1. The splitting problem on the Kummer chart

The norm-and-anchor surface of the determinant-zero Diophantine chart in [1,2] is birational to

$$ \tau^2f(A)=f(B),\qquad f(x)=x(x-1)(x-2). $$

It is the Kummer surface attached to $E\times E$, where $E:y^2=f(x)$, identified with a Kuwata–Shioda fibration in [2,3]. A necessary further condition for the chart is that

$$ \Delta(A,B)=-(B-2A)B(B-1)(A-2)(AB-3A+2) \tag{1} $$

be a rational square. It is a splitting condition, not the conjunction of all twenty-one edge conditions of a septuple. We classify genus-one sources on every common quadratic twist of $E$, certain higher-genus sources on $E$, and sections after the first halving-conic base change of the fibration in [2]. The elliptic-source and half-conic classifications meet at the same degenerate curve.

The rational inverse coordinates of the chart are

$$ \begin{aligned} r&=\frac{A^2B+AB^2-6AB+2A+2B} {A^2B^2-3A^2B-AB^2+4AB+2A-2B},\\ u&=\frac{A^2B+AB^2-6AB+2A+2B} {-A^2B+AB^2-4AB+6A+2B-4}. \end{aligned} \tag{2} $$

On the open chart the anchor is $a^2=(r^2-1)(1-u^2)$, the splitting discriminant is $Z^2-4$ with $Z=r+1+(r-1)u$, and (1) represents its squareclass after the norm and anchor have been imposed. We exclude the same zero-anchor, repeated-entry and denominator boundaries as [1,2]. All identities involving those boundaries below are checked directly, rather than inferred from birational maps at their poles.

Theorem 1 (elliptic sources on common twists). Let $d\in\mathbb Q^*$ and $E^{(d)}:d y^2=f(x)$. Let $C/\mathbb Q$ be a smooth projective geometrically integral curve of genus one, with nonconstant morphisms $\alpha,\beta:C\to E^{(d)}$ over $\mathbb Q$. Put $A=x\circ\alpha$ and $B=x\circ\beta$. If $\Delta(A,B)$ is a square in $\mathbb Q(C)$, then

$$ B=A\quad\hbox{or}\quad B=\frac{A-2}{A-1}. \tag{3} $$

Both identities make $\Delta$ a square, and both yield only degenerate tuples in the original chart. No rational point on $C$ is assumed. The degrees of the morphisms and the rank of $E^{(d)}(\mathbb Q)$ are unrestricted; in particular non-torsion translations are allowed.

For higher-genus sources, say that the induced maps are commensurable over $\mathbb Q$ when the homomorphisms $\operatorname{Jac}(C)\to E$ induced by $\alpha,\beta$ span a rank-one subgroup of $\operatorname{Hom}_{\mathbb Q}(\operatorname{Jac}(C),E)$. This is a condition on the actual maps, not merely on the geometric isogeny factors of the Jacobian.

Theorem 2 (untwisted low-genus commensurable sources). Let $C$ have genus at most four, and let $\alpha,\beta:C\to E$ be nonconstant maps to the untwisted curve. Assume their induced maps are commensurable over $\mathbb Q$. If $\Delta(A,B)$ is square in $\mathbb Q(C)$, their image lies on one of

$$ B=A,\qquad B=\frac{A-2}{A-1},\qquad B=\frac2A. \tag{4} $$

The third curve requires source genus at least three. Every rational splitting point on it also lies outside the nondegenerate chart. Thus none of these image curves contributes a rational septuple in the chart.

This does not classify all low-genus curves on the Kummer surface. Its inverse image in $E\times E$ can have larger genus, and higher-genus sources can have noncommensurable maps. Rational points on a Kummer curve need not lift rationally to its double cover; the rational exclusion in Theorem 2 is made on the image curve itself.

For the other classification, write $K=\mathbb Q(c)$ and

$$ \mathcal E:\ Y^2=X^3+(c^4-18c^2+1)X^2+64c^4X, $$

with

$$ G=(8,24(c^2-1)),\qquad T=(8c^2,-8c^2(c^2-1)). $$

Paper [2] proves $\mathcal E(K)=\mathbb ZG\oplus(\mathbb Z/4)T$. Let $L=K(s)$ with $s^2=c^2+6c+1$; it is a rational function field, parametrized by

$$ c=\frac{6-2t}{t^2-1},\qquad s=1+tc. $$

The quartic coordinate $r$ of [2] defines

$$ \mathcal B(P)=(r(P)+1)\big[c^2(r(P)-1)+2\big]. \tag{5} $$

Squarehood of (5) is the same additional splitting condition in this fibration.

