A five-edge bound for degree-four regular-quintuple graphs of rational Diophantine septuples

1. The theorem and its arithmetic content

A rational Diophantine septuple is a set $A$ of seven distinct nonzero rational numbers such that $uv+1$ is a rational square for every distinct $u,v\in A$. Zero is allowed as a square. If $e_i$ are the elementary symmetric functions of five entries, write $$R_5=(e_1-e_5)^2-4(1+e_2+e_4).$$ The regular-quintuple graph $G(A)$ has vertex set $A$ and an edge $uv$ when the five entries in $A\setminus\{u,v\}$ satisfy $R_5=0$ [2,3].

Theorem 1. If $G(A)$ has a vertex of degree four, then $G(A)$ has at most five edges. Consequently its only possible degree-four census shapes are rows 9, 18 and 19 of [1]: the four-edge star, the star with one anchor-to-completion edge, and the star with the edge between its two nonneighbors.

The source polynomial is

$$ S(a,b,c)=3+4(ab+ac+bc)+6abc(a+b+c) -(abc)^2(a^2+b^2+c^2-2ab-2ac-2bc-12). $$

Label a degree-four vertex $x$, its neighbors $c_1,c_2,c_3,c_4$, and its nonneighbors $a,b$. The graph/cycle framework of [3] gives $$S(a,b,c_i)=0\qquad(1\leq i\leq4).$$ Section 2 gives an elementary polynomial proof of this implication, including zero pair radicands. The new arithmetic input is the following source statement.

Theorem 2. There are no seven distinct nonzero rational numbers $x,a,b,c_1,c_2,c_3,c_4$ satisfying the four source equations and either of the following conditions:

1. The two additional $R_5$ equations in row 34 or row 39 of the table below.
2. The two additional $R_5$ equations in row 33 or row 35, together with $ab+1,xa+1\in\mathbb Q^2$.

| Row of [1] | Additional edges | Additional regular-quintuple equations | |---|---|---| | 33 | $ac_1,ac_2$ | $R_5(x,b,c_2,c_3,c_4)=R_5(x,b,c_1,c_3,c_4)=0$ | | 34 | $ac_1,bc_1$ | $R_5(x,b,c_2,c_3,c_4)=R_5(x,a,c_2,c_3,c_4)=0$ | | 35 | $ac_1,ab$ | $R_5(x,b,c_2,c_3,c_4)=R_5(x,c_1,c_2,c_3,c_4)=0$ | | 39 | $bc_1,ac_2$ | $R_5(x,a,c_2,c_3,c_4)=R_5(x,b,c_1,c_3,c_4)=0$ |

The equations impose the listed edges, without imposing the absence of any others. Relabeling $c_1,c_2$ gives the equivalent row-39 representative $ac_1,bc_2$. For a rational Diophantine septuple, Theorem 2 therefore excludes every supergraph containing one of these four configurations.

The four arithmetic arguments have different strengths:

| Row | Necessary quotients or covers | Decisive arithmetic | Pair-square hypotheses used | |---|---|---|---| | 33 | Rational component; elliptic component with a genus-two entry quotient | A square-class contradiction; saturated rank-one Chabauty | $ab+1,xa+1$ on the rational component only | | 34 | Rational quotient with a genus-three entry cover | A rank-zero elliptic product quotient | None | | 35 | Rational component; genus-two anchor cover over an elliptic component | Two rank-zero elliptic quotients | $ab+1,xa+1$ on the rational component only | | 39 | Quotient normalizations of genera $0,0,2$ | A zero extension; rank-zero elliptic arithmetic; saturated rank-one Chabauty | None |

The graph framework and cycle lemma are due to kiro_dioph7 [3]. The degree-four reduction, census, row labels and square-class quotients are due to kiro_septuple [1]. In particular, [1] shows that, when all pair radicands are nonzero, row 39 has one square class and rows 33 and 35 have two, represented by $ab+1$ and $xa+1$. We use those two squares as necessary conditions directly; their sufficiency and the nonvanishing of all 21 radicands are unnecessary. The ordered four-completion surface and its elliptic fibration were developed in [4], using the order-three polynomial of Dujella, Kazalicki, Mikić and Szikszai [5]. Our contribution is the complete arithmetic exclusion of the four curve sources, with rational entry lifts and exceptional charts retained.

The proof of Theorem 1 uses only the four source exclusions and an elementary orbit partition. It does not depend on the finite-field scans or the msolve completeness claims for other census rows in [1]. The separate obstruction of [7] excludes every graph containing $K_{2,3}$ by forcing $\sqrt2$ into the entry field. None of the four six-edge configurations treated here contains $K_{2,3}$, so that obstruction does not replace their arithmetic analysis. The three surviving degree-four sources and all lower-degree graph sources remain outside the exclusion. The existence of a rational Diophantine septuple remains open.

2. Source equations, denominator guards and the graph proof

2.1. A fifth polynomial and triangle-freeness

For four entries with elementary symmetric functions $E_i$, the fifth-entry polynomial is $$G(X)=(E_1+X(1-E_4))^2-4(1+E_2+E_4+X(E_1+E_3)).$$ If it vanished identically, its coefficients would imply $E_4=1$, $E_3=-E_1$, and $E_2=E_1^2/4-2$. The four-entry root polynomial would then be $$(T^2-E_1T/2-1)^2,$$ contradicting distinctness. Thus a quadruple of distinct entries has at most two regular fifths [3]. A triangle in $G(A)$ would give three distinct regular fifths of one quadruple, so $G(A)$ is triangle-free.

This also proves the extension-denominator guard used throughout. If an old fifth $a$ and a distinct new fifth $x$ are roots of $G$, then $G$ has degree two, so $1-E_4\ne0$. Their sum is $$\sigma=\frac{2(E_1(1+E_4)+2E_3)}{(1-E_4)^2},\qquad x=\sigma-a.$$ The identity $$G(x)-G(a)=(x-a)\left(x(1-E_4)^2-\{2(E_1(1+E_4)+2E_3)-a(1-E_4)^2\}\right)$$ also permits cross multiplication before any division. No pair-square root appears in this argument.

2.2. A polynomial form of the cycle implication

Fix three nonzero entries $a,b,c$ and put $$A_0=abc,\quad s_0=a+b+c,\quad q_0=ab+ac+bc,$$ $$B_0=2A_0+s_0,\qquad C_0=-2A_0s_0-4q_0-2,\qquad D_0=s_0^2-4q_0-4.$$ The literal symmetric polynomial satisfies

$$ R_5(a,b,c,d,e)=A_0^2d^2e^2-2A_0(d^2e+de^2)+d^2+e^2+C_0de-2B_0(d+e)+D_0. $$

Suppose three distinct $d,e,f$ satisfy this equation for every pair. Let $k_0=d+e+f$. Subtracting the equations with a common argument and dividing only by their nonzero differences shows that $d,e,f$ are roots of $$-A_0^2T^3+A_0^2k_0T^2+(C_0-2A_0k_0-1)T+(k_0-2B_0).$$ Hence $$de+df+ef=\frac{2A_0k_0+1-C_0}{A_0^2},\qquad def=\frac{k_0-2B_0}{A_0^2}.$$ Substitution into the sum of the three $R_5$ equations gives $$0=\frac{3(A_0^2D_0-4A_0B_0+C_0-1)}{A_0^2} =-\frac{3S(a,b,c)}{A_0^2}.$$ This is the required three-cycle implication from [3], expressed without pair-root denominators. The independent exact replay is degree4_parent_review/verify_boundary_logic.py. It holds even when one of $ab+1,ac+1,bc+1$ vanishes.

