# Perfect Cuboid C2 — Two-Turn Square-Preserving Repair

## Dual-Square Sum Closure, Magic-Sum Gates, and the Global Completeness of the Two-Axis Move Space

**Series:** Perfect Cuboid Global Closure  
**Branch:** Conservation–Duality Branch C  
**Phase:** C2 — Two-Turn Repair  
**Date:** 2026-10-02

---

# Abstract

C1 identified the previous fixed-$k$ arithmetic programs as one-axis orbits of a simpler conservation move algebra.

C2 removes the fixed-axis restriction.

Two already closed dual pairs are allowed to undergo independent rational rotations.

The resulting problem admits a particularly small exact reformulation.

Define the rational dual-square set

$$
\boxed{
\mathcal P
=
\left\{
X\in\mathbb Q\cap(0,1):
X\in\mathbb Q^2,\;
1-X\in\mathbb Q^2
\right\}.
}
$$

Then Perfect Cuboid existence is equivalent to the additive closure statement

$$
\boxed{
\exists\,X,Y,Z\in\mathcal P
\quad
X+Y=Z.
}
$$

In other words:

> the classical Perfect Cuboid problem asks whether the dual-square set $\mathcal P$ contains a nondegenerate additive triple.

Equivalently, if

$$
\chi(r)
=
\left(
\frac{1-r^2}{1+r^2}
\right)^2,
$$

then Perfect Cuboid existence is equivalent to

$$
\boxed{
\chi(r)+\chi(s)=\chi(t)
}
$$

for suitable rational

$$
r,s,t.
$$

The two-turn problem can be reduced further to two explicit square gates in only two rational variables.

Define

$$
\Delta_D(r,s)
=
\left[
(1+r^2)(1-s^2)
\right]^2
+
\left[
(1-r^2)(1+s^2)
\right]^2,
$$

and

$$
\Delta_C(r,s)
=
-
(rs-r-s-1)
(rs-r+s+1)
(rs+r-s+1)
(rs+r+s-1).
$$

Then

$$
\boxed{
\Delta_D+\Delta_C
=
\left[
(1+r^2)(1+s^2)
\right]^2.
}
$$

The Perfect Cuboid target is exactly

$$
\boxed{
\Delta_D(r,s)\in\mathbb Q^2,
\qquad
\Delta_C(r,s)\in\mathbb Q^2.
}
$$

The two outputs therefore form another conserved dual pair.

This gives the Conservation–Duality Branch a recursive structure:

$$
\boxed{
\text{closed pair}
+
\text{closed pair}
\longrightarrow
\text{new conserved pair},
}
$$

and the problem is whether that new pair is also square-closed.

For the current seed

$$
r=\frac{25}{39},
\qquad
s=\frac{17}{28},
$$

the second gate already passes.

The first gate has exact squarefree kernel

$$
\boxed{
233\cdot1913.
}
$$

Thus the familiar arithmetic defect reappears directly as the squarefree kernel of the two-turn Magic-Sum Gate.

---

# 1. The dual-square set

Define

$$
\mathcal P
=
\left\{
X\in\mathbb Q\cap(0,1):
X=u^2,\;
1-X=v^2
\text{ for some }
u,v\in\mathbb Q
\right\}.
$$

Every point of

$$
\mathcal P
$$

is one cell of a rationally square-closed dual pair:

$$
(X,1-X).
$$

The rational unit circle parameterization gives

$$
u
=
\frac{1-r^2}{1+r^2},
$$

$$
v
=
\frac{2r}{1+r^2}.
$$

Therefore

$$
\boxed{
X=\chi(r)
=
\left(
\frac{1-r^2}{1+r^2}
\right)^2.
}
$$

Conversely every nondegenerate rational dual-square pair arises from some rational projective parameter $r$.

