# Local obstructions and rank conditions for canonical symmetric Diophantine sextuples The rational Diophantine septuple problem asks for seven distinct nonzero rational entries whose pairwise products plus one are squares. We study a restricted route toward that problem: take the two regular fifths of a quadruple $\{a,-a,b,-b\}$ and require them to be compatible. The resulting candidate has the form $\{\pm a,\pm b,\pm c\}$. We give a complete local criterion for its genus-one source, two uniform rank requirements, an elementary Hasse-principle failure, and an explicit family with independent rational directions satisfying the remaining tests at three local fields. The final two rational square conditions remain in every construction. The regular extension formula is classical: our expression for $c$ is the specialization of Dujella's rational extension theorem [1, Theorem 1]. Dujella constructed mixed-sign sextuples [2], and Dujella, Kazalicki, Mikić and Szikszai proved infinitude for each sign pattern [3, Theorem 1]. A prescribed sign pattern does not impose opposite pairs or the canonical relation used here. We do not claim novelty for regular extensions or mixed-sign sextuples. Within rStation, elliptic and dyadic obstructions to product-one and order-four completion were developed in [4]. That work already contains the elementary rank-zero obstruction associated with $y^2=x(x-1)(x-4)$; we give a self-contained version of the auxiliary arithmetic below. The order-four diagonal family in [5] has its own complete local criterion and positive-rank Kummer exclusions. Its criterion concerns primes congruent to five modulo eight dividing a difference of squares. The circle family here instead has the criterion in Theorem 1, involving primes congruent to three modulo four in $M^4+M^2N^2+N^4$, and the uniform rank increases of Theorem 2 and Proposition 3. The regular-twin genus-21 curve in [6] is everywhere locally soluble on its full arithmetic cover and has no rational point; an explicit Cassels pairing proves its global obstruction without a rank computation. Our Hasse example concerns the intermediate genus-one circle source, before the final two square tests. Its obstruction is proved directly by two integer factorizations and a contradiction modulo eight. The contribution is the arithmetic classification and rank requirements for this specified family, with this explicit certificate, rather than a new general descent method or the first Hasse failure in tuple constructions. The rational two-parameter sextuple source in [7] supplies a complementary construction with all fifteen roots; our independent-point family still retains two rational square conditions. None of the results here settles the septuple problem or excludes arbitrary symmetric sextuples. ## 1. The construction and the retained square classes A rational $D(1)$-tuple is a set of distinct nonzero rational numbers whose pairwise products plus one are rational squares. We study the canonical extension of a symmetric quadruple $\{a,-a,b,-b\}$. Write $$ u^2=1-a^2,\qquad v^2=1-b^2,\qquad p=ab,\qquad k^2=1-p^2. $$ Here both $1+p$ and $1-p$ must individually be squares. Its canonical extensions are $\pm c$, where $$ c=\frac{2uv}{1-p^2}. $$ Each of these extends all four original entries. For example the four roots against $c$ are $(u+av)/k,(u-av)/k,(v+bu)/k,(v-bu)/k$. The remaining opposite-pair condition is $1-c^2$ square. A change of fibration begins with $c=2z/(1+z^2)$, and keeps a square root that can be lost by passing