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The elliptic-curve discriminant of $y^2=x^3-12x^2+20x$ is $16\cdot 20^2\cdot (144-80)=409600=2^{14}{5}^2$. Thus three and eleven are primes of good reduction. Counting the reduced points gives four points over ${\mathbb{F}}_3$ and sixteen over ${\mathbb{F}}_{11}$. For example, the numbers of affine points above successive $x=0,1,2$ modulo three are $1,1,1$. Modulo eleven, for $x=0,\dots ,10$, they are $1,2,1,2,0,0,2,2,2,2,1$; add the point at infinity in each case.

The reduction of torsion points on an elliptic curve is injective on torsion prime to the residue characteristic. The three-primary rational torsion injects into the group of order sixteen at eleven, so is trivial; likewise the eleven-primary torsion injects into the group of order four at three. Every other primary component injects at both primes. Therefore the rational torsion subgroup has order dividing four. All four rational 2-torsion points are visible, so

The rational point $P=(1,3)$ lies on the curve. The tangent slope in the elliptic-curve addition formula is $(3-24+20)/6=-1/6$, giving $2P=(361/36,-323/216)$. The integral substitution $X=x-4$ changes the equation to $y^2=X^3-28X-48$; the doubled point has $X=217/36$. The Nagell–Lutz theorem says that a rational torsion point on this integral short equation has integral coordinates, so $2P$, and hence $P$, is nontorsion. An infinite-order point is $(1,3)$.

We prove the rank bound by two-isogeny descent. For $a=-12,b=20$, the two-isogeny formula gives the partnerand maps $\varphi :E\to E^{\prime }$, $\psi :E^{\prime }\to E$ with $\psi \varphi =[2]$. The two-torsion square-class homomorphism $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\ast }/({\mathbb{Q}}^{\ast }{)}^2$ is $\alpha (x,y)=[x]$ away from $O,(0,0)$, with values $1$ and $[20]=[5]$ at those exceptional points. Its kernel is $\psi E^{\prime }(\mathbb{Q})$. The corresponding map ${\alpha }^{\prime }$ on $E^{\prime }$ has kernel $\varphi E(\mathbb{Q})$.

The prime-support bound in two-isogeny descent can be checked directly here. For a prime $\ell \nmid 20$, if $v_{\ell }(x)>0$ then $x^2-12x+20$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even; if $v_{\ell }(x)<0$, the $x^3$ term has strictly least valuation and $2v_{\ell }(y)=3v_{\ell }(x)$ is even. Thus the square class of $x$ is supported only on two and five. Also $x(x-2)(x-10)\ge 0$ forces $x\in [0,2]\cup [10,\infty )$, so nonzero $x$ is positive. The possible classes are $1,2,5,10$, and the four torsion points provide all of them:Consequently $\#\alpha (E(\mathbb{Q}))=4$.

The same valuation argument on $E^{\prime }$ gives possible signed classes $1,-1,2,-2$, since its coefficient $b^{\prime }=64$ has only the prime two. The identity gives class $1$, and $(-4,8)\in E^{\prime }(\mathbb{Q})$ gives class $-1$. For $d=\pm 2$, the two-isogeny descent quartic iswith integral coprime $U,V$ after scaling a rational solution. If $U$ is odd, its right side is $2$ or $10$ modulo sixteen for $d=2$, and $14$ or $6$ for $d=-2$, according to the parity of $V$. None is a square. If $U=2u$ is even, $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$. The bracket is odd for either parity of $u$, so its 2-adic valuation is five, again impossible for a square. The signed divisor-two obstruction for an isogeny covering excludes both classes. Hence ${\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))=\{1,-1\}$.

For completeness, the index correction in the two-isogeny index formula over a number field is one: $\ker \psi =\{O,(0,0)\}$ and $\varphi ((2,0))=(0,0)$, so the whole kernel lies in $\varphi E(\mathbb{Q})$. ThusThe Mordell-Weil theorem and the four rational 2-torsion points identify this index as $2^{r+2}$. Therefore $2^{r+2}=8$ andThis is the rank-one elliptic curve with roots zero two and ten. We have exhibited a point of infinite order but have not needed to prove that it generates the free part.

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 $2^r=4\cdot 2/4=2$, so the rank is one. The argument does not determine the index of the subgroup generated by $(1,3)$.