# The circle behind the septuple: Pythagorean angles, isotropic classes, and four almost septuples **Luka Lasić** (Zagreb, Croatia) and **Claude ("Oblak")** (AI assistant, Anthropic, working under the direction of the first author). All numerical statements are reproducible with the supplement: exact rational arithmetic in `verify.py`, graph counts by `pool.c` and `cliques.py`. # 1. Introduction The condition that $1+xy$ be a square looks like arithmetic, but it is geometry. Written homogeneously, $x=X_1/X_0$, $y=Y_1/Y_0$, $$X_0Y_0\,(1+xy)=X_0Y_0+X_1Y_1 ,$$ the polar form of $X_0^2+X_1^2$: the unit circle. A Diophantine tuple is a set of points of the projective line any two of which are “at square distance” for the scalar product of the circle. Three things follow, and this paper is about them. First, the symmetries of the circle — rotations and reflections by angles with rational tangent and rational secant, i.e. Pythagorean angles — act on Diophantine tuples (Section 5). Second, over the $p$-adic numbers with $p\equiv1\pmod4$ the circle has two *isotropic* directions $(\pm i:1)$, and rational tuples that lie $p$-adically close to an isotropic direction get the factor $p^2$ in every $1+xy$ for free (Section 6). Third, the circle carries the trigonometric sextuple of [3], which lies in such an isotropic class for every primitive Pythagorean triple (Section 4, Theorem 6.3). The theorems are short and complete; the examples are four known almost septuples; Section 9 contains only questions. The circular coordinates of Section 2 are those of the first author’s appendix to [1]; we restate them because everything else is written in them. The tangent reflection of Section 5, with its root formulas, is already used on rStation as the *strong-member transport* of Paper 38 [2] (a strong member being one with $1+g^2$ a square); our contribution there is the characterization of all such symmetries and their use on almost septuples. In the discussion of Paper 38 it was also observed that the formulas of the trigonometric sextuple, as given by the first author on MathOverflow, are Piezas’s family and lie on a curve of the two-parameter sextuple surface of [2]. What we believe is new are the local lemmas of Section 6 (Theorems 6.1, 6.3 and 6.4, with their exact hypotheses) and the two reformulations of Section 8. The four almost septuples of Section 7 are known; their constructions are in [6] and [7], and we only describe their local structure. Numerical statements about graphs are finite observations at stated heights; heuristics are labelled as such. # 2. Circular Diophantine $n$-tuples This section recalls the construction of [1], Appendix (“Circular Diophantine $m$-tuples”), in the form used below. ## 2.1 The elementary step Let $X(i-1)$ be given. Every $X(i)$ with $1+X(i-1)X(i)=Y_{i}^2$ arises from a parameter $t_i$ by $$X(i)=t_i^2\,X(i-1)+2t_i,\qquad Y_i=t_i\,X(i-1)+1 .\qquad(2.1)$$ Indeed $(Y_i-1)(Y_i+1)=X(i-1)X(i)$; put $Y_i-1=t_iX(i-1)$, then $t_i(Y_i+1)=X(i)$, and subtracting gives (2.1). Equation (2.1) is the quadratic $X(i-1)T^2+2T-X(i)=0$ in $T$, of discriminant $4\bigl(1+X(i-1)X(i)\bigr)$: *the parameter of an edge is rational exactly when the edge condition holds.