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

Forthe negative of $(x,y)$ is $(x,-y-1)$. At $P=(0,1)$ the tangent slope isThe tangent is $y=1-4x/3$, so the addition formulas giveThusThe line through $P=(0,1)$ and $Q=(-2,1)$ is $y=1$. Its third intersection has $x=2$, and reflection under $y\mapsto -y-1$ gives

Direct enumeration of the solutions of $y^2+y=x^3-4x+2$ givesafter adjoining the point at infinity. Therefore

The discriminant is $1909=23\cdot 83$, so $3,5,7$ are primes of good reduction of an elliptic curve. The reduction of torsion points on an elliptic curve injects the prime-to-$p$ torsion into $\tilde{E}({\mathbb{F}}_p)$.

If a prime $\ell$ divided the order of $E(\mathbb{Q}{)}_{\mathrm{tors}}$, its primary subgroup would inject at every one of $3,5,7$ except possibly when $p=\ell$. For $\ell =3$, the counts at $5$ and $7$ have greatest common divisor one; for $\ell =5$ use the counts at $3$ and $7$; for $\ell =7$ use those at $3$ and $5$; every other $\ell$ would divide all three counts. Each possibility is excluded. Hence

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