Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 125 / 1

The Hasse theorem for elliptic curves states that every elliptic curve $E/{\mathbb{F}}_q$ satisfiesLet $\pi$ be the Frobenius isogeny and put $a=\mathrm{tr}(\pi )$. Its fixed points are exactly $E({\mathbb{F}}_q)$, so separability of $1-\pi$ givesThe degree of an isogeny is a nonnegative quadratic form on the endomorphism ring, and for all integers $m,n$,If $a^2>4q$, this quadratic polynomial has two real roots and takes a negative value at some rational $m/n$ between them, hence after clearing denominators at some integer pair $(m,n)$. Therefore $a^2\le 4q$, and substituting $a=q+1-\#E({\mathbb{F}}_q)$ proves the bound. This is the degree-form proof of the Hasse bound.

For $x=0,1,2$, the value of $x^3-x-1$ is the nonsquare $2$ in ${\mathbb{F}}_3$. Thus the point at infinity is the only rational point andThe eigenvalues of the Frobenius isogeny are the rootsof $T^2-3T+3$. The elliptic-curve point count over a finite field givesThe last term vanishes exactly when $r\pi /6\equiv \pi /2(\mathrm{mod}\pi )$. Hence