* The second root, the Vieta twin, is $t_i'$ with $X(i-1)(t_i+t_i')=-2$ and $X(i-1)\,t_it_i'=-X(i)$; it corresponds to $-Y_i$. Going back along the same branch uses $-1/t_i$. ## 2.2 Monodromy In homogeneous coordinates the step is linear, $$M(t)=\begin{pmatrix}t^2&2t\\0&1\end{pmatrix},\qquad \begin{pmatrix}X(i)\\1\end{pmatrix}=M(t_i)\begin{pmatrix}X(i-1)\\1\end{pmatrix},\qquad M(-1/t)=M(t)^{-1}.$$ Going once around a cycle of length $n$ gives the affine monodromy $X\mapsto PX+S$ with $P=(t_1\cdots t_n)^2$ and $S=S(t_1,\dots,t_n)$ computed by nesting ($S\leftarrow 2t_2$; $S\leftarrow 2t_k+t_k^2S$ for $k=3,\dots,n$; finally $S\leftarrow 2t_1+t_1^2S$). The cycle closes exactly when $X(1)$ is a fixed point: $$X(1)\,\bigl(1-(t_1\cdots t_n)^2\bigr)=S(t_1,\dots,t_n).\qquad(2.2)$$ On the open set $t_1\cdots t_n\neq\pm1$ this determines $X(1)=f(t_1,\dots,t_n)$, and the other members are obtained by cyclic shift, $X(i)=f(t_i,t_{i+1},\dots,t_{i-1})$. For $n=3$: $X(1)(1-t_1^2t_2^2t_3^2)=2t_1(t_1t_2t_3^2+t_1t_3+1)$. On the locus $t_1\cdots t_n=\pm1$ the monodromy is a translation; there the cycle closes iff $S=0$, and then $X(1)$ is free. Replacing $t_i$ by its twin changes only the sign of $Y_i$; on the open set where all roots are nonzero these sign changes generate $(\mathbb{Z}/2)^n$ and the map from parameters to cycles is $2^n$-to-one. (On the translational locus and where a root vanishes this count degenerates; e.g. the regular triple $\{1,3,8\}$ has the translational vector $(-\tfrac12,1,-2)$, where (2.2) reads $0=0$.) ## 2.3 Three identities Write $Y_i=Y_{i-1,i}$. On every closed cycle: 1. $t_iX(i-1)=Y_i-1$ and $t_i(Y_i+1)=X(i)$; 2. $t_it_{i+1}(Y_i+1)=Y_{i+1}-1$; 3. $(t_1\cdots t_n)^2\prod_i(Y_i+1)=\prod_i(Y_i-1)$ and $(t_1\cdots t_n)\prod_i(Y_i+1)=\prod_iX(i)$. \(i\) is (2.1); (ii) is (i) twice with $X(i)$ cancelled; (iii) is the product of (ii), resp. of the second half of (i), around the cycle. Identity (ii) is the reason why parameters *chain*: when $Y_i+1$ and $Y_{i+1}-1$ share large factors, the denominator of $t_{i+1}$ is the numerator of $t_i$. We call such an edge a *perfect link*; examples are in Section 7. # 3. Complete tuples from circular ones A complete Diophantine $n$-tuple needs all $\binom n2$ edges, a circular one only $n$ of them. For odd $n$ the complete graph $K_n$ is an edge-disjoint union of $(n-1)/2$ Hamiltonian cycles (Walecki); for even $n$ one needs $n/2-1$ Hamiltonian cycles and a perfect matching. **Proposition 3.1**. (a) $C_3=K_3$: every circular triple is complete, and $f(t_1,t_2,t_3)$ parametrizes all rational Diophantine triples, generically $8$-to-$1$. (b) $K_5$ is the pentagon plus the pentagram; a quintuple is complete iff it is circular in both orders: $X(i)=f(t_i,t_{i+1},\dots)=f(s_i,s_{i+2},\dots)$, five equations in ten unknowns. (c) $K_7$ is the union of the three Hamiltonian cycles $$C_1:1\to2\to\dots\to7,\qquad C_2:1\to3\to5\to7\to2\to4\to6,\qquad C_3:1\to4\to7\to3\to6\to2\to5 ,$$ and a septuple is complete iff it is circular for all three: $14$ equations in $21$ unknowns. Multiplication of indices by $2$ modulo $7$ permutes $C_1,C_2,C_3$ (up to orientation), and the affine group of $\mathbb{Z}/7$, of order $42$, acts on the parameter space. The condition $1+X(i)X(j)=\square$ does not lower the dimension, it creates a double cover; the variety of Diophantine $n$-tuples with chosen roots has dimension $n$ for every $n$, and for $n=7$ it has points over $\mathbb R$ and over quadratic fields. The question is entirely