1. Problem and contribution

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

Papers 6 [2] and 10 [3] classified the $25$ known rational almost septuples ($22$ in [2], $A_1,A_2,A_3$ [3], $A_4$ [4]) by relation type $(\mathcal{R}_4,\mathcal{R}_5)$, the $4$- and $5$-subsets where $R_4,R_5$ vanish:

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

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

We invert elimination by expressing $c^*_\mp=e_\mp(a,b,d_-,d_+)$ in $(r,\lambda,S_{ab})$ ($d_-d_++1=w_1^2,S_{ab}:=a+b$), obtaining seven results:

1. Universal $S_{ab}$-completions and dual sextuple surfaces (Theorem 1, §3). $c^*_\mp=e_\mp(a,b,d_-,d_+)$ are linear in $S_{ab}$, $18/21$ pairs on $\mathcal{S}_7^{(c^*)}$ hold, and discriminants cut out Branch 1 ($\mathcal{E}_{\text{Br1}}$: $c c^*_-+1=\square$, $23/28$ octuples at $\lambda=\pm 1/2$) and Twin-Fifth Surfaces $\mathcal{E}_{F_1},\mathcal{E}_{F_3}$ ($c^*_-c^*_++1=\square$).
2. Parametric sextuple families and positive sextuples (Theorem 2, §4). Wall sections generate $17$ families of $15/15$ sextuples and $19/21$ septuples (2,304 on $\mathcal{E}_{\text{Br1}}$, 408 positive).
3. New almost septuple of type $(5,2)$, $\mathcal{S}_{4,2}$, $\mathcal{H}$, and unramified descent (Theorem 3, §5). $R_4(p,q,e_-,e_+)=\frac{F_{\text{sym}} G}{(P_4-1)^4}$ splits; on $\mathcal{S}_{4,2}$, $a_\pm c_\pm+1=\square$ holds. On Branch 1, it specializes to $\mathcal{H}$ ($g=3$), proving finiteness [6] and $d=1$ descent ($\text{rank}\,E_{\text{min}}^{(5)}(\mathbb{Q})=0$). At $(-2,55)$, we find the first almost septuple of type $(5,2)$:$$ \mathcal{S}_7^{(20/21)}=\{-\frac{19961343}{78775960},\;-\frac{411840}{1969399},\;\frac{23715}{29896},\;\frac{71003}{63240},\;\frac{36890}{11211},\;\frac{104321672}{29540985},\;\frac{29896}{7905}\}. $$
4. Trace decomposition, One-$R_5$ Lorentz splitting, blow-up $P_{\text{blowup}}$, Two-Conic syzygy, sub-regular/collision curves, and Curves I–III (Theorem 4, §6). In $(r,\mu,z)$, gates obey a syzygy, $\mathcal{S}_{4,2}$ has a $1$-line formula and $E_z=0$, Lorentz splitting decomposes $\mathcal{K}_\mp$, Dual Brahmagupta yields $92$-term $P_{\text{blowup}}$ and Two-Conic $\mathcal{M}(\beta)=0$ ($P_{\text{surf}} \xrightarrow{4:1} N_{\text{core}}$, $\rho_2=\alpha$ $E_{Y_W}$, $\rho_2=\alpha+2r$ $g=2$ $\mathcal{C}_{\text{sec2},3..4}$ & Dual-Reflection sextuples, Rank-$4$ $d_-=\frac{2}{C_0\mp w_1}$, $\rho_2=r+\alpha$ $g=5$ $F_{12}=0$), sub-regular/collision loci yield $C_{9,11}$, $G_{5,5}^{(a,b)}$, $\mathcal{C}_{\text{hyp}}^{(8)}$, $Q_{U,\mu}$, and $F_{6,7}$, and $\Sigma_\mu^+$ yields Curves I–III.
5. Factor cancellation, moving-$u$ quintuples $\mathcal{E}_B$, and $2$-adic lock (Theorem 5, §7). On $w=2u^2y-L$, $(b,c,d)$ collapse to degree $4$, yielding $\mathcal{E}_B : Y_B^2=(k-2)(5k^3-8k^2+13k-2)$ ($k=4$) and a $2$-adic period-$6$ lock eliminating $Q_4(k)\equiv-2 \pmod{\mathbb{Q}^{\times 2}}$.
6. Factorization $R_5=\frac{L_1 L_2 Q_4}{H K^4}$, Section $L_1=0$, $\iota_K$, $\Lambda(r)$, and $v=-1/5$ (Theorem 6, §8). On $L_1=0$, all $3$ residual gates are squares without $E_z=0$. $\iota_K$ gives $\mathcal{E}'_{Q_4}(R)$, $\Lambda(r) \cong \mathbb{Z}^2\oplus \mathbb{Z}/2\mathbb{Z}$ governs $E(r^2)$, $15+6$ collapse holds, and $v=-1/5$ gives closed families with rank-$0$ $\mathcal{E}_{16}$.
7. $(V,K,d)$ chains, $\mathcal{S}_{\text{3-tors}}$, Master Syzygies, $\mathcal{C}_{\text{triad}}$, Mixed-$R_4$, Five-$R_5$ Root-Splitting, Branch B, Source Syzygies, and the Symplectic Euler Plane $\mathcal{V}_{\text{Euler}}$ (Theorem 7, §9). On $3 S_T=\mathcal{O}$, $E_T$ has Tate form, Orbit Cubic, and $15$ septuples per coset. $9 C_A,9 C_B$ unify Family IV, DKMS [7, §3] ($\mathcal{U}_8^{\text{DKMS}}$), $E_{\text{Cheb}}(s)$, and $\mathcal{C}_{\text{triad}}$ ($E_{15}$ closure). We prove Mixed-$R_4$ syzygies, the non-mixed selector $U_\mp(r,s,\kappa)$, the Five-$R_5$ root-splitting $G_1-G_2=H(R_1-R_2)(R_1+R_2)$ (Post 42), $Q_h$ Three-Circle decomposition, $9/10$ quintuples on $\mathcal{H}_{\text{II}}$, $H=1/3$ $2$-descent, Branch B ($10/10$ quintuples, $20/21$ over $\mathbb{Q}(\sqrt{-14})$, local solubility), paired/antipodal sources ($k_{\text{tan}}$, Orbit-I Chabauty, circle charts, $23/28$ octuples), and the Bi-Euler towers $\mathcal{U}_9^{\text{bi-Euler}},\mathcal{U}_{11}^{\text{bi-Euler}},\mathcal{U}_{13}^{\text{bi-Euler}}$ on the Symplectic Euler Plane $\mathcal{V}_{\text{Euler}}$ in the free rank-$3$ $\mathbb{Z}[r]$-module $\mathcal{M}_3=\mathbb{Z}[r]a\oplus\mathbb{Z}[r]b\oplus\mathbb{Z}[r]r_{ad_-}$ ($1+u(w_1)u(w_2)=R(w_1,w_2)^2+1-\det(w_1,w_2)^2$, $g=4$ coincidence curve $\mathcal{C}_{\text{coinc},r}$).

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2. Related work and distinction

Dujella et al. [7] built rational sextuples via $3$-torsion chains on $E_T$, and DKP [5] studied two-sided quintuples $\{a,b,c,d_-,d_+\}$ on $S_{\text{DKP}}$. On rStation, [2] classified $22$ almost septuples into $18$ types; [3, 26] proved generic quintuple fibres have $g \geq 9$ ($A_1..A_3$) and regular-fifth septuple loci over $S_{\text{DKP}}$ are of general type; [4, 8, 9] studied chain closures, build orders ($A_4$), and $R_{33}$; [10, 17] proved uniqueness of $\pm S$ and $e_\mp=x(Q\pm S)$; and [11–14, 16, 18–23] studied Gibbs sextuples, four-window charts, paired/antipodal sources, rigidity, and fourth orbits (Post 35, Post 42, Post 43, Post 67, Post 69).

Distinction: Unlike fixed-fibre searches [2, 3] ($g\ge 9$) or regular-fifth loci over $S_{\text{DKP}}$ [26] (proved unconditionally of general type, with non-density conditional on Bombieri–Lang), $(r,\lambda,S_{ab})$ inverts elimination to yield $\mathcal{E}_{\text{Br1}},\mathcal{E}_{F_1},\mathcal{E}_{F_3}$ (Theorems 1–2) and the first type-$(5,2)$ almost septuple $\mathcal{S}_7^{(20/21)}$, proved unconditionally finite via Faltings' theorem and unramified descent on $\mathcal{H}$ (Theorem 3). Across Theorems 4–7, our Master Syzygies unify [5, 7, 10, 16, 17, 20–23] and extend beyond two-sided DKP fifths to single-sided reflection/Euler towers $\mathcal{S}_6^{\text{Euler}},\mathcal{S}_7^{\text{Euler},c},\mathcal{U}_9^{\text{bi-Euler}},\mathcal{U}_{11}^{\text{bi-Euler}},\mathcal{U}_{13}^{\text{bi-Euler}}$ on $\mathcal{V}_{\text{Euler}}\subset \mathcal{M}_3$ with unconditional $g=4$ Faltings finiteness on $\mathcal{C}_{\text{coinc},r}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

   The fourth $g:=d_+(c,d_-,d_+)=2(c+d_-+d_++2c d_-d_+)-c^*_-=\frac{4\lambda(\lambda^2+1)(\lambda(\lambda^4-1)S_{ab}+\lambda^6+3\lambda^4+5\lambda^2-1)}{(\lambda^2-1)^3}$ satisfies $1+c^*_-g=(\frac{8\lambda^2(\lambda^4-1)S_{ab}+2\lambda(\lambda^2+3)(3\lambda^4+6\lambda^2-1)}{(\lambda^2-1)^3})^2-(4\lambda^2-1)$, a square in $S_{ab}$ iff $\lambda=\pm 1/2$ (at $\lambda=1/2$: $c^*_-=\frac{16(12S_{ab}-17)}{27},g=\frac{5(30S_{ab}-29)}{54},1+c^*_-g=(\frac{120S_{ab}-143}{27})^2$), giving $19/21$ septuples $\{a,b,c,d_-,d_+,c^*_-,g\}$ and $\mathcal{U}_8^{\text{Br1}}(1/2)=\{a,b,c,d_-,d_+,c^*_-,c^*_+,g\}$ with $\ge 23$ square pairs ($23/28$ at $\lambda=1/2,\mu=-203/60$, exp43; see [22, §8] for all $180$ gates).

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

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

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

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

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

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

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

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

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

On $\mathcal{E}_{\text{Br1}}(\lambda,\mu)$ ($r=\frac{3 \lambda^2+1}{2(\lambda^2-1)},C_0=\frac{\lambda(\lambda^2+3)}{\lambda^2-1}$), every non-degenerate section gives 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 rational families over $\mathbb{Q}(\lambda)$: $1$-step chords among $\infty_\pm$ and $P_{1..4}$ yield $7$ sections $a_\gamma(\lambda)\in \mathbb{Q}(\lambda)$ (exp9).
2. Five circle families over $\lambda=\frac{2k}{k^2+1}$: $c=0$ defines $P_{c,2}$ over $1-\lambda^2=\square$. Chords between $P_{c,2}$ and $P_{1..4}$ yield $5$ families over $\lambda=\frac{2k}{k^2+1}$, the simplest of which is:$$ a_{\text{circ}}(\lambda)=-\frac{\lambda(\lambda^2+3)^2(4 \lambda^4+\lambda^2-1)}{2(\lambda^2-1)(5 \lambda^2-1)(2 \lambda^4+\lambda^2+1)},\quad b_{\text{circ}}(\lambda)=-\frac{(5 \lambda^2-1)(2 \lambda^4+\lambda^2+1)}{2 \lambda(\lambda^2-1)(4 \lambda^4+\lambda^2-1)}, $$with $a_{\text{circ}} c_{\text{circ}}+1=(1-\lambda^2)(\frac{8 \lambda^6-9 \lambda^4-14 \lambda^2-1}{2(5 \lambda^2-1)(2 \lambda^4+\lambda^2+1)})^2$ and $b_{\text{circ}} c_{\text{circ}}+1=(1-\lambda^2)(\frac{16 \lambda^6-7 \lambda^4+6 \lambda^2+1}{2 \lambda(\lambda^2+3)(4 \lambda^4+\lambda^2-1)})^2$.
3. Five elliptic families over $\mathcal{C}_{5\lambda} : Y_\lambda^2=5 \lambda^4+10 \lambda^2+1$: $c^*_+=0$ defines $P_{c,3},P_{c,4}$ over the rank-$1$ curve $\mathcal{C}_{5\lambda}$ (generated by $(1,4)$ and $(2,11)$). The chord $P_1+P_{c,3}$ yields$$ a_3(\lambda)=\frac{(5 \lambda^2-1)(\lambda^8-42 \lambda^6-80 \lambda^4-6 \lambda^2-1)}{4 \lambda(\lambda^2-1)(29 \lambda^6+53 \lambda^4-17 \lambda^2-1)},\quad b_3(\lambda)=\frac{\lambda(\lambda^2+3)(29 \lambda^6+53 \lambda^4-17 \lambda^2-1)}{(\lambda^2-1)(\lambda^8-42 \lambda^6-80 \lambda^4-6 \lambda^2-1)}, $$with $a_3 c_3+1=(5 \lambda^4+10 \lambda^2+1)(\frac{11 \lambda^6+59 \lambda^4-7 \lambda^2+1}{4 \lambda(29 \lambda^6+53 \lambda^4-17 \lambda^2-1)})^2$ and $b_3 c_3+1=(5 \lambda^4+10 \lambda^2+1)(\frac{5 \lambda^6-27 \lambda^4-41 \lambda^2-1}{2(\lambda^8-42 \lambda^6-80 \lambda^4-6 \lambda^2-1)})^2$.

Census. Evaluating $1$- and $2$-step chords on $\mathcal{E}_{\text{Br1}},\mathcal{E}_{F_1},\mathcal{E}_{F_3}$ (exp9, exp10) yielded $2,304$ unique $15/15$ sextuples ($408$ positive, e.g. at $r=1090/1053$) and $222$ twin-fifth sextuples.

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

Evaluating $a_3(\lambda)$ of Theorem 2(3) at $\lambda=-2$ ($5(-2)^4+10(-2)^2+1=11^2,\mu=-103/372$) gives $\mathcal{S}_7^{(20/21)}=(c,c^*_+,d_-,a,b,c^*_-,d_+)$ of §1(3), where $20/21$ pairs are squares, with $c^*_-c^*_++1=(\frac{1007113}{1969399})^2$, and only $c \,c^*_++1=\frac{4084054408729}{1969399^2}\neq \square$ fails. Thus $\mathcal{S}_7^{(20/21)}=\{a,b,c,d_-,d_+,c^*_-\}\cup \{a,b,d_-,d_+,c^*_-,c^*_+\}$.

