genus of a curve $y^p=-x^2+x$ for odd prime $p$

Question

Let $p$ be a odd prime. I want to prove the genus of a curve $y^p=-x^2+x$ is $(p-1)/2$. I know genus of hyper elliptic curve $y^2=a_{2g+2}x^{2g+2}+a_{2g+1}x^{2g+q}+・・・+a_1x +a_o$・・・① has genus $g$, 　Thus If we could find some transformation into　the form ① i.e. $y^2=a_{p-1}x^{p-1}+・・・+a_0$, we are done. Another approach is also appreciated.

Answer 1

You can rewrite the equation as $$(x-1/2)^2=-y^p+1/4.$$ With an obvious coordinate change, this is $y^2=(p$-th degree poly.$)$, which is a double cover of the projective line branching $p+1$ points (zero of RHS plus infinity). Therefore, $g=(p-1)/2$ by the genus formula of hyperelliptic curves.