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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.

First compute the Mordell-Weil group rank by two-isogeny descent. HereThe two-isogeny formula gives $\varphi :E\to E^{\prime }$. If $u=1+i$, then $u^4=-4$, andis an isomorphism over $K$. Its $x$-coordinate multiplier $u^2$ is a square in $K$.

Use the two-torsion square-class homomorphismFor any prime ideal of the Gaussian integers, if $v(x)>0$, then $x^2+i$ is a unit and $2v(y)=v(x)$; if $v(x)<0$, the term $x^3$ has strictly smallest valuation and $2v(y)=3v(x)$. Hence every valuation of $x$ is even. Since the Gaussian integers form a principal ideal domain, dividing by a square leaves a unit. Their units are $\{1,-1,i,-i\}$, whose square classes are $\{1,i\}$ because $-1=i^2$ and $i$ is not a square in $K$. For the latter assertion, $(a+bi{)}^2=i$ with $a,b\in \mathbb{Q}$ would imply $a^2=b^2$ and $2ab=1$, hence $a^2=1/2$, impossible for rational $a$. ThusBoth classes occur, at $O$ and $T$. This is the unit square-class bound for two-isogeny descent.

Define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly, with ${\alpha }^{\prime }((0,0))=[-4i]=[i]$. Since $\theta$ multiplies nonexceptional $x$-coordinates by a square, and preserves the exceptional classes as well, its image is also $\{1,[i]\}$. The standard kernel identities in two-isogeny descent areHere $\hat{\varphi }$ is the dual isogeny and $\hat{\varphi }\varphi =[2]$. These identities can be checked directly from the formulas: $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$, and conversely a square $x$-coordinate lets the quadratic equation for a preimage be solved using the curve equation. For example, if $x=r^2\ne 0$, the equation for $X$ is $X^2-(4x+2a)X+(a^2-4b)=0$, whose discriminant is $16(x^2+ax+b)=16(y/r{)}^2$; the $Y$-coordinate then follows from the dual formula. The exceptional points satisfy the same completed square-class criterion.

The two-isogeny index formula over a number field keeps track of a small kernel factor:In this case $\ker \hat{\varphi }=\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)$, and $T^{\prime }\notin \varphi E(K)$ because ${\alpha }^{\prime }(T^{\prime })=[i]\ne 1$. Thus $\delta =2$, and the index is $2\cdot 2/2=2$. There is only one nonzero rational 2-torsion point on $E$: the other two would require $x^2=-i$, and $-i$ has the same nonsquare class as $i$. The Mordell-Weil theorem now givesso $r=0$.

It remains to identify all torsion, rather than merely the rank. The elliptic-curve discriminant is $-64i^3$, so the primes $(3)$ and $(2-i)$ have good reduction, with residue characteristics three and five. By the supplied point-count information their reduction groups have orders that are powers of two. The reduction of torsion points on an elliptic curve is injective on prime-to-residue-characteristic torsion. Every odd-primary torsion subgroup therefore injects into a group of two-power order at at least one of these two primes, and must be zero. All torsion is two-primary.

Finally, if a point had order four, its double would be $T$. The elliptic-curve addition formula givesFor $2P=T$ the denominator is nonzero, so $x^2=i$, impossible in $K$. A point of higher two-power order would have a multiple of order four, so it too is excluded. Thus the only torsion points are $O,T$. Together with rank zero,