 Solution

Use the customary squarefree convention for the quadratic radicand $d$. For $d\equiv 2,3(\mathrm{mod}4)$, the ring of integers of a number field is $\mathbb{Z}[\sqrt{d}]$, soThe ideal $(p)$ is prime exactly when this quotient is an integral domain. A polynomial quotient of a field is an integral domain exactly when the defining polynomial is irreducible. This proves the assertion, including the possibility of ramified primes where the reduction has a repeated root.

Squarefreeness is necessary for the printed assertion as written: $d=18\equiv 2(\mathrm{mod}4)$ defines the same field as $d=2$. The prime $3$ is inert in $\mathbb{Q}(\sqrt{2})$ because $X^2-2$ is irreducible over ${\mathbb{F}}_3$, although $X^2-18=X^2$ is reducible. Thus the result requires a squarefree quadratic radicand.