# Rational septuples of the mixed system $1-xy=\pm\square$: infinitely many from Pythagorean–Eisenstein pairs, with exactly six forced minus signs

**Luka Lasić** (Zagreb, Croatia) and **Claude ("Oblak")** (AI, Anthropic, working under the direction of the first author; in the writing the two authors are equal partners, a symbiosis).

# 1. The statement

A rational Diophantine $n$-tuple is a set of $n$ nonzero rationals with $1+xy\in\mathbb{Q}^2$ for every two members; sextuples exist in infinite families [1], no septuple is known, and the question whether one exists is the problem this paper is about. We approach it by releasing the sign. Call a set of $n$ nonzero rationals a *septuple (or $n$-tuple) of the mixed system* if $$1-xy\in\pm\mathbb{Q}^2\qquad\text{for every two members }x\ne y .\qquad(1)$$ Replacing each member $x$ by $-x$ does not change the system, and a set in which every sign is $+$ in (1) is, after $x\mapsto-x$, a rational Diophantine tuple. So the original problem is the all-plus case of (1), and the natural questions about a septuple of the mixed system are: how many minus signs does it have, where are they, and are they forced? For $x=ir$, $y=ir'$ one has $1+xy=1-rr'$, and $1-rr'\in\pm\mathbb{Q}^2$ iff $1+xy\in\mathbb{Q}(i)^2$; so septuples of (1) are the same as Diophantine septuples over $\mathbb{Q}(i)$ with members in $i\mathbb{Q}$, and rStation Paper 21 [6] found such septuples, the smallest being $$i\cdot\Bigl\{\tfrac9{10},\ \tfrac{25}{24},\ \tfrac{16}{15},\ \tfrac{480704}{547215},\ \tfrac{1374025}{1236696},\ \tfrac{450450}{516961},\ \tfrac{3296631}{3433960}\Bigr\},\qquad(2)$$ by a mechanism on the curve of the base triple: a point $S$ of order four whose double $T_1$ is a $2$-torsion point that is a “good difference”. This is the analogue, with a point of order four, of the construction of Dujella, Kazalicki, Mikić and Szikszai [1] with a point of order three, which we described as layers in [4]. Here we start from the other end, with two classical triangles, build infinitely many rational septuples of (1) explicitly, read their signs off the curve, and arrive at (2). The answer to the three questions, for this family, is Theorem 1.1(e): six minus signs, on a fixed four-element subset, each forced.

Throughout, $a,c\in\mathbb{Q}$ with $ac\ne0$, $a\ne\pm c$, and $$a^2+ac+c^2=k^2\qquad\text{and}\qquad c^2-a^2=\varepsilon b^2,\qquad(3)$$ the first being the Eisenstein condition (a $120^\circ$ triangle $(a,c,k)$, or a $60^\circ$ triangle $(|a|,c,k)$ if $ac<0$), the second the Pythagorean one, with $k,b\in\mathbb{Q}$, $\varepsilon=\pm1$. Put $N=a^2+c^2$ and $$\rho_1=\frac{a^2+c^2}{ac},\qquad\rho_2=\frac{c^2}{a(a+c)},\qquad\rho_3=\frac{a^2}{c(a+c)},$$ $$E_{a,c}\colon\ Y^2=X(X+a^4)(X+c^4),\qquad e_1=0,\ e_2=-a^4,\ e_3=-c^4,$$ $$T_1=(0,0),\quad T_2=(-a^4,0),\quad T_3=(-c^4,0),$$ $$S=\bigl(-a^2c^2,\ a^2c^2(c^2-a^2)\bigr),\qquad R=\bigl(ack^2,\ ack(a+c)N\bigr),\qquad Q_0=R+T_1=\Bigl(\frac{a^3c^3}{k^2},\ \cdot\Bigr),$$ $$s=X(Q_0)=\frac{a^3c^3}{k^2},\qquad\lambda=\frac{a^2c^2N}{k^2},\qquad r(P)=\frac{s-X(P)}{\lambda}\quad(P\in E_{a,c}(\mathbb{Q}),\ P\ne O).$$

