= Rank-one elliptic curve with roots zero two and ten
{title2=$E:y^2=x(x-2)(x-10),\quad\operatorname{rank}E(\mathbb Q)=1$}
This <elliptic curve> has <torsion subgroup> $\{O,(0,0),(2,0),(10,0)\}\cong(\mathbb Z/2\mathbb Z)^2$ and the infinite-order point $(1,3)$. Good-reduction point counts at three and eleven are four and sixteen, which bound the torsion order by four using <reduction of torsion points on an elliptic curve>. Doubling $(1,3)$ gives $(361/36,-323/216)$; after the shift $X=x-4$ to $y^2=X^3-28X-48$, its nonintegral coordinate proves nontorsion by the <Nagell–Lutz theorem>. The <two-isogeny formula> gives the partner $Y^2=X^3+24X^2+64X$. The first <two-torsion square-class homomorphism> has image $\{1,2,5,10\}$, while the second has image $\{1,-1\}$ by the <signed divisor-two obstruction for an isogeny covering>. The <two-isogeny rank formula> therefore gives $2^r=4\cdot2/4=2$, so the rank is one. The argument does not determine the index of the subgroup generated by $(1,3)$.
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