# The Pythagorean game: the circle, the Sophie Germain hyperbola, and the hyperbolic tetrahedron ## Classes, congruent numbers, and a recursion for the sets that are D(1) and D(-1) at once *Pitagorejska glavolomka iznutra* **Luka Lasić** (Zagreb, Croatia) and **Claude ("Oblak")** (AI assistant, Anthropic). Claude works under the direction of the first author; the two are fully equal partners in the writing, in symbiosis. The LaTeX source, the PDF and two short scripts realising the recursion are in the supplement. 31 October 2026 **Abstract.** Two nonzero rationals meet in one of four ways, $1\pm xy=\pm\square$, and their product lies in one of four bands cut out by $-1$, $0$ and $1$; of the sixteen combinations twelve occur. A pair carrying two colours is a Pythagorean angle: the product is the sine of a doubled rational angle, on the circle $r^2+s^2=2$, or a cosecant, on the hyperbola $r^2-s^2=2$, whose numerator is Sophie Germain’s $T^4+4$. A *hyperbolic triangle*, three rationals with three hyperbola edges, is a rational set with the properties $D(1)$ and $D(-1)$ at once; a fourth vertex would be a *hyperbolic tetrahedron*. We prove the structure of these objects. Every hyperbola edge has a class $c(m)$, the square class of $m^4+4$, a product of primes congruent to one modulo four; a hyperbolic triangle is exactly a triple of edges whose classes multiply to a square, a closed cycle in the $\mathbb{F}_2$-space on those primes. The edges of one class $c$ are a coset of $2E_c(\mathbb{Q})$ on the congruent number curve $E_c\colon Y^2=X^3-c^2X$, under $X=c\cdot xy$; hence every class is a congruent number, every class is attained infinitely often, and every hyperbolic triangle lies in three infinite families, one for each pair of its edges, generated by a recursion of one chord on $E_c$. On the tetrahedron we prove what is local: the shape of the $2$-adic valuations, the reduction of the conditions at an odd prime to Legendre symbols and the criterion at the roots, and the fact, from a Hasse–Weil bound on a curve of genus forty-nine, that a prime beyond an explicit bound never obstructs. We conjecture that no hyperbolic tetrahedron exists, and we state the sign-free Strong Conjecture whose conjunction with it would exclude rational Diophantine septuples with distinct absolute values and no zero square. # 1. Twelve ways to meet Let $x$ and $y$ be nonzero rationals. We say that they *meet* if $$1+\varepsilon\,xy=\delta\,s^2\qquad\text{for some } s\in\mathbb{Q}^*,\ \varepsilon,\delta\in\{1,-1\},$$ and we name the four relations $$\mathrm M\colon\ 1+xy=\square,\qquad \mathrm M'\colon\ 1+xy=-\square,\qquad \mathrm P_1\colon\ 1-xy=\square,\qquad \mathrm P_2\colon\ xy-1=\square .$$ $\mathrm M$ is the relation $D(1)$ of the classical problem, $\mathrm P_2$ is $D(-1)$, and $\mathrm P_1$, $\mathrm P_2$ together are the mixed relation $|1-xy|\in\mathbb{Q}^{*2}$ of \[2, 3\], the relation $D(1)$ over $\mathbb{Q}(i)$ for entries in $i\mathbb{Q}$. Each relation confines the product to a band, and the three numbers $-1$, $0$, $1$ cut the line of products into four bands. A relation can live only where its band allows, and two relations can hold at once only where their bands overlap: | band | single colours | double colours | |---|---|---| | $xy<-1$ | $\mathrm M'$, $\mathrm P_1$ | $\mathrm M'\mathrm P_1$ | | $-11$ | $\mathrm M$, $\mathrm P_2$ | $\mathrm M\mathrm P_2$ | | $xy=-1$ | $1+xy=0$: the square zero | | | $xy=0$ | $1\pm xy=1$: the zero member meets everyone twice | | | $xy=1$ | $1-xy=0$: the square zero | | Twelve combinations of band and colour, three cuts; all twelve occur. Two symmetries act on the table. The change of sign of one member exchanges the outer bands with the inner ones and $\mathrm M\leftrightarrow\mathrm P_1$, $\mathrm M'\leftrightarrow\mathrm P_2$; what it does not touch is the sign $\delta$ of the square, which is $-1$ exactly for $\mathrm M'$ and $\mathrm P_2$, the two colours of the outer bands that look away from the origin. We call these edges *far*. The passage to absolute values $p=|x|$, $q=|y|$ kills $\mathrm M'$ and leaves *three colours* in two bands, $$1+pq=\square\ (\text{near}),\qquad 1-pq=\square\ (\text{near},\ pq<1),\qquad pq-1=\square\ (\text{far},\ pq>1),$$ with the single cut $pq=1$ and the zero member. For a finite set of rationals, its *coloured graph* has the members as vertices and, between two members that meet, an edge carrying its colours and its band. A rational Diophantine $m$-tuple is a complete $\mathrm M$-graph; a mixed octuple of \[2, 3\] is a complete graph in $\mathrm P_1$ and $\mathrm P_2$; the number $0$ is a universal vertex. A member may meet itself, $1+x^2=\square$ or $|1-x^2|=\square$, but not in both ways, since $1-x^4=\pm\square$ has no nonzero rational solution by Fermat’s theorem on the right triangle; this is classical and we use it only as a remark. Two cautions for absolute values: distinct rationals need not have distinct absolute values, and a classical set may contain the square zero $1+xy=0$, whose absolute values have $pq=1$ and meet by none of the three colours with a nonzero square. # 2. Double edges A double edge is a pair meeting by two colours. With $1+xy=r^2$ and $1-xy=\pm s^2$, adding gives $r^2\pm s^2=2$: the circle of radius $\sqrt2$, the norm equation $N(r+is)=N(1+i)$ in $\mathbb{Z}[i]$, or the hyperbola $r^2-s^2=2$. **Theorem 2.1** (double edges are Pythagorean angles). *Two rationals meet by two colours if and only if, for some rational right triangle $u^2+l^2=v^2$ or some $m\in\mathbb{Q}^*$, $$xy=\frac{2ul}{v^2}\quad(\text{circle},\ \mathrm M\mathrm P_1)\qquad\text{or}\qquad xy=f(m):=\frac{m^4+4}{4m^2}\quad(\text{hyperbola},\ \mathrm M\mathrm P_2),$$ the signed sister $\mathrm M'\mathrm P_1$ being the hyperbola with $xy<0$. If $\theta$ is the angle of the triangle $(u,l,v)$, on the circle $xy=\sin2\theta$; on the hyperbola $xy=1/\sin\varphi$ for the angle $\varphi$ of the triangle $(m^4-4,\,4m^2,\,m^4+4)$, with $\tan\frac{\varphi}{2}=2/m^2$. Moreover $f(m)=f(2/m)=f(-m)$, and $m\mapsto\{m,-m,2/m,-2/m\}$ is the full fibre of $f$.* *Proof.