**Theorem 1.1**. Assume (3).

**(a)** $\{\rho_1,\rho_2,\rho_3\}$ solves (1): $1-\rho_2\rho_3=\bigl(\tfrac{k}{a+c}\bigr)^2$, $1-\rho_1\rho_2=-\varepsilon\bigl(\tfrac{bk}{a(a+c)}\bigr)^2$, $1-\rho_1\rho_3=\varepsilon\bigl(\tfrac{bk}{c(a+c)}\bigr)^2$. Moreover $\rho_\ell=\lambda/(s-e_\ell)$, and $E_{a,c}$ is the curve induced by the triple $i\cdot\{\rho_1,\rho_2,\rho_3\}$ in the coordinate $r$: $\prod_\ell(1-\rho_\ell r)=\bigl(Y/Y(Q_0)\bigr)^2$.

**(b)** $S$ has order four with $2S=T_1$; $E_{a,c}(\mathbb{Q})_{\mathrm{tors}}=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$; $R$ has infinite order.

**(c)** For every $P\in E_{a,c}(\mathbb{Q})$ with $X(P)\notin\{0,-a^2c^2\}$: $$\begin{aligned}
1-r(P)\,r(P\pm S)&=\Bigl(\frac{k\bigl((a+c)Y\pm(c-a)(k^2X-a^3c^3)\bigr)}{acN\,(X+a^2c^2)}\Bigr)^2,\tag{A}\\
1-r(P)\,r(P+T_1)&=ac\,X\cdot\Bigl(\frac{k\,(X-a^2c^2)}{acN\,X}\Bigr)^2,\tag{B}\\
1-\rho_\ell\,r(P)&=\frac{X-e_\ell}{s-e_\ell}\qquad(\ell=1,2,3).\tag{C}
\end{aligned}$$ Identities (A) and (B) use only the Eisenstein condition; (C) is unconditional.

**(d)** For every integer $m$ outside a finite set, the eight rationals $$0,\quad\rho_1,\ \rho_2,\ \rho_3,\quad r(Q_0+2mR+jS)\ \ (j=0,1,2,3)$$ are distinct and all $28$ values $1-xy$ lie in $\pm\mathbb{Q}^2$; so $i\cdot\bigl(\{\rho_1,\rho_2,\rho_3\}\cup\{r(Q_0+2mR+jS)\}\bigr)$ is a Diophantine septuple over $\mathbb{Q}(i)$, and each pair $(a,c)$ carries infinitely many of them (different $m$ give different sets, since $r$ determines the point up to sign). The sign pattern is: $+$ on the pairs $(P,P\pm S)$; on $(P,P+2S)$ it is $+$ for $P\in Q_0+2E(\mathbb{Q})$ and $-$ for $P\in Q_0+S+2E(\mathbb{Q})$; on the base pairs $(\rho_2\rho_3,\rho_1\rho_2,\rho_1\rho_3)$ it is $(+,-\varepsilon,\varepsilon)$ as in (a); and a layer point $r(Q_0+2mR+jS)$ is joined to $(\rho_1,\rho_2,\rho_3)$ with signs $(+,+,+)$ for $j$ even and $(-,-\varepsilon,\varepsilon)$ for $j$ odd.

**(e)** Consequently every septuple of (d) has exactly fifteen plus signs and six minus signs, the minus signs forming the complete graph $K_4$ on the four members $\rho_{\ell}$, $\rho_{\ell'}$, $r(Q_0+2mR+S)$, $r(Q_0+2mR+3S)$, where $\{\ell,\ell'\}=\{1,2\}$ if $\varepsilon=1$ and $\{1,3\}$ if $\varepsilon=-1$; these four form a rational $D(-1)$-quadruple and the other three a rational $D(1)$-triple, and all twelve cross pairs are $D(1)$. The six minus signs cannot be removed within the construction: the sign of the join $(P,P+2S)$ is the square class of $ac\,X(P)$, which takes both values on the two cosets $Q_0+2E(\mathbb{Q})$ and $Q_0+S+2E(\mathbb{Q})$ that a layer necessarily meets, and the base triple itself has a minus sign for every Pythagorean–Eisenstein pair. Replacing members by their negatives does not help, since (1) is invariant under $x\mapsto-x$.