directly to an elliptic quotient. Put $$ \delta=1+z^2,\qquad F=z^4+z^2+1, $$ and consider the genus-one curve $$ C_z:\quad s^2=\delta r^4-2(z^4+1)r^2+\delta^3. \tag{1} $$ For rational $z\ne0$ it is smooth: the discriminant of its quartic is $4096z^4\delta^4F^2$. Our main results give a complete local-solubility criterion and a uniform rank condition for a rational point on (1). They do not assert that a rational point alone is a sextuple. To make that distinction explicit, set $$ L=r^2+\delta,\quad M=r^2-\delta,\quad A_d=L^2-4r^2,\quad A_e=L^2-4z^2r^2. $$ The rational formulas $$ \begin{aligned} d&=\frac{z(Ls-2rzM)}{\delta A_d},& g&=\frac{Ms+2rz^3L}{\delta A_d},\\ e&=\frac{Ls+2rzM}{\delta A_e},& f&=\frac{2rL-zMs}{\delta A_e} \end{aligned} \tag{2} $$ satisfy $$ d^2+g^2=e^2+f^2=1,\qquad de=\frac z\delta. $$ The denominators have no rational affine exceptions, since $A_d=(r^2+z^2-1)^2+4z^2>0$ and $A_e=(r^2+1-z^2)^2+4z^2>0$. Set $$ a=g/e,\quad b=f/d,\quad c=2de,\quad p=gf/(de),\quad w=(1-z^2)/\delta. $$ Then $$ 1\pm ac=(d\pm g)^2,\quad 1\pm bc=(e\pm f)^2,\quad 1-c^2=w^2, $$ and $$ 1-a^2=(1-p^2)d^2,\qquad1-b^2=(1-p^2)e^2. $$ Thus precisely the two further tests $1+p$ and $1-p$ square give all fifteen square conditions for $\{\pm a,\pm b,\pm c\}$. Nonzero and distinct entries remain necessary. Conversely a canonical symmetric sextuple gives $d=u/k,e=v/k,g=ae,f=bd$ in this two-circle presentation. Here $k\ne0$: otherwise $|ab|=1$ and the opposite-pair conditions force $|a|=|b|=1$, contradicting distinctness. Put $$ D=d-zg+z^2f+z^3e. $$ The inverse on $D\ne0$ is $$ r=\frac{\delta(df+z^2ge)}D,\qquad s=(r^2+\delta)e-z(r^2-\delta)f. \tag{2a} $$ For a direct verification, these expressions satisfy $Md-zLg=-2rz^2$, $zMe+Lf=2r$, and $s=zLd+Mg$ in the two-circle coordinate ring. Together with $s=Le-zMf$, they recover (2) by solving two linear systems. They also give $s^2+4r^2=\delta A_e(e^2+f^2)=\delta A_e$, which is (1). The supplement checks these identities and the opposite composition. The fixed-product two-circle curve is geometrically integral of genus one: its function field over an algebraic closure is $\overline{\mathbb Q}(d)(\sqrt{1-d^2},\sqrt{d^2-(z/\delta)^2})$. The two branch pairs are disjoint because $0<|z/\delta|\le1/2$. The birational maps therefore extend to inverse isomorphisms of the smooth projective curves. There are no omitted finite real source points, since after substituting (2), $$ D(r,s)D(r,-s)= -\frac{4z^2\bigl[\delta^3(r^2-\delta)^2 +4(z^4+1)^2r^2\bigr]}{A_dA_e}<0. $$ If a rational target point has $D=0$, its inverse is at quartic infinity. Writing $\kappa^2=\delta$, those two images are $\pm(z,1,1,-z)/\kappa$ in coordinate order $(d,e,g,f)$, giving $(a,b)=(1,-1)$. The finite points with $r=0$ similarly give $(a,b)=(-1,1)$. These boundary points are rational exactly when $\delta$ is square, and all fail distinctness. Thus every nondegenerate rational canonical sextuple is covered by the displayed finite chart. ## 2. The Jacobian and the relevant descent class Let $$ J_z:\quad y^2=x\bigl(x^2-2(z^4+1)x+\delta^4\bigr), $$ and let its two-isogenous companion be $$ J'_z:\quad y^2=x(x-4z^2)(x+4F). \tag{3} $$ The map $$ C_z\longrightarrow J_z,\qquad (x,y)=(\delta r^2,\delta rs) \tag{4} $$ requires the abscissa square class $\delta$. Its degree is two. The Jacobian of $C_z$ is $J'_z$, even when $C_z$ has no rational origin. For completeness, write $A=-2(z^4+1)$ and work temporarily over $\mathbb Q(\kappa)$ with $\kappa^2=\delta$ and $d_0=\kappa^3$. An explicit birational chart is $$ \xi=A+\frac{2d_0(s+d_0)}{r^2},\qquad \eta=\frac{2d_0\xi}{r}, $$ with inverse $$ r=\frac{2d_0\xi}{\eta},\qquad s=\frac{(\xi-A)r^2}{2d_0}-d_0. $$ It satisfies $\eta^2=\xi(\xi^2-2A\xi+A^2-4\delta^4)$, which is (3). Changing $\kappa$ to $-\kappa$ changes this chart by translation by $(0,0)$: $\xi\mapsto(A^2-4\delta^4)/\xi$ and $\eta\mapsto-(A^2-4\delta^4)\eta/\xi^2$. Translation acts trivially on $\operatorname{Pic}^0$, giving the asserted descent of the Jacobian. ## 3. Complete local-solubility criterion **Theorem 1.