Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2008 / B / 6 / iv

Write the equation as $y^2=x^3-5x^2+4x$. Its duplication formula uses slope $m=(3x^2-10x+4)/(2y)$ and $x(2P)=m^2+5-2x$. At the three listed points the slopes are respectively $i$, $-1+i\sqrt{3}$, and $2+\sqrt{3}$. Substitution givesThese are nonidentity points of order two, so each listed point has exactly order four.

For $p\ne 2,3$ the three distinct roots zero, one and four give the full two-torsion subgroup of order four over ${\mathbb{F}}_p$. At least one of $-1,3,-3$ is a square: if neither of the first two is, their product is. If $-1$ is a square, the first listed point exists over the field. If three is a square, the third exists. If $-3$ is a square, the second can be written $(1-\sqrt{-3},3+\sqrt{-3})$ and exists without requiring the two separate square roots. The duplication computations remain valid after reduction; their ordinates do not vanish unless $p=2$ or three. Thus a rational two-torsion point has a half over the field. Together with a different two-torsion point this generates a subgroup of order eight. Consequently $\boxed{8\mid \#E({\mathbb{F}}_p)}$ for every stated prime.

Over the rationals there are already four two-torsion points, including the identity. Counting at five gives eight points: the abscissas zero, one and four give one ordinate each, two and three give two each, and there is the identity. Thus rational torsion has order dividing eight.

There are no rational points of order four. For a curve $y^2={\prod }_j(x-e_j)$, translating $e_i$ to zero and using the duplication formula shows that a half of $(e_i,0)$ must satisfy $(x-e_i{)}^2=(e_i-e_j)(e_i-e_k)$. For the three roots here, the candidate abscissas are $\pm 2$, $1\pm \sqrt{-3}$, and $4\pm 2\sqrt{3}$. The last two pairs are nonrational; at $x=2,-2$ the ordinate squares are $-4,-36$, also impossible over $\mathbb{Q}$. A group of order eight containing this two-torsion subgroup would contain an element of order four, because an elliptic curve has only four points killed by two. Hence