**Theorem 1.2**. Up to scaling, the pairs $(a,c)$ satisfying (3) are $(a,c)=(m^2-n^2,\,m^2+n^2)$ with $k^2=3m^4+n^4$ (and the same with $a,c$ exchanged). The quartic $k^2=3m^4+n^4$ has the rational points $(m,n)=(0,1),(1,1),(2,1),(3,11),(28,47),(475,13)$ (the first two are degenerate, $a=-c$ and $a=0$), its Jacobian is $Y^2=X^3-12X$, and it has infinitely many rational points; so there are infinitely many Pythagorean–Eisenstein pairs, the first being $$\begin{gathered}
(a,c;\,b,k)=(3,5;\,4,7),\qquad(-56,65;\,33,61),\qquad(-1425,2993;\,2632,2593),\\(112728,112897;\,6175,195397).
\end{gathered}$$ In the range $0<a<c<3000$, $\gcd(a,c)=1$, the pair $(3,5)$ is the only one; the second and third have $ac<0$, i.e. a $60^\circ$ triangle $(|a|,c,k)$, the fourth, from $(m,n)=(475,13)$, is again a $120^\circ$ triangle.

**Theorem 1.3**. For $(a,c)=(3,5)$: $E_{3,5}\colon Y^2=X(X+81)(X+625)$, $S=(-225,3600)$, $R=(735,28560)$, $Q_0=(\tfrac{3375}{49},\cdot)$, $\lambda=\tfrac{7650}{49}$, the base is $\{\tfrac{34}{15},\tfrac{25}{24},\tfrac9{40}\}$, and the first layer ($m=1$) is $$\Bigl\{\tfrac{34}{15},\ \tfrac{25}{24},\ \tfrac9{40},\ \tfrac{2143479}{1115560},\ \tfrac{2059200}{4669921},\ \tfrac{2113225}{1145814},\ -\tfrac{137344}{15}\Bigr\}.\qquad(4)$$ The member $\tfrac{25}{24}$ is strong, $1-(\tfrac{25}{24})^2=-(\tfrac7{24})^2$, and the reflection $\varphi(r)=(\tfrac{25}{24}-r)/(1-\tfrac{25}{24}r)$ preserves (1); it exchanges $\tfrac{25}{24}$ with the hidden member $0$ and carries the eight-element set $\{0\}\cup(4)$ onto $\{0\}\cup\{\tfrac9{10},\tfrac{25}{24},\tfrac{16}{15},\tfrac{480704}{547215},\tfrac{1374025}{1236696},\tfrac{450450}{516961},\tfrac{3296631}{3433960}\}$: the septuple (2) of [6]. Consequently (2) divided by $i$ is a rational septuple of (1), with fifteen plus signs and six minus signs, the minus signs forming the complete graph on $\{\tfrac{25}{24},\tfrac{16}{15},\tfrac{1374025}{1236696},\tfrac{3296631}{3433960}\}$, a rational $D(-1)$-quadruple. With $P_0=(15^3,60^3)$ one has $R=-P_0+S+T_2$, and the septuple (2) itself is $\{r(3P_0+jS)\}$ for $r=(15^3-X)/60^2$, the construction of [6] with $T=3P_0$.

The next sections prove these statements. The algebra is short; the only computer-assisted step is the expansion behind (A).