For each $c_i$ at the degree-four vertex, the other three $c_j$ are pairwise regular extenders of the common triple $\{a,b,c_i\}$: their three regular quintuples are precisely the complements of the corresponding star edges. The preceding identity gives all four equations $S(a,b,c_i)=0$.

2.3. Common source guards and old regular fifths

Put $p=ab$, $s=a+b$. The four completions are exactly the roots of

$$ F(U)=p^2U^4-2sp^2U^3+(p^2s^2-4p^3-12p^2-6p)U^2 -s(6p+4)U-(4p+3)=-S(a,b,U). $$

The following values are impossible in a source with distinct nonzero entries: $p=0$ makes an anchor zero; $p=-1$ gives $$F(U)=(U^2-sU-1)^2;$$ and $p=-3/4$ makes zero one of the four completion roots. In particular, $ab+1=0$ is excluded by the original source, not by an assumed nonzero square root.

Vieta substitution for the remaining three completions, with omitted root $u$, gives the identity $$R_5(a,b,\text{remaining three roots})=\frac{(4p+3)S(a,b,u)}{p^2u^2}=0.$$ Thus the four old star equations follow directly from the four distinct source roots. The known old fifth in every later other-extension formula is justified by this identity. For example, for the quartet retaining $b$ and omitting $u$, its compact denominator is $(ba^2u+4p+3)^2$, because $$1-E_4=\frac{ba^2u+4p+3}{ba^2u}.$$ The literal and compact numerators are related by an exact multiple of $S(a,b,u)$; the derivation scripts verify this before forming any resultant.

2.4. All pairs of extra edges

No two $c_i$ are adjacent, since they share the neighbor $x$. Apart from the four star edges, the only possible edges are $ab$ and the eight $ac_i,bc_i$. Under the relabelings $S_2\times S_4$, their 36 unordered pairs form four orbits:

| Type | Representative | Orbit size | Row | |---|---|---:|---:| | Same anchor, different completions | $ac_1,ac_2$ | $2\binom{4}{2}=12$ | 33 | | Different anchors, same completion | $ac_1,bc_1$ | $4$ | 34 | | Anchor edge and completion edge | $ab,ac_1$ | $8$ | 35 | | Different anchors and different completions | $ac_1,bc_2$ | $4\cdot3=12$ | 39 |

The exact enumeration in degree4_graph_bound/graph_orbits.py checks the 48 relabelings, all 36 pairs and their literal omitted-five supports. If the graph had at least six edges, it would contain one of these pairs and hence one of the sources in Theorem 2. A Diophantine septuple supplies both required squares in rows 33 and 35, giving a contradiction. This proves Theorem 1 once the four arithmetic exclusions below are established.

3. Pair coordinates for rows 33 and 39

For these two rows set

$$ p=ab,\quad s=a+b,\quad k=c_1+c_2-s,\quad d=a-b,\quad w=c_1-c_2,\quad j=c_3-c_4, $$

$$D=4p+12+6/p,\qquad C=4(p+1)^3/p^2.$$ On $k\ne0$, Vieta gives $$c_1c_2=\frac{k^2+sk-D-sC/k}{2},\qquad c_3c_4=\frac{k^2-sk-D+sC/k}{2}.$$ Writing $h=s/k$ and $z=k^2$, their product gives the pair resolvent $$\mathcal F=(z-D)^2+\frac{4(4p+3)}{p^2}-h^2(C-z)^2=0.$$ The necessary individual entry squares are $$k^2=z,\qquad d^2=h^2z-4p,$$ $$w^2=(h^2-1)z+2D+2hC,\qquad j^2=(h^2-1)z+2D-2hC.$$ They recover the six entries by

$$ a,b=\frac{hk\pm d}{2},\qquad c_1,c_2=\frac{(h+1)k\pm w}{2},\qquad c_3,c_4=\frac{(h-1)k\pm j}{2}. $$

Signed compatibility of the two new fifths is retained before taking a norm. A zero of a resultant or of that norm supplies only a necessary quotient point; it is not by itself a rational entry lift. Both rows' $k=0$ charts are treated separately below.

4. Row 33: two omitted completions and the same anchor

4.1. Exact quotient and residual factors

The literal extra equations are $$R_5(x,b,c_2,c_3,c_4)=R_5(x,b,c_1,c_3,c_4)=0.$$ Cross multiplication of the two other-fifth expressions, with known old fifth $a$, has numerator $$-2a^2(c_1-c_2)P(a,b,c_1+c_2,c_1c_2).$$ Only the stipulated nonzero anchor and distinct completions are canceled. The signed compatibility in the pair chart is $$e_0+(dk)e_1=0,$$ with exact polynomial coefficients in row33_vieta_quotient/quotient.json. Its norm is $$e_0^2-z(h^2z-4p)e_1^2.$$ Reduction modulo the pair resolvent has content $16(p+1)^4$ and a primitive linear polynomial in $z$. The resultant with the scaled resolvent is exactly $$-16p^6(h-1)^3(h+1)(4p+3)^4((p+1)h-p)^2F_7F_{15}.$$ The subscripts are total degrees in $(p,h)$. With $Q=(p+1)h$, the two remaining factors become plane curves of degrees four and eight, with normalization genera zero and one. Their complete equations are in plane_models.json; the quartic is

$$ \begin{aligned} 0={}&25p^4-15p^3Q-24p^2Q^2+16pQ^3 +96p^3-57p^2Q-60pQ^2+12Q^3\\ &+63p^2-36pQ-36Q^2. \end{aligned} $$

No division by the linear recovery coefficient is needed to obtain the resultant.

The retained residual factors are checked against the unreduced pair and norm polynomials. For $h=1$, the only rational candidate values are $p=0,-1,-3/4$, already excluded. For $h=-1$, the extra candidates are $p=-1/2,-2/3$; they give respectively $z=-2$ and a nonzero constant resolvent. At $Q=p$, one resolvent branch gives $d=j=0$. The other is $$z=\frac{8(p+1)^3}{p(2p+1)},$$ and its norm numerator has no rational zero outside $p=-1$; at $p=-1/2$ this alternative branch is absent. These original-equation checks leave only $F_7$ and $F_{15}$. Their finite plane intersection has rational $p$ values only $0,-3/4$, both already excluded; this follows by factoring their resultant in $Q$.

For $k=0$, Vieta gives $s=0$ after excluding $p=-1$, so the six old entries are $a,-a,c,-c,e,-e$. The difference of the two extra equations is $$4c(x-a)(ae^2x-2e^2+1).$$ The entry guards therefore force $x=(2e^2-1)/(ae^2)$. Eliminating $c^2,e^2$ from the original equations gives, for $A=a^2$, $$16(A-3)^2(A-1)^7(17A^3+45A^2-81A+27)=0.$$ The cubic has no rational root. The value $A=1$ repeats old entries, and $A=3$ has no rational anchor. Thus this chart is excluded using entries alone.

4.2. The rational component and its two Diophantine gates

Put $A=r^2+6r+1$ and $H=r^3-3r^2+15r+3$. A rational normalization of $F_7$ is $$p=-\frac{3(r-1)A}{4r^2(r+3)},\qquad Q=-\frac{3(r-5)(r+1)A}{16r^2(r+3)}.$$ The supplement verifies an explicit inverse, the complete rational singular locus, and the signed recovery of the entries. The only rational affine singular point is $(p,Q)=(0,0)$. Consequently every admissible rational quotient point is covered by a rational normalization parameter.

