Help me understand this proof from Artin's Algebra about algebraic integers in $\mathbb Q[\sqrt d]$.

Question

In Case $1$, where does he use $d \equiv 2$ or $3$ mod $4$?

In Case $2$, why is $b^2 d \in \mathbb Z +\frac{1}{4}$ if and only if $d \equiv 1 \pmod 4$?

Answer 1

In case 1, he's not using the fact that $d\equiv 2, 3\mod 4$. What he's doing is saying that case 2 can only happen if $d\equiv1\mod 4$. Case 1 can still happen; in fact, $a=b=0$ is an integer no matter what ring you're in.

Now case 2 can only happen for $d\equiv1\mod 4$ because (as he shows) you need $b^2d$ to be $\frac{1}{4}$ plus an integer. Setting $b=\frac{1}{2}+u$ for integer $u$, squaring and multiplying by $d$ shows that the fractional part of $b^2d$ is the same as the fractional part of $\frac{d}{4}$. Since we want it to be $\frac{1}{4}$, we need $d\equiv1\mod 4$.