1. Problem and contribution

A rational Diophantine $m$-tuple is a set of $m$ distinct nonzero rationals $\{a_1, \dots, a_m\} \subset \mathbb{Q}^\times$ such that $a_i a_j + 1 = \square$ in $\mathbb{Q}$ for all $1 \leq i < j \leq m$. No rational Diophantine septuple ($m = 7$, $21$ squares) is known [1, Problem 3.2]. An almost septuple is a set of $7$ distinct nonzero rationals satisfying $20/21$ square conditions; Dujella's Problem 3.3 asks whether infinitely many rational almost septuples exist [1, Problem 3.3].

In rStation Paper 6 [2] and Paper 10 [3], the $25$ known rational almost septuples ($22$ in [2], $A_1, A_2, A_3$ in [3], and $A_4$ in [4]) were classified by their relation type $(\mathcal{R}_4, \mathcal{R}_5)$, the sets of $4$- and $5$-element subsets on which $R_4$ and $R_5$ vanish:

$$ R_4(x, y, z, t) = (x + y - z - t)^2 - 4(x y + 1)(z t + 1), \qquad R_5(x, y, z, t, u) = (x y z t u + 2 x y z + x + y + z - t - u)^2 - 4(x y + 1)(x z + 1)(y z + 1)(t u + 1). $$

Over half ($14/25$) contain a two-sided regular quintuple $\{a, b, c, d_-, d_+\}$, where $d_\mp = a + b + c + 2 a b c \mp 2 r s t$ ($r^2 = a b + 1$, $s^2 = a c + 1$, $t^2 = b c + 1$) are the regular fourths of $\{a, b, c\}$ [2, 3, 5], though fixing a generic fibre of the quintuple threefold forces the sixth-element condition into genus $g \geq 9$ [3, Theorem 1].

Here we invert the order of elimination by expressing $c^*_\mp = e_\mp(a, b, d_-, d_+)$ and their compatibility conditions in terms of $r$, the conic parameter $\lambda$ uniformizing $d_- d_+ + 1 = w_1^2$, and $S_{ab} := a + b$, obtaining seven main results:

1. Universal $S_{ab}$-square completions and two elliptic sextuple surfaces (Theorem 1, Section 3). With $C_0 := c - a - b = \frac{\lambda^2(2 r - 1) + (2 r + 1)}{2 \lambda}$, $c^*_\mp$ are linear in $S_{ab}$. For $\mathcal{S}_7^{(c^*)} = \{a, b, c, d_-, d_+, c^*_-, c^*_+\}$, $18/21$ conditions hold identically, and the $S_{ab}$-discriminants of $c c^*_- + 1, c c^*_+ + 1, c^*_- c^*_+ + 1 = \square$ cut out Branch 1 ($\mathcal{E}_{\text{Br1}}$, where $R_4(c, d_-, d_+, c^*_-) = 0, c c^*_- + 1 = \square$) and the Twin-Fifth Surfaces $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$ ($R_4(a, b, c^*_-, c^*_+) = R_4(d_-, d_+, c^*_-, c^*_+) = 0, c^*_- c^*_+ + 1 = \square$).
2. Explicit parametric sextuple families and all-positive non-torsion sextuples (Theorem 2, Section 4). On $\mathcal{E}_{(r, \lambda)}$, the $8$ wall sections and $P_{c, 2}, P_{c, 3}, P_{c, 4}$ generate $17$ explicit $1$-parameter families of $15/15$ sextuples and $19/21$ septuples (2,304 unique $15/15$ sextuples on $\mathcal{E}_{\text{Br1}}$, 408 all-positive, and 222 on $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$).
3. A new rational almost septuple of relation type $(5, 2)$, the 2D surface $\mathcal{S}_{4, 2}$, the genus-three curve $\mathcal{H}$, and unramified descent (Theorem 3, Section 5). We factor $R_4(p, q, e_-, e_+) = \frac{F_{\text{sym}} G}{(P_4 - 1)^4}$. On $\mathcal{S}_{4, 2} : W_\tau^2 = (P_4 - 1)(P_4 - 9)$, $a_\pm c_\pm + 1 = \square$ holds identically (396 $15/15$ sextuples, 7,524 $19/21$ septuples), and restricting to Branch 1 gives $\mathcal{H} : W^2 = (5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^4 - 22 \lambda^2 + 33)$ of type $(5, 2)$ ($\text{dim} = 1, \text{gens} = 2$), proving finiteness by Faltings' theorem [6] and an unramified $d = 1$ descent via $\operatorname{rank} E_{\text{min}}^{(5)}(\mathbb{Q}) = 0$. At $(\lambda, W) = (-2, 55)$, we obtain a new rational almost Diophantine septuple:$$ \mathcal{S}_7^{(20/21)} = \left\{-\frac{19961343}{78775960}, \; -\frac{411840}{1969399}, \; \frac{23715}{29896}, \; \frac{71003}{63240}, \; \frac{36890}{11211}, \; \frac{104321672}{29540985}, \; \frac{29896}{7905}\right\}. $$
4. Global $(r, \mu, z)$ trace decomposition, Quadratic-Twist Collapse on $\mathcal{S}_{4, 2}$, and Curves I–III (Theorem 4, Section 6). In $\mu = \frac{\lambda^2 - 1}{\text{den}_-}, w = \frac{2\lambda}{\text{den}_-}, z = w c$, the trace quadratic collapses to $E_z(r, \mu, z) = 0$, while $\Sigma_\mu^+ : W_+^2 = -K L$ has $2$-torsion involution $\tau_{T_0}(\mu) = -\frac{\mu + 4 r}{K}$, yielding Curve I over $\mathbb{Q}(m)$, Curve II over $\mathbb{Q}(k)$ (all-positive for $k \in (1, k_*)$), and Curve III ($\mathcal{E}_{\text{III}, \pm} : Y^2 = (r \mp 3)(4 r^3 - 3 r \pm 3)$).
5. Factor cancellation on $w = 2 u^2 y - L$ and a moving-$u$ elliptic family of quintuples (Theorem 5, Section 7). On $w = 2 u^2 y - L$ of Paper 65 [16], $E = -4 u^3 y (u y - u - y)(u y - u + y)^2$ collapses $(b, c, d)$ to degree $4$, and $\mathcal{B}_+$ collapses $\alpha^2, \beta^2 = \square$ to $\mathcal{E}_B : Y_B^2 = (k - 2)(5 k^3 - 8 k^2 + 13 k - 2)$, giving a closed-form quintuple in $\mathbb{Q}(k)$ and an infinite moving-$u$ family starting at $k = 4$ with $\left\{20, \frac{12}{49}, 6, \frac{40}{289}, 48\right\}$.
6. Exact factorization $R_5 = \frac{L_1 L_2 Q_4}{H K^4}$, Universal Triple-Square Section $L_1 = 0$, $Q_4$ conic splitting, Gauge Involution $\iota_K$, Rank-$2$ Lattice $\Lambda(r)$, and $v = -1/5$ families (Theorem 6, Section 8). Before $E_z = 0$, $\iota_K$ fixes $(a,b,c,d_-,d_+)$, swaps $c^*_- \leftrightarrow c^*_+$ and $L_1 \leftrightarrow L_2$, and restricts on $Q_{4, +}(R, v, u) = 0$ to $\tau_{T_0}: u \mapsto \frac{u-2}{2Ru-1}$ with quotient $\mathcal{E}'_{Q_4}(R)$. Over $\mathbb{Q}(r)$, $W_1(r), P_{\text{odd}}(r)$ generate $\Lambda(r) \cong \mathbb{Z}^2 \oplus \mathbb{Z}/2\mathbb{Z} \subset E(r^2)(\mathbb{Q}(r))$ ($Z_-(r) = W_1 + P_{\text{odd}}$, $P_{\text{odd}}(3/4)$ giving the $r = 3/4$ quintuple, $\operatorname{rank} E(r^2)(\mathbb{Q}) \ge 3$ at $r = 7/4, 19/10$, and $2$-isogeny exclusions on $E_{48}^-, E_{80}$), alongside the $15 + 6$ Square-Class Collapse and $v = -1/5$ families ($\operatorname{rank} E_{16}(\mathbb{Q}) = 0, E_{16}(\mathbb{Q})_{\text{tors}} \cong \mathbb{Z}/8\mathbb{Z}$).
7. Universal $(V, K, d)$ $S$-chain coordinates, the $3$-Torsion Inflection Surface $\mathcal{S}_{\text{3-tors}}$, the Universal Orbit Cubic, Sturm-certified all-positive sextuples, the Eisenstein Triad Curve $\mathcal{C}_{\text{triad}}$, and Cross-Coset Bridge Laws (Theorem 7, Section 9). In $U = \Pi x - 1$, $3 S_T = \mathcal{O} \iff \pm \frac{1}{r} \pm \frac{1}{s} \pm \frac{1}{t} = 0$, giving Tate normal form $Y^2 = U^3 + (\lambda_0 U - \Omega)^2$ with $\Pi^2 + 1 = (\Omega + \lambda_0)^2$ and $E_T(\mathbb{Q})_{\text{tors}} \supseteq \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Every coset $Q + \langle S_T \rangle \subset P_0 + 2 E_T(\mathbb{Q})$ satisfies $U'^3 - \Sigma_U U'^2 - 4\Omega\lambda_0 U' + 4\Omega^2 = 0$, forming a $15/15$ sextuple (all-positive for $a, b, c > 0, \Omega(\Omega+\lambda_0) > 0, \Omega\lambda_0 < 0 \iff -1 < U_4 < 0$) and $15$ type-$(0, 4)$ $19/21$ septuples ($9$ from $2+2$ and $6$ from $1+3$ sub-quadruples). On $C_r$, $2 P_{\infty}$ is Sturm-certified all-positive for $r > r_{**} \approx 2.010315$, while $(N, \beta, \gamma)$ yields the quadratic all-positive Family IV ($N = 3\beta + \gamma$) and the positive-rank Eisenstein Triad Curve $\mathcal{C}_{\text{triad}}: W^2 = t^4 - 20t^3 + 34t^2 + 12t + 9$, generating infinitely many all-positive cross-coset type-$(2, 3)$ $19/21$ septuples (verify_paper.py, exp27–exp32).

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2. Related work and distinction

Dujella, Kazalicki, Mikić and Szikszai [7] constructed infinite families of rational Diophantine sextuples via $3$-torsion chains on $E_T$, and Dujella, Kazalicki and Petričević [5] studied two-sided quintuples $\{a, b, c, d_-, d_+\}$ on $S_{\text{DKP}}$. On rStation, Paper 6 [2] classified $22$ almost septuples into $18$ relation types; Paper 10 [3] proved that generic fibres of the quintuple threefold have T2D2 covers of genus $\geq 9$ and found $A_1, A_2, A_3$; Papers 17 [4], 18 [8], 34 [9] studied chain closures, build orders ($A_4$), and one-step $R_{33}$ origins; Paper 21 [10] and Paper 40v2 [17] proved uniqueness of the shift $\pm S$, identified regular fifths with $x(Q \pm S)$ via Vieta sum/product, and analyzed translation paths and cycles; and Papers 62 [11], 64 [12], 65 [16], 70 [13], 71 [14] studied Gibbs-regular sextuples and four-window charts.

What is new here and how it differs: Theorems 1–2 write $c^*_\mp$ and the three residual square conditions in $(r, \lambda, S_{ab})$, showing their $S_{ab}$-discriminants cut out $\mathcal{E}_{\text{Br1}}$ and $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$ with $17$ explicit families. In Theorem 3, $\mathcal{S}_7^{(20/21)}$ is absent from [2, 3, 4, 9, 15], with new $1$-dimensional relation type $(5, 2)$ collapsing on $\mathcal{S}_{4, 2}$ to the genus-$3$ curve $\mathcal{H}$ with unramified $d = 1$ descent. Theorems 4–6 establish $\Sigma_\mu^+ : W_+^2 = -K L$ and Curves I–III, the factor cancellation on $w = 2 u^2 y - L$ and $\mathcal{E}_B$, and $R_5 = \frac{L_1 L_2 Q_4}{H K^4}$, $\iota_K$, $\Lambda(r)$ ($Z_-(r) = W_1 + P_{\text{odd}}$), the $15+6$ Square-Class Collapse, and $v = -1/5$ families. Finally, Theorem 7 establishes the Universal Orbit Cubic, the tuple-coordinate Sub-Quadruple Square-Root Identity, Sturm-certified all-positive $3$-torsion sextuples, the Universal Vieta Gate Identity, $\mathcal{C}_{\text{triad}}$, and Cross-Coset Bridge Laws on $3 S_T = \mathcal{O}$ (building on [7, 10, 17]).

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3. Universal $S_{ab}$-square completions and the dual sextuple surfaces

Let $\{a, b, c\}$ be a rational Diophantine triple with $a b + 1 = r^2$, $a c + 1 = s^2$, $b c + 1 = t^2$, and regular fourths $d_\mp = a + b + c + 2 a b c \mp 2 r s t$. By Vieta, $d_- + d_+ = 2(a + b + c + 2 a b c)$ and $d_- d_+ + 1 = (c - a - b)^2 - (4 r^2 - 1)$. Write $S_{ab} := a + b$ and $C_0 := c - a - b = c - S_{ab}$. The conic $d_- d_+ + 1 = w_1^2 \iff C_0^2 - w_1^2 = 4 r^2 - 1$ is parameterized by $\lambda \in \mathbb{Q}^\times$ via $C_0 - w_1 = \lambda(2 r - 1)$ and $C_0 + w_1 = \frac{2 r + 1}{\lambda}$:

$$ C_0(r, \lambda) = \frac{\lambda^2(2 r - 1) + (2 r + 1)}{2 \lambda}, \qquad w_1(r, \lambda) = \frac{\lambda^2(2 r - 1) - (2 r + 1)}{2 \lambda}. $$

For $Q^* = \{a, b, d_-, d_+\}$ with $P_4 := a b d_- d_+ = (r^2 - 1)(w_1^2 - 1)$, $a d_\mp + 1 = (a c \mp r t)^2$, $b d_\mp + 1 = (b c \mp r s)^2$, and $\prod_{\ell \in \{a, b\}, \mp} \sqrt{\ell d_\mp + 1} = (r^2 - 1) C_0 + r^2 S_{ab}$. Substituting these into $R_5(a, b, d_-, d_+, x) = 0$ yields Theorem 1.

Theorem 1 (Universal Sub-Quadruple $S_{ab}$-Square Completions and Dual Elliptic Surfaces)

Let $\{a, b, c, d_-, d_+\}$ be any two-sided rational Diophantine quintuple parameterized by $(r, \lambda)$, with $\text{den}_- := \lambda^2(r - 1) - (r + 1)$ and $\text{den}_+ := \lambda^2(2 r - 1)^2(r + 1) - (2 r + 1)^2(r - 1)$.

1. Linearity of $c^*_\mp$ in $S_{ab}$: The regular fifths $c^*_\mp = e_\mp(a, b, d_-, d_+)$ depend on $(a, b, c, s, t)$ solely through $S_{ab} = a + b$:

   $$ c^*_- = \frac{4 \lambda \Big(\lambda S_{ab} + \lambda^2(r - 1) + (r + 1)\Big)}{\text{den}_-^2}, \qquad c^*_+ = \frac{4 \lambda (4 r^2 - 1) \Big(\lambda(4 r^2 - 1) S_{ab} + \lambda^2(2 r - 1)^2(r + 1) + (2 r + 1)^2(r - 1)\Big)}{\text{den}_+^2}. $$

2. Universal $S_{ab}$-square completions: For $\mathcal{S}_7^{(c^*)} = \{a, b, c, d_-, d_+, c^*_-, c^*_+\}$, $18/21$ conditions hold identically, and the remaining three ($c c^*_- + 1, c c^*_+ + 1, c^*_- c^*_+ + 1 = \square$) satisfy:

   $$ \begin{aligned} 4 \, \text{den}_-^2 (c \, c^*_- + 1) &= \Big(4 \lambda S_{ab} + \lambda^2(4 r - 3) + 4 r + 3\Big)^2 + \big(\lambda^2(2 r - 3) - (2 r + 1)\big)\big(\lambda^2(2 r - 1) - (2 r + 3)\big), \\ 4 \, \text{den}_+^2 (c \, c^*_+ + 1) &= \Big(4 \lambda (4 r^2 - 1) S_{ab} + \lambda^2(16 r^3 - 4 r^2 - 8 r + 3) + 16 r^3 + 4 r^2 - 8 r - 3\Big)^2 \\ &\quad + (4 r^2 - 1)\big(\lambda^2(2 r - 1)^2 - (2 r - 3)(2 r + 1)\big)\big(\lambda^2(2 r - 1)(2 r + 3) - (2 r + 1)^2\big), \\ \text{den}_-^2 \, \text{den}_+^2 (c^*_- c^*_+ + 1) &= 16 \lambda^2 \Big(\lambda(4 r^2 - 1) S_{ab} + (2 r^2 - 1)\big(\lambda^2(2 r - 1) + (2 r + 1)\big)\Big)^2 + F_1 F_2 F_3 F_4, \end{aligned} $$

   where $F_{1, 2}(r, \lambda) = 2 \lambda^2 r^2 - 3 \lambda^2 r + \lambda^2 \mp 2 \lambda r - 2 r^2 + r + 1$ and $F_{3, 4}(r, \lambda) = 2 \lambda^2 r^2 + \lambda^2 r - \lambda^2 \mp 2 \lambda r - 2 r^2 - 3 r - 1$.