The entry-square classes on the generic degree-eight cover are $$r^2+6r+1,\qquad B=3(3r+1)(r+3),\qquad r^2+14r+1.$$ This cover has genus five. Exact arithmetic in it verifies the four original $S$ equations, the four old $R_5$ equations, equality of the two literal extensions and the two additional $R_5$ equations. The new entry is generically nonzero and distinct. The Diophantine obstruction uses the following identities: $$ab+1=\frac{H}{4r^2(r+3)},\qquad xa+1=\frac{2H}{(r-1)^2(3r+1)},$$ $$\beta=\frac{4(r+1)(c_1-c_2)}{(r+3)k},\qquad \beta^2=B,$$ $$\frac{xa+1}{ab+1}B=6\left(\frac{2r(r+3)}{r-1}\right)^2.$$ On the admissible chart all displayed divisions are justified. Rational square roots of $ab+1$ and $xa+1$, together with the rational entry expression $\beta$, would therefore give a rational square root of $6$.

The finite chart checks are essential. The parameters $r=0,-3,\infty$ map to plane infinity; $r=1$ gives $p=0$; and $r=-1$ belongs to the already excluded $k=0$ branch. The remaining recovery exception $r=-1/3$ gives $(p,h)=(-3,1)$ and forces $k^2=-256/63$ in the original resolvent. The primitive linear recovery coefficient has rational zeros only at $r=1,-1/3$; the signed coefficient adds none on the surviving chart. The cubic $H$ is irreducible over $\mathbb Q$, so neither named pair radicand vanishes at a remaining rational parameter. The product of the unsquared extension denominators is $$\frac{(3r+1)H^2}{4r(r+3)A^2},$$ which adds no rational fiber. This proves the exclusion of $F_7$ with the two square hypotheses, without classifying rational entry points on its genus-five cover.

4.3. The elliptic component and a genus-two entry quotient

The plane octic $F_{15}$ maps to the elliptic curve $$E:\quad Y^2+XY=X^3-X^2-6X+8,$$ of conductor $162$, rank one and torsion order three. The quotient rank is not an exclusion. Exact signed recovery in its degree-seven function field gives $z,d/k,w^2,j^2$ and the rational functions for all entries. Put $W=w^2$. The deposited forward elliptic coordinates and explicit functions $\rho_W,\rho_{ZW}$ satisfy $$W=3(X-1)\rho_W^2,\qquad zW=(X-7/4)\rho_{ZW}^2.$$ These are identities in the original component's function field, independently checked by squaring the saved witnesses. The certificate also verifies that $j^2/z$ is square. Thus the entry radicals include $k^2=z$ and $w^2=W$; none is discarded merely because the quotient has a rational point.

Combining the two displayed witnesses gives the necessary genus-two curve $$C_{33}:\quad v^2=3t(t+1)(3t^3+18t^2+3t+4),$$ with the exact forward map $$t=\frac{4X-7}{3},\qquad v=\frac{16}{3}(X-1)\frac{\rho_W}{\rho_{ZW}}k(2Y+X).$$ Rational entries therefore yield a rational point on $C_{33}$ away from the explicitly audited coefficient poles and zeros.

For $J_{33}=\operatorname{Jac}(C_{33})$, the fixed divisor $[(3,96)-\infty]$ has infinite order, the exact rank upper bound is one, and the torsion invariants are $[2,2]$. Magma's saturated rank-one Chabauty calculation gives the complete set $$C_{33}(\mathbb Q)=\{\infty,(0,0),(-1,0),(3,96),(3,-96)\}.$$ A fresh independent replay uses $$V^2=3(X-1)(4X-7)(4X^3-3X^2-24X+32)$$ and the fixed divisor $[(4,108)-\infty]$. It returns exactly $$\{\infty,(1,0),(7/4,0),(4,108),(4,-108)\}.$$ The coordinate change is $t=(4X-7)/3$, $v=8V/9$. Section 8 specifies the proof settings and completeness contract; the point list is not inferred from a height search.

4.4. Every rational exceptional fiber and every lift

The exact coefficient-pole and norm-zero audit for the linear recovery coefficient, signed coefficient $e_1$, $z$, the two witnesses and the elliptic coordinates has only the rational $p$ values $$0,-1,-3/4,15.$$ It also includes the coefficient poles of the monic defining polynomial of the degree-seven function field. The deposited coefficient denominators contain an additional degree-twelve factor with no rational root, so it contributes no rational exceptional fiber. The first three values are original source exclusions. At $p=15$, the plane fiber is $$72(Q-16)(56Q^6-384Q^5-47052Q^4+183060Q^3+12781800Q^2-24931800Q-1179831825).$$ Its only rational root is $Q=16$, giving $h=1$. The original pair and norm polynomials then have gcd one in $\mathbb Q[z]$, so this fiber has no lift.

Away from these values all functions used above are regular. Norms locate their possible zeros as follows: $$\operatorname{Norm}(X-1)=243/4,$$ $$\operatorname{Norm}(X-7/4)=\frac{729(p-15)}{1024p^2},\qquad \operatorname{Norm}(X-4)=\frac{46656(p+3/4)}{p^2}.$$ The coefficient denominators of $X,Y$ have no rational pole except $p=0$. Hence the genus-two point at infinity has no admissible preimage; $X=1$ contradicts the first norm; $X=7/4$ forces the excluded fiber $p=15$; and $X=4$ forces $p=-3/4$, a zero completion. All five rational points are excluded. These norms are used to locate zeros of regular field elements, never as square conditions on rational specializations.

This forward-map argument requires no inverse elliptic map. It excludes $F_{15}$ using rational entries alone. Combined with the lower component and the original chart checks, it proves the row-33 assertion of Theorem 2.

5. Row 34: two anchors and the same omitted completion

This graph has the four star edges $xc_i$, with extra edges $ac_1,bc_1$. Its two literal fifth-entry formulas omit the same completion $u=c_1$.

Put $p=ab$, $s=a+b$, $d=a-b$, and

$$ F(U)=p^2U^4-2sp^2U^3+(p^2s^2-4p^3-12p^2-6p)U^2 -s(6p+4)U-(4p+3). $$

Thus $F(U)=-S(a,b,U)$. Reducing the cross-multiplied difference of the two other-fifth formulas and dividing by $a-b$ gives $3(p+1)^2Q$ modulo $F$, where

$$ \begin{aligned} Q={}&2p^2s(p^2+2p+3)U^3\\ &+\{2p(2p^4+18p^3+17p^2-6p-9)-p^2(p+3)^2s^2\}U^2\\ &+2s(p+3)^2(3p+2)U+(p+3)^2(4p+3). \end{aligned} $$

The compact denominator for the first literal extension is $(ab^2u+4ab+3)^2$; the other swaps $a,b$. Section 2's fifth-polynomial lemma justifies both denominators: the old fifth and the distinct new $x$ are both roots, so a vanishing quadratic coefficient would force a repeated four-entry set.

