 Solution

For $f=y^2-x^3-17$, the partials are $f_x=-3x^2$ and $f_y=2y$. If $p\notin \{2,3\}$ their simultaneous vanishing forces $x=y=0$, and that point lies on the curve exactly when $p=17$. In characteristic two, the equations force $x=0$ and then $y^2=1$, so $y=1$. In characteristic three they force $y=0$ and then $x^3=1$, so $x=1$. Consequentlyand there are no singular points for any other prime.