## 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 $a_i a_j + 1$ is a rational square for all $1 \leq i < j \leq m$. No rational Diophantine septuple ($m = 7$, requiring $21$ pairwise squares) is known [1, Problem 3.2]. An *almost septuple* is a set of seven distinct nonzero rationals satisfying $20$ of the $21$ 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] and $A_1, A_2, A_3$ in [3], plus $A_4$ in [4]) were classified by their *relation type* $(\mathcal{R}_4, \mathcal{R}_5)$, where $\mathcal{R}_4$ and $\mathcal{R}_5$ denote the sets of $4$-element and $5$-element subsets on which the regular-quadruple polynomial $R_4$ and regular-quintuple polynomial $R_5$ vanish: $$ \begin{aligned} R_4(x, y, z, t) &= (x + y - z - t)^2 - 4(x y + 1)(z t + 1), \\ R_5(x, y, z, t, u) &= (x y z t u + 2 x y z + x + y + z - t - u)^2 - 4(x y + 1)(x z + 1)(y z + 1)(t u + 1). \end{aligned} $$ More than half ($14$ of $25$) of the known almost septuples 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 two regular fourth elements of $\{a, b, c\}$ [2, 3, 5]. However, as proved in Paper 10 [3, Theorem 1], fixing a generic base fibre of the quintuple threefold forces the sub-quadruple sixth-element condition into a double cover of geometric genus $g \geq 9$. In this paper, we invert the order of elimination by expressing the two regular fifth elements $c^*_\mp = e_\mp(a, b, d_-, d_+)$ of $\{a, b, d_-, d_+\}$ and their compatibility conditions directly in terms of $r$, the conic parameter $\lambda$ uniformizing $d_- d_+ + 1 = w_1^2$, and the trace $S_{ab} := a + b$. This yields seven main results: 1. **Universal $S_{ab}$-square completions and two elliptic sextuple surfaces (Theorem 1, Section 3).** For any two-sided quintuple $\{a, b, c, d_-, d_+\}$ with $C_0 := c - a - b = \frac{\lambda^2(2 r - 1) + (2 r + 1)}{2 \lambda}$, $c^*_\mp = e_\mp(a, b, d_-, d_+)$ are linear in $S_{ab} = a + b$. For $\mathcal{S}_7^{(c^*)} = \{a, b, c, d_-, d_+, c^*_-, c^*_+\}$, $18/21$ conditions hold identically, and the remaining three ($c c^*_- + 1, c c^*_+ + 1, c^*_- c^*_+ + 1 = \square$) are quadratics in $S_{ab}$ whose discriminants in $(r, \lambda)$ cut out **Branch 1** ($\mathcal{E}_{\text{Br1}}$, where $R_4(c, d_-, d_+, c^*_-) = 0$ and $c c^*_- + 1 = \square$) and the **Twin-Fifth Surfaces** $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$ (where $R_4(a, b, c^*_-, c^*_+) = R_4(d_-, d_+, c^*_-, c^*_+) = 0$ and $c^*_- c^*_+ + 1 = \square$). 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 $\mathcal{E}_{\text{Br1}}$, including `408` all-positive $15/15$ sextuples, 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 $R_4(p, q, e_-, e_+) = \frac{F_{\text{sym}} G}{(P_4 - 1)^4}$ with $F_{\text{sym}} = P_4^2 - 6 P_4 - 4(p q + u v) - 3$. 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 in `exp12`), and restricting to Branch 1 gives the genus-$3$ curve $\mathcal{H} : W^2 = (5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^4 - 22 \lambda^2 + 33)$ of relation 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 the trace-lock section $w = 2 u^2 y - L$ 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).** In $\mathbb{Q}(r, \mu, z)$ before $E_z = 0$, $\iota_K: (r, \mu, z, w) \mapsto \left(r, -\frac{\mu+4r}{K}, \frac{4r^2-1}{K}z, \frac{4r^2-1}{K}w\right)$ fixes $(a,b,c,d_-,d_+)$, swaps $c^*_- \leftrightarrow c^*_+$, exchanges $L_1 \leftrightarrow L_2$, and restricts on $Q_{4, +}(R, v, u) = 0$ to $\tau_{T_0}: u \mapsto \frac{u-2}{2Ru-1}$ with $2$-isogenous quotient $\mathcal{E}'_{Q_4}(R)$. Over $\mathbb{Q}(r)$, $W_1(r)$ and $P_{\text{odd}}(r)$ generate $\Lambda(r) \cong \mathbb{Z}^2 \oplus \mathbb{Z}/2\mathbb{Z} \subset E(r^2)(\mathbb{Q}(r))$ ($P_{\text{odd}}(3/4)$ giving the $r = 3/4$ quintuple, $\operatorname{rank} E(r^2)(\mathbb{Q}) \ge 3$ certified at $r = 7/4, 19/10$, while at $u = 2$ the branch $v_+(P_{\text{even}}) = -4r^2$ lies on $2Ru+v=0$ and $v_+(T_0) = v_-(P_{\text{even}}), v_-(2W_1), v_+(3W_1), v_+(3W_1+T_0)$ are excluded by complete $2$-isogeny descent 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 $(V, K, d)$ coordinates for canonical $S$-chains, the $3$-Torsion Inflection Criterion, and Vieta-to-Elliptic $S_T$-Progressions (Theorem 7, Section 9).** For any Diophantine triple $T = \{a, b, c\}$, we prove a universal square completion of $F_2(z)$, reduce $4$-term chain conditions in $(V, K, d)$, and apply the $S_T$-translation dictionary [7, 10, 17] to characterize the zero-root locus $R_4(T, x_4) = 0 \iff x_4 = d_\mp(T)$ and prove in $U = X - 1$ coordinates that $3 S_T = \mathcal{O} \iff \pm \frac{1}{r} \pm \frac{1}{s} \pm \frac{1}{t} = 0$, on which every non-base coset $Q + \langle S_T \rangle \subset P_0 + 2 E_T(\mathbb{Q})$ automatically forms a $15/15$ rational Diophantine sextuple (`verify_paper.py`). --- ## 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 periodic 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:** - Theorem 1 writes $c^*_\mp$ and $c c^*_- + 1, c c^*_+ + 1, c^*_- c^*_+ + 1$ in $(r, \lambda, S_{ab})$, showing that their $S_{ab}$-discriminants are independent of $S_{ab}$ and cutting out Branch 1 ($\mathcal{E}_{\text{Br1}}$) and the Twin-Fifth surfaces ($\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$). - $\mathcal{S}_7^{(20/21)}$ is absent from Dujella's list [15] and [2, 3, 4, 9]; its relation type $(5, 2)$ is a new $1$-dimensional relation type whose full-root cover collapses to the genus-$3$ curve $\mathcal{H}$ with unramified $d = 1$ descent. - Theorem 4 puts $\Sigma_\mu$ into cubic Weierstrass form $\Sigma_\mu^+ : W_+^2 = -K L$ and constructs Curves I, II (all-positive for $k \in (1, k_*)$), and III ($\mathcal{E}_{\text{III}, \pm}$); Theorem 5 proves the factor cancellation on $w = 2 u^2 y - L$ and constructs $\mathcal{E}_B$; and Theorems 6–7 establish the $R_5 = \frac{L_1 L_2 Q_4}{H K^4}$ factorization, gauge involution $\iota_K$, rank-$2$ subgroup $\Lambda(r)$ on $Q_{4,+} = 0$, $15+6$ Square-Class Collapse, $v = -1/5$ families, and the $(V, K, d)$ / $3$-torsion inflection analysis of $S_T$-chains. --- ## 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 fourth elements $d_\mp = a + b + c + 2 a b c \mp 2 r s t$. By Vieta's formulas, $$ d_- + d_+ = 2(a + b + c + 2 a b c), \qquad d_- d_+ + 1 = (c - a - b)^2 - (4 r^2 - 1). $$ Write $S_{ab} := a + b$ and $C_0 := c - a - b = c - S_{ab}$. The two-sided quintuple condition $d_- d_+ + 1 = w_1^2$ is the conic $C_0^2 - w_1^2 = 4 r^2 - 1$, which is rationally parameterized by $\lambda \in \mathbb{Q}^\times$ via $C_0 - w_1 = \lambda(2 r - 1)$ and $C_0 + w_1 = \frac{2 r + 1}{\lambda}$: $$ C_0(r, \lambda) = \frac{\lambda^2(2 r - 1) + (2 r + 1)}{2 \lambda}, \qquad w_1(r, \lambda) = \frac{\lambda^2(2 r - 1) - (2 r + 1)}{2 \lambda}. $$ Now consider the sub-quadruple $Q^* = \{a, b, d_-, d_+\}$. Its product is $P_4 := a b d_- d_+ = (r^2 - 1)(w_1^2 - 1)$, and its four pairwise square roots are rational in $(a, b, c, r, s, t)$: $$ a d_\mp + 1 = (a c \mp r t)^2, \qquad b d_\mp + 1 = (b c \mp r s)^2, \qquad \prod_{\ell \in \{a, b\}, \mp} \sqrt{\ell d_\mp + 1} = (r^2 - 1) C_0 + r^2 S_{ab}. $$ Substituting these symmetric invariants into the regular quintuple quadratic $R_5(a, b, d_-, d_+, x) = 0$ yields our first main theorem. ### 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 $(r, \lambda)$ as above, and set $\text{den}_- := \lambda^2(r - 1) - (r + 1)$ 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 two regular fifth elements $c^*_\mp = e_\mp(a, b, d_-, d_+)$ depend on $(a, b, c, s, t)$ solely through $S_{ab} = a + b$: $$ \begin{aligned} c^*_- &= \frac{4 \lambda \Big(\lambda S_{ab} + \lambda^2(r - 1) + (r + 1)\Big)}{\text{den}_-^2}, \\ c^*_+ &= \frac{4 \lambda (4 r^2 - 1) \Big(\lambda(4 r^2 - 1) S_{ab} + \lambda^2(2 r - 1)^2(r + 1) + (2 r + 1)^2(r - 1)\Big)}{\text{den}_+^2}. \end{aligned} $$ 2. **Universal $S_{ab}$-square completions:** For the $7$-tuple $\mathcal{S}_7^{(c^*)} = \{a, b, c, d_-, d_+, c^*_-, c^*_+\}$, $18$ of the $21$ conditions hold identically, and the remaining three conditions ($c c^*_- + 1 = \square$, $c c^*_+ + 1 = \square$, $c^*_- c^*_+ + 1 = \square$) satisfy the exact polynomial identities: $$ \begin{aligned} 4 \, \text{den}_-^2 (c \, c^*_- + 1) &= \Big(4 \lambda S_{ab} + \lambda^2(4 r - 3) + 4 r + 3\Big)^2 + \big(\lambda^2(2 r - 3) - (2 r + 1)\big)\big(\lambda^2(2 r - 1) - (2 r + 3)\big), \\ 4 \, \text{den}_+^2 (c \, c^*_+ + 1) &= \Big(4 \lambda (4 r^2 - 1) S_{ab} + \lambda^2(16 r^3 - 4 r^2 - 8 r + 3) + 16 r^3 + 4 r^2 - 8 r - 3\Big)^2 \\ &\quad + (4 r^2 - 1)\big(\lambda^2(2 r - 1)^2 - (2 r - 3)(2 r + 1)\big)\big(\lambda^2(2 r - 1)(2 r + 3) - (2 r + 1)^2\big), \\ \text{den}_-^2 \, \text{den}_+^2 (c^*_- c^*_+ + 1) &= 16 \lambda^2 \Big(\lambda(4 r^2 - 1) S_{ab} + (2 r^2 - 1)\big(\lambda^2(2 r - 1) + (2 r + 1)\big)\Big)^2 + F_1 F_2 F_3 F_4, \end{aligned} $$ 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 (The Branch 1 Sextuple Surface $\mathcal{E}_{\text{Br1}}$):** The discriminant of $c c^*_- + 1$ with respect to $S_{ab}$ vanishes identically on **Branch 1**: $$ \lambda^2(2 r - 3) - (2 r + 1) = 0 \iff r(\lambda) = \frac{3 \lambda^2 + 1}{2(\lambda^2 - 1)}, \qquad C_0(\lambda) = \frac{\lambda(\lambda^2 + 3)}{\lambda^2 - 1}, \qquad w_1(\lambda) = \frac{\lambda^4 + 6 \lambda^2 - 3}{2 \lambda(\lambda^2 - 1)}. $$ (The second factor $\lambda^2(2 r - 1) - (2 r + 3) = 0$ is isomorphic under $\lambda \mapsto 1/\lambda$.) On Branch 1, the regular quadruple relation $R_4(c, d_-, d_+, c^*_-) = 0$ and the square condition $$ c \, c^*_- + 1 = \left(\frac{8 \lambda(\lambda^2 - 1) S_{ab} + 3 \lambda^4 + 14 \lambda^2 - 1}{(5 \lambda^2 - 1)(\lambda^2 + 3)}\right)^2 $$ 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 (The Twin-Fifth Sextuple Surfaces $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$):** The discriminant $F_1 F_2 F_3 F_4$ of $c^*_- c^*_+ + 1$ with respect to $S_{ab}$ vanishes identically on the genus-$1$ curves $F_1(r, \lambda) = 0$ 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 $\mathcal{E}_{F_3}$, the regular quadruple relations $R_4(a, b, c^*_-, c^*_+) = R_4(d_-, d_+, c^*_-, c^*_+) = 0$ 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.* In $R_5(a, b, d_-, d_+, x) = 0$, 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 $\prod \sqrt{\ell d_\mp + 1} = (r^2 - 1) C_0 + r^2 S_{ab}$ factors the quadratic in $x$ over $\mathbb{Q}(r, \lambda, S_{ab})$ into the two roots $c^*_-$ and $c^*_+$ of Part 1. Substituting $c = S_{ab} + C_0(r, \lambda)$ into $4 \, \text{den}_-^2(c c^*_- + 1)$, $4 \, \text{den}_+^2(c c^*_+ + 1)$, and $\text{den}_-^2 \text{den}_+^2(c^*_- c^*_+ + 1)$ and completing the square in $S_{ab}$ yields the exact polynomial identities of Part 2 (verified symbolically in `verify_paper.py`). On Branch 1, direct substitution of $r(\lambda)$ and $C_0(\lambda)$ into $(c + c^*_- - (d_- + d_+))^2 - 4(c c^*_- + 1)(d_- d_+ + 1)$ gives $R_4(c, d_-, d_+, c^*_-) = 0$ identically, which by Rule (Q) explains why $d_- d_+ + 1 = w_1^2$ forces $c c^*_- + 1 = \square$. Similarly, on $F_i(r, \lambda) = 0$, $P_4^2 - 6 P_4 - 4(a b + d_- d_+) - 3 = \frac{F_1 F_2 F_3 F_4}{4 \lambda^2 (r^2 - 1)^2} = 0$, which forces $R_4(a, b, c^*_-, c^*_+) = R_4(d_-, d_+, c^*_-, c^*_+) = 0$ and $c^*_- c^*_+ + 1 = \square$. $\square$ --- ## 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 $c - a - b = C_0$, eliminate $s$ by setting $s = a + \mu$. Since $c = a + \frac{r^2 - 1}{a} + C_0$, the condition $a c + 1 = (a + \mu)^2$ is linear in $a$: $$ a(r, C_0, \mu) = \frac{\mu^2 - r^2}{C_0 - 2 \mu}, \qquad b(r, C_0, \mu) = \frac{(r^2 - 1)(C_0 - 2 \mu)}{\mu^2 - r^2}, \qquad s = \frac{C_0 \mu - (\mu^2 + r^2)}{C_0 - 2 \mu}. $$ The remaining condition $b c + 1 = t^2$ then reduces to the quartic elliptic surface $\mathcal{E}_{(r, \lambda)}$: $$ Y_\mu^2 = \mathcal{Q}(r, C_0, \mu) := (r^2 - 1)^2(C_0 - 2 \mu)^2 + C_0(r^2 - 1)(C_0 - 2 \mu)(\mu^2 - r^2) + r^2(\mu^2 - r^2)^2, $$ with $t = \frac{Y_\mu}{\mu^2 - r^2}$. Since the coefficient of $\mu^4$ in $\mathcal{Q}(r, C_0, \mu)$ is $r^2 = \square$, $\mathcal{E}_{(r, \lambda)}$ has two rational points at infinity $\infty_\pm$, whose tangent chord gives the universal rational section $$ \mu_{\text{inf}}(r, C_0) = \frac{C_0^2 + 8 r^4 - 4 r^2}{4 C_0 r^2}, \qquad a_{\text{inf}}(r, C_0) = \frac{C_0^2 - (4 r^3 - 2 r)^2}{8 C_0 r^2 (2 r^2 - 1)}. $$ Furthermore, the boundary conditions $a \in \{d_-, d_+\}$ and $b \in \{d_-, d_+\}$ (along with $\mu = \pm r$ and $\mu = C_0 / 2$) provide **eight built-in rational wall sections** $P_{a, 1..4}(r, \lambda)$ and $P_{b, 1..4}(r, \lambda)$ on $\mathcal{E}_{(r, \lambda)}$, corresponding in the $a$-chart to $a_{P_1} = (r - 1)\lambda$, $a_{P_2} = -\frac{r + 1}{\lambda}$, $a_{P_3} = \frac{r + 1}{\lambda}$, and $a_{P_4} = -(r - 1)\lambda$. ### Theorem 2 (Explicit Parametric Sextuple Curves on $\mathcal{E}_{\text{Br1}}$ and Positive Sextuples) On the Branch 1 surface $\mathcal{E}_{\text{Br1}}(\lambda, \mu)$ ($r = \frac{3 \lambda^2 + 1}{2(\lambda^2 - 1)}, C_0 = \frac{\lambda(\lambda^2 + 3)}{\lambda^2 - 1}$), 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 $\mathbb{Q}(\lambda)$:** Taking $1$-step Mordell–Weil chords among $\infty_\pm$ and $P_{1..4}$ yields $7$ distinct non-degenerate sections $a_\gamma(\lambda) \in \mathbb{Q}(\lambda)$, including the chord $P_{a, 1}^+ + P_{a, 2}^+$: $$ a_1(\lambda) = -\frac{\lambda(\lambda^2 + 3)(5 \lambda^2 - 1)(\lambda^8 + 38 \lambda^6 + 16 \lambda^4 + 22 \lambda^2 - 13)}{4(\lambda^2 - 1)(\lambda^4 - 14 \lambda^2 - 3)(6 \lambda^6 + 5 \lambda^4 + 4 \lambda^2 + 1)}. $$ 2. **Five rational circle families over $\lambda = \frac{2 k}{k^2 + 1}$:** The locus $c = 0$ defines a point $P_{c, 2}$ over $1 - \lambda^2 = \square$. Chords between $P_{c, 2}$ and $P_{1..4}$ yield $5$ rational families over the circle $\lambda = \frac{2 k}{k^2 + 1}$, the simplest of which has degree $8$ in $\lambda$: $$ a_{\text{circ}}(\lambda) = -\frac{\lambda(\lambda^2 + 3)^2(4 \lambda^4 + \lambda^2 - 1)}{2(\lambda^2 - 1)(5 \lambda^2 - 1)(2 \lambda^4 + \lambda^2 + 1)}, \qquad b_{\text{circ}}(\lambda) = -\frac{(5 \lambda^2 - 1)(2 \lambda^4 + \lambda^2 + 1)}{2 \lambda(\lambda^2 - 1)(4 \lambda^4 + \lambda^2 - 1)}, $$ 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 sections $P_{c, 3}, P_{c, 4}$ over the rank-$1$ elliptic curve $\mathcal{C}_{5\lambda}$ (generated by $(\lambda, Y_\lambda) = (1, 4)$ and $(2, 11)$). The chord $P_1 + P_{c, 3}$ yields the section $$ a_3(\lambda) = \frac{(5 \lambda^2 - 1)(\lambda^8 - 42 \lambda^6 - 80 \lambda^4 - 6 \lambda^2 - 1)}{4 \lambda(\lambda^2 - 1)(29 \lambda^6 + 53 \lambda^4 - 17 \lambda^2 - 1)}, \qquad b_3(\lambda) = \frac{\lambda(\lambda^2 + 3)(29 \lambda^6 + 53 \lambda^4 - 17 \lambda^2 - 1)}{(\lambda^2 - 1)(\lambda^8 - 42 \lambda^6 - 80 \lambda^4 - 6 \lambda^2 - 1)}, $$ on which $a_3 c_3 + 1 = (5 \lambda^4 + 10 \lambda^2 + 1)\left(\frac{11 \lambda^6 + 59 \lambda^4 - 7 \lambda^2 + 1}{4 \lambda(29 \lambda^6 + 53 \lambda^4 - 17 \lambda^2 - 1)}\right)^2$ 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 $1$-step and $2$-step Mordell–Weil chords across $1,414$ rational $\lambda$-fibres in `exp9` and $2,879$ rational $\lambda$-fibres in `exp10` on $\mathcal{E}_{\text{Br1}}$, along with $38$ base fibres on $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$, produced **$1,508$ unique $15/15$ non-torsion sextuples** (`33,162` verified $19/21$ septuples and $\pm \mathcal{S}_7^{(20/21)}$) in `exp9`—of which **$408$ are all-positive**—and **$2,304$ unique $15/15$ non-torsion sextuples** (`66,758` verified $19/21$ septuples and $\pm \mathcal{S}_7^{(20/21)}$) in `exp10`, plus **$222$ unique $15/15$ twin-fifth sextuples** on $\mathcal{E}_{F_1}, \mathcal{E}_{F_3}$ (`430` verified $19/21$ septuples) and **$10$ unique $15/15$ sextuples** (`220` verified $19/21$ septuples) on the general $(r, \lambda)$ grid. Examples of all-positive $15/15$ non-torsion sextuples $\{a, b, c, d_-, d_+, c^*_-\}$ include: $$ \left\{\frac{343308}{1229881}, \; \frac{23851}{78732}, \; \frac{1696}{1647}, \; \frac{3796}{1647}, \; \frac{69559}{6588}, \; \frac{115959357}{907924}\right\} $$ (found on the universal surface $\mathcal{E}_{(r, \lambda)}$ at $r = 1090/1053$) and, on Branch 1 ($\mathcal{E}_{\text{Br1}}$), $$ \left\{\frac{8887028761}{164339899920}, \; \frac{8887028761}{17666208000}, \; \frac{4357061596629}{177740575220}, \; \frac{22079269824716984551921}{1460493417168901599120}, \; \frac{85711055036402974977363}{21636939513613357024000}, \; \frac{1413296640000}{8887028761}\right\}. $$ --- ## 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 the $\mathcal{C}_{5\lambda}$-section $a_3(\lambda)$ of Theorem 2(3) at $\lambda = -2$ (where $5(-2)^4 + 10(-2)^2 + 1 = 121 = 11^2$ and $\mu = -103/372$) produces seven distinct nonzero rationals: $$ \mathcal{S}_7^{(20/21)} = \left(c, \; c^*_+, \; d_-, \; a, \; b, \; c^*_-, \; d_+\right) = \left(-\frac{19961343}{78775960}, \; -\frac{411840}{1969399}, \; \frac{23715}{29896}, \; \frac{71003}{63240}, \; \frac{36890}{11211}, \; \frac{104321672}{29540985}, \; \frac{29896}{7905}\right). $$ Exact arithmetic (`verify_paper.py`) confirms that **20 of the 21 pairwise products-plus-one are rational squares**, with $c^*_- c^*_+ + 1 = \left(\frac{1007113}{1969399}\right)^2$, and only the single pair $(c, c^*_+)$ fails: $$ c \, c^*_+ + 1 = \left(-\frac{19961343}{78775960}\right)\left(-\frac{411840}{1969399}\right) + 1 = \frac{4084054408729}{1969399^2} \neq \square. $$ Thus $\mathcal{S}_7^{(20/21)}$ is the union of two overlapping $15/15$ rational Diophantine sextuples $\{a, b, c, d_-, d_+, c^*_-\}$ and $\{a, b, d_-, d_+, c^*_-, c^*_+\}$ sharing the $5$-tuple $\{a, b, d_-, d_+, c^*_-\}$. Why does Condition 21 ($c^*_- c^*_+ + 1 = \square$) hold simultaneously with Condition 19 at $\lambda = -2$? We give a complete algebraic explanation: a **universal regularity factorization** that identifies the 2-dimensional type-$(4, 2)$ surface $\mathcal{S}_{4, 2}$, its Branch 1 genus-$3$ full-root curve $\mathcal{H}$ of relation type $(5, 2)$, an unramified $2$-cover descent on $\mathcal{H}$, and a **transverse Mordell–Weil 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 $R_4(p, q, e_-, e_+)$:** Let $\{p, q, u, v\}$ be any quadruple with product $P_4 = p q u v$, and let $e_-, e_+$ be the two roots of $R_5(p, q, u, v, x) = 0$. Then 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 invariant under exchanging $\{p, q\} \leftrightarrow \{u, v\}$, so $F_{\text{sym}} = 0$ simultaneously forces **both** $R_4(p, q, e_-, e_+) = 0$ and $R_4(u, v, e_-, e_+) = 0$ (and forces $e_- e_+ + 1 = \square$ whenever $p q + 1 = \square$ or $u v + 1 = \square$); - $G(p, q; u, v) := (P_4 - 1)^2(p - q)^2 - 4 P_4^2 + 8 P_4 - 4 u v(p + q)^2 - 4 p q(u + v)^2 - 4(P_4 + 1)(p + q)(u + v) - 4$ is asymmetric and quadratic in $v$ with discriminant $\Delta_v(G) = 16(p - q)^2(p q + 1)(p u + 1)^2(q u + 1)^2$, which splits over $p q + 1 = r^2$ into $v_\pm(p, q, r, u) = \frac{p(r \pm 1) - q(r \mp 1) + 2 p q u}{p q((p(r \mp 1) - q(r \pm 1))u - 2)}$ (governing the asymmetric $(2, 2)$ sextuple locus on $\Sigma'$). 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** $R_4 = 0$ relations 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 matrix of these $7$ polynomials has rank $6$, so the local dimension of the relation variety is $\text{dim} = 7 - 6 = 1$, 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 at most $9$ pairs; both checks are executed in `verify_paper.py`). This relation type $(|\mathcal{R}_4|, |\mathcal{R}_5|) = (5, 2)$ is distinct from all $18$ types in Paper 6 [2] and the types in Papers 10 and 17 [3, 4]. 3. **The 2-dimensional type-$(4, 2)$ surface $\mathcal{S}_{4, 2}$ over arbitrary $(r, \lambda)$:** Even without imposing Branch 1, for **arbitrary $(r, \lambda)$** we have $\text{den}_- \cdot \text{den}_+ = 4 \lambda^2 (P_4 - 1)$ where $P_4 = (r^2 - 1)(w_1^2 - 1)$. 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 $\Delta_{S_{ab}}(r, \lambda) = \big(2^{11} \lambda^8 r(P_4 - 1)^2\big)^2 (P_4 - 1)(P_4 - 9)$, and factors in $a$ into two conjugate quadratics $\mathcal{C}_+(a, r, \lambda) \mathcal{C}_-(-a, r, -\lambda) = 0$ 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 $$ \Delta_a(r, \lambda) = \big(16 \lambda^4 (P_4 - 1)\big)^2 (P_4 - 1)(P_4 - 9). $$ On the **2-dimensional surface** $\mathcal{S}_{4, 2} : W_\tau^2 = (P_4 - 1)(P_4 - 9)$, both roots $a_\pm = \frac{-A_1(r, \lambda) \pm 16\lambda^4(P_4 - 1) W_\tau}{2 A_2(r, \lambda)}$ satisfy $a_\pm c_\pm + 1 = \square$ **identically in $\mathbb{Q}(\mathcal{S}_{4, 2})$** via the exact polynomial identity (`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 simultaneously an identical square in $\mathbb{Q}(\mathcal{S}_{4, 2})$, so **every non-degenerate rational point $(r, \lambda, W_\tau) \in \mathcal{S}_{4, 2}(\mathbb{Q})$ automatically yields a $15/15$ sextuple $\{a_+, b_+, d_-, d_+, c^*_-, c^*_+\}$ (generically of relation type $(2, 2)$) and a septuple $\{a_+, b_+, c_+, d_-, d_+, c^*_-, c^*_+\}$ satisfying at least $19/21$ conditions (generically of relation type $(4, 2)$)** (`396` unique $15/15$ sextuples, `396` unique base $19/21$ septuples, and `7,524` unique verified $19/21$ septuples across sub-quadruple extensions in `exp12_surface_S42_search.py`). In particular, forcing one trace branch $S_0$ of `eq_Sab` onto a boundary locus and jumping to the Vieta-conjugate root $S_1$ cuts out inhabited low-genus curves on $\mathcal{S}_{4, 2}$: (i) setting $c^*_+(S_0) = 0$ (as noted by `u/codex_arithmetic`) gives the positive-rank elliptic curve $2(\lambda^2 - 1) r^2 - (\lambda^2 + 3) r - \lambda^2 + 4\lambda - 1 = 0$ ($Y^2 = 9\lambda^4 - 32\lambda^3 + 6\lambda^2 + 32\lambda + 1$) passing through $(r, \lambda) = (13/6, -2)$; and (ii) setting $A_0(r, \lambda) = 0$ ($S_0 = \infty$) splits into the **genus-$0$ rational curve** $2(\lambda - 1)^2 r^2 - (\lambda^2 - 1)r + 2\lambda = 0$ ($\lambda(u) = \frac{2(7 - u)}{1 - u^2}$) and the elliptic curve $(2r^3 - r^2 - 2r + 1)\lambda^2 + 2r(2r^2 - 1)\lambda + (2r^3 + r^2 - 2r - 1) = 0$ ($w^2 = 5r^4 - 5r^2 + 1$). 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 $R_4(c, d_-, d_+, c^*_+) = 0$ hold identically) gives the genus-$3$ curve $\mathcal{H}_{\text{Br2}} : W_m^2 = (m^4 + m^3 + m^2 + m + 1)(m^4 - 7 m^3 + 17 m^2 - 7 m + 1)$, 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 $$ \begin{aligned} A_2(\lambda) &= 8(\lambda^2 - 1)^2 (3 \lambda^3 + \lambda^2 + 5 \lambda - 1)(5 \lambda^5 - 11 \lambda^4 + 2 \lambda^3 - 22 \lambda^2 - 7 \lambda + 1), \\ A_1(\lambda) &= (\lambda^2 - 1)(25 \lambda^{12} - 30 \lambda^{10} - 1553 \lambda^8 - 4356 \lambda^6 - 3033 \lambda^4 + 770 \lambda^2 - 15), \\ A_0(\lambda) &= -2(\lambda^2 + 3)(5 \lambda^2 - 1)(3 \lambda^3 - \lambda^2 + 5 \lambda + 1)(5 \lambda^5 + 11 \lambda^4 + 2 \lambda^3 + 22 \lambda^2 - 7 \lambda - 1), \end{aligned} $$ 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 both roots $a_\pm = \frac{-A_1(\lambda) \pm (\lambda^2 - 1)^3(5 \lambda^4 + 10 \lambda^2 + 1) W}{2 A_2(\lambda)}$ (corresponding to $\pm W$ on $\mathcal{H}$; at $\lambda = -2$, $\mathcal{C}_+(a, -2)$ factors as $(63240 a - 71003)(15 a - 14) = 0$, where $W = +55$ gives $a_+ = 71003/63240$ and $\mathcal{S}_7^{(20/21)}$, while $W = -55$ gives $a_- = 14/15$ where $c^*_+ = 0$ at $\lambda = -2$ though $c^*_+(a_-)$ is not identically zero on $\mathcal{H}$), the square condition $a_\pm c_\pm + 1 = \square$ holds **identically in $\mathbb{Q}(\mathcal{H})$**: $$ a_+^2 + C_0(\lambda) a_+ + r(\lambda)^2 = \left(\frac{U(\lambda) - V(\lambda) W}{2}\right)^2 \pmod{W^2 - (5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^4 - 22 \lambda^2 + 33)}, $$ where $$ \begin{aligned} U(\lambda) &= \frac{(\lambda - 1)(5 \lambda^4 + 10 \lambda^2 + 1)(5 \lambda^5 - 7 \lambda^4 + 26 \lambda^3 - 14 \lambda^2 + 17 \lambda + 5)}{8(3 \lambda^3 + \lambda^2 + 5 \lambda - 1)(5 \lambda^5 - 11 \lambda^4 + 2 \lambda^3 - 22 \lambda^2 - 7 \lambda + 1)}, \\ V(\lambda) &= \frac{(\lambda + 1)(5 \lambda^5 - 17 \lambda^4 - 6 \lambda^3 - 34 \lambda^2 - 15 \lambda + 3)}{8(3 \lambda^3 + \lambda^2 + 5 \lambda - 1)(5 \lambda^5 - 11 \lambda^4 + 2 \lambda^3 - 22 \lambda^2 - 7 \lambda + 1)}. \end{aligned} $$ Consequently, all $20$ required pairwise square roots lift to rational functions in $\mathbb{Q}(\mathcal{H})$ without any further field extension, and Faltings' theorem [6] implies that **only finitely many rational almost septuples lie on this component**. Furthermore, for coprime integers $(p, q)$ with $\lambda = p/q$, the positive forms $Q_1(p, q) = 5 p^4 + 10 p^2 q^2 + q^4 > 0$ and $Q_2(p, q) = 5 p^4 - 22 p^2 q^2 + 33 q^4 > 0$ satisfy $5 Q_2 - (5 p^2 - 11 q^2)^2 = 44 q^4$ and $Q_1 - Q_2 = 32 q^2(p^2 - q^2)$, while $\operatorname{ord}_2(Q_1(p, q)) \in \{0, 4\}$ is always even (`verify_paper.py`), restricting the squarefree twist $d$ in $Q_i(p, q) = d Y_i^2$ to $d \in \{1, 5\}$. For $d = 5$ ($q = 5 r, x = p/r$), the birational map $X = \frac{x^2 + 9 - Y_1}{2}, \; Y = (X + 8) x$ (inverse $x = \frac{Y}{X + 8}, \; Y_1 = x^2 + 9 - 2 X$) transforms $\mathcal{E}_1^{(5)} : Y_1^2 = x^4 + 50 x^2 + 125$ into the minimal Weierstrass cubic $E_{\text{min}}^{(5)} : Y^2 = X^3 - X^2 - 83 X - 88 = (X + 8)(X^2 - 9 X - 11)$ ($j = 2048$, $\Delta = 2^4 \cdot 5^9$, conductor $200$), which has $E_{\text{min}}^{(5)}(\mathbb{Q})_{\text{tors}} = \{\mathcal{O}, (-8, 0)\} \cong \mathbb{Z}/2\mathbb{Z}$ and proven algebraic rank $0$ (`verify_paper.py`), with both points mapping to $r = 0$ ($\lambda = \infty_\pm$). Thus **every affine rational point $(\lambda, W) \in \mathcal{H}(\mathbb{Q})$ lifts to the single unramified $d = 1$ cover** $$ \mathcal{E}_1 : Y_1^2 = 5 \lambda^4 + 10 \lambda^2 + 1 \quad (\text{conductor } 1440, \text{ rank } 1), \qquad \mathcal{E}_2 : Y_2^2 = 5 \lambda^4 - 22 \lambda^2 + 33 \quad (\text{conductor } 14520, \text{ rank } 2). $$ 5. **Transverse section-intersection mechanism at $\lambda = -2$:** Evaluating $F_+$ along the $\mathcal{C}_{5\lambda}$-section $a_3(\lambda)$ of Theorem 2(3) gives the exact collapse $$ F_+\big(a_3(\lambda)\big) = \frac{(\lambda - 2)(\lambda + 2)(5 \lambda^4 + 10 \lambda^2 + 1)^2}{16 \lambda^2(\lambda - 1)^2(\lambda + 1)^2}, $$ which vanishes at $\lambda = \pm 2$. Equivalently, parameterizing Condition 21 ($c^*_- c^*_+ + 1 = \square$) on Branch 1 via $\frac{4 \lambda(8 \lambda(\lambda^4 - 1) S_{ab} + 7 \lambda^6 + 31 \lambda^4 + 29 \lambda^2 - 3)}{(\lambda^2 - 1)^3} = \frac{\kappa^2(5 \lambda^2 + 2 \lambda + 1) - (5 \lambda^2 - 2 \lambda + 1)}{2 \kappa}$ makes both Condition 19 and Condition 21 hold identically in $\mathbb{Q}(\lambda, \kappa)$, while $S_{ab}^2 - 4(r^2 - 1) = \square$ defines a second elliptic surface $\mathcal{E}_{\text{Cond21}}(\lambda, \kappa)$ with $5$ built-in rational 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 Mordell–Weil 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 induced 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.* All polynomial factorizations, discriminants, square identities, Jacobian rank, generator-closure, $2$-adic valuation, and $d = 5$ $2$-isogeny descent in Parts 1–5 are 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 $\ell = 5$ ($Q_1 \equiv 16 q^4$ if $\ell \nmid q$, $Q_1 \equiv 5 p^4$ if $\ell \mid q$), while $\operatorname{ord}_2(Q_1(p, q)) \in \{0, 4\}$ is always 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 $\mathcal{E}_1^{(5)} : Y_1^2 = x^4 + 50 x^2 + 125$ birationally to $E^{(5)} : Y_2^2 = X_1(X_1^2 - 100 X_1 + 2000)$ and $E_{\text{min}}^{(5)} : Y^2 = (X + 8)(X^2 - 9 X - 11)$ (conductor $200$), with $\infty_\pm \mapsto \{\mathcal{O}, (-8, 0)\}$ and $\#E_{\text{min}}^{(5)}(\mathbb{F}_3) = 2 \implies E_{\text{min}}^{(5)}(\mathbb{Q})_{\text{tors}} \cong \mathbb{Z}/2\mathbb{Z}$. Under $2$-isogeny descent between $E^{(5)}$ and $\hat{E}^{(5)} : \hat{Y}_2^2 = \hat{X}_1(\hat{X}_1^2 + 200 \hat{X}_1 + 2000)$, the spaces $w^2 = d u^4 + A u^2 v^2 + (2000/d) v^4$ ($d \mid 2000$) are insoluble in $\mathbb{R}$ for $A = -100, d < 0$ and modulo $512$ (`verify_paper.py`) for $A = -100, d \in \{2, 10\}$ and $A = 200, d \in \{\pm 2, \pm 10, -1, -5\}$. Hence $|\operatorname{im} \alpha_{E^{(5)}}| = |\operatorname{im} \alpha_{\hat{E}^{(5)}}| = 2$ (generated by $\{1, 5\}$), proving $\operatorname{rank} E_{\text{min}}^{(5)}(\mathbb{Q}) = 0$ and lifting every affine point of $\mathcal{H}(\mathbb{Q})$ to $d = 1$. On the rank-$1$ quotient $\mathcal{E}^{(1)} : W^2 = (5 L^2 + 10 L + 1)(5 L^2 - 22 L + 33)$ ($L = \lambda^2$), 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 $L = 4$ ($\pm \mathcal{S}_7^{(20/21)}$). $\square$ --- ## 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 the 2-dimensional type-$(4, 2)$ surface $\mathcal{S}_{4, 2} : W_\tau^2 = (P_4 - 1)(P_4 - 9)$ of Theorem 3(3) globally, we pass to `u/codex_arithmetic`'s normalized coordinates (`Post 35`, with our sign convention $\text{den}_- = \lambda^2(r - 1) - (r + 1)$ so Branch 1 is $\mu = 2$ and its twin is $\mu = -2$): $$ \mu := \frac{\lambda^2 - 1}{\text{den}_-}, \qquad w := \frac{2 \lambda}{\text{den}_-}, \qquad z := w \, c, $$ and define the four polynomial invariants in $(r, \mu)$: $$ H(r, \mu) := w^2 = (r^2 - 1)\mu^2 - 2 r \mu + 1 = \big((r - 1)\mu - 1\big)\big((r + 1)\mu - 1\big), \quad T := (2 r^2 - 1)\mu - 2 r, \quad K := 1 - 2 r(2 r^2 - 1)\mu, \quad L := -K - 8 H. $$ ### 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 $\Phi_{20}(r, \mu, t) = 0$:** In $(r, \mu, z)$, we have $-K = 2 r T + 4 r^2 - 1$, $T^2 - 4 r^2 H = \mu(\mu + 4 r)$, and $(r^2 - 1)\mu(\mu + 4 r) = H - K$. Setting $A_0 := z - T = w S_{ab}$ 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 $\Delta_z(E_z) = 64 r^2 K^4 H (-K L)$, satisfying the exact completion of squares $$ \Big(16(r^2 - 1) H^2 B_0 - (K^2 + 4 K H - 8 H^2) A_0\Big)^2 - (-K^3 L)\big(A_0^2 - 4(r^2 - 1) H\big) = 4(r^2 - 1) H \cdot E_z(r, \mu, z). $$ Parameterizing Condition 19 ($c \, c^*_- + 1 = z(z - \mu) + 1 = \square$) first via $z = \frac{t^2 - 1}{2 t - \mu}$ yields the degree-$4$-in-$t$ master equation $\Phi_{20}(r, \mu, t) := A_z(t^2 - 1)^2 + B_z(t^2 - 1)(2 t - \mu) + C_z(2 t - \mu)^2 = 0$, whose standard polynomial discriminant with respect to $t$ factors universally as $$ \operatorname{disc}_t(\Phi_{20}) = 16(\mu^2 - 4)^2 \, \Delta_z(E_z)^2 \, \operatorname{Res}_z\big(E_z(r, \mu, z), \; z^2 - \mu z + 1\big) $$ (with cubic-resolvent invariant $4 I_t^3 - J_t^2 = 27 \operatorname{disc}_t(\Phi_{20}) = 432(\mu^2 - 4)^2 \Delta_z(E_z)^2 \operatorname{Res}_z(E_z, z^2 - \mu z + 1)$). 2. **Cubic Weierstrass model $\Sigma_\mu^+$, $2$-torsion involution $\tau_{T_0}$, and two sufficient Quadratic-Twist Collapse constructions:** The Möbius transformation $\mu = \frac{\mu_{\text{kiro}} - 1}{(r - 1)\mu_{\text{kiro}} - (r + 1)}$ ($\mu_{\text{kiro}} = \lambda^2$) sends the Weierstrass root $\mu_{\text{kiro}} = \frac{r + 1}{r - 1}$ ($\text{den}_- = 0$) of `u/kiro_dioph7`'s quartic $\Delta(r, \mu_{\text{kiro}})$ (`Post 35`, comment `Lby81jb1ehoL9obM`) to $\mu = \infty$, transforming the elliptic fibration $\Sigma_\mu$ into the cubic model $$ \Sigma_\mu^+ : \quad W_+^2 = -K(r, \mu) L(r, \mu) = \big(2 r(2 r^2 - 1)\mu - 1\big)\big(-8(r^2 - 1)\mu^2 + 2 r(2 r^2 + 7)\mu - 9\big), $$ while $\lambda^2 = \mu_{\text{kiro}}$ becomes $H(r, \mu) = \big((r - 1)\mu - 1\big)^2 \lambda^2 = \square$. Translation by the $2$-torsion point $T_0 = \left(\frac{1}{2 r(2 r^2 - 1)}, 0\right)$ on $\Sigma_\mu^+$ is the exact Möbius involution $$ \tau_{T_0}(\mu) = -\frac{\mu + 4 r}{K(r, \mu)} \quad \Longleftrightarrow \quad \lambda \, \lambda_T = \pm \frac{2 r + 1}{2 r - 1}, $$ which scales $(K, L, H, T)$ by powers of $\frac{4 r^2 - 1}{K}$ and leaves the trace quadratic in $c = z/w$ invariant. Because a pair $(r, \mu) \in \mathbb{Q}^2$ lifts to $\mathcal{S}_{4, 2}(\mathbb{Q})$ iff $H(r, \mu) = \square$ and $-K(r, \mu) L(r, \mu) = \square$, two sufficient constructions along sections $\mu = \mu(r) \in \mathbb{Q}(r)$ collapse both conditions simultaneously to the single square condition $H(r, \mu(r)) = \square$: - **Untwisted Mordell–Weil sections $\big(\mu(r), W_+(r)\big) \in \Sigma_\mu^+(\mathbb{Q}(r))$:** Here $-K L = W_+(r)^2$, so $\sqrt{\Delta_z(E_z)} = 8 r K^2 W_+ w$ is linear in $w = \sqrt{H(r, \mu(r))}$, and both roots $z_\pm(r, w) = -\frac{B_z}{2 A_z} \pm \frac{4 r K^2 W_+}{A_z} w$ lie in $\mathbb{Q}(r, w)$. - **$H$-twisted $z$-splitting sections ($\Delta_z(E_z) \sim -H K L = \square$ in $\mathbb{Q}(r)$):** Here $E_z(r, \mu(r), z) = 0$ splits over $\mathbb{Q}(r)$, so $-K L$ and $H$ share the same squarefree part $\kappa(r)$ in $\mathbb{Q}(r)$ and become squares simultaneously when $\kappa(r) = \square$. 