= Solution
Let $\pi$ be the <Frobenius isogeny of an elliptic curve> and put
$$
a=p+1-\#E(\mathbb F_p).
$$
Part (a) gives
$$
\pi^2-[a]\pi+[p]=0.
$$
Let $\alpha,\beta$ be the two roots of $T^2-aT+p$. The points over $\mathbb F_{p^r}$ are exactly $\ker(1-\pi^r)$. Since the differential of $1-\pi^r$ is the identity, this isogeny is separable, and therefore
$$
\#E(\mathbb F_{p^r})=\deg(1-\pi^r).
$$
Using $\deg\psi=(1-\psi)(1-\widehat\psi)$ in the endomorphism algebra gives the <elliptic-curve point count over a finite field>
$$
\#E(\mathbb F_{p^r})
=p^r+1-\alpha^r-\beta^r.
$$
Equivalently, if $a_r=\alpha^r+\beta^r$, then
$$
a_0=2,qquad a_1=a,qquad a_r=aa_{r-1}-pa_{r-2},
$$
and $\#E(\mathbb F_{p^r})=p^r+1-a_r$.
Back to article page
