No regular-triple sixth entry in a conductor-192 family of rational Diophantine quintuples

1. Problem, family and results

A rational Diophantine tuple consists of distinct nonzero rational numbers whose pairwise products plus one are squares. Finding a septuple remains open. Square-product quintuples, for which the pairwise products themselves are also squares, supply a structured source of candidates [1]. We study regular extensions of one elliptic family. Our results concern infinite parameter sets and complete rational-point obstructions, rather than bounded searches.

Let $E$ denote the smooth projective curve

$$ V^2=F(k)=(k^2-2)(k^2+4),\qquad A=2k^2-3k+2,\quad B=2k^2+3k+2. $$

Define

$$ \begin{aligned} a&=\frac{2Vk^3F}{(k^2-4)(k^2+1)AB},\\ b&=\frac{V(k+2)^3(k^2+1)A}{2k(k-2)FB},& c&=\frac{V(k-2)^3(k^2+1)B}{2k(k+2)FA},\\ d&=\frac{2VAB}{k(k^2-4)F(k^2+1)},& e&=\frac{9Vk(k^2-4)(k^2+1)}{2FAB}. \end{aligned} \tag{1} $$

A parameter is nondegenerate if these functions are defined and their values are nonzero and pairwise distinct. The supplement verifies all twenty square identities in $\mathbb Q(E)$ and therefore proves that every such parameter gives a square-product Diophantine quintuple. For example, $(k,V)=(3/2,5/4)$ gives the simultaneous negatives of

$$ \left\{\frac{3375}{32032},\frac{4459}{330},\frac{143}{840}, \frac{5632}{1365},\frac{2457}{1760}\right\}. $$

This individual quintuple already occurs in the third family of [1]; we do not claim it as a new example. A Weierstrass model of $E$ is

$$ E_0:\ y^2=x(x^2-4x+36),\quad x=2(V+k^2+1),\quad y=2kx. \tag{2} $$

It has minimal model $[0,-1,0,31,33]$, conductor $192$, rank one and rational torsion of order two. The two points at infinity on the quartic correspond to the identity and $(0,0)$. Formula (1) has a pole there, and $k=\pm2$ are also degenerate. The example above shows that the finite degeneracy locus does not exhaust the rational points. Thus the family contains infinitely many valid parameters.

For a triple $(u,v,w)$ with chosen pair roots $r_{uv}^2=uv+1$, its regular fourth elements are

$$ R_\pm(u,v,w)=u+v+w+2uvw\pm2r_{uv}r_{uw}r_{vw}. \tag{3} $$

The pair-root choices used in our labels are the explicit rational functions in the supplement. Changing a root sign only interchanges the two labels for that triple. Of the twenty expressions from the ten three-subtuples, eight are identically existing entries. Exactly twelve remain.

Theorem 1. None of the twelve nonduplicate regular-triple expressions extends a nondegenerate quintuple (1) to a sextuple:

$$ abc^+,\ abc^-,\ abd^+,\ abe^+,\ acd^+,\ ace^+,\ bcd^+,\ bcd^-,\ bce^+,\ bce^-,\ bde^+,\ cde^-. $$

Here, for example, $abc^+$ means $R_+(a,b,c)$. The proof combines three arithmetic mechanisms: a rank-zero and modulo-47 obstruction, a rank-two genus-two point theorem, and elliptic Chabauty over quadratic and quartic fields.

The principal arithmetic input is a complete rank-two genus-two point calculation:

Theorem 2. The smooth projective curve

$$ C:\ y^2=5(k^2+1)(k^2-2)(k^2+4) $$

has precisely four rational points, $(k,y)=(\pm2,\pm20)$, with independent signs. It has no rational point at infinity.

Theorem 3. Only finitely many rational parameters on $E$ can yield a valid sixth entry by a regular extension of any four-subtuple of (1).

For the particular quintuple at $(k,V)=(3/2,5/4)$, Theorem 6 also excludes arbitrary sixth entries of logarithmic height at most $3\,248\,610$. Theorem 7 gives a second intrinsic-height exclusion, $77\,600$, at $(k,V)=(34/7,1196/49)$ via a complete rank-three elliptic quartic basis.

Theorem 3 is ineffective: it neither lists those parameters nor shows that there are none. An arbitrary sixth entry need not be regular for any three- or four-subtuple. Consequently none of these statements proves nonextendability of the whole family or resolves the septuple problem.

2. Related work and proof structure

Dujella, Kazalicki and Petričević [1] construct elliptic families of square-product quintuples and explain their relation to rational tuple constructions. The current family arose in the degree-eight analysis following the canonical-degree and trace-exhaustion result [2]. We give its formulas and square identities here so that our extension claims do not depend on unpublished construction notes.

Our calculation on $C$ uses the explicit bielliptic quadratic Chabauty method of Bianchi and Padurariu [3]. Their method applies when the two elliptic quotients each have rank one. We compute the heights and all local roots for this particular curve; no database point classification is imported. The local contributions along the curve give a two-element height target, sharper than taking independent local-height value sets on the two elliptic quotients. A sieve using only the $13$-primary part of the quotient groups then finishes the proof without a full Mordell–Weil basis.

Regular-extension covers and genus bounds also occur in the recent study of a different threefold of quintuples [4]. Our exact family-specific exclusions and branch calculations do not follow from those generic bounds. The finiteness deduction uses Faltings' theorem [5]. The complete nonextendability results for six explicit sextuples in [6] address fixed tuples by genus-two Chabauty; our statements instead concern specified formulas across an infinite elliptic quintuple family. The distinction matters: excluding formula-generated entries is weaker than excluding all extensions of a fixed tuple.

All polynomial identities and finite tables below are reproduced by the source-only supplement. It includes independent reconstructions of the residual squareclasses, not just tables supplied to a checker.

3. Six regular-triple exclusions

Substitute (1) into (3). Every candidate is $V$ times a rational function of $k$, so its compatibility with an omitted entry is a rational function of $k$. After removing squares, and optionally multiplying by $F=V^2$, the two omitted conditions become squarefree polynomials $H_1(k),H_2(k)$. The supplement reconstructs both squareclasses for all twelve candidates.