Hence

$$
\boxed{
\mathcal P
=
\chi
\left(
\mathbb P^1(\mathbb Q)
\right)
\cap
(0,1).
}
$$

---

# 2. Perfect Cuboid as an additive closure problem

Use the conservation-square simplex variables:

$$
A,B,C,D,E,F.
$$

Because

$$
D=A+B,
$$

$$
C=1-D,
$$

a Perfect Cuboid is equivalent to

$$
A,B,D\in\mathcal P
$$

with

$$
D=A+B.
$$

The other three cells follow automatically:

$$
F=1-A,
$$

$$
E=1-B,
$$

$$
C=1-D.
$$

Therefore:

## Theorem — Dual-Square Sum Closure

Perfect Cuboid exists if and only if

$$
\boxed{
\exists X,Y,Z\in\mathcal P
:
X+Y=Z
}
$$

with

$$
0<X,Y,Z<1.
$$

Equivalently:

$$
\boxed{
(\mathcal P+\mathcal P)\cap\mathcal P
\neq\varnothing
}
$$

inside the nondegenerate unit interval.

Thus the classical question becomes:

> Is $\mathcal P$ sum-free, or does it contain a nontrivial additive closure triple?

---

# 3. Rational parameter form

Write:

$$
X=\chi(r),
$$

$$
Y=\chi(s),
$$

$$
Z=\chi(t),
$$

with

$$
r,s,t\in\mathbb Q.
$$

Then the Perfect Cuboid condition is:

$$
\boxed{
\chi(r)+\chi(s)=\chi(t).
}
$$

Explicitly:

$$
\boxed{
\left(
\frac{1-r^2}{1+r^2}
\right)^2
+
\left(
\frac{1-s^2}{1+s^2}
\right)^2
=
\left(
\frac{1-t^2}{1+t^2}
\right)^2.
}
$$

This is the global two-turn closure equation.

---

# 4. Global completeness of the two-turn orbit

C1 introduced rational rotations of a closed dual pair.

Suppose a seed dual pair has rational unit-circle root vector

$$
(u_0,v_0).
$$

Let a target closed pair have root vector

$$
(u_1,v_1).
$$

Both satisfy

$$
u_i^2+v_i^2=1.
$$

Define

$$
c=u_0u_1+v_0v_1,
$$

and

$$
s=u_0v_1-v_0u_1.
$$

Then:

$$
c,s\in\mathbb Q,
$$

and:

$$
c^2+s^2=1.
$$

The rational rotation

$$
\begin{pmatrix}
c&-s\\
s&c
\end{pmatrix}
$$

maps the seed vector to the target vector.

Therefore:

$$
\boxed{
\text{the rational rotation group acts transitively on rational closed dual pairs}.
}
$$

Since the current seed already contains two square-closed dual pairs,

$$
(A,F)
$$

and

$$
(B,E),
$$

independent rational rotations of those two pairs can reach any ordered pair

$$
(X,Y)\in\mathcal P^2.
$$

Hence:

$$
\boxed{
\text{the two-turn repair space is globally complete}.
}
$$

It is not merely a local neighborhood of the current seed.

---

# 5. Seed-centered rotation coordinates

The seed pair

$$
(A,F)
$$

has half-angle parameter

$$
p_A=\frac{25}{39}.
$$

The seed pair

$$
(B,E)
$$

has half-angle parameter

$$
p_B=\frac{17}{28}.
$$

Let the two rational turn parameters be

$$
\rho,\sigma.
$$

Under the rational tangent group law,

$$
p\oplus r
=
\frac{p+r}{1-pr},
$$

the moved pair parameters are:

$$
\boxed{
r
=
p_A\oplus\rho
=
\frac{25+39\rho}{39-25\rho},
}
$$

and:

$$
\boxed{
s
=
p_B\oplus\sigma
=
\frac{17+28\sigma}{28-17\sigma}.
}
$$

Because the group action is transitive, allowing arbitrary rational

$$
\rho,\sigma
$$

is equivalent to allowing arbitrary rational

$$
r,s.
$$

Thus the seed-centered and global formulations are exact coordinate changes of the same two-dimensional search space.

