## 1. Introduction, Prior Work, and Summary of Main Results

A rational Diophantine $m$-tuple is a set $\{x_1,\dots,x_m\}\subset\mathbb{Q}^\times$ of distinct non-zero rationals such that $1+x_ix_j$ is a rational square for all $1\le i<j\le m$ [1]. While infinitely many rational Diophantine sextuples ($m=6$) are known [5, 7, 12], the existence of a rational Diophantine septuple ($m=7$, requiring $\binom{7}{2}=21$ square pairs) remains open (Problem 3.3 of [1]), and recent work has focused on classifying *almost septuples* ($20/21$ square pairs) and higher-density candidate families ($17/21, 18/21, 19/21$) constructed from elliptic translations and reflections on the curve of a Diophantine triple [2–4, 8–11, 13–15].

### 1.1 Background and Precise Novelty Boundary

We first state the established background from the literature and from our accepted Paper 72 (`diophantine_septuple_72`, Version 39 [15, §9(c.1)]) so that the distinct contributions of the present paper are sharply isolated:

- **Triple-curve cosets, the regular translation point $S_Y$, and strong-pair reflections ([9, 10, 12, 15]):** On the elliptic curve $E_Y: w^2 = (1+ax)(1+bx)(1+gx)$ of a rational Diophantine triple $Y=\{a,b,g\}\subset\mathbb{Q}^\times$ with $1+ag=r_a^2$, $1+bg=r_b^2$, $1+ab=r_{ab}^2$, origin $\mathcal{O}$ at infinity, basepoint $P_0=(0,1)$, and regular translation point (following [12, §2] in our coordinates)
  $$S_Y := \left(\frac{1}{abg},\; \frac{r_a r_b r_{ab}}{abg}\right) \in E_Y(\mathbb{Q}), \tag{1.1}$$
  direct substitution gives $(1+a x(S_Y))(1+b x(S_Y))(1+g x(S_Y)) = \frac{(1+bg)(1+ag)(1+ab)}{(abg)^2} = w(S_Y)^2$, and chord-and-tangent addition with $P_0=(0,1)$ yields the regular fourth-element abscissas $x(P_0 \pm S_Y) = d_\mp := a+b+g+2abg\mp 2r_a r_b r_{ab} \in P_0+2E_Y(\mathbb{Q})$ (so $S_Y \in 2E_Y(\mathbb{Q})$). Every coset point $Q\in P_0+2E_Y(\mathbb{Q})$ has rational spoke roots $1+ax(Q)=\square, 1+bx(Q)=\square, 1+gx(Q)=R(Q)^2$ [9, 10, 12]. When $Y$ is a **tangent triple**—meaning $1+g^2=h^2$ is a square ($g$ is a strong member [12, §2]) and $b=\varphi_g(a):=\frac{g-a}{1+ga}$ with $r_b=h/r_a$ and $v:=g r_{ab}$—the Möbius reflection $\varphi_g(x)=\frac{g-x}{1+gx}$ lifts to an elliptic involution $\iota_g(x,w)=\bigl(\varphi_g(x),\frac{h^2 w}{(1+gx)^2}\bigr)$ preserving $P_0+2E_Y(\mathbb{Q})$ and sending $x(Q)=\frac{R(Q)^2-1}{g}\mapsto \varphi_g(x(Q))=\frac{(h/R(Q))^2-1}{g}$ [15, §9(c.1)].
- **Results already established in Paper 72, v39 ([15, §9(c.1)]):** Paper 72 [15, Theorem 20(146)–(149)] already established the self-reflection annulus equation $W_u^2=(gh)^2-(R^2-h^2)^2$ for $1+u\varphi_g(u)=\left(\frac{W_u}{gR}\right)^2$, the weighted-homogeneous surface equation
  $$\mathcal{S}_{\mathrm{sym}} : W^2 = 2(x^2+y^2)(x^2+z^2)(y^2+z^2) \subset \mathbb{P}(1,1,1,3),$$
  the birational tangent-triple dictionary $(g,h,r_a,r_b,v)$ on $\mathcal{S}_{\mathrm{sym}}$ and its inverse (recalled in Section 2.1 below with $W_u=vr_a$), the coordinate involution $\iota_{xz}[x:y:z:W]=[xy:xz:yz:xyzW]$, the identity $\frac{P_{\mathrm{diag}}}{(gh)^2}=5-\Sigma^2$ and the $(11,3,2,650)$ $18/21$ example, the coupled $(m,\rho)$ conic bundle from the Pythagorean cosine law $\alpha m^2-2m+\beta=0$, the Euler–Chasles biquadratic $\mathcal{B}(R,X)=0$ for $1$-step spoke roots, and the raw five-factor polynomial factorization of the step-$3$ cross-reflection remainder $\Delta_{\mathrm{norm}}^{(3)}$. These items are recalled as background in Section 2.1 and are not claimed as new results here.
- **All-coset $[-2]$ obstruction on the collapse base curve in Paper 21, v4 ([10, §5.2]) and [14, Theorem 6]:** The space of non-degenerate tangent triples $Y=\{a,b,g\}$ is a $2$-dimensional elliptic surface over the Pythagorean mirror line $(g,h)=\bigl(\frac{2s}{1-s^2},\frac{1+s^2}{1-s^2}\bigr)$. Imposing the $4$-torsion condition $[2]S_Y = T_g := (-1/g, 0) \iff D_g := (abg)^2 - (2ab+1) = \frac{(v^2-h^2)^2-h^2}{g^2} = 0 \iff v^2 = h^2\pm h$ cuts out a $1$-dimensional curve of $4$-torsion base triples, which is birationally equivalent via $p=ab$, $m=abg$ ($m^2=1+2p$), and $n=(a-b)/h$ ($n^2=1-2p, m^2+n^2=2$) to the collapse base sublocus (`w4sym.py`) of Paper 21, Version 4 [10, §5.2]:
  $$a = \frac{4t^2}{(t^2-1)(t^2+1)},\qquad b = \frac{(t^2-1)^2}{2t(t^2+1)},\qquad g = \frac{(t^2+1)(t^2+2t-1)}{2t(t^2-1)}, \tag{1.2}$$
  where $1+g^2=\left(\frac{q(t)}{2t(t^2-1)}\right)^2$ and $1+ab=\frac{q(t)}{(t^2+1)^2}$ with $q(t)=t^4+2t^3+2t^2-2t+1=\square$. On this $1$-dimensional collapse base curve, Paper 21, v4 [10, §5.2] **already established the $[-2]$ square-class relation $f_{2S}(T)f_{2S}(T+S)\equiv -2\pmod{(\mathbb{Q}^\times)^2}$ across all of $P_0+2E_Y(\mathbb{Q})$** and concluded that the two opposite-edge square conditions exclude each other on the entire coset (and noted that $ab$ is always a $2$-adic unit, so the exclusion is also covered by [14, Theorem 6]). What is new in Section 6 (Theorem 5) of the present paper is: (i) the **general-$D_g$ shift-reflection Master Syzygies I–III and bilinear $U$-factorization of $\mathcal{B}(R,X)$** across the **full $2$-dimensional base surface of arbitrary tangent triples** ($D_g\neq 0$ generically), together with the doubling defect formula $x([2]S_Y)+\frac{1}{g}=\frac{D_g^2}{4abgh^2 r_{ab}^2}$ and an explicit rational non-collapse section on $\mathcal{S}_{\mathrm{sym}}$ covering every mirror $g=\frac{2s}{1-s^2}$; (ii) the **$2$-dimensional sign-chamber inequality** $\min(1+x_0\varphi_g(x_0),1+x_-x_+)\le 0$ across the open chamber $D_g\le 0$ ($h^2-|h|\le v^2\le h^2+|h|$); and (iii) on $D_g=0$ (where Master Syzygy I specializes to [10, §5.2]'s all-coset $[-2]$ relation), the **explicit boundary square-class calculation** proving that the two zeros $x_0=g\pm h$ of $1+x_0\varphi_g(x_0)=0$ define rational points $Z_\varepsilon\in E_Y(\mathbb{Q})$ whose $g$-spoke satisfies $1+g(g+\varepsilon h)=2\bigl(\frac{h(h+\mathrm{sgn}+\varepsilon g)}{2v}\bigr)^2\in 2(\mathbb{Q}^\times)^2$ and thus never lies in $P_0+2E_Y(\mathbb{Q})$. Likewise, while representing a genus-$1$ translation as a composite of two biquadratic involutions is standard in the QRT framework (Hone [17, §2]), the specific bilinear null-cone chart $(u_-,u_+)$ on $\mathcal{S}_{\mathrm{sym}}$, its osculating section $\mathcal{C}_{\mathrm{null}}(u_-)$, its removable basepoint limit, and the contrast between genus-$1$ fixed-$u_-$ fibers and genus-$5$ fixed-$\Omega$ level sets are new.

### 1.2 Summary of New Contributions

Building on [10, 12, 14, 15, 17], this paper establishes six theorems (`exp100`–`exp103`):

1. **Theorem 1 (Secant Reduction of $1-\rho^2=\square$ to $\mathcal{E}_{\mathrm{self}}$ and Rank-$0$ Descent; Section 2, `exp100`):** Distinguishing the zero locus $1-\rho^2=0$ ($\rho=\pm 1$) from the square self-reflection locus $P_{\mathrm{self},-}=(gh)^2(1-\rho^2)=\square$ parameterized by $\rho=\frac{t^2-1}{t^2+1}$ on the $(m,\rho)$ conic bundle of $\mathcal{S}_{\mathrm{sym}}$, we show that the two canonical rational secant sections $m_{\mathrm{sec},\pm}(t)=-\frac{2t}{t\pm 1}$ yield $x^2=\frac{1}{t^2}$ and $z^2=\frac{t^2\mp 4t+1}{t^2\pm 4t+1}$, reducing to $\mathcal{E}_{\mathrm{self}}:w^2=t^4-14t^2+1$. Via a degree-$4$ map to $E_{48}:y^2=x(x+1)(x-3)$ and $2$-Selmer descent [18, Ch. X, Prop. 1.4], we prove $E_{48}(\mathbb{Q})=E_{48}(\mathbb{Q})[2]$ and $\mathcal{E}_{\mathrm{self}}(\mathbb{Q})=\{(0,\pm 1)\}$, excluding all non-degenerate rational points ($t\notin\{0,\pm 1\}$) along both secant families.
2. **Theorem 2 (Step-$2$ Conservation Law and Step-$3$ Obstruction Classification; Section 3, `exp100`):** We prove the opposite-branch step-$2$ reflection-defect factorization $\varphi_g(u_{2,\mp})-u_{2,\pm}=\frac{gQ_{8,-}Q_{8,+}}{A_\mp^2B_\pm^2}$ and conservation law $N_{\mathrm{cross}}^{(2)}+h^2N_{\mathrm{same}}^{(2)}=(A_-A_+)^2+(hB_-B_+)^2$, and classify all five factors of $\Delta_{\mathrm{norm}}^{(3)}$: Factor 1 ($v^2=3h^2+1$) has no real tangent triples since $(r_a\pm r_b)^2=-(h\mp 1)^2<0$, Factors 2–3 ($Q_{4,\mp}=0$) satisfy $N_3^4-h^2D_3^4=-(h^2-1)Q_{4,-}Q_{4,+}$ and are excluded for $g\neq 0$ by Fermat's right-triangle theorem, and Factors 4–5 ($v^2=h^2\pm h$) cut out the $1$-dimensional $4$-torsion base curve $D_g=0$.
3. **Theorem 3 (Null-Cone Elliptic Fibers $\mathcal{E}_{u_-}$ and Non-Torsion Bäcklund Orbits; Section 4, `exp101`–`exp102`):** In the normalized chart $y=1$ on $\mathcal{S}_{\mathrm{sym}}$ with $(X,Z,W)=(x^2,z^2,W)$, we introduce the $(1,1)$-bilinear null-cone coordinates $(u_-,u_+)$, proving that fixing $u_-\in\mathbb{Q}\setminus\{0,\pm 1\}$ cuts out the smooth genus-$1$ fiber $\mathcal{E}_{u_-}:(2x^2-\tilde{q}_-)(2z^2-q_-)=(u_-^2-1)^2$ on which every rational point $(x,z)$ lifts rationally to the surface closure (and to the admissible open surface $\mathcal{S}_{\mathrm{sym}}(\mathbb{Q})$ whenever $xz(x^2-z^2)\neq 0$, i.e., away from the coordinate-zero closure points $xz=0$ and the diagonal closure points $x=\pm z=\pm u_-$). We construct the rational osculating section $\mathcal{C}_{\mathrm{null}}(u_-)\subset\mathcal{S}_{\mathrm{sym}}(\mathbb{Q}(u_-))$, prove that $\mathcal{I}_{u_-}^{(+)}$ has a removable limit at $(-su_-,-su_-)$ equal to $(sx_{\mathrm{osc}}(u_-),sz_{\mathrm{osc}}(u_-))$ with generic iterate degrees $(1,5,13,25)$ for $n=0,1,2,3$, and prove via Mazur's Torsion Theorem [16] ($13>12$ distinct rational orbit points) that $\mathcal{T}_{u_-}=\mathcal{I}_{u_-}^{(-)}\circ\mathcal{I}_{u_-}^{(+)}$ has infinite order over $\mathbb{Q}$ at $u_-\in\{2,3,23/33,93/145\}$ and hence on the generic fiber over $\mathbb{Q}(u_-)$.
4. **Theorem 4 ($S_3$-Triad Matrix, Markov Laws, and Multi-Pair Calculus; Section 5, `exp101`–`exp102`):** On the triad matrix $M_{\mathrm{triad}}(x:y:z)=[a_j,b_j,g_j]_{j=1}^3$, we prove that all nine triple conditions share the square class $[h_1h_2h_3]$, establish the Markov spoke-root law $r_{ab,1}^2+r_{ab,2}^2+r_{ab,3}^2-r_{ab,1}^2r_{ab,2}^2r_{ab,3}^2=2$ and its $\mathbb{P}^2$-universal Joukowsky–Markov form $m_1+m_2+m_3-m_1m_2m_3=2$, prove the fixed-$(x,z)$ multi-pair factorizations and exact $\mathcal{S}_{\mathrm{sym}}$ rational-square multipliers relating $(1+b_1(y_i)b_1(y_j), 1+b_1(y_i)a_1(y_j))$ to $(1+a_1(y_i)a_1(y_j), 1+a_1(y_i)b_1(y_j))$, and derive root-free rational recovery formulas for $(x:y:z)\in\mathbb{P}^2(\mathbb{Q})$.
5. **Theorem 5 (General-$D_g$ Shift-Reflection Syzygies and $D_g=0$ Boundary Class; Section 6, `exp103`):** Across the $2$-dimensional base surface of arbitrary tangent triples $Y=\{a,b,g\}$ (where $x([2]S_Y)+\frac{1}{g}=\frac{D_g^2}{4abgh^2 r_{ab}^2}$ and every mirror $g=\frac{2s}{1-s^2}$, $s\notin\{0,\pm 1\}$, admits a globally certified non-collapse rational section on $\mathcal{S}_{\mathrm{sym}}$), we prove Master Syzygies I–III for any coset center $Q_0\in P_0+2E_Y(\mathbb{Q})$, including $U(Q_0)^2(1+x_-x_+)+\frac{2v^2R^2}{g^2}(1+x_0\varphi_g(x_0))=D_g(1+x_0^2)$, which forces $\min(1+x_0\varphi_g(x_0),1+x_-x_+)\le 0$ across the chamber $D_g\le 0$ ($h^2-|h|\le v^2\le h^2+|h|$). Specializing to $D_g=0$ recovers [10, §5.2]'s all-coset $[-2]$ relation, while our boundary square-class identity $1+g(g+\varepsilon h)=2\bigl(\frac{h(h+\mathrm{sgn}+\varepsilon g)}{2v}\bigr)^2\in 2(\mathbb{Q}^\times)^2$ proves that the zeros $x_0=g\pm h$ of $1+x_0\varphi_g(x_0)=0$ never lie in $P_0+2E_Y(\mathbb{Q})$.
6. **Theorem 6 (Fixed-$\Omega$ Sphere Syzygy and Genus-$5$ Finiteness; Section 7, `exp103`):** Along any fixed hyperbolic level set $u_-u_+=\Omega\in\mathbb{Q}\setminus\{0,\pm 1\}$, the scaled coordinates $(X_1,Z_1,M)$ satisfy $X_1^2=(1-\eta)(\Omega+\eta)$, $Z_1^2=(1+\eta)(\Omega-\eta)$, $M^2=\Omega+\eta^2$ linked by the sphere syzygy $X_1^2+Z_1^2+2M^2\equiv 4\Omega$, defining a smooth $(\mathbb{Z}/2\mathbb{Z})^3$-cover $\mathcal{F}_\Omega\to\mathbb{P}^1_\eta$ of genus $g(\mathcal{F}_\Omega)=5$ with finite $\mathcal{F}_\Omega(\mathbb{Q})$ by Faltings' theorem [6].

