Let $y^2=x^3$ the affine cubic in $\mathbb{A}^2(k)$ where $k$ is a finite field with cardinal $q=\# k$. I'm trying to express the number of points of the cubic in terms of $q=\# k$.
I'm trying this: there is of course $(x,y)= (0,0)$. If $x,y$ are non zero, then there are $1+2 \frac{q-1}{2}=q$ quadratic residues $x$.
To find the number of solutions of the equivalent projective version of the curve $y^2z=x^3 \in \mathbb{P}^2(k)$, I find $q^2+q+1$ coordinate points. Do you think it's correct ?
Do you know how I can find the number of affine points of the curve $y(y-1)=x^3 \in \mathbb{A}^2(k)$ or do you know a method I could use ?
I thank you for any suggestions.