 Solution

If $p\mid D$, square-freeness gives the cuspidal reduction $y^2=x^3$. Its nonsingular group is isomorphic to the additive group ${\mathbb{F}}_p$, so it is cyclic of order $p$.

Suppose $p\nmid D$. The reduction is an elliptic curve. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is a quadratic nonresidue. Pairing $x$ with $-x$ in the quadratic-character sum and usingshows that the two contributions cancel. Therefore $\#\tilde{E}({\mathbb{F}}_p)=p+1$. The three roots $0,D,-D$ are distinct, so the full 2-torsion is rational. A cyclic group has at most two elements killed by $2$, hence $\tilde{E}({\mathbb{F}}_p)$ is noncyclic. Thus