** Write $z=M/N\ne0$ in lowest terms. The smooth projective curve $C_z$ is everywhere locally soluble if and only if $$ |v_2(z)|\ge2 $$ and every prime $q\equiv3\pmod4$ occurs to even exponent in $$ M^4+M^2N^2+N^4. \tag{5} $$ In particular, $C_z(\mathbb Q_3)$ is empty if $v_3(z)=0$. **Proof.** Substitute $r=U/(NV)$ and $s=T/(N^3V^2)$. A primitive integral local point satisfies $$ T^2=(M^2+N^2)U^4-2(M^4+N^4)U^2V^2+(M^2+N^2)^3V^4. \tag{6} $$ Choose primitive local homogeneous coordinates $(U,V)$; the case $V=0$ includes both points at infinity. When $M,N$ are odd, the primitive right sides modulo $16$ lie in $\{2,6,8,10,14\}$, excluding a square. When exactly one of $M,N$ has 2-adic valuation one, unequal $U,V$ parity gives $5\pmod8$, while both odd give $32\pmod{64}$. These exclude every primitive 2-adic point. If $v_2(z)\ge2$, then $\delta$ is a 2-adic square and the point $r=0,s=\delta^{3/2}$ exists. The isomorphism $(z,r,s)\mapsto(1/z,r/z,s/z^3)$ handles negative valuations. For odd primes use the norm identity $$ s^2-\delta(r^2+\delta)^2=-4Fr^2. \tag{7} $$ By inversion assume $z$ integral at the given odd prime $q$. At a prime dividing $F$, $\delta$ is a unit and $-\delta$ is a square in the residue field: at a root of $z^2+z+1$ it equals $(z+1)^2$, and at a root of $z^2-z+1$ it equals $(z-1)^2$. If $q\equiv1\pmod4$, $\delta$ is also a square and $r=0$ gives a local point. If $q\equiv3\pmod4$, $\delta$ is a nonsquare unit. There is no point at infinity, since it would require $\sqrt\delta$. A local point must have $r$ a unit, since otherwise the leading or constant term in (1) has nonsquare unit part. The unramified quadratic norm in (7) has even valuation, so $v_q(F)$ must be even. Conversely, write $F=q^{2k}f$ with $f$ a unit and $k\ge1$. Set $r^2+\delta=q^k u$, $s=q^k v$. Equation (7) becomes $$ v^2-\delta u^2+4q^kfu-4\delta f=0. $$ Its reduction $v^2-\delta u^2=4\delta f$ has a nonsingular solution, since the finite-field norm is onto and its right side is nonzero. Hensel's lemma lifts $u,v$, and then $r$ lifts because $-\delta$ is a square unit. This proves sufficiency at these primes. If $q$ divides $z$, then $\delta\equiv1\pmod q$ and $r=0$ lifts. If $q$ divides $\delta$, the radicand at $r=1$ is $-4=(2z)^2\pmod q$, a nonzero square, and again lifts. At all remaining odd primes the quartic has smooth genus-one reduction. The Hasse-Weil bound for the smooth projective genus-one reduction supplies a finite-field point, and smoothness lifts it. The real place is soluble because the radicand is $\delta(r^2-\delta)^2+4z^2r^2>0$. These cases exhaust all places and prove the criterion. For a 3-adic unit $z$, exactly one of $z^2\pm z+1$ has valuation one, so $v_3(F)=1$. $\square$ ## 4. A uniform rank increase for a nonsquare source **Theorem 2.