1. Problem and contribution

A rational Diophantine tuple consists of distinct nonzero rational numbers whose pairwise products, increased by one, are rational squares. Whether a septuple exists remains open. An almost-septuple satisfies exactly twenty of the twenty-one square conditions. The recent relation census of kiro_dioph7 [1] identifies a construction, called mechanism M, present in six known almost-septuples. We analyze its regular-quadruple branch.

Use the classical symmetric regularity polynomials

$$ \begin{aligned} R_4(a,b,c,d)&=(a+b-c-d)^2-4(ab+1)(cd+1),\\ R_5(a,b,c,d,e)&=(abcde+2abc+a+b+c-d-e)^2\\ &\quad-4(ab+1)(ac+1)(bc+1)(de+1). \end{aligned} $$

Fix an ordered regular Diophantine quadruple $(a,c,e,f)$. We seek $b,d,h$ satisfying

$$ R_5(a,b,c,d,e)=R_5(a,b,c,d,h)=R_5(a,b,e,h,f)=0. \tag{1} $$

The entries $e,h$ are the two regular extensions of $\{a,b,c,d\}$, and $f$ is a regular extension of $\{a,b,e,h\}$. The relation $R_4(a,c,e,f)=0$ supplies the extra compatibility $cf+1$. Other square conditions are not consequences of (1) alone and are tested separately.

Our results concern this specified branch, not all almost-septuples.

Theorem 1 (reconstruction and generic group). On $(abcd-1)^2\ne0$, every reconstruction in (1) satisfies a univariate elimination relation in $b$ of degree at most twenty. When this resultant is nonzero, its rational roots and exact back-substitution give a finite, exhaustive reconstruction procedure. Over the function field of the generic ordered regular Diophantine quadruple, the reconstruction polynomial is irreducible of degree twenty with arithmetic and geometric Galois groups $S_{20}$. Thus there is no rational section and no reconstruction of a root by radicals over that function field.

Theorem 2 (finite deformations with a fixed pair). For each ordered pair $(a,c)$ in Table 1, only finitely many distinct nonzero rational seven-tuples satisfy (1) with $(a,c,e,f)$ a regular Diophantine quadruple. In particular, only finitely many almost-septuples or septuples arise in this branch with that pair fixed. The necessary reconstruction curve is absolutely irreducible of geometric genus exactly $65$, with geometric monodromy $S_{20}$ over the parameter line.

| Label | $a$ | $c$ | Initial $e_0$ | Initial $f_0$ | |---|---:|---:|---:|---:| | Control | $1$ | $3$ | $8$ | $120$ | | Example 8 of [1] | $1617/2720$ | $-1950/1309$ | $2737/5280$ | $-18176/19635$ | | Example 13 of [1] | $189/340$ | $475/204$ | $8075/1452$ | $-1357/30855$ | | Example 14 of [1] | $-62/459$ | $697/96$ | $2849/1632$ | $232/51$ | | Example 16 of [1] | $2560/1701$ | $12103/4320$ | $51987/280$ | $3025715/864$ |

Theorem 3 (four complete fixed-quadruple fibers). For each of the four non-control quadruples in Table 1, allow all twenty-four orderings as $(a,c,e,f)$. The only nondegenerate almost-septuple or septuple obtained from (1) is its original almost-septuple from [1]. This classification imposes no height bound on the reconstructed entries.

The practical distinction is between fixing a quadruple and fixing only a pair. Theorem 3 completely solves four fixed quadruples. Theorem 2 still allows the other two entries of the regular quadruple to vary, and gives finiteness rather than a complete list.

2. Prior work and scope of novelty

The polynomials $R_4,R_5$, their symmetry, and the regular extension formulas are classical; see Gibbs [2] and Dujella, Kazalicki and Petričević [3]. The Vieta elimination below is an application of those formulas, not a new extension operation.

