hyperelliptic curve Please help me to solve this question:
Let $H$ be a hyperelliptic curve over $\mathbb{F}_{103}$ given by the equation $ y^2 = x^5+1$. let $J$ be the jacobian of $H$ defined over $\mathbb{F}_{103}$. Show that  # H($\mathbb{F}_{103^2}$) = # J($\mathbb{F}_{103}$).
 A: Magma says that $\# H(\mathbb{F}_{103})=104$ and $\# J(\mathbb{F}_{103})=10610$.
Here is the code I used:
P<x> := PolynomialRing(GF(103));
C := HyperellipticCurve(x^5+1);
#C;
J:=Jacobian(C);
#J;

A: Let us consider a more general problem. Let $p$ be an odd prime, $p \not\equiv 1$ modulo ${5}$, and let $H$ be the hyperelliptic curve defined by $y^2 = x^5 + a$ over $\mathbb{F}_p$, where $a \not=0$. Suppose that $p \equiv 2,3$ modulo $5$. Then
$$ \#H(\mathbb{F}_{p^2}) = \#J(\mathbb{F}_{p}) = p^2 + 1.$$
See p. 571 in Choie, Jeong, Lee - Supersingular hyperelliptic curves of genus 2 over finite fields.
One can prove this fact as follows. Denote by $f(x) \in \mathbb{F}_p[x]$ the polynomial $x^5+a$, which defines $H$. Put $q=p^i$, $i\in\{1,2\}$. As $f(x)$ is a permutation polynomial of $\mathbb{F}_{q}$, the equation $y^2=f(x)$ has exactly $2 \cdot (q-1)/2+1 = q$ solutions in $\mathbb{F}_{q}$. Also, there is only one rational point at infinity on $H$. Hence, $N_1(H\otimes \mathbb{F}_{q})=q+1$. We have just obtained that $N_1(H)=p+1$ and $N_2(H)=p^2+1$.
Recall than the Weil polynomial $P(t)$ of the curve $H$ is
$$
P(t) = p^2 t^4 + p a_1 t^3 + a_2 t^2 + a_1 t + 1,
$$
where $a_1 = N_1(H)-(p+1)$, $a_2 = (N_2(H)-a_1^2-(p^2+1))/2$. We have
$$
P(t) = p^2 t^4 + 1.
$$
Finally, it suffices to note that the number of rational points on the Jacobian of $H$ equals $P(1)$. Thus, $\# J(\mathbb{F}_{p}) = p^2+1$.