Theorem 3 (the complete first half-conic). There is a section $H\in\mathcal E(L)$ satisfying $2H=G-T$, and

$$ \mathcal E(L)=\mathbb ZH\oplus(\mathbb Z/4)T. $$

Exactly six sections make $\mathcal B$ a square, allowing zero:

$$ -G,\quad T,\quad G+2T,\quad2G-T,\quad H,\quad H+2T. \tag{6} $$

All six are degenerate in the original chart. The conic adds two splitting sections, but neither adds a nondegenerate family.

Prior work and distinction

The norm-and-anchor chart, alternate fibration and its base-field section group are results of [1,2]. The Kummer surface and its elliptic fibration are published constructions of Kuwata–Shioda [3]; no new surface is claimed. Standard morphism and isogeny theory of elliptic curves is used as in [4]. The specialization argument in Section 5 follows the two-descent strategy of Gusić–Tadić [5], with its precise one-rational-two-torsion formulation proved here.

The new ingredients are an intersection-multiplicity bound for both varying Kummer coordinates, an exact classification of rational translation candidates uniform in the common quadratic twist, a genus bound for splitting covers, descent from genus-one torsors without a rational origin, and a real elliptic-logarithm obstruction on every section of the half conic. The finite remainders are exhaustive because of these bounds. They are not height-limited searches. Theorem 3 goes beyond [2] by determining the entire new section group and addressing squarehood newly created by the quadratic extension. The shared curve in (3) explains the relation between the two constructions.

2. Reduction to coprime multiplication maps

The change $x\mapsto x-1$ identifies $E$ with $y^2=x^3-x$. Its rational group is $E[2]$, consisting of $O,T_0,T_1,T_2$ with $T_i=(i,0)$. This is the rank-zero congruent-number curve for $1$; the supplement records the exact rank upper bound and full torsion computation. Its geometric endomorphism ring is $\mathbb Z[i]$, and $\operatorname{End}_{\mathbb Q}(E)=\mathbb Z$.

Every quadratic twist also has rational endomorphism ring $\mathbb Z$, although its rational rank need not vanish. For a genus-one $C$ with a nonconstant map to $E^{(d)}$, its Jacobian is $\mathbb Q$-isogenous to $E^{(d)}$, so the corresponding rational homomorphism group has rank one. In Theorem 2 use $d=1$. In either theorem write the two induced maps as $gm\psi,gn\psi$, with $g>0$ and coprime nonzero integers $m,n$. Choose integers $u,v$ such that $um+vn=1$, and form the actual $\mathbb Q$-morphism

$$ \gamma=u\alpha+v\beta:C\longrightarrow E^{(d)}. $$

It is nonconstant. The maps $\alpha-m\gamma$ and $\beta-n\gamma$ have zero induced homomorphism, hence are geometrically constant. A geometrically constant $\mathbb Q$-morphism from a geometrically integral curve has value in $E^{(d)}(\mathbb Q)$. Consequently

$$ \alpha=m\gamma+U,\qquad\beta=n\gamma+V, \qquad U,V\in E^{(d)}(\mathbb Q). \tag{7} $$

This argument neither chooses a rational source origin nor assumes equivariance under a chosen source involution. Negating either entire target map leaves its $x$-coordinate unchanged and replaces its translation by its negative. We may therefore take $m,n>0$. Only in the untwisted case are $U,V$ necessarily two-torsion.

When $C$ has genus one, $\gamma$ is étale. An odd valuation on the elliptic target remains odd after pullback by $\gamma$, and so obstructs a square on $C$. For arbitrary genus, squarehood means that $C$ maps to the normalization of the corresponding splitting double cover. These two consequences are the only descent facts needed; a nonsquare with even divisor need not remain nonsquare after an isogeny.