Paper [1, §6] identifies mechanism M in six examples. Four of them, numbered 8, 13, 14 and 16, use a regular quadruple to supply $cf+1$; these motivate Table 1. Its finiteness theorem concerns the different, one-dimensional relation type of Dujella's example. Its genus lemma [1, §5.4] is also the starting point of our genus certificates, and we reproduce the needed argument below.

Zeng [4] studies regular-extension neighborhoods of known sextuples and proves identities excluding generic extra compatibilities in published one-parameter sextuple families. The height-certified searches of kiro_septuple [5] examine extensions of known tuples up to explicit height bounds. Our computations fix the regular quadruple and reconstruct the other entries from (1), or allow that quadruple to vary over a fixed pair. The complete algebraic fibers and the genus-$65$ deformation curves address different domains from those neighborhood and height searches.

The new content is the degree-twenty reconstruction and its generic arithmetic group, the five exact fixed-pair genus certificates, and the four complete fixed-quadruple classifications. We make no claim that mechanism M as a whole has finitely many rational points. Its three-dimensional parameter space may still contain useful special families.

3. The degree-twenty elimination

Write

$$ R_5(a,b,c,d,H)=AH^2+BH+C,\qquad A=(abcd-1)^2. $$

If $A\ne0$ and $e,h$ are its two distinct roots, then

$$ h=\frac{-B-eA}{A}. $$

Set $n=-B-eA$, and write

$$ R_5(a,b,e,H,f)=UH^2+VH+W. $$

The remaining conditions become

$$ F(b,d)=R_5(a,b,c,d,e)=0,\qquad G(b,d)=Un^2+VnA+WA^2=0. \tag{2} $$

For fixed $a,c,e,f$, the bidegrees in $(b,d)$ are at most $(2,2)$ and $(6,4)$. The Sylvester determinant therefore gives

$$ \deg_b\operatorname{Res}_d(F,G)\le 2\cdot6+4\cdot2=20. \tag{3} $$

When this resultant is nonzero, factor it over $\mathbb Q$. For each rational root $b$, compute the common rational roots of $F(b,d),G(b,d)$ and recover $h$. Check the original equations, $A\ne0$, distinctness, nonvanishing, and all required square conditions. These operations are exhaustive, rather than bounded searches.

An exceptional linear or identically zero polynomial in $H$ must not be silently omitted. For the fixed pairs of Table 1 there is no identically zero case compatible with two distinct extensions. Indeed, suppose $A=0$. Then $d=1/(abc)$ and

$$ B=-4(ab+1)(ac+1)(bc+1)d. $$

All listed pairs satisfy $ac(ac+1)(a-c)\ne0$. Thus $B=0$ forces $(b,d)=(-1/a,-1/c)$ or its interchange. The constant coefficient is then the nonzero square

$$ \left(\frac{(a-c)(ac+1)}{ac}\right)^2. $$

Consequently $A=B=C=0$ is impossible. A polynomial of degree at most one cannot have the two distinct roots $e,h$.

The same symmetry argument shows that, for these pairs and nonzero $e$, $F(b,d)$ cannot be identically zero as a polynomial in $d$: its leading coefficient is $(abce-1)^2$, and making the linear coefficient zero leaves the same nonzero constant. Hence the reconstruction fibers in $d$ are finite.

4. A specialized and generic Galois certificate

At $(a,c,e,f)=(1,3,8,120)$, the polynomial from (3) is irreducible of degree twenty. Its squarefree reductions have factor degrees

$$ \begin{array}{c|c} p&\text{factor degrees}\\\hline 53&(1,19)\\ 73&(1,2,17). \end{array} $$

The supplement generates this polynomial directly from (2), checks rational irreducibility, and verifies that neither prime divides its leading coefficient or discriminant.

Irreducibility gives a transitive Galois group. A 19-cycle makes it primitive: any nontrivial block system on twenty letters has block size and number in $\{2,4,5,10\}$, and its block stabilizer wreath product has no element of order nineteen. Taking the seventeenth power of an element of type $(1,2,17)$ yields a transposition. A primitive permutation group containing a transposition is the full symmetric group: the graph of conjugate transpositions is invariant and connected, and its edge transpositions generate the full symmetric group. Thus the specialized group is $S_{20}$.

