1. Problem and contribution

A rational Diophantine $m$-tuple is a set of $m$ distinct nonzero rationals $\{a_1..a_m\}$ with $a_ia_j+1=\square$ ($i<j$). No rational septuple ($m=7$, $21$ squares) is known [1, Problem 3.2]; Problem 3.3 asks if infinitely many almost septuples ($20/21$ squares) exist [1].

Papers 6 [2] and 10 [3] classified the $25$ known almost septuples ($22$ in [2], $A_{1..3}$ [3], $A_4$ [4]) by relation type $(\mathcal{R}_4,\mathcal{R}_5)$, the $4$- and $5$-subsets where $R_4,R_5=0$:

$$ R_4(x,y,z,t)=(x+y-z-t)^2-4(xy+1)(zt+1),R_5(x,y,z,t,u)=(xyztu+2xyz+x+y+z-t-u)^2-4(xy+1)(xz+1)(yz+1)(tu+1). $$

$14/25$ contain $\{a,b,c,d_-,d_+\}$ [2, 3, 5] ($g\ge9$ on generic fibres [3]).

Eliminating $c^*_\mp=e_\mp(a,b,d_-,d_+)$ in $(r,\lambda,S_{ab})$ ($d_-d_++1=w_1^2,S_{ab}:=a+b$) yields:

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 ($\mathcal{E}_{Br1}$: $cc^*_-+1=\square$, $23/28$ octuples at $\lambda=\pm1/2$) 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 $\mathcal{E}_{Br1}$, 408 positive).
3. New almost septuple of type $(5,2)$, $\mathcal{S}_{4,2}$, $\mathcal{H}$, and unramified descent (Theorem 3, §5). $R_4(p,q,e_-,e_+)=\frac{F_{sym}G}{(P_4-1)^4}$ 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 ($rkE_{min}^{(5)}(\mathbb{Q})=0$). At $(-2,55)$, we find the first type-$(5,2)$ almost septuple:

$$ \mathcal{S}_7^{(20/21)}=\{-19961343/78775960,-411840/1969399,23715/29896,71003/63240,36890/11211,104321672/29540985,\frac{29896}{7905}\}. $$

4. Trace decomposition, Lorentz splitting, $P_{blowup}$ & 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 gives $P_{blowup}$ and $\mathcal{M}(\beta)=0$ ($P_{surf}\xrightarrow{4:1}N_{core}$, $E_{Y_W}$, $\mathcal{C}_{sec2,3..4}$, $F_{12}=0$), and $\Sigma_\mu^+$ gives 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 $4$, giving $\mathcal{E}_B:Y_B^2=(k-2)(5k^3-8k^2+13k-2)$ ($k=4$) and a $2$-adic period-$6$ lock eliminating $Q_4(k)\equiv-2\pmod{\mathbb{Q}^{\times2}}$.
6. Factorization $R_5=\frac{L_1L_2Q_4}{HK^4}$, $L_1=0$, $\iota_K$, $\Lambda(r)$, 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 rank-$0$ $\mathcal{E}_{16}$.
7. $(V,K,d)$ chains, $\mathcal{S}_{3-tors}$, Master Syzygies, $\mathcal{C}_{triad}$, $\mathcal{V}_{Euler}$ & $\mathcal{E}_W(R^2,Z)$ towers (Theorem 7, §9). On $3S_T=\mathcal{O}$, $E_T$ has Tate form, Orbit Cubic, $15$ septuples/coset, and Master Syzygies $9C_A,9C_B$ unifying Family IV, DKMS [7, §3] ($\mathcal{U}_8^{DKMS}$), $E_{Cheb}(s)$, and $\mathcal{C}_{triad}$ ($E_{15}$ closure), with Mixed-$R_4$, Five-$R_5$ (Post 42), $Q_h$, $\mathcal{H}_{II}$, $H=1/3$ $2$-descent, Branch B, sources [20–22], and towers $\mathcal{U}_{4N+3..5}^{Cheb},\hat{\mathcal{U}}_{8N+2..6}^{lat},Q_5^{Cheb}(q),\tilde{\mathcal{Q}}_{5,j}^{refl},\mathcal{S}_7^{osc,(n)}(R,\kappa),\mathcal{S}_7^{can,(n)}(r),\hat{\mathcal{S}}_7^{mir,(n)}(w)$ on $\mathcal{V}_{Euler}\subset\mathcal{M}_3$ and $\mathcal{E}_W(R^2,Z)$ ($g=4$ $\mathcal{C}_{coinc,r}$).

2. Related work and distinction

Dujella et al. [7] built sextuples via $3$-torsion chains on $E_T$, DKP [5] studied $\{a,b,c,d_-,d_+\}$ on $S_{DKP}$, and Gibbs [27] built Stern–Brocot trees from regular quadruples. On rStation, [2] classified $22$ almost septuples into $18$ types; [3, 26] proved generic quintuple fibres have $g\ge9$ ($A_1..A_3$) and regular-fifth loci over $S_{DKP}$ have general type; [4, 8, 9] studied chain closures, build orders ($A_4$), and $R_{33}$; [10, 17, 32] proved uniqueness of $\pm S$, $e_\mp=x(Q\pm S)$, and that on the twisted edge surface of a fixed quadruple only Gibbs translates carry infinitely many points; and [11–14, 16, 18–23, 28–30] studied Gibbs sextuples, charts, paired/antipodal sources, rigidity, reflections $\varphi_b$, fourth orbits (Post 35, 42, 43, 67, 69), and QRT maps [30].

Distinction: Unlike fixed-fibre analyses [2, 3, 32] ($g\ge9$) or $S_{DKP}$ regular-fifth loci [26], $(r,\lambda,S_{ab})$ varies the base quadruple to give $\mathcal{E}_{Br1},\mathcal{E}_{F_1},\mathcal{E}_{F_3}$ (Theorems 1–2) and the first type-$(5,2)$ almost septuple $\mathcal{S}_7^{(20/21)}$, finite on $\mathcal{H}$ via Faltings and descent (Theorem 3). Across Theorems 4–7, our Master Syzygies unify [5, 7, 10, 16, 17, 20–23, 27–32] and organize chain completions [4, §2; 27] into the towers of §1(7) on $\mathcal{V}_{Euler}\subset\mathcal{M}_3$ and $\mathcal{E}_W(R^2,Z)$, with Faltings finiteness on $g=4$ $\mathcal{C}_{coinc,r}$, on $\mathcal{S}_7^{univ}(q,V_4\cdot w_1(q))$, and on the canonical gate curves ($n\in\{2,3,4,5\}$).

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

For a triple $\{a,b,c\}$ ($ab+1=r^2,ac+1=s^2,bc+1=t^2$), $d_\mp=a+b+c+2abc\mp2rst$ obey $d_-+d_+=2(a+b+c+2abc)$ and $d_-d_++1=(c-a-b)^2-(4r^2-1)$. With $S_{ab}:=a+b,C_0:=c-S_{ab}$, $d_-d_++1=w_1^2\iff C_0^2-w_1^2=4r^2-1$ is given by $C_0-w_1=\lambda(2r-1),C_0+w_1=\frac{2r+1}{\lambda}$:

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

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

Theorem 1 ($S_{ab}$-Square Completions and Dual Sextuple Surfaces)

For $\{a,b,c,d_-,d_+\}$ given by $(r,\lambda)$, let $D_-:=\lambda^2(r-1)-(r+1)$ and $D_+:=\lambda^2(2r-1)^2(r+1)-(2r+1)^2(r-1)$.

1. Linearity of $c^*_\mp$ in $S_{ab}$: $c^*_\mp=e_\mp(a,b,d_-,d_+)$ depend only on $S_{ab}=a+b$:

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

2. Universal $S_{ab}$-completions: On $\mathcal{S}_7^{(c^*)}=\{a,b,c,d_-,d_+,c^*_-,c^*_+\}$ ($18/21$), $cc^*_\mp+1,c^*_-c^*_++1$ obey:

$$ \begin{aligned} 4D_-^2(cc^*_-+1)=(4\lambda S_{ab}+\lambda^2(4r-3)+4r+3)^2+(\lambda^2(2r-3)-(2r+1))(\lambda^2(2r-1)-(2r+3)),\\ 4D_+^2(cc^*_++1)=(4\lambda(4r^2-1)S_{ab}+\lambda^2(16r^3-4r^2-8r+3)+16r^3+4r^2-8r-3)^2\\+(4r^2-1)(\lambda^2(2r-1)^2-(2r-3)(2r+1))(\lambda^2(2r-1)(2r+3)-(2r+1)^2),\\ D_-^2D_+^2(c^*_-c^*_++1)=16\lambda^2(\lambda(4r^2-1)S_{ab}+(2r^2-1)(\lambda^2(2r-1)+(2r+1)))^2+F_1F_2F_3F_4, \end{aligned} $$

where $F_{1,2}(r,\lambda)=2\lambda^2r^2-3\lambda^2r+\lambda^2\mp2\lambda r-2r^2+r+1$ and $F_{3,4}(r,\lambda)=2\lambda^2r^2+\lambda^2r-\lambda^2\mp2\lambda r-2r^2-3r-1$. 3. Surface I (Branch 1 $\mathcal{E}_{Br1}$): $\Delta_{S_{ab}}(cc^*_-+1)=0$ on Branch 1:

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

($\lambda\mapsto1/\lambda$ swaps the other factor). On Branch 1, $R_4(c,d_-,d_+,c^*_-)=0$ and $cc^*_-+1=(\frac{4\lambda(\lambda^2-1)S_{ab}+3\lambda^4+14\lambda^2-1}{(\lambda^2-1)^2})^2$ hold for all $(\lambda,S_{ab})$, while Conditions 20–21 become:

$$ \begin{aligned} (\lambda^2-1)^2(5\lambda^4+10\lambda^2+1)^2(cc^*_++1)=(8\lambda(\lambda^4-1)S_{ab}+9\lambda^6+29\lambda^4+27\lambda^2-1)^2+(2\lambda(\lambda^2-1))^2\cdot2(\lambda^2+1)(\lambda^2+3)(3\lambda^2+1),\\ (\lambda^2-1)^4(5\lambda^4+10\lambda^2+1)^2(c^*_-c^*_++1)=(4\lambda(8\lambda(\lambda^4-1)S_{ab}+7\lambda^6+31\lambda^4+29\lambda^2-3))^2+((\lambda^2-1)^3)^2\cdot(25\lambda^4+6\lambda^2+1). \end{aligned} $$

Here $g:=d_+(c,d_-,d_+)=2(c+d_-+d_++2cd_-d_+)-c^*_-=\frac{4\lambda(\lambda^2+1)(\lambda(\lambda^4-1)S_{ab}+\lambda^6+3\lambda^4+5\lambda^2-1)}{(\lambda^2-1)^3}$ obeys $1+c^*_-g=(\frac{8\lambda^2(\lambda^4-1)S_{ab}+2\lambda(\lambda^2+3)(3\lambda^4+6\lambda^2-1)}{(\lambda^2-1)^3})^2-(4\lambda^2-1)$, a square in $S_{ab}$ iff $\lambda=\pm1/2$ ($\lambda=1/2$: $1+c^*_-g=(\frac{120S_{ab}-143}{27})^2$), giving $19/21$ septuples $\{a,b,c,d_-,d_+,c^*_-,g\}$ and $\mathcal{U}_8^{Br1}(1/2)=\{a,b,c,d_-,d_+,c^*_-,c^*_+,g\}$ ($\ge23/28$ squares; $23/28$ at $\lambda=1/2,\mu=-203/60$, exp43; [22, §8]). 4. Surface II (Twin-Fifth $\mathcal{E}_{F_1},\mathcal{E}_{F_3}$): $F_1=0,F_3=0$ are birational to rank-$1$

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

On $\mathcal{E}_{F_1},\mathcal{E}_{F_3}$, $R_4(a,b,c^*_-,c^*_+)=R_4(d_-,d_+,c^*_-,c^*_+)=0$ and $c^*_-c^*_++1=\square$ for all $S_{ab}$, giving $15/15$ $\{a,b,d_-,d_+,c^*_-,c^*_+\}$ and $19/21$ $\mathcal{S}_7^{(c^*)}$. Proof. Substituting $ab=r^2-1,d_-d_+=w_1^2-1,d_-+d_+=2(2r^2S_{ab}+(2r^2-1)C_0),\prod\sqrt{\ell d_\mp+1}=(r^2-1)C_0+r^2S_{ab}$ into $R_5=0$ and $c=S_{ab}+C_0$ gives Parts 1–3; $P_4^2-6P_4-4(ab+d_-d_+)-3=\frac{F_1F_2F_3F_4}{4\lambda^2(r^2-1)^2}$ gives Part 4 (verify_paper.py). $\square$

4. Elliptic surface $\mathcal{E}_{(r,\lambda)}$, chord families, and positive sextuples

With $s=a+\mu$ on $c=a+\frac{r^2-1}{a}+C_0$, $ac+1=(a+\mu)^2$ gives $a(r,C_0,\mu)=\frac{\mu^2-r^2}{C_0-2\mu}$, $b(r,C_0,\mu)=\frac{(r^2-1)(C_0-2\mu)}{\mu^2-r^2}$, $s=\frac{C_0\mu-(\mu^2+r^2)}{C_0-2\mu}$, reducing $bc+1=t^2$ ($t=\frac{Y_\mu}{\mu^2-r^2}$) to $\mathcal{E}_{(r,\lambda)}$:

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

Here $\infty_\pm$ has tangent chord $\mu_{inf}(r,C_0)=\frac{C_0^2+8r^4-4r^2}{4C_0r^2},a_{inf}(r,C_0)=\frac{C_0^2-(4r^3-2r)^2}{8C_0r^2 (2r^2-1)}$, 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 $\mathcal{E}_{Br1}$ and Positive Sextuples)

On $\mathcal{E}_{Br1}(\lambda,\mu)$ ($r=\frac{3\lambda^2+1}{2(\lambda^2-1)},C_0=\frac{\lambda(\lambda^2+3)}{\lambda^2-1}$), generic sections give $15/15$ $\{a,b,c,d_-,d_+,c^*_-\}$ and $19/21$ $\mathcal{S}_7^{(c^*)}$:

1. Seven $\mathbb{Q}(\lambda)$ families: Chords among $\infty_\pm,P_{1..4}$ give $7$ sections $a_\gamma(\lambda)\in\mathbb{Q}(\lambda)$ (exp9).
2. Five circle families ($\lambda=\frac{2k}{k^2+1}$): $c=0$ gives $P_{c,2}$ over $1-\lambda^2=\square$; chords with $P_{1..4}$ give $5$ families, e.g.:

$$ a_{circ}(\lambda)=-\frac{\lambda(\lambda^2+3)^2(4\lambda^4+\lambda^2-1)}{2(\lambda^2-1)(5\lambda^2-1)(2\lambda^4+\lambda^2+1)},b_{circ}(\lambda)=-\frac{(5\lambda^2-1)(2\lambda^4+\lambda^2+1)}{2\lambda(\lambda^2-1)(4\lambda^4+\lambda^2-1)}, $$

with $a_{circ}c_{circ}+1=(1-\lambda^2)(\frac{8\lambda^6-9\lambda^4-14\lambda^2-1}{2(5\lambda^2-1)(2\lambda^4+\lambda^2+1)})^2$ and $b_{circ}c_{circ}+1=(1-\lambda^2)(\frac{16\lambda^6-7\lambda^4+6\lambda^2+1}{2\lambda(\lambda^2+3)(4\lambda^4+\lambda^2-1)})^2$. 3. Five elliptic families ($\mathcal{C}_{5\lambda}:Y_\lambda^2=5\lambda^4+10\lambda^2+1$): $c^*_+=0$ gives $P_{c,3},P_{c,4}$; $P_1+P_{c,3}$ gives

$$ a_3(\lambda)=\frac{(5\lambda^2-1)(\lambda^8-42\lambda^6-80\lambda^4-6\lambda^2-1)}{4\lambda(\lambda^2-1)(29\lambda^6+53\lambda^4-17\lambda^2-1)},b_3(\lambda)=\frac{\lambda(\lambda^2+3)(29\lambda^6+53\lambda^4-17\lambda^2-1)}{(\lambda^2-1)(\lambda^8-42\lambda^6-80\lambda^4-6\lambda^2-1)}, $$

with $a_3c_3+1=(5\lambda^4+10\lambda^2+1)(\frac{11\lambda^6+59\lambda^4-7\lambda^2+1}{4\lambda(29\lambda^6+53\lambda^4-17\lambda^2-1)})^2$ and $b_3c_3+1=(5\lambda^4+10\lambda^2+1)(\frac{5\lambda^6-27\lambda^4-41\lambda^2-1}{2(\lambda^8-42\lambda^6-80\lambda^4-6\lambda^2-1)})^2$.

Census. Chords on $\mathcal{E}_{Br1},\mathcal{E}_{F_1},\mathcal{E}_{F_3}$ (exp9-10) yield $2,304$ sextuples ($408$ positive, e.g. $r=1090/1053$) and $222$ twin-fifth sextuples.