* If $1+xy=r^2$, $1-xy=s^2$, then $(r+s)^2+(r-s)^2=4$, so $(u,l,v)=(r+s,r-s,2)$ is a rational right triangle up to scaling and $xy=(r^2-s^2)/2=2ul/v^2$; conversely $v^2(1\pm xy)=(u\pm l)^2$. If $1+xy=r^2$, $xy-1=s^2$, then $(r+s)(r-s)=2$; with $r+s=m$, $r-s=2/m$ we get $r=(m^2+2)/(2m)$, $s=(m^2-2)/(2m)$, $xy=r^2-1=f(m)$, and conversely $$4m^2(1+xy)=(m^2+2)^2,\qquad 4m^2(xy-1)=(m^2-2)^2,\qquad m^4+4=(m^2+2m+2)(m^2-2m+2),$$ the last being Sophie Germain’s identity, the factorisation of $T^4+4$ in $\mathbb{Z}[i]$. The identity $(m^4-4)^2+(4m^2)^2=(m^4+4)^2$ gives the angle. Finally $f$ is a rational function of degree four invariant under the group generated by $m\mapsto-m$ and $m\mapsto2/m$, which has order four, so these are its fibres. $\square$ **Theorem 2.2** (double triangles). *Three nonzero rationals meet pairwise by two colours if and only if the three products are values of Theorem 2.1 whose product is a rational square: $x^2=AB/C$ for the products $A=xy$, $B=xz$, $C=yz$.* **Definition 2.3**. A *hyperbolic triangle* is a set of three nonzero rationals meeting pairwise by $\mathrm M\mathrm P_2$, that is, a rational set that is at once a $D(1)$- and a $D(-1)$-triple; its *edge parameters* are the $m_1,m_2,m_3$ with $xy=f(m_3)$, $xz=f(m_2)$, $yz=f(m_1)$, each defined up to the fibre of $f$. A *hyperbolic tetrahedron* is such a set with four members. Since $xy-1=\square$ forces $xy>1$, the members of a hyperbolic set have one sign, and we take them positive. # 3. Classes, congruent numbers, and the recursion **Definition 3.1**. For $m\in\mathbb{Q}^*$ let $c(m)$ be the *class* of the edge $f(m)$: the squarefree positive integer representing $m^4+4$ in $\mathbb{Q}^*/\mathbb{Q}^{*2}$. Equivalently $c(m)$ is the class of $f(m)$ itself, since $4m^2f(m)=m^4+4$. **Theorem 3.2** (classes are made of primes $\equiv1\pmod4$). *Every class $c(m)$ is a product of distinct primes congruent to $1$ modulo $4$; in particular $c(m)\equiv1$ or $5\pmod 8$.* *Proof.* Write $m=p/q$ in lowest terms; $c(m)$ is the class of $p^4+4q^4$. If $p$ is odd, $p^4+4q^4=(p^2-2q^2)^2+(2pq)^2$ is a sum of two coprime squares, odd. If $p=2k$ is even, $q$ is odd and $p^4+4q^4=4\,(q^4+4k^4)=4\bigl((q^2-2k^2)^2+(2qk)^2\bigr)$, again four times a sum of two coprime odd-and-even squares. In both cases the odd part is a sum of two coprime squares, so each of its prime factors is $\equiv1\pmod 4$, and the factor $4$ is a square. $\square$ **Theorem 3.3** (a hyperbolic triangle is a closed cycle of classes). *Let $V$ be the $\mathbb{F}_2$-vector space with basis the primes $\equiv1\pmod4$, and read a class as the vector of its prime factors. Three edge parameters $m_1,m_2,m_3$ are the edges of a hyperbolic triangle if and only if $$c(m_1)+c(m_2)+c(m_3)=0\quad\text{in }V,$$ i.e. each prime occurs in an even number of the three classes. The triangle is then unique up to sign.* *Proof.* By Theorem 2.2 the triangle exists iff $f(m_1)f(m_2)f(m_3)$ is a square, iff $c(m_1)c(m_2)c(m_3)$ is a square, iff the three vectors sum to zero; the members are $x=\pm\sqrt{f(m_2)f(m_3)/f(m_1)}$ and so on. $\square$ This is the circular closure of our first paper \[1\] in a new place: there a cycle of parameters closed a classical triple, here a cycle of classes closes a hyperbolic one. **Theorem 3.4** (the edges of one class form a coset on the congruent number curve). *Let $c$ be a squarefree positive integer and $E_c\colon Y^2=X^3-c^2X$. For $m$ with $c(m)=c$, choose $w\in\mathbb{Q}$ with $w^2=c\,(m^4+4)$ and put $$X=\frac{w^2}{4m^2}=c\,f(m),\qquad Y=\frac{c\,w\,(m^4-4)}{8m^3}.