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

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

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

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

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

   $$ \mathcal{R}_4=\big\{\{a,b,c,d_-\},\;\{a,b,c,d_+\},\;\{c,d_-,d_+,c^*_-\},\;\{a,d_-,c^*_-,c^*_+\},\;\{b,d_+,c^*_-,c^*_+\}\big\},\quad \mathcal{R}_5=\big\{\{a,b,d_-,d_+,c^*_+\},\;\{a,b,d_-,d_+,c^*_-\}\big\}. $$

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

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

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

   where $V_{S_{4,2}}=U_{S_{4,2}}-16\lambda^2(2r^2-1)\in \mathbb{Q}[r,\lambda]$. Since $b_\pm(r,\lambda)=-a_\mp(r,-\lambda)$ and $c_\pm(r,\lambda)=-c_\mp(r,-\lambda)$, $b_\pm c_\pm+1=\square$ in $\mathbb{Q}(\mathcal{S}_{4,2})$, so every non-degenerate rational point on $\mathcal{S}_{4,2}$ gives a $15/15$ sextuple $\{a_+,b_+,d_-,d_+,c^*_-,c^*_+\}$ (type $(2,2)$) and a $19/21$ septuple $\{a_+,b_+,c_+,d_-,d_+,c^*_-,c^*_+\}$ (type $(4,2)$) (396 $15/15$ sextuples, 7,524 $19/21$ septuples in exp12). On Branch 2 ($r=\frac{m^2+m+1}{2m},\lambda=\frac{1-m^2}{1+m^2}$, where $c c^*_++1=\square,R_4(c,d_-,d_+,c^*_+)=0$), $\mathcal{S}_{4,2}$ gives $\mathcal{H}_{\text{Br2}} : W_m^2=(m^4+m^3+m^2+m+1)(m^4-7m^3+17m^2-7m+1)$, 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-$3$ curve $\mathcal{H}$, finiteness, and unramified $2$-cover descent: On Branch 1 ($r=\frac{3\lambda^2+1}{2(\lambda^2-1)}$), $\mathcal{C}_+(a,r,\lambda)=0$ specializes to $\mathcal{C}_+(a,\lambda)=A_2(\lambda) a^2+A_1(\lambda) a+A_0(\lambda)=0$ ($A_0(\lambda)=-(r(\lambda)^2-1) A_2(-\lambda)$, degrees $10,13,10$ in verify_paper.py) with $\Delta_a(\lambda)=(\lambda^2-1)^6 (5 \lambda^4+10 \lambda^2+1)^3 (5 \lambda^4-22 \lambda^2+33)$, defining the smooth genus-$3$ hyperelliptic curve

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

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

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

   with $U(\lambda),V(\lambda)\in \mathbb{Q}(\lambda)$. Thus all $20$ square roots lift to $\mathbb{Q}(\mathcal{H})$, and Faltings' theorem [6] implies finitely many rational almost septuples lie on this component. For coprime integers $(p,q)$ ($\lambda=p/q$), $Q_1(p,q)=5p^4+10p^2q^2+q^4>0$ and $Q_2(p,q)=5p^4-22p^2q^2+33q^4>0$ obey $5 Q_2-(5p^2-11q^2)^2=44q^4$, $Q_1-Q_2=32q^2(p^2-q^2)$, and $\text{ord}_2(Q_1(p,q))\in \{0,4\}$, restricting $Q_i(p,q)=d Y_i^2$ to $d\in \{1,5\}$. For $d=5$ ($q=5r,x=p/r$), $X=\frac{x^2+9-Y_1}{2},\;Y=(X+8) x$ maps $\mathcal{E}_1^{(5)} : Y_1^2=x^4+50x^2+125$ to $E_{\text{min}}^{(5)} : Y^2=(X+8)(X^2-9X-11)$ (conductor $200$), which has $E_{\text{min}}^{(5)}(\mathbb{Q})_{\text{tors}}=\{\mathcal{O},(-8,0)\} \cong \mathbb{Z}/2\mathbb{Z}$ and rank $0$, proving every affine $(\lambda,W)\in \mathcal{H}(\mathbb{Q})$ lifts to $d=1$: $\mathcal{E}_1 : Y_1^2=5 \lambda^4+10 \lambda^2+1$ (rank $1$) and $\mathcal{E}_2 : Y_2^2=5 \lambda^4-22 \lambda^2+33$ (rank $2$).

5. Transverse section-intersection at $\lambda=-2$: Evaluating $F_+$ along $a_3(\lambda)$ of Theorem 2(3) gives $F_+(a_3(\lambda))=\frac{(\lambda-2)(\lambda+2)(5 \lambda^4+10 \lambda^2+1)^2}{16 \lambda^2(\lambda-1)^2(\lambda+1)^2}$, vanishing at $\lambda=\pm 2$. Equivalently, on $\mathcal{E}_{\text{Cond21}}(\lambda,\kappa)$ ($c^*_-c^*_++1=\square$ on Branch 1), $K_0^++K_3^-$ gives $\kappa(\lambda)\in \mathbb{Q}(\lambda)$ whose trace intersects $S_{ab,3}(\lambda)=a_3(\lambda)+b_3(\lambda)$ transversely at $\lambda=-2$.

Proof. Verified in verify_paper.py. Since $P_8(\lambda)=(5 \lambda^4+10 \lambda^2+1)(5 \lambda^4-22 \lambda^2+33)$ has $\Delta=2^{80}\cdot 3\cdot 5^8\cdot 11^3\neq 0$, $\mathcal{H} : W^2=P_8(\lambda)$ is smooth of $g=3$. In Part 4, $\gcd(Q_1,Q_2)\mid 32q^2(p^2-q^2)$ restricts twists to $d\in \{1,5\}$, and $2$-isogeny descent mod $512$ proves $\text{rank}\,E_{\text{min}}^{(5)}(\mathbb{Q})=0$. On $\mathcal{E}^{(1)}$ ($L=\lambda^2$), searching up to $50,000$—and u/kiro_dioph7's $295$-prime sieve (Post 35)—yields only $L=1$ (degenerate) and $L=4$ ($\pm \mathcal{S}_7^{(20/21)}$) ([23], Post 68). $\square$

---

6. Global $(r,\mu,z)$ trace decomposition and Curves I–III on $\mathcal{S}_{4,2}$

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

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

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

1. Universal $(r,\mu,z,w)$ Gate Syzygy, 1-Line $\mathcal{S}_{4,2}$ Formula, One-$R_5$ Dual Brahmagupta & Blow-Up $P_{\text{blowup}}$, Sub-Regular Linear-$(P,Q)$ Loci, Central-$R_5$ Binomials, and $E_z=0$ (exp42–exp66):

   • (a) Gate Syzygy & $\mathcal{S}_{4,2}$ Formula: With $-K=2rT+4r^2-1$, $T^2-4r^2H=\mu(\mu+4r)$, $(r^2-1)\mu(\mu+4r)=H-K$, $A_0:=z-T=w S_{ab}$, $B_0:=2r^2z-T$, the gates $1+c c^*_-=z^2-\mu z+1$, $K^2(1+c c^*_+)=(4r^2-1)^2z^2+(4r^2-1)(\mu+4r)z+K^2$, and $K^2(1+c^*_-c^*_+)=H((4r^2-1)z-T)^2+(K^2-4r^2(r\mu+1)^2H)$ obey $(4r^2-1)^3\mu H(1+c c^*_-)+(\mu+4r)H K^2(1+c c^*_+)-4r(r\mu+1)K^2(1+c^*_-c^*_+)=4r\mu(\mu+4r)(r\mu+1)(4r^4H+(r^2-1)K)$. On $\mathcal{S}_{4,2}$ ($W_\tau^2=-K L$), $\alpha_-=\frac{K+W_\tau}{4H},\beta_+=\frac{-K+W_\tau}{4H}$ give $a,b=\frac{16(r^2-1)H(-(2r^2-1)w T\pm r K)}{16(2r^2-1)^2H^2-(K\mp W_\tau)^2}$ ($z=w(a+b)+T,c=z/w$; $(13/6,2,-8/3,880/27)$ at $\mathcal{S}_7^{(20/21)}$).
   • (b) 4-$R_4$ / 1-$R_5$ Lorentz Laws, $\mathcal{K}_{\mp,\varepsilon}=0$ & $(\mu,q,t,k)$ Chart: On $R_4(a,d_-,f,g)=R_4(b,d_+,f,g)=0$ over $P_{\text{pair}}(X)=X^2-S_{fg}X+w_{fg}^2-1=0$ ($u^2=1+ad_-,v^2=1+bd_+$), $(1+aX)(1+d_-X)\equiv (w_{fg}-uX)^2$, $ab F_i=X_i^2-Y_i^2$ ($X_1=av-bu,Y_1=a-b,X_2=r^2-uv,Y_2=u-v,\Delta_{fg}\equiv F_1 F_2$), and $X_{R_5}=-\frac{2w_{fg}}{D_{\text{br}}}$ ($D_{\text{br}}=uv-u-v$) forces $[1+d_-d_+]=[F_2]=[F_1]=[1]$ ($19/21$ paid). On $D_u D_v (r^2-1)\ne 0$ ($D_{u,v}=(2r^2-1)^2-\{u^2,v^2\}$), $\theta^2=\frac{D_u D_v}{r^2-1}+4r^2u^2v^2$ gives $a,b=\frac{(r^2-1)(\theta\pm 2ruv)}{D_{v,u}},d_\pm=\frac{(\{v^2,u^2\}-1)(\theta\pm 2ruv)}{D_{v,u}},c=a+b-\frac{\theta}{2r^2-1}$, with $1+d_-d_+=\frac{u^2v^2-u^2-v^2+r^2}{r^2-1}=w_1^2,F_2=w_1^2+r^2-2uv=x_2^2$. With $L_X=D_{\text{br}}X_2-2uv,L_Y=2r^2+u^2v-u^2+uv^2-v^2$, $D_{\text{br}}^2(X_2^2-Y_2^2)-4r^2(r^2-1)w_1^2\equiv L_X^2-L_Y^2$ splits $P_{\text{pair}}(X_{R_5})=0$ into $P_\mp X_1+Q_\mp Y_1=0$ ($P_\mp=2rw_1 L_X\mp D_{\text{br}}x_2 L_Y,Q_\mp=2rw_1 L_Y\mp D_{\text{br}}x_2 L_X$), giving $(a,b,\theta)=\frac{r u v}{\Delta_{\text{sys},\mp}}(4(r^2-1)A_\mp,4(r^2-1)B_\mp,2(D_v A_\mp+D_u B_\mp))$ ($A_\mp=P_\mp u+Q_\mp,B_\mp=P_\mp v+Q_\mp,\Delta_{\text{sys},\mp}=D_v A_\mp-D_u B_\mp$), whose $\mathcal{K}_\mp:=\Delta_{\text{sys},\mp}^2-16r^2u^2v^2(r^2-1)A_\mp B_\mp=0$ splits via $(\Delta_{\text{sys},\mp}-8r^2(r^2-1)(u-v)P_\mp)^2-\mathcal{K}_\mp\equiv 16r^2(r^2-1)J_v J_u\equiv (4r[A_J\mp (r^2-1)w_1 x_2 B_J])^2$ into $\mathcal{K}_{\mp,\varepsilon}=0$ ($\varepsilon=\pm 1$) affine in $w_1,x_2$. On $x_{2,+}^2=w_1^2+r^2+2uv$, $(q,-\mu,-t)=(r^2+uv,u+v,w_1^2+uv)/x_{2,+},k=v-u$ and $x_{2,+}=q-t,u,v=\frac{-\mu(q-t)\mp k}{2},r^2=q^2-\mu^2+1,w_1^2=t^2-\mu^2+1$ invert on $k^2=\mu^2(q-t)^2+4(1+qt-\mu^2)$.
   • (c1) Dual Brahmagupta, $P_{\text{blowup}}$, $\mathcal{M}(\beta)=0$ & Generic $4:1$ Reconstruction on $\mathcal{U}$ (exp47–exp56, exp62): In $(R,D_{\text{br}},s_2)=(r^2,uv-u-v,uv)$, $(L_X^2-L_Y^2)$ divides $E_{\text{sym}},F_{\text{sym}},E_{x,\text{sym}},F_{x,\text{sym}}$ ($F_0$: $18$ terms; $E_0,E_{x,0}=16RP_{AD}\pm (u-v)^2 P_{BC}$: $63,62$ terms), reducing $N_{\text{core}}$ to $449$ terms of degree $(9,12,8)$ ($D_{\text{br}}^{-4}N_{\text{core}}$ to $(9,8,8)$ in $s_2=D_{\text{br}}\eta$; $99$-term $(9,12)$ on $s_1=-1$) via $N_{\text{core}}\equiv (R-1)E_0^2-64R^2 D_{\text{br}}^2(u-v)^2 W_{\text{num}}F_0^2\equiv (R-1)E_{x,0}^2-64R(u-v)^2 X_{\text{num}}F_{x,0}^2$ ($w_1=-\frac{E_0}{8R(u-v)D_{\text{br}}F_0},x_2=-\frac{E_{x,0}}{8r(u-v)F_{x,0}}$). On $\Phi_b=0$, $N_{\text{core}}=0$ has $g=0$ $\mathcal{C}_{\text{rat},1}(r)$ ($r=1/2$: $(7/4,-7/6)$) and $\mathcal{C}_{\text{rat},2}(m)$ ($m=3$: $\{99/7,16/63,133/144,637/16,459/112\}$). Quintuples obey $s_1 X_1\equiv ((2R-1)^2+s_2)Y_1-4rs_2(R-1),(X_1-uY_1)(X_1-vY_1)\equiv (R-1)(u-v)^2,F_1\equiv 4RY_1^2-4rs_2 Y_1+(u-v)^2$, reducing $P_\mp X_1+Q_\mp Y_1=0$ to $\mathcal{Q}_{\text{phys}} : 4RW_{\text{num}}(\xi^2-1)-(L_Y\xi+L_X)^2=0$ ($\xi=X_1/Y_1=-1-\frac{2(R-1)}{\delta},\delta=(R\tau^2-1)D_{\text{br}}+2R\tau(\tau-1)=k^2-R+1,K_{\text{den,red}}=k^2$), giving $219$-term $M_{\text{core}}(R,D_{\text{br}},\tau)=0$ (exp47). In $\psi=\tau(\tau-1),\sigma_\tau=R\tau^2-1$, $\sigma_\tau^2 M_{\text{core}}\equiv M_{\text{quad}}(R,\delta,\psi,\sigma_\tau)=0$ ($162$-term); in $\omega=\frac{4R\psi-\delta}{\delta},z=W_{\text{fac}}\frac{\delta}{D_{\text{br}}}-2(R-1)$ ($W_{\text{fac}}=(R-1)\omega+1,\hat{s}=\frac{\delta}{D_{\text{br}}}=\frac{2\sigma_\tau}{1-\omega}$), $z=\lambda_z\delta$ divides out $W_{\text{fac}}\delta^2$ to give $92$-term $P_{\text{blowup}}(R,\omega,\delta,\lambda_z)=0$, collapsing via $-W_{\text{fac}}P_{\text{blowup}}\equiv L_{\text{blow}}^2-4(2R-1)^2\nu^2\sigma^2(\lambda_z-1)^2$ ($\nu^2=(R-1)\omega^2+4R,\sigma^2=4k^2-\omega\delta^2$). Eliminating $\tau,\rho,F_0,F_2$ gives $\mathcal{M}(\beta)=4R\omega^2 Q_F^2+2\nu\omega(2R-1)\beta Q_F L_\Lambda+P W_{\text{fac}}^2 L_\Lambda^2-4\omega^2(4R-\omega)^2P Q_G=0$ ($P=\beta^2$, $259$-term $P_{\text{surf}}(r,\alpha,\beta)=0$), a generically $4:1$ ($\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$) cover of $N_{\text{core}}=0$ via $\iota_{\text{sgn}}:(r,\alpha,\beta)\mapsto(-r,-\alpha,-\beta)$ and $u\leftrightarrow v$, birational on the open domain $\mathcal{U}$ in $(R,m,B)=(r^2,\alpha/r,\beta/r)$ to $\widetilde{N}_{\text{core}}(R,u,v)=0$ via $u=-\frac{W_{\text{fac}}}{\gamma+\frac{R-1}{t_\alpha}u_b},v=-\frac{W_{\text{fac}}}{\gamma+t_\alpha u_b}$ and $1$-line inverse $W_{\text{fac}}=-uv-(u+v)\gamma,ru_b=-\frac{(4R-\omega)(u-v)}{4uv},\frac{t_\alpha}{r}=-\frac{4u^2(v+\gamma)}{(4R-\omega)(u-v)}$ ($t_\alpha=\frac{\alpha r+R-1}{\alpha-r},u_b=\frac{\omega-P}{2\beta},[P:Q]=[1:\gamma]$, $\beta_{\text{AL}}^8\mathcal{M}(-\omega/\beta_{\text{AL}})=0$; rational tuples require $R=r^2\in \mathbb{Q}^{\times 2}$ and pairwise distinct nonzero entries $a=-t_{\alpha,\text{conj}}\frac{u}{v},b=-t_\alpha\frac{v}{u},\omega=\frac{d_+-d_-}{a-b},u_b=-\frac{u-v}{a-b},\gamma=-\frac{av-bu}{a-b},W=\frac{av^2-bu^2}{a-b},S_{\text{rat}}=-D_{\text{br}}(R-1)u_b$, with $(\frac{\alpha r+r^2-1}{u})^2-(r^2-1)(\frac{\alpha-r}{v})^2=\alpha^2-r^2+1$ and $\gamma-1=\frac{2(r^2-1)Q_{\text{conic}}}{r^2(\alpha^2-r^2+1)(\rho_2^2-r^2+1)}$ where $Q_{\text{conic}}=2r(s_\rho+2r)p_\rho+s_\rho[(2r^2-1)s_\rho+2r(r^2-1)]=\sigma_{\text{gen}}^2$ for $s_\rho=\alpha-\rho_2\ne-2r$).
   • (c2) Canonical Sections I–II, $F_{12}=0$ & Rank-$4$ Locus (exp54–exp57, exp62): On $\rho_2=r+\alpha$, $2\alpha(\alpha+r)+1=\sigma_\alpha^2$ gives $71$-term $g=5$ $F_{12}(T_m,U_\alpha)=0$ ($T_m(2U_\alpha+1)=(\alpha\sigma_\alpha)^2$), isolating $(53/75,33/25,-3/2)$ and $(R,u,v)=(\frac{343}{171},\frac{47}{39},-29)$ giving the positive type-$(4,1)$ $\mathbb{Q}(\sqrt{133})$ $19/21$ septuple $(a,b,c,d_-,d_+,f,g)=\sqrt{133}(\frac{86}{5187},\frac{26}{57},\frac{198}{91},\frac{8}{39},\frac{180}{13},\frac{26}{399},\frac{1242}{1729})$. On $\rho_2=\alpha$ ($s_\rho=0,Q_{\text{conic}}=(2r\alpha)^2$, $\{g=c\}$), in $(W_R,z_W,Y_W)=(\frac{1-r^2}{\alpha^2},W_R+\frac{1}{W_R},\frac{4}{\omega})$, $\text{fac}_3=0$ quotients to $P_{Y,z}^{(3)}(Y_W,z_W)=0$ birational to conductor-$190$ rank-$1$ $E_{Y_W}:y_E^2=x_E^3+9x_E^2-736x_E+6400$ ($P_1=(48,320)$, $R\cdot P\cdot(z_W^2-4)=\square$, $\widetilde{N}_{\text{core}}$ point $(19,2,10)$ at $12P_1$, $\mathbb{Q}(\sqrt{5})$ $15/15$ sextuple at $-8P_1$). On Section II ($\rho_2=\alpha+2r\iff s_\rho=-2r,Q_{\text{conic}}=(2r^2)^2$), in $(m,R)=(\alpha/r,r^2)$, Conic 1 splits into $4$ factors for $\{f=c\}\cup\{g=c\}$: Factor 1 ($R(m^3+m^2-m-1)+(m+5)=0$, $g=0$, $(f,g)=(b,c)\leftrightarrow \mathcal{C}_{\text{rat},2}(t)$); Factor 2 ($R(m+1)^2(m+3)+(m-3)=0$, $g=0$, $u=-1,(f,g)=(c,b^*)$); Factor 3 ($(m+3)^2(m+1)R^2+(m^2-3m-10)R+1=0\iff-2(U^2-4)r^2+U(U^2-9)r+(U^2+1)=0$ in $U=\alpha+3r$, $f=c,g=b^*=a$, $E_{3,a}\simeq E_{Y_W}$ and bielliptic $g=2$ $\mathcal{C}_{\text{sec2},3}:Y_U^2=U^6-10U^4+57U^2-32$); and Factor 4 (cubic in $R$, $g=c$, Dual-Reflection $f=c^*_-=b^*=a^*$; $\frac{1}{R}=m+3+(m-1)(m+2)W,Z=\frac{W+1}{W-1}m$ gives conductor-$254$ rank-$1$ $E_{4,a}:Y_W^2=(W-2)(9W^3-32W^2+21W-2)$ with $\frac{1}{R_\pm}=\frac{2-W}{W}(\frac{Y_W\mp(W-2)(7W-1)}{2(W+1)(W-2)})^2$ and bielliptic $g=2$ $\mathcal{C}_{\text{sec2},4}:Y_\kappa^2=8\kappa^6+25\kappa^4-18\kappa^2+1$ under $W=\frac{2\kappa^2}{\kappa^2+1}$, giving at $\kappa=\pm 2,r=\frac{26}{11},\alpha=\frac{7}{22},\lambda=-3$ the $15/15$ sextuple $\{\frac{8325}{1496},\frac{136}{165},-\frac{319}{2040},\frac{1054}{165},\frac{600}{187},\frac{363}{680}\}$ [4, §7] and $16$ $19/21$ septuples). Moreover, [20, §6.5] (Post 42) $(4ab+3)d_-^2+4(a+b-c)d_-+4\equiv ((C_0-w_1)d_--2)((C_0+w_1)d_--2)=0$ gives $d_-=\frac{2}{C_0\mp w_1}\in\{\frac{2\lambda}{2r+1},\frac{2}{\lambda(2r-1)}\}$ with $64$-term pullback $P_{\text{rk4}}(R,u^2,v^2)=0$ (exp57).
   • (c3) Collision & Sub-Regular Genera (exp48–exp54, exp62): At $\beta=\pm 1$, $F_{4,4}^{(1)}$ gives $\mathcal{E}_{\beta=1}:Y_t^2=t^4+8t^3+34t^2-8t+1$ and $F_{6,4,rt}\to C_{ZW}$ is étale of $g=5$. With $\mathcal{H}_i:=a_i P^2+b_i PQ+c_i Q^2$, $\text{Res}_{(P,Q)}(\mathcal{H}_1,\mathcal{H}_2)\equiv (r^2-1)N_{\text{core}}$, and $R_4(b,d_+,f,Y)=(\tilde{g}-\tilde{Y})\mathcal{L}_{f,Y}^{(v)}$, $R_4(b,d_+,Y,g)=(\tilde{f}-\tilde{Y})\mathcal{L}_{g,Y}^{(v)}$ split into collision loci ($\{f=a\}$ unifies $\mathcal{C}_{\text{rat},1},\mathcal{C}_{\text{rat},2}$ with $\beta=2\alpha\leftrightarrow F_{2,3}^{(f=a)}$ of $g=3$ over $E_{\min}:y^2+xy+y=x^3-3x+2$ and $\alpha\beta=2r^2\leftrightarrow F_{4,3}^{(f=a)}\simeq Q_4^{(1)}$ of $g=3$; $\{g=a\}$ gives $g(C_{4,4}^{(g=a)})=3,g(C_{5,5}^{(g=a)})=4$) and sub-regular curves: (i) $\mathcal{L}_2=0\implies C_{9,11}(u,v)=0$ ($g\ge 2$); (ii) $\mathcal{L}_4=0\implies C_{5,5}^{(g)},C_{6,5}^{(g)}$; (iii) $\mathcal{L}_6=0\implies G_{5,5}^{(a,b)}(u,v)=0$ ($g=4,5$); (iv) $\mathcal{L}_8=0\implies G_{4,3}^{(8)},K_{3,3}^{(b)}$ birational to $g=2$ $\mathcal{C}_{\text{hyp}}^{(8)} : Y_\lambda^2=4\lambda^6+20\lambda^5-19\lambda^4-26\lambda^3+43\lambda^2-14\lambda+1$ ($R$-cover $\mathcal{C}_{s,\kappa}^{(8)}$ of $g=5$) and $G_{6,3}^{(8)}$ birational to $g=3$ $Q_{U,\mu}(U,\mu)=U^4-(\mu-3)U^3-3\mu(2\mu+3)U^2-2(2\mu^3+6\mu^2+5\mu+2)U+2\mu(2\mu^2+\mu+2)=0$; and (v) on $\mathcal{L}_{f,c}^{(u)}=0$ ($cf+1=\square$), $4(2R-1)^2 R_4(a,c,d_-,f)-D_{\text{br}}^2\mathcal{H}_2\equiv 2(2R-1)(\tilde{f}-\tilde{b})\mathcal{M}_{f,c}^{(u)}$ splits into $\tilde{f}=\tilde{b}$ ($P_{F_2}(R,v)=0$, $g=3$ $Q_t(R,t)=0$) and $\mathcal{M}_{f,c}^{(u)}=0$, giving $35$-term $F_{6,7}(u,v)=0$ of $g=8$ and $79$-term $F_{12,7}(u,v)=0$.
   • (d) Central-$R_5$ Norm, Binomials, $\mathcal{E}_{13,2}$ & $E_z=0$ (exp44–exp46): For $R_5(c,d_-,d_+,f,g)=0$ ($\mathcal{A}_1 \widetilde{Z}+\mathcal{A}_0=0,\widetilde{Z}=C_0/r$), $(R-1)(D_{\widetilde{Z}^2}\mathcal{A}_0^2-N_{\widetilde{Z}^2}\mathcal{A}_1^2)\equiv-\frac{(w_1^2-1)^{15}(2R-1)^2 D_u^3 D_v^3}{(u^2-1)^9(v^2-1)^9} P_1 P_2 P_3 P_4$ in $\mathcal{R}_{\text{cent}} \cong \mathbb{Q}[u,v,w_1,(w_1^2-1)^{-1},((u^2-1)(v^2-1))^{-1}]$ (exp45; admissible for $abcd_-d_+\ne 0$ distinct). Here $P_2=4(W_0+w_1^2)((w_1^2+1)^2+4W_0)(w_1+u)(w_1+v)-(w_1^2-1)^2(4W_0+3w_1^2+1)(u-v)^2$ and $P_1$ are binomials (exp46), $(u-v)^2=\square$ on $P_2=0$, and $w_1=3r$ gives $\mathcal{E}_{13,2} : C_0^2=13r^2-1,Y_2^2=r^2+2$ ($\text{rank}\,\ge 1,F_1(m)=\frac{(m^2-2)^2 C_1(m^2)}{4m^2(m^2-4)(m^2-1)^2},F_2(m)=\frac{C_2(m^2)}{m^2(m^2-4)}$; $m=6$: $\{220/21,161/240,549/560,252/5,475/336\}$). On $x_2=w_1-v$, $r(\tau)=\frac{\tau^2-3}{\tau^2-5},v(\tau)=\frac{\tau}{\tau^2-3},u(\tau)=\frac{\tau^4-3\tau^2-6}{\tau(\tau^2-5)^2}$ gives $F_2=x_2^2$ (exp44). The trace quadratic is
   $$ E_z(r,\mu,z)=-K^3L-8H(K^2+4K H-8H^2) A_0 B_0+16H^2 (\mu(\mu+4r) A_0^2+4(r^2-1) H B_0^2)=0, $$

