from local squarehood alone.

The local solution set has positive $2$-adic radius.

---

# 15. Formal-group interpretation

Let:

$$
P_\ast
$$

be one of the four R9 boundary centers on:

$$
E_0'.
$$

For sufficiently deep:

$$
Q
$$

in the local elliptic formal group, the functions:

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

and:

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

remain in the same nonzero square classes as:

$$
\mathcal R_1(P_\ast),
$$

and:

$$
\mathcal R_2(P_\ast).
$$

Since the center values are nonzero squares, there is an open formal subgroup neighborhood:

$$
\mathcal U_\ast
$$

such that:

$$
\boxed{
P\in P_\ast+\mathcal U_\ast
\Longrightarrow
\mathcal R_1(P),\mathcal R_2(P)
\in\mathbb Q_2^2.
}
$$

Thus the formal-group square conditions cannot force:

$$
Q=0.
$$

---

# 16. General impossibility theorem

## Theorem — Local Squarehood Cannot Isolate a Nonramified Boundary Point

Let:

$$
E/\mathbb Q_p
$$

be a smooth elliptic curve.

Let:

$$
f_1,\ldots,f_r
\in
\mathbb Q_p(E)^\times
$$

be rational functions defined and nonzero at:

$$
P_\ast\in E(\mathbb Q_p).
$$

Suppose:

$$
f_i(P_\ast)\in\mathbb Q_p^{\times2}
$$

for every:

$$
i.
$$

Then there exists a $p$-adic open neighborhood:

$$
U
$$

of:

$$
P_\ast
$$

such that:

$$
\boxed{
f_i(P)\in\mathbb Q_p^{\times2}
}
$$

for every:

$$
P\in U
$$

and every:

$$
i.
$$

Therefore simultaneous local squarehood cannot isolate:

$$
P_\ast.
$$

R10 is the direct application of this theorem to the two R8 gates.

---

# 17. When local isolation could work

A local square condition can become isolating only if some additional structure is present, for example:

- the square function vanishes at the center;
- the square function has a pole;
- ramification forces an odd valuation;
- two local analytic branches must satisfy an independent equation;
- a global Mordell-Weil condition intersects the local branch discretely;
- a $p$-adic logarithmic condition is imposed in addition to squarehood.

None of the four R9 boundary centers is isolated merely by the two nonzero square values.

Therefore R10 must change strategy.

---

# 18. Why this does not weaken R9

R9 remains a strong global theorem.

It used many finite reductions to force every global rational candidate into four thin $2$-adic tubes.

R10 only proves that the final passage

$$
\text{tube}
\longrightarrow
\text{center}
$$

cannot be obtained from the local square-unit criterion alone.

The correct interpretation is:

$$
\boxed{
\text{R9 performs global concentration;}
}
$$

$$
\boxed{
\text{R10 identifies genuine local branches inside that concentration.}
}
$$

The remaining obstruction is global, not local.

---

# 19. Why adding only higher $2$-adic precision is insufficient

Suppose a boundary point has an open neighborhood

$$
U
$$

in which both gate values are squares.

Then for every sufficiently large:

$$
k,
$$

there are noncentral points satisfying:

$$
P\equiv P_\ast\pmod{2^k}
$$

inside the local cover.

Therefore deeper congruence conditions at the same place do not automatically produce a contradiction.

The problem is not insufficient precision.

It is the existence of an actual $2$-adic analytic branch.

---

# 20. The correct remaining question

The remaining fixed-slice question is now:

> Does any nonboundary point on one of these local $2$-adic branches also belong to the global rational Mordell-Weil group and satisfy all global arithmetic compatibility conditions?

This is fundamentally a local-global intersection problem.

The relevant objects are:

$$
\boxed{
E_0'(\mathbb Q)
}
$$

inside

$$
\boxed{
E_0'(\mathbb Q_2),
}
$$

together with the two quadratic-cover branches.

---

# 21. Preferred global methods after R10

The most appropriate next methods are:

1. **elliptic Chabauty on a genus-$2$ square cover;**
2. **Mordell-Weil sieve using many odd primes;**
3. **simultaneous covering descent;**
4. **$p$-adic elliptic logarithm relations involving global generators;**
5. **height bounds plus local residue restrictions.**

A pure formal-square expansion at:

$$
p=2
$$

is no longer the preferred route.

---

# 22. Return to the two genus-$2$ windows

R7 and R8 produced two complementary genus-$2$ windows.

The new window has:

$$
\boxed{
\operatorname{Jac}(G)
\sim
E_f\times E_0.
}
$$

The historical window has:

$$
\boxed{
\operatorname{Jac}(H_2)
\sim
E_0\times E_-.
}
$$

All three distinct elliptic factors have certified rank:

$$
1.
$$

R10 suggests that global compatibility between these two genus-$2$ windows is a stronger next target than local formal isolation.

---

# 23. Shared-$E_0$ strategy

Both genus-$2$ windows map to the same rank-one:

$$
E_0.
$$

A rational point on the original genus-$5$ fiber must satisfy two different quadratic lifting conditions over a common Mordell-Weil orbit.