---

# 6. Remove the third parameter

Define:

$$
u=r^2,
$$

$$
v=s^2,
$$

$$
w=t^2.
$$

Then:

$$
\chi(r)
=
\frac{(1-u)^2}{(1+u)^2}.
$$

The equation

$$
\chi(r)+\chi(s)=\chi(t)
$$

becomes a reciprocal quadratic in

$$
w.
$$

Define:

$$
N(u,v)
=
(1-u)^2(1+v)^2
+
(1-v)^2(1+u)^2,
$$

and:

$$
H(u,v)
=
(1+u)^2(1+v)^2.
$$

Then:

$$
\chi(r)+\chi(s)
=
\frac{N}{H}.
$$

The target equation is:

$$
\frac{N}{H}
=
\frac{(1-w)^2}{(1+w)^2}.
$$

Clearing denominators gives:

$$
\boxed{
A(u,v)(w^2+1)
+
2B(u,v)w
=
0,
}
$$

where:

$$
A=N-H,
$$

and:

$$
B=N+H.
$$

This equation is reciprocal:

$$
w\longleftrightarrow\frac1w.
$$

That symmetry is the algebraic image of complementary half-angle parameterizations.

---

# 7. First gate: make $D=X+Y$ a rational square

Since:

$$
H
=
\left[
(1+r^2)(1+s^2)
\right]^2
$$

is already a rational square, the quantity

$$
D=X+Y
=
\frac{N}{H}
$$

is a rational square if and only if:

$$
\boxed{
N\in\mathbb Q^2.
}
$$

Now:

$$
N
=
2
\left[
(1-uv)^2+(u-v)^2
\right].
$$

Substituting:

$$
u=r^2,
\qquad
v=s^2,
$$

gives the first Magic-Sum Gate:

$$
\boxed{
\Delta_D(r,s)
=
2
\left[
(1-r^2s^2)^2
+
(r^2-s^2)^2
\right].
}
$$

An equivalent and more geometric form is:

$$
\boxed{
\Delta_D(r,s)
=
\left[
(1+r^2)(1-s^2)
\right]^2
+
\left[
(1-r^2)(1+s^2)
\right]^2.
}
$$

Therefore the first gate is itself a rational Pythagorean closure.

---

# 8. Pythagorean interpretation of the first gate

Define:

$$
U
=
(1+r^2)(1-s^2),
$$

and:

$$
V
=
(1-r^2)(1+s^2).
$$

Then:

$$
\boxed{
\Delta_D=U^2+V^2.
}
$$

Thus:

$$
D\in\mathbb Q^2
$$

if and only if the rational pair:

$$
(U,V)
$$

forms the legs of a rational right triangle.

This gives a new interpretation:

> the sum of two already square-closed conservation cells is square exactly when a derived pair of rational turn expressions satisfies one additional Pythagorean gate.

So the first level of the two-turn repair is still classical rational-circle arithmetic, but now applied to the **move coordinates themselves**.

---

# 9. Parameterizing the first gate

Let:

$$
q\in\mathbb Q.
$$

Parameterize:

$$
\frac{U}{V}
=
\frac{2q}{1-q^2}.
$$

Writing:

$$
u=r^2,
$$

$$
v=s^2,
$$

gives:

$$
\frac{
(1+u)(1-v)
}{
(1-u)(1+v)
}
=
\frac{2q}{1-q^2}.
$$

Solving for:

$$
v
$$

gives:

$$
\boxed{
v
=
\frac{
(1+u)(1-q^2)
-
2q(1-u)
}{
(1+u)(1-q^2)
+
2q(1-u)
}.
}
$$

Therefore the first square gate can be solved rationally at the level of

$$
s^2.
$$

The remaining condition:

$$
s\in\mathbb Q
$$

is itself a square-lift condition.

This gives one possible nested construction for future C2 work.

