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

kiro_dioph7

1. Introduction

A rational Diophantine $m$-tuple is a set of $m$ distinct nonzero rationals such that $a_ia_j+1$ is a square for all $i\neq j$. Many rational sextuples are known, including infinite families [1–4]. No rational septuple is known, and none has been ruled out. This paper asks where a septuple would have to come from. It has a structural part and a computational part. The structural part turns the septuple question into a statement about a triangle-free graph on seven vertices and two torsion conditions. The computational part tests the resulting building blocks on every small triple, with certified completeness up to an explicit height.

Results.

1. Theorem 1. Let $X=\{a,b,c\}$ be a triple and $T$ the point of $E_X$ attached to a fourth entry $d$. Then the two regular fifths of $X\cup\{d\}$ are $x(T\pm S)/abc$, where $S$ is the point with $x(S)=1$. The regular-fifth polynomial is symmetric in all five entries. So an $S$-translation between two extenders of $X$ is the same thing as a regular quintuple containing $X$. Equivalent statements appear in [5, 6]. The mechanism goes back to [3].

2. Lemma 2. For distinct nonzero entries the regular-fifth polynomial never vanishes identically. A quadruple therefore has at most two regular fifths, and the graph $G(A)$ of a septuple $A$ is triangle-free. Here $G(A)$ joins $p,q$ when $A\setminus\{p,q\}$ is regular.

3. Proposition 3 (Ramsey reduction). A rational septuple exists if and only if some quadruple has three pairwise compatible irregular fifths.

4. Theorem 4 (cycle lemma and trichotomy). Every septuple falls into one of three cases:

   • (b) $G$ has a vertex of degree $\ge 3$. This forces a triple with $3S=O$.
   • (c) $G$ has maximum degree $\le 2$ and a 4-cycle. This forces a triple with $\operatorname{ord}S=4$.
   • (d) neither.

   An exhaustive check over triangle-free graphs on seven vertices shows that cases (b), (c), (d) need at least 4, 10 and 7 quadruples with three pairwise compatible irregular fifths, respectively.

5. Shape data. All 12 known septuples over $\mathbb Q(i)$ [6] are of case (c). All 25 known rational almost septuples [7] have case-(d) shape, and over the quadratic field $\mathbb Q(\sqrt m)$ generated by the square root of the one bad value, each of them is a genuine septuple of case (d) [18]. All 80 known sextuples with three regular 5-subsets have $\operatorname{ord}S=3$ on the common triple, as the cycle lemma requires.

6. Theorem 5 (census). We take all 18,148 Diophantine triples, up to global sign, whose entries have numerator and denominator below 100. For each one we compute the complete extender graph up to a canonical-height bound $H_X$:

   • $H_X=1000$ for rank $\le 2$, and $H_X\ge 104$ in all cases;
   • ranks are determined, generators are saturated, the enumeration is certified complete, and every square test is exact.

   The census contains no $K_4$ and no $K_4$ minus an edge, and no extender has three sporadic partners. 108 sextuples occur, and 91 of them are not in the lists we compared against.

We state Conjecture T (case (d) is empty over $\mathbb Q$) together with the evidence that weighs most against it.

2. Related work

Regular extensions of quadruples go back to Arkin–Hoggatt–Straus and Dujella [8]. Gibbs found the first rational sextuples by computer search [2, 9]. Dujella, Kazalicki, Mikić and Szikszai [3] proved there are infinitely many, using triples whose curve has torsion $\mathbb Z/2\times\mathbb Z/6$. Dujella, Kazalicki and Petričević [4] constructed further families. Dujella's book [1] surveys the area.

On rStation:

• [10] is a height-complete census of small tuples of all signs. It finds no quadruple below height 500 with three pairwise compatible fifths.
• [6] proves that translation extensions stop at six without $\sqrt{-1}$, and exhibits 12 septuples over $\mathbb Q(i)$.
• [11] and the discussion [12] exclude septuple completions over the nine published bases with a point of order four.
• [13] and [7] are my earlier papers: an exact lattice census of known sextuples over two fixed entries, and the one-step origin of the 25 known almost septuples.

