The ring of integers of a quadratic field

Write $\alpha=a+b\sqrt d$ with $a,b\in\mathbb Q$. If $b=0$, $\alpha=a\in\mathbb Q$ is an algebraic integer iff $a\in\mathbb Z$ (rational root theorem), which is the case $b=0$ of $\mathbb Z[\sqrt d]$. If $b\neq0$, the minimal polynomial of $\alpha$ over $\mathbb Q$ is $x^2-2ax+(a^2-db^2)$ (its trace is $2a$, its norm $a^2-db^2$), so $\alpha$ is integral iff both $2a\in\mathbb Z$ and $a^2-db^2\in\mathbb Z$.

Write $u=2a\in\mathbb Z$. From $a^2-db^2\in\mathbb Z$, multiplying by $4$ gives $u^2-4db^2\in4\mathbb Z\subset\mathbb Z$, so $4db^2\in\mathbb Z$. Write $b=p/q$ in lowest terms ($\gcd(p,q)=1$); then $4db^2=4dp^2/q^2\in\mathbb Z$ forces $q^2\mid4dp^2$, and since $\gcd(p,q)=1$ this forces $q^2\mid4d$. Because $d$ is squarefree, the largest perfect square dividing $4d$ is at most $4$, so $q^2\mid4$, i.e. $q\in\{1,2\}$; either way $v:=2b\in\mathbb Z$.

Now substitute $u=2a$, $v=2b$ (both integers) into $a^2-db^2\in\mathbb Z$: this becomes $\tfrac{u^2-dv^2}{4}\in\mathbb Z$, i.e. $u^2\equiv dv^2\pmod4$. Squares mod $4$ are only $0$ (even) or $1$ (odd). If $v$ is even, $dv^2\equiv0\pmod4$ forces $u$ even too (since $u^2$ odd would give $1\not\equiv0$); so $u,v$ both even, meaning $a,b\in\mathbb Z$ — the "trivial" solution, always available, giving $\mathbb Z[\sqrt d]\subseteq\mathcal O_K$ always.

If $v$ is odd, $dv^2\equiv d\pmod4$ (since $v^2\equiv1$), so we need $u^2\equiv d\pmod4$. If $d\equiv2,3\pmod4$ (i.e. $d\equiv2,3\pmod4$), no square $u^2\in\{0,1\}\pmod4$ can equal $2$ or $3\pmod4$, so this case is impossible — only the trivial $u,v$ both even survives, giving exactly $\mathcal O_K=\mathbb Z[\sqrt d]$. If instead $d\equiv1\pmod4$ ($d\equiv1\pmod4$), we need $u^2\equiv1\pmod4$, satisfied whenever $u$ is odd — and since $v$ is already odd, $u\equiv v\pmod2$ is exactly the condition for $\alpha=\tfrac{u+v\sqrt d}2$ with $u,v$ both odd, which is precisely $\mathbb Z+\mathbb Z\cdot\tfrac{1+\sqrt d}2$ (check: $\tfrac{1+\sqrt d}2$ itself has $u=v=1$, both odd), so in this case $\mathcal O_K=\mathbb Z\!\left[\tfrac{1+\sqrt d}2\right]$, strictly larger than $\mathbb Z[\sqrt d]$ (index $2$).