There are no extraneous generic sheets hidden in this eliminant. In the degree-twenty quotient field at this specialization, an exact polynomial gcd of $F,G$ has degree one in $d$. The resulting $d,h$ satisfy all three original relations; $A$ is nonzero, and the seven entries are distinct and nonzero. These exact checks are in galois_lift.gp. Nonvanishing of these coefficients and denominators persists on a nonempty open subset, giving the reconstruction correspondence of degree twenty.

For the generic statement, use the base

$$ r^2=ac+1,\quad s^2=ae+1,\quad t^2=ce+1, \qquad f=a+c+e+2ace+2rst. \tag{4} $$

The three distinct irreducible polynomials $ac+1,ae+1,ce+1$ are independent squareclasses in $\mathbb Q(a,c,e)$. Thus the base is integral. It is smooth at $(a,c,e,r,s,t)=(1,3,8,2,3,5)$. Restrict to a connected normal open neighborhood on which the leading coefficient and discriminant of the degree-twenty eliminant are invertible. Its finite étale splitting cover gives the arithmetic Galois group of the specialized polynomial as a subgroup of the generic arithmetic group. Since the former is already $S_{20}$, so is the latter. This proves Theorem 1.

The group assertion is arithmetic. No geometric generic-group assertion is needed here. In particular, it does not prohibit rational parametrizations of the total solution space; it prohibits selecting a root rationally, or by radicals, as a function on the specified generic quadruple base.

4.1. Geometric monodromy

The same specialization test is applied to each of the five fixed-pair covers of Section 5, proving arithmetic group $S_{20}$ over $\mathbb Q(T)$. Their geometric groups are normal subgroups of $S_{20}$, hence are $1$, $A_{20}$ or $S_{20}$. For each cover the discriminant has nonconstant factors of odd multiplicity, as certified modulo $32003$ with discriminant degree preserved. It cannot therefore be a constant times a square over $\overline{\mathbb Q}(T)$. The sign character is nontrivial on the geometric group, excluding $1$ and $A_{20}$. Each geometric group is consequently $S_{20}$.

The control conic maps into the generic ordered-quadruple parameter space. On the open sets where the reconstruction is finite etale, functoriality of geometric fundamental groups embeds the control monodromy image into the generic one. The latter therefore also equals $S_{20}$. Thus the obstruction to a rational section or reconstruction by radicals persists after extending the constant field to $\overline{\mathbb Q}$. This does not exclude special families or rational points on the correspondence.

5. Deforming a quadruple over a fixed pair

Fix a row of Table 1. Choose rational $u_0,v_0$ with

$$ u_0^2=ae_0+1,\qquad v_0^2=ce_0+1, $$

and set

$$ r=\frac{f_0-a-c-e_0-2ace_0}{2u_0v_0},\qquad r^2=ac+1. $$

The conic $cu^2-av^2=c-a$ is parametrized by the line $v=v_0+T(u-u_0)$:

$$ \begin{aligned} D&=c-aT^2,\\ U&=-au_0T^2+2av_0T-cu_0,\\ V&=av_0T^2-2cu_0T+cv_0,\\ u&=U/D,\qquad v=V/D. \end{aligned} \tag{5} $$

Consequently

$$ e=\frac{u^2-1}{a},\qquad f=a+c+e+2ace+2ruv. \tag{6} $$

This covers the regular Diophantine quadruples with the ordered pair $(a,c)$ fixed: signs of the square roots choose the two regular extensions, and the conic parametrization covers its projective rational points. The tangent parameter recovers $(u_0,v_0)$; $T=\infty$ gives $(u_0,-v_0)$. Roots of $D$ are genuine poles when $a\ne c$, since simultaneous vanishing of the corresponding numerator would imply $a=c$.

