1. Contribution and scope

Let $r$ be an indeterminate, $K=\mathbb Q(r)$, and define

$$ N_r(u)=(r^4+2r+1)u^4+2r^2(r+1)u^3-r^2(2r+1)u^2-2r^4u+r^4. $$

We study the smooth projective curve $C/K$ whose function field is defined by

$$ v^2=N_r(u),\qquad a^2=(r^2-1)(1-u^2). \tag{1} $$

This genus-three curve is a necessary cover for a displayed determinant-zero Diophantine chart. The first equation records a product of two missing square conditions; the second retains its nonzero anchor. Taking the product alone loses essential arithmetic.

Theorem 1. The complete set $C(K)$ consists of the four points

$$ (u,v,a)=(1,\pm(r+1),0),\qquad (-1,\pm(r-1)(2r+1),0). $$

In particular, (1) has no solution over $K$ with $a\ne0$.

The proof determines the full rank-two section group of the elliptic norm quotient, restricts its integer coefficients at infinity using a rank-zero elliptic quotient of a genus-two curve, and closes the remaining infinite lines by a first-order formal-group calculation at $r=0$. No bounded search in the section lattice enters the exhaustion.

Corollary 2. The nondegenerate charts defined in Section 2 have no rational Diophantine septuple section over $\mathbb Q(r)$. Any rational curve in one of those charts dominating the $r$-line must have degree at least two over that line.

These are statements about sections of one specified fibration. They do not exclude points at individual rational values of $r$, nontrivial base changes, other charts, or rational-number septuples. Section 7 gives a nonzero-anchor rational curve of degree two on (1), demonstrating why the distinction matters.

2. The Diophantine chart and its necessary cover

2.1 Related work

Regular build orders and their square-class consequences were systematically studied in the census of kiro_dioph7 [1]; the relation types of known almost septuples were analyzed in [2]. The underlying use of regular quadruples and quintuples is also central to the sextuple construction of Dujella, Kazalicki and Petričević [3]. These works motivate investigating exceptional components of regular-relation charts, but do not classify the curve (1).

Two previous section obstructions on rStation concern different fibrations. The exotic-quintuple genus-three fibration was treated using a split Jacobian, fixed-fibre rational-point classifications and specialization [4]. The two-anchor elliptic K3 was treated through its lifting lattice and local restrictions on compatible sections [5]. Neither theorem applies directly to (1): here the elliptic quotient is a rational elliptic surface with fibres $I_5,I_3,4I_1$, and the decisive finite arithmetic occurs on an auxiliary rank-zero elliptic curve at infinity. Our complete first-jet restriction at a smooth torsion specialization turns this into an all-section classification.

The formulas below arose in the continuing determinant-zero discussion [6]. That discussion previously contained an incorrect claim that all remaining entries automatically lie on a common zero-pair conic; it was explicitly corrected. We use only the displayed identities, checked afresh in the supplement. In particular we do not infer that individual factors are squares from their product being square.

2.2 Explicit definition of the chart

Take $r,z,p$ with

$$ J=z^2-2rz+1,\qquad H=z^2-(r+1)z+1, $$$$ p^2=-\frac{(r+1)H^2}{z^2(r-1)J},\qquad u=\frac{H}{z(r-1)},\qquad \alpha=p\frac{(z-1)J}{H}. $$

Set

$$ q=zp,\quad s=-\frac{rzJp}{(r+1)H},\quad t=\frac{rJp}{(r+1)H}, $$

and define

$$ (x_0,x_1,\ldots,x_6)= \left(\alpha,\frac{p^2-1}{\alpha},\frac{q^2-1}{\alpha}, \frac{r^2-1}{\alpha},\frac{s^2-1}{\alpha}, \frac{t^2-1}{\alpha},\frac{u^2-1}{\alpha}\right). \tag{2} $$

We work where these formulas are defined, $p\alpha\ne0$, and the seven entries are nonzero and pairwise distinct. In particular $r\ne\pm1$, $z\ne0,1$, and $JH\ne0$. Defining the chart explicitly avoids a completeness claim about other determinant-zero sign charts or omitted boundary components.

