Given is $\epsilon = x+y\sqrt{m}$, where $m \in \mathbb{Z}_{>0}$ and $\epsilon \in \mathbb{Z}[\sqrt{m}]^*$. I have to show that for $\epsilon > 1$ implies $a>0$ and $b>0$
What I tried: I found out (using the norm) that the inverse of $\epsilon$ could be given by $a-b\sqrt{m}$ or $-a+b\sqrt{m}$. If $\epsilon > 1$, then $\epsilon^{-1} < 1$. Using that, I have to be able to show that both $a > 0$ and $b>0$, but I can't fount out how. Can someone please help me with that?
On
The group of units is given by $\{\pm 1\}\times \langle u_1\}$ with a unique element $u_1$, the fundamental unit. For example, for $m=3$, we have $u_1=2+\sqrt{3}$ and we could have $\epsilon=-2-\sqrt{3}$. So $a,b>0$ need not be true in general. It is true, though, for $\epsilon >1$, in particular for the fundamental unit, see here:
From a more elementary viewpoint:
Assume $N(\epsilon)=1$. Then $\epsilon^{-1}=a-b\sqrt{m}$. Since $\epsilon > 1$, $\epsilon > \epsilon^{-1}$, thus $2b\sqrt{m}>0$, hence $b > 0$. Now $\epsilon^{-1}=a-b\sqrt{m} > 0$ thus $a > 0$.
Similarly, if $N(\epsilon)=-1$, $\epsilon^{-1}=-a+b\sqrt{m}$. Since $\epsilon > 1$, $\epsilon^{-1} < \epsilon$, which entails $a > 0$. Since $\epsilon^{-1} > 0$, $b\sqrt{m} > a > 0$ thus $b > 0$.