Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 1 / b / iv / Solution

Modulo $3$, the affine points areEach equals its own inverse because $-1-1=1$ in ${\mathbb{F}}_3$. Thuswhich is noncyclic. The reductions of $P$ and $Q$ are the distinct nonzero points $(0,1)$ and $(1,1)$, so they form a basis.

If $mP+nQ=O$, reduction modulo $3$ shows that $m$ and $n$ are even. Write $m=2m_1$ and $n=2n_1$. Then $m_{1}P+n_{1}Q$ is a rational point of order dividing two, and part (iii) makes it zero. Repeating the argument shows that $m$ and $n$ are divisible by every power of two, so $m=n=0$. Therefore