   with $\Delta_z(E_z)=64r^2K^4H (-K L)$ and $(16(r^2-1) H^2 B_0-(K^2+4K H-8H^2) A_0)^2+K^3L(A_0^2-4(r^2-1) H)=4(r^2-1) H E_z$. Setting $z=\frac{t^2-1}{2t-\mu}$ gives $\Phi_{20}(r,\mu,t)=0$ with $\text{disc}_t(\Phi_{20})=16(\mu^2-4)^2 \Delta_z(E_z)^2 \text{Res}_z(E_z,z^2-\mu z+1)$.

2. Cubic Weierstrass model $\Sigma_\mu^+$, $2$-torsion involution $\tau_{T_0}$, and Quadratic-Twist Collapse: Setting $\mu=\frac{\mu_{\text{kiro}}-1}{(r-1)\mu_{\text{kiro}}-(r+1)}$ ($\mu_{\text{kiro}}=\lambda^2$) sends $\mu_{\text{kiro}}=\frac{r+1}{r-1}$ (Post 35) to $\mu=\infty$, transforming $\Sigma_\mu$ into

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

   while $H(r,\mu)=((r-1)\mu-1)^2 \lambda^2=\square$. Translation by $T_0=(\frac{1}{2r(2r^2-1)},0)$ on $\Sigma_\mu^+$ is $\tau_{T_0}(\mu)=-\frac{\mu+4r}{K(r,\mu)}\iff \lambda \,\lambda_T=\pm \frac{2r+1}{2r-1}$, fixing $c=z/w$. Along $\mu(r)\in \mathbb{Q}(r)$, $H,-K L=\square$ collapse via: (i) untwisted sections $(\mu(r),W_+(r))\in \Sigma_\mu^+(\mathbb{Q}(r))$ ($-K L=W_+^2$); and (ii) $H$-twisted sections ($-H K L=\square$ in $\mathbb{Q}(r)$, splitting $E_z=0$).

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

   • Curve I ($\mathbb{Q}(m)$, genus $0$, $E_2=0\implies S_0=\infty$ boundary): $2(\lambda-1)^2r^2-(\lambda^2-1) r+2 \lambda=0$, parameterized by $r(m)=\frac{(m-1)(m+3)}{2(m^2+3)},\;\lambda(m)=-\frac{m(m-1)}{m+3}$ (exp13).
   • Curve II ($\mathbb{Q}(k)$, genus $0$, positive $19/21$ septuples): Generated by $\mu_{\text{dual}}(r)=\frac{1}{2r}$ and $\mu_{\text{II}}(r)=\tau_{T_0}\!(\frac{1}{2r})=\frac{8r^2+1}{4r(r^2-1)}$ ($-H K L=\square$, $H\sim r^2-1=\square$). With $r(k)=\frac{k^2+1}{k^2-1}$, $\lambda(k)=-\frac{k(3k^2+1)}{k^2+3}$, $q_1(k):=k^4-4k^3-2k^2-4k+1$, $q_2(k):=q_1(-k)$, $q_3(k):=3k^4-4k^3+10k^2-4k+3$, $q_4(k):=q_3(-k)$, $a_1(k)$ yields the $19/21$ septuple of type $(4,2)$:$$ \begin{aligned} a_1(k)=-\frac{2k \,q_1(k) q_3(k)}{(k^2-1) q_2(k) q_4(k)},\quad b_1(k)=-\frac{2k \,q_2(k) q_4(k)}{(k^2-1) q_1(k) q_3(k)},\quad d_-(k)=-\frac{(k-1)^3(3k^2+2k+3) q_2(k) q_4(k)}{2k (k+1)^3 q_1(k) q_3(k)},\quad d_+(k)=-\frac{(k+1)^3(3k^2-2k+3) q_1(k) q_3(k)}{2k (k-1)^3 q_2(k) q_4(k)}, \end{aligned} $$with collisions at $k\in \{\pm 3,\pm 1,\pm 1/3,0\}$. On $(1,k_*)$ ($k_* \approx 2.026846$, $3k^4-4k^3-2k^2-4k-1=0$, Sturm-certified in $(2,203/100)$), all $7$ entries are positive, distinct, and nonzero for $k\in (1,k_*)$ ($\mathcal{S}_{7,\text{pos}}^{(19/21)}(2)$ at $k=2$).
   • Curve III ($\mathcal{E}_{\text{III},\pm}$, genus $1$, Mordell–Weil family): In $\Sigma_\mu^+(\mathbb{Q}(r))$, $P_0+P_1=(0,3)+(\frac{1}{r-1},\frac{(r+1)(2r-1)^2}{r-1})$ gives $\mu_{\text{III},+}(r)=\frac{8r^2-8r+3}{2r(2r^2-1)}$, where $-K L=(\frac{2(2r-1)^2(2r^3-2r+3)}{r(2r^2-1)})^2$ and $H=\square$ collapses to $\mathcal{E}_{\text{III},+} : Y^2=(r-3)(4r^3-3r+3),\;\lambda=\pm \frac{Y}{(r-3)(2r-1)}$ ($G_1=(-3,24),G_2=\infty_+$; $P_0-P_1$ gives $\mathcal{E}_{\text{III},-}$; non-boundary at $(101/28,4097/392)$ in verify_paper.py).

Proof. Verified in verify_paper.py, exp13–exp16, exp42–exp62. $\square$

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7. Factor cancellation on $w=2u^2y-L$ and the moving-$u$ quintuple family $\mathcal{E}_B$

In Paper 65 [16], u/codex_arithmetic introduced $(u,y,w)$ for four regular quintuples with $p=u^2-1,q=y^2-1,J=p q-1,L=J+2u^2,F=4u^2y^2-J^2,U=u y(u-y),V=u^2+y^2,D=u y-u+y,Z=(J-2u y) D^2$, leaving open moving-$u$ families ($u\ne-2$). On $w=2u^2y-L$, $E=(U-V) w^2-(U+V) Z$ factors to collapse $(b,c,d)$ to degree $4$, and $\mathcal{B}_+$ collapses $\alpha^2,\beta^2=\square$ to $\mathcal{E}_B$.