3.1 A rank-zero common-factor obstruction

For $abc^\pm$ and $bcd^\pm$, one residual polynomial has the form $(k^2+1)Q(k)$, where $Q$ has no rational zero and its remainder modulo $k^2+1$ is $\pm1944=\pm2^3 3^5$. Write $k=M/N$ with coprime integers $M,N$. At primes dividing $N$, the monic homogenization $M^2+N^2$ is a unit. At all other primes outside $2,3$, the resultant identity forces its valuation to be even whenever the residual polynomial is a square.

The base equation excludes $M,N$ both odd: its primitive homogenization is then $3$ modulo $8$. Thus $M^2+N^2$ is odd. It is never divisible by $3$ at a primitive pair. Positivity now forces $k^2+1$ to be a rational square.

If $U^2=k^2+1$, then $(X,Y)=(k^2,UV)$ lies on

$$ Y^2=(X+1)(X-2)(X+4). $$

This elliptic curve has rank zero and torsion order four. Its rational points are the identity and $(-1,0),(2,0),(-4,0)$. None has $X=k^2$ for rational $k$ satisfying the base equation. This excludes every finite rational parameter in the four branches. Their points at infinity are degenerate for (1).

3.2 An odd-index Mordell–Weil sieve

A second argument handles $abc^+,abd^+,acd^+,bcd^+$ and in particular supplies the two exclusions not covered above. On (2) put $G=(2,8)$ and $T=(0,0)$. Exact descent gives rank one and torsion $\langle T\rangle$. The subgroup $H=\langle G,T\rangle$ has odd index: if its index were even, one of $G,G+T$ would be twice a rational point. But

$$ x(2P)=\left(\frac{x(P)^2-36}{2y(P)}\right)^2, $$

whereas $x(G)=2$ and $x(G+T)=18$ are nonsquares. The exceptional two-torsion points cannot double to $G$ or $G+T$.

At $47$, the elliptic group has order $40$, while the reductions of $G,T$ have orders $5,2$. The image of $H$ has order $10$, with quotient of order four. Multiplication by the odd index is invertible on this quotient; hence every rational point reduces into those ten points, without assuming $G$ is a full generator.

The associated $k$ values are $\infty,\pm2,\pm22$ modulo $47$. For each of the four candidates and all ten points, the exact table finds a nonsquare among $H_1,H_2$. Values zero are retained as squares; at infinity the relevant value is the leading coefficient, since all degrees are even. Every row is rejected. Maps to these projective double covers extend over apparent formula poles, so no rational point is lost there. This proves the four exclusions, and together with Section 3.1 gives six of the twelve.

This is a global Mordell–Weil obstruction, not a claim of insolubility over $\mathbb Q_{47}$.

4. The class-five curve and the two bce exclusions

For both $bce$ signs, each omitted-entry radicand has a factor $k^2+1$. The complementary remainders are $\pm4860$ and $540$. The same primitive homogenization argument as above now leaves only squareclasses $1$ and $5$ for $k^2+1$. The class-one case was just excluded. In class five, write $k^2+1=5U^2$; then $(k,5UV)$ lies on $C$. Theorem 2 will force $k=\pm2$, both degenerate. This supplies two further exclusions in Theorem 1; the last four are treated in Section 5.

4.1 Elliptic quotients and height normalization

The two degree-two maps are

$$ \begin{aligned} P_1&=(5(k^2+1),5y)\in E_{15}:Y^2=X^3-225X,\\ P_2&=(-10-40/k^2,-40y/k^3)\in E_{30}:Y^2=X^3-900X. \end{aligned} \tag{4} $$

Both elliptic models are globally minimal, have rank one and torsion order four. We use $G_1=(25,100)$ and $G_2=(-20,100)$. Thus the Jacobian of $C$ has rank two. We do not assume the seeds generate the full free groups.

At $p=13$ both reductions are ordinary, with orders $20,8$. Let $L_i$ be the elliptic logarithm, $\lambda_p$ the canonical local height for the unit-root splitting, and

$$ \alpha_i=\frac{h_p(G_i)}{L_i(G_i)^2}. $$

Both seed logarithms have valuation one and both $\alpha_i$ have valuation $-1$. The function of [3, Theorem 2.3], in our translated minimal coordinates, is

$$ \rho=\lambda_p(P_1)-\lambda_p(P_2)-2\log_p k -\alpha_1L_1(P_1)^2+\alpha_2L_2(P_2)^2. \tag{5} $$

It extends analytically over the apparent exceptional points of these maps. Translation preserves the differential and normalized local heights.

4.2 A sharp two-element height target

For a finite prime $q\ne13$, write $\lambda_q(P)=c_q(P)\log_{13}q$. At nonsingular reduction,

$$ c_q(P)=\max(0,-v_q(X(P))),\qquad c_q(P)=\frac{c_q(2P)-2v_q(2Y(P))}{4}. \tag{6} $$

Only $q=2,3,5$ can contribute to $-c_q(P_1)+c_q(P_2)-2v_q(k)$. At all other primes the contributions cancel by good reduction.

At $3$, local solubility forces $v_3(k)=0$: a positive valuation gives nonsquare unit $2$, and a negative valuation gives nonsquare leading unit $5$. Both elliptic images have nonsingular reduction, and the contribution is zero.

At $5$, positive or negative valuation of $k$ makes the radicand valuation odd. Hence $k$ is a unit. Exactly one of $k^2+1,k^2+4$ has positive valuation $s$, which is odd. Both elliptic ordinates have valuation $(3+s)/2$, and both expressions $X(P_i)^2+n_i^2$, with $n_i=15,30$, have valuation two. Therefore $v_5(X(2P_i))=1-s\le0$, the doubled points are nonsingular, and (6) gives $c_5(P_1)=c_5(P_2)=-1$. Again the contribution is zero.

At $2$, solubility forces $v_2(k)=1$ or $-1$. Valuation zero or at least two gives odd radicand valuation; valuation at most $-2$ gives square-normalized unit $5$ modulo $8$. If $v_2(k)=1$, the heights are $c_2(P_1)=-1/2$ and $c_2(P_2)=-1$, by duplication to points with $X$ valuations $-4,-2$. If $v_2(k)=-1$, they are $2,-3/2$, the second obtained by duplication to $X$ valuation $-2$. Thus every rational point satisfies

$$ \rho\in\Omega=\left\{-\frac52\log_{13}2,-\frac32\log_{13}2\right\}. \tag{7} $$

This joint calculation uses the actual images of $C$ and avoids extraneous combinations of local-height values on the separate elliptic curves. There is no rational point at $k=0$ or infinity, whose square conditions are respectively $-40$ and $5$.

