# Elliptic and dyadic obstructions to constrained Diophantine completion ## 1. Contribution and scope A rational Diophantine tuple consists of distinct nonzero rational numbers whose pairwise products plus one are squares. A square-product tuple additionally has every pairwise product square. A number is strong if its square plus one is square. We study two constrained completion routes relevant to the rational septuple problem: regular fifth completion of product-one square-product quadruples, and four-point translation orbits on elliptic curves induced by triples. The first route is related to square-product tuples with a strong member and their correspondence with product-minus-one pairs [1]. The second attempts to add four points to a triple using an order-four translation. For the first route, we give an explicit cubic necessary quotient, elliptic away from stated degeneracies. On a diagonal subfamily, a rank-zero hypothesis excludes every nondegenerate regular square-product completion. Its torsion analysis is uniform, and elementary descent certifies three parameters, one supporting infinitely many valid starting quadruples. For the orbit route, we derive two diagonal identities and prove a uniform obstruction whenever the base product is a 2-adic unit. A rank-zero diagonal quotient also excludes an infinite nonunit family, all of whose induced curves have positive rank, including infinitely many positive triples. We then turn the dyadic obstruction into a rank-zero filter for entire families with fixed Pythagorean square-root ratios. Finally, a rigorous interval L-value calculation resolves an explicit triple whose initial rank bounds were zero to two and proves it has exactly one nonzero common neighbor. These are structural exclusions of specified routes and a complete fixed-triple result. They do not rule out arbitrary irregular completions, all nonunit translation orbits, or rational septuples. Their practical purpose is to eliminate entire branches before a rational-point search and to distinguish survival of tested local conditions from actual extendibility. ## 2. Prior work Dujella, Kazalicki and Petričević [2, Proposition 4] give the product-one parametrization of rational Diophantine quadruples and observe that their pairwise products are squares. Their exotic square-product quintuples have a product-one subquadruple and two specified regular subquadruples. Our condition concerns a **regular quintuple** completion. These are different regularity requirements; the theorem below does not exclude the exotic quintuples of [2]. The regular extension equation and its rational formulas are classical; we use the conventions of Dujella [3]. Strong-pair constructions and fractional linear transformations also appear in [4]. The correspondence and coordinate exclusions in the concurrent rStation paper [1] concern strong members of the particular families in [2]. Our elliptic obstruction concerns a different, explicitly specified completion operation. The dyadic compatibility analysis in [5] supplies local restrictions for product-minus-one pairs but does not imply the global rank obstruction proved here. Dujella and Mikić [6] study rank-zero induced curves, record the regular-triple relation between the distinguished points, and describe the order-four regular family. We use those classical parametrizations. Our orbit contribution is the simultaneous diagonal obstruction and its fixed-ratio rank-zero consequence; the existence of rank-zero triples itself is not new. The certified L-value application concerns one explicit base left rank-ambiguous by the initial