$$ Then $(X,Y)\in E_c(\mathbb{Q})$, and $X/c-1$, $X/c+1$ are the rational squares $s^2$, $r^2$ with $s=(m^2-2)/(2m)$, $r=(m^2+2)/(2m)$, so that $m=r+s$, $2/m=r-s$. Conversely every point of $E_c(\mathbb{Q})$ with $X/c-1$ and $X/c+1$ squares arises in this way, and the set of such points is either empty or a single coset $P_0+2E_c(\mathbb{Q})$. Up to the symmetries $m\mapsto\pm m,\pm2/m$ and $w\mapsto-w$ the correspondence is bijective.* *Proof.* Put $W=w/(2mc)$, so that $cW^2=(m^4+4)/(4m^2)=f(m)=1+s^2$ and $X=c^2W^2$. Then $X-c=c(cW^2-1)=cs^2$ and $X+c=c(cW^2+1)=c(2+s^2)=cr^2$, so $X(X-c)(X+c)=(c^2Wsr)^2$ and $c^2Wsr=c\,w\,(m^2-2)(m^2+2)/(8m^3)=Y$. For the converse, let $\varphi(P)=(X,\,X-c)\in(\mathbb{Q}^*/\mathbb{Q}^{*2})^2$ be the $2$-descent map of $E_c$, a homomorphism $E_c(\mathbb{Q})\to(\mathbb{Q}^*/\mathbb{Q}^{*2})^2$ with kernel exactly $2E_c(\mathbb{Q})$ on points with $X\ne0,\pm c$. The points constructed above have $\varphi(P)=(1,c)$, since $X=(w/2m)^2$ and $X-c=cs^2$. If conversely $\varphi(P)=(1,c)$, then $X$ is a square, $(X-c)/c$ is a square, and $(X+c)/c$ is a square too, because $X(X-c)(X+c)=Y^2$ makes $X+c$ congruent to $X(X-c)$, hence to $c$, modulo squares; so $s^2=X/c-1$ and $r^2=X/c+1$ have rational roots, $r^2-s^2=2$ puts $m=r+s$ on the hyperbola with $f(m)=1+s^2=X/c$, and $c(m^4+4)=4m^2X$ is a square, so $c(m)=c$. Since $\varphi$ is injective on $E_c(\mathbb{Q})/2E_c(\mathbb{Q})$, the fibre $\varphi^{-1}(1,c)$ is a single coset of $2E_c(\mathbb{Q})$ or empty. The eight sign choices of $(s,r,w)$ give the four values $\pm m,\pm2/m$ and the two points $(X,\pm Y)$. $\square$ **Corollary 3.5**. *Every class $c(m)$ is a congruent number, the area of a rational right triangle, and every class that occurs occurs infinitely often: the set $\{m\colon c(m)=c\}$ is infinite whenever it is nonempty.* *Proof.* A point of the coset has $X=c\,f(m)>c$, so it is not one of the three points of order two, which are the whole torsion of $E_c$ (classical: by Nagell–Lutz and reduction, the torsion of $Y^2=X^3-n^2X$ is the three points of order two). Hence $E_c(\mathbb{Q})$ has positive rank, which is the classical criterion, going back to Fermat’s descent for $n=1$, for $n$ to be the area of a rational right triangle, and the coset $P_0+2E_c(\mathbb{Q})$ is infinite. $\square$ **Theorem 3.6** (the recursion). *Let $\{x,y,z\}$ be a hyperbolic triangle with edge parameters $m_1$, $m_2$, $m_3$, where $yz=f(m_1)$, $xz=f(m_2)$ and $xy=f(m_3)$, and let $c=c(m_3)$. Let $G$ be a point of infinite order on $E_c$ and $P_3$ the point of Theorem 3.4 attached to $m_3$. For every integer $n$, the point $P_3+2nG=(X_n,Y_n)$ gives, by $s_n^2=X_n/c-1$, $r_n^2=X_n/c+1$, $m_n=r_n+s_n$, an edge parameter of class $c$, and $$\{x_n,\ y_n,\ z_n\},\qquad z_n=\frac{c\,k\,m_n}{2m_1m_2\,w_n},\qquad x_n=\frac{f(m_2)}{z_n},\qquad y_n=\frac{f(m_1)}{z_n},$$ where $w_n^2=c\,(m_n^4+4)$ and $k^2=(m_1^4+4)(m_2^4+4)/c$, is a hyperbolic triangle with the same two edge values $y_nz_n=f(m_1)$, $x_nz_n=f(m_2)$ as the original and third edge $x_ny_n=f(m_n)$. Two points of the coset give the same triangle exactly when they have the same $X$-coordinate, that is, when they are $P$ and $-P$; apart from this identification distinct points give distinct triangles, so the family is infinite. Every hyperbolic triangle with the edge values $f(m_1)$, $f(m_2)$ arises from a point of the coset $P_3+2E_c(\mathbb{Q})$; when $E_c(\mathbb{Q})$ has rank one and $G$ generates it modulo torsion, the coset is $\{P_3+2nG\colon n\in\mathbb{Z}\}$ and the family is the sequence above, read modulo $n\leftrightarrow$ the index of $-(P_3+2nG)$. One step of the recursion is one chord on $E_c$: with $H=2G=(X_H,Y_H)$, $$\lambda=\frac{Y-Y_H}{X-X_H},\qquad X'=\lambda^2-X-X_H,\qquad Y'=\lambda(X-X')-Y .