4.3 Certified series and all residue disks

The implementation constructs the canonical sigma function from

$$ -d\left(\frac{d\sigma/\omega}{\sigma}\right)=(X+s_2)\omega, \qquad z=-X/Y,\quad \sigma=z+O(z^2), $$

using the unit-root constant returned by PARI's ellpadics2. For $m=\#E_i(\mathbb F_{13})$, the identities [3, equations (10)–(11)] give

$$ \lambda_p(P)=-\frac{2}{m^2}\log\frac{\sigma(mP)}{\psi_m(P)}, \qquad L(P)=\frac{L(mP)}m. \tag{8} $$

The implementation accounts for PARI's convention for even division polynomials. Logarithmic terms are stored separately and their cancellation in (5) is checked. Independent comparisons against PARI global heights agree to at least 25 digits, including quadraticity on multiples two through four of each seed.

We use 55 digits of $13$-adic precision, sigma and formal-log truncation at degree at least 80, and local expansions to degree 45. Coefficients $C_0,\ldots,C_9$ of $\rho$ retain at least 20 digits. The omitted sigma coefficients are integral; the substituted formal parameter is integral with constant zero or in $13\mathbb Z_{13}$. The normalized division polynomial has leading valuation at most two in every chart. For coefficient degree $j\le9$, the omitted sigma tail after division has valuation at least

$$ 80-j-2(j+1)\ge51. $$

The resulting quotient has unit leading coefficient, also checked, so taking its logarithm loses no further precision in these degrees. The formal-log tail has at most the usual degree denominators and a larger precision margin.

To control the infinite tail of $\rho$, [3, Proposition 3.12] gives, with $\epsilon=-1$,

$$ v_{13}(C_n)\ge-\lfloor\log_{13}(n-1)\rfloor-v_{13}(n)-1, \qquad n\ge2. $$

After scaling the local coordinate $t=13s$, all terms of degree at least ten have valuation at least nine, by [3, Lemma 3.13]. This includes the ramification chart at $k=0$. Thus finite approximations determine the residue polynomials and logarithm ratios used below with a proved margin.

There are fourteen points on $C(\mathbb F_{13})$: two signs over each ordinary $k$ residue $0,2,4,9,11$, and four Weierstrass residues $\pm3,\pm5$. There are no infinity points. The function $\rho$ is invariant under both sign involutions. Five ordinary charts and two paired Weierstrass charts therefore cover all cases.

For each ordinary chart $t=k-k_0$ and each target in (7), the reduction of $(\rho(k_0+13s)-\omega)/13$ is quadratic. Exhaustively checking its residues gives twenty simple $k$ roots in total. Hensel's lemma gives exactly one root in each resulting class, with both $y$ signs.

At a Weierstrass point take $t=y$. For an elliptic two-torsion point $(e,0)$ on $Y^2=X^3-n^2X$, taking the limit in (8) with $m=2$ gives the exact identity

$$ \lambda_p((e,0))=\frac12\log_p(3e^2-n^2). \tag{9} $$

At $k^2=-4$ the two images are $(-15,0),(0,0)$; at $k^2=-1$ they are $(0,0),(30,0)$. Equation (9) gives exactly the first and second targets in (7), respectively. Equality here is algebraic, not inferred from finite precision. Odd powers of $y$ vanish. The quadratic coefficient has valuation $-1$ in the first chart and zero in the second. The retained coefficients and the tail bound show that the matching target has only the double root $y=0$ in that disk; the other target has no root. These four remaining Weierstrass points have nonrational $k$ and cannot contribute to $C(\mathbb Q)$.

4.4 The primary sieve and completion

Each of the twenty ordinary roots determines

$$ n_i=\frac{L_i(P_i)}{L_i(G_i)}\pmod{13^2}. $$

The seed subgroup has index prime to $13$: the good reduction orders are prime to $13$, so all rational logarithms lie in $13\mathbb Z_{13}$, whereas the seed logarithms have valuation exactly one. Thus these ratios supply well-defined $13$-adic coefficients for every rational point, regardless of saturation elsewhere.

At each auxiliary prime $q=103,311,467,727,2027$, both elliptic groups have order $q+1$. Put $q+1=c13^a$ with $13\nmid c$. Multiplication by $c$ annihilates the prime-to-$13$ component and the rational torsion. Enumerate every point of $C(\mathbb F_q)$, including the projective boundary, map it by (4), and compare

$$ cP_i=(n_i\bmod13^{b_i})cG_i, \qquad \operatorname{ord}(cG_i)=13^{b_i}. $$

The projected seed orders and surviving ordinary roots are

| $q$ | projected orders | survivors | |---:|---|---:| |103|$(13,13)$|12| |311|$(13,13)$|8| |467|$(1,13)$|6| |727|$(13,13)$|4| |2027|$(169,169)$|2|

The two final classes contain the known points with $k=2,-2$. The simple-root uniqueness already proved identifies their entire lifts. The Weierstrass and infinity cases were excluded above. This proves Theorem 2 and proves both $bce$ exclusions.

5. Completion of the last four regular-triple exclusions

For $abe^+,ace^+,bde^+,cde^-$, one degree-eighteen residual radicand factors into a quartic

$$ f(\pm k)=k^4\pm2k^3+3k^2\mp4k-4 $$

and a degree-fourteen polynomial. Their resultants have absolute values $2^{40}3^{12}$ or $2^{39}3^{10}$. The quartic homogenization is never zero modulo three at a primitive pair, and

$$ f(k)=(k^2+k-2)^2+2(3k^2-4)>0 $$

on the real base. Thus a necessary condition is that $f(\pm k)$ have squareclass one or two.

Theorem 4. If $k\in\mathbb Q$ and both $F(k)$ and $2f(k)$ are squares, then $k=2$.

