> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification. # Perfect Cuboid Center-Axis Rational Closure Theorem ## An Exact Reformulation of the Perfect Cuboid Problem as Rational Closure on a Low-Dimensional Coupling Surface **Series: Perfect Cuboid Global Closure** **Main line: Geometry Branch G** **Verification sub-branch: Verification Branch A** **Document role: Subsequent main paper / Canonical Research Anchor** **Version: v0.1** **Date: 2026-09-28** --- # Abstract This paper proposes and fully proves an exact equivalent reformulation of the Perfect Cuboid problem. The classical Perfect Cuboid problem requires the existence of positive integers: $$ a,b,c,d,e,f,g $$ satisfying: $$ a^2+b^2=d^2, $$ $$ a^2+c^2=e^2, $$ $$ b^2+c^2=f^2, $$ $$ a^2+b^2+c^2=g^2. $$ This paper no longer takes seven integers as the primary variables. Let: $$ x_a=\frac ag, \qquad x_b=\frac bg, \qquad x_c=\frac cg. $$ For each edge and its opposite face diagonal, construct a center-axis right-triangle coupling and introduce a rational half-angle parameter: $$ t_a, \qquad t_b, \qquad t_c. $$ This paper proves: > **A Perfect Cuboid exists if and only if there exist three positive rational numbers** > > $$ > t_a,t_b,t_c\in\mathbb Q\cap(0,1) > $$ > > **satisfying** > > $$ > \boxed{ > \left( > \frac{1-t_a^2}{1+t_a^2} > \right)^2 > + > \left( > \frac{1-t_b^2}{1+t_b^2} > \right)^2 > + > \left( > \frac{1-t_c^2}{1+t_c^2} > \right)^2 > = > 1. > } > $$ This equation is called: $$ \boxed{ \text{Center-Axis Rational Closure Equation} } $$ or, in abbreviated form: $$ \boxed{ \text{CARCE}. } $$ This theorem compresses: $$ \text{Seven positive integers} + \text{Four square relations} $$ exactly into: $$ \boxed{ \text{Three rational coupling parameters} + \text{One global closure equation}. } $$ If any rational triple: $$ (t_a,t_b,t_c) $$ satisfying CARCE is found, an exact integer Perfect Cuboid witness can be generated mechanically. The canonical existence target for subsequent research can therefore be rewritten as: $$ \boxed{ \mathcal M(\mathbb Q)\neq\varnothing, } $$ where: $$ \mathcal M = \left\{ (t_a,t_b,t_c)\in(0,1)^3: \Phi(t_a,t_b,t_c)=0 \right\}. $$ The currently known near-arithmetic seed: $$ (a,b,c;g) = (448,495,840;1073) $$ corresponds exactly in these coordinates to: $$ \boxed{ T_0 = \left( \frac{25}{39}, \frac{17}{28}, \sqrt{\frac{233}{1913}} \right), } $$ and: $$ \Phi(T_0)=0 $$ exactly. Thus, geometric closure is not what is currently missing. What is truly missing is: $$ \boxed{ T^\ast\in\mathbb Q^3\cap\mathcal M. } $$ This paper also provides: - an exact equivalence proof; - an exact integer witness generator; - a square-defect formulation; - a symmetric polynomial surface; - a smoothness theorem; - the exact arithmetic decomposition of the current seed; - geometry-preserving dynamics; - the positioning of the arithmetic observer; - a future proof pipeline; - a formal certificate interface. --- # 1. The Classical Problem A Perfect Cuboid requires: $$ a,b,c,d,e,f,g\in\mathbb Z_{>0} $$ and: $$ a^2+b^2=d^2, $$ $$ a^2+c^2=e^2, $$ $$ b^2+c^2=f^2, $$ $$ a^2+b^2+c^2=g^2. $$ Using the usual naming convention: - $d$ is the $ab$ face diagonal; - $e$ is the $ac$ face diagonal; - $f$ is the $bc$ face diagonal; - $g$ is the body diagonal. The last equation together with the first three can be rearranged as: $$ a^2+f^2=g^2, $$ $$ b^2+e^2=g^2, $$ $$ c^2+d^2=g^2. $$ A Perfect Cuboid can therefore also be viewed as: > Three right triangles sharing the same hypotenuse $g$, whose three short-axis projections must also