---

# 10. Second gate: make the complement $C=1-D$ a square

The complementary state is:

$$
C
=
1-D.
$$

Since:

$$
D=\frac{\Delta_D}{\Lambda^2},
$$

where:

$$
\boxed{
\Lambda
=
(1+r^2)(1+s^2),
}
$$

we have:

$$
C
=
\frac{
\Lambda^2-\Delta_D
}{
\Lambda^2
}.
$$

Define:

$$
\boxed{
\Delta_C
=
\Lambda^2-\Delta_D.
}
$$

Then:

$$
C\in\mathbb Q^2
$$

if and only if:

$$
\boxed{
\Delta_C\in\mathbb Q^2.
}
$$

---

# 11. Four-wall factorization

Exact factorization gives:

$$
\boxed{
\Delta_C(r,s)
=
-
(rs-r-s-1)
(rs-r+s+1)
}
$$

$$
\boxed{
\qquad\cdot
(rs+r-s+1)
(rs+r+s-1).
}
$$

Thus the second square gate is a product of four bilinear “walls”.

This factorization is one of the main simplifications of C2.

The four walls correspond to special tangent-addition boundaries among the two rational turn parameters.

They are natural objects for:

- local residue analysis;
- squareclass tracking;
- defect-wall dynamics;
- symbolic factor grouping.

---

# 12. Meta-conservation identity

The two gates satisfy:

$$
\boxed{
\Delta_D+\Delta_C=\Lambda^2,
}
$$

where:

$$
\Lambda
=
(1+r^2)(1+s^2).
$$

Therefore the output pair:

$$
(\Delta_D,\Delta_C)
$$

is itself a conserved dual pair with total square:

$$
\Lambda^2.
$$

This is a recursive version of the original cuboid conservation law.

After normalization:

$$
D
=
\frac{\Delta_D}{\Lambda^2},
$$

$$
C
=
\frac{\Delta_C}{\Lambda^2},
$$

and:

$$
D+C=1.
$$

Hence the two-turn operation takes two closed input dual pairs and produces a new conserved output dual pair.

Perfect Cuboid closure occurs exactly when the output pair is also square-closed.

---

# 13. Conservation fusion operator

Define a partial arithmetic fusion operation:

$$
\boxed{
\mathfrak F:
\mathcal P\times\mathcal P
\longrightarrow
\mathbb Q^2
}
$$

by:

$$
\mathfrak F(X,Y)
=
(X+Y,\;1-X-Y)
$$

whenever:

$$
0<X+Y<1.
$$

The inputs are already square-closed dual cells.

The output always satisfies conservation:

$$
(X+Y)+(1-X-Y)=1.
$$

The Perfect Cuboid condition is:

$$
\boxed{
\mathfrak F(X,Y)
\in
\mathbb Q^2\times\mathbb Q^2.
}
$$

Equivalently:

$$
\boxed{
X+Y\in\mathcal P.
}
$$

Thus the problem is the existence of a square-closed fusion event.

---

# 14. Recursive closure interpretation

The fundamental object is no longer seven lengths.

It is the binary process:

$$
\boxed{
\text{closed dual pair}
+
\text{closed dual pair}
\longrightarrow
\text{conserved output dual pair}.
}
$$

There are two possibilities.

## Open output

At least one of:

$$
\Delta_D,
\Delta_C
$$

is nonsquare.

Then the arithmetic defect survives.

## Closed output

Both:

$$
\Delta_D,
\Delta_C
$$

are rational squares.

Then:

$$
\boxed{
\text{Perfect Cuboid closure is achieved}.
}
$$

This gives the Magic-Cube formulation a self-similar algebraic meaning.

---

# 15. Current seed in global two-turn coordinates

The current square-closed input pairs have parameters:

$$
\boxed{
r_0=\frac{25}{39},
}
$$

and:

$$
\boxed{
s_0=\frac{17}{28}.
}
$$

They generate:

$$
A=\chi(r_0)
=
\left(
\frac{448}{1073}
\right)^2,
$$

and:

$$
B=\chi(s_0)
=
\left(
\frac{495}{1073}
\right)^2.
$$

Thus:

$$
D=A+B
=
\frac{445729}{1073^2}.
$$

The complement is:

$$
C=1-D
=
\left(
\frac{840}{1073}
\right)^2.
$$

So the seed is exactly a state where:

$$
\boxed{
\Delta_C
\text{ is square},
}
$$

but:

$$
\boxed{
\Delta_D
\text{ is not square}.
}
$$

---

# 16. Exact first-gate defect of the seed

For:

$$
r_0=\frac{25}{39},
\qquad
s_0=\frac{17}{28},
$$

the first gate is:

$$
\Delta_D
=
\frac{
513180723841
}{
355492597824
}.
$$

The denominator is the exact square:

$$
355492597824
=
596232^2.
$$

The numerator factors as:

$$
\boxed{
513180723841
=
29^2
37^2
233
1913.
}
$$

Hence:

$$
\boxed{
\Delta_D
=
\left(
\frac{29\cdot37}{596232}
\right)^2
(233\cdot1913).
}
$$

Therefore the squarefree kernel of the failed Magic-Sum Gate is:

$$
\boxed{
233\cdot1913.
}
$$

This is exactly the defect already observed in:

- the missing face diagonal;
- the shared-core factorization;
- the CARCE half-angle defect;
- both one-axis conservation slices.

The same arithmetic obstruction has now appeared in all coordinate systems.

---

# 17. Exact second-gate closure of the seed

For the same seed:

$$
\Delta_C
=
\frac{
28783225
}{
12595401
}.
$$

Both numerator and denominator are squares:

$$
28783225=5365^2,
$$

$$
12595401=3549^2.
$$

Thus:

$$
\boxed{
\Delta_C
=
\left(
\frac{5365}{3549}
\right)^2.
}
$$

So the current seed has the exact two-gate status:

$$
\boxed{
(\Delta_D,\Delta_C)
=
(\text{nonsquare},\text{square}).
}
$$

The target is:

$$
\boxed{
(\text{square},\text{square}).
}
$$

---

# 18. Why the previous one-axis search was restrictive

The full C2 state has two independent rational input parameters:

$$
r,s.
$$

The previous fixed-$k$ analysis froze one input dual pair and varied only the other.

Therefore it restricted:

$$
(r,s)
$$

to a one-dimensional rational curve inside the full two-dimensional move space.

C2 removes that restriction.

Thus:

$$
\boxed{
\text{one-axis search}
\subset
\text{two-turn search}.
}
$$

The two-turn search is globally complete because of rational rotation transitivity.

---

# 19. Relation to CARCE

The CARCE equation is:

$$
\chi(r)+\chi(s)+\chi(q)=1.
$$

The C2 form is:

$$
\chi(r)+\chi(s)=\chi(t).
$$

Since:

$$
1-\chi(t)
=
\left(
\frac{2t}{1+t^2}
\right)^2
=
\chi
\left(
\frac{1-t}{1+t}
\right),
$$

the two equations are identical after the rational involution:

$$
q
=
\frac{1-t}{1+t}.
$$

Therefore:

$$
\boxed{
\text{C2 is not a new conjecture}.
}
$$

It is a new algebraic organization of the exact same existence problem.

Its advantage is that conservation and defect transfer are explicit.

---

# 20. Magic-Sum Polynomial

Let:

$$
u=r^2,
$$

$$
v=s^2,
$$

$$
w=t^2.
$$

Define:

$$
A(u,v)
=
(1-u)^2(1-v)^2
-
16uv,
$$

and:

$$
B(u,v)
=
(1-u)^2(1+v)^2
+
(1-v)^2(1+u)^2
+
(1+u)^2(1+v)^2.
$$

Then:

$$
\chi(r)+\chi(s)=\chi(t)
$$

is equivalent to:

$$
\boxed{
A(u,v)(w^2+1)
+
2B(u,v)w
=
0.
}
$$

This is the canonical **Magic-Sum Polynomial**.