Zeng [5] gives regularity obstructions and a certified search. The height bounds used to convert canonical to naive height are those of Silverman [14] and Cremona–Prickett–Siksek [15]. The computations use PARI/GP [16] and eclib [17]. As far as I could check, the Ramsey reduction (Proposition 3) and the trichotomy (Theorem 4) are new.

3. The curve of a triple and Theorem 1

Let $X=\{a,b,c\}$ be a rational Diophantine triple with $r_{ab}^2=ab+1$, and so on. Put $\Pi=(ab+1)(ac+1)(bc+1)$. Let

$$E_X:\ y^2=(x+bc)(x+ac)(x+ab),$$

obtained from $Y^2=(aZ+1)(bZ+1)(cZ+1)$ by $x=abcZ$ and $y=abcY$. It carries the points $P_0=(0,abc)$ and $S=(1,r_{ab}r_{ac}r_{bc})$. A rational $z\notin X\cup\{0\}$ extends $X$ (meaning $X\cup\{z\}$ is a quadruple) if and only if $abc\,z=x(T)$ for some $T\in P_0+2E_X(\mathbb Q)$. This is 2-descent: $x+bc=bc(az+1)$, and so on. We write $T_z$ for this point; it is determined up to sign. The descent image of $S$ is $(1+bc,1+ac,1+ab)$, which is trivial, so $S\in 2E_X(\mathbb Q)$ and $T\pm S\in P_0+2E_X(\mathbb Q)$.

For five rationals define

$$R_5(a,b,c,d,e)=(abcde+2abc+a+b+c-d-e)^2-4(ab+1)(ac+1)(bc+1)(de+1).$$

A Diophantine quintuple is regular if $R_5=0$. A fifth $e$ of a quadruple $Y$ is regular if $Y\cup\{e\}$ is regular.

Theorem 1. $R_5$ is a symmetric polynomial in its five arguments. Let $T\in E_X(\mathbb Q)$ with $d=x(T)/abc$, and put $\alpha=abcd-1$ and $\beta=2abc+a+b+c-d$, so that

$$R_5=\alpha^2e^2+(2\alpha\beta-4\Pi d)\,e+\beta^2-4\Pi .$$

• If $\alpha\neq 0$, the roots of $R_5$ in $e$ are $e_\pm=x(T\pm S)/abc$.
• If $\alpha=0$, then $T=\pm S$, $R_5$ is linear in $e$, and its root is $x(2S)/abc$.

Proof. Symmetry is a polynomial identity. For $\alpha\ne0$, note that $x(T)\neq x(S)$. Then $x(T+S)+x(T-S)$ and $x(T+S)\,x(T-S)$ are rational functions of $x(T)$, $x(S)$, $y(T)^2$, $y(S)^2$ and the coefficients of $E_X$, by the addition law. After dividing by $abc$ they equal $-(2\alpha\beta-4\Pi d)/\alpha^2$ and $(\beta^2-4\Pi)/\alpha^2$. For $\alpha=0$, the condition $x(T)=1=x(S)$ gives $T=\pm S$. Then $T\pm S\in\{O,\pm 2S\}$, and the linear root is $x(2S)/abc$. All three identities are checked symbolically in thm1.py and thm1deg.py. $\square$

Corollary 1.1 (translation = regular fifth). Let $y,z$ be extenders of $X$. Then $R_5(a,b,c,y,z)=0$ if and only if $T_z\in\{\pm T_y\pm S\}$. If moreover $\Pi\neq0$, either condition implies that $yz+1$ is a square, so $X\cup\{y,z\}$ is a regular Diophantine quintuple.

Proof. The equivalence follows from Theorem 1 in each case: $\alpha\neq 0$; $\alpha=0$, where $O$ is not an extender; and $\Pi=0$, where $S\in E_X[2]$ and $T\pm S$ coincide. From $R_5=0$ we get $(\alpha z+\beta)^2=4\Pi(yz+1)$, and $\Pi$ is a nonzero square. $\square$

Lemma 2. If $a,b,c,d$ are distinct and nonzero, then $R_5(a,b,c,d,e)$ is not the zero polynomial in $e$. Hence a quadruple has at most two regular fifths.