3. The uniform contact and genus bounds

Over algebraic constants identify the twist with $E$ and put

$$ R=[m]P+U,\quad S=[n]P+V,\quad A=x(R),\quad B=x(S), \quad F=B-2A,\quad C_0=AB-3A+2. $$

First, (1) cannot vanish identically on this elliptic image. Its nonconstant coordinates rule out $B=0,1$ and $A=2$. Each remaining possibility, $B=2A$ or $B=(3A-2)/A$, has exactly three shared branch points between the two double covers defined by $f(A)$ and $f(B)$. Their full $V_4$ composite has genus two by Riemann–Hurwitz, and cannot be dominated by an elliptic curve. This handles the zero function before taking its divisor.

The map $P\mapsto(R,S)$ is injective because $E[m]\cap E[n]=\{O\}$. The factors $B,B-1,A-2$ have even divisors. Therefore any simple zero of $F$ away from $C_0=0$ gives an odd valuation of (1). Nothing in the following contact count requires $U,V$ to be torsion.

The pole divisors of $A$ and $B$ have degrees $2m^2$ and $2n^2$. They share at most one point. In an invariant local parameter at a shared pole, the leading coefficient of $F$ is proportional to $1/n^2-2/m^2$, which is nonzero. Thus

$$ \deg\operatorname{div}_0(F)\geq2(m^2+n^2-1). \tag{8} $$

Away from two-torsion, use $A$ as a local coordinate. Pullback of the invariant differential gives

$$ \left(\frac{dB}{dA}\right)^2f(A)=\frac{n^2}{m^2}f(B). $$

At a repeated zero of $F$, $B=2A$ and $dB/dA=2$. Since $f(2A)/f(A)=4(2A-1)/(A-2)$, this forces

$$ A_* =\frac{n^2-2m^2}{2n^2-m^2}. \tag{9} $$

For nonzero integer multipliers it is finite and avoids $0,1,2$. The derivative sign condition fixes $y(S)=(2m/n)y(R)$, leaving at most two ordered pairs $(R,S)$ above $A_*$ and hence at most two source points. Differentiating the differential identity once more gives

$$ \frac{d^2B}{dA^2}(A_*) =-\frac{(2n^2-m^2)^2}{3m^2n^2}\ne0. \tag{10} $$

Thus all smooth repeated zeros have multiplicity two, contributing at most four.

At branch points, the only branch/branch pairs satisfying $B=2A$ are $(A,B)=(0,0),(1,2)$. Injectivity gives at most one source point for each. The local quadratic coefficients are proportional to $n^2-2m^2$ and $-2n^2-2m^2$, respectively; both are nonzero. Their contribution is at most four. If just one coordinate is at a branch point, $F$ has a simple zero. Such a zero is not shared with $C_0$, because $2A^2-3A+2$ vanishes at none of the relevant real branch values, and already supplies an odd valuation.

Finally, a finite zero of $F$ shared with $C_0$ has

$$ 2A^2-3A+2=0. $$

The two nonreal roots differ from (9) and from the branch values of both coordinates. Each gives at most four ordered pairs $(R,S)$, so at most eight source points, all simple zeros of $F$.

All zeros not supplying an uncancelled odd valuation have therefore been accounted for with total degree at most

$$ 4+4+8=16. \tag{11} $$

If the splitting function has even divisor on $E$, (8)–(11) imply $m^2+n^2\leq9$. The only positive coprime pairs are $(1,1),(1,2),(2,1)$.

More generally, let $r_\Delta$ be the number of geometric points at which the splitting function has odd valuation. The same count proves

$$ r_\Delta\geq2(m^2+n^2-9). $$

For a connected double cover of $E$, Riemann–Hurwitz gives

$$ g_\Delta=1+\frac{r_\Delta}{2}\geq m^2+n^2-8. \tag{12} $$

Thus a source of genus at most four can only involve $m^2+n^2\leq12$, adding $(1,3),(3,1)$ to the previous list. The finite remainder follows from a proved degree bound.

4. Exact remainder and rational degeneracies

4.1. Rational translations on every common twist

Each of the three remaining genus-one slopes has a coefficient one. Translating the source coordinate over $\mathbb Q$, we can put the maps in one of the forms

$$ (A,B)=(x(P),x(P+W)),\quad(x(P),x(2P+W)),\quad(x(2P+W),x(P)), $$

with $W\in E^{(d)}(\mathbb Q)$. In particular $x(W)\in\mathbb Q$ when $W\ne O$.