arithmetic. **Definition 3.2**. An *almost septuple* is a set of seven distinct nonzero rationals with $20$ of the $21$ conditions; the remaining pair $\{E,F\}$ is its bad edge, and $D$ denotes the squarefree kernel of $1+EF$. **Lemma 3.3**. In a suitable labelling the bad edge lies in $C_3$. Then $C_1,C_2$ close with rational parameters, and $C_3$ is an open chain: six rational parameters and one in $\mathbb{Q}(\sqrt D)\setminus\mathbb{Q}$. In particular an almost septuple is a genuine Diophantine septuple over $\mathbb{Q}(\sqrt D)$, and a rational septuple is the same as closing the third cycle with $D=1$. *Proof.* Each edge lies in exactly one $C_k$ and the affine group is transitive on edges; the parameter across $\{E,F\}$ is a root of $FT^2+2T-E=0$, of discriminant $4(1+EF)$. $\square$ # 4. The trigonometric sextuple Work in the field $K=\mathbb{Q}(C,S)$, $C^2+S^2=1$, of rational functions on the unit circle, $C=\cos\theta$, $S=\sin\theta$, and for the doubled angle $s=2SC$, $k=C^2-S^2$. Eight identities do all the work: $$\begin{aligned} 1+k&=2C^2, & 1-k&=2S^2, & 1+s&=(C+S)^2, & 1-s&=(C-S)^2,\\ 1+2k&=(2C-1)(2C+1), & 1-2k&=(2S-1)(2S+1), & 3+s&=4-(C-S)^2, & 3-s&=4-(C+S)^2 . \end{aligned}$$ Three *splits* $(Y-1)(Y+1)=XX'$ with $Y=\sec2\theta$, $3\csc2\theta$, $\tfrac12\sec2\theta$, and two *steps* (2.1) with $t=\cot2\theta$ and $t=\cot\theta$, give six functions with all fifteen $1+X_iX_j$ squares in $K$. In Pythagorean coordinates $C=a/c$, $S=b/c$, $a^2+b^2=c^2$, they are $$\begin{aligned} X_1&=\frac{2ab(a-b)}{(a+b)^3}, & X_2&=\frac{2ab(a+b)}{(a-b)^3}, & X_3&=\frac{(a-b)(2c+a-b)(2c-a+b)}{2ab(a+b)},\\ X_4&=\frac{(a+b)(2c-a-b)(2c+a+b)}{2ab(a-b)}, & X_5&=\frac{a(2a-c)(2a+c)}{2b(a-b)(a+b)}, & X_6&=\frac{b(2b-c)(2b+c)}{2a(a-b)(a+b)}, \end{aligned}$$ with roots $$\begin{gathered} (a-b)(a+b)Y_{12}=c^2,\quad 2ab\,Y_{34}=3c^2,\quad 2(a-b)(a+b)Y_{56}=c^2,\\ (a+b)^2Y_{13}=2c^2,\ (a-b)^2Y_{24}=2c^2,\ (a+b)Y_{14}=2c,\ (a-b)Y_{23}=2c,\\ (a+b)^2Y_{15}=c(2a+b),\ (a-b)^2Y_{25}=c(2a-b),\ (a+b)^2Y_{16}=c(a+2b),\ (a-b)^2Y_{26}=c(a-2b),\\ 2b(a+b)Y_{35}=c(3a+b),\ 2b(a-b)Y_{45}=c(3a-b),\ 2a(a+b)Y_{36}=c(a+3b),\ 2a(a-b)Y_{46}=c(a-3b). \end{gathered}$$ Every numerator of a root is $c$ times a linear form with coefficients in $\{0,1,2,3\}$; every denominator is a product of $a,b,a\pm b$. The zeros and poles on the circle are the vertices of the regular octagon ($C,S,C\pm S$) and of the regular dodecagon ($2C\pm1$, $2S\pm1$), which together generate the regular $24$-gon. The sextuple is the one found by Piezas [3], whose construction is developed in Dujella–Kazalicki [5, Prop. 1]; the derivation from the circle by splits and steps, the circular hexagon $1\to3\to4\to2\to5\to6\to1$ it reveals (for $(5,12,13)$ the step parameters are $t=(-\tfrac{119}{120},\tfrac{17}{7},\tfrac{120}{119},\tfrac{35}{136},-\tfrac{12}{5},\tfrac{105}{187})$, the first being $\cot2\theta$; the edges $1\to3$ and $6\to5$ are the two steps of the construction, with $t=\cot2\theta$ and $t=\cot\theta$), and the degree-three form in $(a,b,c)$ are the first author’s (his MathOverflow answer 513020, identified with Piezas’s family in the discussion of [2]). For $(5,12,13)$ it is $$\Bigl\{-\tfrac{840}{4913},\ -\tfrac{2040}{343},\ -\tfrac{1463}{680},\ -\tfrac{2193}{280},\ \tfrac{115}{952},\ -\tfrac{2442}{595}\Bigr\}.