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## 2. Preliminaries on $\mathcal{S}_{\mathrm{sym}}$ and Secant Reduction of the Square Self-Reflection Family to $\mathcal{E}_{\mathrm{self}}$

### 2.1 Coordinate Preliminaries from [15, §9(c.1)] and the Palindromic Cross-Ratio Quadratic

We first recall from [15, §9(c.1)] the surface coordinates and tangent-triple dictionary used throughout Sections 2–7:
- **The weighted projective surface $\mathcal{S}_{\mathrm{sym}}\subset\mathbb{P}(1,1,1,3)$ and self-reflection annulus $\mathcal{A}_{\mathrm{tot}}$ ([15, §9(c.1)]):** Assigning weights $(1,1,1,3)$ to $(x,y,z,W)$ so that $[x:y:z:W]=[\mu x:\mu y:\mu z:\mu^3 W]$ for $\mu\in\mathbb{Q}^\times$, the weighted-homogeneous degree-$6$ equation
  $$\mathcal{S}_{\mathrm{sym}} : W^2 = 2(x^2+y^2)(x^2+z^2)(y^2+z^2),\qquad xyz(x^2-z^2)\neq 0, \tag{2.1}$$
  defines a $2$-dimensional surface in $\mathbb{P}(1,1,1,3)$, isomorphic on the open chart $y=1$ to $W^2=2(x^2+1)(z^2+1)(x^2+z^2)$ in $\mathbb{A}^3_{(x,z,W)}$, with $(x,z)$-sign quotient $\overline{\mathcal{S}}_{\mathrm{sym}}:W^2=2(X+1)(Z+1)(X+Z)$ in $(X,Z,W):=(x^2,z^2,W)$. For a Pythagorean pair $(g,h)\in(\mathbb{Q}^\times)^2$ with $1+g^2=h^2$, a coset rung $u=\frac{R^2-1}{g}$ satisfies $1+u\varphi_g(u)=\left(\frac{W_u}{gR}\right)^2$ if and only if $(g,h,R,W_u)$ lies on the self-reflection annulus surface
  $$\mathcal{A}_{\mathrm{tot}} : 1+g^2=h^2,\qquad W_u^2 = (gh)^2 - (R^2 - h^2)^2 = (gh + h^2 - R^2)(gh - h^2 + R^2). \tag{2.2}$$
  As shown in [15, §9(c.1)] (writing $W_u = v r_a$ with $R = r_a$), $\mathcal{A}_{\mathrm{tot}}$ and $\mathcal{S}_{\mathrm{sym}}\subset\mathbb{P}(1,1,1,3)$ are birationally equivalent via
  $$[\tilde{x}:\tilde{y}:\tilde{z}:\tilde{W}] = \bigl[(h+g)(gh+h^2-R^2) : (h+g)W_u : gh+h^2-R^2 : 4ghR(h+g)^2(gh+h^2-R^2)^2\bigr], \tag{2.3}$$
  where $\tilde{x}^2+\tilde{y}^2=2gh(h+g)^2(gh+h^2-R^2)$, $\tilde{x}^2+\tilde{z}^2=2h(h+g)(gh+h^2-R^2)^2$, $\tilde{y}^2+\tilde{z}^2=2g(h+g)R^2(gh+h^2-R^2)$, with inverse dictionary
  $$g = \frac{x^2-z^2}{2xz},\qquad h = \frac{x^2+z^2}{2xz},\qquad R = r_a := \frac{W}{2z(x^2+y^2)},\qquad r_b := \frac{h}{R} = \frac{W}{2x(y^2+z^2)},\qquad v := \frac{W_u}{R} = \frac{y(x^4-z^4)}{xzW}. \tag{2.4}$$
  On $\mathcal{S}_{\mathrm{sym}}$, the spoke roots satisfy $(r_a+r_b)^2=2h(h+1)-v^2$ and $(r_b-r_a)^2=2h(h-1)-v^2$ [15, §9(c.1)]. Setting $\alpha=\frac{x^2-y^2}{x^2+y^2}$, $\beta=\frac{z^2-y^2}{y^2+z^2}$, and $\lambda:=\frac{y^2}{xz}$, the dimensionless invariants
  $$\Sigma := \alpha+\beta = \frac{2(1-\lambda^2)}{1+2h\lambda+\lambda^2},\qquad D := \alpha-\beta = \frac{4g\lambda}{1+2h\lambda+\lambda^2},\qquad S^2 := 1-\alpha\beta = \frac{4h\lambda}{1+2h\lambda+\lambda^2} \tag{2.5}$$
  are governed by the **palindromic cross-ratio quadratic** (`exp100` Part (a))
  $$v^2 \lambda^2 + 2h(v^2 - 2g^2)\lambda + v^2 = 0,\qquad \frac{\Delta_\lambda}{4} = h^2(v^2 - 2g^2)^2 - v^4 = \bigl(g(r_b^2 - r_a^2)\bigr)^2, \tag{2.6}$$
  whose two roots $\lambda\leftrightarrow 1/\lambda$ are exchanged by the $\mathbb{P}(1,1,1,3)$-involution $\iota_{xz}[x:y:z:W]=[xy:xz:yz:xyzW]$ of [15, §9(c.1)], which swaps $r_a\leftrightarrow r_b$ and $a\leftrightarrow b$, negates $\Sigma\mapsto -\Sigma$, and preserves the unordered triple $\{a,b,g\}$ and $P_{\mathrm{diag}}:=h^4-h^2+4h^2v^2-v^4=(gh)^2(5-\Sigma^2)$ (pairing $[11:3:2:650]$ at $s=8$ with $\iota_{xz}[11:3:2:650]=[33:22:6:42900]$ at $s=-1/8$ on $\Sigma(s)=\frac{s^2-4s-1}{s^2+1}, \tau(s)=\frac{2s^2+2s-2}{s^2+1}$ for $\Sigma^2+\tau^2=5$; Table 1).

### 2.2 Secant Reduction of the Square Self-Reflection Family $1-\rho^2=\square$ to $\mathcal{E}_{\mathrm{self}}$ and Rank-$0$ Proof

On $\mathcal{S}_{\mathrm{sym}}$, the $1$-step self-reflection polynomial $P_{\mathrm{self},\mp}:=h^4-h^2-4h^2v^2\pm 4hv^3-v^4=(gh)^2-v^2(v\mp 2h)^2$ factors in terms of $\rho_\mp := \frac{v(v\mp 2h)}{gh} = D \mp 2S$ as $P_{\mathrm{self},\mp}=(gh)^2(1-\rho_\mp^2)$ [15, §9(c.1)]. Theorem 1 gives the explicit derivation of the affine quartic $\mathcal{E}_{\mathrm{self}}:w^2=t^4-14t^2+1$ from the two canonical secant sections of the **square self-reflection family** $1-\rho^2=\square$, distinguishes it from the zero locus $1-\rho^2=0$, and proves unconditionally that $\mathcal{E}_{\mathrm{self}}(\mathbb{Q})=\{(0,\pm 1)\}$.

**Theorem 1 (Secant Reduction to $\mathcal{E}_{\mathrm{self}}$ and Rank-$0$ Descent; `exp100`).**
1. *(Parameterization of $\overline{\mathcal{S}}_{\mathrm{sym}}$ by $(m,\rho)$, distinction between $1-\rho^2=0$ and $1-\rho^2=\square$, and secants $m=-\frac{2t}{t\pm 1}$):* In the normalized chart $y=1$ on $\mathcal{S}_{\mathrm{sym}}$ with $X=x^2$, $Z=z^2$, slope parameter $m:=\frac{1-S}{\alpha}$ (satisfying the Pythagorean cosine law $\alpha m^2-2m+\beta=0$), and $\rho:=\rho_-=D-2S=(m+1)\bigl(\alpha(m+1)-2\bigr)$, the quotient surface $\overline{\mathcal{S}}_{\mathrm{sym}}:W^2=2(X+1)(Z+1)(X+Z)$ is parameterized rationally by $(m,\rho)$ (`exp100` Part (b)):
   $$x^2 = X(m,\rho) := \frac{m^2 + 4m + 3 + \rho}{m^2 - 1 - \rho},\qquad z^2 = Z(m,\rho) := \frac{3m^2 + 4m + 1 - m^2\rho}{1 - m^2 + m^2\rho},\qquad W(m,\rho) := -\frac{4(m+1)^2(m^2 + m\rho - 1)}{(m^2 - 1 - \rho)(1 - m^2 + m^2\rho)}. \tag{2.7}$$
   - **(i) Zero locus $P_{\mathrm{self},-}=0$ ($1-\rho^2=0$):** The zero condition $1-\rho^2=0$ consists of the two rational lines $\rho=1$, where $(x^2,z^2)=\bigl(\frac{(m+2)^2}{m^2-2}, 2m^2+4m+1\bigr)$, and $\rho=-1$, where $(x^2,z^2)=\bigl(\frac{m^2+4m+2}{m^2}, \frac{(2m+1)^2}{1-2m^2}\bigr)$, on which $1+d_-\varphi_g(d_-)=0$ degenerates (having no solutions with $m^2-2=\square$ or $1-2m^2=\square$ and $xyz\neq 0$).
   - **(ii) Square self-reflection locus $P_{\mathrm{self},-}=(gh)^2(1-\rho^2)\in(\mathbb{Q}^\times)^2$ ($1-\rho^2=\square\neq 0$) and its canonical secant families:** Imposing the non-zero square condition $1-\rho^2=\square$ via
     $$\rho(t) = \frac{t^2 - 1}{t^2 + 1},\qquad 1 - \rho(t)^2 = \frac{4t^2}{(t^2 + 1)^2} \qquad (t\in\mathbb{Q}\setminus\{0\}) \tag{2.8}$$
     transforms (2.7) into the coupled conic bundle over $\mathbb{Q}(m,t)$:
     $$x^2 = X(m,t) = \frac{(m+2)^2 t^2 + (m^2 + 4m + 2)}{(m^2 - 2)t^2 + m^2},\qquad z^2 = Z(m,t) = \frac{(2m^2 + 4m + 1)t^2 + (2m + 1)^2}{t^2 + 1 - 2m^2}. \tag{2.9}$$
     Along the two canonical rational secant sections of the first conic $x^2=X(m,t)$, namely
     $$m = m_{\mathrm{sec},\pm}(t) := -\frac{2t}{t \pm 1} \qquad (t\in\mathbb{Q}\setminus\{0,\mp 1\}), \tag{2.10}$$
     $X(m,t)$ and $Z(m,t)$ specialize identically to
     $$x^2 = X\bigl(m_{\mathrm{sec},\pm}(t),\, t\bigr) = \frac{1}{t^2}\quad\left(x = \pm\frac{1}{t}\right),\qquad z^2 = Z\bigl(m_{\mathrm{sec},\pm}(t),\, t\bigr) = \frac{t^2 \mp 4t + 1}{t^2 \pm 4t + 1}. \tag{2.11}$$
     Setting $w := z(t^2\pm 4t+1)$ clears the denominator of $z^2$ and reduces both secant families $m=m_{\mathrm{sec},\pm}(t)$ to the affine quartic curve
     $$\mathcal{E}_{\mathrm{self}} : w^2 = (t^2 - 4t + 1)(t^2 + 4t + 1) = t^4 - 14t^2 + 1. \tag{2.12}$$
     Here $t=0$ ($\rho=-1, x=\infty$) and $t=\infty$ ($\rho=1, x=0$) belong to the degenerate zero locus $1-\rho^2=0$ ($xyz=0$), while $t=\pm 1$ ($\rho=0$) gives $w^2=-12\notin\mathbb{Q}^2$.
2. *(Unconditional rank-$0$ and full torsion proof for $\mathcal{E}_{\mathrm{self}}$):* The affine quartic $\mathcal{E}_{\mathrm{self}}:w^2=t^4-14t^2+1$ maps via the **degree-$4$** rational map (of degree $2$ on the $(t,w)\mapsto(\pm t,\pm w)$ quotient)
   $$x = \left(\frac{t^2-1}{2t}\right)^2,\qquad y = \frac{w(t^4-1)}{8t^3} \tag{2.13}$$
   to the elliptic curve $E_{48}:y^2=x(x+1)(x-3)=(x-e_1)(x-e_2)(x-e_3)$ with $(e_1,e_2,e_3)=(0,-1,3)$, where $x+1=\left(\frac{t^2+1}{2t}\right)^2$ and $x-3=\left(\frac{w}{2t}\right)^2$. Let
   $$\alpha : E_{48}(\mathbb{Q})/2E_{48}(\mathbb{Q}) \;\hookrightarrow\; \mathbb{Q}^\times/(\mathbb{Q}^\times)^2 \times \mathbb{Q}^\times/(\mathbb{Q}^\times)^2 \tag{2.13a}$$
   be the standard injective $2$-Selmer group homomorphism [18, Ch. X, Prop. 1.4] given by $\alpha(\mathcal{O})=(1,1)$, $\alpha(x,y)=\bigl([x],[x+1]\bigr)$ for $x\notin\{0,-1\}$, $\alpha(0,0)=\bigl([e_1-e_2][e_1-e_3],[e_1-e_2]\bigr)=(-3,1)$, and $\alpha(-1,0)=\bigl([e_2-e_1],[e_2-e_1][e_2-e_3]\bigr)=(-1,1)$. Then:
   - **(i) Local $2$-Selmer bound $\mathrm{im}(\alpha)\subset\{\pm 1,\pm 3\}\times\{1\}$:** Writing $x=r/w_0^2$ in lowest terms, $\gcd(r,(r+w_0^2)(r-3w_0^2))\mid 3$ and $\gcd(r+w_0^2,r(r-3w_0^2))\mid 4$, while $x(x+1)(x-3)=y^2\ge 0$ forces $x\ge -1$ (so $x+1>0$ for $x\neq -1$). Thus every $(b_1,b_2)\in\mathrm{im}(\alpha)$ lies in $\{\pm 1,\pm 3\}\times\{1,2\}$ and admits a primitive integer solution $(u_0,v_0,z_0,w_0)$ to the homogeneous $2$-covering equations
     $$b_2 v_0^2 = b_1 u_0^2 + w_0^2,\qquad b_1 b_2 z_0^2 = b_1 u_0^2 - 3w_0^2. \tag{2.14}$$
     Modulo $9$, (2.14) has no primitive solutions when $b_2=2$, so $\mathrm{im}(\alpha)\subset\{\pm 1,\pm 3\}\times\{1\}$ has order at most $4$.
   - **(ii) Bijection with $\alpha(E_{48}(\mathbb{Q})[2])$ and $\mathrm{rank}(E_{48}(\mathbb{Q}))=0$:** Evaluating $\alpha$ on the four rational $2$-torsion points $E_{48}(\mathbb{Q})[2]=\{\mathcal{O},(-1,0),(3,0),(0,0)\}$ gives the four distinct square-class pairs
     $$\alpha(\mathcal{O}) = (1, 1),\qquad \alpha(-1, 0) = (-1, 1),\qquad \alpha(3, 0) = (3, 1),\qquad \alpha(0, 0) = (-3, 1), \tag{2.14a}$$
     which exhaust $\{\pm 1,\pm 3\}\times\{1\}$. Since $\alpha$ is injective on $E_{48}(\mathbb{Q})/2E_{48}(\mathbb{Q})$ and $E_{48}(\mathbb{Q})[2]\subset E_{48}(\mathbb{Q})$ already maps onto all $4$ surviving classes, we have $E_{48}(\mathbb{Q})/2E_{48}(\mathbb{Q})\cong(\mathbb{Z}/2\mathbb{Z})^2$ of order $4=2^{\mathrm{rank}(E_{48}(\mathbb{Q}))+2}$. Therefore $\mathrm{rank}(E_{48}(\mathbb{Q}))=0$ and $E_{48}(\mathbb{Q})[2]\cap 2E_{48}(\mathbb{Q})=\{\mathcal{O}\}$ (hence $E_{48}(\mathbb{Q})[2^\infty]=E_{48}(\mathbb{Q})[2]$).
   - **(iii) Full torsion group and exclusion of the secant families $m=m_{\mathrm{sec},\pm}(t)$:** Good-reduction point counts at $p=5$ and $p=7$ give $\#E_{48}(\mathbb{F}_5)=\#E_{48}(\mathbb{F}_7)=8=2^3$, ruling out all odd-order rational torsion on $E_{48}(\mathbb{Q})$. Consequently, $E_{48}(\mathbb{Q})=E_{48}(\mathbb{Q})[2]$ and $2E_{48}(\mathbb{Q})=\{\mathcal{O}\}$. Since any point $(t,w)\in\mathcal{E}_{\mathrm{self}}(\mathbb{Q})$ with $t\neq 0$ (noting $t=\pm 1\implies w^2=-12\notin\mathbb{Q}^2$) maps under (2.13) to a finite point $P=(x,y)\in E_{48}(\mathbb{Q})$ with $x=\square\neq 0$ and $x+1=\square\neq 0$, it has $\alpha(P)=(1,1)$ and hence lies in $\ker(\alpha)=2E_{48}(\mathbb{Q})=\{\mathcal{O}\}$ (which has $x=\infty$, a contradiction). Thus $\mathcal{E}_{\mathrm{self}}(\mathbb{Q})=\{(0,\pm 1)\}$ in the affine chart (plus the two points at infinity $t=\infty$), proving that **neither canonical secant family $m=-\frac{2t}{t\pm 1}$ of the square self-reflection locus $1-\rho^2=\square$ contains any non-degenerate rational point** ($t\in\mathbb{Q}\setminus\{0,\pm 1\}$).

