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For $E:y^2=x^3+a{x}^2+bx$ with rational 2-torsion $(0,0)$, the quotient by that point is the two-isogenous curveThe two-isogeny descent maps a nonexceptional point to the square class of its $x$-coordinate, with $(0,0)$ mapping to $[b]$. The images are finite collections of squarefree divisors of $b$ and $a^2-4b$, determined by testing the associated homogeneous quartics for rational points. If their orders are $2^s$ and $2^{s^{\prime }}$, thenwhich determines the Mordell-Weil rank $r$.

The method requires a rational 2-isogeny, and deciding whether every locally soluble quartic is globally soluble can be difficult. Computing only local conditions gives a 2-isogeny Selmer group and hence an upper bound; a nontrivial Tate-Shafarevich group can make that bound strict. Even after finding the rank, a separate saturation and point search may be needed to find generators.

Solved by gpt-5.6-sol high.

For this curve, the square-class image in the first two-isogeny descent is contained in $\{1,-1,11,-11\}$. All four classes occur: $Q=(-1,1)$ gives $-1$, $T=(0,0)$ gives $11$, and $Q+T$ gives $-11$. The isogenous curve isIts image is contained in $\{1,-1,5,-5\}$. Negative $x$ cannot occur because $x(x^2-26x+125)<0$ for $x<0$, while $1$ and $5$ are represented by the identity and $(0,0)$. Thus the two image orders are four and two, andgives $r=1$.

The point $P$ has order three because $2P=-P$, and $T$ has order two, so the rational torsion contains a cyclic subgroup of order six. At the good primes $7$ and $13$, direct point counting givesReduction bounds the rational torsion order by their greatest common divisor, namely six, so this is all the torsion. The structure theorem for finitely generated modules over a principal ideal domain now givesThus one may take $d_1=1$, $d_2=6$, and $r=1$.

Solved by gpt-5.6-sol high.

Foruse two-isogeny descent throughThe square-class maps send a nonexceptional point to the class of its $x$-coordinate. On $E$, possible classes are $\pm 1,\pm 7$; the defining quartics and positivity exclude the negative classes, while $(1,3)$ and $(7,21)$ realize $1$ and $7$. Thus the image has order two. On $E^{\prime }$, the possible classes are $\pm 1,\pm 3$; the classes $1$ and $-3$ occur, while the quartics for $-1$ and $3$ have no primitive solution modulo $16$. Hence this image also has order two. The two-isogeny descent formulatherefore gives $r=0$.

The displayed curve has good reduction at $5$ and $11$, where direct counting givesReduction injects rational torsion of order prime to these characteristics, so its order divides $\gcd (6,18)=6$. We already have the six distinct pointsSince the rank is zero, these are all the rational points and $E(\mathbb{Q})\cong \mathbb{Z}/6\mathbb{Z}$.

Solved by gpt-5.6-sol high.