Effective divisor of degree $2$ over finite field and number of points in Jacobian I am given a hyperelliptic genus $2$ curve $C\dots y^2=f(x)$ over the finite field $\mathbb{F}_q$ and I need to prove that 
$$\#J(\mathbb{F}_q)=\frac{1}{2}\#C(\mathbb{F}_q)^2+\frac{1}{2}\#C(\mathbb{F}_{q^2})-q.$$ 
I have the theorem that states that for every point $P$ on Jacobian and fixed canonical divisor $D_a$ on $C$ there is unique effective degree $2$ divisor $D$ such that $P=[D_a]-[D]$. 
So, I need to count the number of effective $\mathbb{F}_q$-rational divisors of degree $2$.

My idea: So I have that divisor of degree $2$ has a form $D=P_1+P_2$  where these are $\mathbb{F}_q$-rational points on $C$ (have degree $1$) or it has a form $D=P+\sigma(P)$ where $P$ is point of degree $2$, ie. point that is not $\mathbb{F}_q$-rational but is $\mathbb{F}_{q^2}$-rational and $\sigma$ is element in $Gal(\mathbb{F}_{q^2}/\mathbb{F}_q)=\{\sigma, id\}$.

Now, number of such divisors is equal to $\frac{\#C(\mathbb{F}_q)^2-\#C(F_q)}{2}+\#C(\mathbb{F}_q)+\frac{\#C(\mathbb{F}_{q^2})-\#C(\mathbb{F}_q)}{2}=\frac{1}{2}(\#C(\mathbb{F}_{q^2})+\#C(\mathbb{F}_{q})^2)$. 
Am I right?
Now, I can show that since $D_a=(a,\sqrt{f(a)})+(a,-\sqrt{f(a)})$, for every other $b\in \mathbb{F}_q$ we have $[D_a]=[D_b]$ so I have to substract $(q-1)$. I  am one point short?

"Missing point" is zero point $0$.