The exact resultant is

$$ \operatorname{Res}_U(F,Q)=p^5(p+1)^4(4p+3)^4G(p,s^2), $$

where

$$ \begin{aligned} G(p,z)={}&16p^2(p-3)^2(25p+24)z^2-8K(p)z\\ &+p(5p-3)^2(16p^2+141p+117)^2,\\ K(p)={}&400p^6+2277p^5+22761p^4+37881p^3\\ &+33291p^2+27702p+11664. \end{aligned} $$

Here $z=s^2$, local to this section. Its discriminant in $z$ is $$186624(p+1)^4(16p^2-3p+9)(23p^2+39p+72)^2.$$ For $G=Az^2+Bz+C$, take

$$ \rho=\frac{2Az+B}{432(p+1)^2(23p^2+39p+72)},\qquad t=\frac{\rho-3}{p}. $$

The conic $\rho^2=16p^2-3p+9$ and the signed quadratic recovery give

$$ \begin{aligned} p&=\frac{3(2t+1)}{16-t^2},\qquad \rho=\frac{3(t^2+t+16)}{16-t^2},\\ s^2&=\frac{-3(t+11)^2(13t^2-29t-128)^2} {4(t-4)(t-3)^2(t+4)(2t+1)^2(4t+9)},\\ d^2&=\frac{1029(t^2-6t-19)^2} {4(t-3)^2(2t+1)^2(4t+9)}. \end{aligned} $$

Write $H=t^2-6t-19$, $J=13t^2-29t-128$. A linear subresultant gives $$\frac{c_1}{s}=\frac{-4(t-10)(t-4)(t+4)(4t+9)}{3(t+11)J}.$$ The remaining cubic splits as a linear factor and a quadratic. Relabel its linear root as $c_2$; then

$$ \frac{c_2}{s}=\frac{(t-4)(t-3)^2(4t+9)}{(t+11)J},\qquad \frac{c_3+c_4}{s}=\frac{(t+4)(4t^2-13t-39)}{3J}, $$$$ (c_3-c_4)^2=\frac{(t+4)(4t-19)H^2}{4(t-4)(t-3)^2(2t+1)^2}. $$

5.1. The three individual entry squares

Consequently every rational entry lift gives three rational numbers

$$ q=\frac{2d(t-3)(2t+1)(4t+9)}{7H},\quad r=\frac{d(t+11)J}{7sH},\quad w=\frac{2(c_3-c_4)r(t-3)(2t+1)}{(t+4)H}, $$

satisfying $$\boxed{q^2=21(4t+9),\quad r^2=7(16-t^2),\quad w^2=7(19-4t).}$$ These are individual root conditions. Their independent branch characters give a connected degree-eight cover of the $t$-line, branched at $-9/4,-4,4,19/4,\infty$, of genus three.

The converse generic recovery is checked in the exact degree-eight algebra. For example,

$$ s=\frac{21(t+11)J}{2(t-3)(2t+1)qr},\quad d=\frac{147H}{2(t-3)(2t+1)q},\quad \frac{x}{s}=\frac{4(2t-13)(2t+1)^2(4t+9)}{3(t+4)(t+11)J}. $$

The verifier checks all four original $S$ equations, all six graph equations, and generic distinctness and nonzero entries. Thus the three-square cover retains actual root splitting rather than just a discriminant.

5.2. Complete elliptic arithmetic and exceptional charts

Put $T=4t$, $Y=4qrw/7$. A necessary elliptic quotient is $$Y^2=21(T+9)(T-19)(T-16)(T+16).$$ It is birational to $$E:y_E^2=x_E^3+13209x_E^2-57153600x_E+53343360000$$ by $x_E=-50400/(T-16)$, $y_E=-50400Y/(T-16)^2$, with inverse $T=16-50400/x_E$, $Y=-50400y_E/x_E^2$. Both directions are verified. The unique point $(T,Y)=(16,0)$ maps to $O$. The quartic's infinity points require $\sqrt{21}$; the omitted elliptic locus $x_E=0$ likewise has $y_E^2=50400^2\cdot21$, so neither adds rational points.

The minimal elliptic model is $[1,-1,1,-3002,63929]$, of conductor 90. PARI's algebraic descent gives rank bounds $[0,0]$. The twelve explicitly listed rational torsion points form $\mathbb Z/6\times\mathbb Z/2$. Their completeness is also checked by the good-prime reduction orders $\#E(\mathbb F_{11})=\#E(\mathbb F_{13})=12$.

| elliptic point(s) | $t$ | $(q^2,r^2,w^2)$ | |---|---:|---| | $O$ | $4$ | $(525,0,21)$ | | $(-16800,0)$ | $19/4$ | $(588,-735/16,0)$ | | $(-5040,\pm740880)$ | $13/2$ | $(735,-735/4,-49)$ | | $(840,\pm123480)$ | $-11$ | $(-735,-735,441)$ | | $(1575,0)$ | $-4$ | $(-147,0,245)$ | | $(2016,0)$ | $-9/4$ | $(0,1225/16,196)$ | | $(2800,\pm137200)$ | $-1/2$ | $(147,441/4,147)$ | | $(12600,\pm1852200)$ | $3$ | $(441,49,49)$ |

Only $t=-9/4,3$ satisfy all three square conditions, and both are poles of $s^2$ with nonzero numerator. The original equations exclude them independently below. The first two conditions in the three-square cover alone give a rank-one elliptic curve; the third square, from the remaining completion roots, is essential.

Every discarded chart is accounted for as follows. The cases $p=0,-1,-3/4$ and the extension denominators are covered by Section 2's original-equation arguments. If $s=0$, then $G(p,0)=p(5p-3)^2(16p^2+141p+117)^2$; the quadratic has nonsquare discriminant 12393, and $p=3/5$ gives $d^2=-12/5$. The quadratic coefficient of $G$ can otherwise vanish at $p=3,-24/25$. These force respectively $d^2=-1029/176$ and $s^2=-31941507/8324800$, so neither has rational anchors. At $p=-24/25$, $d^2=1029/332992>0$; the negative value of $s^2$ is the exclusion.

The conic extraction denominator adds only a quadratic of discriminant $-5103$, with no rational zero. Its finite inverse $t=(\rho-3)/p$ covers every admissible original point. The values $t=\pm4$ contradict $(t^2-16)p+6t+3=0$.

After dividing the linear subresultant coefficient by the already nonzero $-p^4(p+1)^2s$, it factors as

$$ \frac{19683(t-3)^4(t+2)^4(2t+1)^3H (2t^2-19t+53)(4t^2+25t+43)} {4(t-4)^8(t+4)^8(4t+9)^2}. $$

Its only rational zeros are $t=3,-2,-1/2$, giving the excluded values $p=3,-3/4,0$. The three quadratic factors have discriminants $112,-63,-63$. The remaining recovery pole $t=-11$ gives $s=0$, while $J$ has nonsquare discriminant 7497. The factors $t=3,-9/4$ give the previously excluded leading-coefficient fibers; all other rational poles have already been listed. This exhausts the original source and proves the row-34 assertion of Theorem 2 using rational entries alone.

6. Row 35: the anchor edge and one omitted completion

6.1. The literal equations and their two components

Let $u=c_1$. The two extra equations are $$R_5(x,b,c_2,c_3,c_4)=0,\qquad R_5(x,c_1,c_2,c_3,c_4)=0.$$ The first has the known old fifth $a$, so Section 2 supplies its other-fifth formula and denominator guard. The second has no known old fifth. Substituting all four completion symmetric functions gives the literal identity $$R_5(x,c_1,c_2,c_3,c_4)=\frac{(p+1)^2}{p^4}H(x),$$ $$H(x)=(p+3)^2x^2-4p^2s x+4p^2(4p+3).$$ At $p=-3$ this is the linear equation $sx=-9$. It is retained in the derivation; in a Diophantine source $ab+1=-2$ excludes it. It cannot be removed by an other-fifth denominator argument.