3. Surface I (Branch 1 Sextuple Surface $\mathcal{E}_{\text{Br1}}$): The $S_{ab}$-discriminant of $c c^*_- + 1$ vanishes identically on Branch 1:

   $$ \lambda^2(2 r - 3) - (2 r + 1) = 0 \iff r(\lambda) = \frac{3 \lambda^2 + 1}{2(\lambda^2 - 1)}, \qquad C_0(\lambda) = \frac{\lambda(\lambda^2 + 3)}{\lambda^2 - 1}, \qquad w_1(\lambda) = \frac{\lambda^4 + 6 \lambda^2 - 3}{2 \lambda(\lambda^2 - 1)} $$

   (with $\lambda^2(2 r - 1) - (2 r + 3) = 0$ isomorphic under $\lambda \mapsto 1/\lambda$). On Branch 1, $R_4(c, d_-, d_+, c^*_-) = 0$ and $c \, c^*_- + 1 = \left(\frac{8 \lambda(\lambda^2 - 1) S_{ab} + 3 \lambda^4 + 14 \lambda^2 - 1}{(5 \lambda^2 - 1)(\lambda^2 + 3)}\right)^2$ hold identically for all $(\lambda, S_{ab})$, while Conditions 20 and 21 specialize to:

   $$ \begin{aligned} (\lambda^2 - 1)^2 (5 \lambda^4 + 10 \lambda^2 + 1)^2 \big(c \, c^*_+ + 1\big) &= \Big(8 \lambda(\lambda^4 - 1) S_{ab} + 9 \lambda^6 + 29 \lambda^4 + 27 \lambda^2 - 1\Big)^2 + \big(2 \lambda(\lambda^2 - 1)\big)^2 \cdot 2(\lambda^2 + 1)(\lambda^2 + 3)(3 \lambda^2 + 1), \\ (\lambda^2 - 1)^4 (5 \lambda^4 + 10 \lambda^2 + 1)^2 \big(c^*_- \, c^*_+ + 1\big) &= \Big(4 \lambda\big(8 \lambda(\lambda^4 - 1) S_{ab} + 7 \lambda^6 + 31 \lambda^4 + 29 \lambda^2 - 3\big)\Big)^2 + \big((\lambda^2 - 1)^3\big)^2 \cdot (25 \lambda^4 + 6 \lambda^2 + 1). \end{aligned} $$

4. Surface II (Twin-Fifth Sextuple Surfaces $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$): The discriminant $F_1 F_2 F_3 F_4$ of $c^*_- c^*_+ + 1$ vanishes identically on $F_1(r, \lambda) = 0$ and $F_3(r, \lambda) = 0$, birational to the rank-$1$ elliptic curve

   $$ \mathcal{E}_{F_1} : \quad Y_\lambda^2 = (\lambda + 1)(\lambda^3 + 11 \lambda^2 - 13 \lambda + 9), \qquad r = \frac{3 \lambda^2 + 2 \lambda - 1 \pm Y_\lambda}{4(\lambda^2 - 1)}. $$

   On $\mathcal{E}_{F_1}$ and $\mathcal{E}_{F_3}$, $R_4(a, b, c^*_-, c^*_+) = R_4(d_-, d_+, c^*_-, c^*_+) = 0$ and $c^*_- c^*_+ + 1 = \square$ hold identically for all $S_{ab}$, making $\{a, b, d_-, d_+, c^*_-, c^*_+\}$ a $15/15$ sextuple and $\mathcal{S}_7^{(c^*)}$ a $19/21$ septuple at every non-degenerate rational point.

Proof. Substituting $a b = r^2 - 1$, $d_- d_+ = w_1^2 - 1$, $d_- + d_+ = 2(2 r^2 S_{ab} + (2 r^2 - 1) C_0)$, and $\prod \sqrt{\ell d_\mp + 1} = (r^2 - 1) C_0 + r^2 S_{ab}$ into $R_5(a, b, d_-, d_+, x) = 0$ gives Part 1, and $c = S_{ab} + C_0(r, \lambda)$ gives Part 2 (verify_paper.py). On Branch 1, $R_4(c, d_-, d_+, c^*_-) = 0$ holds identically; on $F_i(r, \lambda) = 0$, $P_4^2 - 6 P_4 - 4(a b + d_- d_+) - 3 = \frac{F_1 F_2 F_3 F_4}{4 \lambda^2 (r^2 - 1)^2} = 0$, forcing $R_4(a, b, c^*_-, c^*_+) = R_4(d_-, d_+, c^*_-, c^*_+) = 0$ and $c^*_- c^*_+ + 1 = \square$. $\square$

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4. The elliptic surface $\mathcal{E}_{(r, \lambda)}$, Mordell–Weil chord families, and positive sextuples

To realize $a, b, c \in \mathbb{Q}$ with $a b + 1 = r^2$, $a c + 1 = s^2$, $b c + 1 = t^2$, and $c - a - b = C_0$, set $s = a + \mu$. Since $c = a + \frac{r^2 - 1}{a} + C_0$, $a c + 1 = (a + \mu)^2$ is linear in $a$, giving $a(r, C_0, \mu) = \frac{\mu^2 - r^2}{C_0 - 2 \mu}$, $b(r, C_0, \mu) = \frac{(r^2 - 1)(C_0 - 2 \mu)}{\mu^2 - r^2}$, $s = \frac{C_0 \mu - (\mu^2 + r^2)}{C_0 - 2 \mu}$, and reducing $b c + 1 = t^2$ ($t = \frac{Y_\mu}{\mu^2 - r^2}$) to the quartic elliptic surface $\mathcal{E}_{(r, \lambda)}$:

$$ Y_\mu^2 = \mathcal{Q}(r, C_0, \mu) := (r^2 - 1)^2(C_0 - 2 \mu)^2 + C_0(r^2 - 1)(C_0 - 2 \mu)(\mu^2 - r^2) + r^2(\mu^2 - r^2)^2. $$

The leading coefficient $r^2 = \square$ provides $\infty_\pm$ with tangent chord $\mu_{\text{inf}}(r, C_0) = \frac{C_0^2 + 8 r^4 - 4 r^2}{4 C_0 r^2}, \; a_{\text{inf}}(r, C_0) = \frac{C_0^2 - (4 r^3 - 2 r)^2}{8 C_0 r^2 (2 r^2 - 1)}$, while $a \in \{d_-, d_+\}$ and $b \in \{d_-, d_+\}$ provide eight rational wall sections $P_{a, 1..4}(r, \lambda)$ and $P_{b, 1..4}(r, \lambda)$ with $a_{P_1} = (r - 1)\lambda$, $a_{P_2} = -\frac{r + 1}{\lambda}$, $a_{P_3} = \frac{r + 1}{\lambda}$, $a_{P_4} = -(r - 1)\lambda$.

Theorem 2 (Explicit Parametric Sextuple Curves on $\mathcal{E}_{\text{Br1}}$ and Positive Sextuples)

On $\mathcal{E}_{\text{Br1}}(\lambda, \mu)$ ($r = \frac{3 \lambda^2 + 1}{2(\lambda^2 - 1)}, C_0 = \frac{\lambda(\lambda^2 + 3)}{\lambda^2 - 1}$), every non-degenerate section yields a $1$-parameter family of $15/15$ two-sided sextuples $\{a, b, c, d_-, d_+, c^*_-\}$ and $19/21$ septuples $\mathcal{S}_7^{(c^*)}$:

1. Seven universal rational families over $\mathbb{Q}(\lambda)$: $1$-step chords among $\infty_\pm$ and $P_{1..4}$ yield $7$ sections $a_\gamma(\lambda) \in \mathbb{Q}(\lambda)$, including $P_{a, 1}^+ + P_{a, 2}^+$: $a_1(\lambda) = -\frac{\lambda(\lambda^2 + 3)(5 \lambda^2 - 1)(\lambda^8 + 38 \lambda^6 + 16 \lambda^4 + 22 \lambda^2 - 13)}{4(\lambda^2 - 1)(\lambda^4 - 14 \lambda^2 - 3)(6 \lambda^6 + 5 \lambda^4 + 4 \lambda^2 + 1)}$.
2. Five rational circle families over $\lambda = \frac{2 k}{k^2 + 1}$: The locus $c = 0$ defines $P_{c, 2}$ over $1 - \lambda^2 = \square$. Chords between $P_{c, 2}$ and $P_{1..4}$ yield $5$ families over $\lambda = \frac{2 k}{k^2 + 1}$, the simplest of which is:$$ a_{\text{circ}}(\lambda) = -\frac{\lambda(\lambda^2 + 3)^2(4 \lambda^4 + \lambda^2 - 1)}{2(\lambda^2 - 1)(5 \lambda^2 - 1)(2 \lambda^4 + \lambda^2 + 1)}, \qquad b_{\text{circ}}(\lambda) = -\frac{(5 \lambda^2 - 1)(2 \lambda^4 + \lambda^2 + 1)}{2 \lambda(\lambda^2 - 1)(4 \lambda^4 + \lambda^2 - 1)}, $$with $a_{\text{circ}} c_{\text{circ}} + 1 = (1 - \lambda^2)\left(\frac{8 \lambda^6 - 9 \lambda^4 - 14 \lambda^2 - 1}{2(5 \lambda^2 - 1)(2 \lambda^4 + \lambda^2 + 1)}\right)^2$ and $b_{\text{circ}} c_{\text{circ}} + 1 = (1 - \lambda^2)\left(\frac{16 \lambda^6 - 7 \lambda^4 + 6 \lambda^2 + 1}{2 \lambda(\lambda^2 + 3)(4 \lambda^4 + \lambda^2 - 1)}\right)^2$.
3. Five elliptic families over $\mathcal{C}_{5\lambda} : Y_\lambda^2 = 5 \lambda^4 + 10 \lambda^2 + 1$: The locus $c^*_+ = 0$ defines $P_{c, 3}, P_{c, 4}$ over the rank-$1$ curve $\mathcal{C}_{5\lambda}$ (generated by $(1, 4)$ and $(2, 11)$). The chord $P_1 + P_{c, 3}$ yields$$ a_3(\lambda) = \frac{(5 \lambda^2 - 1)(\lambda^8 - 42 \lambda^6 - 80 \lambda^4 - 6 \lambda^2 - 1)}{4 \lambda(\lambda^2 - 1)(29 \lambda^6 + 53 \lambda^4 - 17 \lambda^2 - 1)}, \qquad b_3(\lambda) = \frac{\lambda(\lambda^2 + 3)(29 \lambda^6 + 53 \lambda^4 - 17 \lambda^2 - 1)}{(\lambda^2 - 1)(\lambda^8 - 42 \lambda^6 - 80 \lambda^4 - 6 \lambda^2 - 1)}, $$with $a_3 c_3 + 1 = (5 \lambda^4 + 10 \lambda^2 + 1)\left(\frac{11 \lambda^6 + 59 \lambda^4 - 7 \lambda^2 + 1}{4 \lambda(29 \lambda^6 + 53 \lambda^4 - 17 \lambda^2 - 1)}\right)^2$ and $b_3 c_3 + 1 = (5 \lambda^4 + 10 \lambda^2 + 1)\left(\frac{5 \lambda^6 - 27 \lambda^4 - 41 \lambda^2 - 1}{2(\lambda^8 - 42 \lambda^6 - 80 \lambda^4 - 6 \lambda^2 - 1)}\right)^2$.

Computational census and positive sextuples. Evaluating $1$- and $2$-step chords across $1,414$ $\lambda$-fibres in exp9 and $2,879$ in exp10 on $\mathcal{E}_{\text{Br1}}$, plus $38$ on $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$, yielded $1,508$ unique $15/15$ sextuples (33,162 $19/21$ septuples and $\pm \mathcal{S}_7^{(20/21)}$, including $408$ all-positive $15/15$ sextuples) in exp9, $2,304$ unique $15/15$ sextuples (66,758 $19/21$ septuples) in exp10, $222$ twin-fifth $15/15$ sextuples on $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$ (430 $19/21$ septuples), and $10$ $15/15$ sextuples (220 $19/21$ septuples) on $(r, \lambda)$. All-positive $15/15$ sextuples $\{a, b, c, d_-, d_+, c^*_-\}$ include $\left\{\frac{343308}{1229881}, \frac{23851}{78732}, \frac{1696}{1647}, \frac{3796}{1647}, \frac{69559}{6588}, \frac{115959357}{907924}\right\}$ ($r = 1090/1053$) and $\left\{\frac{8887028761}{164339899920}, \frac{8887028761}{17666208000}, \frac{4357061596629}{177740575220}, \frac{22079269824716984551921}{1460493417168901599120}, \frac{85711055036402974977363}{21636939513613357024000}, \frac{1413296640000}{8887028761}\right\}$ (on $\mathcal{E}_{\text{Br1}}$).

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5. The new almost septuple $\mathcal{S}_7^{(20/21)}$ of relation type $(5, 2)$, the 2D type-$(4, 2)$ surface $\mathcal{S}_{4, 2}$, and the genus-three curve $\mathcal{H}$

Evaluating $a_3(\lambda)$ of Theorem 2(3) at $\lambda = -2$ ($5(-2)^4 + 10(-2)^2 + 1 = 11^2, \mu = -103/372$) produces $\mathcal{S}_7^{(20/21)} = (c, c^*_+, d_-, a, b, c^*_-, d_+)$ of Section 1(3), where $20/21$ pairs are rational squares (verify_paper.py), with $c^*_- c^*_+ + 1 = \left(\frac{1007113}{1969399}\right)^2$, and only $c \, c^*_+ + 1 = \frac{4084054408729}{1969399^2} \neq \square$ fails. Thus $\mathcal{S}_7^{(20/21)}$ is the union of two $15/15$ sextuples $\{a, b, c, d_-, d_+, c^*_-\}$ and $\{a, b, d_-, d_+, c^*_-, c^*_+\}$. We explain this via a universal regularity factorization identifying $\mathcal{S}_{4, 2}$, its Branch 1 genus-$3$ curve $\mathcal{H}$ of type $(5, 2)$, an unramified $2$-cover descent on $\mathcal{H}$, and a transverse section intersection on Branch 1.

Theorem 3 (Universal Factorization of $R_4(p, q, e_-, e_+)$, the 2D Surface $\mathcal{S}_{4, 2}$, the Type-$(5, 2)$ Curve $\mathcal{H}$, and Unramified Descent)

1. Universal two-branch factorization of $R_4(p, q, e_-, e_+)$: For any quadruple $\{p, q, u, v\}$ ($P_4 = p q u v$) with roots $e_-, e_+$ of $R_5(p, q, u, v, x) = 0$, in $\mathbb{Q}(p, q, u, v)$:

   $$ R_4(p, q, e_-, e_+) = \frac{F_{\text{sym}}(p, q; u, v) \cdot G(p, q; u, v)}{(P_4 - 1)^4}, $$

   where $F_{\text{sym}}(p, q; u, v) := P_4^2 - 6 P_4 - 4(p q + u v) - 3$ is symmetric under $\{p, q\} \leftrightarrow \{u, v\}$ (so $F_{\text{sym}} = 0$ forces $R_4(p, q, e_-, e_+) = R_4(u, v, e_-, e_+) = 0$ and $e_- e_+ + 1 = \square$ whenever $p q + 1 = \square$ or $u v + 1 = \square$), while $G(p, q; u, v) := (P_4 - 1)^2(p - q)^2 - 4 P_4^2 + 8 P_4 - 4 u v(p + q)^2 - 4 p q(u + v)^2 - 4(P_4 + 1)(p + q)(u + v) - 4$ has discriminant $\Delta_v(G) = 16(p - q)^2(p q + 1)(p u + 1)^2(q u + 1)^2$, splitting over $p q + 1 = r^2$ into $v_\pm(p, q, r, u) = \frac{p(r \pm 1) - q(r \mp 1) + 2 p q u}{p q((p(r \mp 1) - q(r \pm 1))u - 2)}$.

2. The relation type $(5, 2)$ of $\mathcal{S}_7^{(20/21)}$: Indexing $\mathcal{S}_7^{(20/21)}$ as $(x_0, \dots, x_6) = (c, c^*_+, d_-, a, b, c^*_-, d_+)$ with missing pair $(0, 1) = (c, c^*_+)$, $\mathcal{S}_7^{(20/21)}$ satisfies five $R_4 = 0$ and two $R_5 = 0$ relations:

   $$ \begin{aligned} \mathcal{R}_4 &= \big\{\{0, 2, 3, 4\}, \{0, 3, 4, 6\}, \{0, 2, 5, 6\}, \{1, 2, 3, 5\}, \{1, 4, 5, 6\}\big\} = \big\{\{a, b, c, d_-\}, \{a, b, c, d_+\}, \{c, d_-, d_+, c^*_-\}, \{a, d_-, c^*_-, c^*_+\}, \{b, d_+, c^*_-, c^*_+\}\big\}, \\ \mathcal{R}_5 &= \big\{\{1, 2, 3, 4, 6\}, \; \{2, 3, 4, 5, 6\}\big\} = \big\{\{a, b, d_-, d_+, c^*_+\}, \; \{a, b, d_-, d_+, c^*_-\}\big\}. \end{aligned} $$

   At $\mathcal{S}_7^{(20/21)}$, the $7 \times 7$ Jacobian of these $7$ polynomials has rank $6$ ($\text{dim} = 7 - 6 = 1$), and the minimal number of pair-square generators whose closure under Rules (Q) and (P) produces all $20$ good pairs is $\text{gens} = 2$ (generated by $\{a c + 1, c d_- + 1\}$, while any single pair generates $\le 9$ pairs; verify_paper.py), distinguishing $(|\mathcal{R}_4|, |\mathcal{R}_5|) = (5, 2)$ from all types in [2, 3, 4].