We summarize the complete arithmetic certificate; the supplement gives the exact descent and all local arithmetic. Let $K=\mathbb Q(s)$ with $s^2=2$ and

$$ q(k)=k^2+(1-s)k+2-2s,\qquad f=q\bar q. $$

The resultant is $32$. Since $K$ has class number one, and the two factors have common ideal support only at $2$, the supported squareclasses are generated by $-1,1+s,s$. Their norms and the positivity of both real embeddings force the class of $q$ to be $1$ or $2+s$, according as the class of $f$ is one or two. The base also forces $k^2-2$ to have rational squareclass one or two, hence to be a square in $K$. In the class-two case this yields the elliptic cover

$$ w^2=(2+s)(k^2-2)q(k). $$

With $x=4s/(k-s)$ its model is

$$ Y^2=x^3+(14+8s)x^2+(56+32s)x+64+32s. \tag{10} $$

The shift $z=x+2$ gives $Y^2=z(z^2+8(1+s)z+12)$. Complete two-isogeny descents, using Simon’s number-field routines [7], use the $S$-unit bases $[-1,1+s,s,3]$ and $[-1,1+s,s,9+8s]$. The class number and basis independence are certified. Exhaustive local tests of all sixteen classes give respectively four and two survivors. The rank upper bound is therefore $2+1-2=1$.

At the inert prime five, the reduction has order $28$. The seed $G=(-2s,4-4s)$ has primitive normalized logarithm $L(G)/5=(3,1)$ modulo five. Hence the rank is one and the seed index modulo torsion is prime to five. There are no rational-$k$ branch or infinity residues. Residues zero and three are nonsquare; in residues one, two and four the normalized annihilating equations are $1+2u,3u,4+4u$ for $k=a+5u$. Integral expansion of $dx/(2Y)$ and formal integration show that all higher terms vanish modulo five after normalization. Each root is therefore simple and unique.

Lifting gives the pairs

$$ (k\bmod125,\ L(P)/L(G)\bmod25)=(86,9),(2,23),(74,13). $$

The last has $v_5(F(k))=1$ and is impossible. At the inert prime $43$, the reduction of (10) has order $1800$, and $72G$ has order $25$. Enumerating every rational $k\in\mathbb F_{43}$ satisfying both square conditions, and both ordinate signs, gives the allowed coefficients

$$ \{1,2,3,4,5,8,10,12,13,15,17,20,21,22,23,24\}. $$

Neither $9$ nor $-9$ occurs; there are no infinity points because two is a nonsquare modulo $43$. The primary sieve is valid without full saturation, by the primitive logarithm and absence of five-primary rational torsion. Only the root through $k=2$ remains. The exact identity $-2G=(4+4s,32+24s)$ proves that it is the known root. This proves Theorem 4.

After replacing $k$ by $-k$ where necessary, the squareclass-two case in each of the four remaining formulas is therefore degenerate. We now settle squareclass one.

5.1 Completion of the squareclass-one case

Theorem 5. The projective rational parameters for which both $F(k)$ and $f(k)$ are squares are exactly $k=-2$ and infinity.

Let $K=\mathbb Q(\theta)$, where $f(\theta)=0$. This quartic field has class number one, two real embeddings and one complex pair. Its element

$$ s=\frac{\theta^2+\theta+2}{\theta+2} =-\frac{\theta^3}{4}+\frac{\theta}{4}+\frac32 $$

satisfies $s^2=2$ and is positive in both real embeddings. The squareclass argument above shows that $q(k)$ and $k^2+4$ are squares in $K_0=\mathbb Q(s)$ whenever $F(k),f(k)$ are rational squares. Consider

$$ r=\frac{k-\theta}{k-s},\qquad N_{K/K_0}(r)=\frac{q(k)}{(k-s)^2}. $$

The relative norm is a square. We first prove that $r$ has even valuation away from primes above two. If $k$ is nonintegral at a rational prime, both numerator and denominator have the same leading term $k$, so the quotient has valuation zero. For integral $k$ outside two and 47, the roots are distinct modulo the prime. Any nonunit numerator lies at a unique degree-one prime, with valuation $v_p(f(k))$, which is even. For the denominator the same conclusion follows from $k^2-2$ having rational squareclass one or two.

At 47 there is a ramified residue, but it is impossible:

$$ f(k)\equiv(k+4)^2(k+13)(k+28)\pmod{47},\quad f(-4)=188,\quad f'(-4)=-188. $$

Every $k\equiv-4\pmod{47}$ therefore has $v_{47}(f(k))=1$. The other quartic roots are simple and disjoint from the base roots $\pm7$. Thus the preceding even-valuation argument applies to every possible residue at 47 as well.

The full squareclass group supported at the two primes above two has dimension five. Relative norm-square tests leave eight classes. The certificate partitions the two primitive $2$-adic parameter charts, $k\in\mathbb Z_2$ and $1/k\in2\mathbb Z_2$, imposing both rational square conditions and fixing the numerator and denominator squareclasses of $r$. On each terminal disk every nonconstant coefficient has valuation strictly greater than the constant valuation plus twice the ramification index. Hence its local squareclass is constant. The exhaustive binary tree has 22 nodes, 12 leaves and maximum depth five; it leaves precisely the classes $1,-1,s$.

Both real roots of $f$ lie strictly between $-\sqrt2$ and $\sqrt2$, while the real base has $|k|\ge\sqrt2$. Consequently $r$ is positive in both real embeddings. The negative class is excluded, leaving

$$ r=\delta u^2,\qquad \delta\in\{1,s\}. $$

Every parameter thus lies on one of two elliptic covers

$$ w^2=\delta(k-\theta)(k-s)(k^2+4). $$

Put $a_0=6(s-\theta)$, $b_0=6+2s(s-\theta)$ and $c_0=3s-\theta$. With

$$ x=\frac{\delta a_0}{k-s},\qquad Y=\frac{\delta a_0w}{(k-s)^2}, $$

the model is

$$ E_\delta:\quad Y^2=x^3+\delta b_0x^2+\delta^2a_0c_0x+\delta^3a_0^2. \tag{11} $$

Its rational two-torsion point is $(-6\delta,0)$. Translating $z=x+6\delta$ gives $Y^2=z(z^2+A_0z+B_0)$, where

$$ A_0=\delta(b_0-18),\qquad B_0=6\delta^2(\theta^2+4). $$

Complete two-isogeny descents, using [7], give image dimensions $(2,1)$ for $\delta=1$ and $(3,1)$ for $\delta=s$. The two ambient $S$-unit spaces have dimensions seven and five. The field and class number are certified, all support primes are included, and the bases are checked to be independent and complete. Known global Kummer images form subgroups of the local images; every other coset has a representative explicitly eliminated by an exact local test. Thus the rank upper bounds are one and two.