5. Almost septuple $\mathcal{S}_7^{(20/21)}$ of type $(5,2)$, surface $\mathcal{S}_{4,2}$, and genus-$3$ curve $\mathcal{H}$

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

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

1. Two-branch factorization of $R_4(p,q,e_-,e_+)$: For $\{p,q,u,v\}$ ($P_4=pquv$) and roots $e_\mp$ of $R_5(p,q,u,v,x)=0$:

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

where $F_{sym}:=P_4^2-6P_4-4(pq+uv)-3$ ($F_{sym}=0\implies R_4(p,q,e_-,e_+)=R_4(u,v,e_-,e_+)=0$, $e_-e_++1=\square$ if $pq+1$ or $uv+1=\square$), while $G(p,q;u,v):=(P_4-1)^2(p-q)^2-4P_4^2+8P_4-4uv(p+q)^2-4pq(u+v)^2-4(P_4+1)(p+q)(u+v)-4$ has $\Delta_v(G)=16(p-q)^2(pq+1)(pu+1)^2(qu+1)^2$, splitting over $pq+1=r^2$ into $v_\pm(p,q,r,u)=\frac{p(r\pm1)-q(r\mp1)+2pqu}{pq((p(r\mp1)-q(r\pm1))u-2)}$. 2. Type $(5,2)$ of $\mathcal{S}_7^{(20/21)}$ & Plane Cubic Addition Law: $\mathcal{S}_7^{(20/21)}=(c,c^*_+,d_-,a,b,c^*_-,d_+)$ (missing $(c,c^*_+)$) obeys five $R_4=0$ and two $R_5=0$ relations:

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

At $\mathcal{S}_7^{(20/21)}$, the Jacobian has rank $6$ ($\dim=1$) and $gens=2$ ($\{ac+1,cd_-+1\}$), distinguishing type $(5,2)$ from [2, 3, 4]. Projecting $\tilde{E}_T:\{u_i^2=1+x_it\}_{i=1}^3$ ($T=\{x_1,x_2,x_3\}$) from $O=(0,1,1,1)$ via $\xi_i=u_i-1$ gives $\mathcal{C}_T:\sum_{cyc}x_1\xi_2\xi_3(\xi_2-\xi_3)=0\subset\mathbb{P}^2$ ($O=[x_1:x_2:x_3],t=-\frac{4\xi_1\xi_2(\xi_1-\xi_2)(x_2\xi_1-x_1\xi_2)}{(x_2\xi_1^2-x_1\xi_2^2)^2}$). Under $\pi:(\mathcal{C}_T,\oplus_O)\to(E_T,\oplus_{P_0})$ ($\ker\pi=K_4\cong(\mathbb{Z}/2\mathbb{Z})^2$), $T\cup\{x_0\}\in\mathcal{R}_4\implies\varepsilon_0Q_0=P_0\mp_\infty S_T$ cancels $P_0$ on $32/64$ signed sums of $(x_0,x_1)$ to give $\varepsilon_1Q_1\mp_\infty S_T$ (roots of $R_5(T\cup\{x_1\},e)=0$, $\ge4$ edges; $6$ edges on $16$ sums iff $T\cup\{x_1,y\}\in\mathcal{R}_5$, Post 76, exp68). 3. The 2D type-$(4,2)$ surface $\mathcal{S}_{4,2}$ over $(r,\lambda)$: For arbitrary $(r,\lambda)$, $D_-\cdot D_+=4\lambda^2 (P_4-1)$ ($P_4=(r^2-1)(w_1^2-1)$). Eliminating $st$ from $F_{sym}(a,d_+;b,d_-)=P_4^2-6P_4-4(ad_++bd_-)-3=0$ via $(st)^2=(ac+1)(bc+1)$ gives $\Delta_{S_{ab}}(r,\lambda)=(2^{11}\lambda^8r(P_4-1)^2)^2 (P_4-1)(P_4-9)$, 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 $\mathcal{S}_{4,2}:W_\tau^2=(P_4-1)(P_4-9)$, $a_\pm=\frac{-A_1(r,\lambda)\pm16\lambda^4(P_4-1)W_\tau}{2A_2(r,\lambda)}$ obey $a_\pm c_\pm+1=\square$ in $\mathbb{Q}(\mathcal{S}_{4,2})$:

$$ a_\pm c_\pm+1=a_\pm^2+C_0(r,\lambda)a_\pm+r^2=(\frac{D_-D_+U_{S_{4,2}}(r,\lambda)\mp4\lambda^2W_\tau V_{S_{4,2}}(r,\lambda)}{2A_2(r,\lambda)})^2, $$

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 generic points on $\mathcal{S}_{4,2}$ give $15/15$ $\{a_+,b_+,d_-,d_+,c^*_-,c^*_+\}$ (type $(2,2)$) and $19/21$ $\{a_+,b_+,c_+,d_-,d_+,c^*_-,c^*_+\}$ (type $(4,2)$) (396 sextuples, 7,524 septuples in exp12). On Branch 2 ($r=\frac{m^2+m+1}{2m},\lambda=\frac{1-m^2}{1+m^2}$: $cc^*_++1=\square$), $\mathcal{S}_{4,2}$ gives $\mathcal{H}_{Br2}:W_m^2=(m^4+m^3+m^2+m+1)(m^4-7m^3+17m^2-7m+1)$, 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 $\mathcal{H}$ ($g=3$), finiteness & $2$-cover descent: On Branch 1, $\mathcal{C}_+(a,r,\lambda)=0$ becomes $\mathcal{C}_+(a,\lambda)=A_2(\lambda)a^2+A_1(\lambda)a+A_0(\lambda)=0$ ($A_0(\lambda)=-(r(\lambda)^2-1)A_2(-\lambda)$, 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-$3$ curve

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

On $a_\pm=\frac{-A_1(\lambda)\pm(\lambda^2-1)^3(5\lambda^4+10\lambda^2+1)W}{2A_2(\lambda)}$ (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 $20$ roots lift to $\mathbb{Q}(\mathcal{H})$, so finitely many almost septuples lie on $\mathcal{H}$ [6]. For $\lambda=p/q$, $Q_1(p,q)=5p^4+10p^2q^2+q^4>0$ and $Q_2(p,q)=5p^4-22p^2q^2+33q^4>0$ obey $5Q_2-(5p^2-11q^2)^2=44q^4$, $Q_1-Q_2=32q^2(p^2-q^2)$, and $ord_2(Q_1(p,q))\in\{0,4\}$, restricting $d\in\{1,5\}$. 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 $E_{min}^{(5)}:Y^2=(X+8)(X^2-9X-11)$ (conductor $200$) of rank $0$ ($E_{min}^{(5)}(\mathbb{Q})_{tors}=\{\mathcal{O},(-8,0)\}$), so 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 $\lambda=-2$: $F_+(a_3(\lambda))=\frac{(\lambda-2)(\lambda+2)(5\lambda^4+10\lambda^2+1)^2}{16\lambda^2(\lambda-1)^2(\lambda+1)^2}$ vanishes at $\lambda=\pm2$; on $\mathcal{E}_{Cond21}(\lambda,\kappa)$, $K_0^++K_3^-$ gives $\kappa(\lambda)\in\mathbb{Q}(\lambda)$ meeting $S_{ab,3}(\lambda)=a_3+b_3$ transversely at $\lambda=-2$. Proof. Verified in verify_paper.py. Since $P_8(\lambda)$ has $\Delta=2^{80}\cdot3\cdot5^8\cdot11^3\neq0$, $\mathcal{H}$ is smooth of $g=3$. In Part 4, $\gcd(Q_1,Q_2)\mid32q^2(p^2-q^2)$ restricts twists to $d\in\{1,5\}$, and $2$-isogeny descent mod $512$ proves $rkE_{min}^{(5)}(\mathbb{Q})=0$. On $\mathcal{E}^{(1)}$ ($L=\lambda^2$), height-$50,000$ search and u/kiro_dioph7's $295$-prime sieve (Post 35) yield 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 $\mathcal{S}_{4,2}$

On $\mathcal{S}_{4,2}$ (Post 35), set $\mu:=\frac{\lambda^2-1}{D_-},w:=\frac{2\lambda}{D_-},z:=wc$ ($D_-=\lambda^2(r-1)-(r+1)$), and define

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

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

1. $(r,\mu,z,w)$ Syzygy, $\mathcal{S}_{4,2}$ Formula, $P_{blowup}$ & $E_z=0$ (exp42-66):

• (a) Gate Syzygy & $\mathcal{S}_{4,2}$: With $-K=2rT+4r^2-1,T^2-4r^2H=\mu(\mu+4r),(r^2-1)\mu(\mu+4r)=H-K,A_0:=z-T=wS_{ab},B_0:=2r^2z-T$, $1+cc^*_-=z^2-\mu z+1$, $K^2(1+cc^*_+)=(4r^2-1)^2z^2+(4r^2-1)(\mu+4r)z+K^2$, and $K^2(1+c^*_-c^*_+)=H((4r^2-1)z-T)^2+(K^2-4r^2(r\mu+1)^2H)$ obey $(4r^2-1)^3\mu H(1+cc^*_-)+(\mu+4r)HK^2(1+cc^*_+)-4r(r\mu+1)K^2(1+c^*_-c^*_+)=4r\mu(\mu+4r)(r\mu+1)(4r^4H+(r^2-1)K)$. On $\mathcal{S}_{4,2}$ ($W_\tau^2=-KL$), $a,b=\frac{16(r^2-1)H(-(2r^2-1)wT\pm rK)}{16(2r^2-1)^2H^2-(K\mp W_\tau)^2}$ ($z=w(a+b)+T,c=z/w$; $(13/6,2,-8/3,880/27)$ at $\mathcal{S}_7^{(20/21)}$).
• (b) Lorentz Laws & $\mathcal{K}_{\mp,\varepsilon}=0$: On $R_4(a,d_-,f,g)=R_4(b,d_+,f,g)=0$ over $P_{pair}(X)=X^2-S_{fg}X+w_{fg}^2-1=0$ ($u^2=1+ad_-,v^2=1+bd_+$), $(1+aX)(1+d_-X)\equiv(w_{fg}-uX)^2$, $ab F_i=X_i^2-Y_i^2$ ($X_1=av-bu,Y_1=a-b,X_2=r^2-uv,Y_2=u-v,\Delta_{fg}\equiv F_1F_2$), and $X_{R_5}=-\frac{2w_{fg}}{D_{br}}$ ($D_{br}=uv-u-v$) forces $[1+d_-d_+]=[F_2]=[F_1]=[1]$ ($19/21$). On $D_uD_v(r^2-1)\ne0$ ($D_{u,v}=(2r^2-1)^2-\{u^2,v^2\}$), $\theta^2=\frac{D_uD_v}{r^2-1}+4r^2u^2v^2$ gives $a,b=\frac{(r^2-1)(\theta\pm2ruv)}{D_{v,u}},d_\pm=\frac{(\{v^2,u^2\}-1)(\theta\pm2ruv)}{D_{v,u}},c=a+b-\frac{\theta}{2r^2-1},w_1^2=\frac{u^2v^2-u^2-v^2+r^2}{r^2-1},F_2=w_1^2+r^2-2uv=x_2^2$. With $L_X=D_{br}X_2-2uv,L_Y=2r^2+u^2v-u^2+uv^2-v^2$, $D_{br}^2(X_2^2-Y_2^2)-4r^2(r^2-1)w_1^2\equiv L_X^2-L_Y^2$ splits $P_{pair}(X_{R_5})=0$ into $P_\mp X_1+Q_\mp Y_1=0$ ($P_\mp,Q_\mp=2rw_1\{L_X,L_Y\}\mp D_{br}x_2\{L_Y,L_X\}$), and $\mathcal{K}_\mp:=\Delta_{sys,\mp}^2-16r^2u^2v^2(r^2-1)A_\mp B_\mp=0$ ($A_\mp,B_\mp=P_\mp\{u,v\}+Q_\mp,\Delta_{sys,\mp}=D_vA_\mp-D_uB_\mp$) splits via $(\Delta_{sys,\mp}-8r^2(r^2-1)(u-v)P_\mp)^2-\mathcal{K}_\mp\equiv16r^2(r^2-1)J_vJ_u\equiv(4r[A_J\mp(r^2-1)w_1x_2B_J])^2$ into $\mathcal{K}_{\mp,\varepsilon}=0$, inverting in $(q,-\mu,-t)=(r^2+uv,u+v,w_1^2+uv)/x_{2,+},k=v-u$ ($x_{2,+}=q-t$) on $k^2=\mu^2(q-t)^2+4(1+qt-\mu^2)$.
• (c1) $P_{blowup}$, $\mathcal{M}(\beta)=0$ & $4:1$ Cover (exp47-56,62): In $(R,D_{br},s_2)=(r^2,uv-u-v,uv)$, $L_X^2-L_Y^2$ divides $E_{sym},F_{sym},E_{x,sym},F_{x,sym}$ ($E_0,E_{x,0}=16RP_{AD}\pm(u-v)^2P_{BC}$), reducing $N_{core}\equiv(R-1)E_0^2-64R^2D_{br}^2(u-v)^2W_{num}F_0^2\equiv(R-1)E_{x,0}^2-64R(u-v)^2X_{num}F_{x,0}^2$ ($w_1=-\frac{E_0}{8R(u-v)D_{br}F_0},x_2=-\frac{E_{x,0}}{8r(u-v)F_{x,0}}$) to $449$ terms, $\deg(9,12,8)$ ($\deg(9,8,8)$ in $s_2=D_{br}\eta$; $\mathcal{C}_{rat,1}(r),\mathcal{C}_{rat,2}(m)$ give $\{99/7,16/63,133/144,637/16,459/112\}$ at $m=3$). Since $s_1X_1\equiv((2R-1)^2+s_2)Y_1-4rs_2(R-1),(X_1-uY_1)(X_1-vY_1)\equiv(R-1)(u-v)^2,F_1\equiv4RY_1^2-4rs_2Y_1+(u-v)^2$, $P_\mp X_1+Q_\mp Y_1=0$ gives $\mathcal{Q}_{phys}:4RW_{num}(\xi^2-1)-(L_Y\xi+L_X)^2=0$ ($\xi=X_1/Y_1=-1-\frac{2(R-1)}{\delta},\delta=(R\tau^2-1)D_{br}+2R\tau(\tau-1)=k^2-R+1$), $M_{core}(R,D_{br},\tau)=0$, $M_{quad}(R,\delta,\psi,\sigma_\tau)=0$ ($\psi=\tau(\tau-1),\sigma_\tau=R\tau^2-1$), and in $\omega=\frac{4R\psi-\delta}{\delta},z=W_{fac}\frac{\delta}{D_{br}}-2(R-1)=\lambda_z\delta$ ($W_{fac}=(R-1)\omega+1$) $P_{blowup}(R,\omega,\delta,\lambda_z)=0$ with $-W_{fac}P_{blowup}\equiv L_{blow}^2-4(2R-1)^2\nu^2\sigma^2(\lambda_z-1)^2$ ($\nu^2=(R-1)\omega^2+4R,\sigma^2=4k^2-\omega\delta^2$). Eliminating $\tau,\rho,F_0,F_2$ gives $\mathcal{M}(\beta)=4R\omega^2Q_F^2+2\nu\omega(2R-1)\beta Q_FL_\Lambda+PW_{fac}^2L_\Lambda^2-4\omega^2(4R-\omega)^2PQ_G=0$ ($P=\beta^2,P_{surf}(r,\alpha,\beta)=0$), a $4:1$ ($(\mathbb{Z}/2\mathbb{Z})^2$) cover of $N_{core}=0$ via $\iota_{sgn}:(r,\alpha,\beta)\mapsto(-r,-\alpha,-\beta)$ and $urightarrow v$, birational on $\mathcal{U}$ in $(R,m,B)=(r^2,\alpha/r,\beta/r)$ to $\tilde{N}_{core}(R,u,v)=0$ via $u,v=-\frac{W_{fac}}{\gamma+\{(R-1)/t_\alpha,t_\alpha\}u_b},W_{fac}=-uv-(u+v)\gamma,ru_b=-\frac{(4R-\omega)(u-v)}{4uv},\frac{t_\alpha}{r}=-\frac{4u^2(v+\gamma)}{(4R-\omega)(u-v)}$ ($t_\alpha=\frac{\alpha r+R-1}{\alpha-r},u_b=\frac{\omega-P}{2\beta}=-\frac{u-v}{a-b},\omega=\frac{d_+-d_-}{a-b},\gamma=-\frac{av-bu}{a-b},Q_{conic}=2r(s_\rho+2r)p_\rho+s_\rho[(2r^2-1)s_\rho+2r(r^2-1)]=\sigma_{gen}^2$ for $s_\rho=\alpha-\rho_2\ne-2r$).
• (c2) Sections I–II & $F_{12}=0$ (exp54-57,62): 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 $(R,u,v)=(343/171,47/39,-29)$ ($\mathbb{Q}(\sqrt{133})$ $19/21$). On $\rho_2=\alpha$ ($s_\rho=0,Q_{conic}=(2r\alpha)^2$), 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 $P_{Y,z}^{(3)}(Y_W,z_W)=0\simeq E_{Y_W}:y_E^2=x_E^3+9x_E^2-736x_E+6400$ ($rk\ge1,P_1=(48,320),(19,2,10)$ at $12P_1$). On Section II ($\rho_2=\alpha+2r\iff s_\rho=-2r,Q_{conic}=(2r^2)^2$), in $(m,R)=(\alpha/r,r^2)$, Conic 1 splits into $4$ factors for $\{f=c\}\cup\{g=c\}$: Factor 1 ($R(m^3+m^2-m-1)+m+5=0,g=0,\mathcal{C}_{rat,2}$); Factor 2 ($R(m+1)^2(m+3)+m-3=0,g=0,u=-1$); Factor 3 ($-2(U^2-4)r^2+U(U^2-9)r+U^2+1=0$ in $U=\alpha+3r,E_{3,a}\simeq E_{Y_W},g=2$ $\mathcal{C}_{sec2,3}:Y_U^2=U^6-10U^4+57U^2-32$); and Factor 4 ($g=c,f=c^*_-$; $\frac{1}{R}=m+3+(m-1)(m+2)W,Z=\frac{W+1}{W-1}m$ gives $E_{4,a}:Y_W^2=(W-2)(9W^3-32W^2+21W-2)$ and $g=2$ $\mathcal{C}_{sec2,4}:Y_\kappa^2=8\kappa^6+25\kappa^4-18\kappa^2+1$ under $W=\frac{2\kappa^2}{\kappa^2+1}$, giving at $\kappa=\pm2,r=26/11,\alpha=7/22,\lambda=-3$ $\{8325/1496,136/165,-319/2040,1054/165,600/187,\frac{363}{680}\}$ [4, §7] and $16$ $19/21$ septuples). Also [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 $d_-\in\{\frac{2\lambda}{2r+1},\frac{2}{\lambda(2r-1)}\}$ with $64$-term $P_{rk4}(R,u^2,v^2)=0$ (exp57).
• (c3) Sub-Regular Genera (exp48-54,62): At $\beta=\pm1$, $F_{4,4}^{(1)}$ gives $\mathcal{E}_{\beta=1}:Y_t^2=t^4+8t^3+34t^2-8t+1$ ($F_{6,4,rt}\to C_{ZW}$ étale, $g=5$). With $\mathcal{H}_i:=a_iP^2+b_iPQ+c_iQ^2$, $\text{Res}(\mathcal{H}_1,\mathcal{H}_2)\equiv(r^2-1)N_{core}$, 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 collisions ($\{f=a\}$: $\mathcal{C}_{rat,1},\mathcal{C}_{rat,2}$, $\beta=2\alpharightarrow F_{2,3}^{(f=a)}$ of $g=3$ over $E_{\min}:y^2+xy+y=x^3-3x+2$, $\alpha\beta=2r^2rightarrow F_{4,3}^{(f=a)}\simeq Q_4^{(1)}$ of $g=3$; $\{g=a\}$: $g(C_{4,4}^{(g=a)})=3,g(C_{5,5}^{(g=a)})=4$) and sub-regular curves: (i) $\mathcal{L}_2=0\implies C_{9,11}=0$ ($g\ge2$); (ii) $\mathcal{L}_4=0\implies C_{5,5}^{(g)},C_{6,5}^{(g)}$; (iii) $\mathcal{L}_6=0\implies G_{5,5}^{(a,b)}=0$ ($g=4,5$); (iv) $\mathcal{L}_8=0\implies G_{4,3}^{(8)},K_{3,3}^{(b)}\simeq\mathcal{C}_{hyp}^{(8)}:Y_\lambda^2=4\lambda^6+20\lambda^5-19\lambda^4-26\lambda^3+43\lambda^2-14\lambda+1$ ($g=2$; $\mathcal{C}_{s,\kappa}^{(8)}$ of $g=5$) and $G_{6,3}^{(8)}\simeq Q_{U,\mu}(U,\mu)=0$ ($g=3$); and (v) on $\mathcal{L}_{f,c}^{(u)}=0$ ($cf+1=\square$), $4(2R-1)^2R_4(a,c,d_-,f)-D_{br}^2\mathcal{H}_2\equiv2(2R-1)(\tilde{f}-\tilde{b})\mathcal{M}_{f,c}^{(u)}$ gives $\tilde{f}=\tilde{b}$ ($P_{F_2}=0,g=3$ $Q_t=0$) and $\mathcal{M}_{f,c}^{(u)}=0$ ($g=8$ $F_{6,7}=0$, $F_{12,7}=0$).
• (d) Central-$R_5$ & $E_z=0$ (exp44-46): For $R_5(c,d_-,d_+,f,g)=0$ ($\mathcal{A}_1\tilde{Z}+\mathcal{A}_0=0,\tilde{Z}=C_0/r$), $(R-1)(D_{\tilde{Z}^2}\mathcal{A}_0^2-N_{\tilde{Z}^2}\mathcal{A}_1^2)\equiv-\frac{(w_1^2-1)^{15}(2R-1)^2D_u^3D_v^3}{(u^2-1)^9(v^2-1)^9}P_1P_2P_3P_4$ in $\mathcal{R}_{cent}$ (exp45). 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 $\mathcal{E}_{13,2}:C_0^2=13r^2-1,Y_2^2=r^2+2$ ($rk\ge1,F_1(m)=\frac{(m^2-2)^2C_1(m^2)}{4m^2(m^2-4)(m^2-1)^2},F_2(m)=\frac{C_2(m^2)}{m^2(m^2-4)}$; $m=6$: $\{220/21,161/240,549/560,252/5,475/336\}$). 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