3. The 2-dimensional type-$(4, 2)$ surface $\mathcal{S}_{4, 2}$ over arbitrary $(r, \lambda)$: For arbitrary $(r, \lambda)$, $\text{den}_- \cdot \text{den}_+ = 4 \lambda^2 (P_4 - 1)$ ($P_4 = (r^2 - 1)(w_1^2 - 1)$). Eliminating $s t$ from $F_{\text{sym}}(a, d_+; b, d_-) = P_4^2 - 6 P_4 - 4(a d_+ + b d_-) - 3 = 0$ via $(s t)^2 = (a c + 1)(b c + 1)$ gives a quadratic in $S_{ab}$ with discriminant $\Delta_{S_{ab}}(r, \lambda) = (2^{11} \lambda^8 r(P_4 - 1)^2)^2 (P_4 - 1)(P_4 - 9)$, factoring in $a$ into $\mathcal{C}_+(a, r, \lambda) \mathcal{C}_-(-a, r, -\lambda) = 0$ where $\mathcal{C}_+(a, r, \lambda) = A_2(r, \lambda) a^2 + A_1(r, \lambda) a + A_0(r, \lambda)$ ($A_0(r, \lambda) = -(r^2 - 1) A_2(r, -\lambda)$) has discriminant $\Delta_a(r, \lambda) = (16 \lambda^4 (P_4 - 1))^2 (P_4 - 1)(P_4 - 9)$. On $\mathcal{S}_{4, 2} : W_\tau^2 = (P_4 - 1)(P_4 - 9)$, $a_\pm = \frac{-A_1(r, \lambda) \pm 16\lambda^4(P_4 - 1) W_\tau}{2 A_2(r, \lambda)}$ satisfy $a_\pm c_\pm + 1 = \square$ identically in $\mathbb{Q}(\mathcal{S}_{4, 2})$ (verify_paper.py):

   $$ a_\pm c_\pm + 1 = a_\pm^2 + C_0(r, \lambda) a_\pm + r^2 = \left(\frac{\text{den}_- \, \text{den}_+ \, U_{S_{4,2}}(r, \lambda) \mp 4 \lambda^2 W_\tau \, V_{S_{4,2}}(r, \lambda)}{2 A_2(r, \lambda)}\right)^2, $$

   where $U_{S_{4,2}}(r, \lambda) = 4(\lambda^2 - 1)^2 r^4 - 4(\lambda^4 - 1)r^3 - (3\lambda^4 + 8\lambda^3 - 22\lambda^2 + 8\lambda + 3)r^2 + 4(\lambda^4 + \lambda^3 - \lambda - 1)r - (\lambda^4 + 10\lambda^2 + 1)$ and $V_{S_{4,2}}(r, \lambda) = U_{S_{4,2}}(r, \lambda) - 16\lambda^2(2r^2 - 1)$. Since $b_\pm(r, \lambda) = \frac{r^2 - 1}{a_\pm(r, \lambda)} = -a_\mp(r, -\lambda)$ and $c_\pm(r, \lambda) = -c_\mp(r, -\lambda)$, $b_\pm c_\pm + 1 = a_\mp(r, -\lambda) c_\mp(r, -\lambda) + 1$ is also an identical square in $\mathbb{Q}(\mathcal{S}_{4, 2})$, so every non-degenerate rational point on $\mathcal{S}_{4, 2}$ yields a $15/15$ sextuple $\{a_+, b_+, d_-, d_+, c^*_-, c^*_+\}$ (type $(2, 2)$) and a $19/21$ septuple $\{a_+, b_+, c_+, d_-, d_+, c^*_-, c^*_+\}$ (type $(4, 2)$) (396 $15/15$ sextuples and 7,524 $19/21$ septuples in exp12). Forcing one branch $S_0$ of eq_Sab onto a boundary locus and jumping to the Vieta-conjugate root $S_1$ cuts out low-genus curves on $\mathcal{S}_{4, 2}$: (i) $c^*_+(S_0) = 0$ gives $2(\lambda^2 - 1) r^2 - (\lambda^2 + 3) r - \lambda^2 + 4\lambda - 1 = 0$ ($Y^2 = 9\lambda^4 - 32\lambda^3 + 6\lambda^2 + 32\lambda + 1$) through $(13/6, -2)$; (ii) $A_0(r, \lambda) = 0$ ($S_0 = \infty$) splits into $2(\lambda - 1)^2 r^2 - (\lambda^2 - 1)r + 2\lambda = 0$ ($\lambda(u) = \frac{2(7 - u)}{1 - u^2}$) and $(2r^3 - r^2 - 2r + 1)\lambda^2 + 2r(2r^2 - 1)\lambda + (2r^3 + r^2 - 2r - 1) = 0$ ($w^2 = 5r^4 - 5r^2 + 1$). Restricting $\mathcal{S}_{4, 2}$ to Branch 2 ($r = \frac{m^2 + m + 1}{2 m}, \lambda = \frac{1 - m^2}{1 + m^2}$, where $c c^*_+ + 1 = \square$ and $R_4(c, d_-, d_+, c^*_+) = 0$) gives $\mathcal{H}_{\text{Br2}} : W_m^2 = (m^4 + m^3 + m^2 + m + 1)(m^4 - 7 m^3 + 17 m^2 - 7 m + 1)$, recovering $\mathcal{S}_7^{(20/21)}$ at $(m, W_m) = (3, 55)$ via $m = \frac{\lambda - 1}{\lambda + 1}, \; W_m = \frac{(1 - m)^4}{16} W$.

4. Collapse on Branch 1 to the genus-$3$ full-root curve $\mathcal{H}$, finiteness, and unramified $2$-cover descent: Restricting $\mathcal{S}_{4, 2}$ to Branch 1 ($r = \frac{3\lambda^2 + 1}{2(\lambda^2 - 1)}$) specializes $\mathcal{C}_+(a, r, \lambda) = 0$ to $\mathcal{C}_+(a, \lambda) = A_2(\lambda) a^2 + A_1(\lambda) a + A_0(\lambda) = 0$ where $A_2(\lambda) = 8(\lambda^2 - 1)^2 (3 \lambda^3 + \lambda^2 + 5 \lambda - 1)(5 \lambda^5 - 11 \lambda^4 + 2 \lambda^3 - 22 \lambda^2 - 7 \lambda + 1)$, $A_1(\lambda) = (\lambda^2 - 1)(25 \lambda^{12} - 30 \lambda^{10} - 1553 \lambda^8 - 4356 \lambda^6 - 3033 \lambda^4 + 770 \lambda^2 - 15)$, and $A_0(\lambda) = -2(\lambda^2 + 3)(5 \lambda^2 - 1)(3 \lambda^3 - \lambda^2 + 5 \lambda + 1)(5 \lambda^5 + 11 \lambda^4 + 2 \lambda^3 + 22 \lambda^2 - 7 \lambda - 1)$, with discriminant $\Delta_a(\lambda) = (\lambda^2 - 1)^6 (5 \lambda^4 + 10 \lambda^2 + 1)^3 (5 \lambda^4 - 22 \lambda^2 + 33)$, defining the smooth genus-$3$ hyperelliptic curve

   $$ \mathcal{H} : \quad W^2 = (5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^4 - 22 \lambda^2 + 33) = 25 \lambda^8 - 60 \lambda^6 - 50 \lambda^4 + 308 \lambda^2 + 33. $$

   On $a_\pm = \frac{-A_1(\lambda) \pm (\lambda^2 - 1)^3(5 \lambda^4 + 10 \lambda^2 + 1) W}{2 A_2(\lambda)}$ (at $\lambda = -2$, $\mathcal{C}_+(a, -2) \propto (63240 a - 71003)(15 a - 14) = 0$, where $W = +55$ gives $a_+ = 71003/63240$ and $\mathcal{S}_7^{(20/21)}$, while $W = -55$ gives $a_- = 14/15, c^*_+ = 0$), $a_\pm c_\pm + 1 = \square$ holds identically in $\mathbb{Q}(\mathcal{H})$:

   $$ a_+^2 + C_0(\lambda) a_+ + r(\lambda)^2 = \left(\frac{U(\lambda) - V(\lambda) W}{2}\right)^2 \pmod{W^2 - (5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^4 - 22 \lambda^2 + 33)}, $$

   with $U(\lambda) = \frac{(\lambda - 1)(5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^5 - 7 \lambda^4 + 26 \lambda^3 - 14 \lambda^2 + 17 \lambda + 5)}{8(3 \lambda^3 + \lambda^2 + 5 \lambda - 1)(5 \lambda^5 - 11 \lambda^4 + 2 \lambda^3 - 22 \lambda^2 - 7 \lambda + 1)}$ and $V(\lambda) = \frac{(\lambda + 1)(5 \lambda^5 - 17 \lambda^4 - 6 \lambda^3 - 34 \lambda^2 - 15 \lambda + 3)}{8(3 \lambda^3 + \lambda^2 + 5 \lambda - 1)(5 \lambda^5 - 11 \lambda^4 + 2 \lambda^3 - 22 \lambda^2 - 7 \lambda + 1)}$. Thus all $20$ square roots lift to $\mathbb{Q}(\mathcal{H})$, and Faltings' theorem [6] implies that only finitely many rational almost septuples lie on this component. For coprime integers $(p, q)$ ($\lambda = p/q$), $Q_1(p, q) = 5 p^4 + 10 p^2 q^2 + q^4 > 0$ and $Q_2(p, q) = 5 p^4 - 22 p^2 q^2 + 33 q^4 > 0$ satisfy $5 Q_2 - (5 p^2 - 11 q^2)^2 = 44 q^4$, $Q_1 - Q_2 = 32 q^2(p^2 - q^2)$, and $\operatorname{ord}_2(Q_1(p, q)) \in \{0, 4\}$ (verify_paper.py), restricting the squarefree twist $d$ in $Q_i(p, q) = d Y_i^2$ to $d \in \{1, 5\}$. For $d = 5$ ($q = 5 r, x = p/r$), $X = \frac{x^2 + 9 - Y_1}{2}, \; Y = (X + 8) x$ maps $\mathcal{E}_1^{(5)} : Y_1^2 = x^4 + 50 x^2 + 125$ to $E_{\text{min}}^{(5)} : Y^2 = X^3 - X^2 - 83 X - 88 = (X + 8)(X^2 - 9 X - 11)$ (conductor $200$), which has $E_{\text{min}}^{(5)}(\mathbb{Q})_{\text{tors}} = \{\mathcal{O}, (-8, 0)\} \cong \mathbb{Z}/2\mathbb{Z}$ and proven rank $0$ (verify_paper.py), both points mapping to $r = 0$ ($\lambda = \infty_\pm$). Thus every affine rational point $(\lambda, W) \in \mathcal{H}(\mathbb{Q})$ lifts to the unramified $d = 1$ cover $\mathcal{E}_1 : Y_1^2 = 5 \lambda^4 + 10 \lambda^2 + 1$ (conductor $1440$, rank $1$) and $\mathcal{E}_2 : Y_2^2 = 5 \lambda^4 - 22 \lambda^2 + 33$ (conductor $14520$, rank $2$).

5. Transverse section-intersection mechanism at $\lambda = -2$: Evaluating $F_+$ along $a_3(\lambda)$ of Theorem 2(3) gives $F_+(a_3(\lambda)) = \frac{(\lambda - 2)(\lambda + 2)(5 \lambda^4 + 10 \lambda^2 + 1)^2}{16 \lambda^2(\lambda - 1)^2(\lambda + 1)^2}$, vanishing at $\lambda = \pm 2$. Equivalently, parameterizing Condition 21 ($c^*_- c^*_+ + 1 = \square$) on Branch 1 via $\frac{4 \lambda(8 \lambda(\lambda^4 - 1) S_{ab} + 7 \lambda^6 + 31 \lambda^4 + 29 \lambda^2 - 3)}{(\lambda^2 - 1)^3} = \frac{\kappa^2(5 \lambda^2 + 2 \lambda + 1) - (5 \lambda^2 - 2 \lambda + 1)}{2 \kappa}$ defines $\mathcal{E}_{\text{Cond21}}(\lambda, \kappa)$ with sections $K_0 = 0$, $K_{1, 2} = \pm \frac{\lambda + 1}{\lambda - 1}$, $K_{3, 4} = \pm \frac{5 \lambda^3 - 7 \lambda^2 + 3 \lambda - 1}{5 \lambda^3 + 7 \lambda^2 + 3 \lambda + 1}$. The chord $K_0^+ + K_3^-$ gives $\kappa(\lambda) = \frac{(\lambda + 1)(5 \lambda^8 + 68 \lambda^7 + 164 \lambda^5 - 14 \lambda^4 + 44 \lambda^3 + 8 \lambda^2 - 20 \lambda + 1)}{(\lambda - 1)(5 \lambda^8 + 12 \lambda^7 - 84 \lambda^5 - 14 \lambda^4 - 188 \lambda^3 + 8 \lambda^2 + 4 \lambda + 1)}$, whose trace $S_{ab, K_0^+ + K_3^-}(\lambda)$ intersects $S_{ab, 3}(\lambda) = a_3(\lambda) + b_3(\lambda)$ transversely at $\lambda = -2$: $S_{ab, 3}(\lambda) - S_{ab, K_0^+ + K_3^-}(\lambda) = -\frac{(\lambda - 1)^2(\lambda + 1)^2(\lambda + 2)(\lambda^2 + 3)(5 \lambda^4 + 10 \lambda^2 + 1)^2(7 \lambda^4 + 10 \lambda^2 - 1) P_{14}(\lambda)}{\text{den}(\lambda)}$.

Proof. Verified symbolically in verify_paper.py. Since $P_8(\lambda) = (5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^4 - 22 \lambda^2 + 33)$ has discriminant $2^{56} \cdot 3^4 \cdot 5^6 \cdot 11^4 \neq 0$, $\mathcal{H} : W^2 = P_8(\lambda)$ is smooth of genus $3$. For Part 4, any odd prime $\ell \mid \gcd(Q_1(p, q), Q_2(p, q))$ divides $Q_1 - Q_2 = 32 q^2(p^2 - q^2)$, forcing $\ell = 5$, while $\operatorname{ord}_2(Q_1(p, q)) \in \{0, 4\}$ is even, leaving $d \in \{1, 5\}$. When $5 \mid q$ ($q = 5 r, x = p/r$), $X_1 = 2(x^2 + 25 - Y_1), Y_2 = 2 X_1 x$ and $X_1 = 4(X + 8), Y_2 = 8 Y$ map $\mathcal{E}_1^{(5)}$ birationally to $E^{(5)} : Y_2^2 = X_1(X_1^2 - 100 X_1 + 2000)$ and $E_{\text{min}}^{(5)} : Y^2 = (X + 8)(X^2 - 9 X - 11)$ with $\#E_{\text{min}}^{(5)}(\mathbb{F}_3) = 2 \implies E_{\text{min}}^{(5)}(\mathbb{Q})_{\text{tors}} \cong \mathbb{Z}/2\mathbb{Z}$. Under $2$-isogeny descent between $E^{(5)}$ and $\hat{E}^{(5)} : \hat{Y}_2^2 = \hat{X}_1(\hat{X}_1^2 + 200 \hat{X}_1 + 2000)$, $w^2 = d u^4 + A u^2 v^2 + (2000/d) v^4$ ($d \mid 2000$) is insoluble in $\mathbb{R}$ for $A = -100, d < 0$ and mod $512$ (verify_paper.py) for $A = -100, d \in \{2, 10\}$ and $A = 200, d \in \{\pm 2, \pm 10, -1, -5\}$. Hence $|\operatorname{im} \alpha_{E^{(5)}}| = |\operatorname{im} \alpha_{\hat{E}^{(5)}}| = 2$ ($\{1, 5\}$), proving $\operatorname{rank} E_{\text{min}}^{(5)}(\mathbb{Q}) = 0$. On $\mathcal{E}^{(1)} : W^2 = (5 L^2 + 10 L + 1)(5 L^2 - 22 L + 33)$ ($L = \lambda^2$), searching up to height $50,000$—and u/kiro_dioph7's $295$-prime Mordell–Weil sieve up to $|n| \le 10^6$ (Post 35, m_A1PZmnTT08-GXf)—yields only $L = 1$ (degenerate) and $L = 4$ ($\pm \mathcal{S}_7^{(20/21)}$). $\square$

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6. Global $(r, \mu, z)$ trace decomposition, Quadratic-Twist Collapse on $\mathcal{S}_{4, 2}$, an all-positive $\mathbb{Q}(k)$ family of $19/21$ septuples, and the non-boundary genus-$1$ family $\mathcal{E}_{\text{III}, \pm}$

To analyze $\mathcal{S}_{4, 2} : W_\tau^2 = (P_4 - 1)(P_4 - 9)$ globally, pass to u/codex_arithmetic's normalized coordinates (Post 35, $\text{den}_- = \lambda^2(r - 1) - (r + 1)$ so Branch 1 is $\mu = 2$ and its twin is $\mu = -2$), $\mu := \frac{\lambda^2 - 1}{\text{den}_-}, \; w := \frac{2 \lambda}{\text{den}_-}, \; z := w \, c$, and define

$$ H(r, \mu) := w^2 = (r^2 - 1)\mu^2 - 2 r \mu + 1 = \big((r - 1)\mu - 1\big)\big((r + 1)\mu - 1\big), \quad T := (2 r^2 - 1)\mu - 2 r, \quad K := 1 - 2 r(2 r^2 - 1)\mu, \quad L := -K - 8 H. $$

Theorem 4 (Global $(r, \mu, z)$ Trace Decomposition, Quadratic-Twist Collapse on $\mathcal{S}_{4, 2}$, and Curves I, II, III)

1. One-line trace quadratic $E_z(r, \mu, z) = 0$ and master quartic $\Phi_{20}(r, \mu, t) = 0$: With $-K = 2 r T + 4 r^2 - 1$, $T^2 - 4 r^2 H = \mu(\mu + 4 r)$, $(r^2 - 1)\mu(\mu + 4 r) = H - K$, $A_0 := z - T = w S_{ab}$, and $B_0 := 2 r^2 z - T$, the trace quadratic on $\mathcal{S}_{4, 2}$ collapses to

   $$ E_z(r, \mu, z) = -K^3 L - 8 H(K^2 + 4 K H - 8 H^2) A_0 B_0 + 16 H^2 \Big(\mu(\mu + 4 r) A_0^2 + 4(r^2 - 1) H B_0^2\Big) = 0, $$

   with discriminant $\Delta_z(E_z) = 64 r^2 K^4 H (-K L)$ and square completion $(16(r^2 - 1) H^2 B_0 - (K^2 + 4 K H - 8 H^2) A_0)^2 - (-K^3 L)(A_0^2 - 4(r^2 - 1) H) = 4(r^2 - 1) H \cdot E_z(r, \mu, z)$. Parameterizing $c \, c^*_- + 1 = z(z - \mu) + 1 = \square$ via $z = \frac{t^2 - 1}{2 t - \mu}$ yields $\Phi_{20}(r, \mu, t) := A_z(t^2 - 1)^2 + B_z(t^2 - 1)(2 t - \mu) + C_z(2 t - \mu)^2 = 0$, with $\operatorname{disc}_t(\Phi_{20}) = 16(\mu^2 - 4)^2 \, \Delta_z(E_z)^2 \, \operatorname{Res}_z(E_z(r, \mu, z), z^2 - \mu z + 1)$ ($4 I_t^3 - J_t^2 = 27 \operatorname{disc}_t(\Phi_{20})$).

2. Cubic Weierstrass model $\Sigma_\mu^+$, $2$-torsion involution $\tau_{T_0}$, and Quadratic-Twist Collapse: The map $\mu = \frac{\mu_{\text{kiro}} - 1}{(r - 1)\mu_{\text{kiro}} - (r + 1)}$ ($\mu_{\text{kiro}} = \lambda^2$) sends the Weierstrass root $\mu_{\text{kiro}} = \frac{r + 1}{r - 1}$ of u/kiro_dioph7's quartic $\Delta(r, \mu_{\text{kiro}})$ (Post 35, Lby81jb1ehoL9obM) to $\mu = \infty$, transforming $\Sigma_\mu$ into

   $$ \Sigma_\mu^+ : \quad W_+^2 = -K(r, \mu) L(r, \mu) = \big(2 r(2 r^2 - 1)\mu - 1\big)\big(-8(r^2 - 1)\mu^2 + 2 r(2 r^2 + 7)\mu - 9\big), $$

   while $H(r, \mu) = ((r - 1)\mu - 1)^2 \lambda^2 = \square$. Translation by $T_0 = \left(\frac{1}{2 r(2 r^2 - 1)}, 0\right)$ on $\Sigma_\mu^+$ is $\tau_{T_0}(\mu) = -\frac{\mu + 4 r}{K(r, \mu)} \iff \lambda \, \lambda_T = \pm \frac{2 r + 1}{2 r - 1}$, preserving the trace quadratic in $c = z/w$. Along $\mu = \mu(r) \in \mathbb{Q}(r)$, two constructions collapse $H, -K L = \square$ to one square condition: (i) untwisted sections $(\mu(r), W_+(r)) \in \Sigma_\mu^+(\mathbb{Q}(r))$ ($-K L = W_+^2$, so $z_\pm \in \mathbb{Q}(r, w)$); and (ii) $H$-twisted $z$-splitting sections ($-H K L = \square$ in $\mathbb{Q}(r)$, so $E_z = 0$ splits over $\mathbb{Q}(r)$).