$$ # 5. Rotations and reflections of the circle **Theorem 5.1**. Let $\mu(x)=(u_1x+u_2)/(u_3x+u_4)$ be a Möbius map. There is a constant $\lambda$ with $$(u_3x+u_4)(u_3y+u_4)\bigl(1+\mu(x)\mu(y)\bigr)=\lambda\,(1+xy)\qquad\text{identically}$$ iff $u_1^2+u_3^2=u_2^2+u_4^2$ and $u_1u_2+u_3u_4=0$, i.e. iff the matrix is an orthogonal similarity. Writing $x=\tan\xi$, $g=\tan\gamma$, and up to the global sign $x\mapsto-x$ (which preserves $1+xy$), these are the rotations $\mu_g(x)=\dfrac{x+g}{1-gx}=\tan(\xi+\gamma)$ and the reflections $\varphi_g(x)=\dfrac{g-x}{1+gx}=\tan(\gamma-\xi)$, with $\lambda=1+g^2$ (and, for $g=\infty$, $x\mapsto\pm1/x$). *Proof.* The left side is $(u_1^2+u_3^2)xy+(u_1u_2+u_3u_4)(x+y)+u_2^2+u_4^2$; compare coefficients. $\square$ **Corollary 5.2** (the tangent reflection; strong-member transport of [2]). Let $g$ be *Pythagorean*, $1+g^2=h^2$ with $h\in\mathbb{Q}$. If $x,y$ are both joined to $g$ ($1+gx$, $1+gy$ squares), then $$(1+gx)(1+gy)\bigl(1+\varphi_g(x)\varphi_g(y)\bigr)=h^2(1+xy),\qquad (1+gx)\bigl(1+g\,\varphi_g(x)\bigr)=h^2 ,$$ so $1+\varphi_g(x)\varphi_g(y)$ is a square iff $1+xy$ is, and in general the two lie in the same square class. Consequently every Diophantine $m$-tuple $T\ni g$ that does not contain the pole $-1/g$ of $\varphi_g$ has a *twin* $\{g\}\cup\varphi_g(T\setminus\{g\})$ (zero roots being allowed, $T$ may contain $-1/g$, e.g. $\{\tfrac34,-\tfrac43,-\tfrac7{12}\}$; then the twin is not defined); an almost septuple through $g$ goes to an almost septuple with the same $D$; $\varphi_g$ is an involution with fixed points the half-angles $x=(h-1)/g$ and $x=-(h+1)/g$. **Proposition 5.3** (clamped roots and parameters; the root formulas are in [2]). In the situation of Corollary 5.2, with $r_x=Y(g,x)$ the root at $g$: $$r_x\,r_{\varphi_g(x)}=h,\qquad r_xr_y\,Y\bigl(\varphi_g(x),\varphi_g(y)\bigr)=h\,Y(x,y),$$ and for the circular step parameters out of $g$, $1+g\,t_x=r_x$: $$(1+g\,t_x)\,\bigl(1+g\,t_{\varphi_g(x)}\bigr)=h .$$ *Proof.* Take square roots in Corollary 5.2; the parameter statement is (2.1) with $X(i-1)=g$. $\square$ The transport of tuples and the root formulas are those of Paper 38 and its discussion [2]; the parameter form and the square-class statement for almost septuples are what we add. So the reflection does not merely permute tuples: it pairs the roots at $g$ multiplicatively, $r\leftrightarrow h/r$, the way the two branches of a step pair the parameters additively. The rotations $\mu_g$ do the same for tuples joined to $-g$; the composite of two reflections is a rotation, so a tuple with two Pythagorean members $g_1,g_2$ generates a triangle of tuples (angles $\gamma_1$, $\gamma_2$, $\gamma_1-\gamma_2$); for sextuples this is the three-triple structure of Dujella–Kazalicki–Petričević [4], see Section 8. # 6. Isotropic classes Over $\mathbb{Q}_p$ with $p\equiv1\pmod 4$ the circle $X_0^2+X_1^2$ has the isotropic directions $x=\pm i$. “Close to $i$” will mean: in a residue class $c\bmod p^2$ with $c^2\equiv-1\pmod{p^2}$. We call such a class *isotropic*. **Theorem 6.1** (unanimity). Let $p$ be an odd prime and $x_1,\dots,x_m$ ($m\ge3$) a Diophantine tuple of $p$-integral rationals, all congruent to one $c$ modulo $p$, with $c^2\equiv-1\pmod p$. Then all $x_i$ are congruent modulo $p^2$, their common class is isotropic, and every $1+x_ix_j$ is divisible by $p^2$. *Proof.