$$ E_z(r,\mu,z)=-K^3L-8H(K^2+4KH-8H^2)A_0B_0+16H^2 (\mu(\mu+4r)A_0^2+4(r^2-1)HB_0^2)=0, $$

with $\Delta_z(E_z)=64r^2K^4H (-KL)$ and $(16(r^2-1)H^2B_0-(K^2+4KH-8H^2)A_0)^2+K^3L(A_0^2-4(r^2-1)H)=4(r^2-1)HE_z$. Setting $z=\frac{t^2-1}{2t-\mu}$ gives $\Phi_{20}(r,\mu,t)=0$ with $disc_t(\Phi_{20})=16(\mu^2-4)^2\Delta_z(E_z)^2\text{Res}_z(E_z,z^2-\mu z+1)$. 2. Cubic Weierstrass model $\Sigma_\mu^+$, $2$-torsion involution $\tau_{T_0}$, and Quadratic-Twist Collapse: Setting $\mu=\frac{\mu_{kiro}-1}{(r-1)\mu_{kiro}-(r+1)}$ ($\mu_{kiro}=\lambda^2$) sends $\mu_{kiro}=\frac{r+1}{r-1}$ (Post 35) to $\mu=\infty$, transforming $\Sigma_\mu$ into

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

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 $\mu(r)\in\mathbb{Q}(r)$, $H,-KL=\square$ collapse via untwisted ($-KL=W_+^2$) or $H$-twisted ($-HKL=\square$, splitting $E_z=0$) sections. 3. Three $1$-parameter families of sextuples and septuples on $\mathcal{S}_{4,2}$ (exp13,16):

• Curve I ($\mathbb{Q}(m)$, $g=0$, $E_2=0\implies S_0=\infty$): $2(\lambda-1)^2r^2-(\lambda^2-1)r+2\lambda=0$, given by $r(m)=\frac{(m-1)(m+3)}{2(m^2+3)},\lambda(m)=-\frac{m(m-1)}{m+3}$ (exp13).
• Curve II ($\mathbb{Q}(k)$, $g=0$, positive $19/21$ septuples): Generated by $\mu_{dual}(r)=\frac{1}{2r}$ and $\mu_{II}(r)=\tau_{T_0}(\frac{1}{2r})=\frac{8r^2+1}{4r(r^2-1)}$ ($-HKL=\square$, $H\sim r^2-1=\square$). 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 $(4,2)$:

$$ a_1(k)=-\frac{2kq_1(k)q_3(k)}{(k^2-1)q_2(k)q_4(k)},b_1(k)=-\frac{2kq_2(k)q_4(k)}{(k^2-1)q_1(k)q_3(k)},d_-(k)=-\frac{(k-1)^3(3k^2+2k+3)q_2(k)q_4(k)}{2k (k+1)^3q_1(k)q_3(k)},d_+(k)=-\frac{(k+1)^3(3k^2-2k+3)q_1(k)q_3(k)}{2k (k-1)^3q_2(k)q_4(k)}, $$

colliding at $k\in\{\pm3,\pm1,\pm1/3,0\}$. On $(1,k_*)$ ($k_*\approx2.026846$, $3k^4-4k^3-2k^2-4k-1=0$, Sturm-certified in $(2,203/100)$), all $7$ entries are $>0$ and distinct ($\mathcal{S}_{7,pos}^{(19/21)}(2)$ at $k=2$).

• Curve III ($\mathcal{E}_{III,\pm}$, $g=1$): 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 $\mu_{III,+}(r)=\frac{8r^2-8r+3}{2r(2r^2-1)}$, where $-KL=(\frac{2(2r-1)^2(2r^3-2r+3)}{r(2r^2-1)})^2$ and $H=\square$ collapses to $\mathcal{E}_{III,+}:Y^2=(r-3)(4r^3-3r+3),\lambda=\pm\frac{Y}{(r-3)(2r-1)}$ ($G_1=(-3,24),G_2=\infty_+$; $P_0-P_1$ gives $\mathcal{E}_{III,-}$; $(101/28,4097/392)$ in verify_paper.py). Proof. Verified in verify_paper.py, exp13-16,42-62. $\square$

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

Paper 65 [16] studied four regular quintuples in $(u,y,w)$ with $p=u^2-1,q=y^2-1,J=pq-1,L=J+2u^2,F=4u^2y^2-J^2,U=uy(u-y),V=u^2+y^2,D=uy-u+y,Z=(J-2uy)D^2$, 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 $\deg4$, and $\mathcal{B}_+$ collapses $\alpha^2,\beta^2=\square$ to $\mathcal{E}_B$.

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

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

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

2. The four quadratics and their universal syzygies: With $C=b+c$, $A=\frac{J^2C+2TL}{4u^2y^2}$, and $N_ac=\frac{T+Lc}{2uy}$, $\alpha^2=2+cA+2N_ac$ and $\beta^2=2+cA-2N_ac$ ($r_{ac}=\frac{\alpha+\beta}{2},r_{ec}=\frac{\alpha-\beta}{2}$) factor in $\mathbb{Q}(u,y)$ as

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

where $J(u,y)=(u^2-1)y^2-u^2$, $Q_0(u,y)=(u^2-1)(y+u)^2-u^4$, and $Q_\mp(u,y)=((u\mp1)y+u^2)^2-u^2(u^2-1)$ obey $Q_0-J=2uy(u^2-1),Q_+-Q_-=4uy(u+y),Q_++Q_--2Q_0=4(u^2+uy+y^2)$, 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 & moving-$u$ family $\mathcal{E}_B$: Along $\mathcal{B}_+:(u+1)y^2+2u^2y+u^2=0$, $u(k)=\frac{k^2+1}{2k-1},y(k)=-\frac{k^2+1}{k+2}$ give $Q_+(u,y)=-u^3$ 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 $\alpha^2,\beta^2=\square$ collapse to the rank-$1$ curve:

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

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 $a,e=\frac{((\alpha\pm\beta)/2)^2-1}{c}$ gives in $\mathbb{Q}(k)$:

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

Pairs in $\{a,e,d\},\{b,c,d\}$ are squares in $\mathbb{Q}(k)$, while $\{a,e\}\times\{b,c\}$ equal $\square\cdot Y_B^2$:

$$ ab+1=\frac{Y_B^2}{(k-3)^2(k-2)^2},ac+1=\frac{(k-1)^2Y_B^2}{(k-2)^2(k^2+1)^2},eb+1=\frac{Y_B^2}{(k-2)^2(2k-1)^2},ec+1=\frac{(k^2-k-1)^2Y_B^2}{(k-2)^2(2k-1)^2(k^2+1)^2}, $$

and $R_4(a(k),e(k),b(k),c(k))\ne0$ factors in $\mathbb{Q}[k]$. Outside $k\in\{-2,-1,-1/3,0,1/2,1,4/3,2,3\}$, $\pm Y_B\in\mathcal{E}_B(\mathbb{Q})$ gives $\{a(k),e(k),b(k),c(k),d(k)\}$ ($d>0$ for $k>3$). On $V^2=U^3+41U^2+704U+5120$ ($\#\mathcal{E}_B(\mathbb{F}_{3,11})=5,16\implies\mathcal{E}_B(\mathbb{Q})_{tors}=\{\mathcal{O}\}$), over $K_3=\mathbb{Q}(\alpha)$ ($\alpha^3-\alpha^2+4\alpha-2=0,h(K_3)=1,\varepsilon=2\alpha-1$), $f_U'(U_0)=\alpha^{14}\pi_{5,2}\pi_{53,1}\varepsilon^{-5}$ ($\pi_{5,2}=\alpha^2+\alpha+1$) and mod-$5$ filtering give $Sel^{(2)}(\mathcal{E}_B/\mathbb{Q})=\langle\pi_{5,2}\rangle\implies rk\mathcal{E}_B(\mathbb{Q})=1$. Since $\delta(P_0)\equiv\pi_{5,2}$ and $ord_{\mathbb{F}_{19}}(P_0)=21$ at $P_0=(-8,-40)\mapsto k=-2$ ($4P_0\mapsto4,6P_0\mapsto-22/5$), $\mathcal{E}_B(\mathbb{Q})\subset\mathbb{Z}_{(6)}P_0$, giving infinitely many moving-$u$ quintuples (exp15; $k=4$ in verify_paper.py). For $G(k)=(k-2)(k^5-4k^4+17k^3-46k^2+64k-56)$ and $H(k)=(k^2-4)Q_4(k)$ ($Q_4=k^4-2k^3+k^2-4k-4$), $Q_4(k)\equiv-2\pmod{\mathbb{Q}_2^{\times2}}$ requires $v_2(U)=0,U\equiv7\pmod8$; on $\mathbb{Z}_{(6)}P_0$, $v_2(U(nP_0))=0\iff n\equiv6\pmod{12}$, where $U(6P_0+12mP_0)\equiv3\pmod8$, proving $G(k(P))\in\mathbb{Q}_2^{\times2}\iff P\in(3+6\mathbb{Z}_{(6)})P_0$ ($k\equiv3\pmod8,Q_4(k)\equiv+5\pmod{\mathbb{Q}_2^{\times2}}$) and eliminating $Q_4(k)\equiv-2\pmod{\mathbb{Q}^{\times2}}$ (exp38; [16, §8]). Proof. Verified in verify_paper.py,exp15,38. $\square$

8. Factorization of $R_5$, section $L_1=0$, $15+6$ collapse, and $v=-1/5$ families

Over $\mathbb{Q}$ in $(r,\mu,z)$ (Post 35), $L_1(r,\mu,z)=0$ forces $cc^*_-+1,cc^*_++1,c^*_-c^*_++1=\square$ in $\mathbb{Q}(r,\mu)$ without $E_z=0$.

Theorem 6 (Section $L_1=0$, $15+6$ Collapse, $(a,b)$-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 $(r,\mu,z)$, $P_4-1=-\frac{K}{H}$, $P_4-9=\frac{L}{H}$, $B_0^2-\mu(\mu+4r)=4r^2((r^2-1)z^2+A_0z+H)$, $w^2(d_-d_++1)=(r\mu+1)^2$, $(cd_-+1)(cd_++1)=(\frac{Tz-H}{H})^2$, $\frac{c^*_-}{w}=z-\mu$, and $\frac{c^*_+}{w}=\frac{(4r^2-1)((4r^2-1)z+\mu+4r)}{K^2}$. In $\mathbb{Q}(r,\mu,z)$, $R_5(d_-,d_+,c,c^*_-,c^*_+)$ factors 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 $\tau_{T_0}(\mu)=-\frac{\mu+4r}{K}$:

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

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 $E_z=0$:

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

Eliminating $z$ from $E_z=R_5=0$ 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 $(R,v,u):=(r^2,r\mu,\frac{t-\mu}{r})$, $(2t-\mu)^4Q_4=F_aF_b$ where $F_a=0$ is $Q_{4,+}(R,v,u)=0$ with $disc_v(Q_{4,+})=4(2Ru-1)^2D_4(R,u)$, $D_4=((R-1)u^2+2u-1)(R(4R^2-8R+1)u^2+8R^2u-(2R-1)^2)$. On $Q_{4,+}=0$, all $3$ residual gates and $\Delta_d^2$ are squares, giving interior $10/10$ quintuples at $(r,u,v)=(3/4,-4,-9/28)$, $(7/4,44/67,2009/3886)$, $(19/10,20/57,19/45)$ (exp24). Furthermore (exp23-27):

• Involution $\iota_K$, $\mathcal{E}'_{Q_4}(R)$ & $3P_0$ Law: $\iota_K:(r,\mu,z,w)\mapsto(r,-\frac{\mu+4r}{K},\frac{4r^2-1}{K}z,\frac{4r^2-1}{K}w)$ fixes $(a,b,c,d_-,d_+)$, swaps $c^*_-rightarrow c^*_+$ and $L_1rightarrow L_2$, 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 $d_\pm=d_{3P_0}(d_\mp,c,c^*_-):=\frac{8(s_1^3-4s_1s_2+8s_3)}{(s_1^2-4s_2)^2}$, so at $(7/4,-4/9,98/33)$, $Q_{d_\pm}+S_T=\pm3P_0$.
• Subgroup $\Lambda(r)$, $Z_-(r)$ & $2$-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 $2W_1(r)=P_{even}(r^2)+T_0=O_-$ ($X(2W_1)=\frac{(4r^2-1)^2}{(2r^2-1)^2}$), while $\hat{E}(r^2)$ gives $P_{odd}(r)=((4r^2-1)^2,4r^3(4r^2-1)^2)\in E(r^2)(\mathbb{Q}(r))$ with $Z_-(r):=(1/r,2(2r-1))=W_1(r)+P_{odd}(r)$. Under $r\mapsto-r$, $\Lambda(r):=\mathbb{Z}W_1(r)\oplus\mathbb{Z}P_{odd}(r)\oplus(\mathbb{Z}/2\mathbb{Z})T_0\cong\mathbb{Z}^2\oplus\mathbb{Z}/2\mathbb{Z}$. 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 $0$ ($r\in\{0,\pm1/2,\pm1\}$); 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 $0$ ($r=\pm1/2$).
• 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 $G_u(R,d_0):=R^2d_0^4+(16-6d_0^2)R+(d_0^2-3)=n^2$ (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 & $(a,b)$-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 $\Delta_{ab}^2,\Delta_d^2$:

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

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 $rv(v^2-1)(4R-1)KH\Delta_d\ne0$ with distinct nonzero entries:

• For $\mathcal{G}_d(R,v)=\square,H=w^2$: $\{c,d_-,d_+,c^*_-,c^*_+\}$ is a $10/10$ quintuple ($R_5=0$), giving at $(r,\lambda,v)=(7/4,3/11,98/163),(5/3,13/11,-8/21),(53/30,2/7,3/5)$ 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 $Q_{e_2}=P_0+2Q_c+S_T$ ($e_2=-4580748/6812155$) forms the $3$-fifth $18/21$ septuple.
• $(r,M)$ completions, $(v,t)$ blow-up & $(a,b)$-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:

$$ \begin{aligned} (r-M)^4\mathcal{F}_{ab}(r^2,\frac{r}{r-M})=(-M^2+8r^3M+4r^4-9r^2+2)^2-4(r^2-1)(4r^2-1)^2(M^2-1),\\ (r-M)^4\mathcal{G}_d(r^2,\frac{r}{r-M})=(rM^2-(6r^2-1)M-\frac{8r^6-10r^4-2r^2+1}{2r})^2+(\frac{4r^2-1}{2r})^2(4rM-(4r^2+1)), \end{aligned} $$

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 $\mathcal{E}_{tan}:Y_t^2=(t^2+30t+253)(9t^2+14t-27)$ (exp20).

• For $\mathcal{F}_{ab}(R,v)=\square,H=w^2$: $\{a,b,c,c^*_-,c^*_+\}\subset\mathbb{Q}$ 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).
• For $\mathcal{F}_{ab},\mathcal{G}_d=\square$: $15/21$ pairs of $\{a,b,c,d_-,d_+,c^*_-,c^*_+\}$ are squares, and $ac+1,bc+1,ad_\mp+1,bd_\mp+1$ share one square class in $\mathbb{Q}^\times/\mathbb{Q}^{\times2}$.

3. Closed-form $\mathbb{Q}(\lambda)$ & $\mathbb{Q}(u)$ families at $v=-1/5$, and rank-$0$ / torsion proof on $\mathcal{E}_{16}$: Since $disc_R(\mathcal{F}_{ab})=256v^2(v-1)^2(v+1)^3(5v+1)$, 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:

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

Here $e_{ab,\mp}(\lambda)\in\mathbb{Q}(\lambda)$:

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

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 $\lambda\notin\{-1,0,1\}$, $\{a,b,c^*_-,c^*_+,e_{ab,-},e_{ab,+}\}$ is a $14/15$ sextuple, positive on $(\sqrt{7/8},1)$ ($\lambda=15/16$ 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 $rkE_{16}(\mathbb{Q})=0$ 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 ($\{(0,\pm4),(\pm1,\pm1),\infty_\pm\}$). Proof. Verified in verify_paper.py, exp17-27. $\square$

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

We study $4$-term chains $x(P_0+nS)$ on $E_T:Y^2=(X+r^2-1)(X+s^2-1)(X+t^2-1)$ ($X=\Pi x,\Pi=abc,\Omega=rst$) via $(V,K,d)$ on $\mathcal{S}_{3-tors}$ ($3S_T=\mathcal{O}$) [7, 10, 17].

Theorem 7 ($(V,K,d)$ Chains, $\mathcal{S}_{3-tors}$, $\mathcal{C}_{triad}$, Mixed-$R_4$ Syzygies, and $\mathcal{E}_W(R^2,Z)$ Towers)

For any triple $T=\{a,b,c\}$, let $d_\mp=x(P_0\pm S)=a+b+c+2\Pi\mp2\Omega$ and $W_\mp:=\Pi d_\mp-1$.