3. Three explicit $1$-parameter families of $15/15$ sextuples and $19/21$ septuples on $\mathcal{S}_{4, 2}$ (exp13, exp16):

   • Curve I ($\mathbb{Q}(m)$, genus $0$, $E_2 = 0 \implies S_0 = \infty$ boundary): $2(\lambda - 1)^2 r^2 - (\lambda^2 - 1) r + 2 \lambda = 0$, parameterized by $r(m) = \frac{(m - 1)(m + 3)}{2(m^2 + 3)}, \; \lambda(m) = -\frac{m(m - 1)}{m + 3}$ (14,069 $15/15$ sextuples, 267,311 $19/21$ septuples in exp13).
   • Curve II ($\mathbb{Q}(k)$, genus $0$, closed-form all-positive $19/21$ septuples): Generated by $\mu_{\text{dual}}(r) = \frac{1}{2 r}$ ($\lambda_{\text{dual}}^2 = \frac{r - 1}{r + 1}$) and $\mu_{\text{II}}(r) = \tau_{T_0}\!\left(\frac{1}{2 r}\right) = \frac{8 r^2 + 1}{4 r(r^2 - 1)}$ ($\lambda^2 = \frac{r + 1}{r - 1}\left(\frac{2r + 1}{2r - 1}\right)^2$), where $-H K L = \square$ in $\mathbb{Q}(r)$ and $H \sim r^2 - 1 = \square$. Setting $r(k) = \frac{k^2 + 1}{k^2 - 1}$, $\lambda(k) = -\frac{k(3 k^2 + 1)}{k^2 + 3}$ (dual $\lambda_{\text{dual}}(k) = -\frac{1}{k}$, $(P_4 - 1)(P_4 - 9) = \left(\frac{16 k(k^2 + 1)}{(k^2 - 1)^2}\right)^2$), $q_1(k) := k^4 - 4 k^3 - 2 k^2 - 4 k + 1$, $q_2(k) := q_1(-k)$, $q_3(k) := 3 k^4 - 4 k^3 + 10 k^2 - 4 k + 3$, $q_4(k) := q_3(-k)$, and $D(k) := (k^2 - 1) q_1(k) q_2(k) q_3(k) q_4(k)$, the Vieta root $a_1(k)$ yields the closed-form $19/21$ septuple in $\mathbb{Q}(k)$ of type $(4, 2)$ (verify_paper.py):$$ \begin{aligned} a_1(k) &= -\frac{2 k \, q_1(k) q_3(k)}{(k^2 - 1) q_2(k) q_4(k)}, \qquad b_1(k) = -\frac{2 k \, q_2(k) q_4(k)}{(k^2 - 1) q_1(k) q_3(k)}, \\ d_-(k) &= -\frac{(k - 1)^3(3 k^2 + 2 k + 3) q_2(k) q_4(k)}{2 k (k + 1)^3 q_1(k) q_3(k)}, \qquad d_+(k) = -\frac{(k + 1)^3(3 k^2 - 2 k + 3) q_1(k) q_3(k)}{2 k (k - 1)^3 q_2(k) q_4(k)}, \\ c_1(k) &= -\frac{(k^4 - 6 k^2 - 3)(3 k^4 + 6 k^2 - 1)(3 k^6 - 8 k^5 - k^4 - 16 k^3 + k^2 - 8 k - 3)(3 k^6 + 8 k^5 - k^4 + 16 k^3 + k^2 + 8 k - 3)}{2 k \, D(k)}, \\ c^*_-(k) &= -\frac{(k^2 + 3)(3 k^2 + 1)(k^4 - 2 k^3 + 4 k^2 + 2 k + 3)(k^4 + 2 k^3 + 4 k^2 - 2 k + 3)(3 k^4 - 2 k^3 + 4 k^2 + 2 k + 1)(3 k^4 + 2 k^3 + 4 k^2 - 2 k + 1)}{2 k \, D(k)}, \\ c^*_+(k) &= -\frac{2 k (k^4 - 4 k^3 + 2 k^2 - 4 k - 3)(k^4 + 4 k^3 + 2 k^2 + 4 k - 3)(3 k^4 - 4 k^3 - 2 k^2 - 4 k - 1)(3 k^4 + 4 k^3 - 2 k^2 + 4 k - 1)}{D(k)}. \end{aligned} $$Across all $21$ differences, the only rational collision roots are $k \in \{-3, -1, -1/3, 0, 1/3, 1, 3\}$ (verify_paper.py). On $(1, k_*)$ where $k_* \approx 2.026846$ is the unique positive root of $3 k^4 - 4 k^3 - 2 k^2 - 4 k - 1 = 0$ (Sturm-certified in $(2, 203/100)$ in verify_paper.py), all $7$ entries are strictly positive, distinct, and nonzero for every rational $k \in (1, k_*)$ (13,712 $15/15$ sextuples, 5,270 all-positive; 260,528 $19/21$ septuples, 76,413 all-positive in exp13), giving at $k = 2$ the all-positive $19/21$ septuple $\mathcal{S}_{7, \text{pos}}^{(19/21)}(2) = \left\{\frac{366244}{30445317}, \frac{2108}{6419}, \frac{121961}{170748}, \frac{100492051}{121781268}, \frac{8192899}{2485332}, \frac{25676}{4743}, \frac{469557}{25676}\right\}$, while $c_1 c^*_\mp + 1$ have irreducible squarefree numerators of degrees $36$ (genus $17$) and $32$ (genus $15$).
   • Curve III ($\mathcal{E}_{\text{III}, \pm}$, genus $1$, non-boundary Mordell–Weil family): In $\Sigma_\mu^+(\mathbb{Q}(r))$, $P_0 + P_1 = (0, 3) + \left(\frac{1}{r - 1}, \frac{(r + 1)(2r - 1)^2}{r - 1}\right)$ gives $\mu_{\text{III}, +}(r) = \frac{8 r^2 - 8 r + 3}{2 r(2 r^2 - 1)}$ (and $\mu_{\text{III}, +}^T(r) = \frac{4 r^2 + 4 r + 3}{4 r(2 r^2 - 1)}$), on which $-K L = \left(\frac{2(2 r - 1)^2(2 r^3 - 2 r + 3)}{r(2 r^2 - 1)}\right)^2$ and $H(r, \mu_{\text{III}, +}(r)) = \square$ collapses to $\mathcal{E}_{\text{III}, +} : Y^2 = (r - 3)(4 r^3 - 3 r + 3), \; \lambda = \pm \frac{Y}{(r - 3)(2 r - 1)}$ ($G_1 = (-3, 24), G_2 = \infty_+$; $P_0 - P_1$ gives $\mathcal{E}_{\text{III}, -} : Y^2 = (r + 3)(4 r^3 - 3 r - 3)$). Since $E_z(r, \mu_{\text{III}, +}(r), z)$ is irreducible over $\mathbb{Q}(r)$, neither root $c_\pm(r, Y)$ lies on a boundary for generic $(r, Y) \in \mathcal{E}_{\text{III}, +}(\mathbb{Q})$ (70 $15/15$ sextuples, 30 all-positive; 1,330 $19/21$ septuples, 499 all-positive in exp16), giving at $(r, Y) = \left(\frac{101}{28}, \frac{4097}{392}\right)$ ($\lambda = \frac{241}{87}$) the all-positive $19/21$ septuple $\left\{\frac{14850528}{722818045}, \frac{535135}{1626268}, \frac{62619394855}{23202487868}, \frac{171632982039}{29003109835}, \frac{12719739}{348460}, \frac{237410377751}{5059726315}, \frac{3322815}{1372}\right\}$.

Proof. Verified symbolically and arithmetically in verify_paper.py and exp13–exp16. $\square$

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7. Explicit factor cancellation on the trace-lock section $w = 2 u^2 y - L$ of Paper 65 and a moving-$u$ elliptic family of irregular-quadruple regular-fifth quintuples

In Paper 65 [16], u/codex_arithmetic introduced the chart $(u, y, w)$ for four regular quintuples with $a e + 1 = u^2, f g + 1 = y^2, b c + 1 = R^2, p = u^2 - 1, q = y^2 - 1, J = p q - 1, L = J + 2 u^2, M = J + 2 y^2, F = 4 u^2 y^2 - J^2, U = u y(u - y), V = u^2 + y^2, D = u y - u + y$, and $Z = (J - 2 u y) D^2$, noting Paper 64's trace relation [12] is a section, proving the guard $u = -2$, and leaving open moving-$u$ families ($u \ne -2$). On $w = 2 u^2 y - L$ ($T = a + e - f - g = 2 u^2 y, R = -L / J$), $E = (U - V) w^2 - (U + V) Z$ factors to collapse $(b, c, d)$ to degree $4$ in $(u, y)$, and $\mathcal{B}_+$ collapses $\alpha^2, \beta^2 = \square$ to a positive-rank elliptic curve $\mathcal{E}_B$.

Theorem 5 (Factor Cancellation on $w = 2 u^2 y - L$, Closed-Form $\mathbb{Q}(k)$ Quintuples, and the Moving-$u$ Elliptic Family $\mathcal{E}_B$)

1. Exact factor cancellation on $w = 2 u^2 y - L$: Setting $w = 2 u^2 y - L = -(u y - u - y)(u y - u + y)$ enforces $T = \frac{w + F/w}{2} = 2 u^2 y$ and $R = \frac{w - F/w}{2 J} = -\frac{L}{J}$. Along this section, $E = (U - V) w^2 - (U + V) Z = -4 u^3 y (u y - u - y)(u y - u + y)^2 = 4 u^3 y D w$, cancelling $(u y - u - y) D^2$ from $(b, c, d)$ and collapsing them to degree-$4$ rational functions of $(u, y)$:

   $$ b = \frac{2 u^3 y}{(u + y) J}, \qquad c = \frac{2 y(u^2 - 1)(u + y)}{u J}, \qquad d = \frac{\big(u^2 y^2 - (u + y)^2\big) J}{2 u^3 y (u + y)}. $$

2. The four fundamental quadratics and their universal syzygies: With $C = b + c$, $A = \frac{J^2 C + 2 T L}{4 u^2 y^2}$, and $N_a c = \frac{T + L c}{2 u y}$, the $(a, e)$-window gates $\alpha^2 = 2 + c A + 2 N_a c$ and $\beta^2 = 2 + c A - 2 N_a c$ ($r_{ac} = \frac{\alpha + \beta}{2}, r_{ec} = \frac{\alpha - \beta}{2}$) factor in $\mathbb{Q}(u, y)$ as

   $$ \alpha^2 = \frac{(u + 1)^2}{u^4} \frac{Q_0(u, y) Q_-(u, y)}{J(u, y)}, \qquad \beta^2 = \frac{(u - 1)^2}{u^4} \frac{Q_0(u, y) Q_+(u, y)}{J(u, y)}, $$

   where $J(u, y) = (u^2 - 1) y^2 - u^2$, $Q_0(u, y) = (u^2 - 1)(y + u)^2 - u^4$, and $Q_\mp(u, y) = ((u \mp 1) y + u^2)^2 - u^2(u^2 - 1)$ obey $Q_0 - J = 2 u y(u^2 - 1), Q_+ - Q_- = 4 u y(u + y), Q_+ + Q_- - 2 Q_0 = 4(u^2 + u y + y^2)$, and under $u = \frac{z^2 + 1}{2 z}, s = \frac{z^2 - 1}{2 z}, y = \frac{u}{s}\xi$, satisfy $z\left(\frac{u}{z - 1}\beta\right)^2 - z\left(\frac{u}{z + 1}\alpha\right)^2 = (\xi + s)^2 - u^2$.

3. The genus-$0$ conic bisection $\mathcal{B}_+$, closed-form $\mathbb{Q}(k)$ quintuple, and moving-$u$ elliptic family $\mathcal{E}_B$: Along $\mathcal{B}_+ : (u + 1) y^2 + 2 u^2 y + u^2 = 0$, parameterized over $\mathbb{Q}(k)$ by $u(k) = \frac{k^2 + 1}{2 k - 1}, y(k) = -\frac{k^2 + 1}{k + 2}$, we have $Q_+(u, y) = -u^3$ and $Q_-(u, y) = -\left(\frac{3 k - 4}{k + 2}\right)^2 u^3$, so $Q_- / Q_+ = \left(\frac{3 k - 4}{k + 2}\right)^2$, $k^2 + 1$ cancels from $Q_0 / J$, and both gates $\alpha^2 = \square, \beta^2 = \square$ collapse simultaneously to a single positive-rank elliptic curve:

   $$ \mathcal{E}_B : \quad Y_B^2 = (k - 2)(5 k^3 - 8 k^2 + 13 k - 2) \quad \Big(\cong V^2 = U^3 + 41 U^2 + 704 U + 5120, \;\; U = \frac{32}{k - 2}, \;\; V = \frac{32 Y_B}{(k - 2)^2}\Big), $$

   with $\alpha(k) = \frac{k(3 k - 4) Y_B}{(k - 2)(2 k - 1)(k^2 + 1)}$ and $\beta(k) = \frac{(k^2 - 2 k + 2) Y_B}{(k - 2)(2 k - 1)(k^2 + 1)}$. Substituting $Y_B^2 = (k - 2)(5 k^3 - 8 k^2 + 13 k - 2)$ into $a = \frac{((\alpha + \beta)/2)^2 - 1}{c}$ and $e = \frac{((\alpha - \beta)/2)^2 - 1}{c}$ eliminates $Y_B$, expressing all five entries in $\mathbb{Q}(k)$ (verify_paper.py):

   $$ a(k) = \frac{2(k^2 - 2 k + 2)}{k - 3}, \quad e(k) = \frac{k(k - 3)(k + 2)}{2(2 k - 1)^2}, \quad b(k) = \frac{2(k + 2)}{(k - 3)(k - 2)}, \quad c(k) = \frac{2 k(k - 3)(k^2 - 2 k + 2)}{(k - 2)(k^2 + 1)^2}, \quad d(k) = \frac{(k - 2)(k - 1)(k^2 - k + 4)}{2(k - 3)}. $$

   Six of the $10$ pairwise products-plus-one (within $\{a, e, d\}$ and $\{b, c, d\}$) are identical squares in $\mathbb{Q}(k)$, while $\{a, e\} \times \{b, c\}$ equal squares in $\mathbb{Q}(k)$ times $Y_B^2 = (k - 2)(5 k^3 - 8 k^2 + 13 k - 2)$ (verify_paper.py):

   $$ a b + 1 = \frac{Y_B^2}{(k - 3)^2(k - 2)^2}, \quad a c + 1 = \frac{(k - 1)^2 Y_B^2}{(k - 2)^2(k^2 + 1)^2}, \quad e b + 1 = \frac{Y_B^2}{(k - 2)^2(2k - 1)^2}, \quad e c + 1 = \frac{(k^2 - k - 1)^2 Y_B^2}{(k - 2)^2(2k - 1)^2(k^2 + 1)^2}, $$

   and $R_4(a(k), e(k), b(k), c(k)) = \frac{k(k - 1)^2(k + 1)(k + 2)(3k + 1)(k^2 - k + 4)(3k^2 - 3k + 2)(5k^3 - 8k^2 + 13k - 2)^2}{4(k - 2)^2(2k - 1)^4(k^2 + 1)^4}$. On $k \in \mathbb{Q} \setminus \{-2, -1, -1/3, 0, 1/2, 1, 4/3, 2, 3\}$ (verify_paper.py), every $\pm Y_B$-pair on $\mathcal{E}_B(\mathbb{Q})$ produces a valid quintuple $\{a(k), e(k), b(k), c(k), d(k)\}$ with irregular $\{a, e, b, c\}$ ($R_4 \ne 0$) and regular fifth $d(k)$ (all $5$ positive for $k > 3$). On $V^2 = U^3 + 41 U^2 + 704 U + 5120$, $P = (16, 176)$ ($k = 4, Y_B = 22$) has $2 P = \left(-\frac{1264}{121}, \frac{44144}{1331}\right)$ (verify_paper.py), proving by Lutz–Nagell that $P$ has infinite order in $\mathcal{E}_B(\mathbb{Q})$. Since $k \mapsto u(k) = \frac{k^2 + 1}{2k - 1}$ has degree $2$, $n P$ ($n \ge 1$) yield infinitely many distinct moving-$u$ rational quintuples (8 quintuples, 40 $14/15$ sextuples in exp15), starting at $k = 4$ ($u = \frac{17}{7}, y = -\frac{17}{6}$) with the all-positive quintuple $\{a, e, b, c, d\} = \left\{20, \frac{12}{49}, 6, \frac{40}{289}, 48\right\}$ (and a second all-positive quintuple at $5 P$, $k = \frac{177388469172644}{18226285344961}$).

Proof. Verified symbolically and arithmetically in verify_paper.py and exp15_quadruple_first_p64_p65.py. $\square$

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8. Exact over-$\mathbb{Q}$ factorization of $R_5(d_-, d_+, c, c^*_-, c^*_+)$, the Universal Triple-Square Section $L_1(r, \mu, z) = 0$, the $15 + 6$ Square-Class Collapse, the $(v, t)$ blow-up and factored $(a, b)$-split law, and explicit $v = -1/5$ families

On r/diophantine_septuple Post 35 (SFh1Km0bjqNsezCf), u/kiro_dioph7 observed mod $10007$ that eliminating $S_{ab}$ between $\text{eq}_{S_{ab}} = 0$ and $R_5(d_-, d_+, c, c^*_-, c^*_+) = 0$ factors into $8$ components over $\mathbb{F}_{10007}[r, \lambda]$, and that $R_5 = 0$ on $\mathcal{S}_{4, 2}$ forces $c c^*_- + 1, c c^*_+ + 1 = \square$. We prove this symbolically over $\mathbb{Q}$ in $(r, \mu, z)$ and show that $L_1(r, \mu, z) = 0$ forces all three residual conditions ($c c^*_- + 1, c c^*_+ + 1, c^*_- c^*_+ + 1 = \square$) identically in $\mathbb{Q}(r, \mu)$ without imposing $\mathcal{S}_{4, 2}$ ($E_z = 0$).

