Exercise 1.33

Proof. Let $\alpha =a+\mathit{bi}\in \mathbb{Z}[i]$.

$\text{∙}$ If $\lambda (\alpha )=1$, then $\alpha \bar{\alpha }=1$, where $\bar{\alpha }=a-\mathit{bi}\in \mathbb{Z}[i]$, so $\alpha$ is an unit.

$\text{∙}$ Conversely, if $\alpha$ is an unit, there exists $\beta \in \mathbb{Z}[i]$ such that $\mathit{\alpha \beta }=1$, then $\lambda (\alpha )\lambda (\beta )=1$, where $\lambda (\alpha ),\lambda (\beta )$ are positive integers, hence $\lambda (\alpha )=1$.

So $\alpha =a+\mathit{ib}$ is an unit of $\mathbb{Z}[i]$ if and only if $a^2+b^2=1$. In this case, $|a{\text{|}}^2\le 1$, $a\in \{0,1,-1\}$. If $a=0,b=\pm 1$, and if $a=\pm 1,b=0$, so the only units of $\mathbb{Z}[i]$ are $1,i,-1,-i$. □