** For rational $z\notin\{0,1,-1\}$, the rank of $J_z$ is at least one. If $C_z(\mathbb Q)$ is nonempty and $1+z^2$ is not square, the rank is at least two. In particular, on a rank-one fiber, $C_z(\mathbb Q)$ is nonempty if and only if $1+z^2$ is square. We give an elementary descent proof. For $E:y^2=x(x^2+Ax+B)$ let $\alpha$ be the standard two-isogeny descent map, with values $x$ modulo squares away from the identity and $(0,0)$, and values $1,B$ there respectively. We use the two-isogeny descent conventions and rank formula of [8, Section 3.6]. On a nonsingular integral model its image is supported on the signed squarefree divisors of $B$. Rational coefficients may first be made integral by scaling $x=u^2X,y=u^3Y$, which does not change the abscissa square classes. For the companion coefficients $-2A,A^2-4B$, the standard formula is $$ |\alpha(E(\mathbb Q))|\,|\alpha'(E'(\mathbb Q))| =2^{\operatorname{rank}E+2}. \tag{8} $$ We first prove three nonsquare facts. On $$ E_-:y^2=x(x^2-x+1),\qquad E_+:y^2=x(x^2+x+1), $$ negative $x$ is impossible over the reals. Since $B=1$, each descent image is exactly $\{1\}$. Their companions have $B'=-3$. For the companion of $E_-$, namely $y^2=x(x-1)(x+3)$, the points $(-3,0),(3,6)$ give independent classes $-3,3$. For the companion of $E_+$, namely $$ E_*:y^2=x(x-3)(x+1), $$ $(-1,0),(3,0)$ give independent classes $-1,3$. In both cases all four possible companion classes occur, and (8) proves rank zero. At the good primes $5,7,11$, the reduction counts of $E_-$ are $8,8,8$; those of $E_+$ and $E_*$ are $8,8,16$. Prime-to-characteristic reduction injectivity bounds torsion by eight. The quadratics $x^2\mp x+1$ have discriminant $-3$, so $E_-$ and $E_+$ each have just one nonzero rational two-torsion point. Any enlargement of the displayed subgroups below must therefore supply the stated halves. On $E_-$, $(1,\pm1)$ have order four. A half of either would have abscissa satisfying $$ (x^2-1)^2=4x(x^2-x+1), $$ or $(x+x^{-1})(x+x^{-1}-4)=0$, which has no rational root. Therefore $E_-(\mathbb Q)=\{O,(0,0),(1,\pm1)\}$. On $E_+$, a half of $(0,0)$ requires $x=\pm1$, giving cubic values $3,-1$. Thus $E_+(\mathbb Q)=\{O,(0,0)\}$. On $E_*$, none of its nonzero two-torsion points can be halved: the three shifted abscissas contain a negative value at $x=-1,0$, and the nonsquare $3$ at $x=3$. The full-two-torsion halving criterion [9, Theorem 2.4] gives $E_*(\mathbb Q)=\{O,(0,0),(3,0),(-1,0)\}$. It follows that $z(z^2-z+1)$ is nonsquare for positive $z\ne1$, and $z(z^2+z+1)$ is nonsquare for every positive $z$. Also $F=z^4+z^2+1$ is nonsquare for every nonzero rational $z$. Indeed, a rational square $h^2=F$ maps to $E_*$ by $$ U=2(h+z^2)+1,\qquad V=2zU. $$ The value $U=0$ is impossible at a finite point, since substituting $h=-z^2-1/2$ gives $h^2-F=-3/4$. The other two finite torsion images have $V=0$ and $U\ne0$, forcing $z=0$. Now replace $z$ by $|z|$, which changes none of our curves. Put $H_-=z^2-z+1$, $H_+=z^2+z+1$. The companion $J'_z$ contains $$ (-4F,0),\quad(-4zH_-,8z\delta H_-),\quad (4zH_-,-8zH_-(1-z^2)). $$ Their descent classes are $-F,-zH_-,zH_-$. Among their seven nonempty products, the positive classes reduce to $zH_-,zH_+,F$, each just proved nonsquare. The negative products cannot be squares. Hence these three classes are independent and $|\alpha'(J'_z(\mathbb Q))|\ge8$. Formula (8) proves rank at least one. A rational point on $C_z$ with $r\ne0$ gives class $\delta$ on $J_z$ by (4). If $\delta$ is nonsquare, the original descent image has at least two elements; (8) then gives $2^{\operatorname{rank}J_z+2}\ge16$. The points at $r=0$ and at infinity require $\sqrt\delta$, so none escapes this argument. Conversely square $\delta$ gives a point at $r=0$. This proves