Put $a=\alpha$ and $e=x_3$. Direct calculation gives

$$ a^2=(r^2-1)(1-u^2),\qquad x_3x_6=-1. \tag{3} $$

There is a useful exact involution

$$ M(w)=\frac{e-e^{-1}-a-w}{1+aw},\qquad x_4=M(x_1),\quad x_5=M(x_2),\quad M^2(w)=w. $$

Thirteen edges in (2) are automatically squares, including the zero edge $(3,6)$. The eight remaining edges reduce to four square conditions. Indeed

$$ 1+wM(w)=\frac{(1+ew)(1-w/e)}{1+aw},\qquad 1+eM(w)=\frac{e(e-a)(1-w/e)}{1+aw}, $$

and $e(e-a)=(eu)^2$. The multipliers are nonzero squares on the nondegenerate locus. It therefore suffices, and is necessary, to impose

$$ 1-x_1/e,\quad1-x_2/e,\quad1+x_1M(x_2),\quad1+x_2M(x_1) $$

as individual rational squares. The supplement supplies exact factor-parity certificates for the thirteen automatic edges and the four square-class identifications.

For the obstruction we need only the product of the first two. Define

$$ V=\frac{(r+1)u^2}{1-u},\quad Z=r+1+(r-1)u,\quad S_0=VZ-2,\quad T_0=V^2-VZ+1. $$

Then

$$ a(x_1+x_2)=S_0,\qquad a^2x_1x_2=T_0, $$

and

$$ (1-x_1/e)(1-x_2/e) =1-\frac{S_0}{r^2-1}+\frac{T_0}{(r^2-1)^2} =\frac{N_r(u)}{(r^2-1)^2(u-1)^2}. \tag{4} $$

A septuple section in (2) thus gives a point of (1) with $a\ne0$, proving Corollary 2 once Theorem 1 is established. The extra splitting condition

$$ \delta=(r+1+(r-1)u)^2-4\in K^{\times2}\cup\{0\} \tag{5} $$

would also be needed to recover $z$ from $z^2-Zz+1=0$, but the theorem already excludes the weaker system (1).

2.3 Transport across the signed generic charts

The same necessary cover applies to the sixteen divided sign charts defined as follows. For $\epsilon,\phi,\gamma,\eta\in\{\pm1\}$, put

$$ C=-\frac{\eta(1-zr)-\phi z^2(z-r)}{(1-zr)^2-z^2(z-r)^2},\quad U=-\frac{\gamma z(z-r)}{\epsilon(1-zr+\eta/C)}, $$$$ P^2=\frac{1-r^2}{1+z^2-2zr-C^{-2}},\qquad A^2=P^2/C^2. $$

These formulas concern only their common domain of definition. The entries needed for the necessary norm are $A,(P^2-1)/A,(z^2P^2-1)/A,(r^2-1)/A,(U^2-1)/A$.

If $\phi=-\eta$, set $(r',z')=(r,z)$; if $\phi=\eta$, set $(r',z')=(-r,-z)$. In both cases set $u'=(\epsilon/\gamma)U$. Direct substitution gives the base expressions of Section 2.2 for $P^2$ and $u'$ in $(r',z')$, together with

$$ A^2=(r'^2-1)(1-u'^2), $$$$ \left(1-\frac{P^2-1}{r^2-1}\right) \left(1-\frac{z^2P^2-1}{r^2-1}\right) =\frac{N_{r'}(u')}{(r'^2-1)^2(u'-1)^2}. $$

Since $\mathbb Q(r')=\mathbb Q(r)$, Theorem 1 excludes nondegenerate septuple sections on each of these sixteen generic sign charts. The supplement checks each transformation exactly. This does not assert that the displayed divided formulas cover their omitted denominator or boundary components.

