Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 2 / b / Solution

For $E:y^2=x^3+3$, direct counting givesNeither group order is divisible by $17$, so neither group contains a point of order $17$.

At $p=17$ the Frobenius trace is zero. The elliptic-curve point count over a finite field has trace recurrenceFor every $n\ge 1$, this order is congruent to one modulo $17$. Consequently $E({\mathbb{F}}_{17^n})$ has no point of order $17$ for any $n$.

At $p=31$, the trace is $a=31+1-43=-11$. On $E[17]$, Frobenius has characteristic polynomialIts discriminant is $14$, a nonsquare in ${\mathbb{F}}_{17}$, so its two distinct eigenvalues lie in ${\mathbb{F}}_{17^2}^{\times }$. Their orders divide $17^2-1=288$, whence ${\pi }^{288}=1$ on $E[17]$. Thus all of $E[17]$ is rational over ${\mathbb{F}}_{31^{288}}$, and in particular a point of order $17$ exists over some extension with $n\ge 2$.