the theorem. $\square$ ## 5. An elementary Hasse-principle failure Take $z=8/3$. Its local conditions hold because $|v_2(z)|=3$ and $$ 8^4+8^2\cdot3^2+3^4=7^2\cdot97. $$ We prove directly that $C_{8/3}(\mathbb Q)$ is empty, without computing an elliptic rank. Its integer quartic is $$ v^2=73u^4-8354u^2+73^3. \tag{9} $$ At $u=0$ or at infinity, a rational point would require $\sqrt{73}$. For $u\ne0$, put $T=v/u$, $Z=u-73/u$, $Q=u+73/u$, and $t=1$. The projective equations become $$ T^2-73Z^2=2304t^2,\qquad Q^2-Z^2=292t^2. \tag{10} $$ Conversely, a rational point on (10) has $t\ne0$ and gives $u=(Q+Z)/(2t)$ and $v=uT/t$. The identity $(Q-Z)(Q+Z)=292t^2$ ensures $u\ne0$. Consider the two linear forms $$ L_1=146t+36Z-37Q,\qquad L_2=-4074t+511Q-52T. \tag{11} $$ They are tangents to the conics $292t^2+Z^2-Q^2=0$ and $-19012t^2+73Q^2-T^2=0$ at the rational points $(t,Z,Q)=(1,72,74)$ and $(t,Q,T)=(3,98,728)$ respectively. Neither vanishes on a rational point of (10): the first would force $T^2=24^2\cdot661$ at its unique point of tangency, and the second would force $Z^2=8^2\cdot109$ at its point of tangency. Write $K=L_1L_2\ne0$. We first show that its odd part is $1$ or $7$ modulo $8$. On (10), the exact identities $$ L_1(L_1-292t)=(37Z-36Q)^2,\qquad L_2(1022Q-L_2)=97(728t-3T)^2 \tag{12} $$ hold. Choose primitive integral coordinates $(t,T,Z,Q)$. Their projections $(t,Z,Q)$ and $(t,Q,T)$ are primitive at every prime: if either vanished modulo a prime, (10) would force the fourth coordinate to vanish as well. Let $\ell$ be an odd prime outside $\{7,73,97\}$. The first pair of factors in (12) cannot both vanish modulo $\ell$. Otherwise $t=0$, and $36Z-37Q=37Z-36Q=0$; their determinant is $73$, contradicting primitivity. The second pair cannot both vanish either: it would give $Q=0$ and $4074t+52T=728t-3T=0$, whose determinant is $-2\cdot7^3\cdot73$. Thus whenever $L_i$ is divisible by $\ell$, its partner is a unit. Equation (12) proves that $v_\ell(L_i)$ is even. Consequently the prime factors of odd exponent in the odd part of $K$ are among $7,73,97$. Their residues are $7,1,1$ modulo $8$, and the sign contributes at most $-1$. This proves the assertion. We next show that every $\mathbb Q_2$-point forces the odd part of $K$ to be $5$ modulo $8$, giving a contradiction. In primitive integral $2$-adic coordinates, at least one of $t,Z$ is odd. If $Z$ is odd, (10) gives $T,Q$ odd and $t$ even. Directly modulo $8$, $$ K\equiv(2t+4Z+3Q)(6t+7Q+4T)\equiv5. $$ If $Z$ is even, then $t$ is odd; normalize $t=1$. The valuations $v_2(Z)=1,2$ are excluded by $Q^2=Z^2+292$, and valuations $3,4,5$ by $T^2=73Z^2+2304$. After extracting the even square factors, the respective unit-square contradictions modulo $8$ are $2,5,5,2,5$. Hence $$ Z=64z_0,\qquad Q=2q_0,\qquad T=16v_0, $$ where $q_0,v_0$ are odd. The first equation for $Q$ gives $q_0^2=73+1024z_0^2$, so $q_0\equiv29$ or $99\pmod{128}$. Substitution in (11) yields | $q_0\pmod{128}$ | $v_2(L_1)$ | Odd part of $L_1\pmod8$ | $v_2(L_2)$ | Odd part of $L_2\pmod8$ | |---:|---:|---:|---:|---:| | $29$ | $4$ | $3$ | $2$ | $7$ | | $99$ | $2$ | $5$ | $4$ | $1$ | In both cases $v_2(K)=6$ and its odd part is $5$ modulo $8$. Common homogeneous rescaling multiplies $K$ by a square, whose odd part is $1$ modulo $8$, so the contradiction is independent of the chosen coordinates. This proves $C_{8/3}(\mathbb Q)=\varnothing$. Together with Theorem 1, it is an everywhere locally soluble genus-one curve with no rational point. For contrast, a known rational point occurs