# 2. The triple and the curve

*Proof of Theorem 1.1(a).* Directly: $1-\rho_2\rho_3=1-\dfrac{ac}{(a+c)^2}=\dfrac{a^2+ac+c^2}{(a+c)^2}$; $1-\rho_1\rho_2=\dfrac{a^2(a+c)-c(a^2+c^2)}{a^2(a+c)}=\dfrac{a^3-c^3}{a^2(a+c)}=-\dfrac{(c-a)(a^2+ac+c^2)}{a^2(a+c)}$, and $(c-a)(a^2+ac+c^2)=\dfrac{(c^2-a^2)k^2}{a+c}$ gives $-\varepsilon\bigl(\tfrac{bk}{a(a+c)}\bigr)^2$; the third is the same with $a\leftrightarrow c$, which changes the sign of $c-a$. For the curve: $s-e_1=\dfrac{a^3c^3}{k^2}$, and since $ck^2+a^3=ak^2+c^3=(a+c)N$ when $k^2=a^2+ac+c^2$, $$s-e_2=\frac{a^3c^3+a^4k^2}{k^2}=\frac{a^3(a+c)N}{k^2},\qquad s-e_3=\frac{c^3(a+c)N}{k^2},$$ so $\lambda/(s-e_1)=N/(ac)=\rho_1$, $\lambda/(s-e_2)=c^2/(a(a+c))=\rho_2$, $\lambda/(s-e_3)=\rho_3$. Then $1-\rho_\ell r=\dfrac{(s-e_\ell)-(s-X)}{s-e_\ell}=\dfrac{X-e_\ell}{s-e_\ell}$, which is (C), and taking the product, $\prod_\ell(1-\rho_\ell r)=Y^2/\prod_\ell(s-e_\ell)=Y^2/Y(Q_0)^2$. $\square$

*Proof of Theorem 1.1(b).* On $Y^2=X(X+\alpha^2)(X+\beta^2)$ the point $S_0=(-\alpha\beta,\alpha\beta(\beta-\alpha))$ has order four with $2S_0=(0,0)$: the tangent slope at $S_0$ is $\bigl(3X^2+2(\alpha^2+\beta^2)X+\alpha^2\beta^2\bigr)/(2Y)=-2\alpha\beta(\alpha-\beta)^2/\bigl(2\alpha\beta(\beta-\alpha)\bigr)=\alpha-\beta$, so $X(2S_0)=(\alpha-\beta)^2-(\alpha^2+\beta^2)+2\alpha\beta=0$. Here $\alpha=a^2$, $\beta=c^2$. Torsion: $E_{a,c}(\mathbb{Q})_{\mathrm{tors}}\supseteq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$, and by Mazur’s theorem [5] it is this or $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$; the latter would need a rational half of a point of order four, i.e. all three coordinates $X-e_\ell$ squares at $X=\pm a^2c^2$: at $-a^2c^2$ the first is $-a^2c^2$, not a square; at $a^2c^2$ they are $a^2c^2,\ a^2N,\ c^2N$, squares iff $N=a^2+c^2$ is a square. But $c^2-a^2=\varepsilon b^2$ and $a^2+c^2=\square$ together give $c^4-a^4=\pm\square$, which by Fermat ($x^4-y^4=z^2$ has only trivial solutions) forces $ac=0$ or $a=\pm c$. So the torsion is exactly $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ under (3).

$R$ lies on the curve because $f(ack^2)=ack^2\cdot a(ck^2+a^3)\cdot c(ak^2+c^3)=a^2c^2k^2(a+c)^2N^2$. It is not torsion because its abscissa is none of $0,-a^4,-c^4,\pm a^2c^2$: $ack^2=a^2c^2$ would give $k^2=ac$, i.e. $a^2+c^2=0$; $ack^2=-a^2c^2$ gives $(a+c)^2=0$; $ack^2=-a^4$ gives $(a+c)N=0$; likewise for $-c^4$. $\square$

*Proof of Theorem 1.1(c).* (C) was shown above. (B): translation by $T_1=(0,0)$ on $Y^2=X(X+a^4)(X+c^4)$ is $X\mapsto a^4c^4/X$, so with $\tilde s=k^2s=a^3c^3$, $\tilde\lambda=k^2\lambda=a^2c^2N$, $$\begin{aligned}
1-r(P)r(P+T_1)&=\frac{\tilde\lambda^2X-(\tilde s-k^2X)(\tilde sX-k^2a^4c^4)}{\tilde\lambda^2X}\\&=\frac{k^2\tilde s\,X^2+(\tilde\lambda^2-\tilde s^2-k^4a^4c^4)X+k^2a^4c^4\tilde s}{\tilde\lambda^2X},
\end{aligned}$$ and $\tilde\lambda^2-\tilde s^2-k^4a^4c^4=a^4c^4\bigl(N^2-a^2c^2-k^4\bigr)=-2a^5c^5k^2$ because $k^2=N+ac$ gives $k^4=N^2+2acN+a^2c^2$, so $N^2-a^2c^2-k^4=-2ac(N+ac)=-2ack^2$. Hence the numerator is $a^3c^3k^2(X^2-2a^2c^2X+a^4c^4)=a^3c^3k^2(X-a^2c^2)^2$, and dividing by $\tilde\lambda^2X=a^4c^4N^2X$ gives (B).

