# 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