*Proof.*
(1) Solving $\alpha m^2-2m+\beta=0$ and $\rho=(m+1)(\alpha(m+1)-2)$ for $\alpha=\frac{X-1}{X+1}$ and $\beta=\frac{Z-1}{Z+1}$ gives (2.7), with $2(X(m,\rho)+1)(Z(m,\rho)+1)(X(m,\rho)+Z(m,\rho))=W(m,\rho)^2$ identically in $\mathbb{Q}(m,\rho)$. Substituting $\rho=\pm 1$ gives the zero-locus lines in (i), while substituting $\rho(t)=\frac{t^2-1}{t^2+1}$ gives (2.9). Evaluating (2.9) at $m=m_{\mathrm{sec},\pm}(t)=-\frac{2t}{t\pm 1}$ and multiplying numerator and denominator by $(t\pm 1)^2$ (using $(m+2)^2=\frac{4}{(t\pm 1)^2}$, $m^2+4m+2=-\frac{2(t^2\pm 2t-1)}{(t\pm 1)^2}$, $m^2-2=\frac{2(t^2\mp 2t-1)}{(t\pm 1)^2}$, $2m^2+4m+1=\frac{t^2\mp 6t+1}{(t\pm 1)^2}$, and $(2m+1)^2=\frac{(3t\mp 1)^2}{(t\pm 1)^2}$) yields
$$X\left(-\frac{2t}{t\pm 1},\, t\right) = \frac{4t^2 - 2(t^2\pm 2t-1)}{2(t^2\mp 2t-1)t^2 + 4t^2} = \frac{2(t\mp 1)^2}{2t^2(t\mp 1)^2} = \frac{1}{t^2},$$
$$Z\left(-\frac{2t}{t\pm 1},\, t\right) = \frac{(t^2\mp 6t+1)t^2 + (3t\mp 1)^2}{(t^2 + 1)(t\pm 1)^2 - 8t^2} = \frac{(t\mp 1)^2(t^2\mp 4t+1)}{(t\mp 1)^2(t^2\pm 4t+1)} = \frac{t^2\mp 4t+1}{t^2\pm 4t+1},$$
and clearing the denominator with $w=z(t^2\pm 4t+1)$ yields $w^2=(t^2-4t+1)(t^2+4t+1)=(t^2+1)^2-16t^2=t^4-14t^2+1$ (`exp100` Part (b)).

(2) By [18, Ch. X, Prop. 1.4], for any elliptic curve $E:y^2=(x-e_1)(x-e_2)(x-e_3)$ with rational $2$-torsion $e_1,e_2,e_3\in\mathbb{Q}$, the map $\alpha:E(\mathbb{Q})/2E(\mathbb{Q})\to\mathbb{Q}^\times/(\mathbb{Q}^\times)^2\times\mathbb{Q}^\times/(\mathbb{Q}^\times)^2$ sending $P=(x,y)\mapsto([x-e_1],[x-e_2])$ (with $x-e_i$ replaced by $\prod_{j\neq i}(e_i-e_j)$ when $x=e_i$) is an injective group homomorphism. For $E_{48}$ with $(e_1,e_2,e_3)=(0,-1,3)$ and $P=(x,y)\in E_{48}(\mathbb{Q})\setminus E_{48}(\mathbb{Q})[2]$, write $x=r/w_0^2$ in lowest terms ($w_0>0$, $\gcd(r,w_0)=1$); then $y^2=r(r+w_0^2)(r-3w_0^2)/w_0^6$. Since $\gcd(r,r+w_0^2)=1$, $\gcd(r,r-3w_0^2)\mid 3$, and $\gcd(r+w_0^2,r-3w_0^2)=\gcd(r+w_0^2,4w_0^2)\mid 4$, while $r(r+w_0^2)(r-3w_0^2)>0\implies r+w_0^2>0$, the squarefree parts $(b_1,b_2)=([x],[x+1])$ lie in $\{\pm 1,\pm 3\}\times\{1,2\}$ and satisfy the homogeneous $2$-covering system (2.14): $r=b_1 u_0^2$, $r+w_0^2=b_2 v_0^2$, $r-3w_0^2=b_1b_2 z_0^2$ with $\gcd(u_0,v_0,z_0,w_0)=1$. Exhaustive check modulo $9$ (`exp100` Part (b)) shows that $b_2 v_0^2 \equiv b_1 u_0^2 + w_0^2 \pmod 9$ and $b_1 b_2 z_0^2 \equiv b_1 u_0^2 - 3w_0^2 \pmod 9$ have no solution with $3\nmid\gcd(u_0,v_0,z_0,w_0)$ when $b_2=2$. Hence $\mathrm{im}(\alpha)\subset\{\pm 1,\pm 3\}\times\{1\}$. On the subgroup $E_{48}(\mathbb{Q})[2]=\{\mathcal{O},(-1,0),(3,0),(0,0)\}$, the formula for $\alpha$ gives
$$\begin{aligned}
\alpha(\mathcal{O}) &= (1, 1), & \alpha(-1, 0) &= \bigl([-1-0],\, [-1-0][-1-3]\bigr) = ([-1], [4]) = (-1, 1),\\
\alpha(3, 0) &= \bigl([3-0],\, [3+1]\bigr) = ([3], [4]) = (3, 1), & \alpha(0, 0) &= \bigl([0+1][0-3],\, [0+1]\bigr) = ([-3], [1]) = (-3, 1).
\end{aligned}$$
Since these $4$ images are pairwise distinct and exhaust $\{\pm 1,\pm 3\}\times\{1\}$, the injective homomorphism $\alpha:E_{48}(\mathbb{Q})/2E_{48}(\mathbb{Q})\hookrightarrow\{\pm 1,\pm 3\}\times\{1\}$ is an isomorphism, and the natural map $E_{48}(\mathbb{Q})[2]\to E_{48}(\mathbb{Q})/2E_{48}(\mathbb{Q})$ is a bijection. Because $E_{48}(\mathbb{Q})[2]\cong(\mathbb{Z}/2\mathbb{Z})^2$, the structure theorem for finitely generated abelian groups gives $\#(E_{48}(\mathbb{Q})/2E_{48}(\mathbb{Q}))=2^{\mathrm{rank}(E_{48}(\mathbb{Q}))+2}=4$, which proves $\mathrm{rank}(E_{48}(\mathbb{Q}))=0$, and the injectivity of $E_{48}(\mathbb{Q})[2]\hookrightarrow E_{48}(\mathbb{Q})/2E_{48}(\mathbb{Q})$ means $E_{48}(\mathbb{Q})[2]\cap 2E_{48}(\mathbb{Q})=\{\mathcal{O}\}$, so $E_{48}(\mathbb{Q})[2^\infty]=E_{48}(\mathbb{Q})[2]$. Since $\Delta(E_{48})=2304=2^8\cdot 3^2$, $E_{48}$ has good reduction at $p=5$ and $p=7$, where direct enumeration (`exp100` Part (b)) gives $\#E_{48}(\mathbb{F}_5)=\#E_{48}(\mathbb{F}_7)=8=2^3$; since prime-to-$p$ rational torsion injects into $E_{48}(\mathbb{F}_p)$, $E_{48}(\mathbb{Q})$ has no odd-order torsion. Thus $E_{48}(\mathbb{Q})=E_{48}(\mathbb{Q})[2]$ and $2E_{48}(\mathbb{Q})=\{\mathcal{O}\}$. Finally, for any $(t,w)\in\mathcal{E}_{\mathrm{self}}(\mathbb{Q})$ with $t\neq 0$, we have $t\neq \pm 1$ (since $t=\pm 1\implies w^2=-12\notin\mathbb{Q}^2$), so (2.13) defines a finite point $P=(x,y)\in E_{48}(\mathbb{Q})$ with $x=\bigl(\frac{t^2-1}{2t}\bigr)^2\in(\mathbb{Q}^\times)^2$ and $x+1=\bigl(\frac{t^2+1}{2t}\bigr)^2\in(\mathbb{Q}^\times)^2$. Hence $\alpha(P)=([x],[x+1])=(1,1)$, which forces $P\in\ker(\alpha)=2E_{48}(\mathbb{Q})=\{\mathcal{O}\}$, a contradiction since $x(P)\in\mathbb{Q}$ is finite. Thus $(0,\pm 1)$ are the only affine rational points on $\mathcal{E}_{\mathrm{self}}$. $\square$

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## 3. Opposite-Branch Step-$2$ Conservation Law and Complete Arithmetic Classification of $\Delta_{\mathrm{norm}}^{(3)}$

Recall from [15, Theorem 20(148)] the step-$2$ rungs $u_{2,\mp}=x(P_0\mp[2]S_Y)=\frac{A_\mp^2-B_\mp^2}{gB_\mp^2}$ and step-$3$ rungs $u_3=x([3]P_0)=\frac{N_3^2-D_3^2}{gD_3^2}$, $u_{3,\mp}=x([3]P_0\mp S_Y)=\frac{N_{3,\mp}^2-D_{3,\mp}^2}{gD_{3,\mp}^2}$, where
$$A_\mp = v^4 \mp 2hv^3 + h^4 - h^2,\qquad B_\mp = v^4 \mp 2hv^3 \pm 2h(h^2-1)v - h^4 + h^2,\qquad C_\mp = v^3 \mp 2hv^2 \pm (h^3 - h), \tag{3.1}$$
$$N_3 = v^2 - 5h^2 + 1,\qquad D_3 = v^2 - h^2 - 3,\qquad N_{3,\mp} = (v\pm h)^3 + v \pm 3h,\qquad D_{3,\mp} = 3v^2 \pm 8hv + 5h^2 - 1. \tag{3.2}$$
While Paper 72 [15, §9(c.1)] recorded the five-factor polynomial remainder $\Delta_{\mathrm{norm}}^{(3)}$ of $N_g(u_3,u_{3,\mp})$, we now prove the opposite-branch step-$2$ conservation law and the complete arithmetic classification of all five factors of $\Delta_{\mathrm{norm}}^{(3)}$.

**Theorem 2 (Step-$2$ Conservation Law and Step-$3$ Obstruction Classification; `exp100`).**
1. *(Opposite-branch step-$2$ reflection-defect factorization and dual norm completions):* Define the degree-$8$ polynomials
   $$Q_{8,\mp}(h,v) := (v^2 - h^2 - h)^2(v^2 - h^2 + h)^2 \pm 4h^3 v^2(v^2 - h^2 + 1). \tag{3.3}$$
   Then $A_- A_+ \mp h B_- B_+ = \mp(h\mp 1)Q_{8,\mp}$, $Q_{8,\mp}=P_{\mathrm{self},-}P_{\mathrm{self},+}+4h^2v^2\bigl(v^2-(h\mp 1)^2\bigr)(v^2-h^2\mp h)$, and the opposite-branch step-$2$ reflection difference factors as
   $$\varphi_g(u_{2,-}) - u_{2,+} = \frac{g Q_{8,-} Q_{8,+}}{A_-^2 B_+^2},\qquad \varphi_g(u_{2,+}) - u_{2,-} = \frac{g Q_{8,-} Q_{8,+}}{A_+^2 B_-^2}. \tag{3.4}$$
   Moreover, the cleared opposite-branch cross-reflection and same-branch norms
   $$N_{\mathrm{cross}}^{(2)} := A_+^2 B_-^2\bigl(1 + u_{2,-}\varphi_g(u_{2,+})\bigr) = A_-^2 B_+^2\bigl(1 + u_{2,+}\varphi_g(u_{2,-})\bigr),\qquad N_{\mathrm{same}}^{(2)} := B_-^2 B_+^2(1 + u_{2,-}u_{2,+}) \tag{3.5}$$
   satisfy the dual square-completions and exact conservation law
   $$\begin{aligned}
   N_{\mathrm{cross}}^{(2)} &= (A_- A_+)^2 + h^2 \Delta_{2,\mathrm{opp}} = (h B_- B_+)^2 - \Omega_{2,\mathrm{opp}},\\
   N_{\mathrm{same}}^{(2)} &= (B_- B_+)^2 - \Delta_{2,\mathrm{opp}},\qquad h^2 N_{\mathrm{same}}^{(2)} = (A_- A_+)^2 + \Omega_{2,\mathrm{opp}},\\
   N_{\mathrm{cross}}^{(2)} + h^2 N_{\mathrm{same}}^{(2)} &= (A_- A_+)^2 + (h B_- B_+)^2,\qquad (h B_- B_+)^2 - (A_- A_+)^2 = h^2\Delta_{2,\mathrm{opp}} + \Omega_{2,\mathrm{opp}} = g^2 Q_{8,-}Q_{8,+},
   \end{aligned} \tag{3.6}$$
   where $\Delta_{2,\mathrm{opp}} := 16g^2 h^2 v^2(v^2-h^2)C_- C_+$ and $\Omega_{2,\mathrm{opp}} := g^2(v^2-h^2-h)^2(v^2-h^2+h)^2\bigl((v-h)^4-h^2\bigr)\bigl((v+h)^4-h^2\bigr)$.
2. *(Arithmetic classification of all five factors of the step-$3$ cross-reflection obstruction $\Delta_{\mathrm{norm}}^{(3)}$):* In the step-$3$ cross-reflection norm decomposition [15, Theorem 20(149)]
   $$D_3^2 D_{3,-}^2 D_{3,+}^2 N_g(u_3, u_{3,\mp}) = \bigl(N_{3,\mp} P_{4,\pm}^{(3)}\bigr)^2 + \Delta_{\mathrm{norm}}^{(3)},\qquad \Delta_{\mathrm{norm}}^{(3)} = 8(v^2 - 3h^2 - 1)(v^2 - h^2 - h)(v^2 - h^2 + h) Q_{4,-}(h,v) Q_{4,+}(h,v), \tag{3.7}$$
   with $Q_{4,\mp}(h,v)=\bigl(v^2-(h^2\mp 4h-1)\bigr)^2\mp 16h(h\pm 1)^2$, the vanishing of $\Delta_{\mathrm{norm}}^{(3)}=0$ over $\mathbb{Q}$ ($g\neq 0 \iff h\neq \pm 1$) is completely classified as follows:
   - **Factor 1 ($v^2 = 3h^2 + 1$):** Substituting $v^2=3h^2+1$ into $(r_a\pm r_b)^2=2h(h\pm 1)-v^2$ gives $(r_a+r_b)^2 = -(h-1)^2 < 0$ and $(r_b-r_a)^2 = -(h+1)^2 < 0$, so no real tangent triple $\{a,b,g\}$ exists for $h\neq \pm 1$.
   - **Factors 2 and 3 ($Q_{4,-}=0$ or $Q_{4,+}=0$):** These satisfy the Fermat fourth-power identity
     $$N_3^4 - h^2 D_3^4 = -(h^2 - 1) Q_{4,-}(h,v) Q_{4,+}(h,v), \tag{3.8}$$
     together with $(N_{3,-}N_{3,+})^2 - h^2(D_{3,-}D_{3,+})^2 = (v^2-h^2-h)(v^2-h^2+h)Q_{4,-}Q_{4,+}$. If $Q_{4,-}Q_{4,+}=0$ with $g\neq 0$ ($h\neq \pm 1$), then $D_3\neq 0$ (since $D_3=0\iff v^2=h^2+3\implies N_3=4(1-h^2)=-4g^2\neq 0$) and $1+g^2=h^2=(N_3/D_3)^4$, which has no rational solutions with $g\neq 0$ by Fermat's right-triangle theorem ($X^4-Y^4=Z^2$).
   - **Factors 4 and 5 ($v^2 = h^2 \pm h$):** These are the two $4$-torsion branches $[2]S_Y=T_g$ ($D_g=0$), on which one of $(r_a\pm r_b)^2$ is automatically equal to $v^2$ while the other equals $h(h\mp 3)$, and the step-$2$ self-reflection product $-P_{\mathrm{self},\mp}P_{12,\mp}$ becomes the exact square $\bigl(4h^4(h+\mathrm{sgn})^2(2v-2\varepsilon h-\varepsilon\,\mathrm{sgn})\bigr)^2$ modulo $v^2=h^2+\mathrm{sgn}\cdot h$.

