Singular reduction of the standard Mordell equation ($y^2=x^3+c$)

For the displayed integral equation, reduce its polynomial modulo a prime number $p$. Singular points satisfy $y^2-x^3-c=-3x^2=2y=0$. For $p>3$, they exist exactly when $p\mid c$, and the sole point is $(0,0)$. At $p=2$, the sole point over ${\mathbb{F}}_2$ is $(0,\overline{c})$; at $p=3$, it is $(-\overline{c},0)$ because cubing is the identity on ${\mathbb{F}}_3$. For $c=17$, this gives primes $2,3,17$ and points $(0,1),(1,0),(0,0)$ respectively. This concerns the specified plane equation; intrinsic good or bad reduction of an elliptic curve can require choosing a different integral model, especially at primes two and three.