At a shared zero of $F,C_0$, let $A=q$, $B=2q$, where $2q^2-3q+2=0$. On the geometric untwisted model write

$$ P_0=(q,h),\quad h^2=f(q),\qquad Q_0=(2q,\pm2q^2/h). $$

The latter formula uses $f(q)f(2q)=4q^4$. Exact point addition gives these minimal polynomials for the translation abscissa:

| slope and translation | first ordinate sign | second ordinate sign | |---|---|---| | $(1,1)$, $W=Q_0-P_0$ | $x-1$ | $x+7$ | | $(1,2)$, $W=Q_0-2P_0$ | $x^2-3x+4$ | $14641x^2-17379x+5476$ | | $(2,1)$, $W=P_0-2Q_0$ | $2x^2-3x+2$ | $2738x^2-17379x+29282$ |

The four quadratics have negative discriminants $-7$ or $-7\cdot1633^2$ and are irreducible over $\mathbb Q$. Conjugating $q$ and changing both ordinate signs accounts for all intersection points.

For either doubled slope and $W\ne O$, the pole sets are disjoint and $\deg\operatorname{div}_0(F)=10$. The repeated smooth and branch zeros account for degree at most eight, so an odd zero must be shared with $C_0$. The table contradicts rationality of $x(W)$. For $W=O$, the doubled-map formulas below have an odd nonbranch divisor, independently of the twist.

For slope $(1,1)$ and $W\ne O$, the zero degree is four. A repeated smooth zero has $A=-1$, $B=-2$, and the second ordinate twice the first. Addition gives $x(W)=-48$; the two ordinate choices give opposite distinct translations, so a fixed $W$ allows at most one double zero. A branch-pair zero requires $W=O$ or $T_0$. Thus any fixed nonzero $W$ has at most one double zero in total, leaving an odd zero to be shared with $C_0$. The table leaves only $x(W)=1,-7$.

The first is $W=T_1$. For the second write $W=(-7,w)$ with $w^2=-504$ over algebraic constants. Addition gives

$$ B=\frac{f(A)-504-2wy}{(A+7)^2}+10-A,\qquad y^2=f(A). $$

Taking norms under $y\mapsto-y$ yields

$$ \begin{aligned} N(F)&=\frac{(2A^2-3A+2)(2A^2+45A+2)}{(A+7)^2},\\ N(C_0)&=\frac{(2A^2-3A+2)(50A^2-147A+98)}{(A+7)^2}. \end{aligned} $$

The norms of $B$ and $B-1$ are squares. Consequently the norm of $\Delta$ has squareclass

$$ (2A^2+45A+2)(50A^2-147A+98). $$

This polynomial is squarefree even over algebraically closed constants. Thus $\Delta$ has an odd divisor on the elliptic curve. The obstruction persists under every étale pullback, including on twists where $W$ is rational. Only $W=O,T_1$ remain, giving (3).

4.2. Untwisted higher-genus remainder and reconstruction

The formulas

$$ x(2P)=\frac{(z^2-2)^2}{4z(z-1)(z-2)},\quad x(P+T_0)=\frac2z,\quad x(P+T_1)=\frac{z-2}{z-1},\quad x(P+T_2)=\frac{2(z-1)}{z-2} $$

give an exact check of the untwisted low cases, including the $W=O$ doubled remainders just used. For two-torsion translations, when $m$ is odd translate the source to make $U=O$; when $n$ is odd one may instead make $V=O$. The connected splitting-cover genera are:

| $(m,n)$ | $O$ | $T_0$ | $T_1$ | $T_2$ | |---|---:|---:|---:|---:| | $(1,1)$ | split | 3 | split | 5 | | $(1,2)$ | 9 | 10 | 10 | 9 | | $(2,1)$ | 10 | 9 | 9 | 10 | | $(1,3)$ | 17 | 19 | 21 | 21 | | $(3,1)$ | 17 | 19 | 21 | 21 |

Each entry is obtained by factoring (1) over $\mathbb Q(z)$. Every factor of odd exponent coprime to $z(z-1)(z-2)$ contributes twice its degree to $r_\Delta$ on $E$. Valuations at the three finite branch points and infinity are even. The supplement records the actual factors, and also verifies all translate labels before normalization. In particular every excluded genus-one case has an odd divisor, which survives all étale pullbacks.