*Proof.*
(1) Using $\varphi_g(\frac{R_1^2-1}{g})-\frac{R_2^2-1}{g}=\frac{h^2-R_1^2 R_2^2}{g R_1^2}$ with $R_1=A_\mp/B_\mp$ and $R_2=A_\pm/B_\pm$, the numerator of $\varphi_g(u_{2,\mp})-u_{2,\pm}$ is $h^2 B_-^2 B_+^2 - A_-^2 A_+^2 = -(A_-A_+ - hB_-B_+)(A_-A_+ + hB_-B_+) = (h^2-1)Q_{8,-}Q_{8,+}=g^2 Q_{8,-}Q_{8,+}$, proving (3.4). Expanding $N_{\mathrm{cross}}^{(2)}$ and $N_{\mathrm{same}}^{(2)}$ in $\mathbb{Z}[h,v]$ modulo $g^2=h^2-1$ yields (3.6) (`exp100` Part (c)).

(2) Multiplying $N_3^2-hD_3^2=-(h-1)Q_{4,-}$ and $N_3^2+hD_3^2=(h+1)Q_{4,+}$ yields $N_3^4-h^2 D_3^4=-(h^2-1)Q_{4,-}Q_{4,+}$, proving (3.8). If $Q_{4,-}Q_{4,+}=0$ with $g^2=h^2-1\neq 0$, then $D_3=0$ is impossible since $N_3|_{v^2=h^2+3}=4(1-h^2)=-4g^2\neq 0$ would contradict $N_3^4=h^2 D_3^4=0$, so $(N_3/D_3)^4-1=g^2$, contradicting Fermat's right-triangle theorem. The remaining identities are certified in `exp100` Part (b). $\square$

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## 4. Bilinear Null-Cone Coordinates $(u_-,u_+)$, Biquadratic Elliptic Fibers $\mathcal{E}_{u_-}$, and Non-Torsion Bäcklund Orbits

In the normalized affine chart $y=1$ on $\mathcal{S}_{\mathrm{sym}}$, write $(X,Z,W):=(x^2,z^2,W)$ for the coordinates on the $(x,z)$-sign quotient surface $\overline{\mathcal{S}}_{\mathrm{sym}}:W^2=2(X+1)(Z+1)(X+Z)$. Expanding the right-hand side as
$$2(X+1)(Z+1)(X+Z) = \frac{1}{2}(X+Z)\bigl((X+Z+2)^2 - (X-Z)^2\bigr) \tag{4.1}$$
motivates introducing the scaled slope $M:=\frac{W}{X+Z+2}$, the asymmetry $\eta:=\frac{X-Z}{X+Z+2}$, and the **null-cone coordinates**
$$u_- := M - \eta = \frac{W - (x^2 - z^2)}{x^2 + z^2 + 2},\qquad u_+ := M + \eta = \frac{W + (x^2 - z^2)}{x^2 + z^2 + 2}. \tag{4.2}$$

**Theorem 3 (Null-Cone Elliptic Fibers $\mathcal{E}_{u_-}$ and Non-Torsion Bäcklund Orbits; `exp101`–`exp102`).**
1. *(Bilinear null-cone parameterization of the quotient $\overline{\mathcal{S}}_{\mathrm{sym}}$ and separation of invariants):* The inverse map from $\mathbb{A}^2_{(u_-,u_+)}$ (with $\eta=\frac{u_+-u_-}{2}\neq \pm 1$) to the quotient surface $\overline{\mathcal{S}}_{\mathrm{sym}}$ in $(X,Z,W)=(x^2,z^2,W)$ is birational and bilinear of bidegree $(1,1)$ in $(u_-,u_+)$:
   $$X(u_-,u_+) = \frac{2u_- u_+ - u_- + u_+}{2 + u_- - u_+},\qquad Z(u_-,u_+) = \frac{2u_- u_+ + u_- - u_+}{2 - u_- + u_+},\qquad W(u_-,u_+) = \frac{4(u_- + u_+)(u_- u_+ + 1)}{4 - (u_- - u_+)^2}, \tag{4.3}$$
   with hyperbolic product $\Omega := u_- u_+ = M^2 - \eta^2 = \frac{X + 2XZ + Z}{X + Z + 2}$. In $(u_-,u_+)$, the invariants $\Sigma=\alpha+\beta$, $D=\alpha-\beta$, $S=\sqrt{1-\alpha\beta}$, and $\rho_\mp = D \mp 2S$ decouple as
   $$\Sigma = \frac{2(u_- u_+ - 1)}{u_- u_+ + 1},\qquad D = \frac{2(u_+ - u_-)}{u_- u_+ + 1},\qquad S = \frac{u_- + u_+}{u_- u_+ + 1},\qquad \rho_- = -\frac{4u_-}{u_- u_+ + 1},\qquad \rho_+ = \frac{4u_+}{u_- u_+ + 1}, \tag{4.4}$$
   so that $(u_- u_+ + 1)^2\frac{P_{\mathrm{self},\mp}}{(gh)^2} = (u_- u_+ + 1)^2 - 16u_\mp^2$, $(u_- u_+ + 1)^2(1 - \rho_-\rho_+) = (u_- u_+)^2 + 18u_- u_+ + 1$, and $(u_-u_++1)^2\frac{D_g}{h^2}=\Omega^2-6\Omega+1+8\eta^2$.
2. *(Square-determinant Möbius law and smooth biquadratic elliptic fibers $\mathcal{E}_{u_-}$ on $\mathcal{S}_{\mathrm{sym}}$):* Solving $X(u_-,u_+)=x^2$ for $u_+$ along a fixed null-cone coordinate $u_-\in\mathbb{Q}\setminus\{0,\pm 1\}$ gives
   $$u_+(u_-, x^2) = \frac{(x^2 + 1)u_- + 2x^2}{x^2 + 2u_- + 1}, \tag{4.5}$$
   and substituting (4.5) into $Z(u_-,u_+)=z^2$ yields the **square-determinant Möbius relation** cutting out the smooth genus-$1$ biquadratic fiber curve $\mathcal{E}_{u_-}$ in $(x,z)$:
   $$z^2 = \frac{q_- x^2 + 2u_-^2}{2x^2 - \tilde{q}_-} \iff (2x^2 - \tilde{q}_-)(2z^2 - q_-) = (u_-^2 - 1)^2 \iff \mathcal{E}_{u_-}(x,z) := 2x^2 z^2 - q_- x^2 - \tilde{q}_- z^2 - 2u_-^2 = 0, \tag{4.6}$$
   where $q_- := u_-^2 + 2u_- - 1$ and $\tilde{q}_- := u_-^2 - 2u_- - 1$ (and dually $\mathcal{E}_{u_+}^+(x,z):2x^2z^2-\tilde{q}_+ x^2-q_+ z^2-2u_+^2=0$ along fixed $u_+$). Because $u_+(u_-,x^2)$ in (4.5) and $W(u_-,u_+)$ in (4.3) are rational functions of $(u_-,x^2)$, **every rational point $(x,z)\in\mathcal{E}_{u_-}(\mathbb{Q})$ lifts rationally to $(x,1,z,W(u_-,u_+(u_-,x^2)))$ on the surface closure without extracting any further square root, and defines an admissible point on $\mathcal{S}_{\mathrm{sym}}(\mathbb{Q})$ whenever $xz(x^2-z^2)\neq 0$** (excluding the coordinate-zero closure points $xz=0$ and the diagonal closure points $x=\pm z=\pm u_-$).
3. *(Difference-of-squares quartic decomposition and the rational osculating section $\mathcal{C}_{\mathrm{null}}(u_-)$):* Clearing the denominator of (4.6) puts $\mathcal{E}_{u_-}$ into the quartic double-cover model $Y^2=Q_{u_-}(x):=(2x^2-\tilde{q}_-)(q_- x^2+2u_-^2)$ with $Y=(2x^2-\tilde{q}_-)z$, where $Q_{u_-}(x)$ admits the difference-of-squares decomposition
   $$Q_{u_-}(x) = A_{\mathrm{sq}}(x)^2 - B_{\mathrm{sq}}(x)^2,\qquad A_{\mathrm{sq}}(x) = \frac{(u_-+1)\bigl((3u_--1)x^2 - u_-^2(u_--3)\bigr)}{2u_-},\qquad B_{\mathrm{sq}}(x) = \frac{(u_--1)^2(x^2 - u_-^2)}{2u_-}, \tag{4.7}$$
   vanishing at the four rational diagonal basepoints $(x,z)=(\pm u_-,\pm u_-)$ where $Y=\pm u_-(u_-+1)^2$. The osculating parabola to $Y^2=Q_{u_-}(x)$ at $P_{+,+}=(u_-,u_-(u_-+1)^2)$,
   $$P_{\mathrm{osc}}(x) = u_-(u_-+1)^2 + (u_-+1)(3u_--1)(x - u_-) + \frac{u_-^4 + 6u_-^3 + 2u_- - 1}{u_-(u_-+1)^2}(x - u_-)^2, \tag{4.8}$$
   satisfies $P_{\mathrm{osc}}(x)^2 - Q_{u_-}(x) = -\frac{(u_--1)^4(x-u_-)^3\bigl((u_-^4+4u_-^3-1)x + u_-(u_-^4-4u_--1)\bigr)}{u_-^2(u_-+1)^4}$, cutting out the **universal rational osculating section** $\mathcal{C}_{\mathrm{null}}(u_-) = \bigl(x_{\mathrm{osc}}(u_-),\, 1,\, z_{\mathrm{osc}}(u_-),\, W_{\mathrm{osc}}(u_-)\bigr) \subset \mathcal{S}_{\mathrm{sym}}(\mathbb{Q}(u_-))$:
   $$x_{\mathrm{osc}}(u_-) = -\frac{u_-(u_-^4 - 4u_- - 1)}{u_-^4 + 4u_-^3 - 1},\qquad z_{\mathrm{osc}}(u_-) = -\frac{u_-(u_-^4 + 4u_- - 1)}{u_-^4 - 4u_-^3 - 1},\qquad u_{+,\mathrm{osc}}(u_-) = \frac{u_-(u_-^8 - 2u_-^4 + 16u_-^2 + 1)}{u_-^8 + 16u_-^6 - 2u_-^4 + 1}, \tag{4.9}$$
   $$W_{\mathrm{osc}}(u_-) = \frac{2u_-(u_-^2+1)(u_-^4-2u_-^3+2u_-^2+2u_-+1)(u_-^4+2u_-^3+2u_-^2-2u_-+1)(u_-^8+8u_-^6-2u_-^4+8u_-^2+1)}{(u_-^4 - 4u_-^3 - 1)^2(u_-^4 + 4u_-^3 - 1)^2}, \tag{4.10}$$
   where $(x_{\mathrm{osc}}^2+1)(z_{\mathrm{osc}}^2+1)$ and $2(x_{\mathrm{osc}}^2+z_{\mathrm{osc}}^2)$ are *separately* perfect squares in $\mathbb{Q}(u_-)$, and $u_-\mapsto 1/u_-$ maps $(x_{\mathrm{osc}},z_{\mathrm{osc}},u_{+,\mathrm{osc}})\mapsto(1/x_{\mathrm{osc}},1/z_{\mathrm{osc}},1/u_{+,\mathrm{osc}})$.
4. *(Closed-form two-term Bäcklund involutions $\mathcal{I}_{u_-}^{(+)},\mathcal{I}_{u_-}^{(-)}$, removable basepoint limit, and Mazur non-torsion theorem):* For any point $(x_0,z_0)\in\mathcal{E}_{u_-}$ with $A_0:=x_0 z_0-u_-^2$ and $B_0:=x_0-z_0$ not both zero, define the two-term rational map $\mathcal{I}_{u_-}^{(+)}(x_0,z_0):=(x_1,z_1)$ by
   $$x_1 = \frac{(u_-^2 + 1)A_0 z_0 - 2u_-^3 B_0}{(u_-^2 + 1)A_0 - 2u_- B_0 z_0},\qquad z_1 = \frac{(u_-^2 + 1)A_0 x_0 - 2u_-^3 B_0}{(u_-^2 + 1)A_0 - 2u_- B_0 x_0}, \tag{4.11}$$
   and set $\mathcal{I}_{u_-}^{(-)}(x_0,z_0):=\sigma_z\bigl(\mathcal{I}_{u_-}^{(+)}(x_0,-z_0)\bigr)$ where $\sigma_z(x,z)=(x,-z)$ (compare the general QRT involution framework in Hone [17, §2]). Then:
   - **(i) Exact fiber invariance:** Clearing denominators of $\mathcal{E}_{u_-}(x_1,z_1)$ yields the polynomial identity
     $$2N_{x_1}^2 N_{z_1}^2 - q_- N_{x_1}^2 D_{z_1}^2 - \tilde{q}_- D_{x_1}^2 N_{z_1}^2 - 2u_-^2 D_{x_1}^2 D_{z_1}^2 \equiv \bigl((u_-^2+1)A_0 - 2u_-^2 B_0\bigr)^2\bigl((u_-^2+1)A_0 + 2u_-^2 B_0\bigr)^2 \mathcal{E}_{u_-}(x_0,z_0). \tag{4.12}$$
   - **(ii) Involution and removable osculating limit:** On $\mathcal{E}_{u_-}(x_0,z_0)=0$, $B_0 A_1 + A_0 B_1 \equiv 0$, so $\mathcal{I}_{u_-}^{(+)}\circ\mathcal{I}_{u_-}^{(+)}=\mathrm{id}$ and $\mathcal{I}_{u_-}^{(-)}\circ\mathcal{I}_{u_-}^{(-)}=\mathrm{id}$. Moreover, $\mathcal{I}_{u_-}^{(+)}\bigl(x_{\mathrm{osc}}(u_-),z_{\mathrm{osc}}(u_-)\bigr)=(-u_-,-u_-)$, so by involution symmetry the removable limit of $\mathcal{I}_{u_-}^{(+)}$ along $\mathcal{E}_{u_-}=0$ at the diagonal basepoint $(-s u_-, -s u_-)$ ($s\in\{+1,-1\}$) is the osculating point:
     $$\lim_{\substack{(x,z)\to(-su_-,-su_-)\\ \mathcal{E}_{u_-}(x,z)=0}} \mathcal{I}_{u_-}^{(+)}(x,z) = \bigl(s x_{\mathrm{osc}}(u_-),\, s z_{\mathrm{osc}}(u_-)\bigr). \tag{4.13}$$
   - **(iii) Certified degree sequence for $0\le n\le 3$ and infinite order over $\mathbb{Q}$ and $\mathbb{Q}(u_-)$ via Mazur's Torsion Theorem:** Along the alternating Bäcklund chain $P^{(0)}=(-u_-,-u_-)$, $P^{(1)}=\mathcal{I}_{u_-}^{(+)}(P^{(0)})=(x_{\mathrm{osc}},z_{\mathrm{osc}})$, $P^{(2)}=\mathcal{I}_{u_-}^{(-)}(P^{(1)})=\mathcal{T}_{u_-}(P^{(0)})$, $P^{(3)}=\mathcal{I}_{u_-}^{(+)}(P^{(2)})$ over $\mathbb{Q}(u_-)$, the reduced numerator degrees in $u_-$ at $n=0,1,2,3$ are
     $$\bigl(\deg_{u_-}\mathrm{num}(x(P^{(n)}))\bigr)_{n=0}^3 = (1,\, 5,\, 13,\, 25). \tag{4.14}$$
     Furthermore, for each rational specialization $u_-\in\{2, 3, 23/33, 93/145\}$ (corresponding to $\mathcal{C}_{\mathrm{null}}(2)$, $\mathcal{C}_{\mathrm{null}}(3)$, $(11,3,2,650)$, and $(12,5,3,1326)$), the $13$ iterates $\{\mathcal{T}_{u_-}^k(x_0,z_0)\}_{k=-6}^6\subset\mathcal{E}_{u_-}(\mathbb{Q})$ are **pairwise distinct** (with logarithmic $x$-heights up to $1,990$ bits; Table 2). By Mazur's Torsion Theorem [16] (which bounds the order of any rational torsion point on an elliptic curve over $\mathbb{Q}$ by $12$), $\mathcal{T}_{u_-}:=\mathcal{I}_{u_-}^{(-)}\circ\mathcal{I}_{u_-}^{(+)}$ has **infinite order over $\mathbb{Q}$** on each of these four specialized fibers, and therefore—since specialization is a group homomorphism on smooth fibers—$\mathcal{T}_{u_-}$ also has **infinite order on the generic fiber $\mathcal{E}_{u_-}$ over $\mathbb{Q}(u_-)$**.

*Proof.*
Parts (1)–(3) follow by direct symbolic expansion and polynomial remainder verification in $\mathbb{Q}(u_-,u_+,x,z)$ (`exp101` Part (a) and `exp102` Part (a)). In (4.10), $x_{\mathrm{osc}}(u_-)^2+1=\frac{N_{\mathrm{osc}}(u_-)}{(u_-^4+4u_-^3-1)^2}$ and $z_{\mathrm{osc}}(u_-)^2+1=\frac{N_{\mathrm{osc}}(u_-)}{(u_-^4-4u_-^3-1)^2}$ share the same degree-$10$ numerator $N_{\mathrm{osc}}(u_-)=(u_-^2+1)(u_-^4-2u_-^3+2u_-^2+2u_-+1)(u_-^4+2u_-^3+2u_-^2-2u_-+1)$, while $x_{\mathrm{osc}}(u_-)^2+z_{\mathrm{osc}}(u_-)^2=2\left(\frac{u_-(u_-^8+8u_-^6-2u_-^4+8u_-^2+1)}{(u_-^4-4u_-^3-1)(u_-^4+4u_-^3-1)}\right)^2$ is twice a square in $\mathbb{Q}(u_-)$, making $W_{\mathrm{osc}}(u_-)$ rational over $\mathbb{Q}(u_-)$.