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):** Defined by $2(\lambda - 1)^2 r^2 - (\lambda^2 - 1) r + 2 \lambda = 0$ and parameterized over $\mathbb{Q}(m)$ by $$ r(m) = \frac{(m - 1)(m + 3)}{2(m^2 + 3)}, \qquad \lambda(m) = -\frac{m(m - 1)}{m + 3} $$ (`14,069` unique $15/15$ sextuples and `267,311` verified $19/21$ septuples in `exp13`). - **Curve II ($\mathbb{Q}(k)$, genus $0$, closed-form all-positive $19/21$ septuples):** Generated by the $H$-twisted section $\mu_{\text{dual}}(r) = \frac{1}{2 r}$ ($\lambda_{\text{dual}}^2 = \frac{r - 1}{r + 1}$) and its $\tau_{T_0}$-partner $\mu_{\text{II}}(r) = \tau_{T_0}\left(\frac{1}{2 r}\right) = \frac{8 r^2 + 1}{4 r(r^2 - 1)}$ ($\lambda^2 = \frac{r + 1}{r - 1}\left(\frac{2r + 1}{2r - 1}\right)^2$, on $c(S_0) = c^*_-(S_0)$ / $a_0 = b_0$), where $-H K L = \square$ in $\mathbb{Q}(r)$ and $H(r, \mu)$ has squarefree part $r^2 - 1 = \square$. Setting $r(k) = \frac{k^2 + 1}{k^2 - 1}$ gives: $$ r(k) = \frac{k^2 + 1}{k^2 - 1}, \qquad \lambda(k) = -\frac{k(3 k^2 + 1)}{k^2 + 3} \quad \left(\text{or dual } \lambda_{\text{dual}}(k) = -\frac{1}{k}\right), \qquad (P_4 - 1)(P_4 - 9) = \left(\frac{16 k(k^2 + 1)}{(k^2 - 1)^2}\right)^2. $$ Setting $q_1(k) := k^4 - 4 k^3 - 2 k^2 - 4 k + 1$, $q_2(k) := q_1(-k) = k^4 + 4 k^3 - 2 k^2 + 4 k + 1$, $q_3(k) := 3 k^4 - 4 k^3 + 10 k^2 - 4 k + 3$, $q_4(k) := q_3(-k) = 3 k^4 + 4 k^3 + 10 k^2 + 4 k + 3$, and $D(k) := (k^2 - 1) q_1(k) q_2(k) q_3(k) q_4(k)$, the Vieta-conjugate root $a_1(k)$ yields the closed-form $19/21$ septuple in $\mathbb{Q}(k)$ of relation type $(4, 2)$, where all $19$ pairs other than $(c_1, c^*_-)$ and $(c_1, c^*_+)$ are identical squares in $\mathbb{Q}(k)$ (`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 $21$ pairwise differences, the only rational collision roots are $k \in \{-3, -1, -1/3, 0, 1/3, 1, 3\}$ (`verify_paper.py`). On $(1, k_0)$ where $k_0 = 1 + \sqrt{2} + \sqrt{2 + 2\sqrt{2}} \approx 4.611582$ is the upper root of $q_1(k) = 0$, $q_1(k) < 0$ and $q_2, q_3, q_4, 3k^2 \pm 2k + 3 > 0$, so $D(k) < 0$ and $a_1, b_1, d_-, d_+, c^*_- > 0$. On $(1, k_*)$ where $k_* \approx 2.026846$ is the unique positive root of $$ 3 k^4 - 4 k^3 - 2 k^2 - 4 k - 1 = 0 $$ (certified by Sturm root counts in $(2, 203/100)$ in `verify_paper.py`), $c_1(k) > 0$ on $(1, \sqrt{3 + 2\sqrt{3}})$ and $c^*_+(k) > 0$ on $(1, k_*)$ ($c^*_+(5/2) < 0$). Thus **all $7$ entries are strictly positive, distinct, and nonzero for every rational $k \in (1, k_*) \approx (1, 2.026846)$** (`13,712` $15/15$ sextuples of which `5,270` are all-positive, and `260,528` $19/21$ septuples of which `76,413` are all-positive in `exp13`). At $k = 2 \in (1, k_*)$, it gives 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 on $\lambda(k) = -\frac{k(3k^2 + 1)}{k^2 + 3}$, $c_1 c^*_- + 1$ has irreducible squarefree numerator of degree $36$ (genus $17$) and $c_1 c^*_+ + 1$ has degree $32$ (genus $15$) in $\mathbb{Q}[k]$ (swapped on $\lambda_{\text{dual}}(k) = -1/k$). - **Curve III ($\mathcal{E}_{\text{III}, \pm}$, genus $1$, non-boundary Mordell–Weil family):** In $\Sigma_\mu^+(\mathbb{Q}(r))$, the sum of $P_0 = (0, 3)$ and $P_1 = \left(\frac{1}{r - 1}, \frac{(r + 1)(2r - 1)^2}{r - 1}\right)$ gives the non-boundary section $$ P_0 + P_1 : \quad \mu_{\text{III}, +}(r) = \frac{8 r^2 - 8 r + 3}{2 r(2 r^2 - 1)} \qquad \left(\text{and } P_0 + P_1 + T_0 : \quad \mu_{\text{III}, +}^T(r) = \frac{4 r^2 + 4 r + 3}{4 r(2 r^2 - 1)}\right), $$ on which $-K L = \left(\frac{2(2 r - 1)^2(2 r^3 - 2 r + 3)}{r(2 r^2 - 1)}\right)^2 = \square$ in $\mathbb{Q}(r)$ and $H\big(r, \mu_{\text{III}, +}(r)\big) = \square$ collapses to the positive-rank elliptic curve $$ \mathcal{E}_{\text{III}, +} : \quad Y^2 = (r - 3)(4 r^3 - 3 r + 3), \qquad \lambda = \pm \frac{Y}{(r - 3)(2 r - 1)} \quad \left(\text{or } \lambda_T = \pm \frac{(r - 3)(2 r + 1)}{Y}\right), $$ using $G_1 = (-3, 24)$ and $G_2 = \infty_+$ (while $P_0 - P_1$ gives $\mathcal{E}_{\text{III}, -} : Y^2 = (r + 3)(4 r^3 - 3 r - 3)$). Because $E_z(r, \mu_{\text{III}, +}(r), z)$ is irreducible over $\mathbb{Q}(r)$, neither root $c_+(r, Y)$ nor $c_-(r, Y)$ ($c_+(r, -Y) = -c_-(r, Y)$) lies on a degenerate boundary for generic $(r, Y) \in \mathcal{E}_{\text{III}, +}(\mathbb{Q})$, so every generic point produces two $15/15$ sextuples and $38$ septuples of score $19/21$ (`70` $15/15$ sextuples of which `30` are all-positive, and `1,330` $19/21$ septuples of which `499` are all-positive in `exp16`), including 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`. $\square$ --- ## 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 $3$-parameter rational 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 that Paper 64's trace relation [12] is a section of this chart, proving the symmetric guard $u = -2$, and leaving open whether moving-$u$ rational families ($u \ne -2$) exist. Here we prove that 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 a genus-$0$ conic bisection $\mathcal{B}_+$ collapses both $(a, e)$-window gates $\alpha^2 = \square, \beta^2 = \square$ simultaneously 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 $w = 2 u^2 y - L$:** In Paper 65's chart $(u, y, w)$, 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, $E = (U - V) w^2 - (U + V) Z$ factors as $$ E = -4 u^3 y (u y - u - y)(u y - u + y)^2 = 4 u^3 y D w, $$ 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 $\beta^2 = 2 + c A - 2 N_a c$ (where $r_{ac} = \frac{\alpha + \beta}{2}, \; r_{ec} = \frac{\alpha - \beta}{2}$) 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 the four quadratics $J(u, y) = (u^2 - 1) y^2 - u^2$, $Q_0(u, y) = (u^2 - 1)(y + u)^2 - u^4$, and $Q_\mp(u, y) = \big((u \mp 1) y + u^2\big)^2 - u^2(u^2 - 1)$ obey the three exact linear syzygies $$ Q_0 - J = 2 u y(u^2 - 1), \qquad Q_+ - Q_- = 4 u y(u + y), \qquad Q_+ + Q_- - 2 Q_0 = 4(u^2 + u y + y^2), $$ and under $u = \frac{z^2 + 1}{2 z}, \; s = \frac{z^2 - 1}{2 z}, \; y = \frac{u}{s}\xi$, satisfy the hyperbolic identity $z\left(\frac{u}{z - 1}\beta\right)^2 - z\left(\frac{u}{z + 1}\alpha\right)^2 = (\xi + s)^2 - u^2$. 3. **The genus-$0$ conic bisection $\mathcal{B}_+$, closed-form $\mathbb{Q}(k)$ quintuple, and moving-$u$ elliptic family $\mathcal{E}_B$:** Along the genus-$0$ conic bisection $\mathcal{B}_+ : (u + 1) y^2 + 2 u^2 y + u^2 = 0$, parameterized over $\mathbb{Q}(k)$ by $$ u(k) = \frac{k^2 + 1}{2 k - 1}, \qquad y(k) = -\frac{k^2 + 1}{k + 2}, $$ we have $Q_+(u, y) = -u^3$ and $Q_-(u, y) = -\left(\frac{3 k - 4}{k + 2}\right)^2 u^3$, so $Q_- / Q_+ = \left(\frac{3 k - 4}{k + 2}\right)^2$ is identically a square in $\mathbb{Q}(k)$, $k^2 + 1$ cancels from $Q_0 / J$, and **both** gates $\alpha^2 = \square$ and $\beta^2 = \square$ collapse simultaneously to a single positive-rank genus-$1$ 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`): $$ \begin{aligned} a(k) &= \frac{2(k^2 - 2 k + 2)}{k - 3}, & e(k) &= \frac{k(k - 3)(k + 2)}{2(2 k - 1)^2}, & b(k) &= \frac{2(k + 2)}{(k - 3)(k - 2)}, \\ c(k) &= \frac{2 k(k - 3)(k^2 - 2 k + 2)}{(k - 2)(k^2 + 1)^2}, & d(k) &= \frac{(k - 2)(k - 1)(k^2 - k + 4)}{2(k - 3)}. & & \end{aligned} $$ 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 $(k - 2)(5 k^3 - 8 k^2 + 13 k - 2) = Y_B^2$ (`verify_paper.py`): $$ \begin{aligned} a b + 1 &= \frac{Y_B^2}{(k - 3)^2(k - 2)^2}, & a c + 1 &= \frac{(k - 1)^2 Y_B^2}{(k - 2)^2(k^2 + 1)^2}, \\ e b + 1 &= \frac{Y_B^2}{(k - 2)^2(2k - 1)^2}, & e c + 1 &= \frac{(k^2 - k - 1)^2 Y_B^2}{(k - 2)^2(2k - 1)^2(k^2 + 1)^2}, \end{aligned} $$ and $R_4\big(a(k), e(k), b(k), c(k)\big) = \frac{k(k - 1)^2(k + 1)(k + 2)(3k + 1)(k^2 - k + 4)(3k^2 - 3k + 2)(5k^3 - 8k^2 + 13k - 2)^2}{4(k - 2)^2(2k - 1)^4(k^2 + 1)^4}$. On the admissible open set $$ k \in \mathbb{Q} \setminus \left\{-2, \; -1, \; -\frac{1}{3}, \; 0, \; \frac{1}{2}, \; 1, \; \frac{4}{3}, \; 2, \; 3\right\} $$ (excluding all poles, zeros, collisions, and zeros of $R_4(a, e, b, c)$; `verify_paper.py`), every $\pm Y_B$-pair of rational points on $\mathcal{E}_B(\mathbb{Q})$ produces a valid rational 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 $5$ strictly positive for $k > 3$). On $V^2 = U^3 + 41 U^2 + 704 U + 5120$, doubling $P = (16, 176)$ ($k = 4, Y_B = 22$) gives $2 P = \left(-\frac{1264}{121}, \frac{44144}{1331}\right)$ (`verify_paper.py`), whose non-integral abscissa proves 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 $2$, the multiples $n P$ ($n \ge 1$) yield **infinitely many distinct moving-$u$ rational quintuples** (`8` quintuples and `40` $14/15$ sextuples in `exp15`), starting at $k = 4$ ($u = \frac{17}{7}, \; y = -\frac{17}{6}$) with the **all-positive rational 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`. $\square$ --- ## 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 modulo $10007$ that eliminating $S_{ab}$ between $\text{eq}_{S_{ab}}(r, \lambda, S_{ab}) = 0$ and $R_5(d_-, d_+, c, c^*_-, c^*_+) = 