jointly satisfy a three-dimensional sphere closure. This is the starting point of the Center-Axis formulation. --- # 2. Normalized Center-Axis Coordinates Divide all lengths by: $$ g. $$ Define: $$ x_a=\frac ag, $$ $$ x_b=\frac bg, $$ $$ x_c=\frac cg. $$ and the opposite face-diagonal ratios: $$ y_a=\frac fg, $$ $$ y_b=\frac eg, $$ $$ y_c=\frac dg. $$ Then: $$ x_a^2+y_a^2=1, $$ $$ x_b^2+y_b^2=1, $$ $$ x_c^2+y_c^2=1. $$ At the same time, the body closure is: $$ \boxed{ x_a^2+x_b^2+x_c^2=1. } $$ Since all lengths are positive: $$ 00}^7. $$ From: $$ a^2+f^2=g^2, $$ define: $$ x_a=\frac ag, $$ $$ y_a=\frac fg. $$ Then: $$ x_a^2+y_a^2=1. $$ Define: $$ t_a = \frac{y_a}{1+x_a}. $$ Thus: $$ t_a = \frac{f/g}{1+a/g} = \frac{f}{g+a}. $$ Hence: $$ t_a\in\mathbb Q. $$ And since: $$ 0<\alpha<\frac{\pi}{2}, $$ we have: $$ 00}. $$ Multiplying: $$ P_a^2+R_a^2=Q_a^2 $$ by: $$ Q_b^2Q_c^2 $$ gives: $$ A^2+F^2=G^2. $$ Similarly: $$ B^2+E^2=G^2, $$ and: $$ C^2+D^2=G^2. $$ From CARCE: $$ A^2+B^2+C^2=G^2. $$ Therefore: $$ B^2+C^2 = G^2-A^2 = F^2, $$ $$ A^2+C^2 = G^2-B^2 = E^2, $$ $$ A^2+B^2 = G^2-C^2 = D^2. $$ Thus: $$ \boxed{ (A,B,C,D,E,F,G) } $$ is a Perfect Cuboid. Hence: $$ \boxed{ \mathcal M(\mathbb Q)\neq\varnothing \Rightarrow \text{Perfect Cuboid exists}. } $$ Combining both directions proves the theorem. --- # 10. Exact Witness Generator Thus, once we obtain: $$ t_i=\frac{m_i}{n_i}, $$ we can directly generate: $$ P_i=n_i^2-m_i^2, $$ $$ R_i=2m_in_i, $$ $$ Q_i=n_i^2+m_i^2. $$ Then let: $$ G=Q_aQ_bQ_c. $$ Finally: $$ A=P_aQ_bQ_c, $$ $$ B=P_bQ_aQ_c, $$ $$ C=P_cQ_aQ_b, $$ $$ D=R_cQ_aQ_b, $$ $$ E=R_bQ_aQ_c, $$ $$ F=R_aQ_bQ_c. $$ The rational point itself is therefore a compact certificate. A subsequent verifier need not even trust the Geometry Engine. It only needs to recompute: $$ A^2+B^2-D^2, $$ $$ A^2+C^2-E^2, $$ $$ B^2+C^2-F^2, $$ $$ A^2+B^2+C^2-G^2 $$ and check whether all are exactly equal to: $$ 0. $$ --- # 11. Integer Polynomial Certificate The rational denominators in CARCE can also be cleared directly. Multiplying: $$ \frac{P_a^2}{Q_a^2} + \frac{P_b^2}{Q_b^2} + \frac{P_c^2}{Q_c^2} = 1 $$ by: $$ Q_a^2Q_b^2Q_c^2 $$ gives: $$ \boxed{ P_a^2Q_b^2Q_c^2 + P_b^2Q_a^2Q_c^2 + P_c^2Q_a^2Q_b^2 = Q_a^2Q_b^2Q_c^2. } $$ Thus, Perfect Cuboid existence is also equivalent to the existence of: $$ m_a,n_a, m_b,n_b, m_c,n_c \in \mathbb Z_{>0} $$ satisfying: $$ 0 Geometric closure is already complete; among the three half-angle arithmetic channels, two already have rational closure, and only the third remains to be rationalized. --- # 19. CARCE Surface Define: $$ \mathcal M = \left\{ (t_a,t_b,t_c)\in(0,1)^3: \Phi(t_a,t_b,t_c)=0 \right\}. $$ Perfect Cuboid existence is now equivalent to: $$ \boxed{ \mathcal M\cap\mathbb Q^3\neq\varnothing. } $$ Thus, the object of research is no longer the unbounded: $$ \mathbb Z_{>0}^7. $$ Instead, within the bounded cube: $$ (0,1)^3 $$ it is an exact algebraic surface. --- # 20. Smoothness of the Interior Surface Let: $$ c(t) = \frac{1-t^2}{1+t^2}. $$ Then: $$ \Phi = c(t_a)^2 + c(t_b)^2 + c(t_c)^2 -1. $$ We have: $$ c'(t) = -\frac{4t}{(1+t^2)^2}. $$ Hence: $$ \frac{\partial\Phi}{\partial t} = 2c(t)c'(t) = -\frac{ 8t(1-t^2) }{ (1+t^2)^3 }. $$ For: $$ 00, $$ $$ 1-t^2>0. $$ Therefore: $$ \frac{\partial\Phi}{\partial t}<0. $$ Thus, within: $$ (0,1)^3 $$ we have: $$ \nabla\Phi\neq0. $$ By the implicit function theorem: $$ \boxed{ \mathcal M \text{ is a smooth two-dimensional manifold}. } $$ This provides a formal foundation for Geometry Dynamics. --- # 21. Symmetric Polynomial