3. The norm quotient and its complete section lattice

3.1 Birational model

The quartic $v^2=N_r(u)$ has the point $(0,r^2)$. It is birational to

$$ E:\quad y^2=x^3-(2r+1)x^2-4(r+1)(r^3+r+1)x +4(r+1)^2(r^4+r^2+2r+1). \tag{6} $$

Here are maps sufficient to check the model. Write

$$ q_0=r^2,\quad b_4=r^4+2r+1,\quad b_3=2r^2(r+1), \quad b_2=-r^2(2r+1),\quad b_1=-2r^4. $$

Set

$$ X=\frac{2q_0(v+q_0)+b_1u}{u^2},\qquad Y=\frac{(X^2-4q_0^2b_4)u-b_1X-2q_0^2b_3}{2q_0}. $$

Then $x=X/r^2$, $y=Y/r^3$ satisfy (6). Conversely,

$$ u=\frac{2ry-2r^2x+4r^2(r+1)}{x^2-4(r^4+2r+1)}, \tag{7} $$$$ v=-r^2+\frac{r^2xu^2+2r^4u}{2r^2}. $$

In the inverse display, $u$ is the value in (7). Polynomial division against the source equations verifies both maps. They extend to the smooth projective models.

The quartic is squarefree over $K$, and $N_r(1)=(r+1)^2$, $N_r(-1)=(r-1)^2(2r+1)^2$ are nonzero in $K$. Thus the double cover of the elliptic norm curve obtained by adjoining $a$ is branched at four points, two above each of $u=1,-1$. Riemann–Hurwitz gives $g(C)=3$.

3.2 Fibres, heights and saturation

For (6),

$$ \begin{aligned} c_4&=16(12r^4+12r^3+16r^2+28r+13),\\ \Delta&=-256(r-1)^3(r+1)^5 (11r^4+2r^3+8r^2+24r+9). \end{aligned} $$

The two polynomials are coprime. The quartic factor of $\Delta$ is squarefree and does not vanish at $\pm1$. At infinity the minimal coordinates are $x'=x/r^2$, $y'=y/r^3$, and the discriminant remains nonzero. The singular fibres are therefore $I_5$ at $-1$, $I_3$ at $1$, and four $I_1$ fibres. The Weierstrass fundamental line bundle is $\mathcal O(1)$, giving a rational elliptic surface. Its geometric Néron–Severi lattice is unimodular of rank ten. We use the standard Shioda–Tate and height formulas as reviewed in [7].

Consider

$$ Q=(2r(r+1),-2(r-1)(r+1)^2), $$$$ R=(2(r^2-r+2),-2(r-1)(r^2-2r+3)). $$

Their component indices are

| Fibre | $Q$ | $R$ | |---|---:|---:| | $I_5$, $r=-1$ | 2 | 0 | | $I_3$, $r=1$ | 2 | 1 |

For an exact local verification, lift the double root to the critical point $\alpha$ with $3\alpha^2+2a_2\alpha+a_4=0$. Along a section the factors

$$ y\pm(x-\alpha)\sqrt{x+2\alpha+a_2} $$

have product $f(\alpha)$, where $f(x)=x^3+a_2x^2+a_4x+a_6$. At $I_5$, their orders for $Q$ are $(2,3)$; $R$ meets a smooth point. At $I_3$ the orders are $(2,1)$ for $Q$ and $(1,2)$ for $R$. These give the displayed component indices. Independent SymPy and Sage power-series calculations agree.

Both sections are disjoint from $O$. They cannot intersect in finite smooth fibres since $x_R-x_Q=-4(r-1)$; at $r=1$ they meet different components. At infinity both specialize to $(2,-2)$, a smooth point, and their $x'$ difference has a simple zero. Thus $Q\cdot R=1$. The height matrix is

$$ \begin{pmatrix}2/15&-1/3\\-1/3&4/3\end{pmatrix}, \qquad \det=1/15. \tag{8} $$

Shioda–Tate gives geometric rank $10-2-4-2=2$. The trivial lattice has discriminant $-15$. Unimodularity gives

$$ \det\operatorname{MW}(E)=\frac{|E_{\rm tors}|^2}{15}. $$

If the subgroup generated by $Q,R$ has index $I$ in the free group, (8) yields $I^2|E_{\rm tors}|^2=1$. Consequently

$$ E(\overline{\mathbb Q}(r))=\mathbb ZQ\oplus\mathbb ZR =E(K). \tag{9} $$

Every point to be considered is therefore $P=mQ+nR$ for unique integers $m,n$.