(A): the chord through $P=(X,Y)$ and $S$ has slope $(Y-Y_S)/(X+a^2c^2)$, and $$X(P+S)=\frac{(Y-Y_S)^2-\bigl(a^4+c^4+X-a^2c^2\bigr)(X+a^2c^2)^2}{(X+a^2c^2)^2},\qquad Y_S=a^2c^2(c^2-a^2).$$ Substituting into $1-r(P)r(P+S)$, clearing $\tilde\lambda^2(X+a^2c^2)^2$ and reducing $Y^2$ by the curve, the numerator becomes $A(X)+B(X)Y$ with $\deg_XA=3$, $\deg_XB=1$; it equals $(uY+v)^2$ with $u=ac(a+c)k$ and $v=(c-a)ack(k^2X-a^3c^3)$, i.e. $A=u^2f+v^2$ and $B=2uv$. This is a polynomial identity in $a,c,k$ subject to $k^2=a^2+ac+c^2$; we verified it by exact expansion in $\mathbb{Q}[m,n][X,Y]/(Y^2-f)$ after the substitution $a=m^2-n^2$, $c=2mn+n^2$, $k=m^2+mn+n^2$, which parametrises the conic $a^2+ac+c^2=k^2$ birationally, so that an identity in $\mathbb{Q}[m,n]$ is an identity on the conic. Replacing $S$ by $-S$ changes $Y_S$ to $-Y_S$ and gives $(uY-v)^2$. Dividing by $\tilde\lambda^2(X+a^2c^2)^2=a^4c^4N^2(X+a^2c^2)^2$ gives (A). $\square$

*Proof of Theorem 1.1(d).* Fix $m$ and let $L=\{P_j=Q_0+2mR+jS\colon j=0,\dots,3\}$. Pairs inside $L$: $P_j,P_{j'}$ differ by $\pm S$ (four pairs) or by $2S=T_1$ (two pairs), so (A) and (B) apply; in (B) the class of $ac\,X(P)$ is that of $ac\,X(Q_0)=a^4c^4/k^2$, a square, for $P\in Q_0+2E(\mathbb{Q})$ ($j$ even), and that times the class of $X(S)=-a^2c^2$, i.e. $-1$, for $j$ odd. Pairs base–layer: by (C) we need $(X(P)-e_\ell)/(s-e_\ell)\in\pm\mathbb{Q}^2$. The $2$-descent map $P\mapsto\bigl(X-e_1,X-e_2,X-e_3\bigr)\in(\mathbb{Q}^\times/\mathbb{Q}^{\times2})^3$ is a homomorphism with kernel $2E(\mathbb{Q})$ [7, X.1] (this is Dujella’s description of the extenders of a triple as the coset $P_0+2E$ [2]); for $j$ even the image of $P_j$ equals that of $Q_0$, giving three squares; for $j$ odd it is multiplied by the image of $S$, which is $\bigl(-a^2c^2,\ a^2(a^2-c^2),\ c^2(c^2-a^2)\bigr)=(-1,-\varepsilon,\varepsilon)$ by the Pythagorean condition — this is the only place where it enters beyond (a). Pairs inside the base are (a), and $0$ is joined to everything. Distinctness: $r(P)=r(P')$ iff $P'=\pm P$; within $L$ this needs $4mR=-2Q_0-(j+j')S$, true for at most one $m$ as $R$ is not torsion; $r(P)=\rho_\ell$ iff $X(P)=s-\lambda\rho_\ell$, at most two points of $E(\mathbb{Q})$ for each $\ell$; $r(P)=0$ iff $P=\pm Q_0$, which excludes at most finitely many $(m,j)$ (and always $m=0$). So all but finitely many $m$ give eight distinct values, and the seven nonzero ones form a rational septuple of (1) (times $i$, a Diophantine septuple over $\mathbb{Q}(i)$).