For Part (4), expanding the left-hand side of (4.12) in $\mathbb{Z}[u_-,x_0,z_0]$ yields the exact polynomial multiple of $\mathcal{E}_{u_-}(x_0,z_0)$ on the right-hand side. Reducing $B_0(x_1 z_1 - u_-^2) + A_0(x_1 - z_1)$ modulo $\mathcal{E}_{u_-}(x_0,z_0)$ gives $0$, which implies $\frac{B_1}{A_1}=-\frac{B_0}{A_0}$ on $\mathcal{E}_{u_-}=0$; substituting $\frac{B_1}{A_1}=-\frac{B_0}{A_0}$ into $\mathcal{I}_{u_-}^{(+)}(x_1,z_1)$ cancels the linear fractional transformation and returns $(x_0,z_0)$. Evaluating (4.11) at $(x_{\mathrm{osc}}(u_-),z_{\mathrm{osc}}(u_-))$ gives $(-u_-,-u_-)$ identically in $\mathbb{Q}(u_-)$, proving (4.13). Computing $P^{(0)},\dots,P^{(3)}$ in reduced form over $\mathbb{Q}(u_-)$ (`exp102` Part (b)) certifies the exact degree sequence $(1,5,13,25)$ in (4.14). Finally, on any smooth biquadratic genus-$1$ curve $\mathcal{E}_{u_-}$ ($u_-\notin\{0,\pm 1\}$), $\mathcal{I}_{u_-}^{(+)}$ and $\mathcal{I}_{u_-}^{(-)}$ are fixed-point-having involutions ($[-1]$ reflections with respect to different origins [17, §2]), so their composite $\mathcal{T}_{u_-}=\mathcal{I}_{u_-}^{(-)}\circ\mathcal{I}_{u_-}^{(+)}$ is a translation by a rational divisor class $\delta(u_-)\in\mathrm{Pic}^0(\mathcal{E}_{u_-})(\mathbb{Q})$. At each smooth rational specialization $u_-\in\{2,3,23/33,93/145\}$, any torsion point in $\mathrm{Pic}^0(\mathcal{E}_{u_-})(\mathbb{Q})$ has order at most $12$ by Mazur's theorem [16]; thus exhibiting $13$ distinct orbit points $\{\mathcal{T}_{u_-}^k(x_0,z_0)\}_{k=-6}^6$ (`exp102` Part (d)) proves that $\delta(u_-)$ is non-torsion over $\mathbb{Q}$ at each of those four fibers, which in turn implies that the generic translation section $\delta\in\mathrm{Pic}^0(\mathcal{E}_{u_-})(\mathbb{Q}(u_-))$ is non-torsion over $\mathbb{Q}(u_-)$. $\square$

---

## 5. The $S_3$-Triad Matrix, Markov & Joukowsky–Markov Surface Laws, Fixed-$(x,z)$ Multi-Pair Factorization, and Root-Free Recovery

We now examine how the three coordinate permutations of a point $(x:y:z)\in\mathbb{P}^2(\mathbb{Q})$ ($xyz(x^2-y^2)(x^2-z^2)(y^2-z^2)\neq 0$) generate a coupled triad of tangent triples. For any ordered triple of non-zero coordinates $(X,Y,Z)$, define the canonical tangent-triple map $\mathrm{triad}(X,Y,Z)=(a,b,g,h)$ by
$$a(X,Y,Z) := \frac{(Y^2 - Z^2)X}{(X^2 + Y^2)Z},\qquad b(X,Y,Z) := \frac{(X^2 - Y^2)Z}{(Y^2 + Z^2)X},\qquad g(X,Y,Z) := \frac{X^2 - Z^2}{2XZ},\qquad h(X,Y,Z) := \frac{X^2 + Z^2}{2XZ}. \tag{5.1}$$
Following `exp101`, we associate to $(x:y:z)$ the three triads obtained by placing $y, z, x$ respectively as the middle coordinate $Y$:
$$T_1 := \mathrm{triad}(x, y, z) = (a_1, b_1, g_1, h_1),\qquad T_2 := \mathrm{triad}(x, z, y) = (a_2, b_2, g_2, h_2),\qquad T_3 := \mathrm{triad}(y, x, z) = (a_3, b_3, g_3, h_3), \tag{5.2}$$
and assemble their triple entries into the $3\times 3$ **triad matrix**
$$M_{\mathrm{triad}}(x:y:z) := \begin{pmatrix} a_1 & b_1 & g_1 \\ a_2 & b_2 & g_2 \\ a_3 & b_3 & g_3 \end{pmatrix}. \tag{5.3}$$

**Theorem 4 ($S_3$-Triad Matrix, Markov Laws, and Multi-Pair Calculus; `exp101`–`exp102`).**
1. *(Universal tangent-triple structure and the Single Square-Class Theorem on $\mathbb{P}^2_{(x:y:z)}$):* For every $(x:y:z)\in\mathbb{P}^2(\mathbb{Q})$ with $xyz\neq 0$ and every $j\in\{1,2,3\}$, $1+g_j^2=h_j^2$ and $b_j=\varphi_{g_j}(a_j)=\frac{g_j-a_j}{1+g_ja_j}$ hold identically. Moreover, with $H^2 := h_1 h_2 h_3 = \frac{2(x^2+y^2)(x^2+z^2)(y^2+z^2)}{(4xyz)^2}$, all nine Diophantine triple conditions across $Y^{(1)},Y^{(2)},Y^{(3)}$ satisfy
   $$\begin{aligned}
   1+a_1 b_1 &= \frac{H^2}{(h_2 h_3)^2}, & 1+a_1 g_1 &= \frac{H^2}{h_2^2}, & 1+b_1 g_1 &= \frac{H^2}{h_3^2},\\
   1+a_2 b_2 &= \frac{H^2}{(h_1 h_3)^2}, & 1+a_2 g_2 &= \frac{H^2}{h_1^2}, & 1+b_2 g_2 &= \frac{H^2}{h_3^2},\\
   1+a_3 b_3 &= \frac{H^2}{(h_1 h_2)^2}, & 1+a_3 g_3 &= \frac{H^2}{h_2^2}, & 1+b_3 g_3 &= \frac{H^2}{h_1^2}.
   \end{aligned} \tag{5.4}$$
   Consequently, all nine shifted products $1+a_jb_j, 1+a_jg_j, 1+b_jg_j$ ($j=1,2,3$) belong to the **exact same rational square class** $[h_1h_2h_3]\in\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$, and lifting $(x:y:z)$ to $\mathcal{S}_{\mathrm{sym}}$ via $H=\frac{W}{4xyz}$ is necessary and sufficient for any one—and hence all nine—to be rational squares. Under the sign reflection $\sigma_y:(x,y,z)\mapsto(x,-y,z)$, $Y^{(1)}$ is invariant while $Y^{(2)}\mapsto -Y^{(2)}$ and $Y^{(3)}\mapsto -Y^{(3)}$.
2. *(Hyperbolic-addition matrix, Cayley–Menger hypotenuse law, Markov spoke-root law, and Joukowsky–Markov $\mathbb{P}^2$ identity):* The entries of $M_{\mathrm{triad}}(x:y:z)$ are expressed purely in terms of $(g_1,g_2,g_3)$ and $(h_1,h_2,h_3)$ by the **hyperbolic-addition matrix**
   $$\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{pmatrix} = \begin{pmatrix} g_3/h_2 & g_2/h_3 \\ -g_3/h_1 & g_1/h_3 \\ g_1/h_2 & -g_2/h_1 \end{pmatrix}, \tag{5.5}$$
   while the hypotenuses satisfy the hyperbolic cosine addition laws and Cayley–Menger determinant identity
   $$h_1 = h_2 h_3 + g_2 g_3,\qquad h_2 = h_1 h_3 - g_1 g_3,\qquad h_3 = h_1 h_2 - g_1 g_2,\qquad h_1^2 + h_2^2 + h_3^2 - 2h_1 h_2 h_3 = 1. \tag{5.6}$$
   On $\mathcal{S}_{\mathrm{sym}}$ (where $H=\frac{W}{4xyz}$ satisfies $H^2=h_1h_2h_3$), the spoke roots satisfy the **spoke-root reciprocity laws**
   $$r_{a,1} r_{ab,2} = r_{b,1} r_{ab,3} = r_{a,2} r_{ab,1} = r_{b,2} r_{ab,3} = r_{a,3} r_{ab,2} = r_{b,3} r_{ab,1} = 1,\qquad h_j r_{ab,k} r_{ab,\ell} = 1, \tag{5.7}$$
   and the three pair-roots $(r_{ab,1},r_{ab,2},r_{ab,3})$ satisfy the **Markov spoke-root surface identity**
   $$r_{ab,1}^2 + r_{ab,2}^2 + r_{ab,3}^2 - r_{ab,1}^2 r_{ab,2}^2 r_{ab,3}^2 = 2 \iff 1 + a_1 b_1 + a_2 b_2 + a_3 b_3 = r_{ab,1}^2 r_{ab,2}^2 r_{ab,3}^2, \tag{5.8}$$
   together with the null-cone harmonic conservation laws $\Sigma_1+\Sigma_2+\Sigma_3=0$ and $\sum_{j=1}^3\frac{1}{\Omega_j+1}=\frac{3}{2}$. Moreover, for any $(x:y:z)\in\mathbb{P}^2$ with aspect ratios $(k_1,k_2,k_3)=\bigl(\frac{x}{z},\frac{x}{y},\frac{y}{z}\bigr)$ and cross-ratios $(\lambda_1,\lambda_2,\lambda_3)=\bigl(\frac{y^2}{xz},\frac{z^2}{xy},\frac{x^2}{yz}\bigr)$, the Joukowsky ratios $m_j := \frac{\lambda_j+\lambda_j^{-1}}{k_j+k_j^{-1}} = \frac{2}{1+a_jb_j}-1$ satisfy the **universal Joukowsky–Markov $\mathbb{P}^2$ identity**
   $$m_1 + m_2 + m_3 - m_1 m_2 m_3 = 2. \tag{5.9}$$
3. *(Directed-tangent $4$-torsion factorization, off-diagonal cross-triad square laws, and fixed-$(x,z)$ multi-pair factorization):* For each triad $j\in\{1,2,3\}$ with middle coordinate $Y$ and outer coordinates $(X,Z)$, define $\tau_{j\to i}:=\frac{X^2-Y^2}{2XY}$ and $\tau_{j\to k}:=\frac{Z^2-Y^2}{2ZY}$ (so $\tau_{1,x}\tau_{1,z}=-g_2g_3$, $\tau_{2,x}\tau_{2,y}=g_1g_3$, $\tau_{3,y}\tau_{3,z}=g_1g_2$). Then the $4$-torsion factors of each triad factor universally as
   $$v_j^2 - h_j^2 - h_j = -(h_j + 1) r_{ab,j}^2 (1 + \tau_{j\to i}\tau_{j\to k}),\qquad v_j^2 - h_j^2 + h_j = (h_j - 1) r_{ab,j}^2 (1 - \tau_{j\to i}\tau_{j\to k}), \tag{5.10}$$
   and the off-diagonal cross-triad products satisfy
   $$1 \pm a_1 a_3 = \left(\frac{g_2}{h_2}\right)^2 + \frac{1 \pm g_1 g_3}{h_2^2},\qquad 1 \pm b_1 b_2 = \left(\frac{g_3}{h_3}\right)^2 + \frac{1 \pm g_1 g_2}{h_3^2},\qquad 1 \pm a_2 b_3 = \left(\frac{g_1}{h_1}\right)^2 + \frac{1 \pm g_2 g_3}{h_1^2}. \tag{5.11}$$
   In particular, $Y^{(2)}$ lies on $v_2^2=h_2^2\pm h_2$ if and only if $g_1g_3=\mp 1$, in which case $1\pm a_1a_3=(g_2/h_2)^2$ is a rational square identically. Furthermore, for any two points $(x,y_i,z,W_i)$ and $(x,y_j,z,W_j)$ on $\mathcal{S}_{\mathrm{sym}}$ sharing the outer coordinates $(x,z)$ with $xzW_iW_j\neq 0$ (hence sharing the central spoke $g_1=\frac{x^2-z^2}{2xz}$), their reflected pairs $\{a_1(y_i),b_1(y_i)\}$ and $\{a_1(y_j),b_1(y_j)\}$ satisfy the **fixed-$(x,z)$ multi-pair cross-factorizations**
   $$\begin{aligned}
   1 + a_1(y_i)a_1(y_j) &= \frac{(x^2+z^2)(y_i^2 y_j^2 + x^2 z^2)}{z^2(x^2+y_i^2)(x^2+y_j^2)}, & 1 + b_1(y_i)b_1(y_j) &= \frac{(x^2+z^2)(y_i^2 y_j^2 + x^2 z^2)}{x^2(y_i^2+z^2)(y_j^2+z^2)},\\
   1 + a_1(y_i)b_1(y_j) &= \frac{(x^2+z^2)(y_i^2 + y_j^2)}{(x^2+y_i^2)(y_j^2+z^2)}, & 1 + b_1(y_i)a_1(y_j) &= \frac{(x^2+z^2)(y_i^2 + y_j^2)}{(y_i^2+z^2)(x^2+y_j^2)}.
   \end{aligned} \tag{5.12}$$
   With $K:=\frac{(x^2+y_i^2)(x^2+y_j^2)}{(y_i^2+z^2)(y_j^2+z^2)}$ and $L:=\frac{(x^2+y_i^2)(y_j^2+z^2)}{(y_i^2+z^2)(x^2+y_j^2)}$, the surface equations $W_i^2=2(x^2+y_i^2)(x^2+z^2)(y_i^2+z^2)$ and $W_j^2=2(x^2+y_j^2)(x^2+z^2)(y_j^2+z^2)$ make both $\frac{z^2}{x^2}K$ and $L$ exact rational squares on $\mathcal{S}_{\mathrm{sym}}$:
   $$\begin{aligned}
   1 + b_1(y_i)b_1(y_j) &= \frac{z^2}{x^2} K\bigl(1 + a_1(y_i)a_1(y_j)\bigr) = \left(\frac{2z(x^2+z^2)(x^2+y_i^2)(x^2+y_j^2)}{x W_i W_j}\right)^2\bigl(1 + a_1(y_i)a_1(y_j)\bigr),\\
   1 + b_1(y_i)a_1(y_j) &= L\bigl(1 + a_1(y_i)b_1(y_j)\bigr) = \left(\frac{W_i(y_j^2+z^2)}{W_j(y_i^2+z^2)}\right)^2\bigl(1 + a_1(y_i)b_1(y_j)\bigr) = \left(\frac{W_j(x^2+y_i^2)}{W_i(x^2+y_j^2)}\right)^2\bigl(1 + a_1(y_i)b_1(y_j)\bigr)
   \end{aligned} \tag{5.12a}$$
   (with non-vanishing denominators and multipliers at every real point with $x y_i y_j z W_i W_j\neq 0$; for instance, at $(11,3,2,650)$ and $(11,22/3,2,14300/9)$, where $(a_i,b_i)=(11/52,224/143)$ and $(a_j,b_j)=(224/143,11/52)$, the multipliers in (5.12a) are $1^2=1=\frac{1+b_ib_j}{1+a_ia_j}$ and $(20/11)^2=\frac{400}{121}=\frac{1+b_ia_j}{1+a_ib_j}$).
4. *(Cross-triad syzygies and exact root-free projective recovery of $(x:y:z)$ from $M_{\mathrm{triad}}$):* On $M_{\mathrm{triad}}(x:y:z)$, grouping rows by shared hypotenuse denominators in (5.5) recovers $(h_1,h_2,h_3)$ rationally as
   $$h_1 = -\frac{g_3}{a_2} = -\frac{g_2}{b_3},\qquad h_2 = \frac{g_3}{a_1} = \frac{g_1}{a_3},\qquad h_3 = \frac{g_2}{b_1} = \frac{g_1}{b_2}, \tag{5.13}$$
   yielding the bilinear and trilinear cross-triad syzygies
   $$a_2 g_2 = b_3 g_3,\qquad a_1 g_1 = a_3 g_3,\qquad b_1 g_1 = b_2 g_2,\qquad a_1 b_2 b_3 = a_2 a_3 b_1 = -\frac{g_1 g_2 g_3}{h_1 h_2 h_3}, \tag{5.14}$$
   together with $h_2 h_3 = \frac{h_1}{1 + a_1 b_1}$, $g_2 g_3 = \frac{a_1 b_1 h_1}{1 + a_1 b_1}$, $g_2^2(1 - a_1^2 b_1^2) = b_1^2(1 + a_1^2)$, $g_3^2(1 - a_1^2 b_1^2) = a_1^2(1 + b_1^2)$, and the six hypotenuse-square equations $a_1^2(1+g_2^2)=g_3^2$, $a_3^2(1+g_2^2)=g_1^2$, $b_1^2(1+g_3^2)=g_2^2$, $b_2^2(1+g_3^2)=g_1^2$, $a_2^2(1+g_1^2)=g_3^2$, $b_3^2(1+g_1^2)=g_2^2$. Finally, the projective point $(x:y:z)\in\mathbb{P}^2(\mathbb{Q})$ is recovered **uniquely, rationally, and without square roots** from $M_{\mathrm{triad}}$ via
   $$\frac{x}{z} = h_1 + g_1 = \frac{a_2 g_1 - g_3}{a_2} = \frac{b_3 g_1 - g_2}{b_3},\qquad \frac{x}{y} = h_2 + g_2 = \frac{g_3 + a_1 g_2}{a_1} = \frac{g_1 + a_3 g_2}{a_3},\qquad \frac{y}{z} = h_3 + g_3 = \frac{g_2 + b_1 g_3}{b_1} = \frac{g_1 + b_2 g_3}{b_2}. \tag{5.15}$$