descent. We do not claim novelty for elliptic descent, torsion classification, or the product-one and regular-family parametrizations. The contributions are the completion quotients, the uniform lift and dyadic obstructions, and their exact nonvacuous applications. ## 3. Product-one coordinates and the completion quotient Let $z,a,b,c$ be positive rational numbers satisfying $$ ab=zc, $$ with $a,b,c$ strong. Then $$ Q=\left\{z,\frac{a^2}{z},\frac{b^2}{z},\frac1{zc^2}\right\} \tag{1} $$ has product one and both $D(0)$ and $D(1)$, provided its four entries are distinct. Its six products are $a^2,b^2,c^2,a^{-2},b^{-2},c^{-2}$. Conversely every positive product-one square-product quadruple has this form: distinguish its first member $z$ and take $a=\sqrt{zB}$, $b=\sqrt{zC}$ and $c=\sqrt{BC}$ for its second and third members $B,C$. In fact the square-product hypothesis is automatic for a positive product-one Diophantine quadruple: if $ABCD=1$, then $AB=(AB+1)/(CD+1)$ is a square. The displayed coordinates are a rearrangement of the parametrization in [2]. No strong hypothesis on the distinguished member $z$ is required in this section. For a positive product-one quadruple $\{A,B,C,D\}$, its finite regular fifth candidate is $$ e=\frac{(A+B-C-D)^2-4(AB+1)(CD+1)} {4D(AB+1)(AC+1)(BC+1)}. \tag{2} $$ This follows by imposing the standard regularity equation $$ (ABCDe+2ABC+A+B+C-D-e)^2 =4(AB+1)(AC+1)(BC+1)(De+1). $$ Here $ABCDE=e$, so the left side is independent of $e$. Solving the resulting linear equation gives (2), after using $ABCD=1$. For completeness, direct substitution also gives $$ 1+De=\frac{(2ABC+A+B+C-D)^2}{4(AB+1)(AC+1)(BC+1)}, $$ and the three analogous identities obtained by permuting $A,B,C,D$. Each denominator is a rational square. Thus the candidate is compatible with all four entries whenever it is nonzero and distinct. We always exclude zero or repeated entries. Substituting (1), with $b=zc/a$, into (2) gives $$ \frac ez=\frac{F_{z,c}(a^2)} {4c^2a^2z^2(c^2+1)(a^2+1)(c^2z^2+a^2)}, \tag{3} $$ where $$ \begin{aligned} F_{z,c}(X) ={}&\big[c^2X^2+(-c^2z^2-1-2cz(c^2+1))X+c^4z^2\big]\\ &\cdot\big[c^2X^2+(-c^2z^2-1+2cz(c^2+1))X+c^4z^2\big]. \end{aligned} \tag{4} $$ The denominator in (3) is square: its last factor is $a^2(b^2+1)$, and $a,b,c$ are strong. The completed tuple retains square products precisely when $e/z$ is square, since each original member divided by $z$ is square. Put $s=a-zc/a$. Direct expansion gives the further identity $$ \frac{F_{z,c}(a^2)}{c^4a^4} =\big[s^2-(z+1/c)^2\big] \big[s^2+4cz-(z-1/c)^2\big]. \tag{5} $$ ### Proposition 1 If the regular fifth candidate of (1) retains square products, then $$ \mathcal E_{z,c}:\quad Y^2=X\big[X-(z+1/c)^2\big] \big[X+4cz-(z-1/c)^2\big] \tag{6} $$ has a rational point for which $X$ and $X+4cz$ are rational squares. **Proof.** By (3)–(5), choose $R$ with $R^2=F_{z,c}(a^2)/(c^4a^4)$ and set $X=s^2$, $Y=sR$. Also $X+4cz=(a+zc/a)^2$. ∎ These are necessary conditions only. They do not recover the individual strong conditions on $a,b,c$. At $s=0$, the map forgets $R$, so a lift must be checked in (5). If the cubic in (6) degenerates, it remains a necessary algebraic equation, but no elliptic rank statement is asserted for that specialization. ## 4. Two fixed auxiliary obstructions We record two elementary consequences of rank-zero elliptic curves. The proofs are included to make the uniform argument independent of computer-reported ranks. ### Lemma 2 There are no rational $u,v,w$ with $uv\ne0$ satisfying $$ v^2=1-u^2,\qquad w^2=1+2u. $$ The same holds with $1-2u$ in place of $1+2u$. **Proof.