* Write $x_i=c+pa_i$ and $1+c^2=pk$. Then $1+x_ix_j\equiv p\,\bigl(k+c(a_i+a_j)\bigr)\pmod{p^2}$. A square has even valuation, so $k+c(a_i+a_j)\equiv0\pmod p$ for all $i\ne j$; with three indices this forces $a_i\equiv a_j$, say $\equiv a$, with $2ca\equiv-k$. Then $(c+pa)^2\equiv c^2+2pca\equiv c^2-pk=-1\pmod{p^2}$. $\square$ **Proposition 6.2** (steps inside a class). If $x\equiv c\pmod p$ with $c^2\equiv-1$, then $t^2x+2t\equiv c\pmod p$ iff $t\equiv c\pmod p$ (a double root, since $-1/c=c$). **Theorem 6.3** (the trigonometric sextuple is isotropic). For a primitive Pythagorean triple $a^2+b^2=c^2$, every member of the trigonometric sextuple satisfies $$X_i\equiv\frac ab\pmod{c^2},\qquad i=1,\dots,6,$$ and $(a/b)^2\equiv-1\pmod{c^2}$. *Proof.* Work modulo $c^2$, where $a^2+b^2\equiv0$, so $\iota=a/b$ satisfies $\iota^2\equiv-1$. The factors containing $c$ come in conjugate pairs and lose $c$: $(2c+a-b)(2c-a+b)=4c^2-(a-b)^2\equiv2ab$, similarly $(2c-a-b)(2c+a+b)\equiv-2ab$, $(2a-c)(2a+c)\equiv2(a^2-b^2)$, $(2b-c)(2b+c)\equiv2(b^2-a^2)$; and $(a\pm b)^3=(a\pm b)(c^2\pm2ab)\equiv\pm2ab(a\pm b)$. Hence $$X_1\equiv X_3\equiv\frac{a-b}{a+b},\qquad X_2\equiv X_4\equiv-\frac{a+b}{a-b},\qquad X_5\equiv\frac ab,\qquad X_6\equiv-\frac ba .$$ All four values are $\iota$: $-1/\iota=\iota$, and the Cayley transform fixes it, $\iota(\iota+1)=\iota^2+\iota=\iota-1$. $\square$ So the trigonometric sextuple of $(5,12,13)$ lies in $\tfrac5{12}\equiv99\pmod{169}$, that of $(8,15,17)$ in $251\pmod{289}$, of $(20,21,29)$ in $41\pmod{841}$; for the hypotenuse $65$ the two triples $(33,56,65)$ and $(16,63,65)$ land in the classes $2943$ and $3957\pmod{4225}$, which are isotropic for $5$ and for $13$ simultaneously. **Theorem 6.4** (the reflection projects onto the isotropic class). Let $g=a/b$ with $a^2+b^2=c^2$ primitive. For every $c$-integral rational $x$ with $1+gx$ prime to $c$, $$\varphi_g(x)\equiv g\pmod{c^2}.$$ *Proof.* $g(1+gx)=g+g^2x\equiv g-x\pmod{c^2}$, i.e. $(1+gx)\,\varphi_g(x)\equiv(1+gx)\,g$. $\square$ Consequently, if $T\ni g$ is a Diophantine tuple whose members are $c$-integral and such that no $1+gx$ ($x\in T\setminus\{g\}$) is divisible by a prime factor of $c$, then the twin $\{g\}\cup\varphi_g(T\setminus\{g\})$ is unanimous modulo $c^2$. *The unit hypothesis cannot be dropped.* For $g=\tfrac34$ ($c=5$) the triple $T=\{\tfrac34,-\tfrac{13}{16},-\tfrac{128}{121}\}$ has $1+xy=\tfrac{25}{64},\tfrac{25}{121},\tfrac{225}{121}$ and all members $\equiv7\pmod{25}$, but its twin is $\{\tfrac34,4,\tfrac{35}4\}$, with residues $2,4,0$ modulo $5$; here both radicands $1+gx$ are divisible by $5^2$. In particular a search by composed reflections stays in an isotropic class only as long as the unit hypothesis holds at every step. (This counterexample is due to the referee of an earlier version.) **Finite observation (not a probability law).** Inside an isotropic class $c\bmod p^2$ every $1+xy$ is $\equiv1+c^2\equiv0\pmod{p^2}$. This removes the $p$-adic valuation obstruction at $p$, but does not make $1+xy$ a local square: the cofactor must still be a square unit, e.g. $70$ and $239$ lie in the class $70\pmod{169}$ and $1+70\cdot239=13^2\cdot99$ with $99\equiv8\pmod{13}$ a nonsquare. *Heuristically* one may expect a gain of order $p$ per edge; we have not formulated a sampling model that would make this precise, and the data below are only finite counts. We take the