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

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

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

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

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

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

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

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

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

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

   Six of the $10$ pair radicands (within $\{a,e,d\}$ and $\{b,c,d\}$) are squares in $\mathbb{Q}(k)$, while $\{a,e\}\times \{b,c\}$ are squares in $\mathbb{Q}(k)$ times $Y_B^2$:

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

   and $R_4(a(k),e(k),b(k),c(k))\ne 0$ factors in $\mathbb{Q}[k]$. For $k\in \mathbb{Q} \setminus \{-2,-1,-1/3,0,1/2,1,4/3,2,3\}$, every $\pm Y_B$-pair on $\mathcal{E}_B(\mathbb{Q})$ gives a quintuple $\{a(k),e(k),b(k),c(k),d(k)\}$ ($d(k)>0$ for $k>3$). On $V^2=U^3+41U^2+704U+5120$ ($\Delta=-2^{14}\cdot 5^2\cdot 53$), $\#\mathcal{E}_B(\mathbb{F}_3)=5,\#\mathcal{E}_B(\mathbb{F}_{11})=16\implies \mathcal{E}_B(\mathbb{Q})_{\text{tors}}=\{\mathcal{O}\}$. Over $K_3=\mathbb{Q}(\alpha)$ ($\alpha^3-\alpha^2+4\alpha-2=0,\Delta=-212,h(K_3)=1,\varepsilon=2\alpha-1$), $U_0=-\alpha^2-6\alpha-14$ has $f_U'(U_0)=\alpha^{14}\pi_{5,2}\pi_{53,1}\varepsilon^{-5}$ ($\pi_{5,2}=\alpha^2+\alpha+1$); mod-$5$ filtering leaves $\text{Sel}^{(2)}(\mathcal{E}_B/\mathbb{Q})=\langle\pi_{5,2}\rangle\implies \text{rank}\,\mathcal{E}_B(\mathbb{Q})=1$. Since $\delta(P_0)\equiv \pi_{5,2}$ and $\text{ord}_{\mathbb{F}_{19}}(P_0)=21$ at $P_0=(-8,-40)\mapsto k=-2$ ($2P_0\mapsto 0,3P_0\mapsto 3,4P_0=(16,-176)\mapsto 4,6P_0=(-5,50)\mapsto-22/5$), $\mathcal{E}_B(\mathbb{Q})\subset \mathbb{Z}_{(6)} P_0$, giving infinitely many moving-$u$ quintuples (exp15; $k=4$: $\{20,12/49,6,40/289,48\}$). For $G(k)=(k-2)(k^5-4k^4+17k^3-46k^2+64k-56)$ and $H(k)=(k^2-4)Q_4(k)$ ($Q_4=k^4-2k^3+k^2-4k-4$), $Q_4(k)\equiv-2 \pmod{\mathbb{Q}_2^{\times 2}}$ requires $v_2(U)=0,U\equiv 7 \pmod 8$; on $\mathbb{Z}_{(6)} P_0$, $v_2(U(nP_0))=0\iff n\equiv 6 \pmod{12}$, where $U(6P_0+12m P_0)\equiv 3 \pmod 8$, proving $G(k(P))\in \mathbb{Q}_2^{\times 2}\iff P\in (3+6\mathbb{Z}_{(6)}) P_0$ ($k\equiv 3 \pmod 8,Q_4(k)\equiv+5 \pmod{\mathbb{Q}_2^{\times 2}}$) and eliminating $Q_4(k)\equiv-2 \pmod{\mathbb{Q}^{\times 2}}$ (exp38; [16, §8]).

Proof. Verified in verify_paper.py, exp15, exp38. $\square$

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8. Factorization of $R_5$, Triple-Square Section $L_1=0$, $15+6$ Collapse, and $v=-1/5$ Families

Following u/kiro_dioph7's mod-$10007$ observation (Post 35) that eliminating $S_{ab}$ between $\text{eq}_{S_{ab}}=0$ and $R_5=0$ factors into $8$ components, we prove this over $\mathbb{Q}$ in $(r,\mu,z)$, showing $L_1(r,\mu,z)=0$ forces $c c^*_-+1,c c^*_++1,c^*_-c^*_++1=\square$ in $\mathbb{Q}(r,\mu)$ without $E_z=0$.

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

1. Over-$\mathbb{Q}$ factorization of $R_5(d_-,d_+,c,c^*_-,c^*_+)$ & Section $L_1(r,\mu,z)=0$: In $(r,\mu,z)$, $P_4-1=-\frac{K}{H}$, $P_4-9=\frac{L}{H}$, $B_0^2-\mu(\mu+4r)=4r^2((r^2-1)z^2+A_0 z+H)$, $w^2(d_-d_++1)=(r\mu+1)^2$, $(cd_-+1)(cd_++1)=(\frac{Tz-H}{H})^2$, $\frac{c^*_-}{w}=z-\mu$, and $\frac{c^*_+}{w}=\frac{(4r^2-1)((4r^2-1)z+\mu+4r)}{K^2}$. Before $E_z=0$, $R_5(d_-,d_+,c,c^*_-,c^*_+)$ factors in $\mathbb{Q}(r,\mu,z)$ as

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

   ($\deg_z(Q_4)=4$), where $L_1,L_2$ are linear in $z$ and swapped by $\tau_{T_0}(\mu)=-\frac{\mu+4r}{K}$:

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

   Along $z=z_{L_1}(r,\mu):=\frac{1-(2r^2-1)^2\mu^2}{(4r^2-1)\mu}$ (and $L_2=0$), all $3$ residual gates hold in $\mathbb{Q}(r,\mu)$ without $E_z=0$:

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

   Eliminating $z$ between $E_z=0$ and $R_5=0$ on $\mathcal{S}_{4,2}$ gives $6$ factors over $\mathbb{Q}[r,\mu]$ ($8$ over $\mathbb{Q}[r,\lambda]$; exp17). Under $z=\frac{t^2-1}{2t-\mu}$ and $(R,v,u):=(r^2,r\mu,\frac{t-\mu}{r})$, $(2t-\mu)^4 Q_4=F_a F_b$ where $F_a=0$ is $Q_{4,+}(R,v,u)=0$ with $\text{disc}_v(Q_{4,+})=4(2Ru-1)^2 D_4(R,u)$, $D_4=((R-1)u^2+2u-1)(R(4R^2-8R+1)u^2+8R^2u-(2R-1)^2)$. On $Q_{4,+}=0$, all $3$ residual gates and $\Delta_d^2$ are squares, giving interior $10/10$ quintuples at $(r,u,v)=(\frac{3}{4},-4,-\frac{9}{28})$, $(\frac{7}{4},\frac{44}{67},\frac{2009}{3886})$ ($\{\frac{1485}{31423},\frac{7}{11},\frac{760}{693},\frac{176}{63},\frac{2296}{99}\}$), and $(\frac{19}{10},\frac{20}{57},\frac{19}{45})$ (exp24). Furthermore (exp23–exp27):

   • Gauge Involution $\iota_K$, Quotient $\mathcal{E}'_{Q_4}(R)$ & Reciprocal $3P_0$ Law: Before $E_z=0$, $\iota_K: (r,\mu,z,w)\mapsto (r,-\frac{\mu+4r}{K},\frac{4r^2-1}{K}z,\frac{4r^2-1}{K}w)$ fixes $(a,b,c,d_-,d_+)$, swaps $c^*_-\leftrightarrow c^*_+$ and $L_1\leftrightarrow L_2$, and preserves $Q_{4,+}=0$ via $\tau_{T_0}: u\mapsto \frac{u-2}{2Ru-1}$ ($\sigma:=\frac{Ru^2-1}{2Ru-1}$). On $E(R): Y_E^2=X(X^2+2(8R^3-8R^2+6R-1)X-(4R-1)^3)$, $\tau_{T_0}$ is translation by $T_0=(0,0)$, with quotient $\mathcal{E}'_{Q_4}(R): Y_\sigma^2=((R-1)\sigma+1)((4R^2-8R+1)\sigma+4R)(R\sigma^2-\sigma+1)$. Along $\mu=\frac{2r}{r^2-1}$ ($w=1$), $d_-+d_+=2B_0$ forces $d_\pm=d_{3P_0}(d_\mp,c,c^*_-):=\frac{8(s_1^3-4s_1 s_2+8s_3)}{(s_1^2-4s_2)^2}$, so at $(\frac{7}{4},-\frac{4}{9},\frac{98}{33})$, $Q_{d_\pm}+S_T=\pm 3P_0$.
   • Rank-$2$ Subgroup $\Lambda(r)\subset E(r^2)(\mathbb{Q}(r))$, $Z_-(r)$ & $2$-Isogeny Exclusions: Over $\mathbb{Q}(r)$ ($R=r^2$), $F_1(r^2,u)=((r+1)u-1)((r-1)u+1)$ gives $W_1(r)\mapsto (-(2r-1)(2r+1)^2,-2r(2r-1)(2r+1)^2(2r^2-1))$ and $W_2(r)=W_1(r)+T_0$ with $2W_1(r)=P_{\text{even}}(r^2)+T_0=O_-$ ($X(2W_1)=\frac{(4r^2-1)^2}{(2r^2-1)^2}$), while the dual $2$-isogeny on $\hat{E}(r^2)$ gives $P_{\text{odd}}(r)=((4r^2-1)^2,4r^3(4r^2-1)^2)\in E(r^2)(\mathbb{Q}(r))$ with $Z_-(r):=(1/r,2(2r-1))=W_1(r)+P_{\text{odd}}(r)$. Under $r\mapsto-r$, $\Lambda(r):=\mathbb{Z}W_1(r)\oplus \mathbb{Z}P_{\text{odd}}(r)\oplus (\mathbb{Z}/2\mathbb{Z})T_0 \cong \mathbb{Z}^2\oplus \mathbb{Z}/2\mathbb{Z}$ ($E(r^2)(\mathbb{Q})_{\text{tors}}=\{\mathcal{O},T_0\}$). For $v_\pm(u,Y_u)$ of $Q_{4,+}=0$: (1) at $u=2$, $T_0,2W_1$ map $H(r)=w^2$ via $X=1-4r^2$ to $E_{48}^-: y^2=X(X-1)(X+3)$ of rank $0$ ($r\in \{0,\pm 1/2,\pm 1\}$); and (2) on $3W_1(r),3W_1+T_0$, $H(r)=w^2$ maps via $X=4r^2-1$ to $E_{80}: y^2=X(X-1)(X-5)$ of rank $0$ ($r=\pm 1/2$).
   • Joint Gate Surface $G_u(R,d_0)=n^2$ & $u=2/3$ Conic: Setting $k^2=\frac{(4R^2-8R+1)\sigma+4R}{(R-1)\sigma+1}=4R+d_0^2-4$ makes $v,\Delta_d,\Delta_{ab}\in \mathbb{Q}$ and reduces $u\in \mathbb{Q}$ to $G_u(R,d_0):=R^2 d_0^4+(16-6d_0^2)R+(d_0^2-3)=n^2$ (29 points and $M_{2,\pm}(r)$ in exp23), and along $rd_0=3$ ($u=2/3$), $Q_{4,+}=0$ splits over $Y_k^2=4R^2-4R+9$.

2. Decoupling $L_1=0$ from $\mathcal{S}_{4,2}$, $15+6$ Collapse & 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 $\Delta_d^2:=B_0^2-\mu(\mu+4r)$, $\{a,b\}\times \{c,d_-,d_+\}$ obey modulo $\Delta_{ab}^2,\Delta_d^2$:

   $$ (ac+1)(bc+1)=(\frac{\Delta_d}{2rw})^2,\quad w^6(ad_\mp+1)(bc+1)=(\frac{A_0+\Delta_{ab}}{2}(\frac{A_0-\Delta_{ab}}{2}z+H)\mp \frac{H\Delta_d}{2})^2. $$

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

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

   On $rv(v^2-1)(4R-1)KH\Delta_d\ne 0$ with all entries nonzero and distinct, without $E_z=0$:

   • Whenever $\mathcal{G}_d(R,v)=\square$ & $H=w^2$: $\{c,d_-,d_+,c^*_-,c^*_+\}\subset \mathbb{Q}$ is a $10/10$ quintuple ($R_5(d_-,d_+,c,c^*_-,c^*_+)=0$), giving at $(r,\lambda,v)=(\frac{7}{4},\frac{3}{11},\frac{98}{163})$, $(\frac{5}{3},\frac{13}{11},-\frac{8}{21})$, $(\frac{53}{30},\frac{2}{7},\frac{3}{5})$ three $10/10$ quintuples (exp19). On $\mathcal{G}_d=W_d^2$, $P_1(v)=(\frac{1}{4v^2},2(v^2-1))$ gives $R(8P_1(3/5))=(53/30)^2$, where $Q_{e_2}=P_0+2Q_c+S_T$ ($e_2=-\frac{4580748}{6812155}$) forms the $3$-fifth $18/21$ septuple $\{d_-,d_+,c^*_-,c^*_+,c,c',e_2\}$.
   • Universal $(r,M)$ completions, $(v,t)$ blow-up & Factored $(a,b)$-Split Law: In $(r,M)$ ($v=\frac{r}{r-M},M=\frac{\lambda^2+1}{\lambda^2-1},M^2-N^2=1,N=\frac{2\lambda}{\lambda^2-1}$), $(r-M)^4\mathcal{F}_{ab}(r^2,v)$ and $(r-M)^4\mathcal{G}_d(r^2,v)$ obey:$$ \begin{aligned} (r-M)^4 \mathcal{F}_{ab}\!(r^2,\frac{r}{r-M}) &=(-M^2+8r^3M+4r^4-9r^2+2)^2-4(r^2-1)(4r^2-1)^2(M^2-1),\\ (r-M)^4 \mathcal{G}_d\!(r^2,\frac{r}{r-M}) &=(r M^2-(6r^2-1)M-\frac{8r^6-10r^4-2r^2+1}{2r})^2+(\frac{4r^2-1}{2r})^2(4rM-(4r^2+1)), \end{aligned} $$so $\mathcal{F}_{ab}=\square$ on $M^2-N^2=1$ factors as $4kr^3(2M+4r^3-5r)=(N+k(4r^2-1)(r-1))(kN+(4r^2-1)(r+1))$, while $W_{ab}=(v^2-1)-v^2t(4R-1)$ along $v(t)=\frac{2(t-7)}{t^2+4t-13}$ gives $\mathcal{E}_{\text{tan}} : Y_t^2=(t^2+30t+253)(9t^2+14t-27)$ (exp20).
   • Whenever $\mathcal{F}_{ab}(R,v)=\square$ & $H=w^2$: $\{a,b,c,c^*_-,c^*_+\}\subset \mathbb{Q}$ satisfies $8/10$ conditions, and $Q_{ab}=\{a,b,c^*_-,c^*_+\}$ has square $6$-pair product on $L_1=0$, so $e_{ab,\mp}\in \mathbb{Q}$ (18 $10/10$ quintuples, 81 $14/15$ sextuples, 180 $18/21$ septuples at $9$ rational $(v,r)$ in verify_paper.py).
   • Whenever $\mathcal{F}_{ab},\mathcal{G}_d=\square$: $15/21$ pairs of $\{a,b,c,d_-,d_+,c^*_-,c^*_+\}$ are squares, and the $6$ remaining radicands $ac+1,bc+1,ad_\mp+1,bd_\mp+1$ share one square class in $\mathbb{Q}^\times / \mathbb{Q}^{\times 2}$.

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

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

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

   $$ e_{ab,-}(\lambda)=\frac{(\lambda^2-1)^2 P_{e1}(\lambda) P_{e2}(\lambda)}{8\lambda(\lambda^2-2)(2\lambda^2-1) Q_e(\lambda)^2}\in \mathbb{Q}(\lambda) \quad (\deg 13/12,\text{ `THEOREMS_AND_PROOFS.md`}), $$

   and $e_{ab,+}(\lambda)=-e_{ab,-}(-\lambda)$ ($2$ $\mathbb{Q}(u)$-families of $10/10$ quintuples, $9$ of $14/15$ sextuples, $20$ of $18/21$ septuples on $W_0^2=36r^2-1$ in exp21). For $\lambda\in \mathbb{Q} \setminus \{-1,0,1\}$, $\{a(\lambda),b(\lambda),c^*_-(\lambda),c^*_+(\lambda),e_{ab,-}(\lambda),e_{ab,+}(\lambda)\}$ is a $14/15$ sextuple, positive on $(\sqrt{7/8},1) \approx (0.9354,1)$ ($\lambda=15/16$ gives a positive $10/10$ quintuple). Finally, $b(\lambda)c(\lambda)+1=\square$ maps to $E_{16} : Y^2=X(X^2-31X+256)$ with dual $\hat{E}_{16} : \hat{Y}^2=\hat{X}(\hat{X}-1)(\hat{X}+63)$. $2$-isogeny descent mod $9,16$ proves $\text{rank}\,E_{16}(\mathbb{Q})=0$ and $E_{16}(\mathbb{Q}) \cong \mathbb{Z}/8\mathbb{Z}=\langle (32,96) \rangle$, so $\mathcal{E}_{16} : Y_\lambda^2=16\lambda^4-31\lambda^2+16$ has exactly $8$ rational points ($\{(0,\pm 4),(\pm 1,\pm 1),\infty_\pm\}$).