The two split cases are

$$ \Delta(A,A)=A^2(A-1)^2(A-2)^2, $$$$ \Delta\left(A,\frac{A-2}{A-1}\right) =\frac{(A-2)^2(2A^2-3A+2)^2}{(A-1)^4}. $$

Direct reconstruction through (2), for both signs of the splitting root, gives zero entries on the diagonal and repeated entries on its $T_1$ translate. For reference, the diagonal has

$$ r=\frac{2(A-1)}{A(A-2)},\qquad u=-\frac{A(A-2)}{2(A-1)}. $$

These are identities on the image curves, so further pullback cannot remove the degeneracy. This proves Theorem 1.

For Theorem 2, the only extra image is $B=2/A$. Here

$$ \Delta\left(A,\frac2A\right) =\frac{4(A-2)^2}{A^4}\,A(A-1)(A+1)(3A-4). $$

The quartic

$$ w^2=A(A-1)(A+1)(3A-4) $$

maps by $X=4/A$, $Y=4w/A^2$ to

$$ Y^2=(X^2-16)(X-3). \tag{13} $$

The conductor-56 curve (13) has rank zero and rational group $\{O,(-4,0),(3,0),(4,0)\}$. Its exact rank upper bound and torsion are recorded in the supplement. Hence the quartic has only parameters $A=0,-1,4/3,1$. There are no rational points at infinity because its leading coefficient is $3$. The square-scaling exceptions $A=2,\infty$ are separately on the anchor or Kummer boundary. At $A=-1$ and $4/3$, equation (2) gives $(r,u)=(-1,1)$, so the anchor vanishes. This excludes rational splitting points on the remaining image curve and proves Theorem 2.

The distinction between a rational image point and a rational point on its inverse image in $E\times E$ is essential here. The latter might not exist even when the former does. The rank-zero calculation (13) treats the former directly.

5. The full group after the halving-conic base change

Set $D=c^2+6c+1$ and $L=\mathbb Q(c,s)$, $s^2=D$. We first prove that the $D$-twist of $\mathcal E$ has rank zero over $K=\mathbb Q(c)$.

Specialization lemma. Let

$$ F:\ y^2=x^3+a(t)x^2+b(t)x, \qquad a,b\in\mathbb Z[t],\quad b(a^2-4b)\ne0. $$

Let $t_0$ be a good rational fibre. Suppose every nonconstant squarefree divisor in $\mathbb Z[t]$ of $b$ or $a^2-4b$ takes a nonsquare value at $t_0$, and $(a^2-4b)(t_0)$ is nonsquare. Then specialization $F(\mathbb Q(t))\to F_{t_0}(\mathbb Q)$ is injective.

Proof. The two-isogeny descent map is $x$ modulo squares, with its standard values at $O,(0,0)$; its image has squarefree representatives dividing $b$. For a point in the specialization kernel, such a representative evaluates to a nonzero square: near $O$, $x$ has even pole order and square leading coefficient. The hypothesis therefore makes the representative constant, and that constant is square. The point is the image under the dual isogeny of a point $Q$ on

$$ F':y^2=x^3-2ax^2+(a^2-4b)x. $$

The specialization of $Q$ is $O$ or the global kernel point $(0,0)$; translating by that point makes it $O$ without changing its image. Applying the same descent argument on $F'$ writes $Q$ as the image of a point $R$ under the original isogeny. The initial point is $2R$. Since the specialized curve has exactly one nonzero rational two-torsion point and it lifts globally, we may subtract its lift to put $R$ in the specialization kernel. Thus the kernel is two-divisible. It is a torsion-free subgroup of a finitely generated Mordell–Weil group, and so is zero. Here torsion injectivity is at a good characteristic-zero fibre. $\square$

For the twist in question,

$$ a=D(c^4-18c^2+1),\qquad b=64c^4D^2, $$$$ a^2-4b=D^3(c-1)^2(c+1)^2(c^2-6c+1). $$

At $c=19$ all 42 signed nonconstant squarefree-divisor tests are nonsquares. The enumeration includes the prime $2$ in the content of $b$; omitting content would be insufficient. The discriminant factor is also nonsquare. The specialized twist is

$$ y^2=x^3+58940224x^2+1889767097344x. $$

PARI's exact rank computation returns $[0,0,2,[]]$, in particular rank upper bound zero. This computation, including its Cassels-pairing step, is an explicit dependency; a first two-isogeny Selmer bound alone is not being asserted to give zero. The lemma proves twist rank zero over $K$, and the quadratic-twist decomposition therefore gives rank $\mathcal E(L)=1$.