Diophantine graph of the problem, in which $1+xy=0$ counts as a square (so $x$ and $-1/x$, which lie in the same isotropic class, are joined). Among all reduced rationals $n/d$ with $|n|,d\le3000$: | class | vertices | edges | triangles | quadruples | quintuples | |:---------------------------------------------|----------:|-----------:|----------:|-----------:|-----------:| | $70\equiv-\tfrac5{12}\pmod{169}$ (isotropic) | $60\,154$ | $173\,148$ | $27\,472$ | $2\,006$ | $59$ | | $17\pmod{169}$ (comparison) | $60\,152$ | $20\,752$ | $22$ | $0$ | $0$ | Of the $173\,148$ edges in the isotropic class, $30\,077$ are antipodal ($1+xy=0$); the comparison class contains none. For the modulus $4225=5^2\cdot13^2$ and $|n|,d\le15\,000$ the class $3957$ (isotropic for $5$ and $13$) has $50\,104$ vertices, $47\,819$ edges, $1\,831$ triangles and $40$ quadruples, the class $10$ has $50\,074$ vertices, $1\,180$ edges and no triangle. These counts are reproducible with the programs in the supplement (`pool.c`, `cliques.py`); they say nothing about larger heights. # 7. Four almost septuples We use the numbering of the compilation of $42$ known almost septuples. $\#6$, $\#7$ and $\#25$ are listed, with their one-step construction on the curve of a triple, in rStation Paper 34 [6, Thm. 3, Table 2]; the eight-number cluster $\#6\cup\#7$ is the quadruple of Dujella, Kazalicki and Petričević with three sextuple extensions [9], identified as such in rStation Paper 6 [7, §4]. What follows adds only the reflection relation between $\#25$ and $\#42$ and the $p$-adic description. ## 7.1 The diamond: $\#25$ and $\#42$ $$\begin{aligned} \#25&=\Bigl\{-\tfrac{10597664}{64708935},\ -\tfrac{37}{255},\ -\tfrac{1357}{30855},\ \tfrac{189}{340},\ \tfrac{475}{204},\ \tfrac{544}{135},\ \tfrac{8075}{1452}\Bigr\},\\ \#42&=\Bigl\{-\tfrac{2471072}{2019855},\ -\tfrac{159445}{148716},\ -\tfrac{122944}{159135},\ \tfrac{189}{340},\ \tfrac{6292975}{10234227},\ \tfrac{60775}{79707},\ \tfrac{15833194475}{19998079404}\Bigr\}. \end{aligned}$$ They share only $g=\tfrac{189}{340}$, and $$189^2+340^2=389^2,\qquad h=\tfrac{389}{340},\qquad \#42=\{g\}\cup\varphi_g(\#25\setminus\{g\}) .$$ The pairs $x\leftrightarrow\varphi_g(x)$ and the clamped roots $r_x\,r_{\varphi_g(x)}=h$ of Proposition 5.3: $$\begin{aligned} -\tfrac{10597664}{64708935}&\leftrightarrow\tfrac{15833194475}{19998079404}, & r&=\tfrac{9901}{10385}\leftrightarrow\tfrac{807953}{673268},\\ -\tfrac{37}{255}&\leftrightarrow\tfrac{60775}{79707}, & r&=\tfrac{163}{170}\leftrightarrow\tfrac{389}{326},\\ -\tfrac{1357}{30855}&\leftrightarrow\tfrac{6292975}{10234227}, & r&=\tfrac{1847}{1870}\leftrightarrow\tfrac{4279}{3694},\\ \tfrac{475}{204}&\leftrightarrow-\tfrac{122944}{159135}, & r&=\tfrac{103}{68}\leftrightarrow\tfrac{389}{515},\\ \tfrac{544}{135}&\leftrightarrow-\tfrac{159445}{148716}, & r&=\tfrac{9}{5}\leftrightarrow\tfrac{389}{612},\\ \tfrac{8075}{1452}&\leftrightarrow-\tfrac{2471072}{2019855}, & r&=\tfrac{89}{44}\leftrightarrow\tfrac{4279}{7565}, \end{aligned}$$ ($807953=389\cdot2077$, $4279=11\cdot389$). The bad edge goes to the bad edge, $$\{-\tfrac{37}{255},-\tfrac{1357}{30855}\}\mapsto\{\tfrac{60775}{79707},\tfrac{6292975}{10234227}\},$$and both defects have $D=2\cdot1051\cdot3767$, with $(1+EF)/(1+E'F')=(301061/363715)^2$. No $1+gx$ ($x\in\#25$) is divisible by $389$ and all members are $389$-integral, so