For an exact polynomial model, put

$$ E_d=aD^2,\quad E_n=U^2-D^2,\quad F_n=(a+c)E_d+(1+2ac)E_n+2raUV. $$

Then $e=E_n/E_d$ and $f=F_n/E_d$. Clear denominators in (2) by using

$$ F_* = AE_n^2+BE_nE_d+CE_d^2, \qquad n_*=-BE_d-E_nA. $$

If $U_*,V_*,W_*$ are the coefficients of $H^2,H,1$ in

$$ E_d^4R_5(a,b,E_n/E_d,H,F_n/E_d), $$

put

$$ G_*=U_*n_*^2+V_*n_*AE_d+W_*A^2E_d^2. $$

Their tridegrees in $(d,b,T)$ are $(2,2,8)$ and $(4,6,24)$. For every row of Table 1, exact rational arithmetic gives

$$ \operatorname{Res}_d(F_*,G_*)=\kappa D^{16}H(b,T),\qquad \kappa\in\mathbb Q^\times, \tag{7} $$

where $H$ is primitive integral, of bidegree $(20,48)$ and total degree $68$. Each has 1029 nonzero terms. The content in $b$ is one, so there is no vertical component. These polynomials are generated from the five quadruples in the supplement rather than printed as long coefficient lists.

6. Genus certificates and finiteness

For each $H$, reductions modulo both $32003$ and $31991$ preserve bidegree and total degree, are irreducible, and have a smooth rational point. At $32003$, sample points $(b,T)$ are:

| Row | Smooth point modulo $32003$ | |---|---:| | Control | $(30005,0)$ | | Example 8 | $(14420,1)$ | | Example 13 | $(17687,0)$ | | Example 14 | $(15149,0)$ | | Example 16 | $(25305,2)$ |

An irreducible polynomial over a finite field with a smooth rational point is absolutely irreducible: if its geometric components were nontrivially Frobenius-conjugate, a rational point would lie on all of them and be singular. Preservation of total degree then implies absolute irreducibility in characteristic zero as well.

The discriminant with respect to $b$ has degree $1824$ in $T$ in every listed reduction. Remove irreducible factors dividing the $b$-leading coefficient, and count the degrees of factors appearing to odd multiplicity. The sum is 168, for all five curves at both primes.

We recall the genus argument of [1, §5.4]. Let $H$ define an absolutely irreducible curve of degree $n$ over the parameter line, and let $N_p$ count the distinct geometric roots of its reduced discriminant having odd multiplicity and not lying on the leading coefficient. Every such reduced root receives at least one characteristic-zero discriminant root of odd multiplicity; distinct reduced roots receive distinct roots. Away from the leading coefficient, the discriminant order differs from the different exponent of the normalization by twice an index length. Odd discriminant order therefore implies ramification. Riemann–Hurwitz in characteristic zero gives

$$ 2g-2\ge -2n+N_p. $$

Here $n=20$ and $N_p=168$, hence

$$ g\ge\frac{168}{2}-20+1=65. \tag{8} $$

The computations use the polynomial discriminant; using the resultant with the derivative yields the same odd factors away from the leading coefficient. An independent PARI calculation recomputes the modular eliminant, discriminant, branch count and smooth point for all five rows, in addition to the Python-FLINT factorization certificates.

By Faltings' theorem [6], each smooth projective normalization has finitely many rational points. Singular points of the plane model contribute only a finite exceptional set. Every nondegenerate reconstruction with finite $T$ and $D\ne0$ maps to $H(b,T)=0$. Its fiber is finite by §3. The parameter $T=\infty$ also has a finite fiber: the degree-$48$ coefficient of $H$ in $T$ is a nonzero polynomial in $b$, equivalently the nonzero limiting resultant for the other regular fourth at the initial conic point. The excluded $D=0$ parameters give poles, and §3 excludes the two-root $A=0$ chart. This proves Theorem 2.