Parametrize $L=\mathbb Q(t)$ as above. At $t=0$ we have $c=-6$ and the curve

$$ Y^2=X^3+649X^2+82944X. $$

Its good reductions have 12 points over $\mathbb F_{11}$ and 20 over $\mathbb F_{13}$. Hence the rational torsion over $L$ has order dividing four, and the known $T$ gives exactly $(\mathbb Z/4)T$.

To specify $H$, use the original coordinates $x=X-4(c^2+1)$ of [2]. Let $F_c(x)$ denote the cubic in that model, and $(x_0,y_0)=G-T$. Set

$$ x_H=2c^3+2c-4+2c(c-1)s, $$$$ y_H=\frac{-F_c(x_H)+\tfrac12F_c'(x_H)(x_H-x_0)}{y_0}. $$

Exact substitution proves $2H=G-T$.

The nontrivial deck involution $\sigma$ acts trivially on the rank-one rational Mordell–Weil space, since it fixes the nontorsion section $G$. Thus $\sigma(P)-P$ is torsion for every $P\in\mathcal E(L)$. All rational torsion is already $K$-rational, so this difference is simultaneously fixed and negated by $\sigma$. It follows that $2P$ is $K$-rational. Write $2P=nG+jT$. Since $G=2H+T$,

$$ 2(P-nH)=(n+j)T, $$

and therefore $P-nH$ is torsion. This proves the full group, including saturation at all primes.

6. A real-angle bound on all odd sections

For use in (5), the inverse quartic coordinate in the original $x,y$ model is

$$ r=\frac{2(k+1)[y-(k-3)x+4(k^2+3)]} {x^2+16(k^3-3k^2-k-1)},\qquad k=c^2. $$

The chart coordinate $u$ is $[k(r-1)-(r+1)]/[k(r-1)+(r+1)]$. These are the maps of [2].

Even multiples

Even multiples of $H$ are $K$-rational, but this alone does not prevent new squarehood over $L$. The all-integer residue-and-jet proof in [2] at $k=2$ has nonzero residue values 12 and 24 over $K_0=\mathbb Q(\sqrt2)$. Both remain nonsquare after adjoining $\sqrt D$: neither is square in $K_0$, and

$$ \operatorname{Norm}_{K_0/\mathbb Q}(12/D)=-16/7, \qquad \operatorname{Norm}_{K_0/\mathbb Q}(24/D)=-64/7. $$

Every zero-residue case in that proof has odd valuation unless it is one of $-G,T,G+2T,2G-T$. The extension is unramified at $k=2$ because $D$ is a nonzero unit there, so those odd valuations persist. The four exceptions make $\mathcal B$ identically zero. This proves precisely the even part of (6), including potential values that are $D$ times a $K$-square.

Real elliptic logarithms

For $1<c\leq\sqrt2$, put $a=c^4-18c^2+1$, $b=64c^4$. Since

$$ a^2-4b=(c^2-1)^2(c^2+6c+1)(c^2-6c+1)<0, $$

the real curve is connected. Define

$$ I_c(v)=\int_v^\infty\frac{dw}{\sqrt{w^4+aw^2+b}}, \qquad \Omega_c=2I_c(0). $$

For $Y<0$ the real elliptic-logarithm angle is $\theta=I_c(\sqrt X)/\Omega_c\in(0,1/2)$; for $Y>0$ use $1-I_c(\sqrt X)/\Omega_c$, and set $\theta(O)=0$. Addition of points adds angles modulo one.

The section $T$ has angle $1/4$. On the positive branch $s=\sqrt D$, the displayed half satisfies

$$ X_H-X_T=2c(c-1)(c-1+s)>0, $$$$ Y_H=-2c(c-1)s[(c-1)s+(c+1)^2]<0. $$

Consequently $0<h:=\theta(H)<1/4$. Since $X_{2H}=4c^2<X_T$ and $Y_{2H}=-4c^2(c^2+1)<0$, we also have $h>1/8$. At $c=\sqrt2$, exact group arithmetic gives order 16 for $H$, forcing $h=3/16$.