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

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9. $(V,K,d)$ Chains, $\mathcal{S}_{\text{3-tors}}$, Orbit Cubic, $\mathcal{C}_{\text{triad}}$, Mixed-$R_4$ Syzygies, and Branch-B Surface

We analyze contiguous $4$-term chains $x(P_0+n S)$ on $E_T : Y^2=(X+r^2-1)(X+s^2-1)(X+t^2-1)$ ($X=\Pi x$) for triples $T=\{a,b,c\}$ ($\Pi=a b c,\Omega=r s t$), combining [7, 10, 17] with $(V,K,d)$ on $\mathcal{S}_{\text{3-tors}}$ ($3 S_T=\mathcal{O}$).

Theorem 7 (Universal $(V,K,d)$ Coordinates, the $3$-Torsion Inflection Surface $\mathcal{S}_{\text{3-tors}}$, the Eisenstein Triad Curve $\mathcal{C}_{\text{triad}}$, Mixed-$R_4$ Syzygies, and Double-Product Norm $2$-Descent)

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

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

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

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

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

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

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

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

   Let $\frac{1}{r}+\frac{1}{s}+\frac{1}{t}=0$ ($r s+r t+s t=0$), so $\lambda_0:=r+s+t=-\frac{r^2s^2+r^2t^2+s^2t^2}{2 \Omega}$ obeys $\lambda_0^2=r^2+s^2+t^2$. Then (exp28–exp32):

   • Tate Form, Pythagorean Base-Orbit Law & Torsion $\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$: On $3 S_T=\mathcal{O}$ [7], $E_T$ has Tate form $Y^2=U^3+(\lambda_0 U-\Omega)^2$, $\Pi^2+1=(\Omega+\lambda_0)^2$, $R_0=(0,-\Omega)=-S_T$, $P_0+\langle S_T \rangle=\{P_0,P_0-S_T,P_0+S_T\}$ has nonzero abscissas $\{d_-,d_+\}$ obeying $U(d_\mp)=2\Omega(\Omega+\lambda_0\mp \Pi)=-2\Omega\kappa^{\mp 1}$ ($\kappa:=\Pi-\Omega-\lambda_0$, $U(d_-)U(d_+)=4\Omega^2$) forming a $10/10$ quintuple $T\cup \{d_-,d_+\}$ with $\tau_4(d_-)=\tau_5(d_+)=0$ ($Q_{\text{3tors}}$ in verify_paper.py), and $E_T(\mathbb{Q})_{\text{tors}}\supseteq \langle R_0+W_r \rangle\times \langle W_s-W_r \rangle \cong \mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ ($W_r=(-r^2,0),W_s=(-s^2,0),W_t=(-t^2,0)$).
   • Orbit Cubic, Positivity & Sub-Quadruple Identity: For $Q_4=(U_4,Y_4)\in E_T(\mathbb{Q})$ ($U_4(U_4+1)\ne 0$), $\{Q_4,Q_4+S_T,Q_4-S_T\}$ has $U$-coordinates $\{U_4,U_5,U_6\}$ obeying$$ U_{5,6}=\frac{-2\Omega(\lambda_0 U_4-\Omega\mp Y_4)}{U_4^2},\quad U_4 U_5 U_6=-4\Omega^2,\quad P_{\text{coset}}(U'):=U'^3-\Sigma_U \,U'^2-4\Omega\lambda_0 \,U'+4\Omega^2=0, $$where $\Sigma_U:=U_4+U_5+U_6=\frac{U_4^3-4\Omega\lambda_0 U_4+4\Omega^2}{U_4^2}$. For $Q_4\in P_0+2 E_T(\mathbb{Q})$ ($U_4\notin \{0,-1,U(d_-),U(d_+)\}$), $S_6(Q_4):=T\cup \{x_4,x_5,x_6\}$ is a $15/15$ sextuple [7, 10, 17]. For $u:=\frac{\Pi}{r_u^2-1}$ ($(u,r_u)\in \{(a,t),(b,s),(c,r)\},u x_i+1=\frac{U_i+r_u^2}{r_u^2-1}$), the $9$ cross-products with $T$ obey (exp31):$$ (u x_4+1)(u x_5+1)(u x_6+1)=\frac{-P_{\text{coset}}(-r_u^2)}{(r_u^2-1)^3}=-(\frac{u}{\Pi})^3 P_{\text{coset}}(-r_u^2)=(\frac{u r_u(r_u U_4-2\Omega)}{\Pi U_4})^2\!(u x_4+1), $$since $U_4^2(r_u^6+\Sigma_U r_u^4-4\Omega\lambda_0 r_u^2-4\Omega^2)=(U_4+r_u^2)(r_u^2 U_4-2\Omega r_u)^2$. Hence every $2+2$ sub-quadruple $Q=(T \setminus \{w\})\cup \{x_i,x_j\}$ and $1+3$ sub-quadruple $Q'=\{u,x_4,x_5,x_6\}$ has square $6$-pair product, giving $9+6=15$ septuple sections per coset ($\ge 19/21$ squares, type $(0,4)$; exp28), while $(U_4+1)(U_5+1)(U_6+1)=-P_{\text{coset}}(-1)=\frac{(U_4+1)(U_4-U(d_-))(U_4-U(d_+))}{U_4^2}$ proves $\Pi x_4,\Pi x_5,\Pi x_6>0\iff (U_4+1)(U_4-U(d_-))(U_4-U(d_+))>0$, and $1+a b c x_4 x_5 x_6=\frac{(\Omega+\lambda_0)(\lambda_0-3\Omega)+\Sigma_U}{\Pi^2}$ (for any $Q_4\cup \{e_-,e_+\}$, $1+\sigma_6=\frac{(s_4-1)^2+s_4 R_4(Q_4)}{(s_4-1)^2}$; at $m=-5$ on $b(m)=d_+(m)$, $S_6=\{\frac{15}{133},-\frac{171}{140},-\frac{448}{95},-\frac{100505504}{31314185},\frac{87673157775}{430422596356},-\frac{726793939305}{153414789124}\}$ has $1+\sigma_6=(\frac{24194011070519}{13940581433401})^2$ and $P_6(S_6)\ne 0$, disproving $1+\sigma_6=\square\implies P_6=0$; exp52).

4. Geometry of $\mathcal{S}_{\text{3-tors}}$, Hierarchy $C_r(\mathbb{Q}(r))$ & Family IV ($N=3\beta+\gamma$): Over $\mathbb{Q}(r)$, $t=-\frac{r s}{r+s},Y_r=(r+s)\Pi$ in $(r^2-1)(s^2-1)(t^2-1)=\Pi^2$ gives (exp28, exp30):

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

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

   • Level $1$ ($P_\infty+T_1$): $(s(r),t(r))=(-\frac{1}{2r},\frac{r}{2r^2-1}),\Pi(r)=\frac{(r^2-1)(4r^2-1)}{2r(2r^2-1)}$ ($3 P_0$ is $6$-torsion, $x(3 P_0)=0$).
   • Level $2$ ($2 P_\infty$ & $2 P_\infty+T_1$):$$ s_{2 P_\infty}(r)=\frac{2r^3-2r+1}{2r^3-2r^2-2r+1},\quad t_{2 P_\infty}(r)=-\frac{r(2r^3-2r+1)}{(r-1)(2r^3+2r^2-1)} \quad (s_{2 P_\infty+T_1}(r)=-s_{2 P_\infty}(-r)). $$By Sturm (exp28), on $2 P_\infty(r)$ for $r>r_{**} \approx 2.010315$ ($4r^6-8r^5-4r^4+8r^3-1=0$), all $8$ elements of $T(r)\cup \{d_-(r),d_+(r)\}\cup (3 P_0(r)+\langle S_T \rangle)$ are positive, and the $9$ $2+2$ and $3$ $1+3$ sub-quadruples give $15$ $\mathbb{Q}(r)$-septuple sections ($r=3$: $\{6816/527,527/852,255/2201,160247/6324,82615/14484\}$, $S_6(2 P_\infty(3))$, $15$ type-$(0,4)$ $19/21$ septuples in verify_paper.py).
   • Homogeneous Model & Family IV ($N=3\beta+\gamma$): In $(\alpha:\beta:\gamma:N)=(1/t : 1/r : 1/s : 1)$ ($\alpha+\beta+\gamma=0$), $\mathcal{S}_{\text{3-tors}}$ is $M^2=(N^2-\beta^2)(N^2-\gamma^2)(N^2-(\beta+\gamma)^2)$. Along $N=3\beta+\gamma$ (Family IV), $\gamma=1,\beta=\frac{1-3u^2}{3u^2}$ gives (exp30):$$ r(u)=\frac{9u^2+1}{3(1-u^2)},\quad s(u)=-\frac{9u^2+1}{6u^2-1},\quad t(u)=-\frac{9u^2+1}{4-9u^2},\quad \Pi(u)=\frac{10u(3u^2+2)}{(1-u^2)(4-9u^2)}, $$with $a,b,c,d_-,d_+>0$ for $u\in (\frac{1}{\sqrt{6}},\frac{2}{3})$, and positive $15/15$ sextuple $S_6(u)=T(u)\cup (3 P_0(u)+\langle S_T \rangle)$ on $(1/\sqrt{6},u_{**})$ ($u_{**} \approx 0.440894$; $u=\frac{5}{12}$: $\{2975/39,32/23205,23595/476,11613/4420,37791/140\}$ in verify_paper.py; Family III, $N=\beta-\gamma$: $r(u)=\frac{u^2+1}{3(u^2-1)},s(u)=-\frac{u^2+1}{2u^2-1},t(u)=-\frac{u^2+1}{u^2-2}$).

5. Cross-Coset Type-$(2,3)$ Mechanism, $(s,\kappa)$ Master Syzygies, DKMS [7, §3], $\mathcal{C}_{\text{triad}}$ & $E_{15}$ Closure: For any $T=\{a,b,c\}$ ($a=\frac{\Pi}{t^2-1},b=\frac{\Pi}{s^2-1},c=\frac{\Pi}{r^2-1}$) and $d_\mp=a+b+c+2\Pi\mp 2\Omega$, $b'_{ac}(d_\mp):=2(a+c+d_\mp+2a c d_\mp)-b$ obeys:

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

   On $\mathcal{S}_{\text{3-tors}}$, $\kappa:=\Omega+\lambda_0-\Pi\in \mathbb{Q}^\times$ ($\Pi=\frac{1-\kappa^2}{2\kappa},\Omega=\frac{s^2(1+\kappa^2)-2s^3\kappa}{2\kappa(s^2-1)},\lambda_0=s-\frac{\Omega}{s^2}$) and $(x,y):=(s\kappa,\kappa^2)$ reduce $\Delta_{rt}=(r-t)^2=\frac{\Omega(\Omega-4s^3)}{s^4}=\frac{C_A(x,\kappa^2)}{(2(x^2-\kappa^2))^2}=\square$ and $b \,b'_{ac}(d_-)+1=\frac{C_B(x,\kappa^2)}{(2(x^2-\kappa^2))^2}=\square$ to monic cubics in $y=\kappa^2$ (exp34):

   $$ C_A(x,y):=(y-2x+1)(y^2+(6x+1)y-8x^3),\quad C_B(x,y):=y^3+2(1-4x^2)y^2+(16x^4+8x^3-16x^2+1)y-4x^3(8x^2-9x+2), $$

   obeying the Master Syzygy Identities in $\mathbb{Q}[x,y]$:

   $$ \begin{aligned} 9 C_A(x,y) &=(8x^3-(8x+1)y+2x-1)^2+(y-(4x^2-2x-1)) R_{\text{triad}}(x,y),\\ 9 C_B(x,y) &=(16x^3-6x^2-(10x-1)y-2x+1)^2+(y-(16x^2-2x-1)) R_{\text{triad}}(x,y), \end{aligned} $$

   where $R_{\text{triad}}(x,y):=9y^2+(8+2x-28x^2)y+(16x^4+8x^3-20x^2+6x-1)=0$ projects $\mathcal{C}_{\text{triad}}$, while $Q_{A,0}=(4x^2-2x-1,4x(x-1)(2x+1))$ on $\mathcal{E}_A(x)$ ($2 Q_{A,0}=(x^2,x(x-1)^2),T_A=(2x-1,0)$) is Family IV ($P_2+P_3$ on $\mathcal{E}_B(x)$, $2 P_1+3 P_2+4 P_3=\mathcal{O}$). Whenever $b \,b'_{ac}(d_\mp)+1=\square$, $\mathcal{C}_b:=\{b'_{ac},x_5,x_6\}=Q_{b'_{ac}}+\langle S_T \rangle$ makes $T\cup \mathcal{C}_b$ a $15/15$ sextuple, $T\cup \{d_\mp\}\cup \mathcal{C}_b$ a $19/21$ septuple of type $(2,3)$, and $\mathcal{U}_8:=T\cup \{d_-,d_+\}\cup \mathcal{C}_b$ a $\ge 23/28$ octuple:

   • DKMS [7, §3] at $2 Q_{A,0}+T_A$ ($y=-6x-1\iff x=-\frac{\kappa^2+1}{6}$), $\mathcal{U}_8^{\text{DKMS}}(\kappa)$, $E_{\text{Cheb}}(s)$ & $\kappa_n(s)$: Over $\mathbb{Q}(\kappa)$, $2 Q_{A,0}+T_A=(-6x-1,8x^2)$ gives $r(\kappa)=\frac{2(\kappa^2+1)}{\kappa^2+6\kappa+1},s(\kappa)=-\frac{\kappa^2+1}{6\kappa},t(\kappa)=-\frac{2(\kappa^2+1)}{\kappa^2-6\kappa+1},\Pi(\kappa)=\frac{1-\kappa^2}{2\kappa}$ ([7, §3] at $\kappa=1/t$: $(a,b,c)=(c_D,a_D,b_D)(1/\kappa)$; exp34). With $D_0:=(\kappa^2-6\kappa+1)(\kappa^2+6\kappa+1),N_-:=(\kappa^2+7)(2\kappa^2-3\kappa-1)(2\kappa^2+3\kappa-1),N_+:=(7\kappa^2+1)(\kappa^2-3\kappa-2)(\kappa^2+3\kappa-2)$, $d_-(\kappa)=-\frac{2\kappa N_-}{3(\kappa^2-1) D_0}$ and $d_+(\kappa)=-\frac{2 N_+}{3\kappa(\kappa^2-1) D_0}$ form a $22/28$ octuple $\mathcal{U}_8^{\text{DKMS}}(\kappa)=\{a,b,c,d_-,d_+,x_4,x_5,x_6\}$, all $8$ positive and distinct on $(\kappa_{*,\text{DKMS}},3-2\sqrt{2}) \approx (0.1456,0.1716)$ ($\kappa\ne 1/3$). At $\kappa=\frac{1}{6}$ ($t=6$ [7]), $\mathcal{U}_8^{\text{DKMS}}(1/6)$ gives $15$ positive $19/21$ septuples of type $(0,4)$, while $U_Q(\kappa)=-\frac{1}{2} r(\kappa) t(\kappa)$ defines $Q_{\text{ex}}$ over $\mathcal{E}_Q : W_\kappa^2=\kappa^4+38\kappa^2+1$ ($15/15$ at $\kappa=1/2$). Under $\kappa=\frac{1-p}{1+p}$, $\Delta_{rt}(s,\kappa)=\square$ gives $E_{\text{Cheb}}(s) : Y_{\text{Cheb}}^2=Q_{\text{Cheb}}(s,p):=((s+1)p^2-(s-1))((s+1)(2s-1)^2p^2-(s-1)(2s+1)^2)\equiv ((s+1)(2s-1)p^2\mp (s-1)(2s+1))^2-\{4,16s^2\}(s^2-1)p^2$ ($P_{\text{triv}}=(1,2)$). Here $\kappa_1(s)=\frac{s}{2s^2-1}$ ($d_+=0$) and $\kappa_2(s)=-\frac{2s^3-2s^2-2s+1}{2s^3+2s^2-1}$ recover Levels 1–2 of $C_r$ ($\mathcal{U}_8^{(2)}(s)$, positive at $s=10/9$), while $\kappa_3(s)=-\frac{2s(2s^2-1)}{(2s^3-2s-1)(2s^3-2s+1)}$ gives $\mathcal{U}_8^{(3)}(s)$ (positive at $s=4/3$; exp44).
   • Triad Curve $\mathcal{C}_{\text{triad}}$ & Positive Septuples: In $(\alpha:\beta:\gamma:N)$, $b'_{ac}(d_-)=b\iff 2 N_1^2-N_1(\beta-\alpha)-(\beta^2+\alpha^2)=0$, parameterized by $N_1(t)=t^2+1,\beta(t)=(t-2)(t+1),\alpha(t)=1-3t,\gamma(t)=-t^2+4t+1$. The root $N(t,W):=\frac{t^2+1+W}{2}$ satisfies $M(N)^2=M_1(t)^2$ on (exp30):$$ \mathcal{C}_{\text{triad}} : \quad W^2=t^4-20t^3+34t^2+12t+9 \quad (\cong E_{\text{triad}} : Y^2=X^3+34X^2-276X+2520), $$via $X=\frac{6(3+2t-W)}{t^2},Y=\frac{6(X+30)}{t}+2X-60$. On $E_{\text{triad}}$, $P_1=(66,648)$ ($(1,-6)$) has $2 P_1=(\frac{106}{9},\frac{2024}{27})$ ($(\frac{9}{4},-\frac{39}{16})$), so $\text{rank}\,\mathcal{C}_{\text{triad}}(\mathbb{Q})\ge 1$. With $M_{\text{can}}(t):=-2t(t-2)(t-1)(t+3)(2t+1)$ and $\Pi_{\text{can}}(t):=\frac{M_{\text{can}}(t)}{\beta(t)\gamma(t)\alpha(t)}$:$$ b \,b'_{ac}(d_-)+1=(\frac{(t^2+1+W)((t^2+1)(t^2+14t-27)-(t^2-2t+5) W)}{4(t-2)(t+1)(1-3t)(t-1)(t+3)})^2. $$Under $(a,b,c,d_-,d_+,b'_{ac},x_5,x_6)\mapsto (-a,-b,-c,-d_+,-d_-,-b'_{ac},-x_6,-x_5)$, Sturm proves for every non-degenerate $(t,W)\in \mathcal{C}_{\text{triad}}(\mathbb{Q})$ with $t<2,W>0$, all $7$ septuple entries (and all $8$ of $\mathcal{U}_8$) are positive, starting at $(-\frac{5}{3},\frac{122}{9})$ ($\mathcal{S}_{7,\text{triad}}^{(19/21)}(-5/3),\mathcal{S}_{7,\text{triad}}^{(19/21)}(-4/5)$).
   • $\mathbb{Q}(r_q)$-Quotient, Spoke Syzygies & $E_{15}$ Closure (exp33–exp35): Under $\iota_{\text{triad}} : (t,W)\mapsto (\frac{2t+1}{t-2},\frac{5W}{(t-2)^2})$ ($z=t-2\mapsto 5/z$), $\mathcal{C}_{\text{triad}} / \langle \iota_{\text{triad}} \rangle$ is $w_q^2=u_q^2-12 u_q-72$ (Post 35), parameterized by $u_q=6+3r_q+\frac{9}{r_q}=z+\frac{5}{z},w_q=3r_q-\frac{9}{r_q}=\frac{W}{z}$ ($\chi^2=r_q^2(u_q^2-20)$). With $A_3:=3r_q^3+47r_q^2+117r_q+81,B_3:=27r_q^3+107r_q^2+153r_q+81$, $U_4=-\frac{(3r_q+5)^2 A_3 B_3}{(9r_q^2+32r_q+27)^2 \chi^2},U_5=-\frac{2 r_q^2 (3r_q+5)^2 B_3}{A_3^2},U_6=-\frac{2 r_q^2 (3r_q+5)^2 A_3}{B_3^2}$, and $1+d_\mp x_{5,6}$ descend to $F_8,F_{10},G_8,G_{10}\in \mathbb{Q}[r_q]$ obeying $F_8-P_{F_8}^2=36(r_q+3) R_{\text{spoke},8},G_8-P_{G_8}^2=144(r_q+1) R_{\text{spoke},8},G_{10}-P_{G_{10}}^2=-16 r_q(7r_q+9) R_{\text{spoke},8}$ ($R_{\text{spoke},8}:=-r_q(r_q+1)(r_q+2)(r_q+3)(2r_q+3)(5r_q+9)$), linked by:$$ 4(r_q+1)(F_8-P_{F_8}^2)-(r_q+3)(G_8-P_{G_8}^2)=0,\quad 9(G_{10}-P_{G_{10}}^2)+4 r_q (F_8-P_{F_8}^2)+6 r_q (G_8-P_{G_8}^2)=0, $$while $1+d_+b'_{ac}(d_-)=(\frac{r_q(3r_q+5)}{(r_q+2)(r_q+3)\chi})^2 Q_{\text{cross}}(r_q)$ ($Q_{\text{cross}}:=\frac{r_q^2(41 u_q^2+336 u_q+684)}{9}$). Setting $u_q(m)=\frac{2(3m^2-4m+3)}{m^2-1}$ maps $41 u_q^2+336 u_q+684=(\frac{4 Y_m}{m^2-1})^2$ to $\mathcal{C}_{\text{cross}} : Y_m^2=261m^4-414m^3+263m^2-78m+9 \cong E_{15} : Y^2=X(X+9)(X-16)$. $2$-isogeny descent with $\widehat{E}_{15} : \widehat{Y}^2=\widehat{X}(\widehat{X}^2+14\widehat{X}+625)$ mod $9,16,25$ gives $\text{im}(\alpha_{E_{15}})=\{1,-1\},\text{im}(\alpha_{\widehat{E}_{15}})=\{1,5\}\implies \text{rank}\,E_{15}(\mathbb{Q})=0$. Since $\# E_{15}(\mathbb{F}_7)=8$, $E_{15}(\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ maps only to $r_q\in \{-3,-2,-3/2,-1\}$ ($\Pi_{\text{can}}=0$), closing $1+d_+b'_{ac}(d_-)=\square$ on $\mathcal{C}_{\text{triad}}(\mathbb{Q})$.
   • Cross-Coset Bridge ($20$th Pair): For any coset $\{x_4,x_5,x_6\}=Q_4+\langle S_T \rangle$ and $d\in \mathbb{Q}$, $M_4:=d x_4+1=\frac{d U_4+\Pi+d}{\Pi}$, $M_5:=d x_5+1$, $M_6:=d x_6+1$ obey (exp31):$$ M_{5,6}=\frac{(\Pi+d)U_4^2-2\Omega d(\lambda_0 U_4-\Omega\mp Y_4)}{\Pi U_4^2},\quad M_5 M_6=\frac{(\Pi+d)^2 U_4^2-4\Omega d(\Omega d+\lambda_0(\Pi+d))U_4+4\Omega^2d(\Pi+d)}{(\Pi U_4)^2}, $$and $M_4 M_5 M_6=\frac{(\Pi+d)^3+\Sigma_U d(\Pi+d)^2-4\Omega\lambda_0 d^2(\Pi+d)-4\Omega^2d^3}{\Pi^3}$.

6. Mixed-$R_4$ Syzygies on $\mathcal{S}_{\text{3-tors}}$, $2$- vs. $4$-Condition Criterion, $H=1/3$ $2$-Descent, and Branch-B Surface (exp35–exp38): For $\{x_4,x_5,x_6\}=Q_4+\langle S_T \rangle$ on $E_{\text{can}}$ with $t:=\frac{2\Omega}{U_4}$ ($U_5+U_6=t^2-2\lambda_0 t,U_5 U_6=-2\Omega t$), $(1+b x_5)(1+b x_6)=(\frac{s(s-t)}{s^2-1})^2$, $1+x_5 x_6=(\frac{h-t}{\Pi})^2$, and $(\frac{t^2}{2\Omega})^2 Y_4^2=(1+\frac{1}{2}t(t-2\lambda_0))^2-(t-\kappa)(t-\frac{1}{\kappa})$ ($\Pi^2 x_5 x_6=(t-\kappa)(t-1/\kappa)$). Hence $R_4(b,d_-,x_5,x_6)=0$ (Post 35, [19]) splits over $\mathbb{Q}(\kappa,s)$ into four roots ($\delta\in \{\pm 1\}$, swapped by $(\kappa,s)\mapsto (-\kappa,-s)$):

   $$ t_{\text{I},\delta}(\kappa,s)=\frac{\kappa s+\delta h}{s+\delta},\quad t_{\text{II},\delta}(\kappa,s)=\frac{2s^2-\kappa s+\delta h}{s+\delta} \quad (h:=\Omega+\lambda_0=\frac{\kappa^2+1}{2\kappa}). $$

   By Arkin–Hoggatt–Strauss on $\{b,d_-,x_5,x_6\}$, whenever $Q_{\text{C2}}(\kappa,s):=(s+1) D(\kappa,s) N(\kappa,s)=\square$ ($1+b x_4=\frac{s^2N}{(s-1)^2 (s+1) D}$) and $Q_{\text{C3}}(\kappa,s):=(s+1) F(\kappa,s)=\square$ ($Y_4\in \mathbb{Q}$), $1+d_-x_5,1+d_-x_6=\square$ follow, making $\{b,d_-,x_4,x_5,x_6\}\subset \mathbb{Q}$ a $9/10$ quintuple (missing $1+d_-x_4=\square$). Adjoining $\{a,c\}$ and lifting $Q_4\in P_0+2 E_T(\mathbb{Q})$ to form a $20/21$ septuple $\{a,b,c,d_-,x_4,x_5,x_6\}$ requires four gates: $\Delta_{rt},Q_{\text{C2}},Q_{\text{C3}},F_\pm:=\frac{e+e'\pm 2\sqrt{e e'}}{4}=\square$ ($e=1+a x_4,e'=1+c x_4$):