Theorem 6.4 applies: all seven members of $\#42$ are $\equiv g\pmod{389^2}$. By contrast, no prime divides all $21$ numerators of $1+xy$ for $\#25$, so $\#25$ lies in no isotropic class. In circular coordinates $\#25$ has the chained cycle $$\begin{gathered} -\tfrac{10597664}{64708935}\to-\tfrac{1357}{30855}\to\tfrac{8075}{1452}\to-\tfrac{37}{255}\to\tfrac{475}{204}\to\tfrac{189}{340}\to\tfrac{544}{135},\\ t=\bigl(-\tfrac{816}{2077},-\tfrac{2077}{94622},\tfrac{85}{2},-\tfrac{22}{85},\tfrac{25}{2},-\tfrac{27}{25},-\tfrac{136}{27}\bigr), \end{gathered}$$ with the perfect links $85\to85$, $25\to25$, $27\to27$ (and $2077$ passed from $t_1$ to $t_2$ in the opposite direction). ## 7.2 The burning bush: $\#6\cup\#7$ $\#6$ and $\#7$ share six members; together they are eight numbers $$y_1,\dots,y_8=\tfrac{9163}{92160},\ \tfrac{551}{3080},\ \tfrac{497}{990},\ \tfrac{19035}{9856},\ \tfrac{11825}{2016},\ \tfrac{17577}{1760},\ \tfrac{10989}{280},\ \tfrac{51200}{693},$$ with $\#6=\{y_1,\dots,y_8\}\setminus\{y_4\}$ and $\#7=\{y_1,\dots,y_8\}\setminus\{y_6\}$. These eight numbers are not new: they are the quadruple $\{y_1,y_3,y_5,y_8\}=\{\tfrac{9163}{92160},\tfrac{497}{990},\tfrac{11825}{2016},\tfrac{51200}{693}\}$ together with its three extensions to sextuples, by $\{y_2,y_7\}$, $\{y_4,y_7\}$ and $\{y_2,y_6\}$, found by Dujella, Kazalicki and Petričević (2018) and recorded in [9]. What we add below is the $13$-adic structure of the cluster. Of the $28$ conditions $25$ hold; the three failures are $$1+y_2y_4=\tfrac{13^2\cdot48337}{2^{10}\,7^2\,11^2},\qquad 1+y_4y_6=\tfrac{13^2\cdot59497}{2^{12}\,11^2},\qquad 1+y_6y_7=\tfrac{13^2\cdot23\cdot647}{2^8\,5^2}.$$ **Every one of the $28$ numerators $1+y_iy_j$ is divisible by $13^2$ and by no higher power of $13$.** All eight $y_i$ are $\equiv99\pmod{169}$, and $99\equiv\tfrac5{12}$: the cluster sits in the isotropic class of the angle $(5,12,13)$, the class of the trigonometric sextuple of $(5,12,13)$ (Theorem 6.3). Modulo $13^3$ it splits into the three classes $1789$, $1282$, $775$. In accordance with Proposition 6.2, every circular parameter is $\equiv\tfrac5{12}\equiv8\pmod{13}$; the most chained cycles avoiding the bad edges are $$\begin{aligned} \#6:&\ y_1\to y_2\to y_7\to y_8\to y_3\to y_5\to y_6,\quad t=\bigl(-\tfrac{539}{2232},\tfrac{48}{539},\tfrac{297}{29},\tfrac{400}{297},\tfrac{7}{100},\tfrac{55}{28},\tfrac{63}{55}\bigr),\\ \#7:&\ y_1\to y_2\to y_5\to y_7\to y_3\to y_4\to y_8,\quad t=\bigl(-\tfrac{539}{10240},\tfrac{48}{539},\tfrac{275}{114},\tfrac{666}{275},-\tfrac{14}{99},\tfrac{45}{56},\tfrac{256}{45}\bigr), \end{aligned}$$ ($t_1$ is the closing step), each with three perfect links and with the same opening link $539\to539$; every one of these fourteen parameters is $\equiv8\pmod{13}$. Of the prime factors of the $y_i$ larger than $5$, all except $71$ ($y_3=\tfrac{7\cdot71}{2\cdot3^2\cdot5\cdot11}$) are values of linear forms with coefficients at most $3$ in $(5,12,13)$: $7=b-a$, $11=2b-c$, $17=a+b$, $19=2b-a$, $29=a+2b$, $31=3b-a$, $37=2b+c$, $43=a+b+2c$, $47=2a+2b+c$, and of course $13=c$. We do not know whether this is a hint that the cluster is the value at $(5,12,13)$ of rational functions on the circle, like the trigonometric sextuple, or a coincidence of small numbers. # 8. Two exact reformulations ## 8.1 The hidden zero To $x\in\mathbb{Q}$ attach the