** If $w=0$, then $u=-1/2$ and $v^2=3/4$, impossible. Otherwise $$ (X,Y)=(-w^2,2wv) $$ lies on $E_1:Y^2=X(X-1)(X+3)$. We determine its rational points by 2-isogeny descent. For $E:y^2=x^3+Ax^2+Bx$, the descent classes are represented by signed squarefree divisors $d$ of $B$; a class requires a primitive solution of $$ N^2=dU^4+AU^2V^2+(B/d)V^4. \tag{7} $$ If the descent-image sizes for $E$ and its 2-isogenous curve are $h,h'$, then $2^r=hh'/4$ [8, Chapter X]. For $(A,B)=(2,-3)$ there are at most four classes. The dual has $(A',B')=(-4,16)$. Its cubic is negative for $x<0$, excluding negative classes; class $2$ is impossible modulo $16$ in (7). Thus $h'\le1$ and $r=0$. Good reduction at $5$ and $7$ gives eight points at each prime. Rational torsion therefore has order at most eight, and the points $(-1,2)$ and $(0,0)$ generate eight points. The finite $X$-coordinates are $$ 0,1,-3,-1,3. $$ Of these, only $-1$ is a negative rational square, and $X=-1$ forces $u=0$. Replacing $u$ by $-u$ proves the second assertion. ∎ ### Lemma 3 There are no rational $u,v,w$ with $u\ne0$ satisfying $$ v^2=1-u^2,\qquad w^2=1-4u^2. $$ **Proof.** The point $(X,Y)=(1/u^2,vw/u^3)$ would lie on $E_2:Y^2=X(X-1)(X-4)$. In its 2-isogeny descent, $(A,B)=(-5,4)$ has only class $1$: negative classes are impossible by sign, and class $2$ fails modulo $4$. The dual has $(A',B')=(10,9)$ and at most four classes, hence the rank is zero. Reduction at $5$ and $7$ bounds torsion order by eight. None of the three nonzero rational 2-torsion points is twice a rational point: the standard full-2-torsion halving criterion requires all three differences from the roots $0,1,4$ to be squares, which fails at each root. Therefore the torsion group consists of the four 2-torsion points. Its finite $X$-coordinates are $0,1,4$. The value $0$ is impossible for $1/u^2$; the other two give $u^2=1$ or $1/4$, contradicting one of the required squares. ∎ ## 5. The uniform diagonal criterion Let $z=p/q>0$ be strong, with coprime positive integers $p,q$, and let $r>0$ satisfy $p^2+q^2=r^2$. Consider $$ E_z:\quad y^2=x(x-r^2)(x+3p^2-q^2). \tag{8} $$ Its roots are distinct: $p,q\ne0$ and $q^2=3p^2$ is impossible for nonzero rational $p/q$. ### Theorem 4 If $E_z(\mathbb Q)$ has rank zero, then no nondegenerate quadruple $$ \left\{z,\frac{a^2}{z},\frac{z^3}{a^2},\frac1{z^3}\right\}, \tag{9} $$ with $a,z^2/a$ strong, has a regular fifth completion retaining square products. **Proof.** Apply Proposition 1 with $c=z$. Scaling $X$ by $(pq/r)^2$ and $Y$ by $(pq/r)^3$ gives (8). Its necessary square conditions become $$ x\in\mathbb Q^2,\qquad x+\frac{4p^4}{r^2}\in\mathbb Q^2. \tag{10} $$ The curve contains $\mathbb Z/2\mathbb Z\times\mathbb Z/4\mathbb Z$: it has full rational 2-torsion, and $$ \big(r(r+2p),\,2pr(r+2p)\big) $$ has order four and doubles to $(r^2,0)$. The finite $x$-coordinates of this subgroup are $$ 0,\ r^2,\ q^2-3p^2,\ r(r-2p),\ r(r+2p). \tag{11} $$ Neither $r(r-2p)$ nor $r(r+2p)$ is square. Otherwise $1\pm2p/r$ and $1-(p/r)^2=(q/r)^2$ would both be squares, contrary to Lemma 2. The doubling formula on $y^2=x^3+Ax^2+Bx$ is $$ x(2P)=\left(\frac{x(P)^2-B}{2y(P)}\right)^2. $$ Consequently there is no point of order eight, since it would double to an order-four point with square $x$-coordinate. Mazur's rational torsion classification [7] now shows that the displayed subgroup is the full torsion group. Also $q^2-3p^2$ is not square: after division by $r^2$, it would give $1-4(p/r)^2$ square alongside $1-(p/r)^2$, contradicting Lemma 3. Thus at rank zero only $x=0$ and $x=r^2$ can satisfy the first condition in (10). At $x=r^2$, the second condition asks