Surface Since: $$ \Phi $$ depends only on: $$ t_a^2, t_b^2,t_c^2, $$ let: $$ s_1=t_a^2, $$ $$ s_2=t_b^2, $$ $$ s_3=t_c^2. $$ Then: $$ \left( \frac{1-s_1}{1+s_1} \right)^2 + \left( \frac{1-s_2}{1+s_2} \right)^2 + \left( \frac{1-s_3}{1+s_3} \right)^2 = 1. $$ Clearing the denominators gives the symmetric polynomial: $$ \Psi(s_1,s_2,s_3)=0. $$ Its explicit form is: $$ \Psi = s_1^2s_2^2s_3^2 + s_1^2s_2^2 + s_1^2s_3^2 + s_2^2s_3^2 + s_1^2+s_2^2+s_3^2 $$ $$ \quad -4s_1^2s_2s_3 -4s_1s_2^2s_3 -4s_1s_2s_3^2 -16s_1s_2s_3 $$ $$ \quad -4s_1s_2 -4s_1s_3 -4s_2s_3 +1. $$ This is a fully symmetric polynomial of degree $6$ in: $$ (s_1,s_2,s_3). $$ --- # 22. Elementary-Symmetric Compression Let: $$ \sigma_1=s_1+s_2+s_3, $$ $$ \sigma_2=s_1s_2+s_1s_3+s_2s_3, $$ $$ \sigma_3=s_1s_2s_3. $$ The polynomial above can then be compressed into: $$ \boxed{ \Psi = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 - 6\sigma_2 - 6\sigma_1\sigma_3 - 16\sigma_3 + 1. } $$ Thus, the CARCE surface has an explicit permutation symmetry: $$ S_3. $$ This representation is suitable for future: - symbolic factorization; - singularity analysis; - rational-point search; - birational classification; - modular obstruction; - invariant construction. --- # 23. The Seed Lies Exactly on the Polynomial Surface For the current seed: $$ s_1=\frac{625}{1521}, $$ $$ s_2=\frac{289}{784}, $$ $$ s_3=\frac{233}{1913}. $$ Exact substitution gives: $$ \boxed{ \Psi \left( \frac{625}{1521}, \frac{289}{784}, \frac{233}{1913} \right) = 0. } $$ Note: $$ s_3\in\mathbb Q $$ but: $$ s_3 $$ is not a rational square. Therefore: $$ t_c=\sqrt{s_3}\notin\mathbb Q. $$ This is precisely the only current half-angle arithmetic defect. --- # 24. The Distinction Between Geometry Closure and Rational Closure The current seed already satisfies: $$ \Phi(T_0)=0. $$ Hence: $$ \boxed{ \text{There is no need to search for cuboid geometry again}. } $$ What is truly needed is: $$ \boxed{ \text{Search on }\mathcal M\text{ for a rational point}. } $$ That is: $$ T_0 \longrightarrow T^\ast $$ where: $$ T^\ast\in\mathcal M\cap\mathbb Q^3. $$ This is the core objective of all subsequent dynamics, visualization, and arithmetic observers. --- # 25. Geometry-Preserving Dynamics Since we are currently already on: $$ \Phi=0 $$ this surface, the primary action of the dynamics should no longer be: $$ -\nabla\Phi. $$ That would merely solve geometric closure again. The new dynamics should move along: $$ \mathcal M $$ the tangent space of this surface. Let the arithmetic heuristic: $$ A(T) $$ measure rationality complexity. Then the unconstrained arithmetic descent: $$ -\nabla A $$ can be projected onto the tangent plane. Define: $$ P_T = I - \frac{ \nabla\Phi \nabla\Phi^{\mathsf T} }{ \|\nabla\Phi\|^2 }. $$ Then: $$ \boxed{ \dot T = - P_T\nabla A(T). } $$ The design objective of this flow is: $$ \frac{d}{dt}\Phi(T(t))=0 $$ while reducing, as much as possible: $$ A(T). $$ This is: $$ \boxed{ \text{geometry-preserving arithmetic search}. } $$ --- # 26. Numerical Constraint Correction In an actual floating-point runtime, even if the tangent flow theoretically satisfies: $$ \dot\Phi=0, $$ finite precision may still cause drift. A projection correction can therefore be added at each step: $$ T' = T+\Delta T. $$ Then use: $$ T_{\mathrm{corr}} = T' - \frac{ \Phi(T') }{ \|\nabla\Phi(T')\|^2 } \nabla\Phi(T') $$ to perform one Newton-style normal correction. Repeat until: $$ |\Phi| < \varepsilon_G. $$ However: $$ \varepsilon_G $$ belongs only to the geometric numerical runtime. The final