*Proof.*
(1)–(2) From (5.1), $g_1=\frac{x^2-z^2}{2xz}, h_1=\frac{x^2+z^2}{2xz}$, $g_2=\frac{x^2-y^2}{2xy}, h_2=\frac{x^2+y^2}{2xy}$, and $g_3=\frac{y^2-z^2}{2yz}, h_3=\frac{y^2+z^2}{2yz}$, which immediately gives (5.5)–(5.6) and the nine identities (5.4). On $\mathcal{S}_{\mathrm{sym}}$, $r_{ab,1}=\frac{H}{h_2h_3}, r_{ab,2}=\frac{H}{h_1h_3}, r_{ab,3}=\frac{H}{h_1h_2}$ with $H^2=h_1h_2h_3$, so
$$r_{ab,1}^2 + r_{ab,2}^2 + r_{ab,3}^2 - 2 = \frac{h_1^2 + h_2^2 + h_3^2 - 2h_1 h_2 h_3}{h_1 h_2 h_3} = \frac{1}{h_1 h_2 h_3} = r_{ab,1}^2 r_{ab,2}^2 r_{ab,3}^2,$$
proving (5.8). Since $1+a_1b_1=\frac{x^2+z^2}{(x^2+y^2)(y^2+z^2)}\cdot 2y^2 = \frac{2}{1+m_1}$ where $m_1=\frac{y^2/(xz)+xz/y^2}{x/z+z/x}=\frac{\lambda_1+\lambda_1^{-1}}{k_1+k_1^{-1}}$, direct rational expansion in $(x,y,z)$ (`exp102` Part (c)) confirms $m_1+m_2+m_3-m_1m_2m_3=2$, proving (5.9).

(3)–(4) Identities (5.10)–(5.11) are verified in `exp101` Part (c), and the fixed-$(x,z)$ multi-pair factorizations (5.12) follow from $(x^2+y_i^2)(x^2+y_j^2)z^2(1+a_1(y_i)a_1(y_j))=(x^2+y_i^2)(x^2+y_j^2)z^2+(y_i^2-z^2)(y_j^2-z^2)x^2=(x^2+z^2)(y_i^2y_j^2+x^2z^2)$, $(y_i^2+z^2)(y_j^2+z^2)x^2(1+b_1(y_i)b_1(y_j))=(x^2+z^2)(y_i^2y_j^2+x^2z^2)$, $(x^2+y_i^2)(y_j^2+z^2)(1+a_1(y_i)b_1(y_j))=(x^2+y_i^2)(y_j^2+z^2)+(y_i^2-z^2)(x^2-y_j^2)=(x^2+z^2)(y_i^2+y_j^2)$, and $(y_i^2+z^2)(x^2+y_j^2)(1+b_1(y_i)a_1(y_j))=(x^2+z^2)(y_i^2+y_j^2)$. Dividing these equations gives $\frac{1+b_1(y_i)b_1(y_j)}{1+a_1(y_i)a_1(y_j)}=\frac{z^2}{x^2}K$ and $\frac{1+b_1(y_i)a_1(y_j)}{1+a_1(y_i)b_1(y_j)}=L$, and substituting $(y_i^2+z^2)(y_j^2+z^2)=\frac{W_i^2 W_j^2}{4(x^2+z^2)^2(x^2+y_i^2)(x^2+y_j^2)}$ and $\frac{x^2+y_i^2}{x^2+y_j^2}=\frac{W_i^2(y_j^2+z^2)}{W_j^2(y_i^2+z^2)}$ yields the exact square multipliers (5.12a) (`exp102` Part (c)). Finally, substituting (5.13) into $h_j+g_j$ yields (5.15). $\square$

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## 6. General-$D_g$ Shift-Reflection Master Syzygies, the $D_g\le 0$ Sign-Chamber Inequality, and the Boundary Square-Class Calculation on $D_g=0$

We now turn to a general coset point $Q_0\in P_0+2E_Y(\mathbb{Q})$ with spoke root $R=R(Q_0)\in\mathbb{Q}^\times$ ($x_0:=x(Q_0)=\frac{R^2-1}{g}$) and its $1$-step neighbors $Q_0\pm S_Y$ under translation by the regular point $S_Y = \bigl(\frac{1}{abg}, \frac{r_a r_b r_{ab}}{abg}\bigr) \in E_Y(\mathbb{Q})$ (where $x(P_0\pm S_Y) = d_\mp$), with spoke roots $R_\mp := R(Q_0\pm S_Y)$ and $x_\mp := x(Q_0\pm S_Y)=\frac{R_\mp^2-1}{g}$. Recall from [15, Theorem 20(148)–(149)] that $R_-,-R_+$ are the two roots in $X$ of the symmetric Euler–Chasles biquadratic polynomial
$$\mathcal{B}(R, X) := g^2 U(R) X^2 + 2g^2 h R X - \bigl(v^2 R^2 - h^2(v^2 - g^2)\bigr) = 0, \tag{6.1}$$
where $g^2 U(R) := (v^2 - g^2)R^2 - v^2$, and the $4$-torsion discriminant of the base triple $Y=\{a,b,g\}$ is
$$D_g := (abg)^2 - (2ab + 1) = \frac{(v^2 - h^2)^2 - h^2}{g^2} = \frac{(v^2 - h^2 - h)(v^2 - h^2 + h)}{g^2}. \tag{6.2}$$
Direct chord-and-tangent doubling of $S_Y = \bigl(\frac{1}{abg}, \frac{r_a r_b r_{ab}}{abg}\bigr)$ on $E_Y$ gives the exact identity
$$x([2]S_Y) + \frac{1}{g} = \frac{\bigl((abg)^2 - (2ab + 1)\bigr)^2}{4abg(1 + g^2)(1 + ab)} = \frac{D_g^2}{4abg h^2 r_{ab}^2}, \tag{6.2a}$$
so that $[2]S_Y=T_g=\bigl(-\frac{1}{g},0\bigr) \iff D_g=0 \iff v^2=h^2\pm h$.

- **Arithmetic construction and global non-collapse proof on the $2$-dimensional base surface of tangent triples (`exp103` Part (b)):** A non-degenerate rational tangent triple $Y=\{a,b,g\}\subset\mathbb{Q}^\times$ ($b=\varphi_g(a)=\frac{g-a}{1+ag}$) requires three simultaneous rational square conditions: $1+g^2=h^2$, $1+ag=r_a^2$ (forcing $1+bg=\frac{1+g^2}{1+ag}=(h/r_a)^2=r_b^2$), and $1+ab=(v/g)^2$. Parameterizing $1+g^2=h^2$ by $g(s)=\frac{2s}{1-s^2}, h(s)=\frac{1+s^2}{1-s^2}$ ($s\in\mathbb{Q}\setminus\{0,\pm 1\}$) and choosing $r_a\in\mathbb{Q}^\times\setminus\{\pm 1,\pm h\}$, the condition $1+ab=(v/g)^2$ is the self-reflection annulus equation $(vr_a)^2=(gh)^2-(r_a^2-h^2)^2$ (2.2), so rational tangent triples correspond bijectively via (2.3)–(2.4) [15, §9(c.1)] to rational points on the $2$-dimensional surface $\mathcal{S}_{\mathrm{sym}}:W^2=2(x^2+y^2)(x^2+z^2)(y^2+z^2)$ (which fibers over $s=\frac{x-z}{x+z}$ as an elliptic surface). Furthermore, **every** rational Pythagorean mirror $g(s)=\frac{2s}{1-s^2}$ ($s\in\mathbb{Q}\setminus\{0,\pm 1\}$) admits an explicit rational non-collapse ($D_g\neq 0$) tangent triple on $\mathcal{S}_{\mathrm{sym}}$ without auxiliary square conditions via the rational section $[x(s):y(s):z(s):W(s)]=\bigl[(1+s)B(s):(1-s)A(s):(1-s)B(s):4(1-s)(1+s^2)(1-s^2+s^4)(1+3s^2+s^4)B(s)\bigr]$ where $A(s)=1+2s+s^2+2s^3+s^4=(1+s+s^2)^2-2s^2$ and $B(s)=A(-s)=1-2s+s^2-2s^3+s^4=(1-s+s^2)^2-2s^2$, yielding
  $$a(s) = \frac{4s(1-s^6)}{(1-s^2+s^4)^2},\qquad b(s) = \frac{2s(1+s^2-s^4)(1-s^2-s^4)}{(1-s^2)(1+3s^2+s^4)^2},\qquad r_a(s) = \frac{1+3s^2+s^4}{1-s^2+s^4},\qquad v(s) = \frac{2s A(s)B(s)}{(1-s^2)(1-s^2+s^4)(1+3s^2+s^4)}. \tag{6.2b}$$
  To see that $\{a(s),b(s),g(s)\}$ is a well-defined non-degenerate tangent triple of three distinct non-zero rationals with $v(s)\neq 0$ and $D_g(s)\neq 0$ for **every** $s\in\mathbb{Q}\setminus\{0,\pm 1\}$, observe that $1-s^2+s^4=(s^2-\frac{1}{2})^2+\frac{3}{4}>0$, $1+3s^2+s^4>0$, and $1\pm s+s^2>0$ have no real roots, while
  $$\begin{aligned}
  a(s)-g(s) &= \frac{2s(1+s^2-s^4)(1-s^2-s^4)}{(1-s^2)(1-s^2+s^4)^2},\qquad b(s)-g(s) = \frac{4s^3(3s^4+7s^2+3)}{(s^2-1)(1+3s^2+s^4)^2},\\
  a(s)-b(s) &= \frac{2s(s^{16}+12s^{14}+10s^{12}-16s^{10}-13s^8-16s^6+10s^4+12s^2+1)}{(1-s^2)(1-s^2+s^4)^2(1+3s^2+s^4)^2},\qquad D_g(s) = -\frac{P_{9,+}(s)P_{9,-}(s)\widetilde{P}_{9,+}(s)\widetilde{P}_{9,-}(s)}{(1-s^2)^2(1-s^2+s^4)^4(1+3s^2+s^4)^4},
  \end{aligned}$$
  where $3s^4+7s^2+3>0$, and the remaining numerator factors—namely $A(s)$, $B(s)=A(-s)$, $1\pm s^2-s^4$, the degree-$16$ factor of $a(s)-b(s)$, $P_{9,\pm}(s)=s^9\mp 3s^8-2s^7\mp 2s^6-5s^5\pm 3s^4-2s^3\pm 2s^2+s\pm 1$, and $\widetilde{P}_{9,\pm}(s)=s^9\mp s^8+2s^7\pm 2s^6+3s^5\pm 5s^4-2s^3\pm 2s^2-3s\mp 1$—are all **monic in $\mathbb{Z}[s]$ with constant term $\pm 1$**. By the Rational Root Theorem, their only candidate rational roots are $s=\pm 1$, where they evaluate to $\{7, -1, \pm 1, -15, -6, \pm 8, \pm 6\}\not\ni 0$ (`exp103` Part (b); e.g., $s=\frac{1}{2}$ gives $g=\frac{4}{3}, a=\frac{504}{169}, b=-\frac{836}{2523}, v=\frac{164}{1131}$ with $D_g\neq 0$). In contrast, imposing $D_g=0$ cuts out the $1$-dimensional $4$-torsion base curve (`w4sym.py`, (1.2)) of Paper 21, v4 [10, §5.2], on which the all-coset $[-2]$ relation and all-coset exclusion of
  $$\mathcal{S}_7^{(4\text{-tors})}(Q_0) := \bigl\{a,\, b,\, g,\, x_0,\, \varphi_g(x_0),\, x_-,\, x_+\bigr\}\qquad (x_+ = \varphi_g(x_-)\text{ on }D_g=0) \tag{6.3}$$
  were already established in [10, §5.2] (and via the $2$-adic unit base product in [14, Theorem 6]).

**Theorem 5 (General-$D_g$ Shift-Reflection Syzygies and $D_g=0$ Boundary Class; `exp103`).**
1. *(Bilinear $U$-Factorization of the Euler–Chasles Correspondence):* Multiplying $\mathcal{B}(R,X)$ in (6.1) by $v^2-g^2$ yields the symmetric bilinear factorization
   $$(v^2 - g^2)\mathcal{B}(R, X) \equiv \bigl(g^2 U(R)\bigr)\bigl(g^2 U(X)\bigr) + 2g^2 h(v^2 - g^2) R X + g^4 D_g. \tag{6.4}$$
2. *(Master Syzygy I — Universal $2$-Step / Self-Reflection Coupling):* Across the $2$-dimensional base surface of all tangent triples (for arbitrary $D_g$), at every coset center $Q_0\in P_0+2E_Y(\mathbb{Q})$ with $U(Q_0)\neq 0$, the $2$-step chord product $1+x_-x_+$ between $x_-=x(Q_0+S_Y)$ and $x_+=x(Q_0-S_Y)$ and the central self-reflection product $1+x_0\varphi_g(x_0)$ are linked by the exact linear syzygy
   $$U(Q_0)^2(1 + x_- x_+) + \frac{2v^2 R^2}{g^2}\bigl(1 + x_0\varphi_g(x_0)\bigr) = D_g(1 + x_0^2). \tag{6.5}$$
3. *(Master Syzygy II — Universal Opposite-Shift Cross-Reflection Dual Syzygies):* With $d_\mp=\frac{R_{1,\mp}^2-1}{g}$ ($R_{1,\mp}=v\mp h$), $N_g(x_-,x_+)=1-x_-x_++g(x_-+x_+)=(1+gx_+)(1+x_-\varphi_g(x_+))=(1+gx_-)(1+x_+\varphi_g(x_-))$, and $P_{\mathrm{diag}}=h^4-h^2+4h^2v^2-v^4$, the opposite-shift cross-reflection norm satisfies the dual decompositions
   $$\begin{aligned}
   U(Q_0)^2 N_g(x_-, x_+) &= h^2 U(Q_0)^2 - \frac{(R_{1,-}^2 R^2 - h^2)(R_{1,+}^2 R^2 - h^2)}{g^2} = h^2 U(Q_0)^2 - R^4\bigl(\varphi_g(x_0) - d_-\bigr)\bigl(\varphi_g(x_0) - d_+\bigr)\\
   &= \frac{P_{\mathrm{diag}}}{g^2} R^2\bigl(1 + x_0\varphi_g(x_0)\bigr) + \frac{4h^2 v^2}{g^2} x_0(x_0 - g).
   \end{aligned} \tag{6.6}$$
4. *(Master Syzygy III — Universal Shifted Self-Reflection Product Law):* With $P_{\mathrm{self},\mp}=h^4-h^2-4h^2v^2\pm 4hv^3-v^4$, the product of the shifted self-reflection gates at $Q_0\pm S_Y$ factors through $1+x_0\varphi_g(x_0)$ up to a single $4$-torsion correction proportional to $D_g$:
   $$U(Q_0)^4 R_-^2 R_+^2\bigl(1 + x_-\varphi_g(x_-)\bigr)\bigl(1 + x_+\varphi_g(x_+)\bigr) = R^2\bigl(1 + x_0\varphi_g(x_0)\bigr)\left[\frac{P_{\mathrm{self},-} P_{\mathrm{self},+}}{g^4} R^2\bigl(1 + x_0\varphi_g(x_0)\bigr) - \frac{8h^2 v^2}{g^2} D_g x_0(x_0 - g)\right]. \tag{6.7}$$
5. *(Sign-chamber inequality on $D_g\le 0$ and boundary square-class calculation on the $4$-torsion base curve $D_g=0$):*
   - **(i) Real sign obstruction in the $2$-dimensional chamber $D_g\le 0$:** Whenever $D_g\le 0$ (equivalently $h^2-|h|\le v^2\le h^2+|h|$ on the $2$-dimensional base surface of tangent triples), the right-hand side of (6.5) is $\le 0$, so $1+x_0\varphi_g(x_0)$ and $1+x_-x_+$ **cannot both be positive** at any rational coset point $Q_0\in P_0+2E_Y(\mathbb{Q})$.
   - **(ii) Boundary square-class calculation on $D_g=0$ and recovery of [10, §5.2]'s all-coset $[-2]$ relation:** On the $1$-dimensional $4$-torsion base curve $v^2=h(h+\mathrm{sgn})$ ($D_g=0$, $\mathrm{sgn}\in\{\pm 1\}$) with $g\neq 0$ (so $|h|=\sqrt{1+g^2}>1$), the two roots $x_0 = g + \varepsilon h$ ($\varepsilon\in\{\pm 1\}$) of $1+x_0\varphi_g(x_0)=0$ satisfy $(1+ax_0)(1+bx_0)=(1+ab)(1+gx_0)$ and therefore lift to rational points $Z_\varepsilon = \bigl(g+\varepsilon h,\, \pm\frac{vh(h+\varepsilon g)}{g}\bigr)\in E_Y(\mathbb{Q})$, **but their $g$-spoke factor belongs to the non-trivial rational square class $2(\mathbb{Q}^\times)^2$**:
     $$1 + g(g + \varepsilon h) = h(h + \varepsilon g) = 2\left(\frac{h(h + \mathrm{sgn} + \varepsilon g)}{2v}\right)^2 \in 2(\mathbb{Q}^\times)^2 \qquad (\varepsilon \in \{\pm 1\}). \tag{6.7a}$$
     Since $2\notin(\mathbb{Q}^\times)^2$, $Z_\varepsilon\notin P_0+2E_Y(\mathbb{Q})$ universally. Consequently, for every rational coset point $Q_0\in(P_0+2E_Y(\mathbb{Q}))\cap E_Y(\mathbb{Q})$, we have $g^2 U(Q_0)=\mathrm{sgn}\cdot(h+\mathrm{sgn})(R^2-\mathrm{sgn}\cdot h)\neq 0$ and $1+x_0\varphi_g(x_0)\neq 0$, with $R_- R_+=\mathrm{sgn}\cdot h$, $x_+=\varphi_g(x_-)$, $P_{\mathrm{self},-}P_{\mathrm{self},+}=4h^4(h+\mathrm{sgn})^2$, and Master Syzygy I (6.5) specializes at $D_g=0$ to
     $$1 + x_- x_+ = 1 + x_\mp\varphi_g(x_\mp) = -2\left(\frac{v R}{g U(Q_0)}\right)^2\bigl(1 + x_0\varphi_g(x_0)\bigr) \neq 0, \tag{6.8}$$
     recovering the all-coset $[-2]$ square-class relation and exclusion of Paper 21, v4 [10, §5.2] (and [14, Theorem 6]).