at $z=72/35$: $$ r=\frac{3799}{2135},\qquad s=\frac{1390074864}{159537875}. $$ Here $\delta=6409/1225$ is nonsquare. Theorem 2 proves rank at least two without another rank computation. Formula (2) yields $$ (d,e,g,f)=\left(\frac35,\frac{4200}{6409},\frac45, \frac{4841}{6409}\right). $$ Its product $p=4841/3150>1$ fails $1-p$, illustrating the remaining arithmetic cost after rational solubility of the torsor. ## 6. Selecting independent directions in the fixed-circle fibration The preceding torsor fixes the product $de$. An alternative is to fix a rational circle point $d,g\in\mathbb Q^\times$, $d^2+g^2=1$, and vary $$ e^2+f^2=1,\qquad w^2=1-4d^2e^2. $$ With identity $(e,f,w)=(0,1,1)$ this is the elliptic curve $$ E_d:\quad y^2=x(x-1)(x-4d^2). $$ The tuple formulas and residual tests remain $a=g/e$, $b=f/d$, $c=2de$, and $1\pm p$ square for $p=gf/(de)$. The visible boundary point $$ B=(2(1+d),2g(1+2d)) $$ has circle coordinates $(g,d,1-2d^2)$ and gives $a=b=p=1$. The next proposition shows why simply taking more multiples of this point cannot pass the residual tests. **Proposition 3.** If $d,g\in\mathbb Q^\times$, $d^2+g^2=1$, and $k=v_2(d)\ge2$ is finite, then the subgroup generated by $B$ and rational torsion, and its entire closure in $E_d(\mathbb Q_2)$, cannot supply a full lift. If $\operatorname{rank}E_d(\mathbb Q)=1$, every rational point belongs to that closure. Thus a full rational lift on this fixed-circle fiber requires rank at least two. **Proof.** Put $a_0=4d^2$ and choose $h=\sqrt{1-a_0}\in\mathbb Q_2$ with $h\equiv1\pmod4$. In circle coordinates the addition law is $$ (e,f,w)+(E,F,W)=\left( \frac{eFW+Efw}{1-a_0e^2E^2}, \frac{fF-eEwW}{1-a_0e^2E^2}, \frac{wW-a_0eEfF}{1-a_0e^2E^2}\right). $$ The map $x=(w+1)/(1-f)$, $y=(w+1)(w+f)/(e(1-f))$ identifies this law with the Weierstrass law. Let $R=(1,0,h)$ and $T=(0,1,-1)$. They generate $H\cong\mathbb Z/4\oplus\mathbb Z/2$, and $$ U=\{(e,f,w):e\in4\mathbb Z_2,\ f\equiv w\equiv1\pmod4\} $$ is a subgroup parametrized by $e\in4\mathbb Z_2$. Every local circle point has integral $e,f$, with one a unit and the other divisible by four, and $w$ is a unit. The translations $$ P+R=(f/w,-he/w,h/w),\quad P+T=(-e,f,-w),\quad P+2R=(-e,-f,w) $$ therefore give $E_d(\mathbb Q_2)=H\oplus U$. There are no local points at infinity in the circle model, since they would require $\sqrt{-1}\in\mathbb Q_2$. For $P\in U\setminus\{O\}$, addition on the successive neighborhoods $v_2(e)\ge j$ reduces to addition of $e$ modulo $2^{j+1}$, while $$ e(2P)=\frac{2efw}{1-a_0e^4} $$ has valuation $v_2(e(P))+1$. Thus $v_2(e(nP))=v_2(e(P))+v_2(n)$ for every nonzero integer $n$. In particular $U$ is torsion-free and pro-$2$. The complete rational torsion is $E_d[2]$. Indeed, local torsion lies in $H$, whose order-four points require rational $h$. Rational $g,h$ would give a point $(X,Y)=(1/d^2,gh/d^3)$ on $Y^2=X(X-1)(X-4)$. After $U_0=X-1$ this is the rank-zero curve $E_*$ from Section 4. Its complete rational point set forces $d^2=1$ or $1/4$, contrary to $v_2(d)\ge2$. Changing the sign of $g$ if necessary, assume $g\equiv1\pmod4$. Then $B=R+Q$ with $$ Q=\left(-\frac d{1-2d^2},\frac{gh}{1-2d^2}, \frac h{1-2d^2}\right)\in U, \qquad v_2(e(Q))=k. $$ For $n$ odd and $\tau\in E_d[2]$, the point $nB+\tau$ has $e$ a unit and $v_2(f)=k$, so $v_2(p)=0$. For nonzero even $n$, it has $v_2(e)=k+v_2(n)$ and $f$ a unit, so $v_2(p)=-2k-v_2(n)<0$. For $n=0$ it has $e=0$, outside the tuple chart. But $1+p$ and $1-p$ are simultaneously $2$-adic squares exactly when $p\in8\mathbb