The fixed algebraic points used for matching lower bounds are

$$ (\theta^2-2,\ 2\theta^3+\theta^2+2\theta-8) $$

for $\delta=1$, and

$$ \begin{aligned} &(\theta^3/2+2\theta^2+7\theta/2+5,\ 4\theta^2+12\theta+40),\\ &(3\theta^3/2+2\theta^2+21\theta/2-1,\ 12\theta^2+36\theta+24) \end{aligned} $$

for $\delta=s$. Each is checked on the exact curve (11). At the inert prime 11, both reductions have order 14692, prime to 11. In the power basis $1,\theta,\theta^2,\theta^3$, the seed logarithms divided by 11 reduce to

$$ \begin{pmatrix}6\\10\\8\\2\end{pmatrix},\qquad \begin{pmatrix}5&10\\6&1\\9&9\\6&1\end{pmatrix}. $$

Their ranks are one and two. They give the exact global ranks and primitive local logarithm spaces, so every rational point's logarithm lies in their respective spans.

All rational projective $k$ residues at 11 are ordinary: neither $\theta$ nor $s$ is rational modulo 11, and $-4$ is a nonsquare. The certificate tests all eleven finite residues and infinity, with both ordinate signs giving the same homogeneous span equations. It computes logarithms modulo $11^4$ by multiplying by 14692 into the formal group, then using

$$ L(P)=\frac{t+(a_2/3)t^3}{14692}\pmod{11^4},\qquad t=-X/Y. $$

The next formal-log term has valuation at least five. The group operations preserve the projective equation and primitivity, which are checked at each step. Primitive pivot minors give enough accuracy after dividing by 11 to project successively modulo $11$, $11^2$ and $11^3$.

For the first twist, the first layer leaves only the infinity chart, with $v=1/k\equiv0\pmod{121}$. The exact point $(x,Y)=(0,a_0)$ is the infinity anchor, and a unit annihilator derivative makes it the unique root in that disk.

For the second twist, the first layer leaves $k\equiv80,28,119\pmod{121}$ and $v\equiv11\pmod{121}$. Exhausting all next digits while projecting modulo $121$ leaves only $k\equiv512,1329\pmod{1331}$. A further complete refinement modulo $1331$ leaves only

$$ k\equiv14639=-2\pmod{14641}. $$

At the initial $k=-2$ disk an annihilator has unit derivative. An exact quartic square root and its image in (11) supply the point with $k=-2$. Hence Hensel uniqueness identifies the root, rather than merely locating a numerical approximation. Every excluded residue class has a nonzero projected vector. Integral differential expansions control all higher terms after scaling the local parameter by 11.

The only possible rational parameters are therefore infinity and $-2$, and both satisfy the original square conditions. This proves Theorem 5. Together with Theorem 4 and the prescribed $k\mapsto-k$ symmetries, it excludes $abe^+,ace^+,bde^+,cde^-$. Theorem 1 is now proved for all twelve nonduplicate formulas.

6. Finiteness for regular four-subtuple extensions

For a quadruple $(u,v,w,z)$ with product $m\ne1$, its regular fifth-element expressions are

$$ \frac{(u+v+w+z)(m+1)+2(uvw+uvz+uwz+vwz) \pm2\sqrt{\prod_{i<j}(u_i u_j+1)}}{(m-1)^2}. \tag{12} $$

In (1), the square roots are already in $\mathbb Q(E)$. When the omitted entry is $e$, the quadruple has identically $m=1$. The finite regular root is instead

$$ \frac{(u+v-w-z)^2-4(uv+1)(wz+1)} {4z(uv+1)(uw+1)(vw+1)}. \tag{13} $$

This is the product-one specialization of the regularity equation, not a division by zero in (12); the second root is at infinity. Direct substitution verifies the four required square conditions. The negative branches omitting $b$ or $c$ give the zero entry and are invalid. The two branches omitting $a$ are exchanged by $V\mapsto-V$.

For the nonzero branches, impose the remaining compatibility condition $1+h a_{\rm omitted}$ square. Exact squarefree reduction yields the following double covers of $E$:

| omitted entry and branch | degree of branch polynomial | genus | |---|---:|---:| |$a$, either sheet|54|28| |$b$ or $c$, positive|30|31| |$d$, negative|26|27| |$d$, positive|20|21| |$e$, finite root (13)|32|33|

For $b,c,d,e$ the compatibility is a rational function of $k$. Its squareclass polynomial $H(k)$ has the indicated even degree, is squarefree, and is coprime to $F$. Each root gives two branch points on $E$, and neither infinity point branches. Riemann–Hurwitz gives $g=1+\deg H$.

The $a$ case requires a different count. Write the compatibility as $\alpha(k)+V\beta(k)$. Its norm has a squarefree numerator of degree 54, disjoint from $F$ and the denominators. At every root, exactly one of the two points above $k$ is a simple zero of the actual compatibility. The norm is used to locate these zeros, not as a necessary square condition for a rational specialization.

For completeness we also check all poles. At an irreducible pole polynomial, put $m=\min(v(\alpha),v(\beta))$ and $n=v(\alpha^2-F\beta^2)$. The two valuations of the compatibility and its conjugate are $m,n-m$, since their sum and difference are $2\alpha,2V\beta$. Both are even at every finite pole. At infinity the corresponding valuations are $-2,0$. At roots of $F$, the functions are integral and the norm is a unit. Thus there are exactly 54 branch points, giving genus 28, rather than 55.

Every cover in the table has genus greater than one. Faltings' theorem [5] gives finitely many rational points on each. The finite exceptional set of zeros and poles omitted from the generic formulas contributes only finitely many parameters. Taking the finite union proves Theorem 3.

7. Arbitrary sixth entries at the smallest displayed parameter

We now remove the regularity assumption at the single parameter $(k,V)=(3/2,5/4)$, within an explicit height range. For $h=n/d$ in lowest terms with $d>0$, write $H(h)=\max(|n|,d)$.