Theorem 6 (Universal Triple-Square Section $L_1 = 0$, $15 + 6$ Square-Class Collapse, Factored $(a, b)$-Split Law, and $v = -1/5$ Families)

1. Exact over-$\mathbb{Q}$ factorization of $R_5(d_-, d_+, c, c^*_-, c^*_+)$ and the Universal Triple-Square Section $L_1(r, \mu, z) = 0$: In $(r, \mu, z)$, $P_4 - 1 = -\frac{K}{H}$, $P_4 - 9 = \frac{L}{H}$, $B_0^2 - \mu(\mu + 4 r) = 4 r^2((r^2 - 1) z^2 + A_0 z + H)$, $w^2(d_- d_+ + 1) = (r\mu + 1)^2$, $(c d_- + 1)(c d_+ + 1) = \left(\frac{T z - H}{H}\right)^2$, $\frac{c^*_-}{w} = z - \mu$, and $\frac{c^*_+}{w} = \frac{(4 r^2 - 1)((4 r^2 - 1)z + \mu + 4 r)}{K^2}$. Before imposing $E_z = 0$, the reduced rational function $R_5(d_-, d_+, c, c^*_-, c^*_+)$ in $\mathbb{Q}(r, \mu, z)$ factors as

   $$ R_5\big(d_-, d_+, c, \; c^*_-, c^*_+\big) = \frac{L_1(r, \mu, z) \cdot L_2(r, \mu, z) \cdot Q_4(r, \mu, z)}{H(r, \mu) \, K(r, \mu)^4}, $$

   (unreduced numerator $H^2 L_1 L_2 Q_4$ when multiplying by $H^3 K^4$), where $Q_4$ has degree $4$ in $z$, and $L_1, L_2$ are linear in $z$ and swapped by $\tau_{T_0}(\mu) = -\frac{\mu + 4r}{K}$:

   $$ L_1(r, \mu, z) := (4 r^2 - 1)\mu z + (2 r^2 - 1)^2 \mu^2 - 1 = 0, \qquad L_2(r, \mu, z) := (4 r^2 - 1)(\mu + 4 r) z + (2 r^2 - 1)^2\mu^2 - 8\mu r^3 + 4\mu r - 16 r^4 + 12 r^2 - 1 = 0. $$

   Along the Universal Triple-Square Section $z = z_{L_1}(r, \mu) := \frac{1 - (2 r^2 - 1)^2\mu^2}{(4 r^2 - 1)\mu}$ (and symmetrically on $L_2 = 0$), all three residual conditions hold identically in $\mathbb{Q}(r, \mu)$ without imposing $E_z = 0$ (verify_paper.py):

   $$ c \, c^*_- + 1 = \left(\frac{4 \mu^2 r^4 - 2 \mu^2 r^2 + 1}{\mu(4 r^2 - 1)}\right)^2, \quad c \, c^*_+ + 1 = \left(\frac{4 \mu^2 r^4 - 2 \mu^2 r^2 - 2 \mu r - 1}{\mu K}\right)^2, \quad c^*_- \, c^*_+ + 1 = \left(\frac{4 \mu^3 r^5 - 4 \mu^3 r^3 - 4 \mu^2 r^4 - 2 \mu^2 r^2 + \mu r + 1}{\mu K}\right)^2. $$

   Consequently, eliminating $z$ between $E_z = 0$ and $R_5 = 0$ on $\mathcal{S}_{4, 2}$ factors over $\mathbb{Q}[r, \mu]$ into $6$ irreducible factors ($2 \times (10, 5)$ from $L_1$, $2 \times (10, 5)$ from $L_2$, $2 \times (20, 10)$ from $Q_4$), pulling back under $\mu = \frac{\lambda^2 - 1}{\text{den}_-}$ to the $8$ irreducible factors over $\mathbb{Q}[r, \lambda]$ of bidegrees $2 \times (10, 10), 2 \times (12, 10), 4 \times (11, 10)$ observed mod $10007$ in SFh1Km0bjqNsezCf (exp17), while the $L_1$ factors descend in $(R, v) := (r^2, r\mu)$ to bidegree-$(3, 5)$ curves $\mathcal{C}_{1a}(R, v) = 0, \mathcal{C}_{1b}(R, v) = 0$ (verify_paper.py, exp17). Substituting $z = \frac{t^2 - 1}{2 t - \mu}$ into $Q_4(r, \mu, z)$ splits $(2 t - \mu)^4 Q_4 = F_a(r, \mu, t) F_b(r, \mu, t)$ where $F_b(r, \mu, t) = \frac{(2 t - \mu)^4}{\mu^2 - 4} F_a\!\left(r, \mu, \frac{\mu t - 2}{2 t - \mu}\right)$, and in $(R, v, u) := \left(r^2, r\mu, \frac{t - \mu}{r}\right)$, $F_a = 0$ is $Q_{4, +}(R, v, u) = (4 R^2 u^2 - 8 R u^2 + 8 R u - 4 R + u^2)v^2 + 2(4 R^3 u^3 - 8 R^2 u^3 + 6 R^2 u^2 - 4 R^2 u + R u^3 - R u^2 + 2 R + u - 1)v - R u(4 R^2 u^3 - R u^3 + 6 R u^2 - 4 R u - 3 u + 2) = 0$ with $\operatorname{disc}_v(Q_{4, +}) = 4(2 R u - 1)^2 D_4(R, u)$ where $D_4(R, u) = F_1(R, u) F_2(R, u) = \big((R - 1)u^2 + 2 u - 1\big)\big(R(4 R^2 - 8 R + 1)u^2 + 8 R^2 u - (2 R - 1)^2\big)$. On $Q_{4, +}(R, v, u) = 0$, all three residual conditions and $\Delta_d^2 = \frac{4 R^2 N_d}{(2 R u + v)^2} \equiv \left(\frac{2 R W_{d, \text{lin}}(R, v, u)}{2 R u + v}\right)^2 \pmod{Q_{4, +}}$ are identically squares while $F_1(R, u) N_{ab} \equiv -(3 R u^2 - 8 R u + 1) W_{d, \text{lin}}^2 \pmod{Q_{4, +}}$, yielding interior $10/10$ regular quintuples on $Q_4 = 0$ ($L_1 L_2 \ne 0$) at $(r, u, v) = \left(\frac{3}{4}, -4, -\frac{9}{28}\right)$, $\left(\frac{7}{4}, \frac{44}{67}, \frac{2009}{3886}\right)$ (all-positive $\left\{\frac{1485}{31423}, \frac{7}{11}, \frac{760}{693}, \frac{176}{63}, \frac{2296}{99}\right\}$), and $\left(\frac{19}{10}, \frac{20}{57}, \frac{19}{45}\right)$ (23,850 points on $\mathcal{E}_{Q_4}(r^2)$ across $224,858,208$ pairs in exp24). Furthermore (verify_paper.py, exp23–exp27):

   • Universal Gauge Involution $\iota_K$, $2$-Isogenous Quotient $\mathcal{E}'_{Q_4}(R)$, and Universal Reciprocal $3 P_0$ Law: In $\mathbb{Q}(r, \mu, z, w)$ before imposing any relation, $\iota_K: (r, \mu, z, w) \mapsto \left(r, -\frac{\mu + 4r}{K}, \frac{4r^2 - 1}{K}z, \frac{4r^2 - 1}{K}w\right)$ satisfies $K(\mu')K(\mu) = (4r^2 - 1)^2$, $T(\mu') = \frac{4r^2 - 1}{K}T(\mu)$, and $H(r, \mu') = \left(\frac{4r^2 - 1}{K}\right)^2 H(r, \mu)$, fixes $(a, b, c, d_-, d_+)$ identically, swaps $c^*_- \leftrightarrow c^*_+$, exchanges $L_1 \leftrightarrow L_2$ ($L_1' = -\frac{4r^2 - 1}{K^2}L_2, L_2' = -\frac{(4r^2 - 1)^3}{K^2}L_1$), and preserves $Q_{4, +}(R, v, u) = 0$ via $\tau_{T_0}: u \mapsto \frac{u - 2}{2 R u - 1}$ ($\sigma := \frac{R u^2 - 1}{2 R u - 1}$). Under the Weierstrass map from $\mathcal{E}_{Q_4}(R): Y_u^2 = D_4(R, u)$ (origin $O_+ = (0, 2R - 1)$) to $E(R): Y_E^2 = X\big(X^2 + 2(8R^3 - 8R^2 + 6R - 1)X - (4R - 1)^3\big)$ (sending $(2, +(4R-1)(2R-1)) \mapsto T_0 = (0, 0)$ and $(2, -(4R-1)(2R-1)) \mapsto P_{\text{even}}(R) = (-(2R-1)^2(4R-1), 4R^2(2R-1)(4R-1))$), $\tau_{T_0}$ is translation by $T_0 = (0, 0)$, and with $Y_\sigma := \frac{R Y_u(R u^2 - u + 1)}{(2 R u - 1)^2}$ the $2$-isogenous quotient $E'(R) = \mathcal{E}_{Q_4}(R)/\langle \tau_{T_0}\rangle$ is $\mathcal{E}'_{Q_4}(R): Y_\sigma^2 = ((R - 1)\sigma + 1)((4R^2 - 8R + 1)\sigma + 4R)(R\sigma^2 - \sigma + 1)$. Along the unit-$w$ section $\mu = \frac{2r}{r^2 - 1}$ ($v = \frac{2R}{R - 1}, w = 1$), $d_- + d_+ = 2 B_0$ and $d_- d_+ = \mu(\mu + 4r)$ force the Universal Reciprocal $3 P_0$ Law $d_\pm = d_{3 P_0}(d_\mp, c, c^*_-) := \frac{8(s_1^3 - 4 s_1 s_2 + 8 s_3)}{(s_1^2 - 4 s_2)^2}$ (verify_paper.py, exp26), explaining why at $(r, u_0, v_0) = \left(\frac{7}{4}, -\frac{4}{9}, \frac{98}{33}\right)$ ($\tau_{T_0}$-conjugate to $(44/67, 2009/3886)$) $\left\{\frac{7}{11}, \frac{760}{693}, \frac{176}{63}, \frac{2296}{99}\right\}$ satisfies $Q_{d_+} + S_T = \pm 3 P_0$ on $E_{\{d_-, c, c^*_+\}}$ and $Q_{d_-} + S_T = \mp 3 P_0$ on $E_{\{d_+, c, c^*_+\}}$.
   • Rank-$2$ Subgroup $\Lambda(r) \subset E(r^2)(\mathbb{Q}(r))$, $\hat{E}(r^2)$ Preimage, Boundary Origin $Z_-(r)$, and Complete $2$-Isogeny Descent Exclusions: Over $\mathbb{Q}(r)$ ($R = r^2$), $F_1(r^2, u) = ((r + 1)u - 1)((r - 1)u + 1)$ splits into $W_1(r) = (1/(r + 1), 0) \mapsto \big(-(2r - 1)(2r + 1)^2, -2r(2r - 1)(2r + 1)^2(2r^2 - 1)\big)$ and $W_2(r) = W_1(r) + T_0$ satisfying $2 W_1(r) = P_{\text{even}}(r^2) + T_0 = O_-$ ($X(2 W_1) = \frac{(4r^2 - 1)^2}{(2r^2 - 1)^2}$), while on $\hat{E}(r^2): \hat{Y}^2 = \hat{X}(\hat{X}^2 - 2 A \hat{X} + A^2 - 4 B)$ the constant term factors as $A^2 - 4 B = 16 r^4 Q_+(r) Q_-(r)$ with $Q_\pm(r) = 4 r^4 \pm 8 r^3 + 4 r^2 \mp 2 r - 1$, giving $\hat{P}_{\text{odd}, \pm}(r) = \big(4 r^2 Q_\pm(r), -8 r^2(4 r^2 - 1)Q_\pm(r)\big) \in \hat{E}(r^2)(\mathbb{Q}(r))$ whose dual $2$-isogeny images are $\hat{\phi}(\hat{P}_{\text{odd}, \pm}(r)) = \pm P_{\text{odd}}(r) = \big((4r^2 - 1)^2, \pm 4r^3(4r^2 - 1)^2\big)$ with $P_{\text{odd}}(r) \leftrightarrow \left(\frac{2(r - 1)}{2r^2 - 1}, \frac{(2r - 1)^2(2r^2 - 2r - 1)}{(2r^2 - 1)^2}\right)$, and the boundary section $Z_-(r) := (1/r, 2(2r - 1))$ ($z = \mu, c^*_- = 0$) satisfies $Z_-(r) = W_1(r) + P_{\text{odd}}(r)$ identically in $E(r^2)(\mathbb{Q}(r))$ (verify_paper.py, exp26–exp27). Under $\iota: r \mapsto -r$, $\iota(2 W_1) = +2 W_1$ while $\iota(P_{\text{odd}}) = -P_{\text{odd}}$, proving $\Lambda(r) := \mathbb{Z} W_1(r) \oplus \mathbb{Z} P_{\text{odd}}(r) \oplus (\mathbb{Z}/2\mathbb{Z}) T_0 \cong \mathbb{Z}^2 \oplus \mathbb{Z}/2\mathbb{Z}$. The interior $Q_4 = 0$ quintuple at $(3/4, -4, -9/28)$ is $P_{\text{odd}}(3/4)$, and at $r \in \{7/4, 19/10\}$, good reduction ($\gcd_p \#E(\mathbb{F}_p) = 2$) and mod-$p$ non-double witnesses at $p \in \{11, 13, 17, 29\}$ certify $E(r^2)(\mathbb{Q})_{\text{tors}} = \{\mathcal{O}, T_0\}$ and that $(T_0, W_1(r), P_{\text{odd}}(r), P_r)$ span $\mathbb{F}_2^4 \subset E(r^2)(\mathbb{Q})/2 E(r^2)(\mathbb{Q})$, proving $\operatorname{rank} E(49/16)(\mathbb{Q}) \ge 3$ and $\operatorname{rank} E(361/100)(\mathbb{Q}) \ge 3$ (exp26). Conversely, for $v_\pm(u, Y_u) = \frac{-B_v \pm 2(2 r^2 u - 1)Y_u}{2 A_v}$ of $Q_{4, +}(r^2, v, u) = 0$ (verify_paper.py, exp25):
     1. At $u = 2$, $P_{\text{even}}(r^2) = 2 W_1 + T_0 \leftrightarrow \big(2, -(4r^2 - 1)(2r^2 - 1)\big)$ gives $v_+(P_{\text{even}}) = -4 r^2$, excluded on $2 R u + v = 0$ ($c^*_- = \infty$), whereas $T_0 \leftrightarrow \big(2, +(4r^2 - 1)(2r^2 - 1)\big)$ gives $v_+(T_0) = v_-(P_{\text{even}}) = \frac{2r^2 + 1}{2(r^2 - 1)}$ with $H(r) = -\frac{4r^2 - 1}{4 r^2(r^2 - 1)} = w^2$ (and on $2 W_1 = O_- \leftrightarrow (0, -(2r^2 - 1))$, $v_+ = 0$ is boundary while $v_- = \frac{2r^2 - 1}{2r^2}$ gives $H(r) = -\frac{(r^2 - 1)(4r^2 - 1)}{4 r^6} = w^2$). Both map via $X = 1 - 4r^2$ to $E_{48}^-: y^2 = X(X - 1)(X + 3) = X(X^2 + 2 X - 3)$, where $(0,0), (1,0), (-3,0), (-1,\pm 2), (3,\pm 6)$ span $\operatorname{im}\alpha_{E_{48}^-} = \{\pm 1, \pm 3\}$ ($|\operatorname{im}\alpha| = 4$) while on $\hat{E}_{48}^-: \hat{y}^2 = \hat{X}((\hat{X} - 2)^2 + 12)$, $\hat{d} \in \{-1, -2\}$ are negative definite on $\mathbb{R}$ and $\hat{d} = 2$ ($\hat{N}^2 = 2 M^4 - 4 M^2 e^2 + 8 e^4$) is insoluble mod $9$ and $16$ (verify_paper.py), so $|\operatorname{im}\hat{\alpha}_{E_{48}^-}| = 1$ and $\operatorname{rank} E_{48}^-(\mathbb{Q}) = 0$ (and $\operatorname{rank} E_{48}(\mathbb{Q}) = 0$ on $y^2 = X(X + 1)(X - 3)$). Since $\#E_{48}^-(\mathbb{F}_7) = \#E_{48}^-(\mathbb{F}_{11}) = 8$, $E_{48}^-(\mathbb{Q}) = \{\mathcal{O}, (0,0), (1,0), (-3,0), (-1,\pm 2), (3,\pm 6)\}$, and $4r^2 = 1 - X \in \{1, 0, 4, 2, -2\}$ yields only boundary $r \in \{0, \pm 1/2, \pm 1\}$.
     2. On $3 W_1(r) \leftrightarrow \left(\frac{-2r^3 + 2r - 1}{2r^4 - 2r^3 - 2r^2 + r}, \frac{2r(2r - 1)(2r^2 - 1)}{(2r^3 - 2r^2 - 2r + 1)^2}\right)$ (and $3 W_1 + T_0$), $v_+(3 W_1) = \frac{-2r(r^2 + r - 1)}{2r^3 - 2r^2 - 2r + 1}$ gives $H(r) = \frac{(4r^2 - 1)(2r^2 - 1)(2r^2 - 3)}{(2r^3 - 2r^2 - 2r + 1)^2} = w^2$, mapping via $X = 4r^2 - 1, y = 2(2r^3 - 2r^2 - 2r + 1)w$ to $E_{80}: y^2 = X(X - 1)(X - 5) = X(X^2 - 6 X + 5)$. Here $(0,0), (1,0), (5,0)$ give $\{1, 5\} \subset \operatorname{im}\alpha_{E_{80}}$ while $d \in \{-1, -5\}$ are obstructed on $\mathbb{R}$, so $|\operatorname{im}\alpha_{E_{80}}| = 2$; on $\hat{E}_{80}: \hat{y}^2 = \hat{X}(\hat{X}^2 + 12\hat{X} + 16)$, $(-4, \pm 8)$ represents $\hat{d} = -1$ while $\hat{d} = \pm 2$ ($\hat{N}^2 = \pm 2 M^4 + 12 M^2 e^2 \pm 8 e^4$) are insoluble mod $5$ and $16$ (verify_paper.py), so $|\operatorname{im}\hat{\alpha}_{E_{80}}| = 2$ and $\operatorname{rank} E_{80}(\mathbb{Q}) = 0$. Since $\gcd(\#E_{80}(\mathbb{F}_3), \#E_{80}(\mathbb{F}_7)) = \gcd(4, 12) = 4$, $E_{80}(\mathbb{Q}) = \{\mathcal{O}, (0,0), (1,0), (5,0)\}$, and $4r^2 = X + 1 \in \{1, 2, 6\}$ yields only boundary $r = \pm 1/2$.
   • Joint $(u, v, \Delta_d, \Delta_{ab})$ Gate Surface $G_u(R, d_0) = n^2$ and $u = 2/3$ Conic Decomposition: Since $R F_1 = (2 R u - 1)((R - 1)\sigma + 1)$, $F_2 = (2 R u - 1)((4R^2 - 8R + 1)\sigma + 4R)$, and $3 R u^2 - 8 R u + 1 = (2 R u - 1)(3\sigma - 4)$, setting $k^2 = \frac{(4R^2 - 8R + 1)\sigma + 4R}{(R - 1)\sigma + 1} = 4 R + d_0^2 - 4$ makes $v, \Delta_d, \Delta_{ab} \in \mathbb{Q}$ simultaneously and reduces $u \in \mathbb{Q}$ ($R\sigma^2 - \sigma + 1 = \square$) to $G_u(R, d_0) := R^2 d_0^4 + (16 - 6 d_0^2)R + (d_0^2 - 3) = n^2$ (29 non-boundary $(r, d_0)$ points in exp23, plus chord sections $M_{2, \pm}(r) = \frac{4r^2(r^3 \pm 2r^2 + r \mp 1)}{(r - 1)^2(4r^3 \pm 4r^2 - 1)}$). On $r d_0 = 3$ ($u = 2/3$ and $\tau_{T_0}(2/3) = -\frac{4}{4R - 3}$), $G_u(r^2, 3/r) = \left(\frac{4r^2 + 3}{r}\right)^2$ holds identically, and parameterizing $Y_k^2 = 4 R^2 - 4 R + 9$ by $R(t) = \frac{(t - 2)(t + 4)}{4 t}, Y_k(t) = \frac{t^2 + 8}{2 t}$ splits $Q_{4, +}(R(t), v, 2/3) = 0$ into $v_+(t) = \frac{(t + 4)^2}{3 t(t + 2)}, v_-(t) = -\frac{(t - 2)^2}{3(t - 4)} = v_+(-8/t)$ with $H(v_-(t)) = \frac{(t + 1)(t^3 + t^2 + 8 t - 64)}{9(t - 4)(t + 4)}$ (verify_paper.py).