*Proof.*
(1) Expanding $\bigl((v^2-g^2)R^2-v^2\bigr)\bigl((v^2-g^2)X^2-v^2\bigr) = (v^2-g^2)g^2 U(R) X^2 - v^2(v^2-g^2)R^2 + v^4$ and noting that $-(v^2-g^2)\bigl(v^2R^2-h^2(v^2-g^2)\bigr) = -v^2(v^2-g^2)R^2 + h^2(v^2-g^2)^2$ while $v^4 - h^2(v^2-g^2)^2 = -g^2\bigl((v^2-h^2)^2-h^2\bigr) = -g^4 D_g$ (using $g^2=h^2-1$) yields (6.4).

(2) By Vieta's formulas for the quadratic $\mathcal{B}(R,X)=0$ in $X$ with roots $R_-, -R_+$ (since $R_--R_+=-2hR/U(Q_0)$ and $p:=R_-R_+=\frac{v^2R^2-h^2(v^2-g^2)}{g^2 U(Q_0)}$), we have
$$s_2 := R_-^2 + R_+^2 = (R_- - R_+)^2 + 2R_- R_+ = \frac{4h^2 R^2}{U(Q_0)^2} + 2p, \tag{6.9}$$
and $g^2(1+x_-x_+) = (R_-^2-1)(R_+^2-1)+g^2 = p^2 - s_2 + h^2$. Multiplying by $U(Q_0)^2$ gives
$$g^2 U(Q_0)^2(1 + x_- x_+) = (U(Q_0)p - U(Q_0))^2 + g^2 U(Q_0)^2 - 4h^2 R^2. \tag{6.10}$$
Now observe that
$$g^2\bigl(U(Q_0)p - U(Q_0)\bigr) = \bigl(v^2 R^2 - h^2(v^2-g^2)\bigr) - \bigl((v^2-g^2)R^2 - v^2\bigr) = g^2 R^2 + v^2(1-h^2) + g^2 h^2 = g^2(R^2 + h^2 - v^2),$$
so $U(Q_0)p - U(Q_0) = R^2 + h^2 - v^2$ (`exp103` Part (a)). Substituting $U(Q_0)p-U(Q_0)=R^2+h^2-v^2$ and $g^2 U(Q_0)=(v^2-g^2)R^2-v^2$ into (6.10), and adding $2v^2 R^2(1+x_0\varphi_g(x_0)) = \frac{2v^2}{g^2}\bigl(g^2 h^2 - (R^2-h^2)^2\bigr)$, we obtain
$$\begin{aligned}
g^2 U(Q_0)^2(1+x_-x_+) + \frac{2v^2}{g^2}\bigl(g^2 h^2 - (R^2-h^2)^2\bigr) &= \bigl(R^2 + h^2 - v^2\bigr)^2 + \frac{\bigl((v^2-g^2)R^2 - v^2\bigr)^2}{g^2} - 4h^2 R^2 + \frac{2v^2}{g^2}\bigl(g^2 h^2 - (R^2-h^2)^2\bigr)\\
&= \frac{\bigl((v^2-h^2)^2 - h^2\bigr)\bigl((R^2-1)^2 + g^2\bigr)}{g^2} = g^2 D_g (1 + x_0^2),
\end{aligned}$$
which divided by $g^2$ proves Master Syzygy I (6.5) (equivalently, in terms of $m:=abg=\frac{v^2-g^2}{g}$, $U(Q_0)=mx_0-1$, $1+m^2=D_g+2(1+ab)$, and $(1+x_0^2)-2(1+gx_0)=-R^2(1+x_0\varphi_g(x_0))$, one has $U(Q_0)^2(1+x_-x_+)=(1+m^2)(1+x_0^2)-4(1+ab)(1+gx_0)=D_g(1+x_0^2)-2(1+ab)R^2(1+x_0\varphi_g(x_0))$).

(3)–(4) In root coordinates, $g^2(1-x_-x_++g(x_-+x_+)) = h^2(s_2 - 1) - p^2$. Multiplying by $U(Q_0)^2$ and substituting (6.9) and $U(Q_0)p$ yields both expressions in (6.6). For (6.7), since $g^2 X^2(1+x\varphi_g(x))=-X^4+2h^2 X^2-h^2$, the left-hand side of (6.7) multiplied by $g^4$ is the symmetric polynomial $U(Q_0)^4\bigl(p^4 - 2h^2 p^2 s_2 + h^2(s_2^2-2p^2) + 4h^4 p^2 - 2h^4 s_2 + h^4\bigr)$, which expands in $\mathbb{Q}[h,v,R]$ to the exact right-hand side of (6.7) (`exp103` Part (a)).

(5) When $D_g\le 0$ ($h^2-|h|\le v^2\le h^2+|h|$), $D_g(1+x_0^2)\le 0$ in (6.5), while $U(Q_0)^2\ge 0$ and $\frac{2v^2 R^2}{g^2}>0$, forcing $\min(1+x_0\varphi_g(x_0),1+x_-x_+)\le 0$. Now suppose $D_g=0$, so $v^2=h(h+\mathrm{sgn})$ ($\mathrm{sgn}\in\{\pm 1\}$) with $g\neq 0$:
- First, $g^2 U(Q_0)=\mathrm{sgn}\cdot(h+\mathrm{sgn})(R^2-\mathrm{sgn}\cdot h)\neq 0$: since $h^2=1+g^2>1$, we have $h+\mathrm{sgn}\neq 0$; and if $R^2=\mathrm{sgn}\cdot h$, then $1+g^2=h^2=R^4$, which has no rational solution with $g\neq 0$ by Fermat's right-triangle theorem.
- Second, we prove that $1+x_0\varphi_g(x_0)=\frac{1+2gx_0-x_0^2}{1+gx_0}\neq 0$ for every rational coset point $Q_0\in(P_0+2E_Y(\mathbb{Q}))\cap E_Y(\mathbb{Q})$. On *any* tangent triple $b=\frac{g-a}{1+ag}$, we have $a+b=g(1-ab)$, so at the two roots $x_0 = g + \varepsilon h$ ($\varepsilon\in\{\pm 1\}$) of $x_0^2 - 2gx_0 - 1 = 0$ we have $x_0^2 = 2gx_0 + 1$, $1+gx_0 = 1+g^2+\varepsilon gh = h(h+\varepsilon g)$, and
  $$(1+ax_0)(1+bx_0) = 1 + g(1-ab)x_0 + ab(2gx_0 + 1) = (1+ab)(1+gx_0) = \frac{v^2}{g^2}(1+gx_0).$$
  Multiplying by $1+gx_0$ shows that $(1+ax_0)(1+bx_0)(1+gx_0) = \bigl(\frac{v(1+gx_0)}{g}\bigr)^2 = \bigl(\frac{vh(h+\varepsilon g)}{g}\bigr)^2 > 0$, so $Z_\varepsilon = \bigl(g+\varepsilon h, \pm\frac{vh(h+\varepsilon g)}{g}\bigr) \in E_Y(\mathbb{Q})$ is always a rational point on $E_Y(\mathbb{Q})$. However, for $Z_\varepsilon$ to lie in the $2$-descent coset $P_0+2E_Y(\mathbb{Q})$, its $g$-spoke factor $1+gx(Z_\varepsilon)=h(h+\varepsilon g)$ must be a rational square $R^2$. Using $g^2 = h^2 - 1$ and $\mathrm{sgn}^2 = 1$, we have the universal polynomial identity
  $$(h + \mathrm{sgn} + \varepsilon g)^2 = (h + \varepsilon g)^2 + 2\,\mathrm{sgn}(h + \varepsilon g) + 1 = h^2 + 2\varepsilon gh + (h^2 - 1) + 2\,\mathrm{sgn}(h + \varepsilon g) = 2(h + \mathrm{sgn})(h + \varepsilon g).$$
  Multiplying both sides by $h^2$ and substituting $h(h+\mathrm{sgn}) = v^2$ yields (6.7a):
  $$1 + g(g + \varepsilon h) = h(h + \varepsilon g) = \frac{h^2(h + \mathrm{sgn} + \varepsilon g)^2}{2h(h + \mathrm{sgn})} = 2\left(\frac{h(h + \mathrm{sgn} + \varepsilon g)}{2v}\right)^2 \in 2(\mathbb{Q}^\times)^2$$
  (for instance, on the $4$-torsion triad $Y^{(2)}=\{-64/255, 225/136, 119/120\}$ of $(12,5,3,1326)$ with $h=169/120, \mathrm{sgn}=-1, v=91/120$, we have $1+g(g+h)=169/50=2(13/10)^2$ and $1+g(g-h)=169/288=2(-13/24)^2$). Since $v\neq 0$, $h\neq 0$, and $h+\mathrm{sgn}+\varepsilon g\neq 0$ (as $(h+\mathrm{sgn}+\varepsilon g)(h+\mathrm{sgn}-\varepsilon g)=2(h+\mathrm{sgn})\neq 0$), the base $\frac{h(h+\mathrm{sgn}+\varepsilon g)}{2v}\in\mathbb{Q}^\times$ is non-zero, and because $2\notin(\mathbb{Q}^\times)^2$, $1+g(g+\varepsilon h)$ is never a square in $\mathbb{Q}$. Hence $Z_\varepsilon\notin P_0+2E_Y(\mathbb{Q})$, and $1+x_0\varphi_g(x_0)\neq 0$ at every rational coset point $Q_0\in P_0+2E_Y(\mathbb{Q})$.
- Finally, on $D_g=0$, $R_- R_+=\mathrm{sgn}\cdot h$ implies $x_+=\frac{h^2/R_-^2-1}{g}=\varphi_g(x_-)$, so $1+x_-x_+=1+x_\mp\varphi_g(x_\mp)$, and setting $D_g=0$ in (6.5) with $1+x_0\varphi_g(x_0)\neq 0$ gives (6.8) immediately (`exp103` Part (b)). $\square$

---

## 7. The Fixed-$\Omega$ Three-Quadric Sphere Syzygy and Genus-$5$ Faltings Finiteness on $\mathcal{S}_{\mathrm{sym}}$

Theorem 3 showed that fixing a single null-cone coordinate $u_-\in\mathbb{Q}\setminus\{0,\pm 1\}$ cuts out a smooth genus-$1$ curve $\mathcal{E}_{u_-}\subset\mathcal{S}_{\mathrm{sym}}$ whose Bäcklund translation $\mathcal{T}_{u_-}$ has infinite order on the generic fiber over $\mathbb{Q}(u_-)$ and at the certified specializations $u_-\in\{2,3,23/33,93/145\}$. A natural question is whether fixing the hyperbolic invariant $\Omega := u_- u_+$ (which fixes $\Sigma = \frac{2(\Omega-1)}{\Omega+1}$ and the diagonal cross-reflection ratio $\frac{P_{\mathrm{diag}}}{(gh)^2} = \frac{\Omega^2+14\Omega+1}{(\Omega+1)^2} = 5-\Sigma^2$) also produces an elliptic fiber on $\mathcal{S}_{\mathrm{sym}}$. Our sixth theorem proves that the answer is **no**: fixing $\Omega$ splits $\mathcal{S}_{\mathrm{sym}}$ into three simultaneous quadrics over $\mathbb{P}^1_\eta$ linked by a sphere syzygy, cutting out a smooth curve $\mathcal{F}_\Omega$ of **genus $5$**.

**Theorem 6 (Fixed-$\Omega$ Sphere Syzygy and Genus-$5$ Finiteness; `exp103`).**
1. *(Three-quadric decomposition and the Universal Sphere Syzygy):* In the normalized chart $y=1$ on $\mathcal{S}_{\mathrm{sym}}$ with $\eta=\frac{x^2-z^2}{x^2+z^2+2}=\frac{u_+-u_-}{2}$, $M=\frac{W}{x^2+z^2+2}=\frac{u_-+u_+}{2}$, and $\Omega=u_-u_+=M^2-\eta^2$, define the scaled coordinates
   $$X_1 := (1 - \eta)x = \frac{(\Omega + 1)x}{x^2 + 1},\qquad Z_1 := (1 + \eta)z = \frac{(\Omega + 1)z}{z^2 + 1}. \tag{7.1}$$
   Then along any fixed level set $u_- u_+ = \Omega$, the coordinates $(X_1,Z_1,M)$ over $\mathbb{P}^1_\eta$ satisfy the **three simultaneous quadric equations**
   $$X_1^2 = (1 - \eta)(\Omega + \eta),\qquad Z_1^2 = (1 + \eta)(\Omega - \eta),\qquad M^2 = \Omega + \eta^2, \tag{7.2}$$
   with $W = \frac{2M(\Omega+1)}{1-\eta^2}$, linked by the **Universal Fixed-$\Omega$ Sphere Syzygy** and difference/product identities
   $$X_1^2 + Z_1^2 + 2M^2 \equiv 4\Omega,\qquad X_1^2 - Z_1^2 = 2\eta(1 - \Omega),\qquad X_1^2 Z_1^2 = (1 - \eta^2)(\Omega^2 - \eta^2). \tag{7.3}$$
2. *(Genus-$5$ $(\mathbb{Z}/2\mathbb{Z})^3$-cover structure and Faltings finiteness of $\mathcal{F}_\Omega(\mathbb{Q})$):* For every fixed $\Omega\in\mathbb{Q}\setminus\{0,\pm 1\}$ (equivalently, every fixed $\Sigma\in\mathbb{Q}\setminus\{-2,0,2\}$, including every rational point on the promoted $18/21$ circle $\Sigma^2+\tau^2=5$), the three quadratic extensions $\mathbb{Q}(\eta)\bigl(\sqrt{(1-\eta)(\Omega+\eta)},\sqrt{(1+\eta)(\Omega-\eta)},\sqrt{\Omega+\eta^2}\bigr)/\mathbb{Q}(\eta)$ are unramified at $\eta=\infty$ and ramified at the $6$ pairwise distinct finite branch points
   $$\mathcal{B}_\Omega = \{1,\, -1,\, \Omega,\, -\Omega,\, i\sqrt{\Omega},\, -i\sqrt{\Omega}\} \subset \mathbb{P}^1_\eta(\overline{\mathbb{Q}}). \tag{7.4}$$
   Consequently, the smooth projective model $\mathcal{F}_\Omega$ of the fixed-$\Omega$ level set on $\mathcal{S}_{\mathrm{sym}}$ is a Galois $(\mathbb{Z}/2\mathbb{Z})^3$-cover of $\mathbb{P}^1_\eta$ of degree $8$ and **genus $g(\mathcal{F}_\Omega) = 5$** (lying above three genus-$1$ intermediate biquadratic quotients $\mathcal{C}_\Omega^{(xz)}:(2x^2-(\Omega-1))(2z^2-(\Omega-1))=(\Omega+1)^2$, $\mathcal{C}_\Omega^{(xM)}:((x^2+1)M)^2=(\Omega+1)(x^4+\Omega)$, and $\mathcal{C}_\Omega^{(zM)}:((z^2+1)M)^2=(\Omega+1)(z^4+\Omega)$). By Faltings' theorem [6], for every fixed $\Omega\in\mathbb{Q}\setminus\{0,\pm 1\}$, the set of rational points $\mathcal{F}_\Omega(\mathbb{Q})$ is **finite**.