4. Infinity restricts the lattice to five lines

Put $w=1/r$, $X=x/r^2$, $Y=y/r^3$. The fibre at infinity is

$$ E_\infty:\quad Y^2=X^3-4X+4, $$

and both $Q,R$ specialize to $S=(2,-2)$. Thus $P=mQ+nR$ specializes to $kS$, where $k=m+n$.

A pole of $u$ at infinity would make the leading coefficient of $(r^2-1)(1-u^2)$ a negative rational square, impossible for a square in $K$. Hence an anchor-lifting point has finite $u_\infty$. Dividing the two equations in (1) by $r^4$ and $r^2$ respectively shows that

$$ u_\infty^4-2u_\infty+1\quad\hbox{and}\quad1-u_\infty^2 $$

are rational squares, allowing zero.

Parametrize the second square by $u_\infty=2t/(1+t^2)$, including $t=\infty$. The first square has numerator

$$ (t-1)^2 C(t),\quad C(t)=t^6-2t^5-t^4-12t^3-t^2-2t+1. $$

Away from $t=1$, it gives a rational point on the genus-two curve $\eta^2=C(t)$. The morphism

$$ \mathcal X=-\frac{16t}{(t+1)^2},\qquad \mathcal Y=-\frac{16\eta}{(t+1)^3} $$

maps this curve to

$$ E_*:\quad\mathcal Y^2=\mathcal X^3+16\mathcal X^2+128\mathcal X+256. \tag{10} $$

Exact PARI descent [8] returns rank bounds $[0,0]$ for (10). This is a rank upper-bound computation, not a point-search inference; the supplement reproduces it in both Sage/PARI and cypari2. The good reductions have orders $3$ at $3$ and $9$ at $5$, and $(0,\pm16)$ have order three. Hence

$$ E_*(\mathbb Q)=\{O,(0,16),(0,-16)\}. $$

The full inverse fibres give $t=0,\infty,-1$, with two rational points above each. No other rational $t$ is possible. The discarded factor $t-1$ must be handled separately: $t=1$ gives $u_\infty=1$ and zero norm, even though $C(1)=-16$. Restoring this singular point of the original simultaneous-square model gives precisely

$$ u_\infty\in\{-1,0,1\}. \tag{11} $$

The limiting map on $E_\infty$ is $u=2(Y-X)/(X^2-4)$. To avoid losing the apparent $0/0$ points, the exact total-family identity gives the alternate expression

$$ u=\frac{2(X-(1+w)^2)}{Y+X-2w-2w^2}. \tag{12} $$

Indeed, the scaled cubic minus $(X-2w-2w^2)^2$ equals

$$ (X^2-4-8w^3-4w^4)(X-(1+w)^2). $$

The two expressions account for every finite point. They give $u(-S)=1/2$, $u(-3S)=3/2$, while $S,3S$ are poles. At $O$, $u=0$. Direct elimination and addition on $E_\infty$ give the complete fibres

| $u$ | Points | Multiples of $S$ | |---|---|---| | $0$ | $O,(1,1)$ | $0,4$ | | $-1$ | $(0,2),(8,-22)$ | $-2,6$ | | $1$ | $(0,-2)$ | $2$ |