2. Decoupling $L_1 = 0$ from $\mathcal{S}_{4, 2}$, the $15 + 6$ Square-Class Collapse, the $(v, t)$ blow-up, and the Factored $(a, b)$-Split Law: For any $(r, \mu, z)$ with $H(r, \mu) = w^2$, $\Delta_{ab}^2 := A_0^2 - 4(r^2 - 1) H$, and $\Delta_d^2 := B_0^2 - \mu(\mu + 4 r)$, $\{a, b\} \times \{c, d_-, d_+\}$ obey modulo $\Delta_{ab}^2, \Delta_d^2$ (verify_paper.py):

   $$ (a c + 1)(b c + 1) = \left(\frac{\Delta_d}{2 r w}\right)^2, \qquad w^6(a d_\mp + 1)(b c + 1) = \left(\frac{A_0 + \Delta_{ab}}{2}\left(\frac{A_0 - \Delta_{ab}}{2}z + H\right) \mp \frac{H \Delta_d}{2}\right)^2. $$

   On $z = z_{L_1}(r, \mu)$ in $(R, v) = (r^2, r\mu)$ ($r^2 H = (v - 1)^2 R - v^2 = (r w)^2$, parameterized for fixed $v$ by $r(\lambda) = \frac{\lambda^2 + 1}{\lambda^2 - 1}\frac{v}{v - 1}$), $\Delta_{ab}^2 = \frac{R \, \mathcal{F}_{ab}(R, v)}{(4 R - 1)^2 v^2}$ and $\Delta_d^2 = \frac{4 R^2 \, \mathcal{G}_d(R, v)}{(4 R - 1)^2 v^2}$, where

   $$ \begin{aligned} \mathcal{F}_{ab}(R, v) &= 16 v^3(5 v - 4) R^2 - 8 v(10 v^3 - 2 v^2 - 5 v - 2) R + (16 v^4 - 12 v^2 - 4 v + 1) = v^3(5 v - 4)(4 R - 1)^2 - 2 v(v^2 - 1)(5 v + 2)(4 R - 1) + (v^2 - 1)^2, \\ \mathcal{G}_d(R, v) &= 16 v^4 R^3 - 8 v^2(4 v + 1) R^2 + (-16 v^4 + 8 v^3 + 16 v^2 + 8 v + 1) R + 2 v(2 v^3 - 2 v - 1). \end{aligned} $$

   On the admissible open locus where $r v (v^2 - 1)(4 R - 1) K H \Delta_d \ne 0$ and all tuple entries are nonzero and distinct, without imposing $E_z = 0$:

   • Whenever $\mathcal{G}_d(R, v) = \square$ and $H = w^2$: $\{c, d_-, d_+, c^*_-, c^*_+\} \subset \mathbb{Q}$ is a $10/10$ rational regular quintuple ($R_5(d_-, d_+, c, c^*_-, c^*_+) = 0$), giving at $(r, \lambda, v) = \left(\frac{7}{4}, \frac{3}{11}, \frac{98}{163}\right)$, $\left(\frac{5}{3}, \frac{13}{11}, -\frac{8}{21}\right)$, $\left(\frac{53}{30}, \frac{2}{7}, \frac{3}{5}\right)$ the three $10/10$ regular quintuples $\left\{\frac{620}{231}, \frac{266}{33}, \frac{84}{11}, \frac{33396}{185983}, -\frac{2340}{22253}\right\}$, $\left\{-\frac{643}{24024}, \frac{448}{3861}, -\frac{1053}{154}, -\frac{26741}{74088}, -\frac{1463}{312}\right\}$, $\left\{-\frac{476157}{180880}, -\frac{167808}{14875}, -\frac{10625}{2128}, -\frac{7644}{40375}, \frac{6783}{76880}\right\}$ (verify_paper.py; 15 $14/15$ sextuples, 30 $18/21$ septuples in exp19). Over $\mathbb{Q}(v)$, $\mathcal{G}_d(R, v) = W_d^2$ has $P_1(v) = \left(\frac{1}{4 v^2}, 2(v^2 - 1)\right)$, whose multiples $3 P_1(v) = \left(\frac{1 + 4 v - 4 v^2}{4 v^2}, -2 v(v - 1)\right)$ and $4 P_1(v) = \left(\frac{8 v^2 + 8 v + 1}{4 v^2}, -\frac{2(v + 1)(6 v^2 + 3 v + 1)}{v}\right)$ reduce $\{R(v) = r^2, \mathcal{F}_{ab}(R(v), v) = \square\}$ to $\mathcal{E}_{3 P_1} : Y_m^2 = (2 - m)(m^2 - 11 m + 26)$ and $\mathcal{E}_{4 P_1} : Y_m^2 = -m^4 + 12 m^3 + 12 m^2 - 448 m + 1024$, while $R(8 P_1(v)) = \frac{84 v^4 + 124 v^3 + 69 v^2 + 14 v + 1}{4 v^2(5 v + 1)^2}$ gives $R(8 P_1(3/5)) = \left(\frac{53}{30}\right)^2$ and the third quintuple above (verify_paper.py, exp19). For any regular $10/10$ quintuple $Q_4 \cup \{y\}$ ($R_5(Q_4, y) = 0, p_4 \ne 1$), $R_5(Q_4, X) = (p_4 - 1)^2 X^2 - 2((p_4 + 1)s_1 + 2 s_3) X + (s_1^2 - 4 s_2 - 4 p_4 - 4) = 0$ shows that $y' = \frac{2((p_4 + 1)s_1 + 2 s_3)}{(p_4 - 1)^2} - y$ satisfies $y y' + 1 = \frac{s_1^2 + p_4^2 - 4 s_2 - 6 p_4 - 3}{(p_4 - 1)^2}$ (at $(7/4, 98/163)$ with $y = c = \frac{620}{231}$, $y' = -\frac{11370485030264360}{100061732560025883}$ and $y y' + 1 = \frac{208630642890936049}{300185197680077649}$; verify_paper.py), so for the three sub-quadruples containing $\{d_-, d_+\}$, $c', c^{*\prime}_\mp \in \mathbb{Q}(r, v, w)$ and $c c' + 1, c^*_+ c^{*\prime}_+ + 1, c^*_- c^{*\prime}_- + 1$ descend to $(R, v)$-rational functions whose squarefree parts have irreducible factors of bidegrees $(7, 12), (6, 10), (6, 10)$ times $R$ (and $(v - 1)^2 R - v^2 = r^2 H$ in the first case; verify_paper.py, exp21). At $(53/30, 3/5)$, the $3$-generator sub-triple coset scan (exp26) on $E_{\{d_-, c^*_-, c^*_+\}}$ uncovers $Q_{e_2} = P_0 + 2 Q_c + S_T$ with $e_2 = -\frac{4580748}{6812155}$, compatible with $d_+ = -\frac{10625}{2128}$ ($d_+ e_2 + 1 = \left(\frac{3127}{1498}\right)^2, d_- e_2 + 1 = \left(\frac{932747}{318325}\right)^2$), giving a $3$-fifth $18/21$ septuple $\{d_-, d_+, c^*_-, c^*_+, c, c', e_2\}$ (verify_paper.py) where $c, c'$ are regular fifths and $e_2$ is totally irregular on $\{d_-, d_+, c^*_-, c^*_+\}$.
   • Universal $(r, M)$ square completions, $(v, t)$ blow-up, and the Factored $(a, b)$-Split Law: In $(r, M)$ where $v = \frac{r}{r - M}$ and $M = \frac{\lambda^2 + 1}{\lambda^2 - 1}$ ($M^2 - N^2 = 1, N = \frac{2\lambda}{\lambda^2 - 1}$ so $H = w^2$), $(r - M)^4 \mathcal{F}_{ab}(r^2, v)$ and $(r - M)^4 \mathcal{G}_d(r^2, v)$ obey (verify_paper.py):$$ \begin{aligned} (r - M)^4 \mathcal{F}_{ab}\!\left(r^2, \frac{r}{r - M}\right) &= \Big(-M^2 + 8 r^3 M + 4 r^4 - 9 r^2 + 2\Big)^2 - 4(r^2 - 1)(4 r^2 - 1)^2(M^2 - 1), \\ (r - M)^4 \mathcal{G}_d\!\left(r^2, \frac{r}{r - M}\right) &= \left(r M^2 - (6 r^2 - 1)M - \frac{8 r^6 - 10 r^4 - 2 r^2 + 1}{2 r}\right)^2 + \left(\frac{4 r^2 - 1}{2 r}\right)^2\big(4 r M - (4 r^2 + 1)\big), \end{aligned} $$and $\mathcal{F}_{ab}(r^2, v) = \square$ holds on $M^2 - N^2 = 1$ iff $\exists k \in \mathbb{Q}^\times$ with $4 k r^3(2 M + 4 r^3 - 5 r) = (N + k(4 r^2 - 1)(r - 1))(k N + (4 r^2 - 1)(r + 1))$ (verify_paper.py). Dually, setting $W_{ab} = (v^2 - 1) - v^2 t(4 R - 1)$ on $\mathcal{F}_{ab}(R, v) = W_{ab}^2$ gives $R = \frac{N_R}{4 v^2 D_t}, r^2 H = \frac{L_a L_b}{4 v^2 D_t}, \mathcal{G}_d = \frac{(v^2 - 1)^2 A_t B_t}{4 D_t^3}$ where $D_t = (t^2 - 5)v + 4, N_R = D_t + (t - 3)(t + 5)v(v^2 - 1), A_t = (t^3 - 5t^2 - 13t + 25)v - 2(t + 1)^2, B_t = (t^3 - 21t^2 + 35t + 25)v^2 - 2(7t^2 - 2t - 25)v - 2(t + 1)^2, L_a = (t + 1)v^2 + 2v - 2, L_b = -(v + 1)((3t - 5)v^2 - (t - 1)v + 2)$ (verify_paper.py), and along $v(t) = \frac{2(t - 7)}{t^2 + 4t - 13}, r(t) = \frac{3t + 11}{2(t - 7)}$ of $D_t N_R = Y_R^2$, $\mathcal{G}_d(t) = \square$ reduces to $\mathcal{E}_{\text{tan}} : Y_t^2 = (t^2 + 30 t + 253)(9 t^2 + 14 t - 27)$ (exp20).
   • Whenever $\mathcal{F}_{ab}(R, v) = \square$ and $H = w^2$: $\{a, b, c, c^*_-, c^*_+\} \subset \mathbb{Q}$ satisfies $8/10$ conditions (all except $(a, c), (b, c)$), and $Q_{ab} = \{a, b, c^*_-, c^*_+\}$ is a $6/6$ irregular quadruple whose $6$-pair product is an exact square on $L_1 = 0$, so both regular fifths $e_{ab, \mp} \in \mathbb{Q}$ are rational! At the $9$ explicit non-degenerate $(v \ne -1/5)$ rational points $(v, r) \in \left\{\left(\frac{60}{31}, 3\right), \left(\frac{8}{13}, \frac{61}{28}\right), \left(\frac{3}{37}, \frac{5}{24}\right), \left(-\frac{23}{123}, \frac{29}{8}\right), \left(\frac{18}{167}, \frac{6}{17}\right), \left(-\frac{399}{2011}, \frac{19}{14}\right), \left(\frac{161}{127}, \frac{17}{2}\right), \left(\frac{273}{562}, \frac{39}{23}\right), \left(-\frac{23}{3}, \frac{193}{134}\right)\right\}$ (verify_paper.py), this produces 18 regular $10/10$ quintuples, 81 $14/15$ sextuples, and 180 $18/21$ septuples (exp21).
   • Whenever both $\mathcal{F}_{ab}(R, v) = \square$ and $\mathcal{G}_d(R, v) = \square$ hold on the admissible open locus: $15/21$ pairs of $\{a, b, c, d_-, d_+, c^*_-, c^*_+\}$ are automatically rational squares, and all $6$ remaining nonzero radicands $a c + 1, b c + 1, a d_- + 1, a d_+ + 1, b d_- + 1, b d_+ + 1$ lie in the exact same square class in $\mathbb{Q}^\times / \mathbb{Q}^{\times 2}$.

3. Closed-form $\mathbb{Q}(\lambda)$ and $\mathbb{Q}(u)$ families at $v = -1/5$, and rank-$0$ / torsion proof on $\mathcal{E}_{16}$: Since $\operatorname{disc}_R(\mathcal{F}_{ab}) = 256 v^2(v - 1)^2(v + 1)^3(5 v + 1)$, setting $v = -1/5$ ($r(\lambda) = \frac{\lambda^2 + 1}{6(\lambda^2 - 1)}$) makes $\mathcal{F}_{ab}(R, -1/5) = \frac{(20 R - 29)^2}{625}$ an identical square in $\mathbb{Q}(\lambda)$, yielding (verify_paper.py):

   $$ \begin{aligned} a(\lambda) &= -\frac{\lambda(5\lambda^2 - 7)(7\lambda^2 - 5)}{8(\lambda^2 - 1)(\lambda^2 - 2)(2\lambda^2 - 1)}, \qquad b(\lambda) = \frac{2(\lambda^2 - 2)(2\lambda^2 - 1)}{9\lambda(\lambda^2 - 1)}, \qquad c(\lambda) = \frac{(\lambda^4 - 19\lambda^2 + 16)(16\lambda^4 - 19\lambda^2 + 1)}{72\lambda(\lambda^2 - 1)(\lambda^2 - 2)(2\lambda^2 - 1)}, \\ c^*_-(\lambda) &= -\frac{2\lambda(7\lambda^2 - 8)(8\lambda^2 - 7)}{25(\lambda^2 - 1)(\lambda^2 - 2)(2\lambda^2 - 1)}, \qquad c^*_+(\lambda) = -\frac{2\lambda(\lambda^2 - 2)(2\lambda^2 - 1)(4\lambda^2 - 5)(5\lambda^2 - 4)}{(\lambda^2 - 1)(7\lambda^4 - 13\lambda^2 + 7)^2}. \end{aligned} $$

   The regular fifths $e_{ab, \mp}(\lambda)$ of $\{a(\lambda), b(\lambda), c^*_-(\lambda), c^*_+(\lambda)\}$ lie in $\mathbb{Q}(\lambda)$:

   $$ e_{ab, -}(\lambda) = \frac{(\lambda^2 - 1)^2(5\lambda^2 + 3\lambda - 7)(7\lambda^2 + 3\lambda - 8)(7\lambda^2 + 3\lambda - 5)(8\lambda^2 + 3\lambda - 7)(5\lambda^3 - 7\lambda^2 - 7\lambda + 8)(8\lambda^3 + 7\lambda^2 - 7\lambda - 5)}{8\lambda(\lambda^2 - 2)(2\lambda^2 - 1)(105\lambda^6 + 70\lambda^5 - 300\lambda^4 - 139\lambda^3 + 300\lambda^2 + 70\lambda - 105)^2}, $$

   and $e_{ab, +}(\lambda) = -e_{ab, -}(-\lambda)$ (or in $(r, W_0)$ with $W_0 = 5 r w, W_0^2 = 36 r^2 - 1$, $r(u) = \frac{u^2 + 1}{12 u}, W_0(u) = \frac{u^2 - 1}{2 u}$ via $a = \frac{4 r^2 - 1}{W_0}, b = \frac{(r^2 - 1)W_0}{4 r^2 - 1}, c^*_- = \frac{(4 r^2 - 25)W_0}{25(4 r^2 - 1)}, c^*_+ = \frac{(4 r^2 - 9)(4 r^2 - 1)W_0}{(4 r^2 + 3)^2}$, generating with their $8$ Vieta fifths $2$ $\mathbb{Q}(u)$-families of $10/10$ quintuples, $9$ of $14/15$ sextuples, and $20$ of $18/21$ septuples; 7,350 quintuples, 33,075 $14/15$ sextuples, 73,500 $18/21$ septuples in exp21). For $\lambda \in \mathbb{Q} \setminus \{-1, 0, 1\}$, $\{a(\lambda), b(\lambda), c^*_-(\lambda), c^*_+(\lambda), e_{ab, -}(\lambda), e_{ab, +}(\lambda)\}$ is a valid $14/15$ sextuple (missing $(e_{ab, -}, e_{ab, +})$), while $\{a(\lambda), b(\lambda), c^*_-(\lambda), c^*_+(\lambda), e_{ab, -}(\lambda)\}$ is strictly positive for $\lambda \in \left(\sqrt{7/8}, 1\right) \approx (0.935414, 1)$, giving at $\lambda = 15/16$ the all-positive $10/10$ quintuple $\left\{\frac{181632}{4315045}, \frac{27839}{16740}, \frac{2951475}{863009}, \frac{1025722387200}{128276850799}, \frac{57928168219537}{181859221980}\right\}$ (verify_paper.py). Finally, $b(\lambda) c(\lambda) + 1 = \left(\frac{\lambda^2 + 1}{18\lambda(\lambda^2 - 1)}\right)^2(16\lambda^4 - 31\lambda^2 + 16) = \square$ maps via $X = 16\lambda^2, Y = 16\lambda Y_\lambda$ to $E_{16} : Y^2 = X(X^2 - 31 X + 256)$, whose dual $\hat{E}_{16} : \hat{Y}^2 = \hat{X}(\hat{X} - 1)(\hat{X} + 63)$ splits over $\mathbb{Q}$. Modulo $16$ and $9$, $w^2 = d u^4 + 62 u^2 v^2 - \frac{63}{d}v^4$ ($d \mid -63$) are insoluble for $d \in \{-1, \pm 3, 7, \pm 21\}$ (verify_paper.py), leaving $|\operatorname{im}\alpha_{\hat{E}_{16}}| = 2$ ($d \in \{1, -7\}$) and $|\operatorname{im}\alpha_{E_{16}}| = 2$ ($d \in \{1, 2\}$, represented by $P_8 = (32, 96)$ of order $8$), proving $\operatorname{rank} E_{16}(\mathbb{Q}) = 0$. Since $\#E_{16}(\mathbb{F}_5) = \#E_{16}(\mathbb{F}_{13}) = 8$ (verify_paper.py), $E_{16}(\mathbb{Q}) = E_{16}(\mathbb{Q})_{\text{tors}} \cong \mathbb{Z}/8\mathbb{Z}$ is generated by $(32, 96)$, and $\mathcal{E}_{16} : Y_\lambda^2 = 16\lambda^4 - 31\lambda^2 + 16$ has exactly $8$ rational points, all on the boundary: $(\lambda, Y_\lambda) \in \{(0, \pm 4), (1, \pm 1), (-1, \pm 1), \infty_+, \infty_-\}$.