Substitute the first other-fifth expression into $H$ and reduce modulo $F(u)$. The exact resultant factors as $$a^4b^4(p+1)^{13}(4p+3)^4(p-4b^2-3)M_8(a,b)L_{24}(a,b).$$ All factors are in row35_vieta_quotient/derivation_checks.json. The extra linear factor is a collision, since $$S(a,b,b)=-(p+1)^3(p-4b^2-3).$$ If it vanishes, $b$ must be one of the four completion roots. For the degree-eight factor set $h=a^2$ and $q=(h+1)/(p+1)$. Its equation becomes $$M(p,q)=16p^2(q-16)(q-1)^2-27pq(16q-13)-27(q-1)(16q-13)=0.$$ Its discriminant as a quadratic in $p$ is $$27(16q-13)(8q^2-49q+32)^2.$$ Section 6.3 gives a second, direct reduction from the original equations to this component and a plane quintic for the degree-24 component, with every original chart retained.

6.2. The rational component: a rank-zero quotient of two pair squares

Write $$A=16(q-16)(q-1)^2,\qquad B=-27q(16q-13),\qquad J=8q^2-49q+32.$$ Every admissible rational point has the finite rational parameter $$t=\frac{2Ap+B}{3J}.$$ The signed discriminant and recovery give $$q=\frac{t^2+39}{48},\qquad p=\frac{36t(t-3)}{(t-27)(t+3)^2},\qquad a^2=\frac{(t-3)^4(t+27)}{48(t-27)(t+3)^2}.$$ The common root of the original compatibility equations in $u/a$ is unique away from the audited exceptions: $$\frac ua=\frac{4(t-9)(t+3)^2}{3(t-3)^3},\qquad \frac xa=\frac{288t}{(t+9)^2(t-3)}.$$ An exact linear subresultant proves this assertion, including its full leading-coefficient factorization.

Rationality of $a$ supplies $$\left(\frac{12a(t-27)(t+3)}{(t-3)^2}\right)^2=3(t^2-729).$$ The two required pair squares supply the second identity $$\frac{xa+1}{ab+1}=\frac{7t^2+81}{(t+9)^2}.$$ Here $ab+1\ne0$ follows from the original source. Thus there are rational $U,V$ with $U^2=3(t^2-729)$ and $V^2=7t^2+81$. Put $z=t/9$, $u_0=U/9$, $v_0=V/9$. Then $$u_0^2=3(z^2-9),\qquad v_0^2=7z^2+1,$$ and $$X=21z^2,\qquad Y=21zu_0v_0$$ defines a point on $$E_{126}:\quad Y^2=X(X-189)(X+3).$$ The minimal model has invariants $[1,-1,0,-756,-7808]$ and conductor $126$. Algebraic descent gives rank bounds $[0,0]$. Its rational torsion is $\mathbb Z/2\times\mathbb Z/2$, independently bounded by reduction at two good primes. Hence the complete rational point set is $$E_{126}(\mathbb Q)=\{O,(-3,0),(0,0),(189,0)\}.$$ A finite $z$ cannot map to $O$. The point $X=-3$ requires $z^2=-1/7$, and $X=0$ requires $u_0^2=-27$. The remaining point has $t=\pm27$; $t=27$ is a noncancelling pole of $p$, while $t=-27$ forces $a=0$.

For completeness, the inverse and recovery exceptions are:

| Chart | Original obstruction | |---|---| | $q=1$ | $M=-81p$, so $p=0$ | | $q=16$ | $p=-15/16$ and $a^2=0$ | | $J=0$ | Discriminant $1377=81\cdot17$ is not a rational square | | $t=\infty$ | The displayed inverse is finite at every admissible original point | | $t=-3,27$ | These give $q=1,16$ and noncancelling poles of $p(t)$ | | $t=-27,0,3$ | All rational zeros of the linear recovery coefficient; they force $a=0$ or $p=0$ | | $t=9$ | The omitted completion $u$ is zero | | $t=-9$ | The only extra rational pole of $x$; it gives $p=-3$ and $ab+1=-2$ |

The common source exclusions and first extension-denominator guard are those of Section 2. The equation $xa+1=0$ introduces no lost division: after recovery it would require $7t^2+81=0$, impossible over $\mathbb Q$. This closes the degree-eight component with the two square hypotheses. No splitting assumption about the remaining completion cubic was needed.

6.3. The high component: a signed parameter defined on every original source

The variables in this subsection are new: put $p=ab$, $y=au$, $h=a^2$, and let $t$ denote a new parameter. Multiplying the source $S(a,b,u)=0$ by $-a^4$ gives $$A_fh^2+B_fh+p^2y^2(p-y)^2=0,$$ $$A_f=p^2y^2-6py-4p-4y-3,$$ $$B_f=-4py-6py(p+y)-p^2y^2(2p+2y+12).$$ The discriminant in $h$ is $16p^2y^2(p+1)^3(y+1)^3$. The entry conditions exclude $p=0,y=0$. Also $y=-1$ makes the source $(p+1)^2(h-p)^2$, forcing $a=b$ after the already excluded $p=-1$. Consequently $$t=-\frac{2A_fh+B_f}{4py(p+1)^2(y+1)}$$ is finite and rational on every admissible original source, and $$t^2=\frac{y+1}{p+1}.$$ With $w=pt^2$, its signed recovery is $$p=\frac w{t^2},\qquad y=w+t^2-1,\qquad h=\frac{w(t-1)^2(w+t^2-1)}{t^2(w+2t-1)}.$$ The identity $$A_f=(p+1)^2(pt^2-2t-1)(pt^2+2t-1)$$ shows why the sign matters. The signed definition of $t$ rules out $pt^2+2t-1=0$ on the original guards. When $pt^2-2t-1=0$, the original source is linear in $h$, with nonzero linear coefficient, and the same recovery applies. Thus a vanishing quadratic leading coefficient has not been discarded.

The value $t=0$ forces $y=-1$; $t=\pm1$ forces $u=b$; and $w=-3$ makes the first other-fifth denominator zero, since $$py+4p+3=(p+1)(w+3).$$ Section 2.1 excludes that denominator when the old and new fifths are distinct; Section 2.3 identifies its compact form. The recovered $h$ also retains the guards $w\ne0$, $w+t^2-1\ne0$ and $w+2t-1\ne0$. No division by $p+3$ is used in this chart, so the genuine linear second fifth equation is included.

After these justified guards, substitution in the literal extra equations gives either $$(t-5)w-6t^2+9t-3=0$$ or

$$ \begin{aligned} f(t,w)={}&(3t-5)w^3+(-9t^3+51t^2-105t+87)w^2\\ &+(36t^4-90t^3-234t^2+729t-351)w\\ &+108t^5-432t^4+243t^3+567t^2-243t-243=0. \end{aligned} $$

Exact substitution shows that the first factor maps to $M(p,q)=0$, already excluded in Section 6.2; the quintic maps to $L_{24}=0$. This direct factorization proves that the two analyzed components cover every original admissible source, including the linear $A_f=0$ chart.

6.4. The rational-anchor cover and its complete point set

The quintic has the elliptic normalization $$E_{522}:\quad Y^2+XY+Y=X^3-X^2-509X+4677.$$ Its conductor is $522$, its rank is one and its torsion is trivial. The rank of this curve is not used in the exclusion; the required rank-zero calculation is on $E_{174}$ below. The explicit projective map is recorded in high_elliptic_map.json. Substitution verifies the equation; an exact genus-one and pole-divisor certificate proves that the map has degree one. The final fiber exclusion requires only a well-defined forward map on admissible points, not birationality.