1. Shift product & $F_2(z)$ completion: For $Q\in E_T$ ($z=x(Q)$), $x(Q-S)x(Q+S)=\frac{(z-d_+)(z-d_-)}{(\Pi z-1)^2}$, so $x(Q-S)x(Q+S)+1=\frac{F_2(z)}{(\Pi z-1)^2}$ where $F_2(z):=(\Pi z-1)^2+(z-d_+)(z-d_-)$ obeys $W_-^2F_2(z)=(W_-^2+(\Pi W_--2\Omega)(z-d_-))^2+(z-d_-)^2D_4$ with $D_4=4\Omega^2d$ 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 & $4$-term chains: In $u_0:=\frac{W_-}{2\Omega}$ ($V:=u_0^2$), $K:=W_--d=\Pi d_--X(2S)$, $d=X(2S)-1$, $\Pi=\frac{d+1-u_0^2}{2u_0}$, $\Omega=\frac{K+d}{2u_0}$, and for all $m,n\in\mathbb{Z}\setminus\{0\}$, $x_mx_n+1=\frac{(d+1-V)^2+4VX_mX_n}{(d+1-V)^2}$, where $X_n=\Pi x(P_0+nS)$ are $X_1=K+d+1$, $X_2=\frac{K+V}{V}$, $X_3=\frac{d(K(V-1)+Vd)}{K^2}$, $X_4=\frac{(K^2+K-Vd)(K(d+1)-Vd)}{(K-Vd)^2}$. On $\mathcal{S}_7^{(1,2,3,4)}=\{a,b,c,d_-,x_2,x_3,x_4\}$ ($18/21$), $d_-x_3+1,x_2x_4+1=\square\iff num_{19},num_{20}=\square$.
3. Recalled Halving $2H=S_T$ [7, §2; 17, §3], $2P_0$ Classification, Antipodal $R_5$ & $11$ $S_5$ Laws, $3S_T=\mathcal{O}$, Tate Form & Orbit Cubic: By [7, 10, 17], for $Q=(\Pi x_4,Y_4)\in E_T$ ($p_4=\Pi x_4\ne1$), $(p_4-1)^2(X-x(Q-S_T))(X-x(Q+S_T))\equiv R_5(a,b,c,x_4,X)$ on $E_T$, where $H=(1+rs+rt+st,(r+s)(r+t)(s+t))\in E_T(\mathbb{Q})$ [7, §2; 17, §3] has $\lambda_H=r+s+t$ and $2H=(1,\Omega)=S_T$ ($t=0\implies H\in E_T[4](\mathbb{Q})$), while $2P_0=(\frac{(c-a-b)^2-4ab}{4},-\frac{(a-b-c)(a-b+c)(a+b-c)}{8})$ obeys $2P_0\in\{\pm S_T\}\iff R_3(a,b,c)=0\iff c=a+b\pm2r\iff b=a+c\pm2s\iff a=b+c\pm2t$ ($2P_0=-S_T$ at $(1,3,8)$) and $2P_0\in E_T[2]\setminus\{\mathcal{O}\}\iff(a+b-c)(a-b+c)(-a+b+c)=0$ (e.g. $(1,0)$ at $(3/2,2,-1/2)$), and on regular $10/10$ $Q_5=(x_1..x_5)$ ($T_{ij}=Q_5\setminus\{x_i,x_j\},Q_j=Q_i+S_{T_{ij}}$), $\tau_j(x_j)=x(Q_i-S_{T_{ij}})=0\iff x_i\in\{d_\mp(T_{ij})\}$. For any $(x_1..x_4,y)$ ($P_4=\prod x_i,\sigma_1=\sum x_i$) and roots $w_\pm$ of $R_5(x_1..x_4,w)=0$, $y^2R_5(x_1..x_4,y)\equiv((P_4+1)y^2-\sigma_1y-2)^2-4\prod(1+x_iy)$, and under $y\mapsto-1/y$, $(P_4-1)^2(1+yw_+)(1+yw_-)\equiv y^2R_5(x_1..x_4,-1/y)$ reorganizes [4, 5, 17] into $1+6+4=11$ $S_5$ completions: $(2y^2-\sigma_1y-(P_4+1))^2-4\prod(y-x_i)$, six $2+2$ laws $A_{ij;kl}(y)^2-4(1+x_ix_j)(1+x_kx_l)(y-x_i)(y-x_j)$ ($A_{ij;kl}=(P_4+1)+2x_ix_j+(x_k+x_l-x_i-x_j)y$), and four $3+1$ laws $((P_4-1)-(2x_ix_jx_k+x_i+x_j+x_k-x_m)y)^2-4(1+x_ix_j)(1+x_ix_k)(1+x_jx_k)y(y-x_m)$. In $U:=X-1=\Pi x-1$ ($\Sigma_2:=r^2s^2+r^2t^2+s^2t^2$), $E_T$ is $Y^2=(U+r^2)(U+s^2)(U+t^2)$ with $S_T=(0,\Omega)$, $\lambda_{0,gen}:=\frac{\Sigma_2}{2\Omega}$ ($\Omega\ne0$), $W_\mp=U(d_\mp)=2\Omega(\Omega\mp\Pi)-\Sigma_2$, and $U(Q\pm S_T)=\frac{2\Omega(\lambda_{0,gen}U+\Omega\mp Y)}{U^2}$ ($U\Omega\ne0$), so $(\Pi(\Pi+y)U)^2(1+yw_+)(1+yw_-)\equiv L_U^2-4\Pi^2\Omega^2y(y-a)(y-b)(y-c)$ ($L_U=(\Pi+y)^2U+\Sigma_2y(\Pi+y)-2y^2\Omega^2$, genus-$0$ conic in $x$ for $\Pi(\Pi+y)\Omega\tau\ne0$ and $y(y-a)(y-b)(y-c)\ne0$; exp89), while $3S_T=\mathcal{O}\iff2S_T=-S_T\iff d=0$ iff $\lambda_{0,gen}^2-(r^2+s^2+t^2)=0\iff\pm\frac{1}{r}\pm\frac{1}{s}\pm\frac{1}{t}=0$. For $\frac{1}{r}+\frac{1}{s}+\frac{1}{t}=0$ ($\lambda_0:=r+s+t=-\lambda_{0,gen},\lambda_0^2=r^2+s^2+t^2$; exp28-32):

• Tate Form & Torsion $\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$: On $3S_T=\mathcal{O}$ [7], $E_T$ has Tate form $Y^2=U^3+(\lambda_0U-\Omega)^2$, $\Pi^2+1=(\Omega+\lambda_0)^2$, $R_0=(0,-\Omega)=-S_T$, $P_0+\langle S_T\rangle=\{P_0,P_0-S_T,P_0+S_T\}$ has nonzero abscissas $\{d_-,d_+\}$ obeying $U(d_\mp)=2\Omega(\Omega+\lambda_0\mp\Pi)=-2\Omega\kappa^{\mp1}$ ($\kappa:=\Pi-\Omega-\lambda_0$, $U(d_-)U(d_+)=4\Omega^2$) giving $10/10$ $T\cup\{d_-,d_+\}$ with $\tau_4(d_-)=\tau_5(d_+)=0$, and $E_T(\mathbb{Q})_{tors}\supseteq\langle R_0+W_r\rangle\times\langle W_s-W_r\rangle\cong\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ ($W_r=(-r^2,0),W_s=(-s^2,0)$).
• Orbit Cubic & Sub-Quadruple Identity: For $Q_4=(U_4,Y_4)\in E_T(\mathbb{Q})$ ($U_4(U_4+1)\ne0$), $\{Q_4,Q_4+S_T,Q_4-S_T\}$ has $U$-coordinates $\{U_4,U_5,U_6\}$ obeying $U_{5,6}=\frac{-2\Omega(\lambda_0U_4-\Omega\mp Y_4)}{U_4^2}$, $U_4U_5U_6=-4\Omega^2$, and $P_{coset}(U'):=U'^3-\Sigma_UU'^2-4\Omega\lambda_0U'+4\Omega^2=0$, where $\Sigma_U:=U_4+U_5+U_6=\frac{U_4^3-4\Omega\lambda_0U_4+4\Omega^2}{U_4^2}$. For $Q_4\in P_0+2E_T(\mathbb{Q})$ ($U_4\notin\{0,-1,U(d_-),U(d_+)\}$), $S_6(Q_4):=T\cup\{x_4,x_5,x_6\}$ is a $15/15$ sextuple [7, 10, 17]. For $(u,r_u)\in\{(a,t),(b,s),(c,r)\}$ ($ux_i+1=\frac{U_i+r_u^2}{r_u^2-1}$), $(ux_4+1)(ux_5+1)(ux_6+1)=\frac{-P_{coset}(-r_u^2)}{(r_u^2-1)^3}=(\frac{ur_u(r_uU_4-2\Omega)}{\Pi U_4})^2(ux_4+1)$ (exp31), making every $2+2$ $(T\setminus\{w\})\cup\{x_i,x_j\}$ and $1+3$ $\{u,x_4,x_5,x_6\}$ a square-product quadruple and giving $9+6=15$ $\ge19/21$ type-$(0,4)$ septuples/coset (exp28), while $\prod_{i=4}^6(U_i+1)=-P_{coset}(-1)=\frac{(U_4+1)(U_4-U(d_-))(U_4-U(d_+))}{U_4^2}$ gives $\Pi x_4,\Pi x_5,\Pi x_6>0\iff(U_4+1)(U_4-U(d_-))(U_4-U(d_+))>0$, and $1+abcx_4x_5x_6=\frac{(\Omega+\lambda_0)(\lambda_0-3\Omega)+\Sigma_U}{\Pi^2}$ ($1+\sigma_6=\frac{(s_4-1)^2+s_4R_4(Q_4)}{(s_4-1)^2}$; $m=-5$ has $1+\sigma_6=\square,P_6\ne0$, exp52).

4. Geometry of $\mathcal{S}_{3-tors}$, $C_r(\mathbb{Q}(r))$ & Family IV ($N=3\beta+\gamma$): Over $\mathbb{Q}(r)$, $t=-\frac{rs}{r+s},Y_r=(r+s)\Pi$ gives $C_r:Y_r^2=(r^2-1)(s^2-1)(rs-r-s)(rs+r+s)$ (exp28,30), with $C_r(\mathbb{Q}(r))[2]=\{W_2=(-1,0),T_1=(1,0),T_2=(-\frac{r}{r+1},0),T_3=(\frac{r}{r-1},0)\}\cong(\mathbb{Z}/2\mathbb{Z})^2$, and $P_\infty$ ($s=\infty$) generating $\mathbb{Z}P_\infty\oplus(\mathbb{Z}/2\mathbb{Z})^2\subset C_r(\mathbb{Q}(r))$:

• Level $1$ ($P_\infty+T_1$): $(s(r),t(r))=(-\frac{1}{2r},\frac{r}{2r^2-1}),\Pi(r)=\frac{(r^2-1)(4r^2-1)}{2r(2r^2-1)}$ ($3P_0$ is $6$-torsion, $x(3P_0)=0$).
• Level $2$ ($2P_\infty$ & $2P_\infty+T_1$): $s_{2P_\infty}(r)=\frac{2r^3-2r+1}{2r^3-2r^2-2r+1}$, $t_{2P_\infty}(r)=-\frac{r(2r^3-2r+1)}{(r-1)(2r^3+2r^2-1)}$ ($s_{2P_\infty+T_1}(r)=-s_{2P_\infty}(-r)$). By Sturm (exp28), on $2P_\infty(r)$ for $r>r_{**}\approx2.010315$ ($4r^6-8r^5-4r^4+8r^3-1=0$), all $8$ of $T(r)\cup\{d_\mp(r)\}\cup(3P_0(r)+\langle S_T\rangle)$ are positive, giving $15$ $19/21$ $\mathbb{Q}(r)$-septuple sections ($15$ type-$(0,4)$ $19/21$ septuples at $r=3$ in verify_paper.py).
• Family IV ($N=3\beta+\gamma$): In $(\alpha:\beta:\gamma:N)=(1/t:1/r:1/s:1)$ ($\alpha+\beta+\gamma=0$), $\mathcal{S}_{3-tors}$ 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 $r(u)=\frac{9u^2+1}{3(1-u^2)}$, $s(u)=-\frac{9u^2+1}{6u^2-1}$, $t(u)=-\frac{9u^2+1}{4-9u^2}$, $\Pi(u)=\frac{10u(3u^2+2)}{(1-u^2)(4-9u^2)}$ (exp30), with $a,b,c,d_-,d_+>0$ for $u\in(\frac{1}{\sqrt{6}},2/3)$, and positive $15/15$ sextuple $S_6(u)=T(u)\cup(3P_0(u)+\langle S_T\rangle)$ on $(1/\sqrt{6},u_{**})$ ($u_{**}\approx0.440894$; $u=5/12$; Family III, $N=\beta-\gamma$: $(r,s,t)=\frac{u^2+1}{(3(u^2-1),1-2u^2,2-u^2)}$).

5. Type-$(2,3)$ Mechanism, $(s,\kappa)$ Master Syzygies, DKMS [7, §3], $\mathcal{C}_{triad}$ & $E_{15}$ 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 $d_\mp=a+b+c+2\Pi\mp2\Omega$, $b'_{ac}(d_\mp):=2(a+c+d_\mp+2acd_\mp)-b$ obeys $bb'_{ac}(d_\mp)+1=(a+c-d_\mp)^2-(4s^2-1)=(b+2(\Pi\mp\Omega))^2-(4s^2-1)=\frac{((2s^2-1)\Pi\mp2(s^2-1)\Omega)^2-(s^2-1)^2(4s^2-1)}{(s^2-1)^2}$. On $\mathcal{S}_{3-tors}$, $\kappa:=\Omega+\lambda_0-\Pi\in\mathbb{Q}^\times$ ($\Pi=\frac{1-\kappa^2}{2\kappa},\Omega=\frac{s^2(1+\kappa^2)-2s^3\kappa}{2\kappa(s^2-1)},\lambda_0=s-\frac{\Omega}{s^2}$) 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 $bb'_{ac}(d_-)+1=\frac{C_B(x,\kappa^2)}{(2(x^2-\kappa^2))^2}=\square$ to monic cubics $C_A(x,y):=(y-2x+1)(y^2+(6x+1)y-8x^3)$ and $C_B(x,y):=y^3+2(1-4x^2)y^2+(16x^4+8x^3-16x^2+1)y-4x^3(8x^2-9x+2)$ in $y=\kappa^2$ (exp34), obeying in $\mathbb{Q}[x,y]$:

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

where $R_{triad}(x,y):=9y^2+(8+2x-28x^2)y+(16x^4+8x^3-20x^2+6x-1)=0$ projects $\mathcal{C}_{triad}$, while $Q_{A,0}=(4x^2-2x-1,4x(x-1)(2x+1))$ on $\mathcal{E}_A(x)$ ($2Q_{A,0}=(x^2,x(x-1)^2),T_A=(2x-1,0)$) is Family IV ($P_2+P_3$ on $\mathcal{E}_B(x)$, $2P_1+3P_2+4P_3=\mathcal{O}$). Whenever $bb'_{ac}(d_\mp)+1=\square$, $\mathcal{C}_b:=\{b'_{ac},x_5,x_6\}=Q_{b'_{ac}}+\langle S_T\rangle$ 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 $\ge23/28$ octuple:

• DKMS [7, §3] at $2Q_{A,0}+T_A$ ($y=-6x-1$), $\mathcal{U}_8^{DKMS}(\kappa)$, $E_{Cheb}(s)$ & $\kappa_n(s)$: Over $\mathbb{Q}(\kappa)$, $2Q_{A,0}+T_A=(-6x-1,8x^2)$ 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 $\kappa=1/t$; 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 $d_+(\kappa)=-\frac{2N_+}{3\kappa(\kappa^2-1)D_0}$ form a $22/28$ octuple $\mathcal{U}_8^{DKMS}(\kappa)=\{a,b,c,d_-,d_+,x_4,x_5,x_6\}$, all $8$ $>0$ on $(\kappa_{*,DKMS},3-2\sqrt{2})\approx(0.1456,0.1716)$ ($\kappa\ne1/3$). At $\kappa=1/6$ ($t=6$ [7]), $\mathcal{U}_8^{DKMS}(1/6)$ gives $15$ positive $19/21$ septuples, while $U_Q(\kappa)=-\frac{1}{2}r(\kappa)t(\kappa)$ defines $Q_{ex}$ 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 $E_{Cheb}(s):Y_{Cheb}^2=Q_{Cheb}(s,p):=((s+1)p^2-(s-1))((s+1)(2s-1)^2p^2-(s-1)(2s+1)^2)\equiv((s+1)(2s-1)p^2\mp(s-1)(2s+1))^2-\{4,16s^2\}(s^2-1)p^2$ ($P_{triv}=(1,2)$). 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 $C_r$ ($\mathcal{U}_8^{(2)}(s)$, $>0$ at $s=10/9$), and $\kappa_3(s)=-\frac{2s(2s^2-1)}{(2s^3-2s-1)(2s^3-2s+1)}$ gives $\mathcal{U}_8^{(3)}(s)$ ($>0$ at $s=4/3$, exp44).
• Triad Curve $\mathcal{C}_{triad}$ & Positive Septuples: In $(\alpha:\beta:\gamma:N)$, $b'_{ac}(d_-)=b\iff2N_1^2-N_1(\beta-\alpha)-(\beta^2+\alpha^2)=0$, given 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 $N(t,W):=\frac{t^2+1+W}{2}$ gives $\mathcal{C}_{triad}:W^2=t^4-20t^3+34t^2+12t+9\cong E_{triad}:Y^2=X^3+34X^2-276X+2520$ (exp30) via $X=\frac{6(3+2t-W)}{t^2},Y=\frac{6(X+30)}{t}+2X-60$. On $E_{triad}$, $P_1=(66,648)$ ($(1,-6)$) has $2P_1=(106/9,2024/27)$ ($(9/4,-39/16)$), so $rk\mathcal{C}_{triad}(\mathbb{Q})\ge1$. With $M_{can}(t):=-2t(t-2)(t-1)(t+3)(2t+1)$ and $\Pi_{can}(t):=\frac{M_{can}(t)}{\beta(t)\gamma(t)\alpha(t)}$, $bb'_{ac}(d_-)+1=(\frac{(t^2+1+W)((t^2+1)(t^2+14t-27)-(t^2-2t+5)W)}{4(t-2)(t+1)(1-3t)(t-1)(t+3)})^2$. Negating entries, Sturm proves on $\mathcal{C}_{triad}(\mathbb{Q})$ ($t<2,W>0$) all $8$ of $\mathcal{U}_8$ are $>0$ (from $(-5/3,122/9)$).
• $\mathbb{Q}(r_q)$-Quotient, Spoke Syzygies & $E_{15}$ Closure (exp33-35): Under $\iota_{triad}:(t,W)\mapsto(\frac{2t+1}{t-2},\frac{5W}{(t-2)^2})$ ($z=t-2\mapsto5/z$), $\mathcal{C}_{triad}/\langle\iota_{triad}\rangle$ is $w_q^2=u_q^2-12u_q-72$ (Post 35; $u_q=6+3r_q+\frac{9}{r_q}=z+\frac{5}{z},w_q=3r_q-\frac{9}{r_q}=\frac{W}{z},\chi^2=r_q^2(u_q^2-20)$). With $A_3=3r_q^3+47r_q^2+117r_q+81,B_3=27r_q^3+107r_q^2+153r_q+81$, $U_4=-\frac{(3r_q+5)^2A_3B_3}{(9r_q^2+32r_q+27)^2\chi^2},U_{5,6}=-\frac{2r_q^2(3r_q+5)^2\{B_3,A_3\}}{\{A_3,B_3\}^2}$, and $1+d_\mp x_{5,6}$ descend to $F_8,F_{10},G_8,G_{10}\in\mathbb{Q}[r_q]$ with $(F_8-P_{F_8}^2,G_8-P_{G_8}^2,G_{10}-P_{G_{10}}^2)=(36(r_q+3),144(r_q+1),-16r_q(7r_q+9))R_{spoke,8}$ ($R_{spoke,8}=-r_q(r_q+1)(r_q+2)(r_q+3)(2r_q+3)(5r_q+9)$), while $1+d_+b'_{ac}(d_-)=(\frac{r_q(3r_q+5)}{(r_q+2)(r_q+3)\chi})^2Q_{cross}(r_q)$ ($Q_{cross}=\frac{r_q^2(41u_q^2+336u_q+684)}{9}$). Setting $u_q(m)=\frac{2(3m^2-4m+3)}{m^2-1}$ maps $41u_q^2+336u_q+684=(\frac{4Y_m}{m^2-1})^2$ to $\mathcal{C}_{cross}:Y_m^2=261m^4-414m^3+263m^2-78m+9\cong E_{15}:Y^2=X(X+9)(X-16)$. $2$-isogeny descent with $\hat{E}_{15}:\hat{Y}^2=\hat{X}(\hat{X}^2+14\hat{X}+625)$ mod $9,16,25$ gives $im(\alpha_{E_{15}})=\{1,-1\},im(\alpha_{\hat{E}_{15}})=\{1,5\}\implies rkE_{15}(\mathbb{Q})=0$. 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 $r_q\in\{-3,-2,-3/2,-1\}$ ($\Pi_{can}=0$), closing $1+d_+b'_{ac}(d_-)=\square$ on $\mathcal{C}_{triad}(\mathbb{Q})$.
• Cross-Coset Bridge ($20$th Pair): For any $\{x_4,x_5,x_6\}=Q_4+\langle S_T\rangle$ and $d\in\mathbb{Q}$, $M_4:=dx_4+1=\frac{dU_4+\Pi+d}{\Pi}$ and $M_{5,6}:=dx_{5,6}+1=\frac{(\Pi+d)U_4^2-2\Omega d(\lambda_0U_4-\Omega\mp Y_4)}{\Pi U_4^2}$ obey $M_5M_6=\frac{(\Pi+d)^2U_4^2-4\Omega d(\Omega d+\lambda_0(\Pi+d))U_4+4\Omega^2d(\Pi+d)}{(\Pi U_4)^2}$ and $M_4M_5M_6=\frac{(\Pi+d)^3+\Sigma_Ud(\Pi+d)^2-4\Omega\lambda_0d^2(\Pi+d)-4\Omega^2d^3}{\Pi^3}$ (exp31).

