Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 26 / 1 / c

Use $P=(0,0)$ in the family from part (b). For order four, take $a=b=1$, givingIts discriminant is $-15$, and $2P=(-1,0)$ is a nonzero point equal to its own inverse. Thus $4P=O$ and $2P\ne O$.

For order five, take $a=2,b=1$:The discriminant is $-11$, and $a=b+1$ gives exact order five by part (b).

For order six, take $a=0,b=-2$:Its discriminant is $80$. Here $2P=(2,2)$ and $3P=(1,1)$; negation is $(x,y)\mapsto (x,2-y)$, so $3P$ is nonzero 2-torsion. Neither $2P$ nor $3P$ is $O$, excluding orders two and three, while $6P=O$. All coefficients and points are rational, and the nonzero discriminants certify genuine elliptic curves.