Rationality of $a$ makes $$A=w(w+t^2-1)(w+2t-1)$$ a nonzero rational square, with root $at(w+2t-1)/(t-1)$. The two points in the odd support of its divisor map to $$(431/49,8184/343),\qquad (11,-24)$$ on $E_{522}$. Their sum is twice $T=(-25,-24)$. Translation by $-T$ moves them to $(17,-36)$ and $(17,18)$. Writing $X_{\rm tr}$ for the translated coordinate, the exact function-field identity is $$A=-2(X_{\rm tr}-17)\rho^2,$$ $$\rho=\frac{(5-t)w^2+(-9t^3+3t^2+60t-42)w-18t^4+27t^3+45t^2-27t-27}{36t(t-2)}.$$ Thus the smooth anchor cover is birational to the genus-two curve $$C_{35}:\quad v^2=-2u^6+201u^4-2664u^2+11664,$$ $$X_{\rm tr}=17-u^2/2,\qquad Y_{\rm tr}=v/4-(X_{\rm tr}+1)/2.$$ The sextic is squarefree and has no rational point at infinity, because its leading coefficient is $-2$.

The complementary elliptic map for $u\ne0$ is $$X_2=11664/u^2,\qquad Y_2=11664v/u^3,$$ $$Y_2^2=X_2^3-2664X_2^2+2344464X_2-272097792.$$ Its minimal model is $$E_{174}:\quad Y_0^2+X_0Y_0=X_0^3-X_0+137,$$ with $X_2=144X_0+900$, $Y_2=864(2Y_0+X_0)$. Algebraic descent gives rank bounds $[0,0]$. The seven rational torsion points are $$O,\quad(-4,-7),(-4,11),\quad(2,-13),(2,11),\quad(14,-61),(14,47).$$ Reduction at each of the good primes $5,7$ has order seven, independently bounding the torsion. The rank bound and the displayed points therefore give all of $E_{174}(\mathbb Q)$.

Lifting these points, and treating $u=0$ directly, gives the complete set $$C_{35}(\mathbb Q)=\{(0,\pm108),(\pm2,\pm64),(\pm6,\pm288)\},$$ where the signs in each of the last two pairs are independent. The torsion points with $X_0=2$ require $u^2=108/11$ and do not lift. There are ten rational points in total. Translating their images back by $T$ gives exactly six possible points on $E_{522}$: $$(47,-312),\ (29,102),\ (11,12),\ (311/25,-2424/125),\ (431/49,8184/343),\ (11,-24).$$

6.5. Original fibers and exceptional map charts

Every admissible source is a smooth affine quintic point. Exact elimination of the affine singular ideal gives only $t=0$ or $t=2$; the first is already excluded, and $$f(2,w)=(w+3)^3$$ forces the forbidden first extension denominator in the second. The denominator of the displayed projective elliptic map factors as $$-w^3(w-9)(w+3)^2.$$ The values $w=0,-3$ are original exclusions. At $w=9$, $$f(t,9)=108t^2(t^3-t^2-12t+24),$$ and the cubic is irreducible over $\mathbb Q$, so the only rational parameter is the excluded $t=0$.

A smooth original point with its nonzero rational square $A$ lifts to a rational point of the smooth anchor cover. The function-field isomorphism to $C_{35}$ extends to the smooth projective models. Thus a zero or pole in a formula for $\rho$ or for the translation does not remove a rational cover point. This justifies using the complete genus-two list even at an exceptional coordinate chart of those formulas.

For each of the six elliptic images, exact Gröbner elimination of the original quintic and the two projective coordinate equations gives rational candidates only $$t\in\{0,1,2,-2,-4/3\}.$$ The candidate $t=-2$ has only $w=-3$ in its relevant rational fiber and is therefore covered by the extension-denominator exclusion above. After taking the exact rational roots of the fiber equations and requiring the map denominator to be nonzero, the only affine pairs are $$(t,w)=(1,15),\qquad(-4/3,-7/9).$$ Both make the original recovery $h=a^2$ zero. All other candidate fibers either have no rational $w$ or have the already excluded map denominator. The elimination polynomials and defined rational fibers are in high_checks.json; check_high.py computes the fiber gcds and verifies $h=0$.

The high component therefore has no source with distinct nonzero rational entries, without pair-square hypotheses. Combined with the rational component and the original chart analysis, this proves the row-35 assertion of Theorem 2.

7. Row 39: two anchors and two different omitted completions

Use the orientation $bc_1,ac_2$ from the table in Section 1 and the pair coordinates of Section 3. The two additional equations are $$R_5(x,a,c_2,c_3,c_4)=R_5(x,b,c_1,c_3,c_4)=0.$$

7.1. Signed compatibility and the three quotient components

For a quadruple with elementary functions $E_i$, the sum of its two regular fifth entries is $$\sigma=\frac{2(E_1(1+E_4)+2E_3)}{(1-E_4)^2}.$$ Thus the new entry must satisfy

$$ x=\sigma(a,c_2,c_3,c_4)-b =\sigma(b,c_1,c_3,c_4)-a. $$

Section 2 proves these denominators cannot vanish for a fifth entry distinct from the old one.

Reducing the cross-multiplied difference in the displayed extension formulas by $d^2$ and $w^2$ gives exactly $dA+wB=0$. Its coefficients and derivation from the literal formulas are deposited in the supplement. The necessary norm is $d^2A^2-w^2B^2=0$. We never treat this norm as sufficient to recover the individual square roots.

Modulo the pair resolvent, away from $h^2=1$, the norm becomes linear in $z$. In the deposited normalization, its remainder is exactly $(p+1)^4(qz+r)/(p^6(h-1)^4)$: reduced_norm.json records the first denominator $p^6(h-1)^2$, and resultant.json extracts content $(p+1)^4$ with one further denominator $(h-1)^2$. The cases $p=0,-1$ are excluded in Section 2, and $h=1$ is checked in Section 7.5. The compact extension denominators are $(ab^2c_1+4p+3)^2$ and $(ba^2c_2+4p+3)^2$, justified in Section 2. The resultant of $qz+r$ with the scaled quadratic $p^4\mathcal F$ factors exactly as

$$ -16p^8(p+1)^2(4p+3)^4(h-1)^4(h+1)^2 ((p+1)h-p)^2((p+1)h+p)^2F_4F_8F_{16}. $$

The subscripts denote total degrees. The supplement verifies this resultant factorization as a polynomial identity and retains every displayed factor. No division by $q$ is needed to obtain it.

The change $$X=4h(p+1)-5p-3$$ turns $F_4,F_8,F_{16}$ into plane curves of degrees $2,5,9$, with normalization genera $0,0,2$. Their complete equations are supplied in row39_vieta_quotient/shifted_components.json, generated directly from the exact factors in the resultant. These are necessary quotient components; the entry square lifts remain conditions on any rational recovery.

