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The Hasse theorem for elliptic curves states that, for an elliptic curve over ${\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve and put $a=q+1-\#E({\mathbb{F}}_q)$. The fixed points of $\pi$ are $E({\mathbb{F}}_q)$, and $1-\pi$ is separable, soHence the trace of an elliptic-curve endomorphism iswhile $\mathrm{deg}\pi =q$.

The degree on $\mathrm{End}(E)$ is a nonnegative quadratic form. Polarization and the identities for the dual isogeny give, for integers $m,n$,If $a^2>4q$, this real binary quadratic form is indefinite. An open cone on which it is negative contains a nonzero rational point and therefore a nonzero integer point, contradicting nonnegativity of isogeny degree. Thus $a^2\le 4q$, which is exactly the claimed inequality.

The Hasse theorem for elliptic curves states thatLet $\pi$ be the Frobenius isogeny of an elliptic curve and put $a=p+1-\#E({\mathbb{F}}_p)$. The degree on $\mathrm{End}(E)$ is a positive-definite quadratic form, its associated bilinear form gives $\mathrm{tr}(\pi )=a$, and $\mathrm{deg}\pi =p$. Consequentlyfor all integers $m,n$. If $a^2-4p>0$, this real quadratic form is indefinite, so by density of rational slopes it is negative at some nonzero integer pair $(m,n)$, contradicting nonnegativity of the degree. Hence $a^2\le 4p$, which is the claimed bound.

Equality of the two point groups implies equality of their orders. Since ${\alpha }^2+{\beta }^2=a^2-2p$,soIts two integral solutions are $a=p+1$ and $a=-p$. The Hasse theorem for elliptic curves excludes the first for every prime and permits the second only when $p\le 4$. Thus $p=2$ or $p=3$.

Both occur. Over ${\mathbb{F}}_2$, the smooth curve $y^2+y=x^3+x$ has five rational points and trace $-2$. Over ${\mathbb{F}}_3$, the smooth curve $y^2=x^3-x+1$ has seven rational points and trace $-3$. In either case the point-count formula gives $\#E({\mathbb{F}}_{p^2})=\#E({\mathbb{F}}_p)$. Since $E({\mathbb{F}}_p)\subseteq E({\mathbb{F}}_{p^2})$, equal orders give equality of groups.