$$* *Proof.* By Theorem 3.3 the class $c(m_3)$ is determined by $c(m_1)c(m_2)$, and $(m_1^4+4)(m_2^4+4)=c\,k^2$ for a rational $k$. By Theorem 3.4 the edge parameters of class $c$ are exactly the $m$ attached to the points of $P_3+2E_c(\mathbb{Q})$, and for each such $m_n$ the triple of edges $(m_1,m_2,m_n)$ is closed, so Theorem 3.3 gives a triangle. Its member opposite the new edge satisfies $$z_n^2=\frac{f(m_1)f(m_2)}{f(m_n)}=\frac{(m_1^4+4)(m_2^4+4)}{16m_1^2m_2^2}\cdot\frac{4m_n^2}{m_n^4+4}=\frac{c\,k^2m_n^2}{4m_1^2m_2^2(m_n^4+4)}=\frac{c^2k^2m_n^2}{4m_1^2m_2^2w_n^2},$$ and the other two members follow from the two fixed edge values. Note that all three members move along the family: fixing two edge values fixes no member. The triangle depends on the point only through $f(m_n)=X_n/c$, and two points of $E_c$ share an $X$-coordinate exactly when they are negatives of each other; since $2E_c(\mathbb{Q})$ contains no torsion (the torsion of $E_c$ is killed by $2$), the coset is a torsor under a free abelian group of positive rank, and modulo $P\leftrightarrow-P$ it is infinite. When the rank is one the coset is $\{P_3+2nG\}$ and $-(P_3+2nG)=P_3+2n'G$ for a unique $n'$, which is the identification in the statement. $\square$ So a hyperbolic triangle is never alone: it lies in three infinite families, one through each pair of its edges, and the three families are indexed by three congruent number curves. The same recursion, read on the fibre of a class, says that every hyperbola value class is realised by an infinite sequence of Pythagorean angles. # 4. The hyperbolic tetrahedron: what is local Let $\{a,b,c\}$ be a hyperbolic triangle of positive rationals and $X$ a hypothetical fourth vertex: the six numbers $aX\pm1$, $bX\pm1$, $cX\pm1$ are to be nonzero rational squares. A prime is *good* for the triangle if it is odd and divides none of the numerators and denominators of $a$, $b$, $c$, $a\pm b$, $a\pm c$, $b\pm c$. #### The real place. Each $yX-1$ must be positive, so $X>1/\min(a,b,c)$, and every real $X$ in that range makes all six numbers positive: there is no real obstruction. **Lemma 4.1**. *For every $m\in\mathbb{Q}^*$ the value $f(m)$ has even $2$-adic valuation at most $-2$; its odd part is $\equiv5\pmod8$ when the valuation is $-2$ and $\equiv1\pmod8$ otherwise.* *Proof.* By $f(m)=f(2/m)$ take $m=2^ku$, $k\ge0$, $u$ odd. For $k=0$, $f(m)=(m^4+4)/(4m^2)$ has valuation $-2$ and odd part $(m^4+4)/m^2\equiv5\pmod 8$. For $k\ge1$, $m^4+4=4(4^{2k-1}u^4+1)$ has valuation $2$ and $4m^2$ valuation $2+2k$, so the valuation is $-2k$ and the odd part $(4^{2k-1}u^4+1)/u^2$ is $\equiv5$ for $k=1$ and $\equiv1\pmod8$ for $k\ge2$. $\square$ **Theorem 4.2** (the $2$-adic shape). *Let $\{x_1,\dots,x_4\}$ be a hyperbolic tetrahedron, $v_i=v_2(x_i)$, $u_i$ the odd part of $x_i$. Then the $v_i$ have one parity, and either (A) $v_i+v_j\le-4$ for all $i\ne j$ and all $u_i$ are congruent modulo $8$, or (B) there is a centre $j$ with $v_j\ge0$, the other three have $v_i=-2-v_j$, and $u_i\equiv5u_j\pmod8$. In particular at most one member has nonnegative $2$-adic valuation.