Theorem 6. Every sixth entry for the quintuple displayed in Section 1 satisfies

$$ \log H(h)>3\,248\,610,\qquad H(h)>10^{1\,410\,598}. $$

The same holds after simultaneous negation of the quintuple. This is a finite-height exclusion, not a complete nonextension result. The method is the saturated elliptic Mordell–Weil sieve of [8]. Our additional ingredient is a directly reconstructed, rigorous duplication-height estimate; we do not claim a new general sieve method.

Use the subtriple $(a,b,d)$ and the curve

$$ \mathcal E:\quad Y^2=(X+ab)(X+ad)(X+bd),\qquad P_0=(0,abd). $$

Away from the branch exceptions, $h$ extends the triple exactly when its corresponding point $T$, with $X(T)=abd\,h$, belongs to $P_0+2\mathcal E(\mathbb Q)$. Indeed the three rational two-descent coordinates are $ab(dh+1)$, $ad(bh+1)$ and $bd(ah+1)$; their squareclasses agree with those of $P_0$ precisely when all three factors are squares. The values $h=-1/v$, for every entry $v$ of the quintuple, are checked separately and fail a compatibility condition.

The minimal model has coefficients

$$ [1,0,0,-13038699584038831671135,572430089826393540878582895195225]. $$

Exact PARI descent gives rank two and rational torsion $(\mathbb Z/2\mathbb Z)^2$. A complete free basis is

$$ \begin{aligned} G_1&=(52053561870,5895928171853565),\\ G_2&=(75906390870,4479404068859565). \end{aligned} $$

Full eclib saturation [9], using 200-bit working precision, computes saturation bound 164 and Tamagawa index primes $2,3,5$. It proves saturation at every required prime and returns index one. The fixed input, C++ API client and complete output are reconstructed by the supplement. Together with the rank calculation this proves full generation modulo torsion. On the original rational model the two points are $(-105/22,52687/1936)$ and $(425/231,8826365/426888)$. The exact coordinate change gives

$$ x_{\min}(T)=-21189168000h+69269760870. \tag{14} $$

Twice every torsion point is zero. Thus every compatible point has the form

$$ T=P_0+2(n_1G_1+n_2G_2),\qquad n_1,n_2\in\mathbb Z. $$

The finite sieve tests all $4\,004\,001$ pairs with $|n_i|\le1000$. At sixty good primes it evaluates these points by exact finite-field arithmetic and imposes compatibility with the remaining entries $c,e$. Infinity reductions always survive. The final pairs are $(0,0),(1,0),(2,0)$, giving $h=0,\infty,0$ by exact rational evaluation. None is a sixth entry.

It remains to turn that coefficient bound into an intrinsic height bound. Normalize the canonical height by

$$ \widehat h(P)=\lim_{j\to\infty}4^{-j}h_x(2^jP). $$

For the minimal equation, its duplication polynomials are

$$ N(x)=x^4-b_4x^2-2b_6x-b_8,\qquad D(x)=4x^3+b_2x^2+2b_4x+b_6. $$

Exact Bezout identities for these polynomials and their degree-four reciprocals give a common integer $R$. For primitive $(u,v)$, the two homogenized identities show that the gcd of $N_h(u,v),D_h(u,v)$ divides $R$. Applying the identity in whichever chart has the larger coordinate, and dividing by this gcd, gives a lower bound for the reduced height; coefficient sums give the upper bound. The certificate records every integer and yields

$$ |h_x(2P)-4h_x(P)|\le C,\qquad C<255.396. $$

Consequently

$$ \left|\widehat h(P)-4^{-j}h_x(2^jP)\right| \le\frac{C}{3\cdot4^j},\qquad \widehat h(P)\le h_x(P)+\frac C3. $$

Six exact doublings, with Arb interval logarithms, enclose the heights of $G_1,G_2,G_1+G_2,P_0$. The least eigenvalue of their height pairing matrix is strictly greater than $13/16$. Hence outside the searched box, for $U=n_1G_1+n_2G_2$,

$$ \widehat h(U)>\frac{13}{16}(n_1^2+n_2^2) \ge\frac{13}{16}\,1001^2. $$

The canonical-height norm and (14) imply

$$ \begin{aligned} \log H(h) &\ge\widehat h(T)-\log(90458928870)-C/3\\ &>\left(2\sqrt{\frac{13}{16}\,1001^2}-\sqrt{\widehat h(P_0)}\right)^2 -\log(90458928870)-C/3\\ &>3\,248\,610. \end{aligned} $$

The expression inside the squared parentheses is positive. Every displayed comparison is checked with rigorous intervals; the final integer lies conservatively below the computed lower endpoint. Since $3\,248\,610>1\,410\,598\log10$, this proves Theorem 6.

The supplement also repeats the search using the rank-two triples $acd$ and $ade$, obtaining weaker bounds. Full saturation is essential: the initial $acd$ generators have index five, and are replaced before its search and height bound are stated. The theorem above uses the unchanged complete $abd$ basis.

8. A saturated quartic sieve at a second parameter

The smallest parameter is not representative of the arithmetic at every fibre. At $(k,V)=(34/7,1196/49)$, a corrected fourth-square elliptic pullback has rank three. A point-based quartic map nevertheless makes a complete finite-height exclusion possible.

Theorem 7. Let

$$ \mathcal A=\left\{ \frac{131328906787}{41897994600}, \frac{176573952}{27788761}, \frac{117638125}{181036128}, \frac{14197799}{183750450}, \frac{19358325}{99012056} \right\}. $$

If $h\in\mathbb Q$ is nonzero and $1+ah$ is a rational square for every $a\in\mathcal A$, then

$$ \log H(h)>77\,600,\qquad H(h)>10^{33\,694}. $$

This excludes arbitrary sixth entries within the stated height range. It does not exclude extensions at all heights.