that $1+4(p/r)^4$ be square. Then a rational right triangle with legs $1$ and $2(p/r)^2$ would have square area $(p/r)^2$, contradicting Fermat's square-area theorem. Here is a short descent verification of the exact special case used: if $w^2=1+4u^4$ with $u\ne0$, then $(4u^2,4uw)$ lies on $V^2=U^3+4U$. Its 2-isogenous curve is isomorphic to $V^2=U^3-U$. The latter has at most two descent classes, and the former also has at most two, since its nonzero $U$-coordinates are positive. Thus both curves have rank zero. Reduction of $V^2=U^3+4U$ at $3$ and $5$ gives orders $4$ and $8$, so its rational points are $O,(0,0),(2,\pm4)$. Neither finite nonzero coordinate is $4u^2$ for rational $u$. At $x=0$, the original expression $s=a-z^2/a$ vanishes, giving $a^2=z^2$. The second member of (9) then equals its first member. Hence neither case gives a nondegenerate completion. ∎ The theorem is conditional only on the rank of the explicit curve (8). Its torsion calculation and lift obstruction are uniform, and its proof does not assume any unproved rank-parity or distribution conjecture. ## 6. Explicit rank-zero applications and nonvacuity ### Corollary 5 The regular square-product completion in Theorem 4 is impossible for $$ z=\frac{15}{8},\qquad\frac{21}{20},\qquad\frac9{40}. $$ **Proof.** The corresponding $(A,B)$ in $y^2=x^3+Ax^2+Bx$ are | $z$ | $A$ | $B$ | Original image upper bound | Dual image upper bound | |---|---:|---:|---:|---:| | $15/8$ | $322$ | $-176579$ | $4$ | $1$ | | $21/20$ | $82$ | $-776243$ | $4$ | $1$ | | $9/40$ | $-3038$ | $2281117$ | $4$ | $1$ | For each curve the supplement lists every signed squarefree divisor in the two descents and a real or finite congruence obstruction for each excluded class. A standalone integer checker verifies these complete lists. Equation (7) and the formula $2^r=hh'/4$ therefore give rank zero in all three cases. For example, at $z=15/8$, the original classes surviving reduction modulo $3$ and $5$ are $1,-221,-611,799$. The dual has $(A',B')=(-644,810000)$; its negative classes are excluded by sign. Modulo $13$ excludes $2,5,6,15$, modulo $17$ excludes $3,10$, and modulo $16$ excludes $30$, leaving only $1$. The remaining two rows are certified in exactly the same way, with the largest modulus $23^2$. ∎ There are infinitely many nondegenerate starting quadruples in (9) at $z=15/8$. To prove this without a software rank assertion, consider $$ H_z:\quad v^2=u(u+1)(u+z^4). $$ The point $$ T=\left(\frac{11025}{43264},\frac{72294075}{35995648}\right) $$ has $u,u+1,u+z^4$ all rational squares, so lies in $2H_z(\mathbb Q)$ by the full-2-torsion descent criterion. Reduction at $11$ and $13$ gives order $16$ at both primes, bounding rational torsion by $16$. Exact duplication verifies $16T\ne O$, so $T$ has infinite order. Every multiple of $T$ remains in $2H_z(\mathbb Q)$. Except at finitely many exceptional points, put $a=\sqrt u>0$ and $b=z^2/a$. Then $a$ and $b$ are strong and $ab=z^2$. Infinitely many different $u$-coordinates occur, while zero and coincidence conditions in (9) exclude only finitely many values. This produces infinitely many valid quadruples to which Corollary 5 applies. The proof excludes their regular square-product fifth completion; it makes no assertion about an irregular completion. ## 7. A dyadic obstruction for order-four translation orbits Let $\{a,b,c\}$ be a rational Diophantine triple, and write $$ A=ab,\qquad B=(abc)^2,\qquad r^2=ab+1,\quad s^2=ac+1,\quad t^2=bc+1. $$ On the induced curve $$ E:\quad y^2=(x+ab)(x+ac)(x+bc), $$ the distinguished points are $P=(0,abc)$ and $S=(1,rst)$. The full rational