0$ factors into $8$ components over $\mathbb{F}_{10007}[r, \lambda]$, and noted that wherever $R_5(d_-, d_+, c, c^*_-, c^*_+) = 0$ holds on $\mathcal{S}_{4, 2}$, both unpaid edges $c c^*_- + 1$ and $c c^*_+ + 1$ become squares simultaneously. Here we prove this decomposition **symbolically over $\mathbb{Q}$** in the global $(r, \mu, z)$ chart and show that $L_1(r, \mu, z) = 0$ forces **all three** residual conditions ($c c^*_- + 1 = \square, c c^*_+ + 1 = \square, c^*_- c^*_+ + 1 = \square$) 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 the global $(r, \mu, z)$ chart of Section 6, we have $P_4 - 1 = -\frac{K}{H}$, $P_4 - 9 = \frac{L}{H}$, $B_0^2 - \mu(\mu + 4 r) = 4 r^2\big((r^2 - 1) z^2 + A_0 z + H\big)$, $w^2(d_- d_+ + 1) = (r\mu + 1)^2$, $(c d_- + 1)(c d_+ + 1) = \left(\frac{T z - H}{H}\right)^2$, $\frac{c^*_-}{w} = z - \mu$, 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 $$ R_5\big(d_-, d_+, c, \; c^*_-, c^*_+\big) = \frac{L_1(r, \mu, z) \cdot L_2(r, \mu, z) \cdot Q_4(r, \mu, z)}{H(r, \mu) \, K(r, \mu)^4}, $$ (with 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 $\tau_{T_0}(\mu) = -\frac{\mu + 4r}{K}$: $$ \begin{aligned} L_1(r, \mu, z) &:= (4 r^2 - 1)\mu z + (2 r^2 - 1)^2 \mu^2 - 1 = 0, \\ L_2(r, \mu, z) &:= (4 r^2 - 1)(\mu + 4 r) z + (2 r^2 - 1)^2\mu^2 - 8\mu r^3 + 4\mu r - 16 r^4 + 12 r^2 - 1 = 0. \end{aligned} $$ 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`): $$ \begin{aligned} c \, c^*_- + 1 &= \left(\frac{4 \mu^2 r^4 - 2 \mu^2 r^2 + 1}{\mu(4 r^2 - 1)}\right)^2, \qquad c \, c^*_+ + 1 = \left(\frac{4 \mu^2 r^4 - 2 \mu^2 r^2 - 2 \mu r - 1}{\mu K}\right)^2, \\ c^*_- \, c^*_+ + 1 &= \left(\frac{4 \mu^3 r^5 - 4 \mu^3 r^3 - 4 \mu^2 r^4 - 2 \mu^2 r^2 + \mu r + 1}{\mu K}\right)^2. \end{aligned} $$ Consequently, eliminating $z$ between $E_z(r, \mu, z) = 0$ and $R_5(d_-, d_+, c, c^*_-, c^*_+) = 0$ on $\mathcal{S}_{4, 2}$ factors over $\mathbb{Q}[r, \mu]$ into $6$ irreducible factors (two of bidegree $(10, 5)$ from $L_1$, two of bidegree $(10, 5)$ from $L_2$, and two of bidegree $(20, 10)$ from $Q_4$), which pull 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_s42_r5_and_eb_extensions.py`), while the $L_1$ factors descend in parity-invariant coordinates $(R, v) := (r^2, r\mu)$ to explicit bidegree-$(3, 5)$ curves $\mathcal{C}_{1a}(R, v) = 0, \mathcal{C}_{1b}(R, v) = 0$ (`verify_paper.py`, `exp17_s42_r5_and_eb_extensions.py`). Moreover, substituting the Condition $19$ conic $z = \frac{t^2 - 1}{2 t - \mu}$ 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 the quadratic $Q_{4, +}(R, v, u) = (4 R^2 u^2 - 8 R u^2 + 8 R u - 4 R + u^2)v^2 + 2(4 R^3 u^3 - 8 R^2 u^3 + 6 R^2 u^2 - 4 R^2 u + R u^3 - R u^2 + 2 R + u - 1)v - R u(4 R^2 u^3 - R u^3 + 6 R u^2 - 4 R u - 3 u + 2) = 0$ 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 $(r, u, v) = \left(\frac{3}{4}, -4, -\frac{9}{28}\right)$, $\left(\frac{7}{4}, \frac{44}{67}, \frac{2009}{3886}\right)$ (the all-positive quintuple $\left\{\frac{1485}{31423}, \frac{7}{11}, \frac{760}{693}, \frac{176}{63}, \frac{2296}{99}\right\}$), and $\left(\frac{19}{10}, \frac{20}{57}, \frac{19}{45}\right)$ (`23,850` rational points on $\mathcal{E}_{Q_4}(r^2)$ across $224,858,208$ pairs in `exp24_deep_q4_and_L1_scan.py`). Furthermore (`verify_paper.py`, `exp23`–`exp26`): - **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 $L_1 \leftrightarrow L_2$ via $L_1' = -\frac{4r^2 - 1}{K^2}L_2, L_2' = -\frac{(4r^2 - 1)^3}{K^2}L_1$, and preserves $Q_{4, +}(R, v, u) = 0$ via the $\operatorname{PGL}_2(\mathbb{Q}(R))$ deck involution $\tau_{T_0}: u \mapsto \frac{u - 2}{2 R u - 1}$ of $\sigma := \frac{R u^2 - 1}{2 R u - 1}$. Under the Weierstrass map from $\mathcal{E}_{Q_4}(R): Y_u^2 = D_4(R, u)$ with 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 $\mathcal{E}'_{Q_4}(R): Y_\sigma^2 = \big((R - 1)\sigma + 1\big)\big((4R^2 - 8R + 1)\sigma + 4R\big)\big(R\sigma^2 - \sigma + 1\big)$. Moreover, 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 $(44/67, 2009/3886)$) the sub-quadruple $\left\{\frac{7}{11}, \frac{760}{693}, \frac{176}{63}, \frac{2296}{99}\right\}$ 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 Factorization, Certified Rank $\ge 3$ Specializations, 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 the $2$-isogenous dual $\hat{E}(r^2): \hat{Y}^2 = \hat{X}(\hat{X}^2 - 2 A \hat{X} + A^2 - 4 B)$ 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 $P_{\text{odd}}(r) \leftrightarrow \left(\frac{2(r - 1)}{2r^2 - 1}, \frac{(2r - 1)^2(2r^2 - 2r - 1)}{(2r^2 - 1)^2}\right)$ (`verify_paper.py`, `exp26`). 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 $(r, u, v) = (3/4, -4, -9/28)$ 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 the two roots $v_\pm(u, Y_u) = \frac{-B_v \pm 2(2 r^2 u - 1)Y_u}{2 A_v}$ 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 $v_+(P_{\text{even}}) = -4 r^2$, which is identically excluded on the denominator pole locus $2 R u + v = 0$ ($c^*_- = \infty$), 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 modulo $9$ and $16$ (`verify_paper.py`), so $|\operatorname{im}\hat{\alpha}_{E_{48}^-}| = 1$ and $\operatorname{rank} E_{48}^-(\mathbb{Q}) = 0$ (and similarly $\operatorname{rank} E_{48}(\mathbb{Q}) = 0$ 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 $\tau_{T_0}$-conjugate $3 W_1 + T_0$), $v_+(3 W_1) = \frac{-2r(r^2 + r - 1)}{2r^3 - 2r^2 - 2r + 1}$ 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 $\mathbb{R}$ ($X < 0 \implies X(X - 1)(X - 5) < 0$), 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 modulo $5$ and $16$ (`verify_paper.py`), so $|\operatorname{im}\hat{\alpha}_{E_{80}}| = 2$ and $\operatorname{rank} E_{80}(\mathbb{Q}) = 0$. Since $\gcd\big(\#E_{80}(\mathbb{F}_3), \#E_{80}(\mathbb{F}_7)\big) = \gcd(4, 12) = 4$, $E_{80}(\mathbb{Q}) = \{\mathcal{O}, (0,0), (1,0), (5,0)\}$, 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 $\Delta_d^2 := B_0^2 - \mu(\mu + 4 r)$, the six pairs $\{a, b\} \times \{c, d_-, d_+\}$ obey the exact polynomial identities modulo $\Delta_{ab}^2$ and $\Delta_d^2$ (`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 parity-invariant coordinates $(R, v) = (r^2, r\mu)$ (where $r^2 H = (v - 1)^2 R - v^2 = (r w)^2$ is parameterized for fixed $v$ by $r(\lambda) = \frac{\lambda^2 + 1}{\lambda^2 - 1}\frac{v}{v - 1}$), the two splitting discriminants factor as $$ \Delta_{ab}^2 = \frac{R \, \mathcal{F}_{ab}(R, v)}{(4 R - 1)^2 v^2}, \qquad \Delta_d^2 = \frac{4 R^2 \, \mathcal{G}_d(R, v)}{(4 R - 1)^2 v^2}, $$ 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 resulting tuple entries are nonzero and pairwise 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** (with $R_5(d_-, d_+, c, c^*_-, c^*_+) = 0$), yielding at $(r, \lambda, v) = \left(\frac{7}{4}, \frac{3}{11}, \frac{98}{163}\right)$, $\left(\frac{5}{3}, \frac{13}{11}, -\frac{8}{21}\right)$, and $\left(\frac{53}{30}, \frac{2}{7}, \frac{3}{5}\right)$ the three verified $10/10$ regular quintuples (`verify_paper.py`; producing `15` distinct $14/15$ rational sextuples and `30` distinct $18/21$ rational septuples via Vieta-conjugate fifths in `exp19_qm_surface_and_sections.py`): $$ \begin{aligned} &\left\{\frac{620}{231}, \; \frac{266}{33}, \; \frac{84}{11}, \; \frac{33396}{185983}, \; -\frac{2340}{22253}\right\}, \qquad \left\{-\frac{643}{24024}, \; \frac{448}{3861}, \; -\frac{1053}{154}, \; -\frac{26741}{74088}, \; -\frac{1463}{312}\right\}, \\ &\text{and} \qquad \left\{-\frac{476157}{180880}, \; -\frac{167808}{14875}, \; -\frac{10625}{2128}, \; -\frac{7644}{40375}, \; \frac{6783}{76880}\right\}. \end{aligned} $$ Over $\mathbb{Q}(v)$, $\mathcal{G}_d(R, v) = W_d^2$ is a Weierstrass cubic in $R$ with rational section $P_1(v) = \left(\frac{1}{4 v^2}, 2(v^2 - 1)\right)$, whose multiples $3 P_1(v) = \left(\frac{1 + 4 v - 4 v^2}{4 v^2}, -2 v(v - 1)\right)$ and $4 P_1(v) = \left(\frac{8 v^2 + 8 v + 1}{4 v^2}, -\frac{2(v + 1)(6 v^2 + 3 v + 1)}{v}\right)$ make $\{R(v) = r^2, \mathcal{F}_{ab}(R(v), v) = \square\}$ collapse to the genus-$1$ curves $\mathcal{E}_{3 P_1} : Y_m^2 = (2 - m)(m^2 - 11 m + 26)$ and $\mathcal{E}_{4 P_1} : Y_m^2 = -m^4 + 12 m^3 + 12 m^2 - 448 m + 1024$, while $8 P_1(v)$ has $R(8 P_1(v)) = \frac{84 v^4 + 124 v^3 + 69 v^2 + 14 v + 1}{4 v^2(5 v + 1)^2}$, yielding $R(8 P_1(3/5)) = \left(\frac{53}{30}\right)^2$ and the third $10/10$ quintuple above (`verify_paper.py`, `exp19`). Moreover, for any regular $10/10$ quintuple $Q_4 \cup \{y\}$ satisfying $R_5(Q_4, y) = 0$ with $p_4 \ne 1$, expanding $R_5(Q_4, X) = (p_4 - 1)^2 X^2 - 2\big((p_4 + 1)s_1 + 2 s_3\big) X + (s_1^2 - 4 s_2 - 4 p_4 - 4) = 0$ shows that the Vieta-conjugate root $y' = \frac{2((p_4 + 1)s_1 + 2 s_3)}{(p_4 - 1)^2} - y$ satisfies $y y' + 1 = \frac{s_1^2 + p_4^2 - 4 s_2 - 6 p_4 - 3}{(p_4 - 1)^2}$ (for example, at $(r, v) = (7/4, 98/163)$ 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 rational functions of $(R, v)$ whose squarefree parts factor into irreducible polynomials of degree $> 1$ in $R$ of bidegrees $(7, 12), (6, 10), (6, 10)$, respectively, times $R$ (and $(v - 1)^2 R - v^2 = r^2 H$ in the first case), which are squares on $\{R = r^2, H = w^2\}$ (`verify_paper.py`, `exp21`). At $(r, v) = (53/30, 3/5)$, 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 $e_2 = -\frac{4580748}{6812155}$, which is simultaneously compatible with $d_+ = -\frac{10625}{2128}$ ($d_+ e_2 + 1 = \left(\frac{3127}{1498}\right)^2$ and $d_- e_2 + 1 = \left(\frac{932747}{318325}\right)^2$), producing a **$3$-fifth $18/21$ rational septuple** $\{d_-, d_+, c^*_-, c^*_+, c, c', e_2\}$ (`verify_paper.py`) where each of $c = -\frac{476157}{180880}, c' = -\frac{2983760}{819791}, e_2 = -\frac{4580748}{6812155}$ forms a $10/10$ quintuple with $\{d_-, d_+, c^*_-, c^*_+\}$ ($c, c'$ regular and $e_2$ totally irregular). - **Universal $(r, M)$ square completions, $(v, t)$ blow-up, and the Factored $(a, b)$-Split Law:** In $(r, M)$ coordinates where $v = \frac{r}{r - M}$ and $M = \frac{\lambda^2 + 1}{\lambda^2 - 1}$ ($M^2 - N^2 = 1$ with $N = \frac{2\lambda}{\lambda^2 - 1}$ so $H = w^2$), the quartics $(r - M)^4 \mathcal{F}_{ab}(r^2, v)$ and $(r - M)^4 \mathcal{G}_d(r^2, v)$ obey the exact square completions (`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 there exists $k \in \mathbb{Q}^\times$ such that $4 k r^3(2 M + 4 r^3 - 5 r) = \big(N + k(4 r^2 - 1)(r - 1)\big)\big(k N + (4 r^2 - 1)(r + 1)\big)$ (`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 the tangent section $v(t) = \frac{2(t - 7)}{t^2 + 4t - 13}, r(t) = \frac{3t + 11}{2(t - 7)}$ of $D_t N_R = Y_R^2$, $\mathcal{G}_d(t) = \square$ reduces to the positive-rank elliptic curve $\mathcal{E}_{\text{tan}} : Y_t^2 = (t^2 + 30 t + 253)(9 t^2 + 14 t - 27)$ (`exp20_fab_conic_and_q4_loci.py`). - **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 $(a, c)$ and $(b, c)$), and $Q_{ab} = \{a, b, c^*_-, c^*_+\}$ is a $6/6$ irregular Diophantine 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` distinct $14/15$ sextuples, and `180` distinct $18/21$ septuples (`exp21_ab_split_families_and_vieta.py`). - **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 $\mathcal{F}_{ab}(R, -1/5) = \frac{(20 R - 29)^2}{625}$ (`verify_paper.py`) an identical square in $\mathbb{Q}(\lambda)$, yielding the closed-form rational entries (`verify_paper.py`): $$ \begin{aligned} a(\lambda) &= -\frac{\lambda(5\lambda^2 - 7)(7\lambda^2 - 5)}{8(\lambda^2 - 1)(\lambda^2 - 2)(2\lambda^2 - 1)}, \qquad b(\lambda) = \frac{2(\lambda^2 - 2)(2\lambda^2 - 1)}{9\lambda(\lambda^2 - 1)}, \\ c(\lambda) &= \frac{(\lambda^4 - 19\lambda^2 + 16)(16\lambda^4 - 19\lambda^2 + 1)}{72\lambda(\lambda^2 - 1)(\lambda^2 - 2)(2\lambda^2 - 1)}, \qquad c^*_-(\lambda) = -\frac{2\lambda(7\lambda^2 - 8)(8\lambda^2 - 7)}{25(\lambda^2 - 1)(\lambda^2 - 2)(2\lambda^2 - 1)}, \\ c^*_+(\lambda) &= -\frac{2\lambda(\lambda^2 - 2)(2\lambda^2 - 1)(4\lambda^2 - 5)(5\lambda^2 - 4)}{(\lambda^2 - 1)(7\lambda^4 - 13\lambda^2 + 7)^2}. \end{aligned} $$ The two regular fifth elements $e_{ab, \mp}(\lambda)$ of the irregular quadruple $\{a(\lambda), b(\lambda), c^*_-(\lambda), c^*_+(\lambda)\}$ 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 equivalently in $(r, W_0)$ with $W_0 = 5 r w, W_0^2 = 36 r^2 - 1$ parameterized by $r(u) = \frac{u^2 + 1}{12 u}, W_0(u) = \frac{u^2 - 1}{2 u}$ via $a = \frac{4 r^2 - 1}{W_0}, b = \frac{(r^2 - 1)W_0}{4 r^2 - 1}, c^*_- = \frac{(4 r^2 - 25)W_0}{25(4 r^2 - 1)}, c^*_+ = \frac{(4 r^2 - 9)(4 r^2 - 1)W_0}{(4 r^2 + 3)^2}$, which together with their $8$ Vieta-conjugate fifths generate $2$ explicit $\mathbb{Q}(u)$-families of $10/10$ quintuples, $9$ of $14/15$ sextuples, and $20$ of $18/21$ septuples; `7,350` quintuples, `33,075` $14/15$ sextuples, and `73,500` $18/21$ septuples across $3,675$ values of $u$ in `exp21_ab_split_families_and_vieta.py`). For every rational $\lambda \in \mathbb{Q} \setminus \{-1, 0, 1\}$ (the unique rational zeros, poles, and collision roots of these $6$ functions; `verify_paper.py`), $\{a(\lambda), b(\lambda), c^*_-(\lambda), c^*_+(\lambda), e_{ab, -}(\lambda), e_{ab, +}(\lambda)\}$ is a valid **$14/15$ rational Diophantine sextuple** (missing only $(e_{ab, -}, e_{ab, +})$), while $\{a(\lambda), b(\lambda), c^*_-(\lambda), c^*_+(\lambda), e_{ab, -}(\lambda)\}$ is **strictly positive** for all rational $\lambda \in \left(\sqrt{7/8}, 1\right) \approx (0.935414, 1)$, giving at $\lambda = 15/16$ the all-positive $10/10$ quintuple (`verify_paper.py`): $$ \left\{\frac{181632}{4315045}, \; \frac{27839}{16740}, \; \frac{2951475}{863009}, \; \frac{1025722387200}{128276850799}, \; \frac{57928168219537}{181859221980}\right\}. $$ Finally, $b(\lambda) c(\lambda) + 1 = \left(\frac{\lambda^2 + 1}{18\lambda(\lambda^2 - 1)}\right)^2(16\lambda^4 - 31\lambda^2 + 16) = \square$ reduces via $X = 16\lambda^2, Y = 16\lambda Y_\lambda$ to $E_{16} : Y^2 = X(X^2 - 31 X + 256)$, whose $2$-isogenous dual $\hat{E}_{16} : \hat{Y}^2 = \hat{X}(\hat{X} - 1)(\hat{X} + 63)$ splits completely over $\mathbb{Q}$. Modulo $16$ and modulo $9$, the homogeneous spaces $w^2 = d u^4 + 62 u^2 v^2 - \frac{63}{d}v^4$ for squarefree $d \mid -63$ are insoluble for $d \in \{-1, 3, -3, 7, 21, -21\}$ (`verify_paper.py`), leaving $|\operatorname{im}\alpha_{\hat{E}_{16}}| = 2$ ($d \in \{1, -7\}$) and $|\operatorname{im}\alpha_{E_{16}}| = 2$ ($d \in \{1, 2\}$, with class $2$ represented by $P_8 = (32, 96)$ of order $8$), which proves $\operatorname{rank} E_{16}(\mathbb{Q}) = 0$. Since good reduction at $p = 5$ and $p = 13$ gives $\#E_{16}(\mathbb{F}_5) = \#E_{16}(\mathbb{F}_{13}) = 8$ (`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 the smooth quartic $\mathcal{E}_{16} : Y_\lambda^2 = 16\lambda^4 - 31\lambda^2 + 16$ has **exactly $8$ rational points**, all on the degenerate boundary: $$ (\lambda, Y_\lambda) \in \big\{(0, \pm 4), \; (1, \pm 1), \; (-1, \pm 1), \; \infty_+, \; \infty_-\big\}. $$ *Proof.* All polynomial factorizations, square-class identities, collision sets, $2$-isogeny local solubility obstructions, and mod-$p$ torsion counts are verified symbolically in `verify_paper.py` and `exp17`–`exp26`. $\square$ --- ## 9. Universal $(V, K, d)$ coordinates for canonical $S$-chains, the $3$-Torsion Inflection Criterion, and Vieta-to-Elliptic $S_T$-Progressions To complement the two-sided sub-quadruple, four-window, and triple-square theory of Sections 3–8, we record the exact algebraic structure of contiguous $4$-term translation chains $x(P_0 + n S)$ on the elliptic curve $E_T : Y^2 = (X + r^2 - 1)(X + s^2 - 1)(X + t^2 - 1)$ ($X = a b c \, x$) of an arbitrary Diophantine triple $T = \{a, b, c\}$ with $\Pi = a b c$ and $\Omega = r s t$, where $P_0 = (0, \Pi)$ and $S = (1, \Omega) = 2 R_0$ with $R_0 = \big(1 + r s + r t + s t, \; (r + s)(r + t)(s + t)\big) \in E_T(\mathbb{Q})$, and combine the $S_T$-translation dictionary for regular fifths [7, 10, 17] with our $(V, K, d)$ invariants to characterize the zero-root / $3$-torsion inflection locus $3 S_T = \mathcal{O}$. ### Theorem 7 (Universal $(V, K, d)$ Coordinates for $S$-Chains, the $3$-Torsion Inflection Criterion, and Vieta-to-Elliptic $S_T$-Progressions) 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 point $Q \in E_T$ with $z = x(Q)$, the product of its two $S$-translates $x(Q \mp S) = e_\mp(a, b, c, z)$ is $x(Q - S) x(Q + S) = \frac{(z - d_+)(z - d_-)}{(\Pi z - 1)^2}$. Thus $x(Q - S) x(Q + S) + 1 = \frac{F_2(z)}{(\Pi z - 1)^2}$ where $F_2(z) := (\Pi z - 1)^2 + (z - d_+)(z - d_-)$ satisfies the universal square completion $$ W_-^2 \, F_2(z) = \Big(W_-^2 + (\Pi W_- - 2 \Omega)(z - d_-)\Big)^2 + (z - d_-)^2 D_4, $$ with $D_4 = W_- W_+ - 4 \Omega^2 = 4 \Omega^2 d$, where $d := X(2 S) - 1 = \frac{W_- W_+}{4 \Omega^2} - 1$ is given in terms of $(\alpha, \beta, \gamma) = (1/r, 1/s, 1/t)$ by Heron's formula: $$ d = -\frac{(\alpha + \beta + \gamma)(\alpha + \beta - \gamma)(\alpha - \beta + \gamma)(-\alpha + \beta + \gamma)}{4 \alpha^2 \beta^2 \gamma^2}. $$ 2. **Closed-form chain coordinates and contiguous $4$-term chains:** In terms of the three dimensionless invariants $u_0 := \frac{W_-}{2 \Omega}$ ($V := u_0^2$), $K := W_- - d = \Pi d_- - X(2S)$, 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 rational functions of $(V, K, d)$ alone: $$ X_1 = K + d + 1, \quad X_2 = \frac{K + V}{V}, \quad X_3 = \frac{d\big(K(V - 1) + V d\big)}{K^2}, \quad X_4 = \frac{(K^2 + K - V d)\big(K(d + 1) - V d\big)}{(K - V d)^2}. $$ For the contiguous $4$-term chain $\mathcal{S}_7^{(1, 2, 3, 4)} = \{a, b, c, d_-, x_2, x_3, x_4\}$, $18$ of the $21$ conditions hold identically, and the $19$th ($d_- x_3 + 1 = \square$) and $20$th ($x_2 x_4 + 1 = \square$) conditions reduce to: $$ \begin{aligned} \text{num}_{19}(V, K, d) &= \Big(K(V + d - 1) + 2 V d\Big)^2 + 4 d \Big(K(V - 1) + V d\Big)^2 = \square, \\ \text{num}_{20}(V, K, d) &= \Big(2 K^2 + (K - V d)(V - d + 1)\Big)^2 + 4 d \Big(K^2 + K - V d\Big)^2 = \square. \end{aligned} $$ 3. **Application of the Regular-Fifth Translation Dictionary [7, 10, 17], Zero-Root Degeneracy, and the $3$-Torsion Inflection Criterion ($3 S_T = \mathcal{O}$):** By the translation dictionary of [7, 10] and Paper 40v2 [17, Theorem 1, Corollary 1.1, and Cycle Lemma] (verified symbolically over $\mathbb{Q}(a,b,c,x_4)[Y_4, \Omega]$ in `verify_paper.py`), for any point $Q = (\Pi x_4, Y_4) \in E_T$ with $p_4 = \Pi x_4 \ne 1$ and $Q_4 = \{a, b, c, x_4\}$, the $S_T$-translates $x(Q \mp S_T) = \frac{X(Q \mp S_T)}{\Pi}$ satisfy $$ x(Q - S_T) + x(Q + S_T) = \frac{2\big((p_4 + 1)s_1 + 2 s_3\big)}{(p_4 - 1)^2}, \qquad x(Q - S_T) x(Q + S_T) = \frac{(x_4 - d_+)(x_4 - d_-)}{(p_4 - 1)^2} = \frac{s_1^2 - 4 s_2 - 4 p_4 - 4}{(p_4 - 1)^2}, $$ 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 $Q_j = Q_i + S_{T_{ij}}$ on $E_{T_{ij}}$, the Vieta flips $\tau_k(x_k) = \frac{2((p_4 + 1)s_1 + 2 s_3)}{(p_4 - 1)^2} - x_k$ walk along $Q_i + \mathbb{Z} S_{T_{ij}}$ via $\tau_j(x_j) = x(Q_i - S_{T_{ij}})$ and $\tau_i(x_i) = x(Q_i + 2 S_{T_{ij}})$. Combining this with Part 1 yields: - **Zero-Root & $3$-Torsion Inflection Criterion:** $\tau_j(x_j) = 0 \iff s_1^2 - 4 s_2 - 4 p_4 - 4 = R_4(T_{ij}, x_i) = 0 \iff x_i \in \{d_-(T_{ij}), d_+(T_{ij})\}$. In the shifted coordinate $U := X - 1$, $E_T$ takes the symmetric form $Y^2 = (U + r^2)(U + s^2)(U + t^2)$ with $S_T = (U=0, Y=rst)$ and tangent slope $\lambda_S = \frac{r^2 s^2 + r^2 t^2 + s^2 t^2}{2 rst}$, so $S_T$ is an inflection point ($3 S_T = \mathcal{O} \iff 2 S_T = -S_T \iff d = 0$) **if and only if** (`verify_paper.py`): $$ \lambda_S^2 - (r^2 + s^2 + t^2) = -\frac{(r s + r t + s t)(r s + r t - s t)(r s - r t + s t)(-r s + r t + s t)}{4 r^2 s^2 t^2} = 0 \iff \pm \frac{1}{r} \pm \frac{1}{s} \pm \frac{1}{t} = 0. $$ Whenever $3 S_T = \mathcal{O}$ on $E_T$, $2 S_T = -S_T$ closes the base coset $P_0 + \langle S_T \rangle = \{P_0, P_0 - S_T, P_0 + S_T\}$ into a $3$-cycle whose nonzero abscissas $\{d_-(T), d_+(T)\}$ differ by $-S_T$ and thus form a $10/10$ quintuple $T \cup \{d_-, d_+\}$ with $\tau_4(d_-) = \tau_5(d_+) = 0$ (as in $Q_{\text{3tors}} = \left\{-\frac{289}{252}, -\frac{108}{119}, \frac{49}{68}, -\frac{572}{1071}, \frac{928}{1071}\right\}$, where $(r, s, t) = \left(\frac{10}{7}, -\frac{5}{12}, \frac{10}{17}\right)$ satisfies $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = 0$), while **every non-base coset $Q + \langle S_T \rangle \subset P_0 + 2 E_T(\mathbb{Q})$ automatically forms a $15/15$ rational Diophantine sextuple** $T \cup \{x(Q - S_T), x(Q), x(Q + S_T)\}$ (yielding at $Q = 3 P_0$ on $T_{45}$ the verified $15/15$ sextuple $\left\{-\frac{289}{252}, -\frac{108}{119}, \frac{49}{68}, \frac{57469402289728}{145393246858671}, \frac{843393140604}{1028879549569}, -\frac{83594902600352}{77453118042111}\right\}$ and $18/21$ septuples with $d_\mp(T_{45})$; `verify_paper.py`). - **Open Non-Degenerate Progression Locus:** Where $p_4 \ne 1$, $R_4(Q_4) \ne 0$, and the translates are distinct and outside $T_{ij}$, the Sibling Union $S_{i,j}(Q_5) = T_{ij} \cup \{x(Q_i - S), x(Q_i), x(Q_i + S), x(Q_i + 2S)\}$ and Two-Step Trajectory $C_{j,i}(Q_5) = T_{ij} \cup \{x(Q_i - 2S), x(Q_i - S), x(Q_i), x(Q_i + S)\}$ are $4$-term $S_{T_{ij}}$-progressions with $18/21$ automatic squares obstructed at $2 S_{T_{ij}}$ and $3 S_{T_{ij}}$ by Part 1 (`2,556` quintuples, `12,780` $14/15$ sextuples, and `76,680` $18/21$ septuples in `exp25`, plus `2,659` extenders across $60$ sub-triple curves in `exp26`). *Proof.* Verified symbolically and arithmetically in `verify_paper.py`, `exp25_mw_lattice_gauge_and_coxeter.py`, `exp26_kummer_rank_and_q4_extensions.py`, and `THEOREMS_AND_PROOFS.md`. $\square$ --- ## 10. Limitations and scope - **No rational Diophantine septuple ($21/21$) is claimed.** Our explicit tuple $\mathcal{S}_7^{(20/21)}$ 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 census of 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-$3$ full-root cover $\mathcal{H}$, so by Faltings' theorem it carries only finitely many rational almost septuples, ruling out deforming $\mathcal{S}_7^{(20/21)}$ inside its full $7$-relation type $(5, 2)$ while leaving open $2$-dimensional sub-systems such as $\mathcal{S}_{4, 2}$ or $\mathcal{E}_{\text{Br1}}(\lambda, \mu)$. - **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 Mordell–Weil sieve up to $|n| \le 10^6$ (`Post 35`, `m_A1PZmnTT08-GXf`) are bounded searches, not a complete Chabauty–Coleman determination of $\mathcal{H}(\mathbb{Q})$. On Curve II ($\mathbb{Q}(k)$), the two unpaid edges $c_1 c^*_- + 1$ and $c_1 c^*_+ + 1$ have irreducible squarefree numerators of degrees $36$ (genus $17$) and $32$ (genus $15$); on $w = 2 u^2 y - L$ (Theorem 5), the $(f, g)$-window gates $\gamma^2, \delta^2$ remain bidegree-$(4, 4)$ in $(u, y)$; and on $z = z_{L_1}(r, \mu)$ (Theorem 6), trivializing the common $6$-pair square class ($a c + 1 = \square$) simultaneously with $\mathcal{F}_{ab}, \mathcal{G}_d, H = \square$ remains open. --- ## 11. Reproducibility The standalone verifier `verify_paper.py` in `supplement.zip` requires only Python 3 and SymPy (`python3 verify_paper.py`, $\sim 70\text{ s}$) 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`–`exp26` reproduce `results_exp9_*.json`–`results_exp26_*.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 $\mathcal{H}$, $\mathcal{H}_{\text{Br2}}$, the $2$-isogeny descent proof for $\operatorname{rank} E_{\text{min}}^{(5)}(\mathbb{Q}) = 0$, 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 (all $19$ 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 three interior quintuples on $Q_4 = 0$, $\iota_K$, $\mathcal{E}'_{Q_4}(R)$, the Universal Reciprocal $3 P_0$ Law on $\mu = \frac{2r}{r^2 - 1}$, $\Lambda(r) \subset E(r^2)(\mathbb{Q}(r))$ ($2 W_1 = O_-$, $A^2 - 4 B = 16 r^4 Q_+ Q_-$ with $\hat{\phi}(\hat{P}_{\text{odd}, \pm}) = \pm P_{\text{odd}}$, $P_{\text{odd}}(3/4)$, the $u = 2$ branches $v_+(P_{\text{even}}) = -4r^2$ and $v_+(T_0) = v_-(P_{\text{even}})$, and the complete $2$-isogeny descent and torsion proofs on $E_{48}^-, E_{48}, E_{80}$), $G_u(R, d_0) = n^2$ and $u = 2/3$, the $15 + 6$ Square-Class Collapse, the $(r, M)$ / $(v, t)$ identities and $9$ points, the three quintuples on $L_1 = 0$ and $3$-fifth $18/21$ septuple at $(53/30, 3/5)$, the Vieta-conjugate formula, the $v = -1/5$ family ($\lambda = 15/16$, $\operatorname{rank} E_{16}(\mathbb{Q}) = 0, E_{16}(\mathbb{Q})_{\text{tors}} \cong \mathbb{Z}/8\mathbb{Z}$), and $3 P_1, 4 P_1, 8 P_1$ on $\mathcal{G}_d = W_d^2$; 5. **Theorem 7:** the $(V, K, d)$ chain identities, the symbolic $S_T$-translation sum and product identities on $E_T$, the $U = X - 1$ inflection identity ($3 S_T = \mathcal{O} \iff \pm \frac{1}{r} \pm \frac{1}{s} \pm \frac{1}{t} = 0$), and the $Q_{\text{3tors}}$ zero-root and $3 P_0 + \langle S_T \rangle$ $15/15$ sextuple checks, together with `THEOREMS_AND_PROOFS.md` and `exp9`–`exp26`. --- ## References [1] A. 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