Z_2$, as follows by considering square residues modulo eight and ruling out negative valuations. Thus all these points fail. The same valuation formula extends by continuity to $n\in\mathbb Z_2$, proving the assertion for the closure as well. Finally, if the rational rank is one, write $B=mG+\tau$ for a primitive generator $G$ modulo torsion. The first two-isogeny descent character is trivial on rational torsion, but its value $x(B)=2(1+d)$ has odd $2$-adic valuation. Consequently $m$ is odd. Multiplication by an odd integer is an automorphism of the pro-$2$ group $H\oplus U$, and its inverse is a limit of integer multiplications. Hence $G$ and every rational point lie in the excluded closure. $\square$ This condition applies after exchanging the two circles to every full canonical symmetric sextuple. Over $\mathbb Q_2$, $d,e$ cannot both be units, since $1-4d^2e^2\equiv5\pmod8$, and they cannot both be even, since then $p$ has negative valuation. Exactly one has valuation at least two and may be chosen as the fixed coordinate. There is a rational family with two independent directions on this fibration. Start with $t,h_0\in\mathbb Q^\times$, $t^2+h_0^2=1$, $v_2(t)=0$, and finite $v_2(h_0)=j\ge2$. Double the boundary point on $E_t$, then exchange the two circles. Put $$ D_0=1-4t^2h_0^4,\quad w_0=1-2t^2, $$ $$ d=\frac{2th_0w_0}{D_0},\quad g=\frac{t^2-h_0^2w_0^2}{D_0},\quad w_P=\frac{w_0^2-4t^4h_0^2}{D_0}. $$ Then $d^2+g^2=1$ and $P=(e,f,w)=(t,h_0,w_P)$ belongs to $E_d$. The denominator is a $2$-adic unit and $v_2(d)=j+1$. In the local splitting from Proposition 3, $B$ has an odd order-four component with its $U$ component of depth $j+1$, whereas $P$ has an odd order-four component with depth $j$. A primitive integral relation between $B$ and $P$ modulo rational torsion would force both coefficients odd by projection to $H/H[2]$. Their $U$ components would then have different valuations and could not cancel. Thus $B$ and $P$ are independent for every rational parameter satisfying these conditions. For example $t=3/5$, $h_0=4/5$ gives $$ d=\frac{4200}{6409},\quad g=\frac{4841}{6409},\quad P=\left(\frac35,\frac45,-\frac{3959}{6409}\right) $$ in circle coordinates. The independence proof requires no rank computation. The point $P$ itself has $v_2(p(P))=-1$ and fails both residual squares. However, the combination $S=B+2P$ reaches a permitted local neighborhood uniformly. Put $u=t^2$, $K=4u^3-4u^2+2u-1$, and $H=12u^3-16u^2+4u+1$. Its first two circle coordinates simplify to $$ e(S)=-\frac{(2u-1)K}{H},\qquad f(S)=\frac{4th_0(u-1)(2u^2-1)}H. $$ Consequently $v_2(f(S))=3j+2$ and $v_2(p(S))=2j+1\ge5$, so both residuals are $2$-adic squares. Their exact remaining rational conditions are $$ 1+p(S)=\frac{(4u^3-4u+1)(4u^3-8u^2+8u-3)}{(2u-1)^2K}, $$ $$ 1-p(S)=-\frac{(4u^3-4u^2+1)(4u^3-12u^2+8u-1)}{(2u-1)^2K}. $$ Also $p(S)\to0$ as $u\to1$, so the family reaches the real region where both residuals are positive. Each displayed square condition defines a genus-four double cover of the $u$-line; rational construction additionally requires $u$ and $1-u$ to be squares. This is an explicit remaining arithmetic problem, not a claim that either residual is automatically a rational square or that a full sextuple has been constructed. The same short section can satisfy all fifteen edge conditions over $\mathbb R$, $\mathbb Q_2$, and $\mathbb Q_3$ simultaneously. Set $t=(1-q^2)/(1+q^2)$, $h_0=2q/(1+q^2)$ and choose $$ q=\frac6{48n+1},\qquad n\in\mathbb Z_{\ge1}. $$ Then $v_2(h_0)=2$, $v_3(h_0)=1$, and $9/10