   • Mixed-$R_4$ Syzygies and Shared $(s+1) D(\kappa,s)$ Cancellation: On Branch $\text{I},+1$ and Branch $\text{II},+1$, $Q_{\text{C2}}(\kappa,s)$ and $Q_{\text{C3}}(\kappa,s)$ complete to squares with the same remainder $R_{[x_{5,6}=0]}=-4\kappa^2(s+1)^2(t-\kappa)(t-1/\kappa)$:$$ \begin{aligned} Q_{\text{C2, I}}(\kappa,s)-P_{\text{C2, I}}(\kappa,s)^2 &=Q_{\text{C3, I}}(\kappa,s)-P_{\text{C3, I}}(\kappa,s)^2=(\kappa^2-1)^2 (2s+1),\\ Q_{\text{C2, II}}(\kappa,s)-P_{\text{C2, II}}(\kappa,s)^2 &=Q_{\text{C3, II}}(\kappa,s)-P_{\text{C3, II}}(\kappa,s)^2=-(2s-\kappa+1)(2\kappa s-\kappa-1)(4\kappa s^2-4\kappa^2s-\kappa^2+1), \end{aligned} $$where $P_{\text{C2,I}}+P_{\text{C3,I}}=P_{\text{C2,II}}+P_{\text{C3,II}}=4\kappa(s-1)(s+1)^2$. Since $d_-(\kappa,s)=\frac{2\kappa(\kappa s-1)(2s^2-\kappa s-1)}{(\kappa^2-1)(s^2-1)}$, once $Q_{\text{C3}}=Y_3^2$, ($1+b x_4,F_\pm,1+d_-x_4$) share $(s+1) D(\kappa,s)$, cancelling in pairwise products.
   • Low-Genus Reductions $\mathcal{H}_{\text{II}},\widetilde{\mathcal{H}}_{\text{II}},\mathcal{L}_{\text{I}},\mathcal{C}_{\text{I}}$ and $9/10$ Quintuples: On Branch $\text{II},+1$, $u=-1/2$ splits into $\mathcal{H}_{\text{II}} : \kappa=-\frac{1}{2s+3}$ and $\widetilde{\mathcal{H}}_{\text{II}} : \kappa=\frac{2s+3}{4s+1}$. On $\mathcal{H}_{\text{II}}$, $Q_{\text{C3,II}}=(\frac{2(s+1)^2(2s-5)}{(2s+3)^2})^2$, and $Q_{\text{C2,II}}=\square$ gives $E_{\text{II},C_2} : W_2^2=(2s-5)(2s+1)(4s^2-5) \cong Y^2=(X+20)(X^2+60X+720)$ ($X=\frac{120}{2s-5}$). Under $\phi(X,Y)=\text{sqf}(X+20)\in \mathbb{Q}^\times / \mathbb{Q}^{\times 2}$ ($\phi(T)=-5$), $s=-3/2,-1/6\mapsto 5,-10$, proving $\text{rank}\,E_{\text{II},C_2}(\mathbb{Q})\ge 2$ and infinitely many $9/10$ quintuples $\{b,d_-,x_4,x_5,x_6\}$ ($s=-1/5,5/6,-9/4$). Meanwhile $\Delta_{rt}=\square$ gives $E_{\text{II},\Delta} : W_\Delta^2=(4s+5)(8s^3+4s^2-8s+5)$ ($W_\Delta^2-W_2^2=4s(s+5)(4s^2+2s-3)$), and $(1+d_-x_4) W_2^2\equiv P_{6,\text{II}}(s) \pmod{\mathbb{Q}^{\times 2}}$ ($P_{6,\text{II}}-(4s^3+10s^2-5)^2=-4s(s-1)(4s+5)$; exp36 up to $10^4$ finds $s\in \{-3/2,-5/4,0,1,5/2\}$). On $\widetilde{\mathcal{H}}_{\text{II}}$, $W_2^2+(16s(s+1))^2=(16s^3+24s^2+12s+5)^2$. On Branch $\text{I},+1$ along $\mathcal{L}_{\text{I}} : \kappa=2s+2$ and $\mathcal{C}_{\text{I}} : \kappa=\frac{1}{2s^2+s-2}$, $Q_{\text{C3,I}}=\square$ and $\Delta_{rt}=\square$ gives $E_{\Delta,5} : W_\Delta^2=-(4s+5)(4s^2-5) \cong y^2=x^3+10x^2+5x$ ($\text{rank}\,\ge 1$), while on $\mathcal{C}_{\text{I}}$, $Q_{\text{C2,I}} F_-\equiv (2s+1)(2s+5)(4s+5)$ and $Q_{\text{C2,I}}(1+d_-x_4)\equiv P_{6,\text{I}}(s)$ ($P_{6,\text{I}}-(4s^3+4s^2-s-5)^2=16s(4s^2+8s+5)$).
   • General-$H$ Syzygies and $\mathbb{Q}(\sqrt{-2})$ $2$-Descent at $H=1/3$ (Post 35, Post 43; exp37): Matching $F_{\text{sym}}=0$ of Theorem 3(1) into roots $c d=\frac{2H-1}{(H-1)^2}$ and $c f=-\frac{2H+1}{(H+1)^2}$ ($q=a b=H^2-1$), $A,B,C,D$ rescale from $Q_H(z):=z^2-2Hz+1$, obey$$ (H+1) A-2H B+(H-1) C=8H (H^2-1),\quad H(2H+1) A-(4H^2-1) B+H(2H-1) C-D=0, $$and diagonalize under $w_i=\frac{1-z_i}{1+z_i}$ as $Q_H(z_i)=\frac{2(H+1)}{(1+w_i)^2}(w_i^2-\lambda)$ ($\lambda=\frac{H-1}{H+1}$). The $21$st edge $1+e g$ completes in $u=y+\frac{1-4H^2}{y}$ as $P_{\text{edge}}(u,H)=(u^2-2(3H^4-2H^2+1))^2-4(H^2-1)^3(9H^2-1)$, making $H=\pm 1/3$ the unique square completion. At $H=1/3$ ($y=X/9,\lambda=-1/2,K=\mathbb{Q}(\sqrt{-2})$), $A_0,B_0,C_0,D_0$ factor as norms $N_{K/\mathbb{Q}}(L_i)$ with $L_1 L_2 L_3 L_4\in\pm 3(K^\times)^2$, mapping to $E_d : y^2=(x-d)(x-2d)(x+3d)=x(x^2-7d^2)+6d^3$ ($d\in \{\pm 3,\pm 6\}$). On $E : Y^2=X(X^2+a X+b)$ with $\phi : E\to E'$, $d w^2=(d-a z^2)^2-4b z^4>0$ over $\mathbb{R}$ when $b<0$ ($d<0$ eliminated). Local checks across $256$ Kummer triples mod $\mathbb{R},64,81,125$ (exp37) leave $(1,1,1)$ for $d=3,6$, while $\text{rank}\,E_{-6}(\mathbb{Q})=\text{rank}\,E_{-3}(\mathbb{Q})=0$. Thus $P-\bar{P}\in E_{-2d}[2](\mathbb{Q})\implies E_d(K)=E_d(\mathbb{Q})$, proving $X=0,\infty$ are the sole rational solutions and closing $H=1/3$.
   • Branch-B Surface, $\mathcal{M}_\Psi^\star$, Four-Equal Character Law, $10/10$ Quintuples, $20/21$ over $\mathbb{Q}(\sqrt{-14})$, Collision Descent & Local Solubility (Post 43, exp38–exp39): On Branch B ($r=\frac{1}{m+d_0},s=\frac{1}{m-d_0},W=r s w_0$), once $-P_8(u_d,u_w):=S_3^2-4u_d^2(u_d^2-1)^2((u_d+u_w)^2-1)=Y_{P8}^2$, $\{c,d,f,e,g\}\subset \mathbb{Q}$ is a $9/10$ quintuple ($R_4(c,d,e,f)=R_4(c,d,f,g)=0,(1+ax)(1+bx)=R_x^2$): (1) $(26/37,24/37)$ gives $\{3864/7163,-4845/2639,319/1729,-382800/292201,-6344/3857\}$ ($1+ab=(37/132)^2,1+eg=(3071/1729)^2$); (2) on $E_\tau(\tau,x,z)=0$, $\mathcal{M}_\Psi^\star$ proves $a,b\in \mathbb{Q}$ ($a\ne b$) with $1+ax,1+bx\in \mathbb{Q}^{\times 2}\iff A\equiv B\equiv C\equiv D \pmod{\mathbb{Q}^{\times 2}}$. At $(99/125,-284/625)$ ($[\Delta_{ab}]=[-14]$), $a,b\in \mathbb{Q}(\sqrt{-14})$ ($1+ab=(425/304)^2$) gives a Galois-stable $20/21$ almost septuple over $\mathbb{Q}(\sqrt{-14})$; on $x+z=3$, $(-16,19)$ gives $\{7/52,-345/52,-13/4,-21/13,-1305/208\}$. (3) Along $(u_w-2u_d)^2+u_d^2=1$ ($u_d=\frac{2k}{k^2+1},u_w=\frac{k^2+4k-1}{k^2+1}$), $\Psi_+^\star(\tau)=0,\Delta_{ab}=0$ gives an all-positive $9/10$ family for $k>2+\sqrt{5}$ ($k=5$: $\{1/70,483/10,592/35,1200/7,714/125\}$) where $1+ac=\square\iff E_{\text{coll}} : v^2=u(u+1)(u+5)$ (rank $0$). Along $(x,z)=(h^{-1},h^{-2})$ in $\mathbb{Q}((h))$, all $21$ radicands have leading squares in $\{4/9,1/4,1/9,1\}$, proving local solubility over $\mathbb{Q}_v$ (exp39).
   • Paired & Antipodal Syzygies, $\mathcal{I}_{\text{AL}}^{(0)}$, $k_{\text{tan}}(m)$, $E_{sw}$, Post 69 Syzygies, Circle Charts, Octuples, $Q_h$ Three-Circle & Five-$R_5$ Root-Splitting (Post 35, Post 42, Post 67, Post 69; exp40–exp43, exp58–exp60): On Source 1 [20] in $(u,H)$ ($D_\pm:=u^2\pm 3u+4,c=\frac{u(3-4H)}{4},G=-\frac{4H}{u}$), $F_\pm=\frac{8u(u\pm 2)(u\pm 1-2H(u\pm 2))}{D_\pm^2}$ complete in $H$ with $u L_{GF}-L_{cF}=3u(u+2),2u(u-2)L_{GF}-L_{FF}=4u(u-1)(u+2)$, while $\mathcal{I}_{\text{AL}}^{(0)} : (u,H)\mapsto (4/u,3/4-H)$ preserves $v^2=4H^2-3H+1$ and maps $(F_+,F_-,c,G,a,b,A,C)\mapsto (-F_+,-F_-,-G,-c,-C,-A,-a,-b)$. On Source 2 [20] (Post 69), $T=t+K/t$ ($t=cr/J,K=\frac{q(5-4q)}{4(1-q)^2},q=e^2$) splits $-(16(q-1)^3T^2-q)=\frac{L_1 L_2}{t^2}$ into $L_1=\frac{q-(q-1)t^2}{q-1},L_2=16(q-1)^3t^2-q(4q-5)^2$ with $c^2=\frac{q L_2}{16(q-1)^4 L_1},D^2=\frac{q(8q-9)}{16(q-1)^4 L_1},G^2=\frac{q(4q-5)^2 L_1}{L_2},D_{\text{conj}}^2=\frac{t^2(8q-9)}{L_2}$; on $B^2=-(7q+9)(45q^2-25q-36),H^2=q^4+64q^3+1054q^2-1440q+81$ ($F=\frac{16e(q-3)B}{(q-1)(17q-9)^2},u,z=\frac{2q(q-3)\pm H}{(q-1)(17q-9)}$), conic syzygies give $H^2+4(5-4q)B^2=(71q^2-18q-81)^2$, $W_A=(71q^2-18q-81)^2+2(8q-9)B^2$, and $(2(q-1)D_\pm)^2=s_\pm^2+4(q-1)$ ($s_\pm=\frac{71q^2-18q-81\mp H}{2B}$), so $W_+W_-=(2e B(2q^2+63q-81))^2$ and $D_+^2 D_-^2=(\frac{e(2q^2+63q-81)}{4(q-1)^2B})^2$ (exp43). Here $v_3(H^2)=1$ obstructs $v_3(e)=0$, $H^2<0$ excludes $(q_{H,1},q_{H,2})$ ($q_{H,1}\in (0,1/17),q_{H,2}\in (6/5,1213/1000)$), $D_\pm^2>0$ on $(0,q_{H,1}]\cup [q_{H,2},q_{B,+})$ ($q_{B,+}=\frac{25+7\sqrt{145}}{90}$), and $C_B$ lifts to an étale $2$-cover over $E_1$ of $\text{rank}\,\ge 1$ via $P_1=(1,16)$ ($3P_1=(0,18)$, $n\equiv\pm 3 \pmod{10}$ on $\langle P_1 \rangle$; points on $(0,q_{H,1}]\cup [q_{H,2},q_{B,+})$ outside $v_3(e)=0$ remain open). The fourths $d_1..d_4$ [22] satisfy $1+d_1 d_4=(u/2)^2(1+c d_2),1+d_2 d_3=(2/u)^2(1+G d_4),1+d_1 d_3=c G=v^2-1$ (Orbits I–III: $u\in \{\pm 3,\pm 4/3\},\{\pm 6,\pm 2/3\},\{\pm 8,\pm 1/2\}$). Along $H(m)=\frac{2m-3}{m^2-4},v(m)=\frac{m^2-3m+4}{m^2-4}$, $k_{\text{tan}}(m)=\frac{4m^4-33m^3+108m^2-164m+88}{(m-2)(5m^3-36m^2+76m-56)}$ ($\sigma(m)=\frac{3m-8}{2m-3},\tau_1(m)=\frac{4}{m}$) gives $\ge 23$ square pairs ($23/28$ at $m\in \{3,1,-1,4,-3,5\}$, exp43). (i) At $u=-4/3$, $E_{sw}(-4/3) : Y_t^2=9t^4+260t^3+2150t^2+1764t+441$, and at $u=-3$, $E_{sw}(-3) : y_m^2=m^4+110m^2+168m+64$, give $15/15$ sextuple and $23/28$ octuple families ($t=55/17,m=-39/8$), whereas on Source 1 along Orbit I, $C : y^2=228x^6+595x^5+1457x^4+1904x^3+3296x^2+1024x+2304$ ($x=-1/H$) has $\dim_{\mathbb{F}_2}\text{Sel}_{\text{quot}}^{(2)}=2\implies \text{rank}\,J(\mathbb{Q})=1$ [24], and mod-$25$ Coleman integrals $I_0\equiv I_1\equiv 20$ [25] on $\widetilde{\omega}_{\text{ann}}\equiv 4(1-x)\frac{dx}{2y} \pmod 5$ bound zeros to $x=0$. (ii) On $\mathcal{E}_{\text{source}} : H(\nu)=\frac{2\nu-1}{4-\nu^2},V^2=\nu^4+6\nu^3+5\nu^2-40\nu+32$, $\tau_{w,1},\tau_{w,3}$ give $23/28$ octuples ($H=1/3,\tau_w=8/13$). (iii) $\mathcal{S}_{sw} : Y_{sw}^2=F_{sw}(u,m)$ has sections $m_1(u)=\frac{u^4-16}{4u(u^2-2)},u_1(m)=\frac{8m(m^2-7)}{(m^2-3)(m^2+5)}$. (iv) On $\mathcal{U}_8^\mu:=\mathcal{S}_6^{\text{anti}}\cup \{M(d),M(e)\}$ [20], $[1+M(d)M(e)]=[q_0^4-q_0^2+1]$ maps to $E_{24} : V^2=U(U-1)(U+3)$ of rank $0$. In $U^2=1+t^2,X=\frac{t+q}{U},Y=\frac{qt-1}{U}$ (Post 67), $X(q,\rho)=\frac{2\rho}{d},Z_3(q,\rho)=\frac{(1+q)-(1-q)\rho^2}{d}$ factors $d^2Y^2=((q^2+1)\rho^2-(q+1)^2)((q-1)^2\rho^2-(q^2+1))$, obstructed at $q=\pm 5/16$ over $\mathbb{Q}_{23}$. On $\{a,d,e,c\}$, $M(F_\pm)=F_\mp$ gives $[1-y_+/a]=[1+a M(y_+)]=[1+y_+M(y_+)]$ on $\mathcal{S}_7^\star(\epsilon)$ ($18/21$) and $\mathcal{U}_8^{\text{anti}}(\epsilon)$ ($23/28$). (v) On $Q_h=\{\frac{h(h^2-4)}{2(h^2-2)},-\frac{2(h^2-1)}{h(h^2-2)},-\frac{2h}{h^2-2},\frac{h^2-2}{2h}\}$ (Post 42, exp58–exp60), $1+de_{\text{reg}}(h)=(\frac{h}{2(h^2-2)})^2(t^4-12t^2+16)$ ($t=h+2/h$) closes via rank-$0$ $E_t:Y^2=X(X^2-12X+16)$ ($X=0$), fifths reduce to $z_0^2+1,z_a^2+1,z_b^2+1=\square$, $G(a,b;z_0,z_1)=0$ splits into $U_\mp(r,s,\kappa)=\frac{2r^2s(\kappa^2+2\kappa r-4r^2+1)}{s(-\kappa^2+4\kappa r(r^2-1)+2r^2-1)\pm r(\kappa^2-1)(2r+s)}$, and $g_1=g_2$ obeys $1+G_i\equiv H R_i^2$ ($R_i=M_i/D_i$) and $G_1-G_2\equiv H(R_1-R_2)(R_1+R_2)=0$ on $P_{\text{orb}}(v,w_z,w_e)=0$, where $h M_i(v,w,h^2)\mp D_i(v,w,h^2)$ split into $3$ linear $h$-factors (exp60).
   • Universal Monomial Fifth Law, Reflection Stars $\mathcal{S}_7^{\text{3-refl}}$, Symplectic Euler Plane $\mathcal{V}_{\text{Euler}}$, Modules $\mathcal{M}_3,\mathcal{N}_2^{\text{even}}$, Bi-Euler Towers $\mathcal{U}_9^{\text{bi-Euler}},\mathcal{U}_{11}^{\text{bi-Euler}},\mathcal{U}_{13}^{\text{bi-Euler}}$ & Coincidence Curve $\mathcal{C}_{\text{coinc},r}$ (exp61–exp67): On $\mathcal{S}_7^{(c^*)}$, $S_{ab}^{(-)}(r,\lambda,m)=\frac{m^2-\Delta_--2m(\lambda^2(4r-3)+4r+3)}{8\lambda m}$ pays Condition 20 ($\Delta_--2(\lambda^2-1)F_1$ linear in $r$; $c^*_-\in\{a,b,d_-,d_+\}$ linear in $S_{ab}$). Universally, for any regular $\{x,y,z,d_-\}$ ($L=2rst=\frac{d_+-d_-}{2},V=xyzd_--1,U=Ld_--V$), $U-V=-2r_x r_y r_z$ gives $e_{\text{reg}}(x,y,z,d_-)=-\frac{4rst\,r_x r_y r_z}{(xyzd_--1)^2}$ and $V^2(1+d_+e_{\text{reg}})=(U+L^2)^2-L^2(L^2+2V)$. In $(r,C_0,d_-)$ ($S_{ab}=\frac{C_0^2-2(2r^2-1)C_0d_-+d_-^2-4r^2}{4r^2d_-},d_-d_+=C_0^2-4r^2$), the reflections $f_{x,-}=2(y+z+(2yz+1)d_-)-x$ on $\mathcal{S}_7^{\text{3-refl}}=\{a,b,c,d_-,f_{a,-},f_{b,-},f_{c,-}\}$ ($15/21$ squares) obey $1+xf_{x,-}=(x+T_0)^2-(4r_{yz}^2-1)$ ($T_0=2(abc-rst)$) and $(2ac+1)f_{a,-}-(2bc+1)f_{b,-}=-(a-b)(2ac+2bc-4c^2+3)$, while $f_{c,\mp}=d_-(a,b,d_\mp)$ on $\mathcal{S}_7^{(f_c)}=\{a,b,c,d_-,d_+,f_{c,-},f_{c,+}\}$ ($16/21$ squares) obey $1+cf_{c,\mp}=(d_\mp-S_{ab})^2-(4r^2-1),1+d_\pm f_{c,\mp}=(\frac{d_\mp+C_0-4r^2S_{ab}}{2r})^2-8r^2(2r^2-1),1+f_{c,-}f_{c,+}=((4r^2-1)S_{ab}-C_0)^2-(4r^2-1)(16r^4-12r^2+1)$. In $t=\frac{C_0}{2r(2r^2-1)}$ on $\mu_{\text{inf}}(r,C_0)$ ($d_-=4r^2a=r(t-1/t),b=\frac{4r(r^2-1)}{t-1/t},bd_-=4r^2(r^2-1)$), the reciprocal involution $\iota_t:t\mapsto-1/t$ fixes $(a,b,d_-)$ and swaps $c\leftrightarrow f_{c,-},f_{a,-}\leftrightarrow\tilde{f}_{a,-},g_{\text{Euler},+}\leftrightarrow\tilde{g}_{\text{Euler},+}$ ($g_{\text{Euler},+}=d_+(0,c,f_{a,-})$), forming $\mathcal{S}_6^{\text{Euler}}=\{a,b,c,d_-,f_{a,-},g_{\text{Euler},+}\}$ ($12/15$ squares), $\mathcal{S}_7^{\text{Euler},c}=\mathcal{S}_6^{\text{Euler}}\cup\{f_{c,-}\}$ ($15/21$ squares), $\mathcal{U}_8^{\text{Euler}}=\mathcal{S}_7^{\text{Euler},c}\cup\{f_{b,-}\}$ ($18/28$ squares, $1+bf_{b,-}=X_7^2-\frac{t^4+t^2+1}{t^2}$), and the Universal Bi-Euler $9$-Tuple $\mathcal{U}_9^{\text{bi-Euler}}=\mathcal{S}_7^{\text{Euler},c}\cup\{\tilde{f}_{a,-},\tilde{g}_{\text{Euler},+}\}$ ($21/36$ squares; $23/36$ on $t_1(r),\mathcal{C}_{\text{coinc}}$). Here all $9$ entries, the $t$-odd automatic roots $R_{c,a},R_{c,g},R_{a,g}$ ($R_{c,g}-R_{c,a}=c,R_{a,g}-R_{c,a}=f_{a,-}$), and the $10$ $t$-odd open-pair bases $X_1,\tilde{X}_1,X_2,X_4,\tilde{X}_4,X_5,X_6,X_{11},X_{12},X_{13}$ lie in the $t$-odd free rank-$3$ $\mathbb{Z}[r]$-module $\mathcal{M}_3=\mathbb{Z}[r]a\oplus\mathbb{Z}[r]b\oplus\mathbb{Z}[r]r_{ad_-}$ ($r_{ad_-}=\frac{t^2+1}{2t}=\sqrt{1+ad_-},r_{ad_-}^2-(2ra)^2=1$), while the $t$-even base $X_3=r_{bc}+r_{bf_{a,-}}=(4r^2-1)r_{bc}-r=(2r^2-1)^2r+(4r^2-1)b r_{ad_-}$ lies in the complementary $t$-even free rank-$2$ $\mathbb{Z}[r]$-module $\mathcal{N}_2^{\text{even}}=\mathbb{Z}[r]r\oplus\mathbb{Z}[r](b r_{ad_-})$ ($r_{bc}=(2r^2-1)r+b r_{ad_-}$):$$ \begin{aligned} c,\,f_{c,-} &=(8r^4-4r^2+1)a+b\pm J_c(r)r_{ad_-},\quad f_{a,-},\,\tilde{f}_{a,-}=(128r^8-192r^6+96r^4-16r^2+1)a+4(2r^2-1)^2b\pm J_a(r)r_{ad_-},\\ g_{\text{Euler},+},\,\tilde{g}_{\text{Euler},+} &=8r^4(16r^4-16r^2+5)a+(4r^2-1)^2b\pm J_g(r)r_{ad_-},\quad f_{a,-}=(8r^2-1)a+2b+2(8r^4-8r^2+1)c,\\ (2r^2-1)g_{\text{Euler},+} &=2r^2(8r^2-3)a+(4r^2-1)b+2r^2(4r^2-1)(4r^2-3)c,\quad X_2=r_{cd_-}+X_1,\quad X_3=(2r^2-1)^2r+(4r^2-1)b r_{ad_-}, \end{aligned} $$with jump amplitudes $J_c(r)=2r(2r^2-1),J_a(r)=4r(2r^2-1)(8r^4-8r^2+1),J_g(r)=4r^3(4r^2-1)(4r^2-3)$ and all $15$ open pairs of $\mathcal{U}_9^{\text{bi-Euler}}$ equal to rank-$1$ differences of squares with $r$-only shifts: $4r^2(1+af_{a,-})=X_1^2+(4r^2-1)=\tilde{X}_1^2-4r^2(4r^2-1)(4r^2-3)$ ($X_1=r_{d_-f_{a,-}}$), $1+d_-g_{\text{Euler},+}=X_2^2-(4r^2-1)$, $1+bg_{\text{Euler},+}=X_3^2-(4r^2-1)$ ($r_{bc}+r_{bf_{c,-}}=J_c(r),X_3+\tilde{X}_3=4r^3(4r^2-3)$), diagonal norms $1+cf_{c,-}=((4r^2-1)a\mp b)^2-(4r^2-1)(4r^2-3)^{(1\mp 1)/2}$, $1+f_{a,-}\tilde{f}_{a,-}=X_{12}^2-(4r^2-1)(16r^4-12r^2+1)$ ($X_{12},\tilde{X}_{12}=(4r^2-1)(16r^4-12r^2+1)a\mp 4(2r^2-1)^2b$), $1+g_{\text{Euler},+}\tilde{g}_{\text{Euler},+}=(32r^4(2r^2-1)a\mp(4r^2-1)^2b)^2-((4r^2(4r^2-1)(4r^2-3)^{(1\mp 1)/2})^2-1)$, and cross-pairs obeying the Jump-Product Shift Law $1+u_i\tilde{u}_j=X_k^2-\frac{J_i(r)J_j(r)}{r^2}$ ($X_5,X_6,X_{11}\in\mathcal{M}_3$ with $X_k-\iota_t(X_k)=-2r\,r_{ad_-}$ and Zero-Sum Triangle Laws $X_6-X_5=\tilde{X}_4,X_{11}-X_5=\tilde{X}_{12}$). Universally, in $E_1=\frac{t}{r}=2a+\frac{1}{r}r_{ad_-},E_2=-\frac{1}{rt}=2a-\frac{1}{r}r_{ad_-},E_3=b$ ($E_1 E_2=-\frac{1}{r^2},E_1 E_3+E_2 E_3=4(r^2-1)$), on the Symplectic Euler Plane $\mathcal{V}_{\text{Euler}}=\{(p_+,p_-,s):p_++p_-=2r^2s\}=\mathbb{Q}(r)e_1\oplus\mathbb{Q}(r)e_2$ ($w=\xi e_1+\eta e_2$, $p_+=\frac{\xi+(4r^2-1)\eta}{2},p_-=\frac{-\xi+\eta}{2},s=\eta$, $\iota_V(\xi,\eta)=(-\xi-2(2r^2-1)\eta,\eta)$), $u(w):=p_+^2 E_1+p_-^2 E_2+s^2 E_3\in\mathcal{M}_3$ and $R(w_1,w_2):=p_{1,+}p_{2,+}E_1+p_{1,-}p_{2,-}E_2+s_1 s_2 E_3\in\mathcal{M}_3$ obey the Universal Unimodular Determinant & Master Laws ($Q(\xi,\eta):=\xi^2+2(2r^2-1)\xi\eta+\eta^2$, $X_{d_-}=\xi r_{ad_-}+\eta r_{cd_-}\in\mathcal{M}_3,X_b=\xi r+\eta r_{bc}\in\mathcal{N}_2^{\text{even}},\tilde{X}_a=2r(\xi a+\eta r_{ac})\in\mathcal{M}_3$):$$ 1+u(w_1)u(w_2)=R(w_1,w_2)^2+1-(\xi_1\eta_2-\eta_1\xi_2)^2,\quad 1+d_-u(\xi,\eta)=X_{d_-}^2+1-Q(\xi,\eta),\quad 1+bu(\xi,\eta)=X_b^2+1-Q(\xi,\eta),\quad 4r^2(1+au(\xi,\eta))=\tilde{X}_a^2+4r^2(1-\eta^2), $$with universal osculating section $t_{b,(\xi,\eta)}(r)=-\frac{4r^4\eta(\xi+(r^2-1)\eta)+(\xi-\eta)^2-1}{4r^2\eta((r^2-2)\xi+(r^2-1)(3r^2-2)\eta)-(\xi-\eta)^2+1}$ making $1+bu(\xi,\eta)=\square$ (reproducing $t_{\text{III}}(r)=-t_{\text{sec}}(r)=t_{b,(-1,4r^2-1)}(r)$ with $22/36$ squares in $\mathcal{U}_9^{\text{bi-Euler}}$, $t_{b,(1,1)}(r)=-\frac{4r^6-1}{12r^6-16r^4+1}$ for $g_0=d_+(0,a,c)=u(1,1)=4r^3t+b$, and degree-$4$ $t_{b,(\eta\pm 1,\eta)}(r)=-\frac{r^2(r^2\eta\pm 1)}{r^2(3r^2-4)\eta\pm(r^2-2)}$); under $M_{\text{Cheb}}(\xi,\eta)=(-\eta,\xi+2(2r^2-1)\eta)$, $u_n=u(M_{\text{Cheb}}^n(1,0))$ ($\iota_t(u_n)=u_{-n}$) and $g_n=u(M_{\text{Cheb}}^n(1,1))=d_+(0,u_n,u_{n+1})$ ($\iota_t(g_n)=g_{-n-1},1+ag_n=\square$) form $\mathcal{U}_{4N+3}^{\text{bi-Euler}}=\{b,d_-\}\cup\{u_n\}_{-N}^N\cup\{g_n\}_{-N}^{N-1}\subset\mathcal{M}_3$ with $12N+1$ automatic squares ($13/21,25/55,37/105$ for $N=1,2,3$) and $8N^2-2N+2$ $r$-only shifts ($1+d_-g_0=(2r^2t)^2-(4r^2-1)$ parameterized by $t_{\text{conic}}(r,M)$, and adjoining $g_{\text{Euler},-},\tilde{g}_{\text{Euler},-}$ to $\mathcal{U}_{11}^{\text{bi-Euler}}$ gives $\mathcal{U}_{13}^{\text{bi-Euler}}$ with $29/78$ squares and $49$ $r$-only shifts). Finally, $4rt(t^2-1)(g_{\text{Euler},+}-f_{a,-})=4t^2(X_2^2-X_1^2-(4r^2-1))=P_{\text{CaseA}}(r,t^2)$ closes Case A, $t=\pm\frac{1}{4r^2-1}$ gives $1+f_{a,-}f_{c,-}=\square$ ($b=c$), and $R_4(a,d_-,f_{a,-},g_{\text{Euler},+})=0$ splits into $t=\pm t_1(r)=\pm\frac{4r^2-1}{2(2r^2-1)}$ ($g=0$: $\mathcal{Q}_5^\star(r),\mathcal{S}_6^{\text{refl}}(r)$ whose $9$ $6/6$ sub-quadruples split $R_5=0$ into $3$ zero + $15$ distinct $18/21$ septuple branches in $\mathbb{Q}(r)$, certified at $r=2$; infinite-order $P_1(r)$ on $\mathcal{E}_{\text{I}},\mathcal{E}_{\text{II}}$ with $\deg_r([n]P_1)=2,12,26$ and $2,10,22$ and reciprocal involution $\iota_{A,C}(t,Y)=(\frac{C}{At},\frac{CY}{At^2})=\infty_+\ominus P$ giving $\mathcal{Q}_{5,(2,\varepsilon)}^{\text{refl}}(r)$) and smooth $g=4$ $\mathcal{C}_{\text{coinc},r}:F_{\text{coinc},\pm}(r^2,t)=0$ ($r^2=\frac{(Y-2X-4)(Y+2X-4)}{64(X+4)},t=-\frac{4}{X}$) over $E_{\text{coinc}}:Y^2=X^3+8X^2+32X+80$, where $Q_{\text{I}},Q_{\text{II}}=\square$ via explicit polynomial division identities (exp61–exp67).