It is reciprocal in:

$$
w.
$$

---

# 21. Discriminant of the Magic-Sum Polynomial

The discriminant in:

$$
w
$$

is:

$$
\boxed{
\Delta_w
=
32(1+u)^2(1+v)^2
\left[
u^2v^2+u^2-4uv+v^2+1
\right].
}
$$

But:

$$
u^2v^2+u^2-4uv+v^2+1
=
(1-uv)^2+(u-v)^2.
$$

Therefore:

$$
\boxed{
\Delta_w
=
32(1+u)^2(1+v)^2
\left[
(1-uv)^2+(u-v)^2
\right].
}
$$

The condition for:

$$
w\in\mathbb Q
$$

is exactly the first square gate:

$$
\boxed{
2
\left[
(1-uv)^2+(u-v)^2
\right]
\in
\mathbb Q^2.
}
$$

This is precisely:

$$
\Delta_D\in\mathbb Q^2.
$$

Then the additional requirement:

$$
t\in\mathbb Q
$$

is:

$$
\boxed{
w\in\mathbb Q^2,
}
$$

which is equivalent to the complementary gate:

$$
\Delta_C\in\mathbb Q^2.
$$

Thus the reciprocal-quadratic viewpoint and the two-gate conservation viewpoint are exactly equivalent.

---

# 22. Two-stage arithmetic hierarchy

The global two-turn repair therefore has a clean two-stage structure.

## Stage 1 — Sum becomes square

Require:

$$
\boxed{
\Delta_D(r,s)\in\mathbb Q^2.
}
$$

This makes:

$$
D=A+B
$$

a rational square.

## Stage 2 — Complement becomes square

Require:

$$
\boxed{
\Delta_C(r,s)\in\mathbb Q^2.
}
$$

This makes:

$$
C=1-D
$$

a rational square.

Therefore:

$$
\boxed{
\text{Perfect Cuboid}
=
\text{square sum}
+
\text{square complement}.
}
$$

This is the minimal arithmetic meaning of the two-turn repair.

---

# 23. C2 search space

The raw geometric search space is:

$$
(r,s)\in\mathbb Q^2.
$$

Define the two gate loci:

$$
\mathcal G_D
=
\{
(r,s):
\Delta_D(r,s)\in\mathbb Q^2
\},
$$

and:

$$
\mathcal G_C
=
\{
(r,s):
\Delta_C(r,s)\in\mathbb Q^2
\}.
$$

Then the Perfect Cuboid target is:

$$
\boxed{
\mathcal G_D\cap\mathcal G_C
}
$$

inside the nondegenerate positivity region.

The current seed lies on:

$$
\mathcal G_C
$$

but not on:

$$
\mathcal G_D.
$$

---

# 24. Defect motion in two dimensions

The one-axis approach tried to repair the defect while remaining on a fixed rational slice.

C2 instead permits motion in both:

$$
r
$$

and:

$$
s.
$$

Therefore a deformation path can move:

$$
(r_0,s_0)
\longrightarrow
(r,s)
$$

while tracking the squarefree kernels:

$$
\operatorname{sf}(\Delta_D),
$$

and:

$$
\operatorname{sf}(\Delta_C).
$$

The target event is:

$$
\boxed{
\operatorname{sf}(\Delta_D)
=
\operatorname{sf}(\Delta_C)
=
1.
}
$$

This is the arithmetic analogue of simultaneous six-cell closure.

---

# 25. Local residue signatures

For every odd prime:

$$
p,
$$

one can attach quadratic characters:

$$
\left(
\frac{\Delta_D}{p}
\right),
$$

and:

$$
\left(
\frac{\Delta_C}{p}
\right).
$$

A rational solution must satisfy local square compatibility at every place.

Because:

$$
\Delta_C
$$

factors into four bilinear walls, its local character can be decomposed into four elementary character contributions.

This is a promising route for C3:

$$
\boxed{
\text{local wall-sign dynamics}.
}
$$

Instead of performing high-genus descent immediately, one can study how rational conservation turns change the local sign pattern of the four walls.

---

# 26. Algebraic wall structure

Define:

$$
L_1=rs-r-s-1,
$$

$$
L_2=rs-r+s+1,
$$

$$
L_3=rs+r-s+1,
$$

$$
L_4=rs+r+s-1.
$$

Then:

$$
\boxed{
\Delta_C=-L_1L_2L_3L_4.
}
$$

The hypersurfaces:

$$
L_i=0
$$

are degenerate boundaries of the two-turn state.