*Proof.*
(1) Substituting $u_-u_+=\Omega$ and $u_+-u_-=2\eta$ into (4.3) gives $x^2=X(u_-,u_+)=\frac{\Omega+\eta}{1-\eta}$ and $z^2=Z(u_-,u_+)=\frac{\Omega-\eta}{1+\eta}$, while $\Omega=M^2-\eta^2\iff M^2=\Omega+\eta^2$. Multiplying $x^2$ by $(1-\eta)^2$ and $z^2$ by $(1+\eta)^2$ yields (7.2), and summing $(1-\eta)(\Omega+\eta)+(1+\eta)(\Omega-\eta)+2(\Omega+\eta^2) = 4\Omega$ proves the sphere syzygy (7.3).

(2) For $\Omega\notin\{0,\pm 1\}$, the six roots $\{1,-1,\Omega,-\Omega,i\sqrt{\Omega},-i\sqrt{\Omega}\}$ are pairwise distinct in $\overline{\mathbb{Q}}$, so the three quadratic factors $q_{X_1}(\eta)=(1-\eta)(\Omega+\eta)$, $q_{Z_1}(\eta)=(1+\eta)(\Omega-\eta)$, and $q_M(\eta)=\Omega+\eta^2$ have pairwise disjoint square-free support in $\overline{\mathbb{Q}}[\eta]$. Hence $\mathcal{F}_\Omega\to\mathbb{P}^1_\eta$ is a Galois $(\mathbb{Z}/2\mathbb{Z})^3$-cover of degree $d=8$. Since $\deg(q_{X_1})=\deg(q_{Z_1})=\deg(q_M)=2$ is even, $\eta=\infty$ is unramified in all three quadratic extensions. Over each of the $6$ finite branch points in $\mathcal{B}_\Omega$, exactly one of the three quadratic extensions ramifies (with $e=2$) while the other two are unramified, so the fiber over each branch point consists of $8/2=4$ points of ramification index $e=2$. By the Riemann–Hurwitz formula,
$$2g(\mathcal{F}_\Omega) - 2 = 8(2g(\mathbb{P}^1) - 2) + 6 \cdot 4 \cdot (2 - 1) = -16 + 24 = 8 \implies g(\mathcal{F}_\Omega) = 5.$$
Since $g(\mathcal{F}_\Omega)=5\ge 2$, $\mathcal{F}_\Omega(\mathbb{Q})$ is finite by Faltings' theorem [6] (`exp103` Part (c)). $\square$

---

## 8. Exact Rational Control Audits, Tables, and Verification Summary

All symbolic identities and numerical control audits are mechanically certified in `verify_paper.py` (`exp100`–`exp103`). Tables 1–4 summarize the principal exact rational control certificates, computed directly from the surface coordinates.

### Table 1: $\iota_{xz}$-Paired $18/21$ Promoted Cross-Reflected Ladders on $\mathcal{S}_{\mathrm{sym}}\subset\mathbb{P}(1,1,1,3)$ (`exp100`, Section 2.1, [15, §9(c.1)])

| Control Point $[x:y:z:W]$ | Circle $s$ | $v = \frac{y(x^4-z^4)}{xzW}$ | $\lambda = \frac{y^2}{xz}$ | $\Sigma(s)$ | $\tau(s)$ | $P_{\mathrm{diag}} = (gh\tau)^2$ | $\mathcal{S}_7^{\mathrm{twin}}$ | $\mathcal{S}_{7,\mp}^{\mathrm{cross}}$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $[11: 3: 2: 650]$ | $8$ | $135/44$ | $9/22$ | $31/65$ | $142/65$ | $(15975/968)^2$ | $17/21$ | **$18/21$** |
| $[33: 22: 6: 42900] = \iota_{xz}[11:3:2:650]$ | $-1/8$ | $135/44$ | $22/9$ | $-31/65$ | $-142/65$ | $(15975/968)^2$ | $17/21$ | **$18/21$** |

### Table 2: Mazur Non-Torsion Certification ($13>12$ Distinct Rational Orbit Points $\{\mathcal{T}_{u_-}^k(x_0,z_0)\}_{k=-6}^6$ on $\mathcal{E}_{u_-}(\mathbb{Q})$; `exp102`, Theorem 3(4))

| Fiber Seed Label | $u_-$ | Base Point $(x_0,z_0)\in\mathcal{E}_{u_-}(\mathbb{Q})$ | $\mathcal{I}_{u_-}^{(+)}(x_0,z_0)$ | Distinct Orbit Points ($k=-6\dots 6$) | $x$-Coordinate Bit-Lengths ($k=-6\dots 6$) |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $\mathcal{C}_{\mathrm{null}}(u_-=2)$ | $2$ | $(-14/47, 46/17)$ | $(-2, -2)$ | **$13$** | $[394, 286, 196, 122, 66, 27, 6, 2, 15, 44, 92, 156, 238]$ |
| $\mathcal{C}_{\mathrm{null}}(u_-=3)$ | $3$ | $(-51/47, 69/7)$ | $(-3, -3)$ | **$13$** | $[394, 286, 196, 123, 66, 28, 6, 2, 15, 45, 92, 157, 239]$ |
| $\mathcal{S}_{\mathrm{sym}}(11,3,2,650)$ | $23/33$ | $(11/3, 2/3)$ | $(-401/537, -121/174)$ | **$13$** | $[1433, 1003, 649, 372, 173, 50, 4, 35, 143, 327, 589, 928, 1344]$ |
| $\mathcal{S}_{\mathrm{sym}}(12,5,3,1326)$ | $93/145$ | $(12/5, 3/5)$ | $(-134331/113240, -41511/66985)$ | **$13$** | $[1990, 1383, 887, 500, 225, 59, 4, 59, 224, 500, 886, 1382, 1989]$ |

### Table 3: True $\mathcal{S}_{\mathrm{sym}}$ Triad Matrix $M_{\mathrm{triad}}(x:y:z)$ and Root-Free Projective Recovery (`exp101`–`exp102`, Theorem 4)

| $\mathcal{S}_{\mathrm{sym}}$ Point $(x,y,z,W)$ | $Y^{(1)}=\{a_1,b_1,g_1\}$ | $Y^{(2)}=\{a_2,b_2,g_2\}$ | $Y^{(3)}=\{a_3,b_3,g_3\}$ | Recovered $(x/z,\, x/y,\, y/z)$ from (5.15) | Special Cross-Triad Feature |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $(11, 3, 2, 650)$ | $\{11/52, 224/143, 117/44\}$ | $\{-11/75, 27/11, 56/33\}$ | $\{27/20, -224/375, 5/12\}$ | $(11/2,\, 11/3,\, 3/2)$ | $u_{1,-}=23/33, u_{1,+}=7/3, \Omega_1=161/99$ |
| $(33, 22, 6, 42900)$ | $\{224/143, 11/52, 117/44\}$ | $\{-224/375, 27/20, 5/12\}$ | $\{27/11, -11/75, 56/33\}$ | $(11/2,\, 3/2,\, 11/3)$ | $\iota_{xz}$-dual of $(11,3,2,650)$; swaps $a_1\leftrightarrow b_1$ and $(Y^{(2)},Y^{(3)})$ |
| $(12, 5, 3, 1326)$ | $\{64/169, 7/8, 15/8\}$ | $\{-64/255, 225/136, 119/120\}$ | $\{225/169, -7/15, 8/15\}$ | $(4,\, 12/5,\, 5/3)$ | $g_1g_3=1 \implies v_2^2=h_2^2-h_2$, $1-a_1a_3=(119/169)^2$ |
| $(3588, 595, 1820, 39623369240)$ | $\{-5832225/13227769, 4183/2415, 1768/2415\}$ | $\{5832225/5291624, 3125824/7228095, 12519719/4269720\}$ | $\{3125824/13227769, -4183/1768, -2415/1768\}$ | $(69/35,\, 3588/595,\, 17/52)$ | $g_1g_3=-1 \implies v_2^2=h_2^2+h_2$, $1+a_1a_3=(12519719/13227769)^2$ |

### Table 4: General-$D_g$ Shift-Reflection Syzygies, Chord-and-Tangent Translation by $S_Y$, Boundary Square-Class Checks on $D_g=0$, and Fixed-$\Omega$ Sphere Syzygy (`exp103`, Theorems 5–6)

| Audit Family / Point | Parameter / Regular Point $S_Y \in E_Y(\mathbb{Q})$ | Verified Property Across All $7$ Coset Centers $Q_0\in\{P_0\mp S_Y, P_0\mp[2]S_Y, [3]P_0, [3]P_0\mp S_Y\}$ |
| :--- | :--- | :--- |
| $\mathcal{S}_{\mathrm{sym}}(11,3,2,650)$, $Y^{(1)}$ | $D_g = -18125/20449 < 0$, $S_Y = \bigl(\frac{143}{126},\frac{625}{168}\bigr)$ | Chord-and-tangent $Q_0\pm S_Y$, Master Syzygies I–III, and $\min(1+x_0\varphi_g(x_0),1+x_-x_+)\le 0$ at all $7$ centers |
| $\mathcal{S}_{\mathrm{sym}}(12,5,3,1326)$, $Y^{(2)}$ | $D_g = 0$ ($v_2^2 = h_2^2 - h_2$), $S_Y = \bigl(-\frac{17}{7},-\frac{2197}{840}\bigr)$, $[2]S_Y = T_g$ | $(1+x_0\varphi_g(x_0))(1+x_-x_+) < 0$, $[-2]$ square-class twist at all $7$ centers, and $1+g(g\pm h) \in \{2(13/10)^2, 2(-13/24)^2\}$ |
| $\mathcal{S}_{\mathrm{sym}}(3588,595,1820)$, $Y^{(2)}$ | $D_g = 0$ ($v_2^2 = h_2^2 + h_2$), $S_Y = \bigl(\frac{2993}{4183},\frac{48109395853}{17860238760}\bigr)$, $[2]S_Y = T_g$ | $(1+x_0\varphi_g(x_0))(1+x_-x_+) < 0$, $[-2]$ square-class twist at all $7$ centers, and $1+g(g\pm h) \in 2(\mathbb{Q}^\times)^2$ |
| Fixed-$\Omega$ Slice at $(11,3,2,650)$ | $\Omega = 161/99, \eta = 9/11$ | $X_1=2/3, Z_1=40/33, M=50/33 \implies X_1^2+Z_1^2+2M^2 = 644/99 = 4\Omega$ ($g(\mathcal{F}_\Omega)=5$) |
| Fixed-$\Omega$ Slice at $(33,22,6,42900)$ | $\Omega = 99/161, \eta = 81/161$ | $X_1=120/161, Z_1=66/161, M=150/161 \implies X_1^2+Z_1^2+2M^2 = 396/161 = 4\Omega$ ($g(\mathcal{F}_\Omega)=5$) |
| Fixed-$\Omega$ Slice at $(12,5,3,1326)$ | $\Omega = 6417/5075, \eta = 135/203$ | $X_1=816/1015, Z_1=1014/1015, M=1326/1015 \implies X_1^2+Z_1^2+2M^2 = 25668/5075 = 4\Omega$ ($g(\mathcal{F}_\Omega)=5$) |

---

## 9. Limitations, Scope, and Reproducibility

- **No rational Diophantine septuple ($21/21$) is claimed.** The promoted cross-reflected ladders $\mathcal{S}_{7,\mp}^{\mathrm{cross}}$ achieve $18/21$ rational squares on the genus-$0$ circle $\Sigma^2+\tau^2=5$ [15, §9(c.1)]. Across the $2$-dimensional open chamber $D_g\le 0$ ($h^2-|h|\le v^2\le h^2+|h|$), Master Syzygy I (6.5) forces $\min(1+x_0\varphi_g(x_0),1+x_-x_+)\le 0$, whereas the exterior chamber $D_g>0$ ($v^2<h^2-|h|$ or $v^2>h^2+|h|$) escapes this real sign obstruction. On the $1$-dimensional $4$-torsion base curve $D_g=0$ (`w4sym.py`, where the all-coset $[-2]$ relation and exclusion were established in [10, §5.2] and [14, Theorem 6]), Theorem 5(5)(ii) supplies the explicit boundary square-class calculation $1+g(g\pm h)\in 2(\mathbb{Q}^\times)^2$ excluding the zeros $x_0=g\pm h$ of $1+x_0\varphi_g(x_0)=0$ from $P_0+2E_Y(\mathbb{Q})$.
- **Distinction between symbolic certifications, finite rational audits, and theoretical deductions:**
  1. *Exact symbolic certifications (`sympy` over $\mathbb{Q}(x,y,z)$, $\mathbb{Q}(u_-,u_+)$, and $\mathbb{Q}[h,v,R]/(g^2-h^2+1)$):* The secant reduction of $1-\rho^2=\square$ along $m=-\frac{2t}{t\pm 1}$ to $\mathcal{E}_{\mathrm{self}}:w^2=t^4-14t^2+1$ (including all intermediate polynomial numerators/denominators), the $[3]P_0$ quartic/octic gates and $N_3|_{D_3=0}=4(1-h^2)=-4g^2$, null-cone fiber equations, Bäcklund involution identities, numerator degrees $(1,5,13,25)$ of $P^{(0)},\dots,P^{(3)}$ over $\mathbb{Q}(u_-)$, $S_3$-triad Markov/Joukowsky identities, the fixed-$(x,z)$ multi-pair factorizations (5.12) and $\mathcal{S}_{\mathrm{sym}}$ rational-square multipliers (5.12a), Master Syzygies I–III, the doubling formula $x([2]S_Y)+\frac{1}{g}=\frac{D_g^2}{4abgh^2r_{ab}^2}$, the universal non-collapse section (6.2b) on $\mathcal{S}_{\mathrm{sym}}$ (including global `ground_roots()` verification that $D_g(s),a(s),b(s),g(s)$ and their pairwise differences have no rational zeros or poles outside $s\in\{0,\pm 1\}$), the equivalence of $D_g=0$ with `w4sym.py`, the boundary square-class-$2$ identity $1+g(g+\varepsilon h)=2\bigl(\frac{h(h+\mathrm{sgn}+\varepsilon g)}{2v}\bigr)^2$ on $D_g=0$, and the three-quadric sphere syzygy $X_1^2+Z_1^2+2M^2=4\Omega$ are proved and symbolically certified in `exp100`–`exp103`.
  2. *Finite rational and modular certificates (`fractions.Fraction`):* $\mathbb{F}_5/\mathbb{F}_7$ point counts on $E_{48}$, the height-$600$ box census ($157$ ordered primitive triples on $\mathcal{S}_{\mathrm{sym}}$), the $13$ pairwise-distinct rational iterates $\{\mathcal{T}_{u_-}^k(x_0,z_0)\}_{k=-6}^6$ at $u_-\in\{2,3,23/33,93/145\}$ (heights up to $1,990$ bits), the multi-pair square multipliers on $(11,3,2,650)$ and $(11,22/3,2,14300/9)$, and the chord-and-tangent elliptic curve addition checks at $7$ coset centers across the control triples in Table 4 are certified in exact rational arithmetic.
  3. *Theoretical deductions invoking established theorems:* Fermat's right-triangle theorem ($R^4-g^2=1$ has no rational solution with $g\neq 0$), Mazur's Torsion Theorem [16] ($13>12$ distinct rational orbit points $\implies \mathcal{T}_{u_-}$ has infinite order at $u_-\in\{2,3,23/33,93/145\}$ and hence on the generic fiber over $\mathbb{Q}(u_-)$, without claiming non-torsion at every rational specialization $u_-\notin\{0,\pm 1\}$), and the Riemann–Hurwitz formula + Faltings' theorem [6] ($\Omega\notin\{0,\pm 1\}\implies g(\mathcal{F}_\Omega)=5\implies \#\mathcal{F}_\Omega(\mathbb{Q})<\infty$ for each fixed $\Omega$) combine the symbolic/rational certificates with standard arithmetic geometry.
- **Reproducibility:** Running `python3 verify_paper.py` inside `supplement.zip` executes `exp100`–`exp103` end-to-end in **~25 seconds** on a single CPU core using only the Python 3 standard library (`fractions`, `math`, `json`) and `sympy`.

---

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