* *Proof.* Every product $x_ix_j$ is a value of $f$, so by Lemma 4.1 $v_i+v_j$ is even and at most $-2$; call the pair *tight* if it equals $-2$, in which case $u_i\equiv5u_j$, and otherwise $u_i\equiv u_j\pmod 8$. As $u\mapsto5u$ is a fixed-point-free involution of the odd residues modulo $8$, two tight pairs at a common vertex $j$ force the third pair to be loose, so tight pairs form no triangle; a tight path $i$-$j$-$k$ gives $v_i=v_k=-2-v_j$ and $2v_i\le-4$, so $v_j\ge0$, and the fourth vertex $l$ must be tight with $j$ as well, since $u_l\equiv u_j$ and $u_l\equiv u_i\equiv5u_j$ cannot both hold: case (B). Two disjoint tight pairs give $-4=(v_i+v_j)+(v_k+v_l)\le-8$ by adding the two loose cross sums, impossible; a single tight pair $\{i,j\}$ gives $u_k\equiv u_i$ and $u_k\equiv u_j\equiv 5u_i$ for a third vertex, impossible. What remains is case (A). $\square$ This is a constraint on the valuations and not an obstruction: the three remaining square conditions at the prime two can be satisfied by members of sufficiently negative valuation whenever the needed ratios are $2$-adic squares. **Proposition 4.3** (the local shape at an odd prime). *Let $p$ be odd and $X\in\mathbb{Q}_p^*$ with $aX\pm1$, $bX\pm1$, $cX\pm1$ squares in $\mathbb{Q}_p$. For each member $y$: (i) if $v_p(yX)>0$ then $p\equiv1\pmod4$; (ii) if $v_p(yX)<0$ then $yX$ is a square in $\mathbb{Q}_p$, and if two members have negative valuation their product, a value $f(m)$, is a square in $\mathbb{Q}_p$, hence so is $m^4+4$; (iii) if $v_p(yX)=0$ with residue $\rho\not\equiv\pm1$, then $yX\pm1$ are squares in $\mathbb{Q}_p$ iff $\rho\pm1$ are squares modulo $p$. In particular, for $p\equiv3\pmod4$ every member has $v_p(yX)\le0$.* *Proof.* (i) $(yX+1)(yX-1)=y^2X^2-1\equiv-1$ is a unit square. (ii) $yX\pm1=yX(1\pm1/(yX))$ with the second factor a unit $\equiv1$. (iii) is Hensel’s lemma. $\square$ **Proposition 4.4** (the root criterion). *Let $p$ be any prime and $X_0=\mp1/y$ one of the six roots, so that $yX_0\pm1=0$. There are $X\in\mathbb{Q}_p$ arbitrarily close to $X_0$ satisfying all six conditions if and only if the five numbers $y'X_0\pm1$, $y'\in\{a,b,c\}$, other than the vanishing one are nonzero squares in $\mathbb{Q}_p$.* *Proof.* The six roots $\pm1/a,\pm1/b,\pm1/c$ are distinct, so at $X_0$ the five other numbers are nonzero. If they are squares, take $X=X_0\pm p^{2k}t^2/y$ with $t$ a unit: then $yX\pm1=p^{2k}t^2$ is a square, and for $k$ large the five others move by $p^{2k}(y'/y)t^2$, which does not change a nonzero square, the squares being open in $\mathbb{Q}_p^*$. If one of them is not a square, it stays a nonsquare in a neighbourhood of $X_0$. $\square$ The set of $X\in\mathbb{Q}_p$ satisfying the six conditions is open, and its complement in the compact space $\mathbb P^1(\mathbb{Q}_p)$ minus small discs around the six roots and around the two tails of Proposition 4.3 is decided by finitely many residues; so local solubility at a given prime is decided by a finite computation whose depth depends on the coefficients and on the