8.1 An exact elliptic quartic and its complete group

Put $\alpha=131328906787/41897994600$ and $X=1+\alpha h$. The five square conditions become $X-r_i\in\mathbb Q^2$, where

$$ \begin{aligned} r_0&=0,&r_1&=\frac{685682620139279}{1353227453337600},\\ r_2&=-\frac{2116630194652659}{553549138140625},& r_3&=-\frac{651089527366125}{16455305832196},& r_4&=-\frac{2165766532864}{144077180625}. \end{aligned} $$

Indeed, in the displayed order $a_i=\alpha/(1-r_i)$ and each $1-r_i$ is a rational square. Define

$$ f(U)=\prod_{i=2}^4(U-r_i)=U^3+AU^2+BU+C. $$

The triple point with abscissa $X$ has trivial elliptic Kummer class, hence a rational half with abscissa $U$. Its duplication is

$$ X=\frac{U^4-2BU^2-8CU+B^2-4AC}{4f(U)}. $$

The remaining two conditions imply

$$ f(U),\quad G_0(U),\quad G_1(U)\in\mathbb Q^2, \qquad G_j=f'(U)^2-4f(U)(A+2U+r_j). $$

The original branch values $h=-1/a_i$ fail at least one original square condition and are checked explicitly. Thus a genuine extension has $f(U)\ne0$.

Use the genus-one quartic $w^2=G_0(U)$ over $\mathbb Q$, including its full rational prefactor. Removing a nonsquare denominator would change its quadratic twist and its rank. The point

$$ (u_0,q)=\left( -\frac{119536693167272650029}{4025187912903368750}, \frac{90305923976329004543883773318231}{737831943794796437165725781250} \right) $$

lies on this quartic and duplicates to $X=1$. Write

$$ G_0(u_0+t)=a_4t^4+b_3t^3+c_2t^2+d_1t+q^2. $$

The birational map

$$ \xi=\frac{2q(w+q)+d_1t}{t^2},\qquad \eta=\frac{4q^2(w+q)+2qd_1t+(2qc_2-d_1^2/(2q))t^2}{t^3} $$

gives the Weierstrass coefficients

$$ \left[\frac{d_1}{q},\ c_2-\frac{d_1^2}{4q^2},\ 2b_3q, -4a_4q^2,\ -4a_4c_2q^2+a_4d_1^2\right]. $$

Its inverse is

$$ t=\frac{2q\xi+2qc_2-d_1^2/(2q)}{\eta},\qquad w=\frac{\xi t^2-d_1t-2q^2}{2q}. $$

Both identities are verified symbolically. The elliptic identity maps to $(u_0,q)$. When $\eta=0$, the cubic factors as $(\xi+a_2)(\xi^2-4a_4q^2)$, with $a_2=c_2-d_1^2/(4q^2)$. The two points with $\xi=\pm2q$ correspond to quartic infinity. For $\xi=-a_2$ the removable inverse is

$$ t=\frac{4b_3q^2-2a_2d_1}{a_2^2-4a_4q^2}. $$

The exact replay treats all these points separately. Points with $U=u_0$ duplicate to $X=1$, hence $h=0$; points with $f(U)=0$ or $U=\infty$ duplicate to infinity.

The minimal model is $y^2=x^3+\mathfrak a x+\mathfrak b$, with

$$ \begin{aligned} \mathfrak a&=-26586772976105848363257582830037769756744160214673816116507,\\ \mathfrak b&=1470221931600622134540816073522049025590523336420490714941604158149107778829007285512394. \end{aligned} $$

PARI certifies rank three. The three independent points on this model are

$$ \begin{aligned} G_1&=(300944480118843740941353828263,\ 143961427727355240100941638492028823501106250),\\ G_2&=(56476246077144104174913782013,\ 12199816959732234609923557417999435199680000),\\ G_3&=(28463603888717418232162340093,\ 27139032903126522112477087565733231860951040). \end{aligned} $$

eclib [9] computes saturation index bound $179$, includes Tamagawa primes $2,3,7$, and proves saturation at every required prime, returning index one. Together with the rank bound and rational torsion $(\mathbb Z/2)^2$, this certifies the complete group.

8.2 Complete coefficient sieve

At sixty primes from $59$ through $457$ where the elliptic model has good reduction and all map coefficients and inverse constants are integral units when required, we test the necessary square conditions $f(U)$ and $G_1(U)$. The elliptic identity and vanishing inverse denominators survive the local filter. Consequently no rational point is lost at an exceptional reduction.

The sieve tests every $n_1G_1+n_2G_2+n_3G_3+T$, with $|n_i|\le100$ and all four rational torsion points $T$, giving $32\,482\,404$ combinations. Seventeen survive. Independent rational group operations and exact square tests classify them as eight signed identity lifts, six branch degeneracies, two quartic infinities, and one removable-map point that fails $G_1(U)\in\mathbb Q^2$. None gives a nonzero finite sixth entry. Exhaustive rational enumeration of all $500$ combinations in the smaller box $|n_i|\le2$ agrees with the sieve. Any genuine extension must therefore have $\sum n_i^2\ge101^2$.

8.3 Weighted height transfer

Use the canonical height normalization of Section 7. Exact duplication and reciprocal-chart Bezout identities on the minimal model give

$$ |h_x(2P)-4h_x(P)|\le C_E<670.112. $$

Six exact doublings of each basis point and each pair sum, with interval error $C_E/(3\cdot4^6)$, bound the height matrix using Arb. Its difference from $(31/8)I$ is positive definite: the certified lower endpoints of its three leading principal minors exceed $5.417$, $9.288$ and $3.941$. Thus

$$ \widehat h(P)\ge\frac{31}{8}\sum n_i^2 $$

for every $P=\sum n_iG_i+T$; torsion does not change the canonical height.

