Cubic splitting law: $p$ splits completely in $\mathbb{Q}[x]/(x^3+dx+e)$ iff $(\Delta/p)=1$

Let $D,E\in\mathbb{Z}$ with $m(x)=x^3+Dx+E$ irreducible over $\mathbb{Q}$, let $K=\mathbb{Q}[x]/(m)$, and let $\Delta=-4D^3-27E^2$ be the discriminant. Let $p$ be an odd prime with $p\nmid\Delta$, and suppose $p$ divides the binary form $F(a,b)=a^3+Dab^2+Eb^3$ for some coprime integers $a,b$.

Then $p$ splits completely in $K$ — there are exactly three distinct primes of $\mathcal{O}_K$ above $p$ — if and only if $\Delta$ is a quadratic residue mod $p$, i.e. the Legendre symbol $\left(\frac{\Delta}{p}\right)$ equals $1$.

This is the classical Dedekind splitting criterion for a cubic field, in the exact form the Collapsible Cubics mission needs. The hypothesis $p\mid F(a,b)$ with $\gcd(a,b)=1$ is what supplies a root of $m$ mod $p$; the hypothesis $p\nmid\Delta$ makes $p$ unramified and (via the index-discriminant relation) guarantees $p$ does not divide the index $[\mathcal{O}_K:\mathbb{Z}[\theta]]$, so Kummer-Dedekind applies. The count is stated with Nat.card so that no finiteness instance is needed in the statement.