6. Mixed-$R_4$ Syzygies, $2$- vs. $4$-Condition Criterion, $H=1/3$ Descent & Branch B (exp35-38): For $\{x_4,x_5,x_6\}=Q_4+\langle S_T\rangle$ on $E_{can}$ with $t:=\frac{2\Omega}{U_4}$ ($U_5+U_6=t^2-2\lambda_0t,U_5U_6=-2\Omega t$), $(1+bx_5)(1+bx_6)=(\frac{s(s-t)}{s^2-1})^2$, $1+x_5x_6=(\frac{h-t}{\Pi})^2$, and $(\frac{t^2}{2\Omega})^2Y_4^2=(1+\frac{1}{2}t(t-2\lambda_0))^2-(t-\kappa)(t-\frac{1}{\kappa})$ ($\Pi^2x_5x_6=(t-\kappa)(t-1/\kappa)$). Hence $R_4(b,d_-,x_5,x_6)=0$ (Post 35, [19]) splits over $\mathbb{Q}(\kappa,s)$ into four roots ($\delta\in\{\pm1\}$, swapped by $(\kappa,s)\mapsto(-\kappa,-s)$, $h:=\Omega+\lambda_0=\frac{\kappa^2+1}{2\kappa}$): $t_{I,\delta}(\kappa,s)=\frac{\kappa s+\delta h}{s+\delta}$ and $t_{II,\delta}(\kappa,s)=\frac{2s^2-\kappa s+\delta h}{s+\delta}$. By AHS on $\{b,d_-,x_5,x_6\}$, $Q_{C2}:=(s+1)DN=\square$ ($1+bx_4=\frac{s^2N}{(s-1)^2 (s+1)D}$) and $Q_{C3}:=(s+1)F=\square$ ($Y_4\in\mathbb{Q}$) force $1+d_-x_{5,6}=\square$, giving a $9/10$ quintuple $\{b,d_-,x_4,x_5,x_6\}$ (missing $1+d_-x_4=\square$). Adjoining $\{a,c\}$ ($Q_4\in P_0+2E_T(\mathbb{Q})$) for $20/21$ $\{a,b,c,d_-,x_4,x_5,x_6\}$ needs four gates: $\Delta_{rt},Q_{C2},Q_{C3},F_\pm:=\frac{e+e'\pm2\sqrt{ee'}}{4}=\square$ ($e=1+ax_4,e'=1+cx_4$):

• Mixed-$R_4$ Syzygies & $(s+1)D(\kappa,s)$ Cancellation: On Branches $\text{I},+1$ and $\text{II},+1$, $Q_{C2}$ and $Q_{C3}$ complete with shared remainder $R_{[x_{5,6}=0]}=-4\kappa^2(s+1)^2(t-\kappa)(t-1/\kappa)$: $Q_{C2,I}-P_{C2,I}^2=Q_{C3,I}-P_{C3,I}^2=(\kappa^2-1)^2(2s+1)$ and $Q_{C2,II}-P_{C2,II}^2=Q_{C3,II}-P_{C3,II}^2=-(2s-\kappa+1)(2\kappa s-\kappa-1)(4\kappa s^2-4\kappa^2s-\kappa^2+1)$, where $P_{C2,I}+P_{C3,I}=P_{C2,II}+P_{C3,II}=4\kappa(s-1)(s+1)^2$. Since $d_-(\kappa,s)=\frac{2\kappa(\kappa s-1)(2s^2-\kappa s-1)}{(\kappa^2-1)(s^2-1)}$, once $Q_{C3}=Y_3^2$, ($1+bx_4,F_\pm,1+d_-x_4$) share $(s+1)D(\kappa,s)$, cancelling in pairwise products.
• Reductions $\mathcal{H}_{II},\tilde{\mathcal{H}}_{II},\mathcal{L}_{I},\mathcal{C}_{I}$ & $9/10$ Quintuples: On Branch $\text{II},+1$, $u=-1/2$ gives $\mathcal{H}_{II}:\kappa=-\frac{1}{2s+3}$ and $\tilde{\mathcal{H}}_{II}:\kappa=\frac{2s+3}{4s+1}$. On $\mathcal{H}_{II}$, $Q_{C3,II}=(\frac{2(s+1)^2(2s-5)}{(2s+3)^2})^2$ and $Q_{C2,II}=\square$ gives $E_{II,C_2}:W_2^2=(2s-5)(2s+1)(4s^2-5)\cong Y^2=(X+20)(X^2+60X+720)$ ($X=\frac{120}{2s-5}$); $\phi(X,Y)=sqf(X+20)$ ($\phi(T)=-5$) proves $rkE_{II,C_2}(\mathbb{Q})\ge2$ and infinitely many $9/10$ quintuples $\{b,d_-,x_4,x_5,x_6\}$ ($s=-1/5,5/6$), while $\Delta_{rt}=\square$ gives $E_{II,\Delta}:W_\Delta^2=(4s+5)(8s^3+4s^2-8s+5)$ ($W_\Delta^2-W_2^2=4s(s+5)(4s^2+2s-3)$) and $(1+d_-x_4)W_2^2\equiv P_{6,II}(s)=(4s^3+10s^2-5)^2-4s(s-1)(4s+5)$ (exp36). On $\tilde{\mathcal{H}}_{II}$, $W_2^2+(16s(s+1))^2=(16s^3+24s^2+12s+5)^2$; on Branch $\text{I},+1$ along $\mathcal{L}_{I}:\kappa=2s+2$ and $\mathcal{C}_{I}:\kappa=\frac{1}{2s^2+s-2}$, $Q_{C3,I}=\square$, $\Delta_{rt}=\square$ gives $E_{\Delta,5}:W_\Delta^2=-(4s+5)(4s^2-5)\cong y^2=x^3+10x^2+5x$ ($rk\ge1$), and on $\mathcal{C}_{I}$, $Q_{C2,I}F_-\equiv(2s+1)(2s+5)(4s+5),Q_{C2,I}(1+d_-x_4)\equiv(4s^3+4s^2-s-5)^2+16s(4s^2+8s+5)$.
• $H$-Syzygies & $H=1/3$ Descent (Post 35, 43; exp37): Matching $F_{sym}=0$ into $cd=\frac{2H-1}{(H-1)^2},cf=-\frac{2H+1}{(H+1)^2}$ ($q=ab=H^2-1$), $A,B,C,D$ rescale from $Q_H(z):=z^2-2Hz+1$, obey $(H+1)A-2HB+(H-1)C=8H (H^2-1)$ and $H(2H+1)A-(4H^2-1)B+H(2H-1)C-D=0$, and diagonalize under $w_i=\frac{1-z_i}{1+z_i}$ as $Q_H(z_i)=\frac{2(H+1)}{(1+w_i)^2}(w_i^2-\lambda)$ ($\lambda=\frac{H-1}{H+1}$). In $u=y+\frac{1-4H^2}{y}$, $1+eg$ completes via $P_{edge}(u,H)=(u^2-2(3H^4-2H^2+1))^2-4(H^2-1)^3(9H^2-1)$, selecting $H=\pm1/3$. At $H=1/3$ ($y=X/9,\lambda=-1/2,K=\mathbb{Q}(\sqrt{-2})$), $A_0..D_0$ factor as norms $N_{K/\mathbb{Q}}(L_i)$ ($L_1L_2L_3L_4\in\pm3(K^\times)^2$), mapping to $E_d:y^2=(x-d)(x-2d)(x+3d)=x(x^2-7d^2)+6d^3$ ($d\in\{3,6\}$; $d<0$ eliminated over $\mathbb{R}$). Kummer filtering mod $\mathbb{R},64,81,125$ (exp37) leaves $(1,1,1)$ and $rkE_{-6}(\mathbb{Q})=rkE_{-3}(\mathbb{Q})=0\implies E_d(K)=E_d(\mathbb{Q})$, so $X=0,\infty$ are the only $\mathbb{Q}$-points, closing $H=1/3$.
• Branch B, $\mathcal{M}_\Psi^\star$, $10/10$ Quintuples & $\mathbb{Q}(\sqrt{-14})$ (Post 43, exp38-39): On Branch B ($r=\frac{1}{m+d_0},s=\frac{1}{m-d_0},W=rsw_0$), once $-P_8(u_d,u_w):=S_3^2-4u_d^2(u_d^2-1)^2((u_d+u_w)^2-1)=Y_{P8}^2$, $\{c,d,f,e,g\}$ is a $9/10$ quintuple ($R_4(c,d,e,f)=R_4(c,d,f,g)=0$): (1) $(26/37,24/37)$ gives a $10/10$ quintuple ($1+ab=(37/132)^2,1+eg=(3071/1729)^2$); (2) on $E_\tau(\tau,x,z)=0$, $\mathcal{M}_\Psi^\star$ gives $a\ne b\in\mathbb{Q}$ with $1+ax,1+bx\in\mathbb{Q}^{\times2}\iff A\equiv B\equiv C\equiv D\pmod{\mathbb{Q}^{\times2}}$, yielding at $(99/125,-284/625)$ ($[\Delta_{ab}]=[-14]$) a Galois-stable $20/21$ almost septuple over $\mathbb{Q}(\sqrt{-14})$ and a $10/10$ quintuple at $(-16,19)$; (3) on $(u_w-2u_d)^2+u_d^2=1$ ($u_d=\frac{2k}{k^2+1},u_w=\frac{k^2+4k-1}{k^2+1}$), $\Psi_+^\star(\tau)=0,\Delta_{ab}=0$ gives a positive $9/10$ family ($k>2+\sqrt{5}$) with $1+ac=\square\iff E_{coll}:v^2=u(u+1)(u+5)$ (rank $0$), locally soluble over $\mathbb{Q}_v$ (exp39).
• Paired & Antipodal Syzygies, $\mathcal{I}_{AL}^{(0)}$, $E_{sw}$, $Q_h$ & Five-$R_5$ (Post 35, 42, 67, 69; exp40-43,58-60): On Source 1 [20] in $(u,H)$ ($D_\pm:=u^2\pm3u+4,c=\frac{u(3-4H)}{4},G=-\frac{4H}{u}$), $F_\pm=\frac{8u(u\pm2)(u\pm1-2H(u\pm2))}{D_\pm^2}$ complete in $H$, while $\mathcal{I}_{AL}^{(0)}:(u,H)\mapsto(4/u,3/4-H)$ preserves $v^2=4H^2-3H+1$ and negates/swaps $(c\leftrightarrow G,a\leftrightarrow C,b\leftrightarrow A)$. 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 $-(16(q-1)^3T^2-q)=\frac{L_1L_2}{t^2}$; on $B^2=-(7q+9)(45q^2-25q-36),H^2=q^4+64q^3+1054q^2-1440q+81$, $H^2+4(5-4q)B^2=(71q^2-18q-81)^2$ gives $W_+W_-=(2eB(2q^2+63q-81))^2$ and $D_+^2D_-^2=(\frac{e(2q^2+63q-81)}{4(q-1)^2B})^2$ (exp43), where $v_3(H^2)=1$ obstructs $v_3(e)=0$, $D_\pm^2>0$ on $(0,q_{H,1}]\cup[q_{H,2},q_{B,+})$, and $C_B$ lifts over $E_1$ ($rk\ge1,P_1=(1,16),n\equiv\pm3\pmod{10}$). The fourths $d_1..d_4$ [22] satisfy $1+d_1d_4=(u/2)^2(1+cd_2),1+d_2d_3=(2/u)^2(1+Gd_4),1+d_1d_3=v^2-1$ (Orbits I–III: $u\in\{\pm3,\pm4/3\},\{\pm6,\pm2/3\},\{\pm8,\pm1/2\}$), and $k_{tan}(m)$ along $H(m)=\frac{2m-3}{m^2-4}$ gives $23/28$ squares (exp43). (i) $E_{sw}(-4/3)$ and $E_{sw}(-3)$ give $15/15$ sextuples & $23/28$ octuples ($t=55/17,m=-39/8$); on Source 1 Orbit I, $C:y^2=228x^6+595x^5+1457x^4+1904x^3+3296x^2+1024x+2304$ ($x=-1/H$) has $\dim_{\mathbb{F}_2}Sel_{quot}^{(2)}=2\implies rkJ(\mathbb{Q})=1$ [24], Coleman integrals [25] bounding zeros to $x=0$. (ii) On $\mathcal{E}_{source}:H(\nu)=\frac{2\nu-1}{4-\nu^2},V^2=\nu^4+6\nu^3+5\nu^2-40\nu+32$, $\tau_{w,1},\tau_{w,3}$ 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 $\mathcal{U}_8^\mu:=\mathcal{S}_6^{anti}\cup\{M(d),M(e)\}$ [20], $[1+M(d)M(e)]=[q_0^4-q_0^2+1]$ maps to rank-$0$ $E_{24}:V^2=U(U-1)(U+3)$; in $U^2=1+t^2,X=\frac{t+q}{U},Y=\frac{qt-1}{U}$ (Post 67), $d^2Y^2=((q^2+1)\rho^2-(q+1)^2)((q-1)^2\rho^2-(q^2+1))$ is obstructed at $q=\pm5/16$ over $\mathbb{Q}_{23}$, and $[1-y_+/a]=[1+aM(y_+)]=[1+y_+M(y_+)]$ on $\mathcal{S}_7^\star(\epsilon)$ ($18/21$) and $\mathcal{U}_8^{anti}(\epsilon)$ ($23/28$). (v) On $Q_h$ (Post 42, exp58-60), $1+de_{reg}(h)=(\frac{h}{2(h^2-2)})^2(t^4-12t^2+16)$ ($t=h+2/h$) 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 $U_\mp(r,s,\kappa)$, $g_1=g_2$ obeys $G_1-G_2\equiv H(R_1^2-R_2^2)=0$ on $P_{orb}=0$, and on $E_\rho:Y^2=X(X+\rho^2)(X+1/\rho^2)$ ($T=(1,\rho+1/\rho)$), $X(R+T)X(R-T)=1$ gives $\frac{F_\rho(X(R))}{F_\rho(X(R+T))}=\frac{\rho(X(R)-1)^2}{2(\rho^2+1)X(R)}\in[-2(\rho^2+1)(\rho^2-\rho+1)]$ at $X(U)=-\frac{\rho}{\rho^2-\rho+1}$ (exp60,74).
• Stars $\mathcal{S}_7^{3-refl}$, $\mathcal{V}_{Euler}$, Towers, $\mathcal{E}_W(R^2,Z)$ & $\mathcal{C}_{coinc,r}$ (exp61-88):
• (a) Monomial Fifth, Stars $\mathcal{S}_7^{3-refl},\mathcal{S}_7^{(f_c)}$ & Modules $\mathcal{M}_3,\mathcal{N}_2^{even}$ (exp61-67): 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. For regular $\{x,y,z,d_-\}$ ($L=2rst=\frac{d_+-d_-}{2},V=xyzd_--1,U=Ld_--V$), $U-V=-2r_xr_yr_z$ gives $e_{reg}(x,y,z,d_-)=-\frac{4rstr_xr_yr_z}{(xyzd_--1)^2}$ and $V^2(1+d_+e_{reg})=(U+L^2)^2-L^2(L^2+2V)$. In $(r,C_0,d_-)$ ($S_{ab}=\frac{C_0^2-2(2r^2-1)C_0d_-+d_-^2-4r^2}{4r^2d_-},d_-d_+=C_0^2-4r^2$), $f_{x,-}=2(y+z+(2yz+1)d_-)-x$ on $\mathcal{S}_7^{3-refl}=\{a,b,c,d_-,f_{a,-},f_{b,-},f_{c,-}\}$ ($15/21$) 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 $\mathcal{S}_7^{(f_c)}=\{a,b,c,d_-,d_+,f_{c,-},f_{c,+}\}$ ($16/21$) 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 $\mu_{inf}(r,C_0)$ ($d_-=4r^2a=r(t-1/t),b=\frac{4r(r^2-1)}{t-1/t},bd_-=4r^2(r^2-1)$), $\iota_t:t\mapsto-1/t$ fixes $(a,b,d_-)$ and swaps $crightarrow f_{c,-},f_{a,-}rightarrow\tilde{f}_{a,-},g_{Euler,+}rightarrow\tilde{g}_{Euler,+}$ ($g_{Euler,+}=d_+(0,c,f_{a,-})$), forming $\mathcal{S}_6^{Euler}$ ($12/15$), $\mathcal{S}_7^{Euler,c}$ ($15/21$), $\mathcal{U}_8^{Euler}$ ($18/28$), and $\mathcal{U}_9^{bi-Euler}$ ($21/36$; $23/36$ on $t_1(r),\mathcal{C}_{coinc}$). All $9$ entries, $R_{c,a},R_{c,g},R_{a,g},X_1,\tilde{X}_1,X_2,X_4,\tilde{X}_4,X_5,X_6,X_{11..13}\in$ $\mathcal{M}_3=\mathbb{Z}[r]a\oplus\mathbb{Z}[r]b\oplus\mathbb{Z}[r]r_{ad_-}$ ($r_{ad_-}=\frac{t^2+1}{2t}=\sqrt{1+ad_-},r_{ad_-}^2-(2ra)^2=1$), while $X_3=r_{bc}+r_{bf_{a,-}}=(2r^2-1)^2r+(4r^2-1)br_{ad_-}\in\mathcal{N}_2^{even}=\mathbb{Z}[r]r\oplus\mathbb{Z}[r](br_{ad_-})$ ($r_{bc}=(2r^2-1)r+br_{ad_-},X_2=r_{cd_-}+X_1,J_c=2r(2r^2-1),J_a=4r(2r^2-1)(8r^4-8r^2+1),J_g=4r^3(4r^2-1)(4r^2-3)$):