mathematical certificate must still satisfy: $$ \Phi=0 $$ exactly. --- # 27. Two-Coordinate Surface Explorer By smoothness: $$ \frac{\partial\Phi}{\partial t_c}\neq0 $$ in the interior. Thus, locally we can write: $$ t_c = h(t_a,t_b). $$ The software therefore need not move arbitrarily in three dimensions. It can directly use: $$ (t_a,t_b) $$ as the control plane. Each time, solve: $$ \Phi(t_a,t_b,t_c)=0 $$ to obtain the unique local branch: $$ t_c=h(t_a,t_b). $$ This directly reduces the visualization to: $$ \boxed{ \text{2D arithmetic landscape over a geometric closure surface}. } $$ --- # 28. Explicit Third-Coordinate Formula Let: $$ x_a = \frac{1-t_a^2}{1+t_a^2}, $$ $$ x_b = \frac{1-t_b^2}{1+t_b^2}. $$ Define: $$ u = 1-x_a^2-x_b^2. $$ If: $$ 0 Does the CARCE smooth surface in the open cube $(0,1)^3$ contain a rational point? That is: $$ \boxed{ \exists T\in \mathbb Q^3\cap(0,1)^3 : \Phi(T)=0? } $$ This statement is exactly equivalent to classical Perfect Cuboid existence. --- # 52. What Is Currently Complete and What Remains Incomplete ## Complete - Center-Axis formulation; - half-angle rational compression; - CARCE; - exact equivalence theorem; - integer witness generator; - square-defect bridge; - shared-core bridge; - symmetric polynomial surface; - smoothness proof; - exact embedding of the current seed; - proof completion criterion. ## Incomplete - Find: $$ T^\ast\in\mathbb Q^3\cap\mathcal M; $$ - Prove: $$ \mathcal M(\mathbb Q)\neq\varnothing; $$ - Or prove in the opposite direction: $$ \mathcal M(\mathbb Q)=\varnothing. $$ Thus, the classical Perfect Cuboid problem remains unsolved by this paper. However, the problem has been compressed exactly into an explicit rational-point existence problem. --- # 53. Research Philosophy This research line can now be simplified to: $$ \boxed{ \text{Geometry gives the surface.} } $$ $$ \boxed{ \text{Arithmetic chooses the points.} } $$ $$ \boxed{ \text{Exact verification decides the theorem.} } $$ The Geometry Engine is no longer responsible for proof. The Arithmetic Observer is no longer responsible for generating the world. The Verifier does not participate in heuristics. The three are separated: $$ \boxed{ \text{Generate} \rightarrow \text{Rationalize} \rightarrow \text{Certify}. } $$ --- # 54. Final Conclusion This paper reformulates Perfect Cuboid existence exactly as: $$ \boxed{ \mathcal M(\mathbb Q)\neq\varnothing, } $$ where: $$ \mathcal M = \left\{ (t_a,t_b,t_c)\in(0,1)^3: \left( \frac{1-t_a^2}{1+t_a^2} \right)^2 + \left( \frac{1-t_b^2}{1+t_b^2} \right)^2 + \left( \frac{1-t_c^2}{1+t_c^2} \right)^2 = 1 \right\}. $$ The current seed: $$ \boxed{ T_0 = \left( \frac{25}{39}, \frac{17}{28}, \sqrt{\frac{233}{1913}} \right) } $$ already lies exactly on: $$ \mathcal M. $$ Among the three arithmetic half-angle channels: $$ t_a\in\mathbb Q, $$ $$ t_b\in\mathbb Q, $$ while: $$ t_c^2\in\mathbb Q $$ but currently: $$ t_c\notin\mathbb Q. $$ Thus, the main objective of subsequent research is very clear: $$ \boxed{ \text{On the CARCE surface, bring the third square-defect closure to a rational square, or find any other rational point.} } $$ Once we find: $$ \boxed{ T^\ast\in\mathbb Q^3\cap\mathcal M, } $$ we can automatically generate: $$ \boxed{ (A,B,C,D,E,F,G)\in\mathbb Z_{>0}^7 } $$ and complete an exact constructive proof of Perfect Cuboid existence. --- ## Canonical One-Line Target $$ \boxed{ \exists\, t_a,t_b,t_c\in\mathbb Q\cap(0,1) : \sum_{i\in\{a,b,c\}} \left( \frac{1-t_i^2}{1+t_i^2} \right)^2 = 1. } $$ --- **End of Canonical Paper v0.1**