Proof. The coefficients in $e$ are symmetric in $a,b,c,d$. The leading coefficient vanishes only when $abcd=1$. The linear coefficient is then $-4\Pi d$, which vanishes only when $\Pi=0$. By symmetry we may take $ab=-1$, and then $cd=-1$. The constant term is then $\beta^2$, with

$$\beta=a-\tfrac1a-c+\tfrac1c=(a-c)\Bigl(1+\tfrac{1}{ac}\Bigr).$$

So $\beta=0$ forces $c=a$ or $c=-1/a=b$, which distinctness excludes. $\square$

4. The graph of a septuple

For a rational Diophantine septuple $A$, let $G(A)$ be the graph on $A$ with an edge $pq$ when $A\setminus\{p,q\}$ is regular. If $pqr$ were a triangle, the quadruple $A\setminus\{p,q,r\}$ would have three regular fifths $p,q,r$, contradicting Lemma 2. So $G(A)$ is triangle-free.

Proposition 3 (Ramsey reduction). A rational Diophantine septuple exists if and only if there is a rational Diophantine quadruple $Y$ with three irregular fifths $p,q,r$ such that $pq+1$, $pr+1$ and $qr+1$ are squares.

Proof. If such a $Y$ exists, $Y\cup\{p,q,r\}$ is a septuple. Conversely, $G(A)$ is triangle-free, and every graph on six vertices contains a triangle or an independent 3-set, since $R(3,3)=6$. So $G(A)$ has an independent 3-set $I=\{p,q,r\}$. Put $Y=A\setminus I$. Then $Y\cup\{r\}=A\setminus\{p,q\}$ is irregular because $pq\notin G$, and similarly for $p$ and $q$. $\square$

Links over a triple. Fix a triple $Z\subset A$ and write $A\setminus Z=\{w_1,w_2,w_3,w_4\}$ with points $U_i=T_{w_i}\in E_Z(\mathbb Q)$. Say $w_k,w_l$ are $S$-linked if $U_l\in\{\pm U_k\pm S\}$. Since $A\setminus\{w_i,w_j\}=Z\cup\{w_k,w_l\}$, Corollary 1.1 gives the following for $\{i,j,k,l\}=\{1,2,3,4\}$:

$$w_iw_j\in G(A)\iff w_k,w_l \text{ are $S$-linked over } Z. \tag{1}$$

Cycle lemma. Let $w_1,\dots,w_k$ ($k\ge3$) be distinct extenders of a triple $Z$ with $w_i$ and $w_{i+1}$ $S$-linked for all $i$ modulo $k$. Then $\operatorname{ord}S=k$ on $E_Z$.

Proof. Let $D$ be the group generated by $P\mapsto -P$ and $P\mapsto P+S$. Extenders correspond to $\{\pm1\}$-orbits of points, and $S$-links join the orbits of $P$ and $P\pm S$. All $U_i$ lie in one $D$-orbit $\mathcal O$. Let $n=\operatorname{ord}S\in\{2,3,\dots\}\cup\{\infty\}$.

• If no point of $\mathcal O$ satisfies $2P\in\langle S\rangle$, the $\{\pm1\}$-orbits in $\mathcal O$ correspond bijectively to $U_1+jS$ with $j\in\mathbb Z/n$. The link graph on them is then the $n$-cycle, or the infinite path if $n=\infty$.
• Otherwise the link graph is the quotient of that cycle or path by a reflection, which is a path, possibly with loops.

A path contains no cycle on $k\ge3$ distinct vertices. An $n$-cycle contains one only when $k=n$. $\square$

Theorem 4 (trichotomy). Let $A$ be a rational Diophantine septuple. Exactly one of the following holds.

• (b) $G(A)$ has a vertex of degree $\ge3$. Then some triple $Z\subset A$ has $3S=O$ on $E_Z$, so $E_Z(\mathbb Q)_{\rm tors}\cong\mathbb Z/2\times\mathbb Z/6$.
• (c) $G(A)$ has maximum degree $\le2$ and contains a 4-cycle. Then some triple $Z\subset A$ has $\operatorname{ord}S=4$.
• (d) Neither. Then $G(A)$ is a disjoint union of paths and of cycles of length $\ge5$.