two-torsion descent identifies common-neighbor points with the coset $P+2E(\mathbb Q)$: a finite point with $x=abc\,d$ represents a common neighbor precisely when its three descent components have the same square classes as those of $P$ [8]. Branch points use the usual extended descent map. The point $S$ belongs to $2E(\mathbb Q)$ [6]. Suppose $S$ has order four. Its translations give four candidate entries $$ d_i=\frac{x(T+iS)}{abc},\qquad i=0,1,2,3. $$ A septuple from this construction must satisfy, in particular, the two opposite-edge square conditions. The following theorem excludes a whole valuation branch without determining the Mordell–Weil group. ### Theorem 6 If $v_2(abc)=0$, no order-four translation orbit can supply four finite, distinct, nonzero common neighbors that, together with $a,b,c$, form a Diophantine septuple. **Proof.** Relabel the base so that $2S=(-A,0)$ and set $k=A+1=r^2$. In the coordinate $u=x+A$, the order-four condition and the value of $S$ put the curve in the form $$ y^2=u\bigl(u^2+(L^2-2k)u+k^2\bigr),\qquad S=(k,kL), \tag{12} $$ where $L$ is rational and nonzero. Indeed, doubling $S$ to $(0,0)$ forces the product of the two other shifted roots to be $k^2$, and substitution of $u=k$ determines the quadratic coefficient. At $x=0$, equation (12) gives $$ B=A(AL^2+1). $$ Thus $M=AL$ satisfies $M^2=B-A$. With $N=B+1$, we have $$ k+M^2=N. \tag{13} $$ Put $u=x(T)+A$. The cases $u=0,k$ make the orbit contain infinity, so they cannot supply four finite entries. Base compatibility gives $u/A=cd_0+1=q^2$ for a nonzero rational $q$. Define $$ v=\frac{u-k}{q}\ne0. $$ Translation by $2S$ has $x$-coordinate $k^2/u-A$. Consequently $$ g=x(T)x(T+2S)+B=N-v^2. \tag{14} $$ The addition formula at $S=(k,kL)$ gives $$ x(T\pm S)=\left(\frac{y(T)\mp kL}{u-k}\right)^2-(L^2-2k)-u-k-A. $$ Multiplying these two expressions and replacing $y(T)^2$ using (12) yields $$ h=x(T+S)x(T-S)+B=N-\frac{4kM^2}{v^2}. \tag{15} $$ These identities are also checked symbolically in the supplement. If the two missing diagonals are compatible, $g$ and $h$ are rational squares. Equations (13)–(15) therefore require $$ r^2+M^2=N,\qquad v^2+G^2=N,\qquad W^2=Nv^2-4r^2M^2 \tag{16} $$ for rational $G,W$, with $W=v\sqrt h$. When $abc$ is a 2-adic unit, $N\equiv2\pmod8$. Any two rational squares summing to $N$ have unit roots. To see this, a negative minimum valuation $m$ gives sum valuation $2m$ for unequal root valuations, or $2m+1$ for equal valuations; both contradict $v_2(N)=1$. Once the roots are integral, reduction modulo eight forces both odd. Apply this to the first two equations in (16): $r,M,v,G$ are all units. The last equation then forces $W^2\equiv2-4\equiv6\pmod8$, impossible. ∎ The proof also yields a useful reduction outside the excluded branch. Put $K=4kM^2$. The equations $G^2=N-v^2$ and $W^2=Nv^2-K$ map to $$ \mathcal D:\quad Y^2=X(X+N^2)(X+K),\qquad X=-Nv^2,\quad Y=NvGW. \tag{17} $$ This is elliptic unless $B=2A+1$. Lifting its points still requires additional square conditions; it is not a sufficient construction criterion. For instance, the original group equation also forces $v^2+4Ak$ and $Av^2+M^2$ to be squares. The special real-obstruction relation $B=2A+1$ gives $B+1=2r^2$, which itself forces $v_2(abc)=0$, so Theorem 6 includes that specialization. ### Corollary 6.1 Suppose the diagonal curve (17) is nonsingular and has rank zero. A nondegenerate four-orbit septuple must then contain a product-minus-one pair among opposite orbit entries, and $abc$ must be strong. In particular, if $1+(abc)^2$ is not square, the entire orbit construction is excluded. **Proof.