We have to Show that $\mathbb{Z}[\sqrt{n}]$ is a Euclidean domain for $n = −1,−2, 2, 3$, where $\mathbb{Z}[\sqrt{n}] =\{a + b\sqrt{n} :a,b \in \mathbb{Z}\}$
Now for $ n=-1, \text{the ring} \ D = \mathbb{Z}[i]$ of Gaussian integers is a Euclidean domain with Euclidean norm function $\delta$ defined by $\delta(m+in) = m^2 +n^2 \ \forall m,n (\text{not both zero}) \in \mathbb{Z}$. To define division algorithm let $\alpha,\beta \in D$ and $ \beta \neq 0$. Then $\frac{\alpha}{\beta} = a + ib$, where $a, b$ are rational numbers. Thus there exist integers $m_0, n_0$ such that $|m_0 − a| \leq \frac{1}{2}$ and $|n_0 − b| \leq \frac{1}{2}$ Hence $a +ib = (m_0 +in_0)+\epsilon_1 +i\epsilon_2$, where $|\epsilon_1| \leq \frac{1}{2}$ and $|\epsilon_2| ≤ \frac{1}{2}$ $\implies \alpha = (m_0 +in_0)\beta + (\epsilon_1 + i \epsilon_2)\beta$. Now $ \alpha,\beta \in D$ and $m_0 + in_0 \in D \implies \alpha − (m_0 + in_0)\beta \in D \implies (\epsilon_1 + i\epsilon_2)\beta \in D \implies \alpha = q\beta + r, \text{where} \ r = (\epsilon_1 + i\epsilon_2)\beta \ \text{and} \ q = (m_0 + in_0)$. Now $\delta(r) = \delta((\epsilon_1 +i\epsilon_2)\beta) = |{\epsilon_1}^2+{\epsilon_2}^2|{|\beta|}^2 \leq ( \frac{1}{4} + \frac{1}{4}){|\beta|}^2 = \frac{1}{2} \delta(\beta) \lt \delta(\beta) \implies \delta: D → \mathbb{N}$ is a Euclidean function.
Similarly we can try this procedure Exactly to show that $\mathbb{Z}[\sqrt{n}]$ is Euclidean domain for $n=-2,2,3$.
Now I have a question, is this the procedure to do it or there is general procedure to find n such that $\mathbb{Z}[\sqrt{n}]$ is Euclidean domain for $n=-1,-2,2,3$?