$$ \begin{aligned} c,f_{c,-}=(8r^4-4r^2+1)a+b\pm J_cr_{ad_-},f_{a,-},\tilde{f}_{a,-}=(128r^8-192r^6+96r^4-16r^2+1)a+4(2r^2-1)^2b\pm J_ar_{ad_-},\\ g_{Euler,+},\tilde{g}_{Euler,+}=8r^4(16r^4-16r^2+5)a+(4r^2-1)^2b\pm J_gr_{ad_-}, \end{aligned} $$

with all $15$ open pairs of $\mathcal{U}_9^{bi-Euler}$ having $r$-only shifts: $4r^2(1+af_{a,-})=X_1^2+(4r^2-1)=\tilde{X}_1^2-4r^2(4r^2-1)(4r^2-3)$ ($X_1=r_{d_-f_{a,-}}$), $1+d_-g_{Euler,+}=X_2^2-(4r^2-1)$, $1+bg_{Euler,+}=X_3^2-(4r^2-1)$ ($r_{bc}+r_{bf_{c,-}}=J_c,X_3+\tilde{X}_3=4r^3(4r^2-3)$), $1+cf_{c,-}=((4r^2-1)a\mp b)^2-(4r^2-1)(4r^2-3)^{(1\mp1)/2}$, $1+f_{a,-}\tilde{f}_{a,-}=X_{12}^2-(4r^2-1)(16r^4-12r^2+1)$, $1+g_{Euler,+}\tilde{g}_{Euler,+}=(32r^4(2r^2-1)a\mp(4r^2-1)^2b)^2-((4r^2(4r^2-1)(4r^2-3)^{(1\mp1)/2})^2-1)$, and $1+u_i\tilde{u}_j=X_k^2-\frac{J_iJ_j}{r^2}$ ($X_k-\iota_t(X_k)=-2rr_{ad_-},X_6-X_5=\tilde{X}_4,X_{11}-X_5=\tilde{X}_{12}$).

• (b) Symplectic Euler Plane $\mathcal{V}_{Euler}$, Half-Step $H_{Cheb}$, Scale Duality, Towers & $Q_5^{Cheb}(q)$ (exp67-75): In $E_1=\frac{t}{r}=2a+\frac{1}{r}r_{ad_-},E_2=-\frac{1}{rt}=2a-\frac{1}{r}r_{ad_-},E_3=b$ ($E_1E_2=-\frac{1}{r^2},E_1E_3+E_2E_3=4(r^2-1),b+d_-=r^2E_1+r^2E_2+E_3$), $t$-independence of $R(w_1,w_2)^2-u(w_1)u(w_2)$ ($M_{+s}^2=M_{-s}^2$) splits into $\mathcal{V}_{Euler}=\{(p_+,p_-,s):p_++p_-=2r^2s\}=\mathbb{Q}(r)e_1\oplus\mathbb{Q}(r)e_2$ ($M_{+s}=-M_{-s}$). Here $w=\xi e_1+\eta e_2$ ($p_+=\frac{\xi+(4r^2-1)\eta}{2},p_-=\frac{-\xi+\eta}{2},s=\eta,Q(\xi,\eta)+4p_+p_-=4r^2\eta^2,\iota_V(\xi,\eta)=(-\xi-2R\eta,\eta)$), $u(w):=p_+^2E_1+p_-^2E_2+s^2E_3\in\mathcal{M}_3$, and $R(w_1,w_2):=p_{1,+}p_{2,+}E_1+p_{1,-}p_{2,-}E_2+s_1s_2E_3\in\mathcal{M}_3$ obey ($Q(\xi,\eta):=\xi^2+2R\xi\eta+\eta^2,X_{d_-}=\xi r_{ad_-}+\eta r_{cd_-}=p_+t-\frac{p_-}{t}\in\mathcal{M}_3,X_b=\xi r+\eta r_{bc}\in\mathcal{N}_2^{even},X_a=\xi a+\eta r_{ac}=\frac{p_+t+p_-/t}{2r},\tilde{X}_a=2rX_a\in\mathcal{M}_3$):

$$ \begin{aligned} 1+u(w_1)u(w_2)=R(w_1,w_2)^2+1-(\xi_1\eta_2-\eta_1\xi_2)^2,1+d_-u(\xi,\eta)=X_{d_-}^2+1-Q(\xi,\eta)=(2rX_a)^2+1-4r^2\eta^2,\\ 1+bu(\xi,\eta)=X_b^2+1-Q(\xi,\eta),1+au(\xi,\eta)=X_a^2+1-\eta^2=(\frac{X_{d_-}}{2r})^2+1-\frac{Q(\xi,\eta)}{4r^2},\\ u(w_1\pm w_2)=u(w_1)+u(w_2)\pm2R(w_1,w_2),u(M_{Cheb}^{\pm1}(w))=Q(w)(b+d_-)+(1+2bd_-)u(w)\pm2RX_bX_{d_-}, \end{aligned} $$

with $u(w)u(M_{Cheb}(w))+Q(w)^2=R(w,M_{Cheb}(w))^2$, $(d_--b)(Q(w),u(w),R(w_1,w_2))=(d_-X_b^2-bX_{d_-}^2,X_{d_-}^2-X_b^2,X_{d_-}(w_1)X_{d_-}(w_2)-X_b(w_1)X_b(w_2))$, $\binom{X_b(M_{Cheb}(w))}{X_{d_-}(M_{Cheb}(w))}=(\begin{smallmatrix}R&b\\d_-&R\end{smallmatrix})\binom{X_b(w)}{X_{d_-}(w)}$, $R(w_{n-1},w_{n+1})=u(w_n)-Q(w_0)(b+d_-)$, $1+au(2rw)=1+d_-u(w)$ ($1+au=\square$ on $\eta=\pm1$ or $Q=4r^2$; $1+d_-u=\square$ on $Q=1$ or $\eta=\pm\frac{1}{2r}$), and $t_{b,(\xi,\eta)}(r)=-\frac{4r^4\eta(\xi+(r^2-1)\eta)+(\xi-\eta)^2-1}{4r^2\eta((r^2-2)\xi+(r^2-1)(3r^2-2)\eta)-(\xi-\eta)^2+1}$ making $1+bu(\xi,\eta)=\square$ ($t_{III}=-t_{b,(-1,4r^2-1)}$ gives $22/36$ in $\mathcal{U}_9^{bi-Euler}$; $t_{b,(1,1)}=-\frac{4r^6-1}{12r^6-16r^4+1}$ for $g_0=u(1,1)$; $t_{b,(\eta\pm1,\eta)}=-\frac{r^2(r^2\eta\pm1)}{r^2(3r^2-4)\eta\pm(r^2-2)}$). Moreover, $M_{Cheb}(\xi,\eta)=(-\eta,\xi+2R\eta)$ has square root $M_{1/2}=H_{Cheb}:=\frac{I_2+M_{Cheb}}{2r}=\frac{1}{2r}(\begin{smallmatrix}1&-1\\1&4r^2-1\end{smallmatrix})\in\text{SO}_Q(\mathcal{V}_{Euler})$ ($M_{1/2}^2=M_{Cheb},v_{j/2}(w_0):=M_{1/2}^jw_0,v_{j/2}(e_0)=\frac{(-U_{j-3}(r),U_{j-1}(r))}{2r},S(v_{j/2}(e_0))=v_{(2-j)/2}(e_0),\det(v_{i/2}(w_0),v_{j/2}(w_0))=Q(w_0)\frac{U_{j-i-1}(r)}{2r}$), making $\hat{\mathcal{Q}}_k^{half}(r,t)=\{b,d_-,\hat{g}_k,\hat{g}_{k+1}\}$ ($\hat{g}_k=\frac{g_k}{4r^2},1+\hat{g}_k\hat{g}_{k+1}=(u_{k+1}-a-\tilde{a})^2$) and $\mathcal{Q}_{j/2}^{(u)}(r,t,m)=\{b,d_-,\hat{u}_{j/2}(m),\hat{u}_{(j+2)/2}(m)\}$ ($\hat{u}_{j/2}=B(v_{j/2},v_{j/2}),\hat{G}_{j/2}=4r^2\hat{u}_{j/2},1+\hat{u}_{i/2}\hat{G}_{(i\pm1)/2}=(2rB(v_{i/2},v_{(i\pm1)/2}))^2$) regular $6/6$ quadruple towers ($R_4=0$; exp74-75), obeying ($Q(w_0)=1$):

$$ \begin{aligned} 1+\hat{u}_{i/2}\hat{u}_{j/2}=B(v_{i/2},v_{j/2})^2-(\frac{U_{j-i-1}(r)^2}{4r^2}-1),1+\hat{u}_{i/2}\hat{G}_{j/2}=4r^2B(v_{i/2},v_{j/2})^2-(U_{j-i-1}(r)^2-1),\\ 1+\hat{G}_{i/2}\hat{G}_{j/2}=16r^4B(v_{i/2},v_{j/2})^2-(4r^2U_{j-i-1}(r)^2-1). \end{aligned} $$

On $Q(\xi,\eta)=1$, $w(m)=(\frac{1-m^2}{D_m(m)},\frac{2(m-R)}{D_m(m)})$ ($m=\frac{1-\xi}{\eta},D_m(m)=m^2-2Rm+1,1-\eta(m)^2=\frac{(m^2-1)(m-(4r^2-1))(m-(4r^2-3))}{D_m(m)^2},\Delta_\eta=16r^2r_{ad_-}^2((2ra)^2+(r^2-1)(K^2-1))$ along $X_{d_-}(w)=K$) intertwines $M_{1/2}$ with $\sigma_r(m):=\frac{(2r+1)m+1}{(4r^2+2r-1)-m}$ ($\sigma_r^2=\mu_+,v_n=w_n+w_{n+1}=2rw(\sigma_r^{2n+1}),Q(v_n)\equiv4r^2$) and $S(\xi,\eta)=(\eta,\xi)$ with $\iota_S(m)=\frac{(4r^2-1)-m}{m+1}=\mu_+(m)^{-1}$ ($\iota_S\circ\sigma_r^j=\sigma_r^{-j}\circ\iota_S$, exp75). For distinct nonzero entries (exp69,72-75, $N=1,2,3$), $Q_4(r,t,m)=\{b,d_-,u_0(m),u_1(m)\}$ is a regular $6/6$ quadruple (at $(2,4,3)$), and edge-partitioning ($1+u_kg_k=(u_k+\Pi_k)^2,1+u_{k+1}g_k=(u_{k+1}+\Pi_k)^2$) proves for all $N\ge1$: (i) $\mathcal{U}_{4N+3}^{Cheb}(r,t,m):=\{a,b,d_-,u_{-N}..u_N,g_{-(N-1)}..g_{N-1}\}$ has $\ge12N+2$ squares ($\mathcal{U}_{4N+3}$: $\ge10N+3$; $\mathcal{U}_{4N+3}^{bi-Euler}$: $\ge12N+1$; $\mathcal{U}_{13}^{bi-Euler}$: $29/78$), increasing to $\ge12N+3$ on $t=t_{b,v_0(m)}(r)$ or $C_2\cap L_2:\{Q(v_0)=4r^2,\eta(v_0)=\frac{1}{2r}\}$ over $5r^2-1=\rho^2$ ($r(q)=\frac{q^2+4q+5}{5-q^2},\rho(q)=\frac{2(q^2+5q+5)}{5-q^2},\tilde{m}=6r^2-2r\rho-1,m(q)=\sigma_r^{-1}(\tilde{m}),1+d_-g_0=(2rX_a(v_0))^2$), and to $\ge12N+4$ on both over $\mathbb{Q}(q)$, embedding non-regular $10/10$ $Q_5^{Cheb}(q)=\{b,d_-,u_0,u_1,g_0\}$, $14/15$ $\mathcal{S}_6^{Cheb}(q)=Q_5^{Cheb}(q)\cup\{e_+\}$, and $17/21$-compatible septuples; (ii) adjoining $\tilde{a}:=\frac{b}{4r^2}$ ($1+\tilde{a}d_-=r^2,1+\tilde{a}g_n=(\frac{X_n+X_{n+1}}{2r})^2,1+a\tilde{a}=\frac{5r^2-1}{4r^2}$) and $g_{-N}$ gives $\mathcal{U}_{4N+4}^{Cheb}$ ($\ge14N+2..5$) and $\mathcal{U}_{4N+5}^{Cheb}$ ($\ge14N+6..9$); and (iii) $\hat{\mathcal{U}}_{8N+6}^{lat}(r,t,m)=\{a,\tilde{a},b,d_-,\hat{u}_{(1-2N)/2}..\hat{u}_{(2N+1)/2},\hat{G}_{(1-2N)/2}..\hat{G}_{(2N+1)/2}\}$ has $\ge28N+7..9$ squares. For $G(\xi,\eta):=4r^2u(\xi,\eta)$, Scale Duality gives $1+aG(w)=1+d_-u(w),1+\tilde{a}G(w)=1+bu(w)$, and under $S(\xi,\eta)=(\eta,\xi)$ with $A_w:=(d_--b)\xi\eta+\frac{U_{11}}{2(d_--b)}Q_w$ ($U_{11}=2r_{au_1}(d_--b)+4rr_{bc},s_u=\frac{8r^2r_{ad_-}c_0r_{bc}}{(d_--b)^2},s_G=\frac{4rt(t^2-1)(u_1-\tilde{a})((16r^4-8r^2-1)t^4-2(2r^2+1)t^2-1)}{((t^2+1)^2-4r^2t^2)^2}$):

$$ \begin{aligned} Q_w^2+u(w)u(S(w))=A_w^2-s_uQ_w^2,Q_w^2+G(w)G(S(w))=16r^4A_w^2-s_GQ_w^2,\\ R_{4,hom}(b,d_-,u(w),u(S(w));Q_w)=(d_--b)^2s_u((\xi^2-\eta^2)^2-Q_w^2),R_{4,hom}(a,\tilde{a},G(w),G(S(w));Q_w)=(a-\tilde{a})^2s_G(16r^4(\xi^2-\eta^2)^2-Q_w^2). \end{aligned} $$

On $w_0=(1,0)$, $R_4(a,b,d_-,\tilde{a})=(\tilde{a}-u_1)(\tilde{a}-u_{-1})$ with $u_{-1}-\tilde{a}=\frac{((t-r)^2-(5r^2-1))((t+r)^2-(5r^2-1))}{4rt(t^2-1)}$; on $5r^2-1=\rho^2$ ($r(\lambda)=\frac{\lambda^2+1}{2(1+\lambda-\lambda^2)},\rho(\lambda)=\frac{\lambda^2+4\lambda-1}{2(1+\lambda-\lambda^2)}$), $t_\mp(\lambda)=(r+\rho)^{\pm1}$ forces $u_{\mp1}\equiv\tilde{a}$, $1+\tilde{a}b=(\frac{r(r^2+1+r\rho)}{3r^2-1+r\rho})^2$, and $s_G(r,t_+)=0$, giving $R_4=0$ quads $\{a,\tilde{a},b,d_-\}$ ($\lambda=4/3$), $\mathcal{Q}_4^{swap}(\lambda,m)=\{a,\tilde{a},G(\xi_m,\eta_m),G(\eta_m,\xi_m)\}$, $\mathcal{Q}_n^{(+)}=\{a,\tilde{a},g_{-n},g_n\}$, $\mathcal{Q}_n^{(-)}=\{\tilde{a},a,g_{-n-1},g_{n-1}\}$, $\ge15N+2$ squares in $\mathcal{U}_{4N+4}^{Cheb}(\lambda)$, $\ge30N-9$ squares ($\ge6N-2$ $R_4=0$ quads, $\ge4N-3$ $9/10$-quints) in $\hat{\mathcal{U}}_{8N+2}^{lat}(\lambda)$, and swap pools $\mathcal{S}_6^{swap}$ ($11/15$), $\mathcal{U}_8^{swap}(m)$ ($15..17/28$), $\mathcal{U}_{11}^{swap}$ ($26/55$; exp73-75).