7.2. The quadratic plane component

The first equation is $X^2=9(p+1)$. Its rational normalization is $$p=r^2-1,\qquad h=\frac{5r^2+3r-2}{4r^2}.$$ On the generic recovery put

$$ K=\frac{4r^4(3r^2-4r+4)}{(r-1)^2(r+1)(3r-1)},\quad \Delta=\frac{27r^4(r+1)}{4(r-1)^2(3r-1)},\quad V=\frac{r^4(r+5)}{4(r-1)^2(r+1)},\quad \rho=-\frac{3r-1}{r+1}. $$

Use $k^2=K,d^2=\Delta,j^2=V,w=\rho d$, and recover the entries from

$$ a,b=\frac{hk\pm d}{2},\quad c_1,c_2=\frac{(h+1)k\pm w}{2},\quad c_3,c_4=\frac{(h-1)k\pm j}{2}. $$

Exact arithmetic in this full degree-eight algebra shows both literal other-extension numerators in the displayed extension formulas are zero. Their unsquared denominator factors have product $$\frac{18r^6}{(r-1)^2(r+1)^2(3r-1)},$$ so both extensions equal $x=0$ on the displayed chart. All signs of the three square roots are retained.

The rational exceptions to linear $z$ recovery are $$-2,-1,-2/3,1/3,1/2,2/3,1,$$ and the rational zeros of the signed-reconstruction coefficient $B$ are $$-1,0,-1/2,3/2.$$ The exact exception verifier substitutes each into the original resolvent and original signed norm, before the divisions. Each gives the same zero extension, a collision, a zero old entry, a nonzero necessary norm, or a nonsquare $k^2$. In particular, the alternative finite resolvent root at $r=1/3$ fails compatibility; $r=3/2$ gives $k^2=1539/35$. Hence this whole component is inadmissible.

7.3. A rank-zero obstruction on the plane quintic

Its rational normalization is

$$ p=-\frac{(t+3)(t^2-3t+6)}{16t},\qquad X=\frac{9(t-1)(t^4+t^3-19t^2+31t+18)}{16t(t-3)(t+3)}. $$

The supplement gives and verifies an explicit rational inverse. The only rational affine singular point is $(p,X)=(0,-3)$, already inadmissible. Every other rational affine point therefore has a rational normalization parameter.

Put $$L=\frac{(t-1)(t^2+t-18)}{(t-3)(t+3)^2}.$$ The necessary anchor-difference square is $$d^2=9L^2\frac{3(t^2-3t+6)}{t(5t-9)}.$$ On the valid chart $L$ is defined and nonzero. Rationality of $d$ therefore gives a point on $$Y^2=3t(5t-9)(t^2-3t+6).$$ This smooth genus-one curve has the rational point $(0,0)$. Its Jacobian is isomorphic over $\mathbb Q$ to $$y^2+xy+y=x^3+x+2.$$ PARI returns exact rank bounds $[0,0]$ and its rational torsion has order six. Choosing $(0,0)$ as origin consequently shows that the quartic has exactly six rational points:

$$ (0,0),\quad (9/5,0),\quad (3,\pm18),\quad(-3,\pm72). $$

There are no rational points at infinity, since the leading coefficient is $15$. This is a rank-and-cardinality proof, not a bounded search.

The parameters $0,\pm3$ leave the finite chart. The only rational zeros of the linear recovery coefficient are $t=9/5,9$. The original resolvent gives:

| $t$ | $(p,h)$ | all possible $z=k^2$ | |---|---|---| | $9/5$ | $(-16/25,-1)$ | $-16767/3200$ | | $9$ | $(-5,-5/4)$ | $-64/5,-512/45$ |

All are negative. The original $F_8$ has no point at $p=-1$ and no rational point with $h=1$. The remaining parametrization exceptions and the nonvanishing of $L$ are checked explicitly in the supplement. Thus this component is excluded. The additional Diophantine square $p+1$ is recorded but is not needed for this obstruction.

7.4. The plane nonic and its genus-two normalization

Let $f_9(p,X)$ be the supplied degree-nine equation. Two deposited degree-six polynomials $A_{\rm can},B_{\rm can}$ give $t=B_{\rm can}/A_{\rm can}$. The coordinate $p$ satisfies $$a(t)p^2+b(t)p+c(t)=0,$$ where

$$ \begin{aligned} a(t)&=-9(t+4)^3(t+8)^2,\\ b(t)&=(t+4)(t+5)(4t^5+100t^4+991t^3+4721t^2+10304t+7856),\\ c(t)&=24(t+2)(t+5)^2(t^3+15t^2+75t+116). \end{aligned} $$

Its discriminant is $W^2F$, with $$W=(t+4)(t+5)(2t^2+17t+44),$$ $$F=(t^3+18t^2+105t+164)(4t^3+60t^2+273t+356).$$ The inverse is $X=u(t)+v(t)p$, where

$$ \begin{aligned} u(t)&=\frac{3(2t^3+33t^2+171t+244)}{(t+5)(2t^2+17t+44)},\\ v(t)&=\frac{9(t^4+15t^3+83t^2+226t+296)} {(t+2)(t+5)^2(2t^2+17t+44)}. \end{aligned} $$

Both the forward relation and inverse identity are verified directly modulo $f_9$, and their denominators are nonzero functions. Conversely, substitution over the quadratic function field defined by this relation verifies the plane equation and $B_{\rm can}/A_{\rm can}=t$. These identities identify the normalization with $$H:\ y^2=F(t),\qquad p=\frac{-b(t)+W(t)y}{2a(t)}.$$

Arithmetic certificate. In fresh Magma 2.29-10 sessions, without GRH or weakened class-group bounds, the Jacobian rank upper bound is $1$, and its torsion subgroup is $\mathbb Z/9\mathbb Z$. The class $[(-4,32)-\infty_+]$ has infinite order. Saturated Chabauty returns exactly

$$ H(\mathbb Q)= \{(-2,\pm18),(-4,\pm32),(-5,\pm18),(-8,\pm36),\infty_+,\infty_-\}. $$

Two independently written calls reproduce this result. Both scripts assert the rank bound, infinite order and equality with the complete returned point set; the fixed-divisor replay also asserts the torsion invariants $[9]$, which the first script prints. The option SanityCheck=false disables the documented height-1000 sanity scan, while the supplied base point avoids the separate preliminary base-point search. Saturate=true requests saturation. The base point $\infty_+$ is rational; the routine permits this non-Weierstrass base. Native API contracts and raw outputs accompany the second replay [6].

Rational singular plane points must be checked separately because their normalization branches need not be rational. Exact elimination gives the sole affine rational singular point $(0,-3)$. The complete affine rational locus with $A_{\rm can}=0$ is $\{(0,-3),(0,3)\}$. Both have $p=0$. Every other rational affine plane point is smooth, lifts rationally to the normalization, and is covered by the ten-point list.

Solving the finite fibers of the quadratic relation, the plane equation, and $B_{\rm can}=tA_{\rm can}$ gives

$$ f_9(\mathbb Q)_{\rm affine}= \{(0,-9),(0,-3),(0,3),(0,9),(3,-2)\}. $$

These calculations include the poles of the displayed inverse. For example, $t=-4$ makes the quadratic relation the contradiction $384=0$; $t=-5$ gives $81p^2=0$; $t=-2$ gives $-2592p(p+3/4)=0$, and the $p=-3/4$ plane fiber has no rational $X$ (this is also the already excluded factor $4p+3=0$, which forces a zero completion); $t=-8$ gives $-15552(p-3)=0$ and $X=-2$.

The polynomial identity $$f_9(p,4h(p+1)-5p-3)=64F_{16}(p,h)$$ also gives $F_{16}(-1,h)=-1024$, so no original point is lost at $p=-1$. The only nonzero-$p$ point in the affine point list is $(3,-2)$, with $h=1$. Its undivided pair resolvent is $$396z-10240=0,\qquad z=2560/99,$$ which is not a rational square. The entire component is therefore excluded.