\(e\) Count the signs listed in (d). Inside the layer: four $+$ (differences $\pm S$), one $+$ ($P_0,P_2$) and one $-$ ($P_1,P_3$). Base–layer: $P_0,P_2$ give $+++$, $P_1,P_3$ give $(-,-\varepsilon,\varepsilon)$, i.e. two minus signs each. Base: one minus sign, on $\rho_1\rho_2$ if $\varepsilon=1$ and on $\rho_1\rho_3$ if $\varepsilon=-1$. Total: $1+2+2+1=6$ minus signs, on the pairs among $\{\rho_1,\rho_{2\text{ or }3},r(P_1),r(P_3)\}$: exactly the six pairs of that four-element set. (For $\varepsilon=1$: $\rho_1\rho_2$; $\rho_1r(P_1),\rho_2r(P_1)$; $\rho_1r(P_3),\rho_2r(P_3)$; $r(P_1)r(P_3)$.) The remaining $15$ pairs are $+$. That the signs are forced is the content of (B) and of the descent images computed above: on a layer the class of $ac\,X$ is $+1$ for $j$ even and $-1$ for $j$ odd, and the base sign pattern $(+,-\varepsilon,\varepsilon)$ holds for every pair satisfying (3) by (a). $\square$

# 3. The pairs

*Proof of Theorem 1.2.* The conditions (3) are homogeneous in $(a,c)$, and so are $\rho_1,\rho_2,\rho_3$ (of degree $0$) and the whole construction up to the scaling $X\mapsto t^4X$, $Y\mapsto t^6Y$ of $E_{a,c}$. Every rational point of the conic $c^2-a^2=\varepsilon b^2$ is, up to scaling, $(a,c,b)=(m^2-n^2,m^2+n^2,2mn)$ for $\varepsilon=1$ (and $(m^2+n^2,m^2-n^2,2mn)$ for $\varepsilon=-1$, the same pairs with $a,c$ exchanged, which permutes $\rho_2,\rho_3$). Then $a^2+ac+c^2=(m^2-n^2)^2+(m^4-n^4)+(m^2+n^2)^2=3m^4+n^4$. The quartic $C\colon k^2=3m^4+n^4$ has the point $(m,n,k)=(0,1,1)$, so it is an elliptic curve; by the classical reduction of $y^2=\alpha x^4+\beta$ to $Y^2=X^3-4\alpha\beta X$ its Jacobian is $Y^2=X^3-12X$. The latter has torsion group $\mathbb{Z}/2\mathbb{Z}$: its order divides $\gcd\bigl(\#E(\mathbb{F}_5),\#E(\mathbb{F}_7)\bigr)=\gcd(10,8)=2$, both primes being of good reduction. So $C(\mathbb{Q})$, a torsor under it, is finite only if it has at most two points; but $(m{:}n{:}k)=(0{:}1{:}\pm1),(1{:}1{:}\pm2),(2{:}1{:}\pm7),(3{:}11{:}\pm122),(28{:}47{:}\pm2593)$ are ten. Hence $C(\mathbb{Q})$ is infinite. The points give $(a,c)=(m^2-n^2,m^2+n^2)$: $(2,1)\mapsto(3,5)$, $k=7$; $(3,11)\mapsto(-112,130)\sim(-56,65)$, $k=122/2=61$, $b=66/2=33$; $(28,47)\mapsto(-1425,2993)$, $b=2632$, $k=2593$; $(475,13)\mapsto(112728,112897)$, $b=6175$, $k=195397$. The statement about the range $0<a<c<3000$ is a direct search. $\square$