Septuples of type (b), (c) and (d) contain at least 4, 10 and 7 quadruples, respectively, with three pairwise compatible irregular fifths. They also have at least 14, 26 and 21 triples $Z$ over which some $w\in A\setminus Z$ is $S$-linked to none of the other three.

Proof. (b) Let $v$ have neighbours $p,q,r$ and put $Z=A\setminus\{v,p,q,r\}$. By (1), the edges $vp$, $vq$, $vr$ give $S$-links $qr$, $pr$, $pq$ over $Z$. The cycle lemma with $k=3$ applies. (c) For a 4-cycle $v_1v_2v_3v_4$ put $Z=A\setminus\{v_1,\dots,v_4\}$. By (1), its four edges give the $S$-links $v_3v_4$, $v_4v_1$, $v_1v_2$, $v_2v_3$, which form a 4-cycle; apply the lemma with $k=4$. The 2-torsion of $E_Z$ is full, so by Mazur's theorem a point of order 3 forces torsion $\mathbb Z/2\times\mathbb Z/6$, and a point of order 4 forces $\mathbb Z/2\times\mathbb Z/4$ or $\mathbb Z/2\times\mathbb Z/8$. (d) is the complement. The counts come from an exhaustive enumeration of triangle-free graphs on seven vertices (gcls2.py). Each independent 3-set $I$ gives the quadruple $A\setminus I$ of Proposition 3. Over each of its four triples $Z$, the fourth element is $S$-linked to none of $I$, by (1). $\square$

Shape data.

• Over $\mathbb Q(i)$. For the 12 septuples of [6], $G$ has 4 edges (8 septuples) or 5 edges (4 septuples). In every case it has maximum degree 2 and contains a 4-cycle, so all 12 are of case (c). Computing $\operatorname{ord}S$ over all 35 triples of each, we find $\operatorname{ord}S=4$ on exactly 12 triples, one per septuple. Each is the complement of the 4-cycle.
• Almost septuples. For the 25 rational almost septuples of [7], define $G$ by the same rule (7-sets with exactly one bad pair; edges are pairs whose complement is a regular Diophantine quintuple). It has 1, 2 or 3 edges (5, 14 and 6 cases), with degree sequences $(1,1)$, $(1,1,1,1)$ and $(2,1,1,1,1)$. Every edge has exactly one endpoint in the bad pair. So all 25 have case-(d) shape.
• Sextuples. In the corpus of 1123 known sextuples, the number of regular 5-subsets is 0, 1, 2, 3 for 23, 581, 439 and 80 sextuples respectively, and never more. Three regular 5-subsets $A\setminus\{p\}$, $A\setminus\{q\}$, $A\setminus\{r\}$ give an $S$-triangle over $A\setminus\{p,q,r\}$. All 80 such common triples have torsion $\mathbb Z/2\times\mathbb Z/6$ and $\operatorname{ord}S=3$, as the cycle lemma predicts.

Conjecture T. Every rational Diophantine septuple is of case (b) or (c).

Conjecture T is a statement about $\mathbb Q$ only. Its analogue over number fields is false: if $A$ is a rational almost septuple whose bad value is $m\cdot\square$ with $m$ squarefree, then $A$ is a septuple over $K=\mathbb Q(\sqrt m)$ with seven distinct nonzero rational entries. The graph $G(A)$ is cut out by $R_5=0$, which does not depend on the field, so $A$ is of case (d) over $K$. So the 25 almost septuples of [7] are case-(d) septuples over quadratic fields; for 22 of them kiro_septuple finds 22 pairwise distinct fields $\mathbb Q(\sqrt m)$, with no local pattern in $m$ [18]. An earlier version of this paper said that the only septuples known over any field are those over $\mathbb Q(i)$; that was wrong. The support for T is that all 12 septuples known over $\mathbb Q(i)$ are of case (c), the only case-(d) septuples known live over fields where a rational almost septuple acquires its missing square, and the census below finds no rational septuple. The strongest evidence against is that every rational almost septuple has case-(d) shape. Each is a regular structure plus one coincidence [7], so the nearest misses over $\mathbb Q$ sit exactly in the case the conjecture says is empty. The census below cannot decide T. It bounds both sides.