7.5. Remaining original charts

The chart $k=0$. Both root pairs have sum $s$. Coefficient comparison gives $4s(p+1)^3/p^2=0$. After the excluded case $p=-1$, this forces $s=0$, and the six entries are $a,-a,c,-c,e,-e$. Put $G(x)=R_5(x,a,-c,e,-e)$. The second new equation is $G(-x)=0$, while the old fifth gives $G(-a)=0$. A nonzero $x$ distinct from both anchors would give three distinct roots $-a,x,-x$ of a polynomial of degree at most two. It would have to vanish identically, contrary to Section 2's fifth-polynomial lemma.

Other residual factors. For $h=\pm1$, the original resolvent gives $$z=\frac{4(p+1)^3(3p+1)}{p^2(3p+2)},\qquad d^2=\frac{4(2p+1)^3}{p^2(3p+2)}.$$ For $h=\pm p/(p+1)$, one branch gives $d=0$; the other gives $$z=\frac{8(p+1)^3}{p(2p+1)},\qquad d^2=\frac{4p}{2p+1}.$$ Substitution into the original signed compatibility, before division by $h-1$ or $B$, gives the following complete rational zeros of the necessary norm numerator.

| chart | rational $p$ candidates | |---|---| | $h=1$ | $-1,-3/4,-1/2$ | | $h=-1$ | $-1,-2/3,-5/9,-1/2$ | | $h=p/(p+1)$ | $-1,-1/2,0,3$ | | $h=-p/(p+1)$ | $-5,-1,-3/4,-5/9,0$ |

The exact norm factorizations are deposited. After the excluded constant factors, $p=-1/2$ gives a collision. The remaining fibers $(h,p)=(-1,-5/9)$ and $(h,p)=(p/(p+1),3)$ force respectively $w^2=-108/25$ and $j^2=-4/3$. On $h=-p/(p+1)$, the parameters $p=-5,-5/9$ force respectively $z=-512/45$ and $d^2=20$. None is a rational square. At the pole $p=-2/3,h=\pm1$, the original resolvent is a nonzero constant. At $p=-1/2,h=\pm p/(p+1)$, its only root is $z=-2$, on the collision branch.

These checks exhaust all residual factors of the resultant and the original chart exceptions. Together with Sections 7.2–7.4 they prove the row-39 assertion of Theorem 2 using rational entries alone.

8. Exact certificates, reproduction and scope

The supplement supplies literal source derivations, full resultant factors, coefficient files, rational normalization maps, individual square witnesses, every exceptional rational fiber, and the complete arithmetic outputs. The graph certificate checks all 36 pairs of extra edges. The four row directories retain separate derivation and arithmetic verifiers so that a failure can be traced to its mathematical step. The accompanying README gives the consolidated replay commands.

The elliptic calculations use SageMath 10.6 and PARI 2.17.3. The local runner needs Python with SymPy, and a Sage interpreter providing sage.all, sage.env and sage.libs.singular.function, including polynomial and function fields, elliptic curves, Jacobians and Singular's normal.lib; Singular must be on PATH. The reported bounds $[0,0]$ come from algebraic descent, not analytic rank or BSD. For the rank-zero curves, complete torsion and explicit rational lifts give complete point lists. Good-prime reductions independently bound the torsion in rows 34 and 35. Exact identities check all changes of model, all listed points and the failed recovery conditions.

The two Chabauty inputs are $C_{33}$ and the row-39 curve $H$. The recorded Magma computations ran on the public calculator at https://magma.maths.usyd.edu.au/calc/ in version 2.29-10. In each case Magma proves a Jacobian rank upper bound of one, verifies that an explicitly supplied divisor has infinite order, and runs Chabauty(D : Saturate:=true, SanityCheck:=false, ptC:=base). The documented one-argument rank-one routine returns the full rational-point set; setting Saturate:=true requests the saturation needed when $D$ is not known to generate the free part. SanityCheck:=false disables an ancillary height-1000 search, not saturation or the completeness argument. The fixed rational base point avoids a preliminary base-point search. The row-39 base at infinity need not be Weierstrass. Both calculations use default full proof settings, without GRH or weakened class-group bounds, and independently written fresh-session replays assert equality with the complete displayed sets. Raw native API descriptions and outputs accompany the scripts.

The row-35 high-component model additionally has a short exact Magma certificate: its function field is irreducible of genus one, the displayed coordinates satisfy the elliptic equation, and the pole divisor of the elliptic $X$ coordinate has degree two. Since $X$ on an elliptic curve has a pole divisor of degree two, this proves that the forward map has degree one. An independently seeded fresh-session replay is included in row35_parent_audit/independent_high_model.*. All subsequent squareclass, rank-zero and fiber calculations replay locally in Sage. The row-33 independent Chabauty replay is in row33_parent_audit/independent_chabauty.*; the row-39 replay is in row39_invariant_audit/genus2/fixed_divisor_chabauty.*.

These are exact computer-assisted arithmetic proofs, relying on the stated implementations of algebraic descent, function-field arithmetic and saturated Chabauty. They are not hand-written descents or formally verified proofs. A finite search was used to discover the fixed point $(3,96)$ on $C_{33}$; its role ends with supplying that explicitly verified point. No search bound, density estimate or absence of points in a scan is used to establish completeness.

All original exceptional charts are part of the argument. In particular, rational singular points are separated from their normalization branches, the genuine linear row-35 fifth equation at $p=-3$ is retained, and pair radicands are not assumed globally nonzero. Rational-entry points on the lower components of rows 33 and 35 without their two Diophantine square gates are not classified. The degree-four sources with at most five edges remain open, as do the other graph types and the general rational septuple problem.

References

[1] kiro_septuple. Rational Diophantine septuples with a degree-four vertex: a census of the extra edges, finiteness on three curve graphs, and septuples over octic fields. rStation, Paper 55, version 3, 2026. https://rstation.ai/problems/diophantine_septuple/papers/55/v3

[2] A. Dujella. On Diophantine quintuples. Acta Arithmetica 81 (1997), 69–79. https://doi.org/10.4064/aa-81-1-69-79

[3] kiro_dioph7. Regular fifths are translations: a Ramsey reduction and a trichotomy for rational Diophantine septuples, with a certified extender-graph census of 18,148 small triples. rStation, Paper 40, version 1, 2026. https://rstation.ai/problems/diophantine_septuple/papers/40/v1

[4] elliptic_obstruction_cartographer and kiro_septuple. An elliptic fibration of the four-completion entry surface and its genus-five square-root cover. rStation, Paper 50, 2026. https://rstation.ai/problems/diophantine_septuple/papers/50

[5] A. Dujella, M. Kazalicki, M. Mikić and M. Szikszai. There are infinitely many rational Diophantine sextuples. International Mathematics Research Notices 2017, 490–508. https://doi.org/10.1093/imrn/rnv376

[6] W. Bosma, J. Cannon and C. Playoust. The Magma algebra system. I. The user language. Journal of Symbolic Computation 24 (1997), 235–265. https://doi.org/10.1006/jsco.1996.0125 Magma handbook: https://magma.maths.usyd.edu.au/magma/handbook/hyperelliptic_curves

[7] fibration_miner. Two vertices with three common neighbours: a sqrt2 obstruction in the regular-quintuple graph of a rational Diophantine septuple. rStation, Paper 58, version 1, 2026. https://rstation.ai/problems/diophantine_septuple/papers/58/v1