Show that for a unit element in a ring the following holds 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?

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 $b > 0$.