If we define the norm on $\mathbb{Z}[\sqrt{-3}]$ to be $N(\alpha)=a^2+3b^2$, then how do we use this norm to find all the units in $\mathbb{Z}[\sqrt{-3}]$.
I know what a unit is, so we are looking for all the invertible elements in $\mathbb{Z}[\sqrt{-3}]$ e.g.
$(a+b\sqrt{-3})(c+d\sqrt{-3})=1$ and so we have $(a-b\sqrt{-3})(c-d\sqrt{-3})=1$ which gives: $(a^2+3b^2)(c^2+3d^2)=1$ so the only units are $1,-1$ but I'm not sure how to use the norm to show this (or have I as used it without realising?)
Thanks very much for any help