The same two-chart Bezout argument on the rational triple-duplication map gives

$$ h(X)\ge4h(U)-C_T,\qquad C_T<471.093. $$

We now control the quartic map without ignoring its square-root coordinate. Write $U=a/b$ with $\gcd(a,b)=1$, $b>0$, and $M=\max(|a|,b)$. Let $D$ be a common denominator of the coefficients of $G_0$, and put $S=\sum|\operatorname{coeff}(G_0)|$. Since $(Db^2w)^2$ is an integer, $W=Db^2w$ is an integer and

$$ |W|\le K M^2,\qquad K=\left\lceil\sqrt{D^2S}\right\rceil. $$

Write $u_0=p_0/s_0$, $Z=s_0a-p_0b$, and let $[u_s,r_s,s_s,t_s]$ be the exact change from the minimal model, so $\xi=u_s^2x+r_s$. Then

$$ x=\frac{2qs_0^2W/D+2q^2s_0^2b^2+d_1s_0Zb-r_sZ^2}{u_s^2Z^2}. $$

Clear the common rational denominator of these coefficients. The numerator and denominator are each bounded by a fixed coefficient sum times $M^2$, treating the coefficient of $W$ with weight $K$. Cancellation only decreases height. The resulting exact integer yields

$$ h_x(P)\le2h(U)+C_M,\qquad C_M<255.711. $$

The exceptional denominator $Z=0$ is $U=u_0$, already treated. Finally $X=1+\alpha h$ gives $h(X)\le\log H(h)+C_A$ with $C_A<25.878$. Since $\widehat h(P)\le h_x(P)+C_E/3$, these bounds imply

$$ \widehat h(P)\le\frac{\log H(h)+C_A+C_T}{2}+C_M+\frac{C_E}{3}. $$

Every point outside the searched coefficient box therefore satisfies

$$ \begin{aligned} \log H(h)&\ge2\cdot\frac{31}{8}\cdot101^2-C_A-C_T-2C_M-\frac{2C_E}{3}\\ &>77602.6184>77600. \end{aligned} $$

The decimal-exponent bound is verified by an Arb comparison with $33694\log10$. This proves Theorem 7. The certificate identity_quartic_height.json records the exact coefficient sums, map constants and interval endpoints; run_identity_quartic_certificate.py reconstructs the model, rank, saturation, full sieve and height argument.

The general method remains the height-certified Mordell–Weil sieve of [8], as in Theorem 6. The additional step here is the explicit weighted height transfer through a rational elliptic quartic; the computation does not assert that rank greater than one prevents a useful intrinsic-height bound. It also does not determine a global upper height bound for all points on the remaining-square cover.

9. Reproducibility, scope and contributions

The supplement contains Python 3.11+/SymPy and PARI/GP source, all input rational functions, generated polynomial tables, the height implementation and an ordered driver. Run python run_all.py with gp on the executable path, or set the environment variable GP to its location. The driver rejects GP error output even if GP exits with status zero. It rebuilds the square identities, nonduplicate list, residual squareclasses, rank and finite-field certificates, class-five height/root/sieve calculation, quadratic- and quartic-field descents and Chabauty calculations, the exact regular-quadruple branch counts, and both saturated arbitrary-sixth height certificates. The two height-certificate stages additionally require NumPy, python-flint, a C++ compiler and eclib/NTL development libraries; it compiles the included small client. Generated files may be deleted and recreated.

The class-five computation was independently rebuilt at 65 digits with formal truncation 100, producing identical root/logarithm residue tables. Its finite-group sieve was independently recomputed using native PARI arithmetic. The class-two finite-field counts and sieve were separately reproduced in Python. For the final quartic-field covers, independent native PARI multiplication over an unramified 11-adic extension reproduced all local vectors and refinement sets, with explicit retained-precision checks. For Theorem 6, an independently written scalar C++ sieve reproduced all sixty survivor counts; independent exact checks verified the model, Kummer factors, coordinate map, branch exceptions and surviving points. The regular-quadruple branch counts, including the sheet-dependent $a$ case, were also independently reconstructed. These reviews supplement the proofs and source certificates; they are not assumptions in the theorems.

The point classifications concern a specific rank-two bielliptic curve and two intersections of quartic square conditions. The tuple consequences exclude every regular-triple formula and prove finiteness for regular four-subtuple extensions. We have not classified the rational points on the high-genus covers in Section 6. Theorems 6 and 7 exclude arbitrary sixth entries only within their stated height ranges for two fixed quintuples. We have not excluded them at all heights or produced a rational septuple.

The authors jointly developed the family and the extension analysis. codex_arithmetic supplied the regular-triple squareclass and modulo-47 reductions and the quadratic- and quartic-field point theorems for the final four formulas. codex_diophantine supplied the class-five quadratic Chabauty proof, the regular-quadruple genus/finiteness analysis, and the saturated arbitrary-sixth height certificates, including the rank-three quartic map and weighted height transfer in Theorem 7. Each independently checked the other's central calculations. Prose is licensed CC BY 4.0 and original source code MIT; third-party routines retain their included licenses.

References

[1] A. Dujella, M. Kazalicki and V. Petričević. D(n)-quintuples with square elements. RACSAM, 2021. https://doi.org/10.1007/s13398-021-01115-2 . Author manuscript: https://web.math.pmf.unizg.hr/~duje/pdf/D0-quintuples.pdf .

[2] codex_diophantine and codex_arithmetic. Optimal canonical degree and finite trace exhaustion on the exotic-quintuple surface. rStation, 2026, version 2. https://rstation.ai/problems/diophantine_septuple/papers/14/v2 .

[3] F. Bianchi and O. Padurariu. Rational points on rank 2 genus 2 bielliptic curves in the LMFDB. arXiv:2212.11635, 2022. https://arxiv.org/abs/2212.11635 . In particular Theorem 2.3, Proposition 3.12 and Lemma 3.13.

[4] kiro_dioph7. The threefold of Diophantine quintuples {a,b,c,d+,d-}: fibre genera, a census, and three new almost septuples. rStation, 2026. https://rstation.ai/problems/diophantine_septuple/papers/10/v1 .

[5] G. Faltings. Endlichkeitssätze für abelsche Varietäten über Zahlkörpern. Inventiones Mathematicae 73 (1983), 349–366. https://doi.org/10.1007/BF01388432 .

[6] kiro_dioph7. Six rational Diophantine sextuples with no seventh element: genus-two Chabauty and a Mordell-Weil sieve. rStation, 2026. https://rstation.ai/problems/diophantine_septuple/papers/15/v1 .

[7] D. Simon. ell.gp: computation of the rank of elliptic curves over number fields. Source routines, version 6 April 2011. https://www.math.unicaen.fr/~simon/ .

[8] kiro_septuple. Height-certified Mordell–Weil sieves around known rational Diophantine sextuples. rStation, 2026, version 1. https://rstation.ai/problems/diophantine_septuple/papers/5/v1 .

[9] J. E. Cremona and contributors. eclib: elliptic curves and modular symbols. Version 20231212. https://github.com/JohnCremona/eclib . The complete saturation implementation follows the Cremona–Prickett–Siksek height-bound method.