Thus the natural global problem is:

$$
\boxed{
P\in E_0(\mathbb Q)
}
$$

such that two independent rational functions are simultaneously squares.

This is exactly the global interpretation of the R8 double-square reduction.

The local tubes are the $2$-adic realizations of these covers.

---

# 24. Suggested finite-prime continuation

At an arbitrary good odd prime:

$$
p,
$$

one should not require the entire reduction group to be a power of two.

Instead:

1. enumerate reduced points satisfying both square gates;
2. project those points to the Sylow-$2$ subgroup;
3. project the known odd-index Mordell-Weil surrogate by the same odd cofactor;
4. compare projected candidate images with the projected Mordell-Weil subgroup;
5. intersect the resulting coefficient constraints globally.

This is the proof-safe philosophy already used successfully in the older dual-quotient sieve.

The critical rule is:

$$
\boxed{
\text{project candidate and Mordell-Weil data by the same map.}
}
$$

---

# 25. Why this can improve on R9

R9 restricted itself to primes where:

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

was already a power of two.

That made coefficient tracking especially clean.

However many additional good primes have large Sylow-$2$ subgroups even when their total group order has an odd factor.

Those primes were not used by R9.

A synchronized Sylow-$2$ projection can potentially add independent coefficient restrictions without any assumption about the unknown odd saturation index.

Thus there remains substantial global finite-prime information that has not yet been used in the R8 double-gate representation.

---

# 26. Strategy correction

The original proposed R10 objective was:

$$
\boxed{
\text{prove }u=0
\text{ inside each }2\text{-adic boundary tube}.
}
$$

R10 proves this cannot follow from local squarehood alone.

The corrected objective becomes:

$$
\boxed{
\text{show that no noncentral point on the local branch is globally compatible}.
}
$$

This is a stronger local-global statement and requires information from outside:

$$
\mathbb Q_2.
$$

---

# 27. R10 theorem package

## Theorem A — Local openness

Each of the four boundary points has a nontrivial open:

$$
\mathbb Q_2
$$

neighborhood in which both fixed-fiber gates remain squares.

## Theorem B — Explicit radii

The following are sufficient:

$$
\boxed{
v_2(m)\ge7,
}
$$

$$
\boxed{
v_2(1/m)\ge7,
}
$$

$$
\boxed{
v_2(m-1)\ge8,
}
$$

and:

$$
\boxed{
v_2(m+1)\ge8.
}
$$

## Theorem C — Sharp first transition shells

If:

$$
v_2(m)=6,
$$

the first gate is nonsquare.

If:

$$
v_2(m\mp1)=7,
$$

the second gate is nonsquare.

## Theorem D — Failure of pure local isolation

No argument based only on sufficiently deep $2$-adic square-unit expansions can prove that the R9 tubes contain only their boundary centers.

---

# 28. R11 proposed program

R11 should return to global arithmetic with the R10 local obstruction understood.

## R11-A — General Sylow-$2$ double-gate sieve

Extend R9 from power-of-two reduction groups to arbitrary good primes.

For each prime:

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

with:

$$
m
$$

odd, apply:

$$
[m]
$$

to both candidate points and Mordell-Weil data.

Build the resulting normalized coefficient restrictions.

## R11-B — Search for higher $2$-power coefficient forcing

Target primes whose projected Mordell-Weil direction has large order:

$$
2^s.
$$

Determine whether the R9 condition:

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

can be strengthened globally.

## R11-C — Genus-$2$ Mordell-Weil sieve

Apply local reduction simultaneously to the two genus-$2$ windows:

$$
G,
\qquad
H_2.
$$

Use their common:

$$
E_0
$$

image as the synchronization coordinate.

## R11-D — Elliptic-Chabauty implementation target

On either degree-$2$ cover of:

$$
E_0,
$$

classify rational lifts of the rank-one Mordell-Weil group.

Because the cover has genus:

$$
2,
$$

and the elliptic base has rank:

$$
1,
$$

this is the natural elliptic-Chabauty regime.

## R11-E — Boundary-branch intersection

Use the R10 local radii to identify the exact local disks that must be retained in every $p$-adic implementation.

Do not waste computation on disks already excluded by R9.

---

# 29. Final conclusion

R10 does not produce the hoped-for local nonexistence theorem.

Instead, it proves something strategically important:

$$
\boxed{
\text{the hoped-for theorem cannot come from pure }\mathbb Q_2\text{ squarehood.}
}
$$

The four R9 boundary centers are not isolated $2$-adic solutions.

They sit inside genuine open families of local simultaneous square solutions.

For the historical fixed fiber, explicit neighborhoods are:

$$
m\in2^7\mathbb Z_2,
$$

$$
m^{-1}\in2^7\mathbb Z_2,
$$

$$
m\in1+2^8\mathbb Z_2,
$$

and:

$$
m\in-1+2^8\mathbb Z_2.
$$

Thus the R9 boundary tubes represent real local geometry rather than incomplete numerical precision.

The remaining task is global:

$$
\boxed{
\text{intersect those local branches with the global rank-one Mordell-Weil orbit.}
}
$$

That is the canonical starting point for R11.