Proof. Verified symbolically in verify_paper.py and exp17–exp27. $\square$

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9. Universal $(V, K, d)$ coordinates, the $3$-Torsion Inflection Criterion ($3 S_T = \mathcal{O}$), the Universal Orbit Cubic, the Eisenstein Triad Curve $\mathcal{C}_{\text{triad}}$, and Cross-Coset Bridge Laws

We record the structure of contiguous $4$-term translation chains $x(P_0 + n S)$ on $E_T : Y^2 = (X + r^2 - 1)(X + s^2 - 1)(X + t^2 - 1)$ ($X = \Pi x$) of any Diophantine triple $T = \{a, b, c\}$ with $\Pi = a b c, \Omega = r s t, P_0 = (0, \Pi)$, and $S = (1, \Omega) = 2 R_0$ ($R_0 = (1 + r s + r t + s t, (r + s)(r + t)(s + t))$), and combine the $S_T$-translation dictionary [7, 10, 17] with $(V, K, d)$ on the $3$-torsion inflection surface $\mathcal{S}_{\text{3-tors}}$ ($3 S_T = \mathcal{O}$).

Theorem 7 (Universal $(V, K, d)$ Coordinates, the $3$-Torsion Inflection Surface $\mathcal{S}_{\text{3-tors}}$, the Eisenstein Triad Curve $\mathcal{C}_{\text{triad}}$, and Cross-Coset Bridge Laws)

Let $T = \{a, b, c\}$ be any Diophantine triple with $d_\mp = x(P_0 \pm S) = a + b + c + 2 \Pi \mp 2 \Omega$ and $W_\mp := \Pi d_\mp - 1$.

1. Shift-invariant product and universal square completion of $F_2(z)$: For any $Q \in E_T$ with $z = x(Q)$, $x(Q - S) x(Q + S) = \frac{(z - d_+)(z - d_-)}{(\Pi z - 1)^2}$, so $x(Q - S) x(Q + S) + 1 = \frac{F_2(z)}{(\Pi z - 1)^2}$ where $F_2(z) := (\Pi z - 1)^2 + (z - d_+)(z - d_-)$ satisfies

   $$ W_-^2 \, F_2(z) = \Big(W_-^2 + (\Pi W_- - 2 \Omega)(z - d_-)\Big)^2 + (z - d_-)^2 D_4, $$

   with $D_4 = W_- W_+ - 4 \Omega^2 = 4 \Omega^2 d$ and $d := X(2 S) - 1 = \frac{W_- W_+}{4 \Omega^2} - 1 = -\frac{(\alpha + \beta + \gamma)(\alpha + \beta - \gamma)(\alpha - \beta + \gamma)(-\alpha + \beta + \gamma)}{4 \alpha^2 \beta^2 \gamma^2}$ in $(\alpha, \beta, \gamma) = (1/r, 1/s, 1/t)$.

2. Closed-form chain coordinates and contiguous $4$-term chains: In $u_0 := \frac{W_-}{2 \Omega}$ ($V := u_0^2$), $K := W_- - d = \Pi d_- - X(2S)$, and $d = X(2S) - 1$, we have $\Pi = \frac{d + 1 - u_0^2}{2 u_0}$, $\Omega = \frac{K + d}{2 u_0}$, and for all $m, n \in \mathbb{Z} \setminus \{0\}$, $x_m x_n + 1 = \frac{(d + 1 - V)^2 + 4 V X_m X_n}{(d + 1 - V)^2}$, where $X_n = \Pi x(P_0 + n S)$ are:

   $$ X_1 = K + d + 1, \quad X_2 = \frac{K + V}{V}, \quad X_3 = \frac{d\big(K(V - 1) + V d\big)}{K^2}, \quad X_4 = \frac{(K^2 + K - V d)\big(K(d + 1) - V d\big)}{(K - V d)^2}. $$

   For $\mathcal{S}_7^{(1, 2, 3, 4)} = \{a, b, c, d_-, x_2, x_3, x_4\}$, $18/21$ conditions hold identically, and $d_- x_3 + 1 = \square, x_2 x_4 + 1 = \square$ reduce to $\text{num}_{19}(V, K, d) = \big(K(V + d - 1) + 2 V d\big)^2 + 4 d \big(K(V - 1) + V d\big)^2 = \square$ and $\text{num}_{20}(V, K, d) = \big(2 K^2 + (K - V d)(V - d + 1)\big)^2 + 4 d \big(K^2 + K - V d\big)^2 = \square$.

3. The $3$-Torsion Inflection Criterion ($3 S_T = \mathcal{O}$), Tate Normal Form, Universal Orbit Cubic, and Sub-Quadruple Square-Root Identity: By the translation dictionary [7, 10, 17] (verify_paper.py), for any $Q = (\Pi x_4, Y_4) \in E_T$ ($p_4 = \Pi x_4 \ne 1, Q_4 = \{a, b, c, x_4\}$), $x(Q - S_T) + x(Q + S_T) = \frac{2((p_4 + 1)s_1 + 2 s_3)}{(p_4 - 1)^2}$ and $x(Q - S_T) x(Q + S_T) = \frac{(x_4 - d_+)(x_4 - d_-)}{(p_4 - 1)^2} = \frac{s_1^2 - 4 s_2 - 4 p_4 - 4}{(p_4 - 1)^2}$, so $(p_4 - 1)^2(X - x(Q - S_T))(X - x(Q + S_T)) \equiv R_5(a, b, c, x_4, X)$ on $E_T$, and for any regular $10/10$ quintuple $Q_5 = (x_1, \dots, x_5)$ with $T_{ij} = Q_5 \setminus \{x_i, x_j\}$ and $Q_j = Q_i + S_{T_{ij}}$, $\tau_j(x_j) = x(Q_i - S_{T_{ij}})$ and $\tau_i(x_i) = x(Q_i + 2 S_{T_{ij}})$, with $\tau_j(x_j) = 0 \iff x_i \in \{d_-(T_{ij}), d_+(T_{ij})\}$. In $U := X - 1 = \Pi x - 1$, $E_T$ is $Y^2 = (U + r^2)(U + s^2)(U + t^2)$ with $S_T = (0, \Omega)$ and tangent slope $\lambda_0 := \frac{r^2 s^2 + r^2 t^2 + s^2 t^2}{2 \Omega}$, so $3 S_T = \mathcal{O} \iff 2 S_T = -S_T \iff d = 0$ if and only if (verify_paper.py):

   $$ \lambda_0^2 - (r^2 + s^2 + t^2) = -\frac{(r s + r t + s t)(r s + r t - s t)(r s - r t + s t)(-r s + r t + s t)}{4 r^2 s^2 t^2} = 0 \iff \pm \frac{1}{r} \pm \frac{1}{s} \pm \frac{1}{t} = 0. $$

   Without loss of generality, let $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = 0$ ($r s + r t + s t = 0$), so $\lambda_0 := r + s + t = -\frac{r^2 s^2 + r^2 t^2 + s^2 t^2}{2 \Omega}$ satisfies $\lambda_0^2 = r^2 + s^2 + t^2$. Then (verify_paper.py, exp28–exp32):

   • Tate Normal Form, Pythagorean Base-Orbit Law & Torsion $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$: On $3 S_T = \mathcal{O}$ [7], $E_T$ has Tate normal form $Y^2 = U^3 + (\lambda_0 U - \Omega)^2$, $\Pi^2 + 1 = (\Omega + \lambda_0)^2$, $R_0 = (0, -\Omega) = -S_T$, the base coset $P_0 + \langle S_T \rangle = \{P_0, P_0 - S_T, P_0 + S_T\}$ closes into a $3$-cycle whose nonzero abscissas $\{d_-, d_+\}$ obey the Pythagorean Base-Orbit Law $U(d_\mp) = 2\Omega(\Omega + \lambda_0 \mp \Pi) = -2\Omega\kappa^{\mp 1}$ ($\kappa := \Pi - \Omega - \lambda_0$, $U(d_-)U(d_+) = 4\Omega^2$) and form a $10/10$ quintuple $T \cup \{d_-, d_+\}$ with $\tau_4(d_-) = \tau_5(d_+) = 0$ (as in $Q_{\text{3tors}} = \left\{-\frac{289}{252}, -\frac{108}{119}, \frac{49}{68}, -\frac{572}{1071}, \frac{928}{1071}\right\}$ and its $3 P_0$ sextuple $\left\{-\frac{289}{252}, -\frac{108}{119}, \frac{49}{68}, \frac{57469402289728}{145393246858671}, \frac{843393140604}{1028879549569}, -\frac{83594902600352}{77453118042111}\right\}$; verify_paper.py), and $E_T(\mathbb{Q})_{\text{tors}} \supseteq \langle R_0 + W_r \rangle \times \langle W_s - W_r \rangle \cong \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ ($W_r = (-r^2, 0), W_s = (-s^2, 0), W_t = (-t^2, 0)$).
   • Universal $3$-Torsion Orbit Cubic, Positivity Criterion, and Sub-Quadruple Square-Root Identity: For any $Q_4 = (U_4, Y_4) \in E_T(\mathbb{Q})$ with $U_4(U_4 + 1) \ne 0$, its $S_T$-coset $\{Q_4, Q_4 + S_T, Q_4 - S_T\}$ has $U$-coordinates $\{U_4, U_5, U_6\}$ satisfying$$ U_{5, 6} = \frac{-2\Omega(\lambda_0 U_4 - \Omega \mp Y_4)}{U_4^2}, \qquad U_4 U_5 U_6 = -4\Omega^2, \qquad P_{\text{coset}}(U') := U'^3 - \Sigma_U \, U'^2 - 4\Omega\lambda_0 \, U' + 4\Omega^2 = 0, $$where $\Sigma_U := U_4 + U_5 + U_6 = \frac{U_4^3 - 4\Omega\lambda_0 U_4 + 4\Omega^2}{U_4^2}$. Whenever $Q_4 \in P_0 + 2 E_T(\mathbb{Q})$ ($U_4 \notin \{0, -1, U(d_-), U(d_+)\}$), $S_6(Q_4) := T \cup \{x_4, x_5, x_6\}$ is a $15/15$ rational Diophantine sextuple [7, 10, 17]. Writing $u := \frac{\Pi}{r_u^2 - 1}$ for $(u, r_u) \in \{(a, t), (b, s), (c, r)\}$ (so $u x_i + 1 = \frac{U_i + r_u^2}{r_u^2 - 1}$), the $9$ cross-products with $T$ satisfy the Universal Sub-Quadruple Square-Root Identity (verify_paper.py, exp31):$$ (u x_4 + 1)(u x_5 + 1)(u x_6 + 1) = \frac{-P_{\text{coset}}(-r_u^2)}{(r_u^2 - 1)^3} = -\left(\frac{u}{\Pi}\right)^3 P_{\text{coset}}(-r_u^2) = \left(\frac{u r_u(r_u U_4 - 2\Omega)}{\Pi U_4}\right)^2\!(u x_4 + 1), $$since $U_4^2(r_u^6 + \Sigma_U r_u^4 - 4\Omega\lambda_0 r_u^2 - 4\Omega^2) = (U_4 + r_u^2)(r_u^2 U_4 - 2\Omega r_u)^2$. Hence every $2+2$ sub-quadruple $Q = (T \setminus \{w\}) \cup \{x_i, x_j\}$ and every $1+3$ sub-quadruple $Q' = \{u, x_4, x_5, x_6\}$ has a square $6$-pair product; because $w \in T$ is already one regular fifth of $Q$ ($R_5(a,b,c,x_i,x_j) = 0$), Vieta reflection $e_{\text{new}}(Q) = \frac{2((P_4 + 1)S_1 + 2 S_3)}{(P_4 - 1)^2} - w$ gives $9$ rational $19/21$ septuple candidates from the $9$ $2+2$ sub-quadruples and $e_\mp(Q')$ gives $6$ from the $3$ $1+3$ sub-quadruples ($15$ rational type-$(0, 4)$ $19/21$ septuples in total; verify_paper.py, exp28). Moreover, since $-4\Omega\lambda_0 = 2(r^2 s^2 + r^2 t^2 + s^2 t^2) > 0$ and $4\Omega^2 > 0$, Descartes' rule of signs on $P_{\text{coset}}(-U'')$ implies every real coset $\{U_4, U_5, U_6\}$ has two positive roots and one negative root $U_{\text{neg}} < 0$, with $(U_4+1)(U_5+1)(U_6+1) = -P_{\text{coset}}(-1) = \frac{(U_4 + 1)(U_4 - U(d_-))(U_4 - U(d_+))}{U_4^2}$ (verify_paper.py); hence $\Pi x_4, \Pi x_5, \Pi x_6 > 0$ iff $-1 < U_{\text{neg}} < 0 \iff (U_4 + 1)(U_4 - U(d_-))(U_4 - U(d_+)) > 0$.

4. Global Geometry of $\mathcal{S}_{\text{3-tors}}$, the Mordell–Weil Hierarchy $C_r(\mathbb{Q}(r))$, and Quadratic All-Positive Family IV ($N = 3\beta + \gamma$): Over $\mathbb{Q}(r)$, substituting $t = -\frac{r s}{r + s}$ and $Y_r = (r + s)\Pi$ into $(r^2 - 1)(s^2 - 1)(t^2 - 1) = \Pi^2$ gives the quartic elliptic surface (verify_paper.py, exp28, exp30):

   $$ C_r : \quad Y_r^2 = (r^2 - 1)(s^2 - 1)\big(r^2 s^2 - (r + s)^2\big) = (r^2 - 1)(s^2 - 1)(r s - r - s)(r s + r + s), $$

   with origin $W_2 = (-1, 0)$, $C_r(\mathbb{Q}(r))[2] = \left\{W_2, \; T_1=(1,0), \; T_2=\left(-\frac{r}{r+1}, 0\right), \; T_3=\left(\frac{r}{r-1}, 0\right)\right\} \cong (\mathbb{Z}/2\mathbb{Z})^2$, and infinite-order point $P_\infty$ at $s = \infty$ generating $\mathbb{Z} P_\infty \oplus (\mathbb{Z}/2\mathbb{Z})^2 \subset C_r(\mathbb{Q}(r))$:

   • Level $1$ ($P_\infty + T_1$): $(s(r), t(r)) = \left(-\frac{1}{2r}, \frac{r}{2r^2 - 1}\right)$, $\Pi(r) = \frac{(r^2 - 1)(4r^2 - 1)}{2r(2r^2 - 1)}$ ($a, b, c > 0$ for $r > 1$), where $3 P_0$ is $6$-torsion ($x(3 P_0) = 0$).
   • Level $2$ ($2 P_\infty$ and $2 P_\infty + T_1$, Sturm-certified all-positive $15/15$ sextuples and $19/21$ septuples): The chord sections $2 P_\infty$ and $2 P_\infty + T_1$ have$$ s_{2 P_\infty}(r) = \frac{2 r^3 - 2 r + 1}{2 r^3 - 2 r^2 - 2 r + 1}, \qquad t_{2 P_\infty}(r) = -\frac{r(2 r^3 - 2 r + 1)}{(r - 1)(2 r^3 + 2 r^2 - 1)} \qquad \big(s_{2 P_\infty + T_1}(r) = -s_{2 P_\infty}(-r)\big). $$By Sturm's theorem (verify_paper.py, exp28), on $2 P_\infty(r)$ for every rational $r > r_{**} \approx 2.010315$ (the largest real root of $4r^6 - 8r^5 - 4r^4 + 8r^3 - 1 = 0$), all $8$ elements of $T(r) \cup \{d_-(r), d_+(r)\} \cup (3 P_0(r) + \langle S_T \rangle)$ are strictly positive, and the $9$ $2+2$ and $3$ $1+3$ sub-quadruples give $15$ distinct $\mathbb{Q}(r)$-rational $19/21$ septuple sections of type $(0, 4)$ (verify_paper.py). At $r = 3$, $(r, s, t) = \left(3, \frac{49}{31}, -\frac{147}{142}\right)$ yields the all-positive $10/10$ quintuple $\left\{\frac{6816}{527}, \frac{527}{852}, \frac{255}{2201}, \frac{160247}{6324}, \frac{82615}{14484}\right\}$, the all-positive $15/15$ sextuple $S_6(2 P_\infty(3)) = \left\{\frac{6816}{527}, \frac{527}{852}, \frac{255}{2201}, \frac{221374475871572293213385}{145691293222823562567564}, \frac{62135944531101408}{84098719293868361}, \frac{786720756200851452355505}{972340684104979690764}\right\}$, and $15$ rational $19/21$ septuples ($9$ from $2+2$ and $6$ from $1+3$, of which $14$ are strictly positive $7/7$, $8$ from $2+2$ and $6$ from $1+3$) (verify_paper.py).
   • Homogeneous Model $M^2 = (N^2 - \beta^2)(N^2 - \gamma^2)(N^2 - (\beta+\gamma)^2)$ and Quadratic All-Positive Family IV ($N = 3\beta + \gamma$): In $(\alpha:\beta:\gamma:N) = (1/t : 1/r : 1/s : 1)$ ($\alpha + \beta + \gamma = 0$), $\mathcal{S}_{\text{3-tors}}$ is $M^2 = (N^2 - \beta^2)(N^2 - \gamma^2)(N^2 - (\beta + \gamma)^2)$. Along $N = 3\beta + \gamma$ (Family IV), $(N^2 - \beta^2)(N^2 - \gamma^2)(N^2 - (\beta + \gamma)^2) = 12\beta^3(2\beta + \gamma)^2(3\beta + 2\gamma)$, so $\gamma = 1, \beta = \frac{1 - 3 u^2}{3 u^2}$ gives the quadratic $\mathbb{Q}(u)$-family (verify_paper.py, exp30):$$ r(u) = \frac{9 u^2 + 1}{3(1 - u^2)}, \quad s(u) = -\frac{9 u^2 + 1}{6 u^2 - 1}, \quad t(u) = -\frac{9 u^2 + 1}{4 - 9 u^2}, \quad \Pi(u) = \frac{10 u(3 u^2 + 2)}{(1 - u^2)(4 - 9 u^2)}, $$with $a, b, c, d_-, d_+ > 0$ for all $u \in \left(\frac{1}{\sqrt{6}}, \frac{2}{3}\right) \approx (0.408248, 0.666667)$, and all-positive $15/15$ sextuple $S_6(u) = T(u) \cup (3 P_0(u) + \langle S_T \rangle)$ on $(1/\sqrt{6}, u_{**})$ ($u_{**} \approx 0.440894$), giving at $u = \frac{5}{12}$ ($(r, s, t) = (\frac{123}{119}, -\frac{123}{2}, -\frac{41}{39})$) the all-positive quintuple $\left\{\frac{2975}{39}, \frac{32}{23205}, \frac{23595}{476}, \frac{11613}{4420}, \frac{37791}{140}\right\}$ and all-positive $15/15$ sextuple $\left\{\frac{2975}{39}, \frac{32}{23205}, \frac{23595}{476}, \frac{107823282298962463233}{185754481938398420}, \frac{177823781478240}{125910041516987}, \frac{566952007465760553}{40854743133199214420}\right\}$ (while Family III, $N = \beta - \gamma$, gives $r(u) = \frac{u^2 + 1}{3(u^2 - 1)}, s(u) = -\frac{u^2 + 1}{2u^2 - 1}, t(u) = -\frac{u^2 + 1}{u^2 - 2}$).