• (c.1) Two-Gram-Matrix Theory, Chebyshev Shift Table, Reflection Defect & $\mathcal{S}_{7,j}^{osc}$ (exp63-82): For $\{b,d_-,u_0\}$ with $P=bd_-=R^2-1$ ($R=2r^2-1$) and $\Pi=\Pi_0:=Ru_0+X_0Y_0,\Pi_{-1}:=Ru_0-X_0Y_0$, $V=\mathbb{Q}(R)e_0\oplus\mathbb{Q}(R)e_1$ carries Gram matrices $G_B=(\begin{smallmatrix}u_0&\Pi_0\\\Pi_0&u_1\end{smallmatrix})$ ($\det G_B=-1$) and $G_Q=(\begin{smallmatrix}1&R\\R&1\end{smallmatrix})$ ($\det G_Q=-P,tr(G_BadjG_Q)=u_0+u_1-2R\Pi_0=S_{b,d_-},B(\xi,\eta;\eta,\xi)=Q_{Cheb}(\xi,\eta)\Pi_0+\xi\eta S_{b,d_-}$) with $1+B(v_1,v_1)B(v_2,v_2)=B(v_1,v_2)^2-(\det(v_1,v_2)^2-1),1+bB(v,v)=X(v)^2+1-Q_{Cheb}(v,v),1+d_-B(v,v)=Y(v)^2+1-Q_{Cheb}(v,v)$. For $e_n=M_R^ne_0=(-U_{n-2}(R),U_{n-1}(R)),h_n=e_n+e_{n+1}$, $u_n=U_{n-2}^2u_0+U_{n-1}^2(S_{b,d_-}-u_0)+2U_{n-1}T_n\Pi_0$, $\Pi_n=U_{n-2}U_{n-1}u_0+U_{n-1}U_n(S_{b,d_-}-u_0)+U_{2n}\Pi_0$, and $g_n=2(R+1)\Pi_n+S_{b,d_-}$. Setting $N_R:=M_R-RI_2=2r(M_{1/2}-rI_2)$, $N_R^TG_BN_R=(R^2-1)G_B+S_{b,d_-}G_Q$ and $N_R^TG_B-G_BN_R=S_{b,d_-}\Omega$ give $(M_R^{-n})^TG_BM_R^n=G_B-S_{b,d_-}(U_{n-1}^2G_Q+T_nU_{n-1}\Omega)$ and $B(M_R^{-n}v,M_R^nv)=B(v,v)-U_{n-1}(R)^2Q_{Cheb}(v,v)S_{b,d_-}$ (exp77), proving for all $k\in\mathbb{Z},n\ge1$ (exp73) the All-Step Infinite Chebyshev Shift Table:

$$ \begin{aligned} 1+u_{k-n}u_{k+n}=(u_k-U_{n-1}^2S_{b,d_-})^2-(U_{2n-1}^2-1),1+u_{k+1-n}u_{k+n}=(\Pi_k-U_{n-2}U_{n-1}S_{b,d_-})^2-(U_{2n-2}^2-1),\\ 1+g_{k-n}g_{k+n-1}=(2(R+1)u_k-(U_{n-2}+U_{n-1})^2S_{b,d_-})^2-(4(R+1)^2U_{2n-2}^2-1),\\ 1+g_{k-n}g_{k+n}=(2(R+1)\Pi_k-(U_{n-2}+U_{n-1})(U_{n-1}+U_n)S_{b,d_-})^2-(4(R+1)^2U_{2n-1}^2-1),\\ 1+u_{k\mp n}g_{k\pm n-(1\pm1)/2}=(\Pi_{k-(1\pm1)/2}+u_k-U_{n-1}(U_{n-2}+U_{n-1})S_{b,d_-})^2-((U_{2n-2}+U_{2n-1})^2-1), \end{aligned} $$

giving at $n=1$ the eight universal laws on $\mathcal{W}_7=\{b,d_-,u_{-1},u_0,u_1,g_{-1},g_0\}$ ($13/21$), while $e_+=\frac{4R\Pi(P\Pi^2+RS_{b,d_-}\Pi+R^2)}{(P\Pi^2-R^2)^2}$ and $(P\Pi^2-R^2)^2(1+g_0e_+)=(2R\Pi S_{b,d_-}+(R+1)(3R-1)\Pi^2+R^2)^2-4R(R+1)\Pi^2((R+1)\Pi^2-R)$. In $\kappa:=\frac{d_-(X_0-1)}{Y_0-1}=\frac{d_-}{k}$ ($k=\frac{Y_0-1}{X_0-1};\kappa=48/7$ at $(2,4,3)$), $(u_0,X_0,Y_0)=(-\frac{4\kappa(\kappa-b)(\kappa-d_-)}{(\kappa^2-P)^2},-\frac{\kappa^2-2b\kappa+P}{\kappa^2-P},\frac{\kappa^2-2d_-\kappa+P}{\kappa^2-P})$ gives $1-\det(w(\kappa_1),w(\kappa_2))^2=\frac{\prod_{\varepsilon_1,\varepsilon_2\in\{\pm1\}}(\kappa_1\kappa_2+(\varepsilon_1R+\varepsilon_2)(\kappa_1-\kappa_2)-P)}{(\kappa_1^2-P)^2(\kappa_2^2-P)^2}$, and $\Pi(\tau_\kappa)-1=\frac{\tau_\kappa+2R}{2}u_0(\tau_\kappa)$ descends $(P\Pi^2-R^2)^2(1+g_0e_+)$ to $\deg_{\tau_\kappa}8$ in $\tau_\kappa=\kappa+\frac{P}{\kappa}$, lifting iff $\tau_\kappa^2-4P=\square$ ($\deg_{\tau_\kappa}8,g=3$ at $(2,4)$; on $Q_5^{Cheb}(q)$, $1+g_0e_+=\frac{N_{38}}{D_{19}^2}$, lifting iff $S,5S-1=\square$). In $\sigma_\kappa=\kappa-\frac{P}{\kappa}$, $u_0-(b+d_-)=C_0$ gives $\Pi_{(-1\pm1)/2}=\mp\frac{\tau_\kappa\pm2R+2C_0}{\tau_\kappa\mp2R},u_{\pm1}=-\frac{(C_0\pm2R)(\tau_\kappa\pm2R)}{\tau_\kappa\mp2R}$, $S_{b,d_-}=-\frac{C_0\sigma_\kappa^2+4\tau_\kappa}{\sigma_\kappa^2-4}$, and $S_{b,d_-}^2-4P=(\frac{\sigma_\kappa}{\sigma_\kappa^2-4})^2M(R,C_0,\tau_\kappa)$ ($M=(C_0\tau_\kappa+4R^2)^2-4P(C_0+\tau_\kappa)^2$), where $\tau_{\bar{\kappa}}=I_{C_0}(\tau_\kappa):=-\frac{C_0\tau_\kappa+4R^2}{C_0+\tau_\kappa}$ fixes $S_{b,d_-},u_0$ and swaps $u_1rightarrow u_{-1}$, while $J_R:(R,\Lambda)\mapsto(-R,-\frac{2R-1}{2R+1}\Lambda)$ swaps $c^*_-rightarrow c^*_+$ and $(D_-D_+W)^2(1+c^*_-c^*_+)=16\Lambda^4[(C_0W+4(4R^2-1)(C_0+\tau_\kappa))^2+((R-1)^2D_0^2-R^2)((R+1)^2D_0^2-R^2)W^2]$ ($W=\sigma_\kappa^2-4$). In $T_*=\tau_\kappa+2R$ ($\sigma_\kappa^2=T_*^2-4RT_*+4$), $L_j(r,T_*):=U_{j-1}(r)T_*-2U_{j-3}(r)$ and $T_j:=L_{j+1}-rL_j=T_j(r)T_*-2T_{j-2}(r)$ ($T_j^2-(r^2-1)L_j^2=\sigma_\kappa^2,L_iT_j-L_jT_i=-U_{j-i-1}(r)\sigma_\kappa^2$) factor $\hat{u}_{j/2}=\frac{L_j(L_jS_{b,d_-}-4rT_j)}{4r^2\sigma_\kappa^2}$, $1+\{b,d_-\}\hat{u}_{j/2}=(\frac{\{b,d_-\}L_j-2rT_j}{2r\sigma_\kappa})^2$ ($\varphi_b(x)=\frac{b-x}{1+bx}$ [28, 29] has $b^6R_5=\Phi_5^{(-)}\Phi_5^{(+)}$, mapping $\mathcal{Q}_{5,j}^{reg}$ on $b=\frac{2t_b}{1-t_b^2}$ to $10/10$ $\tilde{\mathcal{Q}}_{5,j}^{refl}$, with $R_4(b,\varphi_b(x_i))=\frac{(1+b^2)^2}{\prod(1+bx_i)^2}R_5(b,x_i,b)$ and $R_5(b,x,y,\varphi_b(x),\varphi_b(y))<0$; exp80,82), and $1+\hat{u}_{i/2}\hat{u}_{j/2}=(\frac{L_iL_jS_{b,d_-}-2r(L_iT_j+L_jT_i)}{4r^2\sigma_\kappa^2})^2+\frac{4r^2-U_{j-i-1}(r)^2}{4r^2}$, while $\iota_Z:(T_*,S)\mapsto(4R-T_*,-S)$, $\iota_{V_*}:(T_*,S)\mapsto(\frac{4}{T_*},-S)$, and $\mathcal{M}_{1/2}(T_*)=\frac{(4R+2)T_*-4}{T_*+2}$ ($\mathcal{M}_{1/2}^m(T_*)=\frac{2L_{m+2}}{L_m}$) shift $\hat{u}_{j/2}(\mathcal{M}_{1/2}^m(T_*))=\hat{u}_{(j+m)/2}$ and unite $\mathcal{S}_{7,j}^{osc}$ ($W_j=4r^2\sigma_\kappa^2-L_j^2,D_j=r^2(4R^2-1)\sigma_\kappa^4-(r^2-1)W_j^2$) into one $\text{PGL}_2(\mathbb{Q}(R))$-orbit (exp78-79): $j=0$ ($D_\sigma:=3\sigma_\kappa^4+32P\sigma_\kappa^2-64P,Z=\sigma_\kappa^2$) and $j=1$ ($\hat{W}=-W_1=(4R+2-T_*)((2R+1)T_*-2),\hat{D}=D_1$), i.e., the $18/21$ osculating surfaces $\mathcal{S}_7^{osc}(R,\kappa)$ and $\hat{\mathcal{S}}_7^{osc}(r,\kappa)$ (exp70-80, graph $(2,2)$: $13$ quads [$2$ $R_4=0$], $3$ quints [$2$ $R_5=0$]):

$$ \begin{aligned} \Lambda_{sol}=-\frac{8(2R+1)\sigma_\kappa^2\tau_\kappa}{D_\sigma},b=\frac{16P(\sigma_\kappa^2-4)\tau_\kappa}{D_\sigma},d_-=\frac{D_\sigma}{16(\sigma_\kappa^2-4)\tau_\kappa},X_0=-\frac{(3Z^2+64P)\tau_\kappa}{\sigma_\kappa D_\sigma},Y_0=\frac{5Z^2-32Z-64P}{8(Z-4)\sigma_\kappa\tau_\kappa},\\ \hat{\Lambda}_{sol}=-\frac{(R+1)(2R+1)(T_*^2-4)\sigma_\kappa^2}{\hat{D}},\hat{b}=-\frac{P(T_*^2-4)\hat{W}}{\hat{D}},\hat{d}_-=-\frac{\hat{D}}{(T_*^2-4)\hat{W}},2\hat{D}\equiv(R+1)(4R^2-1)\sigma_\kappa^4-(R-1)\hat{W}^2. \end{aligned} $$

Here $G_\mp(R,Z),G_{-+}(R,Z)\in\mathbb{Q}(R,Z)$ ($G_+(R,Z)=G_-(-R,Z)$) and $\hat{G}_\mp(R,V_*),\hat{G}_{-+}(R,V_*)\in\mathbb{Q}(R,V_*)$ ($V_*=\frac{\sigma_\kappa^2}{T_*}=T_*+\frac{4}{T_*}-4R$) have degrees $10,10,18$ via $Z=\frac{8(R+1)V_*}{V_*+4(R+1)}$ (exp78), with $\Delta_-=4D_-^2-(\Lambda^2-1)^2,\Delta_+=4D_+^2-((2R-1)^2\Lambda^2-(2R+1)^2)^2,\Delta_{-+}=(D_-D_+)^2-16\Lambda^2R^2((2R-1)\Lambda^2-(2R+1))^2$ (exp85) splitting on $R_{Pell}=\frac{q^2+1}{q^2-1}$ ($Q_{base}=(\frac{\sigma_\kappa}{\kappa})^2F_{Pell,-}F_{Pell,+}$), $R_{Br1}=\frac{3q^2+1}{2(q^2-1)}$, $R_{conj}=\frac{q^2+3}{2(q^2-1)}$. At $\kappa=R-1$ ($Z=4,V_*=\frac{1}{R}$), $\hat{\mathcal{S}}_7^{osc}(r,R-1)$ gives $18/21$ and $\hat{\mathcal{U}}_{15}^{osc}$ ($\ge39..40/105$), $s=\frac{\kappa}{R-1}$ halves degrees, and adjoining $a,\tilde{a},g_k,\hat{g}_k,G_k$ gives $\mathcal{U}_{10,11,13,16}^{osc},\hat{\mathcal{U}}_{16}^{osc}$ ($25/45,28..29/55,33..34/78,43..44/120$).