They divide the rational move plane into arithmetic chambers.

The squareclass of:

$$
\Delta_C
$$

depends only on the combined squareclasses of these four walls.

Thus the magic-cube move plane acquires an explicit wall-and-chamber arithmetic geometry.

---

# 27. Interpretation of a successful move

A successful two-turn move is a rational point:

$$
(r,s)
$$

such that:

$$
\Delta_D=p^2,
$$

$$
\Delta_C=q^2
$$

for:

$$
p,q\in\mathbb Q_{>0}.
$$

Then:

$$
\Lambda^2=p^2+q^2.
$$

Normalize:

$$
d=\frac p\Lambda,
$$

$$
c=\frac q\Lambda.
$$

Together with:

$$
a=
\frac{1-r^2}{1+r^2},
$$

$$
b=
\frac{1-s^2}{1+s^2},
$$

we have:

$$
a^2+b^2=d^2,
$$

and:

$$
a^2+b^2+c^2=1.
$$

The complementary dual pairs automatically give the remaining rational face diagonals.

Thus a successful point gives a rational Perfect Cuboid immediately.

A common denominator then produces an integer witness.

---

# 28. No seed dependence in the final theorem

Although the current seed motivated the move algebra, the final C2 theorem does not depend on it.

The seed only provides:

- a useful initial rational point;
- a one-defect diagnostic;
- a known arithmetic signature:
  $$
  233\cdot1913.
  $$

The two-turn search itself spans the entire rational dual-square state space.

Therefore any solution found in C2 is a global Perfect Cuboid solution, not merely a deformation of the numerical seed in a restricted family.

---

# 29. Recommended C3 direction

The next phase should avoid large integer-height scans.

The strongest algebraic directions are:

## C3-A — Parameterize the first gate

Use:

$$
\Delta_D=R^2
$$

as a rational Pythagorean condition and parameterize it exactly.

Then study whether:

$$
\Delta_C
$$

can become square on that parameterized surface.

## C3-B — Wall-squareclass invariant

Track:

$$
[L_1],
[L_2],
[L_3],
[L_4]
\in
\mathbb Q^\times/\mathbb Q^{\times2}.
$$

Search for algebraic relations forcing or forbidding:

$$
-L_1L_2L_3L_4
$$

to be trivial in the squareclass group.

## C3-C — Seed-defect cancellation

Use the current seed's squarefree kernel

$$
233\cdot1913
$$

as an explicit local target.

Find rational move directions that alter the first-gate kernel at precisely these primes without introducing new defects in the second gate.

## C3-D — Conservation renormalization

Study the binary operator:

$$
\mathfrak F
$$

as a partial arithmetic composition law on:

$$
\mathcal P.
$$

Determine whether:

$$
\mathcal P
$$

is sum-free.

This is the most abstract and potentially theorem-oriented form of the problem.

---

# 30. Canonical C2 theorem target

Perfect Cuboid existence is equivalent to:

$$
\boxed{
\exists r,s\in\mathbb Q
}
$$

such that:

$$
\boxed{
\left[
(1+r^2)(1-s^2)
\right]^2
+
\left[
(1-r^2)(1+s^2)
\right]^2
\in
\mathbb Q^2,
}
$$

and:

$$
\boxed{
-
(rs-r-s-1)
(rs-r+s+1)
(rs+r-s+1)
(rs+r+s-1)
\in
\mathbb Q^2,
}
$$

with the corresponding normalized cells strictly between:

$$
0
$$

and:

$$
1.
$$

This is the canonical two-variable Conservation–Duality formulation.

---

# 31. One-line C2 formulation

$$
\boxed{
\text{Find two rational closed dual pairs whose conserved fusion is again a closed dual pair.}
}
$$

Or, in additive language:

$$
\boxed{
(\mathcal P+\mathcal P)\cap\mathcal P
\neq\varnothing\ ?
}
$$

---

**End of C2**