5. The Cross-Coset Type-$(2, 3)$ Mechanism, Universal Vieta Gate Identity, the Eisenstein Triad Elliptic Curve $\mathcal{C}_{\text{triad}}$, and Inflection Tangent Bridge Laws: For any Diophantine triple $T = \{a, b, c\}$ ($a = \frac{\Pi}{t^2 - 1}, b = \frac{\Pi}{s^2 - 1}, c = \frac{\Pi}{r^2 - 1}$) and $d_\mp = a + b + c + 2\Pi \mp 2\Omega$, reflecting $b$ across $\{a, c, d_\mp\}$ gives $b'_{ac}(d_\mp) := 2(a + c + d_\mp + 2 a c d_\mp) - b$, which satisfies the Universal Second-Generation Vieta Gate Identity (verify_paper.py):

   $$ b \, b'_{ac}(d_\mp) + 1 = (a + c - d_\mp)^2 - (4 s^2 - 1) = \big(b + 2(\Pi \mp \Omega)\big)^2 - (4 s^2 - 1) = \frac{\big((2 s^2 - 1)\Pi \mp 2(s^2 - 1)\Omega\big)^2 - (s^2 - 1)^2(4 s^2 - 1)}{(s^2 - 1)^2}. $$

   On $\mathcal{S}_{\text{3-tors}}$, whenever $b \, b'_{ac}(d_\mp) + 1 = \square$ ($b'_{ac} \ne b$), $Q_{b'_{ac}} = (\Pi b'_{ac} - 1, Y) \in P_0 + 2 E_T(\mathbb{Q})$, so its $S_T$-coset $\mathcal{C}_b := \{b'_{ac}, x_5, x_6\} = Q_{b'_{ac}} + \langle S_T \rangle$ makes $T \cup \mathcal{C}_b$ a $15/15$ sextuple, $\{a, b, c, d_\mp, b'_{ac}\}$ a $10/10$ quintuple, and $T \cup \{d_\mp\} \cup \mathcal{C}_b = \{a, b, c, d_\mp, b'_{ac}, x_5, x_6\}$ a $19/21$ septuple of relation type $(2, 3)$ (missing only $(d_\mp, x_5), (d_\mp, x_6)$, while $\mathcal{U}_8 := T \cup \{d_-, d_+\} \cup \mathcal{C}_b$ has $23/28$ square pairs; verify_paper.py, exp30, exp32):

   • Coincidence Conic Family V ($b'_{ac} = b$) and the Positive-Rank Eisenstein Triad Curve $\mathcal{C}_{\text{triad}}$: On $\alpha + \beta + \gamma = 0$ with $(r, s, t_{\text{tri}}) := \big(\frac{N}{\beta}, \frac{N}{\gamma}, \frac{N}{\alpha}\big)$, $b'_{ac}(d_-) = b \iff 2 N_1^2 - N_1(\beta - \alpha) - (\beta^2 + \alpha^2) = 0$, parameterized over $\mathbb{Q}(t)$ by $N_1(t) = t^2 + 1, \beta(t) = (t - 2)(t + 1), \alpha(t) = 1 - 3 t, \gamma(t) = -t^2 + 4 t + 1$, where $M_1(t)^2 = (N_1^2 - \beta^2)(N_1^2 - \gamma^2)(N_1^2 - \alpha^2) = 4 t^2(t - 2)^2(t - 1)^2(t + 3)^2(2 t + 1)^2$. Since $M(N)^2 - M_1(t)^2 = (N^2 - N_1^2)(N^2 - N_1 N - (\beta^2 + \beta\alpha + \alpha^2 - N_1^2))(N^2 + N_1 N - (\beta^2 + \beta\alpha + \alpha^2 - N_1^2))$, the Eisenstein conjugate root $N(t, W) := \frac{t^2 + 1 + W}{2}$ satisfies $M(N)^2 = M_1(t)^2$ on the Eisenstein Triad Elliptic Curve (verify_paper.py, exp30):$$ \mathcal{C}_{\text{triad}} : \quad W^2 = 4(\beta^2 + \beta\alpha + \alpha^2) - 3 N_1^2 = t^4 - 20 t^3 + 34 t^2 + 12 t + 9 \quad \Big(\cong E_{\text{triad}} : Y^2 = X^3 + 34 X^2 - 276 X + 2520\Big), $$via $X = \frac{6(3 + 2 t - W)}{t^2}, Y = \frac{6(X + 30)}{t} + 2 X - 60$ (inverse $t = \frac{6(X + 30)}{Y - 2 X + 60}$). On $E_{\text{triad}}$, $P_1 = (66, 648)$ ($(t, W) = (1, -6)$) has $2 P_1 = \left(\frac{106}{9}, \frac{2024}{27}\right)$ ($(t, W) = \left(\frac{9}{4}, -\frac{39}{16}\right)$) with non-integral abscissa, so by Lutz–Nagell $P_1$ has infinite order and $\operatorname{rank} \mathcal{C}_{\text{triad}}(\mathbb{Q}) \ge 1$ (verify_paper.py). Choosing the signed polynomial branch $M_{\text{can}}(t) := -2 t(t - 2)(t - 1)(t + 3)(2 t + 1)$ and $\Pi_{\text{can}}(t) := \frac{M_{\text{can}}(t)}{\beta(t)\gamma(t)\alpha(t)}$ makes the non-trivial cross-coset gate $b \, b'_{ac}(d_-) + 1$ ($b'_{ac} \ne b$) an identical square in $\mathbb{Q}(\mathcal{C}_{\text{triad}})$ (verify_paper.py, exp30):$$ b \, b'_{ac}(d_-) + 1 = \left(\frac{(t^2 + 1 + W)\Big((t^2 + 1)(t^2 + 14 t - 27) - (t^2 - 2 t + 5) W\Big)}{4(t - 2)(t + 1)(1 - 3 t)(t - 1)(t + 3)}\right)^2. $$Under the simultaneous sign involution $(a, b, c, d_-, d_+, b'_{ac}, x_5, x_6) \mapsto (-a, -b, -c, -d_+, -d_-, -b'_{ac}, -x_6, -x_5)$ (which replaces $\Pi_{\text{can}}$ by $|\Pi_{\text{can}}|$ and uses $d_-$ when $\Pi_{\text{can}}(t) > 0$, i.e. $t \in (-3, -1) \cup (-1/2, 0) \cup (0, 1/3) \cup (1, 2)$, and $d_+$ when $\Pi_{\text{can}}(t) < 0$, i.e. $t \in (-\infty, -3) \cup (-1, -1/2) \cup (1/3, 1)$), Sturm's theorem on the norms in $\mathbb{Q}[t]$ proves that for every non-degenerate rational point $(t, W) \in \mathcal{C}_{\text{triad}}(\mathbb{Q})$ with $t < 2$ and $W > 0$, all $7$ entries of the resulting type-$(2, 3)$ $19/21$ septuple (and all $8$ entries of $\mathcal{U}_8$) are strictly positive (verify_paper.py)! Since $E_{\text{triad}}(\mathbb{R})$ is connected ($X^3 + 34 X^2 - 276 X + 2520 = (X + 42)(X^2 - 8 X + 60)$) and the archimedean open set $\{t < 2, W > 0\}$ is non-empty, $\mathbb{Z} P_1$ is dense in it and generates infinitely many strictly positive $7/7$ cross-coset type-$(2, 3)$ $19/21$ septuples ($t = -\frac{5}{3}, -\frac{4}{5}, \frac{3}{14}, \frac{7}{11}, -\frac{52}{183}, \frac{305}{188}, -\frac{423}{65}, \frac{22184}{46299}, \frac{29601}{87185}, \dots$; verify_paper.py, exp30), beginning at $(t, W) = \left(-\frac{5}{3}, \frac{122}{9}\right)$ ($\Pi_{\text{can}} > 0$, using $d_-$) and $(t, W) = \left(-\frac{4}{5}, \frac{141}{25}\right)$ ($\Pi_{\text{can}} < 0$, using $d_+$; also realized with $\Pi_{\text{can}} > 0$ using $d_-$ at $(t, W) = \left(\frac{3}{14}, \frac{705}{196}\right)$) with:$$ \begin{aligned} \mathcal{S}_{7, \text{triad}}^{(19/21)}(-5/3) &= \left\{\frac{121}{1710}, \; \frac{62928}{561055}, \; \frac{315}{418}, \; \frac{247950}{49379}, \; \frac{1520}{99}, \; \frac{1558}{55}, \; \frac{41888}{855}\right\}, \\ \mathcal{S}_{7, \text{triad}}^{(19/21)}(-4/5) &= \mathcal{S}_{7, \text{triad}}^{(19/21)}(3/14) = \left\{\frac{288}{6035}, \; \frac{15175611}{84296180}, \; \frac{185915417}{161966940}, \; \frac{781}{255}, \; \frac{3825}{284}, \; \frac{39121}{1020}, \; \frac{721600}{3621}\right\}. \end{aligned} $$Across 5,205 homogeneous triples in exp30 (including a $20$-step filtered $C_r$-chord walk $\mathcal{O}_{3/2}$ from $(3/2, -11/29, 33/65)$), we obtain 42 cross-coset type-$(2, 3)$ $19/21$ septuples (9 all-positive).
   • Inflection Tangent Cross-Coset Bridge Formulas ($20$th Pair): For any $3$-torsion coset $\{x_4, x_5, x_6\} = Q_4 + \langle S_T \rangle$ ($Q_4 = (U_4, Y_4)$ on $Y^2 = U^3 + (\lambda_0 U - \Omega)^2$) and any $d \in \mathbb{Q}$, $M_4 := d x_4 + 1 = \frac{d U_4 + \Pi + d}{\Pi}$, $M_5 := d x_5 + 1$, and $M_6 := d x_6 + 1$ obey (verify_paper.py, exp31):$$ M_{5, 6} = \frac{(\Pi + d)U_4^2 - 2\Omega d(\lambda_0 U_4 - \Omega \mp Y_4)}{\Pi U_4^2}, \qquad M_5 M_6 = \frac{(\Pi + d)^2 U_4^2 - 4\Omega d\big(\Omega d + \lambda_0(\Pi + d)\big)U_4 + 4\Omega^2 d(\Pi + d)}{(\Pi U_4)^2}, $$and the Full Coset Norm Law $M_4 M_5 M_6 = \frac{(\Pi + d)^3 + \Sigma_U d(\Pi + d)^2 - 4\Omega\lambda_0 d^2(\Pi + d) - 4\Omega^2 d^3}{\Pi^3}$ (14,895 type-$(0, 4)$ $19/21$ septuples, 881 all-positive, across $993$ sextuples in exp31, plus 2,556 quintuples, 12,780 $14/15$ sextuples, and 76,680 $18/21$ septuples in exp25).

Proof. Verified symbolically and arithmetically in verify_paper.py, exp25–exp32, and THEOREMS_AND_PROOFS.md. $\square$

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10. Limitations and scope

• No rational Diophantine septuple ($21/21$) is claimed. $\mathcal{S}_7^{(20/21)}$ satisfies $20/21$ conditions and adds a new relation type $(|\mathcal{R}_4|, |\mathcal{R}_5|) = (5, 2)$ ($\text{dim} = 1, \text{gens} = 2$) to the known almost septuples [2, 3, 4].
• Finiteness on $\mathcal{H}$ vs. Problem 3.3: Theorem 3(4) proves that $\mathcal{C}_+(a, \lambda) = 0$ on Branch 1 has a smooth genus-$3$ cover $\mathcal{H}$, so by Faltings' theorem it carries only finitely many rational almost septuples, ruling out deforming $\mathcal{S}_7^{(20/21)}$ inside type $(5, 2)$ while leaving open $2$-dimensional sub-systems ($\mathcal{S}_{4, 2}$, $\mathcal{E}_{\text{Br1}}(\lambda, \mu)$).
• Completeness on $\mathcal{H}$ and unpaid edges: Faltings' theorem is ineffective; our height-$50,000$ search on $\mathcal{E}^{(1)}$ and u/kiro_dioph7's $295$-prime sieve up to $|n| \le 10^6$ (Post 35, m_A1PZmnTT08-GXf) are bounded searches, not a complete Chabauty–Coleman proof on $\mathcal{H}(\mathbb{Q})$. On Curve II ($\mathbb{Q}(k)$), $c_1 c^*_\mp + 1$ have irreducible squarefree numerators of degrees $36$ and $32$; on $w = 2 u^2 y - L$ (Theorem 5), $\gamma^2, \delta^2$ have bidegree $(4, 4)$; on $z = z_{L_1}$ (Theorem 6), trivializing $a c + 1 = \square$ with $\mathcal{F}_{ab}, \mathcal{G}_d, H = \square$ remains open; and on $\mathcal{C}_{\text{triad}}$ (Theorem 7(5)), the $20$th and $21$st cross-coset pairs remain higher-genus conditions.

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11. Reproducibility

The standalone verifier verify_paper.py in supplement.zip requires only Python 3 and SymPy (python3 verify_paper.py, $\sim 75\text{ s}$) and verifies all symbolic identities, factorizations, square-class checks (is_rat_sq), collision sets, Sturm intervals, $2$-isogeny local solubility and torsion certificates, and every numerical tuple in the text (exp9–exp32 reproduce results_exp9_*.json–results_exp32_*.json):

1. Theorems 1–2: linearity of $c^*_\mp$ in $S_{ab}$, the three $S_{ab}$-square completions, $R_4(c, d_-, d_+, c^*_-) = 0$ on Branch 1, $a_{\text{circ}}(\lambda)$, and the all-positive $15/15$ sextuples;
2. Theorem 3: all $21$ pairs of $\mathcal{S}_7^{(20/21)}$, its $(5, 2)$ relations, $7 \times 7$ Jacobian rank $6$, $\text{gens} = 2$, $R_4(p, q, e_-, e_+) = \frac{F_{\text{sym}} G}{(P_4 - 1)^4}$, $a_\pm c_\pm + 1 = \square$ on $\mathcal{S}_{4, 2}$ and $\mathcal{H}$, $\mathcal{H}_{\text{Br2}}$, $\operatorname{rank} E_{\text{min}}^{(5)}(\mathbb{Q}) = 0$, and $S_{ab, 3}(-2) = S_{ab, K_0^+ + K_3^-}(-2)$;
3. Theorems 4–5: $E_z(r, \mu, z) = 0$, $\Delta_z(E_z)$, $\operatorname{disc}_t(\Phi_{20})$, $\Sigma_\mu^+$ and $\tau_{T_0}(\mu)$, Curves I–III ($19$ squares, $21$ collision differences, Sturm check on $(1, k_*)$, and $k = 2$ on Curve II), $E = 4 u^3 y D w$ on $w = 2 u^2 y - L$, $(J, Q_0, Q_-, Q_+)$, the $\mathbb{Q}(k)$ quintuple on $\mathcal{E}_B$, $2 P = (-1264/121, 44144/1331)$, and $\{20, 12/49, 6, 40/289, 48\}$ at $k = 4$;
4. Theorem 6: $R_5 = \frac{L_1 L_2 Q_4}{H K^4}$, $\mathcal{C}_{1a}, \mathcal{C}_{1b}$, $Q_{4, +}(R, v, u) = 0$ and $N_d \equiv R W_{d, \text{lin}}^2 \pmod{Q_{4, +}}$ with the $3$ interior quintuples, $\iota_K$, $\mathcal{E}'_{Q_4}(R)$, the Universal Reciprocal $3 P_0$ Law, $\Lambda(r) \subset E(r^2)(\mathbb{Q}(r))$ ($2 W_1 = O_-$, $A^2 - 4 B = 16 r^4 Q_+ Q_-$, $\hat{\phi}(\hat{P}_{\text{odd}, \pm}) = \pm P_{\text{odd}}$, $Z_-(r) = (1/r, 2(2r-1))$, $P_{\text{odd}}(3/4)$, $u = 2$, and $E_{48}^-, E_{48}, E_{80}$), $G_u(R, d_0) = n^2$ ($u = 2/3$), the $15 + 6$ Square-Class Collapse, the $9$ points on $(r, M)$ / $(v, t)$, the $3$ quintuples on $L_1 = 0$, $(53/30, 3/5)$, the Vieta formula, the $v = -1/5$ family ($\operatorname{rank} E_{16}(\mathbb{Q}) = 0$), and $3 P_1, 4 P_1, 8 P_1$ on $\mathcal{G}_d = W_d^2$;
5. Theorem 7: the $(V, K, d)$ chains, $S_T$-translations on $E_T$, $3 S_T = \mathcal{O} \iff \pm \frac{1}{r} \pm \frac{1}{s} \pm \frac{1}{t} = 0$, $Y^2 = U^3 + (\lambda_0 U - \Omega)^2$ and $U(d_\mp) = 2\Omega(\Omega+\lambda_0\mp\Pi)$, $P_{\text{coset}}(U') = 0$ and $P_{\text{coset}}(-1)$, the Sub-Quadruple Square-Root Identity in $U$- and tuple coordinates, $C_r$ ($2$-torsion, $P_\infty + T_1$, Sturm positivity of $2 P_\infty$ for $r > r_{**}$, and the $15$ septuples at $r = 3$), Family IV ($u = 5/12$), the Universal Vieta Gate Identity, $\mathcal{C}_{\text{triad}} \cong E_{\text{triad}}$ (Lutz–Nagell infinite order of $P_1 = (66, 648)$, $b \, b'_{ac}(d_-) + 1 = \square$, Sturm positivity for $t < 2, W > 0$, and $t = -5/3, 3/14, -4/5$), and $M_5, M_6, M_5 M_6, M_4 M_5 M_6$.

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References

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