 Degree-form proof of the Hasse bound

For Frobenius $\pi$, put $a=\mathrm{tr}(\pi )$ and note that $\mathrm{deg}\pi =q$. Nonnegativity of the quadratic formfor all integers $m,n$ forces its discriminant to be nonpositive, so $a^2\le 4q$. Since $\#E({\mathbb{F}}_q)=\mathrm{deg}(1-\pi )=q+1-a$, this is the Hasse bound.