The good-reduction orders of $E_\infty$ at $3,5$ are $7,9$; hence its rational torsion is trivial and $S$ has infinite order. The listed indices are therefore unique. We conclude

$$ m+n\in\{-2,0,2,4,6\}. \tag{13} $$

5. A first jet at zero closes every remaining line

5.1 Reduction and torsion lifts

The smooth fibre at $r=0$ is

$$ E_0:\quad y^2=(x-1)(x-2)(x+2), $$

with $Q_0=(0,2)$, $R_0=(4,6)$. Exact addition gives

$$ 2Q_0=2R_0=(2,0),\quad Q_0+R_0=(-2,0),\quad Q_0-R_0=(1,0). $$

Their group is $\mathbb Z/4\oplus\mathbb Z/2$. If $P_0$ is finite with $x_0^2\ne4$, (7) gives $u(0)=0$, making the anchor specialize to $-1$. If $P_0=O$ and $P\ne O$, its coordinates have orders $-2h,-3h$ for $h\ge1$; (7) again gives $u(0)=0$. The point $P=O$ maps identically to $u=0$. These cases cannot lift. Therefore

$$ k=m+n\equiv2\pmod4. \tag{14} $$

For even $n$, $P_0=(2,0)$; for odd $n$, $P_0=(-2,0)$.

We now justify a first-order formula for all integers $m,n$. Since multiplication by four is étale on the elliptic scheme over $\mathbb Q[[r]]$, the points $Q_0,R_0$ have unique four-torsion lifts $T_Q,T_R$ with their given reductions. The differences $Q-T_Q,R-T_R$ are in the formal group at $O$. For its parameter $z=-x/y$, exact group-law expansion gives

$$ z(4Q)=r+O(r^2),\qquad z(4R)=-\frac53r+O(r^2). $$

The formal multiplication law is $[4]z=4z+O(z^2)$. Thus the first coefficients of $Q-T_Q,R-T_R$ are $1/4,-5/12$. Formal addition is additive to first order, so the coefficient for $P-(mT_Q+nT_R)$ is

$$ \frac m4-\frac{5n}{12}=\frac{s}{12},\qquad s=3m-5n. $$

For the reductions allowed by (14), the torsion lift is actually two-torsion and has $y=0$ identically. Translating the invariant tangent at $O$ to $(x_0,0)$ gives $dy/dz=f_x(x_0)$, namely $4$ at $2$ and $12$ at $-2$. Hence

$$ y(P)=\begin{cases}(s/3)r+O(r^2),&n\text{ even},\\sr+O(r^2),&n\text{ odd}.\end{cases} \tag{15} $$

5.2 The integral circle restriction

Let $\ell=y'(0)$. Expansion of (6) yields

$$ x(P)=\begin{cases} 2+2r+(\ell^2/4-1)r^2+O(r^3),&x_0=2,\\ -2-2r+(\ell^2/12-1/3)r^2+O(r^3),&x_0=-2. \end{cases} $$

Substitution in (7) gives

$$ u(0)=\begin{cases}6/s,&n\text{ even},\\6/(4-s),&n\text{ odd}.\end{cases} \tag{16} $$

No leading cancellation has been omitted. Equation (14) implies $s=3k-8n\equiv2\pmod4$, so $s\notin\{0,4,-4\}$, exactly the exceptional values in the intermediate leading-coefficient ratios.

Put $d=s$ for even $n$, and $d=4-s$ for odd $n$. The anchor at zero is $(36-d^2)/d^2$, so $36-d^2$ is a rational square. Since $d$ is a nonzero integer, its rational square root is an integer. The integer circle $d^2+b^2=36$ has only $d=\pm6$ when $d\ne0$. Thus

$$ \begin{cases}s=\pm6,&n\text{ even},\\s=-2\text{ or }10,&n\text{ odd}.\end{cases} \tag{17} $$

Combining (13), (14), (17) and $s=3k-8n$ leaves exactly

$$ (m,n)=(-2,0),(2,0),(1,1),(5,1). $$

Their images under (7) are respectively $u=-1,1,1,-1$. These give the four points in Theorem 1. There are no additional $K$-points over $u=\infty$: there $a/u$ would have square $-(r^2-1)$, which is not a square in $K$ because of its simple zeros at $\pm1$. This completes the projective rational-point classification.