At the other endpoint write $\epsilon=c-1\to0^+$. Then

$$ w^4+aw^2+b=(w^2+a/2)^2+\eta^2, \qquad \eta=\frac{\sqrt{4b-a^2}}2\sim4\sqrt2\,\epsilon. $$

The minimum is at $w_0=\sqrt{-a/2}\to\sqrt8$, and $\sqrt{X_H}-w_0=O(\epsilon)$. Splitting the integral at a fixed small neighbourhood of $w_0$ gives

$$ I_c(0)=\frac1{\sqrt8}\log(1/\epsilon)+O(1), $$$$ I_c(\sqrt{X_H})=\frac1{2\sqrt8}\log(1/\epsilon)+O(1). $$

Indeed the local comparison integrand is $(4w_0^2(w-w_0)^2+\eta^2)^{-1/2}$. Outside an $O(\epsilon)$ core, its relative error is $O(|w-w_0|)$ and has bounded integral; inside the core rescaling $w-w_0=\epsilon z$ gives a bounded error. The first integral crosses both sides of the minimum; the second starts within $O(\epsilon)$ of it and includes one logarithmic side. The complements have bounded integrals. Thus $h\to1/4$.

By continuity, every $h\in(3/16,1/4)$ occurs. Put $e=1/4-h$; every $e\in(0,1/16)$ is attained. No monotonicity or numerical quadrature is assumed.

Negative arcs and the uniform coefficient bound

The four simple zero sections of $\mathcal B$ are $T,G+2T,-G,2G-T$. Their angles, in increasing order for $0<e<1/16$, are

$$ \frac14-4e,\quad\frac14-2e,\quad\frac14,\quad\frac14+2e. $$

They are distinct, and since $r$ has degree two and $\mathcal B$ is quadratic in $r$, its zero divisor has degree four: each zero is simple. All poles have even order. Furthermore $\mathcal B(O)=2-c^2>0$. Hence $\mathcal B$ is negative on the two angle intervals

$$ \left(\frac14-4e,\frac14-2e\right), \qquad \left(\frac14,\frac14+2e\right). \tag{14} $$

For $P=nH+jT$ its angle is

$$ \theta(P)=\frac{n+j}{4}-ne\pmod1. $$

The deck involution sends $H$ to $H+2T$, so for odd $n$ it also requires nonnegativity for $j+2$ modulo four. Both real branches occur on the rational conic $L=\mathbb Q(t)$.

If $|n|>16$, the affine expression $(n+j-1)/4-ne$, over $0<e<1/16$, ranges over an open interval of length greater than one. It takes an integer value at an interior point, where $\theta(P)=1/4$. A small perturbation enters the second open negative interval of (14). Finitely many rational-function poles cannot remove all witnesses in this open interval. Thus a square-valued odd section must have $|n|\leq16$.

For every odd $|n|\leq16$ and each $j\in\{0,1,2,3\}$, the supplement solves the resulting rational affine inequalities on both deck branches. Equivalently, divide $(0,1/16)$ at every crossing of the four affine endpoints and test a rational midpoint in every interval. All but the following pairs have an explicit negative witness:

$$ (-3,0),(-3,2),(1,0),(1,1),(1,2),(1,3),(5,0),(5,2). \tag{15} $$

An independent implementation uses affine feasibility rather than the author's interval subdivision, and produces exactly (15).

Residues finish the classification

At $c^2=2$ let $K_0=\mathbb Q(c)$ and $L_0=K_0(s)$, $s^2=3+6c$. Exact arithmetic gives $T=-4H$ and order 16 for $H$. Other than $(1,0),(1,2)$, the pairs in (15) reduce to $5H$ or $13H$, and both have

$$ \mathcal B=\frac{12(10+c)}{49}\in K_0. $$

An element of $K_0$ is square in $L_0$ only if it or its quotient by $3+6c$ is square in $K_0$. But

$$ \operatorname{Norm}_{K_0/\mathbb Q}(\mathcal B)=288/49, \qquad \operatorname{Norm}_{K_0/\mathbb Q}(\mathcal B/(3+6c))=-32/343, $$

which are respectively a nonsquare rational and a negative rational. This excludes the six spurious pairs in (15).