** Choose the signs of $r,M$ positive and put $s=2rM$. The curve $\mathcal D$ has torsion exactly $\mathbb Z/2\times\mathbb Z/4$. The points of order four have abscissas $\pm sN$. Halving the positive one would require both $N$ and $s$ to be squares, since its root differences include $sN$ and $N(N+s)=N(r+M)^2$. The rational right triangle with legs $r,M$ would then have square area $rM/2=s/4$, contrary to Fermat's square-area theorem. The negative abscissa cannot be halved, and Mazur excludes other larger torsion groups. Since $X=-Nv^2<0$, rank zero leaves only $X=-N^2,-s^2,-sN$. The first gives $g=0$; the second gives $h=0$. Thus an opposite pair has product minus one. These cases also force $N$ square, since respectively $v^2=N$ or $v^2=s^2/N$. It remains to exclude $X=-sN$, which gives $v^4=4kM^2$. On (12), duplication yields $$ x_{\mathrm{shift}}(2T)-k =\frac{(u-k)^4-4kL^2u^2}{4y(T)^2}. $$ The numerator vanishes because $v^2=A(u-k)^2/u$ and $M=AL$. Here $y(T)\ne0$: otherwise $u\ne0$ and $(u-k)^2=-L^2u$, which combined with the vanishing numerator forces $L^2=4k$ and makes the induced curve singular. Therefore $x(2T)=1$, so $2T=\pm S$. It follows that $-T=T\mp S$ lies in the same orbit, repeating an $x$-coordinate. This contradicts four distinct entries and proves the corollary. ∎ In the regular order-four subfamily of [6], $N=r^2+1/(4r^2)$ can never be square: multiplying by $4r^2$ would make $1+4r^4$ square, excluded in Section 5. Thus rank zero of the nonsingular diagonal quotient excludes this regular subfamily even when the original induced curve has positive rank. Rank zero of the original induced curve gives a separate obstruction in Section 9. ### Corollary 6.2: an infinite nonunit branch If $M^2=4r^2$, no order-four orbit supplies four distinct compatible new entries. Every valid base in this subfamily has $v_2(abc)=1$ and positive induced rank. There are infinitely many such bases, including infinitely many positive triples. **Proof of exclusion.** Here $N=5r^2$, and (17) scales by $X=r^4x$, $Y=r^6y$ to $$ D_2:\quad y^2=x(x+16)(x+25). $$ This curve has rank zero by elementary two-isogeny descent. In the notation (7), its original pair is $(41,400)$ with signed squarefree classes $\pm1,\pm2,\pm5,\pm10$. The classes $\pm2,\pm10$ fail modulo $16$, leaving an image of size at most four. The dual pair is $(-82,81)$ with classes $\pm1,\pm3$; negative classes fail by sign and class $3$ fails modulo $5$. Thus $2^{\operatorname{rank}}=hh'/4\le1$. As $N=5r^2$ is not a rational square, Corollary 6.1 proves the exclusion. The supplement also checks the full eight-point group, generated by $(-20,20)$ and $(-16,0)$, against good reductions of orders $8,16$ at $7,11$. **Parametrization and valuation.** Write $K_0=abc$, so $K_0^2=5r^2-1$, and $A=r^2-1$. The order-four relation gives $$ a+b=\frac{4r^2-2A^2}{AK_0},\qquad ab=A. $$ Consequently $n=(a-b)AK_0/(4r^3)$ satisfies $n^2=2-r^2$. Conversely, rational solutions of $$ r^2+n^2=2,\qquad K_0^2=5r^2-1 $$ give, away from finitely many degeneracies, $$ a=\frac{2r^2-A^2+2r^3n}{AK_0},\quad b=\frac{2r^2-A^2-2r^3n}{AK_0},\quad c=\frac{K_0}{A}. $$ Indeed, $$ ab=A,\quad ac+1=\left(\frac{r(r+n)}A\right)^2,\quad bc+1=\left(\frac{r(r-n)}A\right)^2,\quad abc=K_0, $$ and $(ac-A)(bc-A)=(A+1)^2$, which is the required order-four condition. The sum $r^2+n^2=2$ forces both roots to be 2-adic units, by the valuation argument in Theorem 6. Hence $K_0^2=5r^2-1\equiv4\pmod8$, proving $v_2(K_0)=1$ for every valid base. **Infinitude.