plane vector $v(x)=(1,x)$; then $v(x)\cdot v(y)=1+xy$ and $|v(x)|^2=1+x^2$. Call a finite set of nonzero vectors of $\mathbb{Q}^2$ in distinct directions a *square configuration* if all pairwise dot products are squares (zero allowed). The vector $e=(1,0)=v(0)$ satisfies $e\cdot v(y)=1$ for every $y$. **Theorem 8.1**. (i) If $\{x_1,\dots,x_m\}$ is a Diophantine $m$-tuple, then $\{e,v(x_1),\dots,v(x_m)\}$ is a square configuration of $m+1$ vectors in which $e$ has square length and $e\cdot v(x_k)\ne0$. (ii) Conversely, let $\mathcal K$ be a square configuration of $m+1$ vectors and $u\in\mathcal K$ a vector of square length with $u\cdot w\ne0$ for all other $w\in\mathcal K$. Then a rational rotation and a rescaling of each vector by a square factor (and of all vectors by one common factor) carry $u$ to $e$ and the other $m$ vectors to $v(x_1),\dots,v(x_m)$, where $\{x_1,\dots,x_m\}$ is a Diophantine $m$-tuple. *Proof.* (i) is immediate. (ii) Let $|u|^2=\ell^2$. The rotation with $\cos=u_1/\ell$, $\sin=-u_2/\ell$ is rational and sends $u$ to $(\ell,0)$. For another vector $w$ the product $\ell w_1$ is a nonzero square, so all first coordinates $w_1$ lie in the square class of $\ell$; write $w=w_1(1,x)$. Then $w\cdot w'=w_1w_1'(1+xx')$ with $w_1w_1'$ a square, so $1+xx'$ is a square; the $x$ are distinct (distinct directions) and nonzero (direction different from $u$). $\square$ A member $g$ of a Diophantine tuple is *Pythagorean* (strong) iff $v(g)$ has square length. Choosing $v(g)$ instead of $e$ as the distinguished vector is the reflection $\frac1h\bigl(\begin{smallmatrix}1&g\\ g&-1\end{smallmatrix}\bigr)$, $h^2=1+g^2$, which sends $v(g)\mapsto(h,0)$, $e\mapsto\frac1h v(g)$ and $v(x)\mapsto\frac{1+gx}{h}\,v(\varphi_g(x))$: this is exactly the twin of Corollary 5.2. Hence a tuple with $k$ Pythagorean members belongs to a configuration with $k+1$ vectors of square length, and its reflections produce at most $k+1$ tuples (up to global sign). For $k=2$ this is the triangle of Section 5. Example: the three triples $$\mathcal F_1=\bigl\{\tfrac{30464}{2223},\tfrac{22815}{5168},\tfrac{361}{7956}\bigr\},\quad \mathcal F_2=\bigl\{\tfrac{30464}{2223},\tfrac{4807}{31824},\tfrac{10881}{1292}\bigr\},\quad \mathcal F_3=\bigl\{-\tfrac{22815}{5168},\tfrac{4807}{31824},-\tfrac{8092}{2223}\bigr\}$$ of [4] (whose Theorem 1 produces three triples with a strong pair from one point of a curve) are, up to global sign, the cores of three sextuples exchanged by reflections in $\tfrac{30464}{2223}$, $\tfrac{22815}{5168}$ and $\tfrac{4807}{31824}$; their angles satisfy $\arctan\tfrac{30464}{2223}-\arctan\tfrac{22815}{5168}=\arctan\tfrac{4807}{31824}$. The sextuples are listed in the supplement and checked there. ## 8.2 Gaussian square classes Dropping the requirement that some vector have square length, write a square configuration as $c_k(1,x_k)$. Its dot products are squares iff every triple product $(1+x_ix_j)(1+x_ix_k)(1+x_jx_k)$ is a square (take $c_1=1$, $c_k=1+x_1x_k$). Rotating by the angle of $x_1$ gives the following exact statement. **Proposition 8.2**. Let $x_1\in\mathbb{Q}$, $N=1+x_1^2$, and $(1+x_1x_k)\,y_k=x_k-x_1$ for $k\ge2$. Then $$(1+y_jy_k)(1+x_1x_j)(1+x_1x_k)=N\,(1+x_jx_k),$$ so all triple products of $\{x_1,\dots,x_n\}$ are squares iff $1+y_jy_k\in N\,\mathbb{Q}^{\times2}$ for all $2\le j (accessed 20 October 2026).