The <zeta function of an elliptic curve over a finite field> is the formal power series
$$
Z_E(T)=\exp\left(\sum_{r\geq1}\#E(\mathbb F_{p^r})\frac{T^r}{r}\right).
$$
The proof of Hasse's theorem gives the characteristic equation $\pi^2-[a]\pi+[p]=0$. If $\alpha,\beta$ are the roots of $X^2-aX+p$, then the <elliptic-curve point count over a finite field> is
$$
\#E(\mathbb F_{p^r})=p^r+1-\alpha^r-\beta^r.
$$
Using $\exp(\sum_{r\geq1}u^rT^r/r)=(1-uT)^{-1}$ therefore gives
$$
Z_E(T)=\frac{(1-\alpha T)(1-\beta T)}{(1-T)(1-pT)}
=\frac{1-aT+pT^2}{(1-T)(1-pT)}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 125 / 1 / b / 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 $\#E(\mathbb F_{p^r})=p^r+1-a_r$.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 / 2 / b / Solution

If $\alpha,\beta$ are the roots of $T^2-aT+q$, where $a=q+1-\#E(\mathbb F_q)$, then
$$
\#E(\mathbb F_{q^r})=q^r+1-\alpha^r-\beta^r.
$$


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