** On $$ H:\quad y^2=x(x+1)(x+10),\qquad Q=(-5,10), $$ good reductions at $7,13$ have orders $12,16$, bounding rational torsion by four; exact duplication gives $4Q\ne O$. Thus $Q$ is non-torsion. Its odd multiples have Kummer classes $(-5,-1,5)$, so give $$ x=-5r^2,\quad x+1=-K_0^2,\quad x+10=5n^2. $$ They supply infinitely many distinct $r^2$, hence infinitely many unordered base triples after the finitely many zeros, poles and coincidences are removed. The odd multiples are dense in the bounded real component: $2Q$ has infinite order in the identity circle, and its cyclic subgroup is dense. The open interval $-100$ gives $A,c>0$, $ab=A>0$, and $a+b=(4r^2-2A^2)/(AK_0)>0$, so all three entries are positive. **Positive induced rank.** No valid base in this subfamily is regular. The parametrization gives $c-a-b=(2r^2-1)/K_0$. Regularity would require $(c-a-b)^2=4r^2$, hence $16r^4=1$, giving $r^2=1/4$ and the impossible rational square $n^2=7/4$. If the induced curve had rank zero, $S\in2E(\mathbb Q)$ of order four would force its torsion to be $\mathbb Z/2\times\mathbb Z/8$. The point $P=(0,abc)$ then has order four or eight. In the latter case the two order-four points in $2E(\mathbb Q)$ give $S=\pm2P$, which is precisely regularity [6]. In the former case, relabel so $2P=(-ab,0)$. The duplication formula gives $x(2P)+ab=(a+b-c)^2/4$, hence $c=a+b$. The points doubling to $(-ab,0)$ have original abscissas $0,-2ab$. Since the doubles of all order-four points in this torsion group agree, $S$ would force $1=-2ab$. This contradicts $ab+1$ being a rational square. Both cases are impossible, proving positive rank. ∎ For a small mixed-sign example, $3Q$ gives $(r,n,K_0)=(41,119,22)/89$ and, up to simultaneous negation, $$ \left\{\frac{12800}{38181},-\frac{16731}{7120},\frac{979}{3120}\right\}. $$ The supplement independently verifies $48P\ne O$ for this base. It also checks an explicit positive triple from $5Q$. Thus neither the unit-product theorem nor the rank-zero induced-curve obstruction accounts for this entire excluded branch. ## 8. A rank-zero filter for fixed Pythagorean ratios The order-four relation can be written $1/r^2=1/s^2+1/t^2$ [6]. Fix positive primitive Pythagorean integers $m,n,h$ with $m^2+n^2=h^2$, and impose $$ \frac{s}{r}=\frac hm,\qquad \frac tr=\frac hn. $$ Signs of the chosen square roots do not change the underlying conditions. ### Theorem 7 If $$ E_{m,n}:\quad V^2=(X-h^2)(X-m^2)(X-n^2) \tag{18} $$ has rank zero, no valid base triple with these fixed ratios yields a septuple by the order-four orbit construction. **Proof.** Existence of a rational base requires $$ Y^2=(r^2-1)(h^2r^2-m^2)(h^2r^2-n^2),\qquad Y=mn\,abc. $$ The map $X=h^2r^2$, $V=hY$ gives a point on (18). Shift $Z=X-h^2$ to obtain $$ V^2=Z(Z+m^2)(Z+n^2). $$ Its full two-torsion and the points with $Z=\pm mn$ give a subgroup $\mathbb Z/2\times\mathbb Z/4$. It has no rational eight-torsion: halving a point with $Z=mn$ would require $mn$ to be a square, hence $m/n$ to be a rational square. Together with $m^2+n^2=h^2$ this would solve $u^4+1=w^2$ with $u\ne0$, contrary to Fermat's descent. For completeness, $w^2=1+u^4$ would give the point $(2(w+1)/u^2,4(w+1)/u^3)$ on $y^2=x^3-4x$, with nonzero $y$. This curve has rank zero: its two-isogeny image has at most four classes, while the dual $y^2=x^3+16x$ has only class $1$ (negative classes fail by sign and class $2$ fails modulo $16$). Good reductions at $3,5$ both have four points, exhausting the curve by $O,(0,0),(2,0),(-2,0)$. This supplies the required contradiction. Points with $Z=-mn<0$ cannot be halved by the rational two-torsion criterion. Mazur's theorem thus makes the torsion exactly $\mathbb Z/2\times\mathbb Z/4$. Rank zero now exhausts the finite $X$-coordinates: $$ m^2,\ n^2,\ h^2,\ h^2-mn,\ h^2+mn. $$ The first three make a base product zero and are invalid. In the two remaining cases, direct substitution of $r^2=(h^2\pm mn)/h^2$ gives $$ (abc)^2=\frac{(m\pm n)^2}{h^2}. $$ The primitive legs have opposite parity and $h$ is odd, so $abc$ is a 2-adic unit. Theorem 6 applies. ∎ For $(m,n,h)=(8,15,17)$, the rank hypothesis admits a short elementary certificate. Shift $X=Z+225$ in (18), obtaining $(A,B)=(97,-10304)$ in the isogeny notation (7). Congruences modulo $3$ or $5$ leave at most the four classes $1,-14,46,-161$. The dual has $(A',B')=(-194,50625)$: negative classes are excluded by sign, classes $3,15$ by modulo $16$, and class $5$ by modulo $49$. Thus the dual image has size at most one and the rank is zero. The complete class lists are supplied in the supplement. Good reductions at $11,13$ have $8,16$ points. Eight explicit rational points therefore exhaust (18), with finite $X$-coordinates $64,169,225,289,409$. It follows that the rational points on $$ y^2=(z^2-1)(289z^2-64)(289z^2-225) $$ are exactly the two infinities, the six branch points $z=\pm1,\pm8/17,\pm15/17$, and $(z,y)=(\pm13/17,\pm840/17)$. This complete reduced-case classification illustrates the uniform filter; it is not needed to assert a broad nonexistence result beyond Theorem 7. ## 9. Rank zero and a certified rigid triple We record a classical torsion consequence to distinguish survival of tested local conditions from actual extendibility. We do not claim novelty for the underlying descent or torsion classification. ### Proposition 8 Every triple contained in a rational Diophantine quintuple induces an elliptic curve of positive Mordell–Weil rank. **Proof.** Suppose the induced curve has rank zero. By Mazur and full rational two-torsion, $E(\mathbb Q)\cong\mathbb Z/2\times\mathbb Z/(2m)$, with $m\in\{1,2,3,4\}$. The point $P=(0,abc)$ is not two-torsion, so $m\ne1$. Extension points lie in $P+2E(\mathbb Q)$, a coset of size $m$, stable under negation and containing the distinct points $P,-P$, both with $x=0$. For $m\le3$ at most one further finite $x$-coordinate remains. For $m=4$, the two remaining points are either a pair $\pm Q$, again giving at most one further coordinate, or two distinct two-torsion points. The latter would give neighbors $-1/a_i,-1/a_j$ for distinct base members. A quintuple containing both would have two disjoint product-minus-one pairs. Such pairs cannot coexist in a distinct real Diophantine tuple: choose their positive members $u,v$; cross-product nonnegativity requires $u/v\le1$ and $v/u\le1$, hence $u=v$. This contradiction proves the proposition. ∎ The regular-family parametrization in [6] supplies the base $$ \left\{\frac{6727}{9840},-\frac{9600}{8897},\frac{32759}{52080}\right\}. \tag{19} $$ Its induced curve has integral minimal model $$ y^2+xy=x^3-562198913575863696451480x +160785133294502034126172196982094400, \tag{20} $$ conductor $N=477491580510$ and root number $+1$. An algebraic descent gave only bounds $[0,2]$. We resolve this example by a rigorous $L$-value calculation and the analytic-rank-zero theorem [9], using modularity of elliptic curves over $\mathbb Q$. For the exact newform coefficients $a_n$, the functional equation gives $$ L(E,1)=2\sum_{n\ge1}\frac{a_n}{n}t^n,\qquad t=e^{-2\pi/\sqrt N}. $$ The bound $|a_n|\le d(n)\sqrt n\le2n$ implies that the absolute tail after $M$ terms is at most $4t^{M+1}/(1-t)$. Exact coefficients through $M=2{,}000{,}000$, evaluated with 128-bit Arb intervals, give partial sum $15.835296483717517\ldots$ and tail bound below $0.005565019931056$. Hence $$ 15.8297314637864