How does the ring of algebraic integers in $\mathbb{Q}[\sqrt{D}]$ depend on $D\mod 4$?

Question

Let $O_K$ be the ring of algebraic integers inside of $\mathbb{Q}[\sqrt{D}]$. Why is it that when $D \equiv 2, 3 \mod 4$ that $O_K = \{x + y\sqrt{D} : x, y \in \mathbb{Z}\}$ and that when $D \equiv 1 \mod 4$ that $O_K = \{x + y\left(\frac{1 + \sqrt{D}}{2}\right) : x, y \in \mathbb{Q}\}$?

Answer 1

The minimal polynomial of $\alpha=a+b\sqrt{D}$ then is given by $$ (x-\alpha)(x-\sigma(\alpha))=x^2-2ax+(a^2-b^2D), $$ where $\sigma(a+b\sqrt{D})=a-b\sqrt{D}$. The coefficients are integers if and only if the above conditions are met. The claim follows now from your previous (deleted) question.