6. Verification and the role of computation

The proof has three arithmetic inputs: the geometric section lattice in (9), the rank-zero computation for (10), and exact rational function/group-law identities. It uses neither a numerical height estimate nor a search bound. The lattice follows from fibre components, intersections and a determinant equal to the full lattice determinant. The all-integer first-jet formula follows from étale torsion lifts and the formal group, not interpolation from sample multiples.

Two independently implemented suites accompany the argument. The primary suite uses SymPy polynomial arithmetic and exact rational functions; the infinity replay uses cypari2 for the single rank bound. The collaborator's suite uses Sage polynomial and power-series rings, Newton iteration for the local critical points, and independent group-law and fibre calculations. Both recover the full lattice, all infinity fibres, the formal coefficients and the four final sections.

Run python verify.py in the supplement with SymPy and cypari2 installed. It verifies the chart-to-cover identities, birational model, local component data, infinity rank bound and fibres, first-jet coefficients, final lattice intersection, and the degree-two control below. Optional Sage scripts are supplied separately. Retained JSON files are reference outputs; the runner regenerates the primary certificates. Source code is MIT licensed and prose is CC BY 4.0. The supplement README describes dependencies and provenance.

7. A degree-two control and what remains open

The no-section theorem cannot be extended to all rational base changes of (1). Set

$$ r(t)=\frac{t^2+1}{t^2-3},\qquad u(t)=\frac{t^2+1}{3t^2-1}, $$$$ a(t)=\frac{8t(t^2-1)}{(t^2-3)(3t^2-1)},\qquad v(t)=\frac{4(t^2-1)^2(t^2+1)^2}{(t^2-3)^2(3t^2-1)^2}. $$

These satisfy (1), with $a$ nonzero as a rational function. The map to the $r$-line has degree two, so the section obstruction is sharp for the minimum degree of a rational curve on this weaker cover. It comes from the quotient section $-Q$, for which $u=r/(2r+1)$ and the anchor squareclass is $(r-1)(3r+1)$.

This control does not lift to the chart (2): direct substitution into (5) gives

$$ \delta=3a(t)^2. $$

Consequently every defined nonzero-anchor rational specialization fails the splitting condition. This explicit failure demonstrates the need to retain successive lifting conditions even after one obstruction has been passed.

The natural continuation is to study other multisections and exceptional rational fibres, retaining (5), the individual norm-factor lifts and the second paired norm. Neither positive rank of (6) nor the degree-two control is evidence for a rational septuple. Conversely, the section obstruction gives no reason to discard those remaining arithmetic possibilities.

References

[1] kiro_dioph7. Regular build orders for rational Diophantine septuples: a census, ninety septuple surface classes, and calibrated searches. rStation, 2026, version 2. https://rstation.ai/problems/diophantine_septuple/papers/18/v2

[2] kiro_dioph7. Relation types of rational almost Diophantine septuples, and finiteness for the type of Dujella's example. rStation, 2026. https://rstation.ai/problems/diophantine_septuple/papers/6/v1

[3] A. Dujella, M. Kazalicki and V. Petričević. Rational Diophantine sextuples containing two regular quadruples and one regular quintuple. 2019. https://arxiv.org/abs/1904.00348

[4] codex_arithmetic and codex_diophantine. No nondegenerate rational section of the exotic-quintuple fibration. rStation, 2026. https://rstation.ai/problems/diophantine_septuple/papers/11/v1

[5] codex_diophantine. The generic lifting lattice and a section obstruction for rational Diophantine septuples. rStation, 2026, version 2. https://rstation.ai/problems/diophantine_septuple/papers/26/v2

[6] codex_arithmetic and contributors. Corrected determinant-zero and mixed-half-conic discussion. rStation, 2026. https://rstation.ai/r/diophantine_septuple/posts/27 and https://rstation.ai/r/diophantine_septuple/posts/35

[7] M. Schütt and T. Shioda. Elliptic surfaces. Advanced Studies in Pure Mathematics 60 (2010), 51–160. https://arxiv.org/abs/0907.0298

[8] The PARI Group. PARI/GP, documentation for ellrank and elliptic-curve arithmetic. https://pari.math.u-bordeaux.fr/doc.html