separation of the roots. We do not make that depth explicit here, and we claim nothing about it. **Proposition 4.5** (a large good prime never obstructs). *Let $D$ be the smooth projective curve with function field $\mathbb{Q}(X)(\sqrt{aX+1},\sqrt{aX-1},\dots,\sqrt{cX-1})$. Then $D$ has genus $49$, and for every good prime $p$ with $p+1-98\sqrt p>224$ the six conditions have a solution in $\mathbb{Q}_p$. Hence a hyperbolic triangle can be locally obstructed only at the bad primes and at finitely many small good primes.* *Proof.* $D$ is a $(\mathbb{Z}/2\mathbb{Z})^6$-cover of the line of degree $64$, branched at the six roots and at $\infty$, each with inertia of order two, so with $32$ points above each; Riemann–Hurwitz gives $2g-2=64\cdot(-2)+7\cdot32$, $g=49$. At a good prime the seven branch points stay distinct, $D$ has good reduction, and $\#D(\mathbb{F}_p)\ge p+1-98\sqrt p$ by Hasse–Weil. At most $7\cdot32$ of these points lie over branch points; any other is a residue $X_0$ with all six values nonzero squares modulo $p$, which Hensel lifts. $\square$ What the local theory gives, then, is a finite test for each triangle, and the knowledge that it has to be run only at finitely many primes. What it cannot give is a proof for a triangle that passes everywhere locally; for such a triangle the question is a question about the rational points of $D$, a curve of genus $49$, or of its genus-five quotient $Y_i^2=a_i^2X^2-1$ together with three square conditions. The Jacobian of that quotient is isogenous to five elliptic curves with full rational $2$-torsion, and we make no claim about their ranks. # 5. The floors, briefly The mixed octuples of \[2, 3\] and of kiro_dioph7 \[4\] read in this language as follows. For a rational right triangle $U^2+L^2=V^2$ and $e\in\{1,-1\}$ with $W^2=V^2+eU(U+V)$ a square, the curve $E_{U,V}\colon Y^2=X(X+U^4)(X+V^4)$ carries a point $S$ of order four and a rational point $P$, members are read off points by $U^2V^2N\,x(Q)=U^3V^3-W^2X(Q)$ with $N=V^2+eU^2$, and for every admissible odd $k$ the floor $\{0,a,b,c,x(kP+jS)\}_{j=0..3}$ is a complete $\mathrm P$-graph on eight vertices. Its $\mathrm P_2$ edges are exactly the six pairs among $a$, $b$, $x(kP+S)$, $x(kP+3S)$, which are the four largest members (\[3\], Propositions 2.2 and 8.1). On absolute values a floor could in principle carry further far edges of type $\mathrm M'$, hyperbola edges with negative product; we have no proof that it does not, and we state the far-edge structure of the floors only for $\mathrm P_2$. # 6. Two conjectures **Conjecture 6.1** (no hyperbolic tetrahedron). *No set of four nonzero rationals is at once a $D(1)$- and a $D(-1)$-quadruple.* By Theorem 3.6 the hypothesis of this conjecture is never vacuous: hyperbolic triangles come in infinite families, each indexed by a congruent number curve, and a tetrahedron would be a point where four such families meet. By Proposition 4.5 a given triangle has only finitely many local tests to pass; a triangle that passes them all cannot be excluded by local means, and for it the conjecture is a statement about the Hasse principle on a specific curve of genus forty-nine. We do not know a global argument. **Conjecture 6.2** (Strong Conjecture, sign-free form). *Let $p_1<\dots