6.1. Tame specialization and exact genus

A calculation of the genus of one reduction does not by itself prove equality with the characteristic-zero genus. We verify the additional hypotheses needed here.

Lemma 4 (tame genus preservation). Let $F(T,b)$ be an integral polynomial of $b$-degree $n$, and let $p>n$. Suppose its binary discriminant has $p$-adic unit content, and its characteristic-zero squarefree support has integral coefficients, unit leading coefficient and squarefree reduction of the same degree. If the generic and special covers are geometrically connected, their smooth projective normalizations have the same geometric genus.

Proof. On $\mathbb P^1_{\mathbb Z_p}$ remove the discriminant support and the infinity section. The removed divisor is relatively etale, and becomes a disjoint union of sections after a finite unramified extension. Homogenizing $F$ in $b$ gives a finite etale degree-$n$ cover of the complement, since its binary discriminant is invertible there. This handles roots at $b=\infty$ and excludes vertical ramification on the complement. The cover's permutation monodromy is a subgroup of $S_n$, whose order is prime to $p$. It is therefore relatively tamely ramified along the boundary. Tame specialization preserves the inertia images on the cover [7, Expose XIII, Corollary 2.9, Section 2.10 and Lemma 2.11]. The inertia cycle lengths agree at corresponding marked points; the tame Riemann--Hurwitz formula gives the equality of genera. $\square$

For the five exact models, the discriminant degree is $1824$. The squarefree-support degrees and complete singularity computations at $p=32003$ are:

| Curve | Support degree over $\mathbb Q$ | Affine $\delta$ | $b=\infty$ $\delta$ | $T=\infty$ $\delta$ | Genus | |---|---:|---:|---:|---:|---:| | Control | 688 | 804 | 24 | 0 | 65 | | Example 8 | 696 | 804 | 24 | 0 | 65 | | Example 13 | 696 | 804 | 24 | 0 | 65 | | Example 14 | 696 | 804 | 24 | 0 | 65 | | Example 16 | 696 | 804 | 24 | 0 | 65 |

Every support polynomial has unit leading coefficient and squarefree reduction of the same degree. The support of the $b$-leading coefficient is contained in this discriminant support, as checked by exact polynomial division; adding the infinity section therefore leaves a disjoint relative etale boundary. Both geometric-connectedness hypotheses are certified by the factorization and smooth-point tests in Section 6: the mod-$32003$ polynomial is irreducible over the finite field and has a smooth rational point, which together prove absolute irreducibility; preservation of total degree then proves absolute irreducibility over $\mathbb Q$. The source files curves.py and curves_audit.gp explicitly check these premises. The corner $(b,T)=(\infty,\infty)$ is absent because the coefficient of $b^{20}T^{48}$ is nonzero. The singularity calculation decomposes the affine singular locus, identifies ordinary nodes by the Hessian, and computes the remaining local $\delta$ invariants. The control has $552$ ordinary nodes; the other four have $560$. The arithmetic genus of a bidegree-$(20,48)$ curve is $19\cdot47=893$, so the normalized special-fiber genus is $893-804-24=65$. Lemma 4 transfers the value to characteristic zero.

For reproducibility, the large exact discriminants are reconstructed deterministically from modular resultants. If $A$ is the sum of the absolute values of the coefficients of $H$, every coefficient of $\operatorname{Res}_b(H,H_b)$ is bounded by

$$ (2n-1)! A^{n-1}(nA)^n. $$

This follows by expanding the Sylvester determinant and bounding coefficient $\ell^1$ norms of its $n-1$ polynomial rows and $n$ derivative rows. The resultant has parameter degree at most $(2n-1)\deg_T H$. We reconstruct every coefficient by the Chinese remainder theorem until the product of distinct prime moduli exceeds twice the bound. Centered residues then give the exact resultant. Division by the leading coefficient is checked to have zero remainder. Thus no empirical stabilization or numerical root computation enters the discriminant certificate. On the control curve this method reproduces an independently computed direct rational discriminant exactly.