• (c.2) $\mathcal{S}_7^{univ}(q,\Lambda)$, Equi-Gate $S_{eq}$, Two-Gate Sections $S_{tan},S_{sec}$ & Off-Branch $19/21$ Loci on $S_{2sq}(R,C_0)$ (exp87-90): Since $C_0^2-4P=w_1^2+3$ and $\binom{2R}{C_0}=(\begin{smallmatrix}V_\Lambda&U_\Lambda\\U_\Lambda&V_\Lambda\end{smallmatrix})\binom{1}{w_1}$, $\kappa_{univ}=-\frac{C_0(w_1^2+3)}{2(w_1^2+1)}$ gives $M(R,C_0,\tau_{\kappa_{univ}})=(\frac{-4C_0^2w_1^2+w_1^6+5w_1^4+7w_1^2+3}{2C_0(w_1^2+1)\sqrt{w_1^2+3}})^2$, giving $18/21$ $\mathcal{S}_7^{univ}(q,\Lambda)$ on $w_1(q)=\frac{q^2-3}{2q}$ ($b(-3/q,\Lambda)=d_-(q,\Lambda)$); $V_4=\{\Lambda,\frac{1}{\Lambda},-\frac{w_1\Lambda+1}{\Lambda+w_1},-\frac{\Lambda+w_1}{w_1\Lambda+1}\}$ spans all $4$ $19/21$ branches $\Lambda\in V_4\cdot w_1$ of $\Delta_-\Delta_+=0$ and $f_{1..4}=2\Lambda((w_1\mp1)R\mp w_1)=0$ (exp87), with $R_{br}-R_{f_j}=\pm\frac{w_1\mp1}{2(w_1\pm1)}$ or $\pm\frac{(2w_1)^2+(w_1\pm1)^2}{2(w_1^2-1)}$ ($\Delta=-16<0$, $\mathbb{R}$-disjoint outside $w_1,\Lambda\in\{0,\pm1\}$). On $S_{eq}=-\frac{2(3R^2-1)}{4R^2-1}C_0$ ($u_{0,eq}=-\frac{2R^2-1}{4R^2-1}C_0$), $(4R^2-1)^2den_-^2(1+u_0c^*_-)-den_+^2(1+u_0c^*_+)\equiv-16\Lambda^2Rw_1((4R^2-1)S+2(3R^2-1)C_0)=0$ gives $[1+u_0c^*_-]=[1+u_0c^*_+]$ and ($Z_w=w_1^2$): $L_{gate}=R^2(4R^2-1)^2+(8R^4-5R^2+1)Z_w>0$, $L_{bd}=R^2(4R^2-1)(5R^2-1)+(3R^2-1)^2Z_w$, $L_\tau=R^2(4R^2-1)(4R^4+4R^2-1)+(2R^2-1)(2R^4+3R^2-1)Z_w$, $Q_{-+}^{eq}=(PZ_w+R^2)^2+4R^4(4R^2-1)$ with $1$-term $Z_w$-eliminants and $L_\tau+(4R^2-1)L_{gate}\equiv4R^2L_{bd}$. Globally, $(4R^2-1)^2\tilde{Q}_-(S),\tilde{Q}_+(S)\equiv(4R^2-1)Y_{par,+}(S)\mp2Rw_1((4R^2-1)S+2(3R^2-1)C_0)\equiv X_u^2-(C_0\pm2Rw_1)X_u+C_\mp^2$ ($X_u=(4R^2-1)u_0,C_\mp=RC_0\mp(2R^2-1)w_1=\mp\frac{(4R^2-1)^{(1\pm1)/2}den_\mp}{2\Lambda}$); on $\Lambda(4R^2-1)C_0den_-den_+D_{coll}\tilde{\nu}\ne0$, doubling $P_{coll,+}=(S_0,A_1(S_0))$ from $\infty_-$ on $F_4=\tilde{Q}_-\tilde{Q}_+\equiv A_1^2-4R^2w_1^2(C_0S+C_0^2+P)$ ($S_0=-\frac{C_0^2+P}{C_0},A_{coll}=(2R^2-1)C_0^2-P(4R^2-1),M_\mp=A_{coll}\mp C_0Rw_1,D_{coll}=\frac{M_-M_+}{4R^2-1}$) gives $S_{tan}(R,C_0)=S_{eq}+\frac{A_{coll}(C_0^2A_{coll}-D_{coll})}{(4R^2-1)C_0D_{coll}}$ ($(4R^2-1)\sqrt{\tilde{Q}_-},\sqrt{\tilde{Q}_+}=\frac{M_\pm}{C_0}\mp\frac{Rw_1C_0^2A_{coll}}{D_{coll}}$), while the $(0,\infty)$-secant $C_+W_--C_-W_+=(C_+-C_-)X_u$ ($\tilde{\mu}=R(R^2C_0^2-(2R^2-1)^2),\tilde{\nu}=(2R^2-1)(PC_0^2-(2R^2-1)^2)$) gives $u_{0,sec}(R,C_0)=\frac{\tilde{\nu}^2-\tilde{\mu}^2}{C_0(4R^2-1)\tilde{\nu}}$ ($S_{sec}=u_{0,sec}-C_0$, $(4R^2-1)\sqrt{\tilde{Q}_-},\sqrt{\tilde{Q}_+}=\frac{\tilde{\nu}^2-\tilde{\mu}^2+C_0C_\mp\tilde{\mu}}{C_0\tilde{\nu}}$), paying both $1+u_0c^*_-,1+u_0c^*_+=\square$ in $\mathbb{Q}(R,\Lambda)$. For $Q_i=(c_iU+m_i)^2+d_i,Q_j=(c_jU+m_j)^2+d_j$ ($\Delta m_{ij}=c_jm_i-c_im_j\ne0,\Lambda_{ij}=\frac{c_j^2d_i-c_i^2d_j}{c_j^2b_i-c_i^2b_j}\ne0$) on generically $g=1$ $\mathcal{E}_{ij}:\{Q_i=W_i^2,Q_j=W_j^2\}$, $c_jW_i-c_iW_j=\Delta m_{ij}$ uniquely selects $U_{2sq}(i,j)=\frac{1}{2}(\frac{d_i}{c_i^2\Lambda_{ij}}-\Lambda_{ij}-\frac{b_i}{c_i^2})$ ($Q_i,Q_j=\square$). On $(D_{bd},D_\tau)$ [5, 17, 28], this gives $a_{2sq}=\frac{C_0^2-4R^2(2R^2-1)^2}{8C_0R^2(2R^2-1)},b_{2sq}=\frac{P}{a_{2sq}},c_{2sq}=\frac{2C_0(C_0^2-4R^2)(2R^2-1)}{C_0^2-4R^2(2R^2-1)^2},d_{2sq}=4R^2a_{2sq},u_{0,2sq}=S_{2sq}+C_0$; substituting $C_0=C_0(R,\Lambda)$ gives $1+c_{2sq}d_{2sq}=C_0^2-4R^2+1=w_1(R,\Lambda)^2$ and an $18/21$ family over $\mathbb{Q}(R,\Lambda)$. By $(P_4-1)^2(1+e_-e_+)\equiv(x_i+x_j-x_k-x_l)^2+\prod_{\epsilon_1,\epsilon_2\in\{\pm1\}}(r_{ij}r_{kl}+\epsilon_1r_{ij}+\epsilon_2r_{kl})$, $q_{-+}(S_{2sq})\equiv M_{base}^2+\text{rem}_4\equiv M_{new}^2+\frac{(C_0^2-(4R^4+4R^2-1))(C_0^2-8R^2(8R^6-8R^4+6R^2-1))}{(C_0^2-4R^2(2R^2-1)^2)^2}\text{rem}_4$ ($M_{base}=a_{2sq}-b_{2sq}+c_{2sq}-d_{2sq},\text{rem}_4=\frac{(R^2C_0^2-(2R^2-1)^2)(P^2C_0^2-R^2(2R^2-1)^2)}{R^2}$) gives off-branch $19/21$ loci ($1+c^*_-c^*_+=\square$): $g=0$ Curve B$_\pm$ ($C_0=\pm\frac{2R^2-1}{R}$: $R=\frac{2m}{m^2+3},\Lambda_1=\frac{m(m+1)}{m-3},\Lambda_2=-\frac{m+3}{m(m-1)}$, $19/21$ at $(4/7,6)$, unpaid $\deg_m36,g_m=17$), $g=1$ Curve A$_\pm$ ($C_0=\pm\frac{R(2R^2-1)}{P}$, $(3,7/4)$), and $g=1$ Curve C ($w_1=\pm2R^2$, $(5/4,1/2)$) (exp89-90).
• (c.3) Root Coordinate $U$ on $\Lambda=w_1$, $10$-Syzygy & $6$-Pair Two-Square Atlases, and $\mathcal{C}_{coinc,r}$ (exp61-65,89-90): On $\Lambda=w_1$ in $U=\frac{4w_1(w_1^2-1)S+3w_1^4+14w_1^2-1}{(w_1^2-1)^2}=\pm\sqrt{1+u_0c^*_-}$ ($D_{V_4}=5w_1^4+10w_1^2+1$; $V_4$-transport swaps $c^*_-rightarrow c^*_+$), $D_{V_4}^2(1+u_0c^*_+)=(2(w_1^4-1)U+(w_1^2-1)(3w_1^2+1))^2+8w_1^2(w_1^2+1)(w_1^2+3)(3w_1^2+1)>0$ and $D_{V_4}^2(1+c^*_-c^*_+)=(4w_1(2(w_1^2+1)U+w_1^2-1))^2+(w_1^2-1)^2(25w_1^4+6w_1^2+1)>0$ ($1+c^*_-c^*_+=\square$ at $U=\frac{(w_1^2-1)(25w_1^4+6w_1^2+1-8w_1t-t^2)}{16w_1(w_1^2+1)t}$), descending at $U_{univ}(Z)$ on $\mathcal{S}_7^{univ}(q,V_4\cdot w_1(q))$ to $Z=w_1^2=\frac{(q^2-3)^2}{4q^2}$ with coprime irreducible $\deg_Z20$ numerators $P_+,P_{-+}\in\mathbb{Z}[Z]$ ($g_Z=9,g_{w_1}=g_{q^2}=19,g_q=39$). For free $S$ on $\Lambda=w_1$, all $\binom{5}{2}=10$ linear $U$-syzygies among $\{\Delta_{bd},\Delta_\tau,Q_+,Q_{-+},Q_-\}$ ($Q_-=((w_1^2-1)^2U)^2$) obey $c_k^2L_{ij}+c_i^2L_{jk}+c_j^2L_{ki}\equiv0$, $\frac{1}{U_{bd,\tau}}-\frac{1}{U_{-,+}}\equiv1$, $(3Z+1)^2D_{bd}(U_{-,+})-(Z-1)^2D_\tau(U_{-,+})\equiv\frac{2Z(11Z^3+27Z^2+25Z+1)^2}{(Z-1)^2(Z+1)}$, splitting into $3$ collisions ($U_{bd,\tau},U_{\tau,+},U_{-,\tau}$), $U_{-,bd}(w_1)=\frac{9w_1^4+22w_1^2+1}{2(3w_1^4+14w_1^2-1)}$ ($1+bu_0=\frac{(1-w_1^2)(15w_1^2+1)}{(8w_1)^2}=\square\mapsto$ rank-$0$ $E_{15,w_1}:Y^2=X(X^2-7X+16)$ via $X=\frac{4(1+w_1)}{1-w_1}$, closing $w_1\in\{0,\pm1\}$), and $6$ noncollision $Z$-sections ($U_{eq}$ [$g_Z=2,g_{w_1}=5,\Delta_{\tau_\kappa}<0$ for $Z>Z_*\approx1.9781$]; $U_{-,+}=U(S_{eq})$ [$g_Z=2,3;g_{w_1}=5,7$]; $U_{bd,-+},U_{\tau,-+},U_{+,-+},U_{-,-+}$ [$g_Z=3,g_{w_1}=7$]). On $U_+(Z,t)=\frac{2Z(Z+3)-(Z-1)(3Z+1)t-(Z+1)(3Z+1)t^2}{2(Z^2-1)t}$ ($Q_+=\square$, giving $U_{-,+}$, $U(S_0)=-\frac{Z(Z+3)}{(Z-1)^2}$), $U(S_{tan}(w_1))=\frac{2Z^2(Z+1)(Z+3)^2}{(Z-1)^2(5Z^3+3Z^2+7Z+1)}$ has $\deg_Z10,12,14$ ($g_Z=4,5,6;g_{w_1}=9,11,13$) while $U_{sec}(Z)=-\frac{4Z(Z-1)(Z+1)(Z+3)}{(5Z^2+10Z+1)(7Z^2+10Z-1)}$ ($t_{sec}=-\frac{7Z^2+10Z-1}{5Z^2+10Z+1}$) has uniform $\deg_Z10$ ($g_Z=4,g_{w_1}=9$). On the open nondegenerate chart, the $6$ $3$-gate sections $U_{2sq}(i,j)(Z)$ pay $Q_-,Q_i,Q_j=\square$ ($U_{2sq}(bd,\tau)=U(S_{2sq}(w_1))$ gives $19/21$ at $w_1=2$; $U_{2sq}(+,-+)$ pays $Q_-,Q_+,Q_{-+}=\square$), while full $21/21$ lifts require unpaid odd-multiplicity kernels of $\deg_Z=(30,18,34,26,30,24)$, $g_Z=(14,8,16,12,14,11)$, $g_{w_1}=(29,17,33,25,29,23)$ (finite by Faltings [6]; syndrome audit on $40$ $U_+$-fibers & $131$ Curve-B$_\pm$ points in exp89-90). Finally, $4rt(t^2-1)(g_{Euler,+}-f_{a,-})=P_{CaseA}(r,t^2)$ closes Case A, $t=\pm\frac{1}{4r^2-1}$ gives $1+f_{a,-}f_{c,-}=\square$ ($b=c$), and $R_4(a,d_-,f_{a,-},g_{Euler,+})=0$ splits into $t=\pm t_1(r)=\pm\frac{4r^2-1}{2(2r^2-1)}$ ($\mathcal{Q}_5^\star(r),\mathcal{S}_6^{refl}(r)$) and smooth $g=4$ $\mathcal{C}_{coinc,r}:F_{coinc,\pm}(r^2,t)=0$ ($r^2=\frac{(Y-2X-4)(Y+2X-4)}{64(X+4)},t=-\frac{4}{X}$) over $E_{coinc}:Y^2=X^3+8X^2+32X+80$, where $Q_{I},Q_{II}=\square$.
• (d) Weierstrass Cubic $\mathcal{E}_W(R^2,Z)$, QRT Translation [30], Halving Laws, Shioda Height $\hat{h}(P_\infty)=1/2$ [31] & Towers $\mathcal{S}_7^{osc,(n)},\mathcal{S}_7^{can,(n)},\hat{\mathcal{S}}_7^{mir,(n)}$ (exp81-88): In $u=\Lambda b$ and $(X_W,Y_W)=(c_3u,c_3Y_u)$ ($c_3=-2Z(2R-1)(Z-4)$), $E_{master}(\Lambda,b,d_-)=0$ is birational to $\mathcal{E}_W(R^2,Z):Y_W^2=X_W^3+A_WX_W^2+B_WX_W$ ($A_W=-Z((8R^2-5)Z-32R^2+16),B_W=4Z^2P(4R^2-1)(Z-4)^2=X_W(P_\infty)X_W(Q_0),T_0=(0,0),P_\infty=(4P(Z-4)^2,16\tau_\kappa P(Z-4)^2),Q_0=((4R^2-1)Z^2,-4\tau_\kappa(4R^2-1)Z^2)$), where $\mathcal{I}_b=\ominus$, $(brightarrow d_-)$ reflects through $P_\infty$, giving QRT translation [30] $Q_{n-1}=[n]P_\infty\oplus T_0$, with $D_Z=X_W(Q_0)-X_W(P_\infty),X_W(Q_1)=\frac{64(Z+4P)X_W(P_\infty)X_W(Q_0)}{D_Z^2},N_{W,1}+D_{Z,2}=64(Z+4P)D_Z,N_{W,1}-D_{Z,2}=2N_{W,2}$, even multiplier $\mu_n(R^2,Z)=-\frac{X_W(P_\infty)}{X_W(Q_{n-2})}\frac{X_W(Q_{n-1})-X_W(Q_0)}{X_W(Q_{n-1})-X_W(P_\infty)}$ ($\mu_2=\frac{N_{W,1}}{D_{Z,2}}$ splitting on $R_{Pell},R_{HPell}$), $2$-descent $\frac{X_W(Q_{m+1})}{X_W(Q_0)}=P^{((m+1)\bmod2)}\mathcal{W}_m^2,\Lambda_n=-\frac{(2R+1)Z}{2(Z-4)}\mathcal{W}_{m_{ev}}\mathcal{W}_{m_{od}},b_n=P\frac{\mathcal{W}_{m_{ev}}}{\mathcal{W}_{m_{od}}},d_{-,n}=\frac{\mathcal{W}_{m_{od}}}{\mathcal{W}_{m_{ev}}},\frac{b_n}{d_{-,n}}=(\frac{X_W(Q_{n-1})}{X_W(Q_n)})^{(-1)^n}$ adding $+1+8$ squares to $\mathcal{U}_{12}^{pre}$, and $T_0$-invariant halving steps $\mathcal{W}_{2k-2}=-\frac{4(Z-4)Y_W(Q_{k-1})}{X_W(Q_{k-1})^2-B_W},\mathcal{W}_{2k-1}=-\frac{X_W(P_\infty)(X_W(Q_k)-X_W(Q_{k-1}))}{X_W(Q_{k-1})X_W(Q_k)-B_W}$ (exp88). At $Z=8(R+1)=16r^2$, $\mathcal{E}_{\min}^{can}(0):y^2=x^3+3x^2+x$ ($\Delta=80$) is smooth and $P_\infty(0)=(-1,-1)$ has order $4$ with $[2]P_\infty(0)=(0,0)=T_0(0)$, so $Q_1(0)=\mathcal{O}$ and $z(\ominus Q_1)=\frac{R}{8(R^2-R-1)}$ ($\omega_u=-64\frac{du}{2Y_u}$) has $\log_{\mathcal{E}}(z(\ominus Q_1))=8R-16R^3-\frac{8}{3}R^4+O(R^5)\ne0$; with $k_n=(-1)^{n-1}\lfloor\frac{n+1}{2}\rfloor$, $Q_{n-1}(0)=[n+2]P_\infty(0)$ cycles mod $4$ through $(-1,1),\mathcal{O},(-1,-1),T_0(0)$ ($Q_{n-1}\in\hat{\mathcal{E}}$ for $n\equiv2\bmod4$, $T_0\oplus\hat{\mathcal{E}}$ for $n\equiv0\bmod4$), giving $\Lambda_n^{can}=-\frac{1}{k_nR}+O(1)$ ($n\equiv1,2\bmod4$) vs. $k_nR+O(R^2)$ ($n\equiv3,0\bmod4$) and exchanging $(c^*_{-,n},c^*_{+,n})=(-8k_n(k_n-1),-8k_n^2)R^2+O(R^3)$ ($n\equiv1,2\bmod4$) with $(-8k_n^2,-8k_n(k_n-1))R^2+O(R^3)$ ($n\equiv3,0\bmod4$) while $\{u_{\pm1,n}-u_{0,n}\}=\{2(1-1/k_n)+O(R),2+O(R)\}$ (exp83,88). Here $n=1$ is degenerate ($b_1^{can}=\frac{P}{R},d_{-,1}^{can}=R,c^*_-=g_{-1}=0$), while every $n\ge2$ gives non-degenerate $\ge18/21$ $\mathcal{S}_7^{can,(n)}(r)$, $\mathcal{U}_{14}^{can,(n)}(r)$ ($\ge34..35/91$ on $\mathbb{Q}(r),5r^2-1=\rho^2$; exact at $n=2,3$), and duals $\hat{\mathcal{S}}_7^{mir,(n)}(w)=-\mathcal{S}_7^{can,(n)}(-R)|_{R=1-2w^2}$ ($T_*=2R,Z_{mir}=16w^2$, $\ge18/21$ & $\ge34..35/91$; exp84). Rescaling $(X_W,Y_W)=(64x,512y)$ gives the minimal K3 surface $\mathcal{E}_{\min}^{can}(R)$ ($\chi=2$, fibers $I_2(1),I_2^*(-1),I_2(1/2),I_6(-1/2),3I_1,I_3(\infty)$) where $\hat{h}(P_\infty)=1/2$ [31] gives for all $n\ge1$ $\deg_R(den(\Lambda_n^{can}))=\binom{n+1}{2}$ and $\deg_R(num(\Lambda_n^{can}))=\binom{n+1}{2}-\mathbf{1}_{n\equiv0,2\pmod3}$ ($1\le n\le12$ in exp88), while at $n\in\{2,3,4,5\}$ gate degrees $(12,16,26),(24,28,50),(36,44,80),(60,64,122)$ have genera $g(\mathcal{C}_{-,n})=(5,11,17,29),g(\mathcal{C}_{+,n})=(7,13,21,31),g(\mathcal{C}_{-+,n})=(12,24,39,60)$, finite by Faltings [6], and $\mathcal{B}_{+,i}^{(n)}-\mathcal{B}_{-,i}^{(n)}=2(X_W(Q_{n-1})X_W(Q_n)+B_W)$ keeps $\Delta_\mp(R,\Lambda_n)=0$ disjoint ($n\ge1$; exp87). Proof. Verified in verify_paper.py, exp25-89, and THEOREMS_AND_PROOFS.md. $\square$

10. Limitations and scope

• No rational septuple ($21/21$) is claimed. $\mathcal{S}_7^{(20/21)}$ adds type $(|\mathcal{R}_4|,|\mathcal{R}_5|)=(5,2)$ ($\dim=1,gens=2$) [2–4], and larger pools ($\mathcal{U}_{4N+3..5}^{Cheb},\hat{\mathcal{U}}_{8N+2..6}^{lat},\mathcal{U}_{10..16}^{osc},\hat{\mathcal{U}}_{15..16}^{osc},\mathcal{U}_{14}^{can,(n)},\hat{\mathcal{U}}_{14}^{mir,(n)},\mathcal{U}_{8,11}^{swap}$) organize squares without raising $m=7$ compatibility.
• Necessary-condition reductions & Faltings vs. Problem 3.3: Theorem 3(4) proves $\mathcal{C}_+(a,\lambda)=0$ on Branch 1 has $g=3$ cover $\mathcal{H}$ (finite by Faltings [6]), leaving $\mathcal{S}_{4,2},\mathcal{E}_{Br1}$ open; Faltings finiteness of the unpaid gates at $U_{univ}(Z)$ on $\mathcal{S}_7^{univ}(q,V_4\cdot w_1(q))$, of Curve B$_\pm(m)$, $U(S_{tan}(w_1))$, $U_{sec}(Z)$, the $6$ $3$-gate sections $U_{2sq}(i,j)(Z)$ ($g_{w_1}=17..33$), and the $6$ linear $U$-syzygy sections (plus rank-$0$ $E_{15,w_1}$-closed $U_{-,bd}$) on $\Lambda=w_1$ (exp89), and of $n\in\{2..5\}$ canonical gate curves are necessary-condition reductions not enumerating points or proving remaining lifts ($\tau_\kappa^2-4P$, $(S,5S-1)$, $(Z,Z+4P)$, $((V_*+4R)^2-16,T_*V_*)=\square$ with distinct nonzero entries).
• Completeness on $\mathcal{H}$ & unpaid edges: Faltings is ineffective; our $50,000$-height search on $\mathcal{E}^{(1)}$ and u/kiro_dioph7's $295$-prime sieve (Post 35) are bounded, while [23] proves $\mathcal{H}(\mathbb{Q})=\{(\pm1,\pm16),(\pm2,\pm55),\infty_\pm\}$ via Quadratic Chabauty. Curve II has $\deg 36,32$; Theorem 5 closes $Q_4\equiv-2$ ($k\in\{2,3\}$ [16]); $z=z_{L_1}$ leaves $ac+1,\mathcal{F}_{ab},\mathcal{G}_d,H=\square$; $\mathcal{C}_{triad}$ closes $1+d_+b'_{ac}(d_-)=\square$ via $E_{15}$; Theorem 7 closes $H=1/3$ via $\mathbb{Q}(\sqrt{-2})$ $2$-descent, $u_w=2u_d$ via Pythagorean Syzygy, $(u_w-2u_d)^2+u_d^2=1$ via $E_{coll}$, and [20–22] give $23/28$ octuples ($E_{24}$, Orbit-I $2$-Selmer/Chabauty [24, 25], $\mathbb{Q}_{23}$ at $q=\pm5/16$, Post 69 on $(0,q_{H,1}]\cup[q_{H,2},q_{B,+})$); $\kappa_2,\kappa_3$ give $22/28$.

11. Reproducibility

verify_paper.py and verify_jac_selmer.py in supplement.zip (Python 3/SymPy) certify Theorems 1–7 (THEOREMS_AND_PROOFS.md, exp9-89):

• Suite: python3 -I verify_paper.py runs check_theorem_1()–check_theorem_7(), evaluate_S42_point, and verify_exp39()–verify_exp89() in $495s$.
• Scripts: python3 -I exp78*.py–exp89*.py run in $5.0,7.8,7.0,2.0,2.9,31.3,8.8,39.6,16.5,2.0,38.7,34.5s$.

References

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