5. Census method

Population. We take all Diophantine triples whose entries are $p/q$ with $|p|<100$ and $0<q<100$, up to the sign change $A\mapsto -A$, which preserves the problem. There are 18,148 of them.

Curve and points. We use an integral model $y^2=(x+bcu^2)(x+acu^2)(x+abu^2)$, in which $z=x(T)/(abcu^2)$, $P_0=(0,abcu^3)$ and $x(S)=u^2$.

• Rank. PARI's ellrank gave equal lower and upper bounds.
• Generators. These were saturated with PARI's ellsaturation up to 100 and then by eclib's mwrank saturation at all primes up to its index bound. On all 18,121 triples of positive rank the index was 1.

Enumeration. Every extender has $T=P_0+\tau+2v$, with $\tau\in 2E_X(\mathbb Q)_{\rm tors}$ and $v$ in the lattice spanned by the generators. The canonical height is a squared norm, so $\hat h(T)\le H$ implies

$$\hat h(v)\le \tfrac14\bigl(\sqrt H+\sqrt{\hat h(P_0+\tau)}\bigr)^2 .$$

For each $\tau$, qfminim lists all $v$ in this ellipsoid, widened by a factor $1+10^{-6}$, and we keep those with $\hat h(T)\le H_X$. The largest count over all triples was 5222, against a cap of $10^7$, so no list was truncated. The bound is

$$H_X=\min\Bigl(1000,\;4\bigl(K\sqrt{\operatorname{reg}}/(|\tau|V_r)\bigr)^{2/r}\Bigr)\quad\text{with } K=2000,$$

a budget of about $K$ lattice points, where $|\tau|$ is the number of classes $\tau$ and $V_r$ is the volume of the unit $r$-ball.

Pairs. For each pair of extenders $y,z$ we decide whether $yz+1$ is a square.

• Prefilter. A quadratic-residue prefilter at the 16 primes from 1009 to 1097 rejects only values that are non-squares modulo some prime, so it never rejects a square.
• Exact test. Survivors get an exact integer square test.

Each compatible pair is then classified:

• translation if $T_z\in\{\pm T_y\pm S\}$, which by Corollary 1.1 means $X\cup\{y,z\}$ is regular;
• $S+E[2]$ if $T_z\in\{\pm T_y\pm S+t\}$ for some $t\in E_X[2]\setminus\{O\}$;
• sporadic otherwise.

We record triangles (sextuples $X\cup\{y,z,w\}$), $K_4$ (septuples) and $K_4^-$, meaning four extenders with five compatible pairs, i.e. almost septuples whose bad pair avoids $X$.

Naive-height coverage. Using the bound of [14, 15] in the form of [10], every extender $z$ with logarithmic height $h(z)\le B_X$ satisfies $\hat h(T_z)\le H_X$, where

$$B_X=H_X-h(abcu^2)-2\mu_X-2.14 .$$

Over the census, $B_X$ has minimum 52.5, 10th percentile 433.7, median 944.0 and maximum 987.2, and it is positive on every triple.

Controls.

• Known sextuples. The census recovers all 17 known sextuples that its range can see.

• Corpus run. A separate run, with the same code and $K$, takes the lowest-height triple of each of 191 randomly chosen known sextuples. It recovers all 191.

• Almost-septuple detection. For each of the 25 almost septuples, we run the census on each of the 10 triples avoiding the bad pair. The other four entries appear as a $K_4^-$ in all 250 runs.

• Calibration. We take the 2978 labellings of known sextuples as (triple, three further entries). In 69 of them some further entry has $\hat h$ above 104, in 21 above 128, and in none above 241.

  So for rank 5, where $H_X$ is as low as 104, about 2% of configurations of the known kind would be missed. For rank $\le 3$ none would be missed once $H_X\ge241$.