Proof. Verified in verify_paper.py, exp25–exp67, and THEOREMS_AND_PROOFS.md. $\square$

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10. Limitations and scope

• No rational septuple ($21/21$) is claimed. $\mathcal{S}_7^{(20/21)}$ adds relation type $(|\mathcal{R}_4|,|\mathcal{R}_5|)=(5,2)$ ($\text{dim}=1,\text{gens}=2$) [2, 3, 4].
• Finiteness on $\mathcal{H}$ vs. Problem 3.3: Theorem 3(4) proves $\mathcal{C}_+(a,\lambda)=0$ on Branch 1 has smooth $g=3$ cover $\mathcal{H}$, so by Faltings [6] it has finitely many rational almost septuples, leaving open $\mathcal{S}_{4,2}$ and $\mathcal{E}_{\text{Br1}}$.
• Completeness on $\mathcal{H}$ and unpaid edges: Faltings' theorem is ineffective; our height-$50,000$ search on $\mathcal{E}^{(1)}$ and u/kiro_dioph7's $295$-prime sieve (Post 35) are bounded, while [23] (Post 68) proves $\mathcal{H}(\mathbb{Q})=\{(\pm 1,\pm 16),(\pm 2,\pm 55),\infty_\pm\}$ via Quadratic Chabauty. On Curve II, $c_1 c^*_\mp+1$ have degrees $36,32$; on Theorem 5, the $2$-adic lock eliminates $Q_4\equiv-2$ while $Y_B^2,G(k)=\square$ is closed at $k\in \{2,3\}$ [16]; on $z=z_{L_1}$ (Theorem 6), $a c+1,\mathcal{F}_{ab},\mathcal{G}_d,H=\square$ remains open; on $\mathcal{C}_{\text{triad}}$, $1+d_+b'_{ac}(d_-)=\square$ is closed by rank-$0$ $E_{15}$; on Theorem 7, $H=1/3$ is closed by $\mathbb{Q}(\sqrt{-2})$ $2$-descent, $u_w=2u_d$ by Pythagorean Syzygy, $(u_w-2u_d)^2+u_d^2=1$ by rank-$0$ $E_{\text{coll}}$, and [20–22] give $23/28$ octuples ($1+M(d)M(e)$ closed by $E_{24}$, Orbit-I Source-1 by $2$-Selmer/Chabauty [24, 25], circle charts obstructed over $\mathbb{Q}_{23}$ at $q=\pm 5/16$, Post 69 open on $(0,q_{H,1}]\cup [q_{H,2},q_{B,+})$ outside $v_3(e)=0$); $\kappa_2,\kappa_3$ give $22/28$ octuples.

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11. Reproducibility

verify_paper.py and verify_jac_selmer.py in supplement.zip use Python 3 and SymPy (python3 -I verify_paper.py, $\sim 90\text{ s}$) to verify all identities, factorizations, square-class checks, Sturm intervals, $2$-descent certificates, and tuples across Theorems 1–7 (exp9–exp66, THEOREMS_AND_PROOFS.md):

1. Theorems 1–3: $c^*_\mp$, $S_{ab}$-completions, $\mathcal{E}_{\text{Br1}},\mathcal{E}_{F_1}$, $a_{\text{circ}}(\lambda)$, positive sextuples, $\mathcal{S}_7^{(20/21)}$, $R_4=\frac{F_{\text{sym}} G}{(P_4-1)^4}$, $\mathcal{S}_{4,2}$, $\mathcal{H}$, $\text{rank}\,E_{\text{min}}^{(5)}(\mathbb{Q})=0$.
2. Theorems 4–6: $(r,\mu,z,w)$ syzygy, $1$-line formula, 4-$R_4$ / 1-$R_5$ Lorentz syzygies, One-$R_5$ surface ($\mathcal{K}_\mp=\mathcal{K}_{\mp,-1}\mathcal{K}_{\mp,+1}$, central $R_5$ norm $P_1 P_2 P_3 P_4$, Dual Brahmagupta, $\mathcal{Q}_{\text{phys}}$, $P_{\text{blowup}}$, $\mathcal{M}(\beta)=0$, $P_{\text{surf}}\xrightarrow{4:1} N_{\text{core}}$, $\rho_2=\alpha$ $E_{Y_W}$, $(19,2,10)$ & $\mathbb{Q}(\sqrt{5})$ sextuple, $\rho_2=\alpha+2r$ $\mathcal{C}_{\text{sec2},3..4}$, $(26/11,7/22)$ sextuple [4] & Rank-$4$ $d_-=\frac{2}{C_0\mp w_1}$, $\mu_{\text{inf}}$, $\mathcal{S}_7^{(f_c)},\mathcal{S}_7^{\text{3-refl}},\mathcal{S}_6^{\text{Euler}},\mathcal{S}_7^{\text{Euler},c},\mathcal{U}_8^{\text{Euler}},\mathcal{U}_9^{\text{bi-Euler}},\mathcal{M}_3,t_{\text{III}}(r),\mathcal{Q}_{5,(n,\varepsilon)}^{\text{refl}},\mathcal{Q}_5^\star,\mathcal{S}_6^{\text{refl}},\mathcal{C}_{\text{coinc},r}$, $L_1$-splitting, $\rho_2=r+\alpha$ $F_{12}=0$, $C_{9,11},G_{5,5}^{(a,b)},\mathcal{C}_{\text{hyp}}^{(8)},Q_{U,\mu},F_{6,7},C_{4,4}^{(g=a)},C_{5,5}^{(g=a)}$, $\mathbb{Q}(\sqrt{133})$ septuple; exp45–exp66), Curves I–III, $\mathcal{E}_B$, $R_5=\frac{L_1 L_2 Q_4}{H K^4}$, $\iota_K$, $\mathcal{E}'_{Q_4}(R)$, $\Lambda(r)$, $15+6$ Collapse, $v=-1/5$ ($\text{rank}\,E_{16}(\mathbb{Q})=0$).
3. Theorem 7: $(V,K,d)$ chains, $\mathcal{S}_{\text{3-tors}}$, Orbit Cubic, $1+\sigma_6$ & $P_6\ne 0$ (exp52), $C_r$, Master Syzygies, $\mathcal{U}_8^{\text{DKMS}}(\kappa)$, $E_{\text{Cheb}}(s)$, $Q_{\text{ex}}$, $\mathcal{C}_{\text{triad}}$, rank-$0$ $E_{15}$, Mixed-$R_4$ Syzygies, $U_\mp(r,s,\kappa)$, Five-$R_5$ $G_1-G_2=H(R_1-R_2)(R_1+R_2)$, $3$-factor $h M_i\mp D_i$ & $6$-torsion $H$-splitting on $P_{\text{orb}}=0$, $Q_h$ Three-Circle & rank-$0$ $E_t$, $(r,H_k)$ $R_3(d,3P_0,g)$ (exp58–exp66), $\mathcal{H}_{\text{II}}$ ($\text{rank}\,E_{\text{II},C_2}(\mathbb{Q})\ge 2$), $H=1/3$ $2$-descent, Branch B ($10/10$ quintuple, $20/21$ over $\mathbb{Q}(\sqrt{-14})$, $\mathcal{M}_\Psi^\star$, rank-$0$ $E_{\text{coll}}$), and [20–22] ($\mathcal{I}_{\text{AL}}^{(0)}$, $d_1..d_4$, $k_{\text{tan}}(m)$, $E_{sw}$, $\mathcal{S}_{sw}$, Orbit-I $2$-Selmer/Chabauty, Post 69 syzygies, circle charts, rank-$0$ $E_{24}$, $23/28$ octuples).

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