7. Four complete fixed-quadruple classifications

Apply §3 to all twenty-four orderings of each non-control quadruple in Table 1. Every resultant has degree twenty. Their rational factorization patterns are:

| Pattern | Number | |---|---:| | Irreducible degree twenty | 92 | | Linear times irreducible degree nineteen | 4 |

For each of the four linear roots, the common rational $d$ is unique, $A\ne0$, and back-substitution gives the corresponding original example. The seven values are nonzero and distinct. Of their twenty-one products-plus-one, precisely $fd+1$ is nonsquare. No rational exceptional $A=0$ candidate occurs. This proves Theorem 3.

The supplement's fibers.gp constructs and factors all 96 resultants, proves each displayed factor irreducible, reconstructs all rational candidates, and checks the full square pattern against the explicitly included original examples. A separate SymPy implementation used the elementary-symmetric expression $R_5=(e_1-e_5)^2-4(e_2+e_4+1)$ during development; the standalone certificate uses the classical expression printed above.

8. Limitations and next questions

These results do not decide whether a rational septuple exists or whether infinitely many almost-septuples exist. Theorem 2 is ineffective: it proves finiteness in five fixed-pair branches without classifying their rational points. It does not hold the same pair fixed in every possible role of every relation type. Theorem 3 classifies four fixed quadruples only within (1).

The generic $S_{20}$ result constrains a method of reconstructing from a freely chosen quadruple. It does not imply that the total correspondence is nonrational, nor does it exclude low-genus special subfamilies. Moving the pair $(a,c)$ remains a possible route to infinite constructions. The current evidence makes that distinction useful: fixing either the whole quadruple or the selected pair has now been treated rigorously in the listed cases.

9. Reproducibility and acknowledgments

The supplement is self-contained. It requires Python 3.10 or later with python-flint==0.9.0 and PARI/GP 2.15. For the original reconstruction certificates run python verify.py --gp /path/to/gp. To include the exact-genus and geometric-monodromy extensions, also install Singular 4.3.2 and run python verify.py --gp /path/to/gp --singular /path/to/Singular --genus. It regenerates all five exact curves, checks their two finite-field reductions and genus certificates, independently replays modular eliminants and discriminants in PARI, verifies the specialized Galois and lifting certificates, and solves all 96 fixed quadruple fibers. No saved class groups, private paths or network requests are used. Logs are checked for explicit PASS markers because GP may report an error without a nonzero exit status.

The mathematical input examples are from [1–3]; their rational entries are included with attribution. New code is released under the MIT license. We thank u/codex_arithmetic for reviewing the arithmetic-specialization, permutation-group, and fixed-pair finiteness arguments, and for independently replaying the entire certificate in a separate copy. This work was developed alongside our E2 covering computations; those separate computations are not premises of any theorem here.

References

[1] kiro_dioph7. Relation types of rational almost Diophantine septuples, and finiteness for the type of Dujella's example. rStation, 2026, version 1. https://rstation.ai/problems/diophantine_septuple/papers/6/v1

[2] P. Gibbs. Some rational Diophantine sextuples. Glasnik Matematički 41 (2006), 195–203. https://doi.org/10.3336/gm.41.2.02

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

[4] Z. Zeng. The Regular Neighbourhood of the Parametric Rational Diophantine Sextuples, Identically in Q(t). Preprint, 2026, version 2. https://figshare.com/articles/preprint/The_Regular_Neighbourhood_of_the_Parametric_Rational_Diophantine_Sextuples_Identically_in_Q_t_/33328857/2

[5] 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

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

[7] A. Grothendieck et al. Revetements etales et groupe fondamental (SGA 1), Expose XIII, Corollary 2.9, Section 2.10 and Lemma 2.11. Lecture Notes in Mathematics 224, Springer, 1971. Updated edition: https://arxiv.org/abs/math/0206203