= Singular reduction of the standard Mordell equation
{title2=$y^2=x^3+c$}
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.
Back to article page
