Let $q=r^2$ with $r$ an odd power of a prime. For the curve $X$ defined over $\mathbb F_q$ by the equation
$$x^{r+1}+y^{r+1}+z^{r+1}=0$$
we have the following $\mathbb F_q$-rational intersection points between the curve and the line $z=0$:
$$P_i=[\beta_i:1:0],\ \ \ i=1,\ldots,r+1$$
with $\beta_i\in\mathbb F_q$ and $\beta_i^{r+1}=-1$.
My question is how to prove that for the divisor $D=P_1+\ldots+P_{r+1}$ on $X$ with $P_i$ an arbitrary point we have
$$L(D-P_i)=2$$
and
$$L(D)=3.$$
Note that the divisor $D$ is the intersection divisor of $z=0$ with $X$. I don't know if this is relevant or I was wrong in some argument.
Thanks all for any suggestion.