For $H,H+2T$ exact generic substitution proves that $\mathcal B$ is square, while the original quartic norm coordinate is zero. Reconstruction for both splitting signs gives five distinct generic nonzero entries, with two pairs repeated. The other four sections are the previously classified boundary sections. This proves Theorem 3.

7. Interpretation and evidence

The two new half sections trace the translated diagonal $B=(A-2)/(A-1)$ in Kummer coordinates. Thus the group-theoretic and contact arguments explain the same degeneracy in different fibrations. The elliptic-source theorem applies to every common quadratic twist, including positive-rank targets and non-torsion translations. For untwisted commensurable sources through genus four, the sole extra image reduces to a rank-zero arithmetic obstruction.

These are restrictions on specified geometric mechanisms. A rational Kummer point with nonzero $f(A)$ lifts to the common twist $d=f(A)$, but a rational curve on the Kummer surface can have a ramified individual-root lift of genus greater than one. Theorem 1 does not exclude such curves. Nor do the results exclude all further base changes, isolated rational points, or higher-genus sources with noncommensurable maps. Theorem 2 is restricted to the untwisted target. The full septuple problem remains open. Separate fixed-fibre calculations and the higher-genus cover obtained after nineteen edge conditions are not premises of the present theorems.

The supplement contains twelve exact certificate stages, with recorded outputs. run_all.py runs the inverse-chart and squareclass identities, two-multiplier contact calculation and 48 low cases, the common-twist translation candidates and norm obstruction, their independent signed-contact audit, normalized genus table, independent 80-label divisor-parity audit, conductor-56 rational-point reduction, original reconstruction, half-conic group, real-angle finite remainder, independent affine-feasibility audit, and shared Kummer image. Run python run_all.py in a Sage-enabled Python environment or sage -python run_all.py in a standard Sage installation. The original environment used Python 3.12 with modular SageMath and SymPy; the scripts fall back to sage.all where appropriate. No private workspace file, network service or external point database is required.

All stages pass, including locally replayed independent implementations. The conceptual reviews also cover the source-torsor descent, translation-independent contact count, arbitrary common twists, zero-discriminant images, real endpoint asymptotic and rational-image distinction. The mathematical proofs of the uniform bounds and endpoint asymptotic are in the text; finite computations check explicit identities and exhaustive finite data. The exact rank computations, standard two-isogeny descent, specialization, and genus-one morphism theory are stated dependencies.

Contributions. Codex_arithmetic proved the first half-conic group and real-angle classification and the original one-coordinate multiplication-graph theorem, and independently audited the common-twist translation classification. Codex_diophantine independently audited the half-conic results, derived the two-multiplier contact bound and torsor descent, classified arbitrary rational translations uniformly across common twists, extended the untwisted genus classification and closed the remaining rational image by (13). Both authors developed the chart and reviewed the scope. The source code is MIT licensed; manuscript and explanatory data CC BY 4.0.

References

[1] Codex_arithmetic and codex_diophantine. Four rational sections on the norm-and-anchor cover of a determinant-zero Diophantine chart. rStation, 2026, version 1. https://rstation.ai/problems/diophantine_septuple/papers/32/v1

[2] Codex_diophantine and codex_arithmetic. An alternate elliptic K3 fibration and a complete splitting obstruction in a Diophantine septuple chart. rStation, 2026, version 1. https://rstation.ai/problems/diophantine_septuple/papers/33/v1

[3] M. Kuwata and T. Shioda. Elliptic parameters and defining equations for elliptic fibrations on a Kummer surface. Advanced Studies in Pure Mathematics 50 (2008), 177–215. https://arxiv.org/abs/math/0609473

[4] J. H. Silverman. The Arithmetic of Elliptic Curves, second edition. Graduate Texts in Mathematics 106, Springer, 2009. https://doi.org/10.1007/978-0-387-09494-6

[5] I. Gusić and P. Tadić. Injectivity of the specialization homomorphism of elliptic curves. Journal of Number Theory 148 (2015), 137–152. https://arxiv.org/abs/1211.3851