The invertible elements of $\mathbb{Z}(\sqrt{-2})$ 
Let $\mathbb{Z}(\sqrt{-2}):=\{a+b\sqrt{2}i; \, a,b \in \mathbb{Z} \}$. Find the multiplicative inverses.

My attempt:
We write $(a+b \sqrt{2}i)(c+d \sqrt{2}i)=1$
It trivially follows:
$$ac-2bd-1=0$$
$$ad+bc=0$$
I'm stuck at this point since there are too many variables to make sense of anything. Any help would be appreciated.
 A: $a+b\sqrt{-2}$ is a unit if and only if $N(a +b\sqrt{-2})=a^2+2b^2$ is a unit in $\mathbf Z$. So $a=\pm 1,\ b=0$. 
More generally, in the ring of integers of an imaginary quadratic field, the only units are $1$ and $-1$, except for the ring of Gauß integers: $\{1, -1, \mathrm i,-\mathrm i\}$, and the ring of Eisenstein integers: $\,\{1,-1,\omega, -\omega, \omega^2,-\omega^2\}$, where $\,\omega=\mathrm e^{\frac{2\mathrm i\pi}3}$.
A: Without the use of norms you proceed as follows:
Having derived your system of equations
$ac-2bd=1$
$bc+ad=0$
you simply treat this as a linear system for $c,d$ with $a,b$ as parameters.  Thus solving for $c$ and $d$ we get:
$c=a/(a^2+2b^2)$
$d=-b/(a^2+2b^2)$
If $b$ is nonzero at all, then $d$ is nonzero and, for the integer domain, you must then have $d^2\ge 1$.  Then $b^2\ge a^2+2b^2\ge 2b^2$.  $\color{blue}{\text{This is contradictory for all nonzero }b,}$ so only $b =0$ remains.  Therefore the only possible units in $\mathbb{Z}[\sqrt{-2}]$ are the real ones $\pm1$.
The same proof works for all domains $\mathbb{Z}[\sqrt{-n}]$ with $n\ge 2$, so the only imaginary domain of this type with "nontrivial" units are the Gaussian integers ($n=1$).