6. Results

Curves. The curves have the following ranks, torsion groups and height bounds.

| rank | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---| | triples | 27 | 3901 | 9507 | 4321 | 386 | 6 | | $H_X$ (min / median / max) | 1000 | 1000 | 1000 | 241 / 548 / 1000 | 128 / 204 / 392 | 104 / 138 / 154 |

Torsion is $\mathbb Z/2\times\mathbb Z/2$ on 17,570 triples, $\mathbb Z/2\times\mathbb Z/4$ on 563, $\mathbb Z/2\times\mathbb Z/6$ on 13 and $\mathbb Z/2\times\mathbb Z/8$ on 2. In 545 triples some product equals $-1$, so $S\in E_X[2]$.

Extenders and edges. There are 6,866,176 extenders, with median 211 and maximum 1022 per triple; the run took 4.96 CPU hours. There are 6,306,187 compatible pairs:

• 6,305,114 translations;
• no $S+E[2]$ edges;
• 1073 sporadic edges.

Of the sporadic edges, 931 lie in a quintuple that contains a regular quadruple and 142 do not. They form 831 distinct sporadic quintuples, or 824 up to sign. The number of sporadic edges per triple is 0, 1, 2, 3 for 17,156, 919, 65 and 8 triples.

Theorem 5 (certified census). For each of the 18,148 triples $X$, the census lists every $z$ such that $X\cup\{z\}$ is a Diophantine quadruple and $\hat h(T_z)\le H_X$. It also lists every pair of these $z$ with product plus one a square. This is subject to the correctness of PARI's rank bounds and of eclib's saturation, and to the $10^{-6}$ floating-point margin. Among these graphs:

1. There is no $K_4$ and no $K_4^-$.
2. No extender has three sporadic partners. The sporadic degree is 1 at 2068 extenders and 2 at 39.
3. The triangles give 108 distinct sextuples. Of these, 94 are of DKMS type (a triple with three regular 5-subsets), 11 are twins (a quadruple together with both regular fifths), and 3 are neither. 91 of the 108 are not in our comparison lists (§7), including 3 of the 11 twins.

Corollary 6. Let $X$ be a triple of the census.

• No rational Diophantine septuple contains $X$ with its other four entries satisfying $\hat h(T_z)\le H_X$. In particular, none has all four entries with $h(z)\le B_X$.
• No quadruple $X\cup\{w\}$ has three irregular fifths with all four points below $H_X$, even without asking the three fifths to be compatible. This is the case-(d) building block of Theorem 4.

Proof. A septuple would be a $K_4$. An irregular fifth $p$ of $X\cup\{w\}$ is compatible with $w$ and not $S$-linked to it, by Corollary 1.1. Since there are no $S+E[2]$ edges, three such fifths would give $w$ sporadic degree 3. $\square$

Torsion and triangles. Every DKMS triangle lies on one of nine rank-one triples with torsion $\mathbb Z/2\times\mathbb Z/6$ and $\operatorname{ord}S=3$. The 563 triples with torsion $\mathbb Z/2\times\mathbb Z/4$ carry no triangle, and on 544 of them $\operatorname{ord}S=2$ because a product equals $-1$.

The 39 vertices of sporadic degree 2. At 7 of them the two sporadic partners are compatible. These are exactly the vertices of the triangles of the 3 "other" sextuples. At the remaining 32 they are not compatible. Each of these 32 is a quadruple with two irregular fifths that are not compatible with each other. We report this as data. A uniform null model, with the 89 pairs of sporadic edges over a common triple placed on about 200 extenders at random, predicts one or two shared vertices. That model ignores height bias, so we draw no conclusion here (see §8).

Two further observations, in brief.

• Relations among points. For each of the 142 sporadic quintuples without a regular quadruple, we test all 10 of its triples. On each curve we look for an unforced linear relation among $P_0$, $S$, $T_y$, $T_z$, using height matrices and LLL and verifying any relation exactly. 92 of the 142 have no such relation on any of the 10 curves.
• Almost septuples. In the 25 almost septuples, 45 of the 50 compatible pairs of irregular fifths are regular over some sub-triple.
• Order four. Case (c) needs a triple with $\operatorname{ord}S=4$. Over $\mathbb Q(i)$ it is the only case that occurs. Over $\mathbb Q$ the nine published order-four bases are excluded in [11, 12].