# 4. The pair $(3,5)$ and Paper 21

*Proof of Theorem 1.3.* With $a=3$, $c=5$, $k=7$, $N=34$: $S=(-225,3600)$, $R=(735,\,15\cdot7\cdot8\cdot34)=(735,28560)$, $Q_0=R+T_1=(50625/735,\cdot)=(3375/49,\cdot)$, $\lambda=225\cdot34/49=7650/49$, $\rho=(\tfrac{34}{15},\tfrac{25}{24},\tfrac9{40})$. The layer $m=1$, $Q_0+2R+jS$, has abscissas $$-\tfrac{25780815}{111556},\qquad\tfrac{165375}{4669921},\qquad-\tfrac{41833500}{190969},\qquad\tfrac{70048815}{49}$$ and $r$-values as in (4); the $21$ conditions can be checked directly. The reflection $\varphi_g(x)=(g-x)/(1+gx)$ in a member $g$ with $1+g^2=h^2$ satisfies $1+\varphi_g(x)\varphi_g(y)=h^2(1+xy)/\bigl((1+gx)(1+gy)\bigr)$ [3, Cor. 5.2]; on $x=ir$ with $g=\tfrac{25}{24}i$ this is $r\mapsto(\tfrac{25}{24}-r)/(1-\tfrac{25}{24}r)$ and preserves (1). It sends $\tfrac{34}{15}\mapsto\tfrac9{10}$, $\tfrac9{40}\mapsto\tfrac{16}{15}$, $\tfrac{25}{24}\mapsto0$, $0\mapsto\tfrac{25}{24}$, and the four layer members to $\tfrac{480704}{547215},\tfrac{1374025}{1236696},\tfrac{450450}{516961},\tfrac{3296631}{3433960}$; this is (2). The sign count follows by evaluating the $21$ values. Finally $R=(735,28560)=-P_0+S+T_2$ by direct addition on $E_{3,5}$, and $\{r(3P_0+jS)\}$ with $r=(3375-X)/3600$ reproduces the four large members of (2). $\square$

**Remark 4.1**. The base of [6] is the analogous construction on $E_{3,5}$ with the point $P_0=((ac)^3,\cdot)$ in place of $Q_0$ and $\lambda=s+a^2c^2$: the triple $\bigl\{\tfrac{ac+1}{ac},\tfrac{c^2(ac+1)}{a(c^3+a)},\tfrac{a^2(ac+1)}{c(a^3+c)}\bigr\}=\{\tfrac{16}{15},\tfrac{25}{24},\tfrac9{10}\}$, which solves (1) because $(a^3+c)(c^3+a)=(ac+1)^3$ and $ac+1=4^2$ for $(3,5)$. The reflection in the strong member identifies the two pictures.

**Remark 4.2**. Theorem 1.1 produces septuples but no octuple: within a layer the four points are pairwise joined, but points of different layers are joined only sporadically if at all. For $E_{3,5}$ no two of the first seven layers share a join. For fixed $D\in2E(\mathbb{Q})+\langle S\rangle\setminus\langle S\rangle$ the condition $1-r(P)r(P+D)=\pm z^2$ is a double cover of $E$ branched at the at most four odd-order zeros of $1-r(P)r(P+D)$ (its poles, at $O$ and $-D$, are double), hence of genus at most three; whether these curves have points beyond the obvious ones is the question of Diophantine octuples over $\mathbb{Q}(i)$ containing a base of this family. We conjecture they do not.

**Remark 4.3**. What this says about the original, all-plus problem. The construction produces rational septuples with every condition in $\pm\mathbb{Q}^2$ and identifies the six obstinate signs with two facts about the curve: the base triple of a Pythagorean–Eisenstein pair is a mixed triple (one minus sign, Theorem 1.1(a)), and the join across $2S=T_1$ changes sign between the two cosets that a layer must use (identity (B)). Over $\mathbb{Q}(i)$ both facts become invisible, which is why [6] found the septuples there; over $\mathbb{Q}$ they are exactly the gap. In the language of [6], Theorem 1 there says that without $\sqrt{-1}$ no translation family reaches seven; our Theorem 1.1(e) says, for this family, that the shortfall is six specific signs on a $K_4$, no more and no less. A rational Diophantine septuple would have to come from a mechanism that keeps the fifteen plus signs and turns these six, and the two facts above show that no choice of pair $(a,c)$ or of layer does so.

# References

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[2] A. Dujella, *Diophantine $m$-tuples and elliptic curves*, J. Théor. Nombres Bordeaux 13 (2001), 111–124.

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