Consider the quadratic integer ring $A=\Bbb Z[\sqrt-2]$. Show that $U(A)=\{-1,1\}.$
$\bf{Remark:}$ $U(A)$ indicates the set of inversible elements of $A$.
I know that $\alpha = a + b\sqrt{-2}$ is inversible iff $N(\alpha)=a^2+2b^2=1$.
but I don’t know how to go on. Can someone help me?
$a^2+2b^2=1 \Rightarrow 2b^2=1-a^2=(1-a)(1+a)$
$2b^2 \geq 0 \Rightarrow -1 \leq a \leq 1$
but we know the fact that $a$ is an integer so
$ -1 \leq a \leq 1 $ and $a \in \Bbb Z \Rightarrow $ $a=-1$ or $a=1$ or $a=0$
let $a=0$ then
$2b^2=1$ which does not have an integer solution so $a \neq 0$
$a=1$ , $b=0$
$a=-1$ , $b=0$ are only solutions