7. Limitations

• The census is bounded. A septuple whose small triples all have their other entries above $H_X$, or whose triples all have entries of height $\ge100$, is invisible to it.
• For rank 4 and 5 the caps are as low as 128 and 104, and calibration shows that about 2% of known configurations would lie above them.
• Certification relies on PARI's ellrank bounds and eclib's saturation, and on floating-point height matrices with a relative margin of $10^{-6}$.
• Theorem 4 is a classification, not an obstruction. Conjecture T is unproved, and the evidence on case (d) cuts both ways. Conjecture T is specific to $\mathbb Q$, since case (d) is inhabited over quadratic fields [18].
• The novelty count of 91 sextuples is relative to the lists we had (the corpus used in [7], the sextuples of [10], and our own families). It may overstate novelty against unpublished lists.

8. Shown, not shown, next

Shown.

• Regular fifths are $S$-translations (Theorem 1), and a quadruple has at most two of them (Lemma 2).
• A septuple exists if and only if some quadruple has three pairwise compatible irregular fifths (Proposition 3).
• Every septuple is of case (b), (c) or (d), with explicit torsion consequences for (b) and (c) (Theorem 4).
• The extender graphs of all 18,148 triples of height below 100 are certified complete up to $H_X$. They contain no septuple, no almost septuple of type $K_4^-$, and no vertex of sporadic degree 3 (Theorem 5).

Not shown.

• That case (d), or either of (b) and (c), is empty over $\mathbb Q$.
• Any uniform statement beyond the height caps.

Next.

1. Explain the 32 incompatible degree-2 vertices. First, a permutation null model that preserves height ranks. Then a test of whether the second irregular fifth comes from a four-term relation $P_0\pm T_y\pm T_{z_1}\pm T_{z_2}\in\{O,\pm S\}$, as for the almost septuples [7].
2. Case (d) contains very structured graphs: $C_7$, $C_6\cup K_1$ and $C_5\cup2K_1$. A $C_7$ is seven equations $R_5=0$ in seven unknowns, so it should have finitely many solutions over $\overline{\mathbb Q}$. Solving that system would decide whether $G\supseteq C_7$ is possible.
3. Over $\mathbb Q(\sqrt m)$ case (d) is inhabited [18]. Find a structural property of the 25 almost septuples, invisible to $G$, that separates them from a putative rational case-(d) septuple; otherwise Conjecture T rests on the census alone.
4. Run a census of triples with $\operatorname{ord}S=3$, organised by pairs, for case (b).

Reproducibility

The supplement contains the census code (census.py) and the analysis scripts:

• thm1.py, thm1deg.py, fifth2.py for Theorem 1 and Lemma 2;
• gcls2.py for Theorem 4;
• gskel.py, rqi2.py, asg.py, sext108.py, torsS.py for the shape data;
• spdeg.py, ctrl100.py, bcov.py, ana.py, calib.py for §§5–6.

It also contains the outputs quoted here (outs.txt, outs2.txt) and the census records with generators stripped (data/h100.tar.xz). The software is Python 3 with cypari2 on PARI/GP 2.17.4, eclib, sympy and gmpy2. The census ran on four processes in about 5 CPU hours.

References

1. A. Dujella, Diophantine m-tuples and Elliptic Curves, Developments in Mathematics 79, Springer (2024). https://doi.org/10.1007/978-3-031-56724-7
2. P. Gibbs, Some rational Diophantine sextuples, arXiv:math/9902081 (1999). https://arxiv.org/abs/math/9902081
3. A. Dujella, M. Kazalicki, M. Mikić, M. Szikszai, There are infinitely many rational Diophantine sextuples, Int. Math. Res. Not. IMRN 2017 (2017) 490–508. https://doi.org/10.1093/imrn/rnv376
4. A. Dujella, M. Kazalicki, V. Petričević, Rational Diophantine sextuples containing two regular quadruples and one regular quintuple, arXiv:1904.00348. https://arxiv.org/abs/1904.00348
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