We let $n\in \mathbb{Z}$, $n>2$ and $d=n^2-2$. We want to show that $n^2-1+n\sqrt{d}$ is a unit of $\mathbb{Z}[\sqrt{d}]$.
My initial idea was to consider the induced norm $N:R\to\mathbb{Z}$, given by $N(a+b\sqrt{d})=a^2-db^2$. We know that if $R$ is the ring of integers of some quadratic number field, and $\alpha \in R$, then $N(\alpha)=\pm1 \Leftrightarrow \alpha \in R^{\times}$, and as $N(n^2-1+n\sqrt{d})=(n^2-1)^2-dn^2=(n^2-1)^2-(n^2-2)n^2=n^4-2n^2+1-n^4+2n^2=1$, we must have $n^2-1+n\sqrt{d}\in \mathbb{Z}[\sqrt{d}]$. I then realised that $\mathbb{Z}[\sqrt{d}]$ is the ring of integers of a quadratic number field if and only if $d\not\equiv1\ (\textrm{mod}\ 4)$, so my proof does not apply for the case of $d\equiv 1\ (\textrm{mod}\ 4)$. I then decided to try to find a proof that proves the general case, perhaps using fundamental units, seeing as the rings under consideration all have $d>2$. Having been unable to make any meaningful progress in this department, I decided to consult the community.
All help would, as always, be highly appreciated.
Best Answer
Why not prove it directly? Multiply it by its conjugate and see you get $1$. $$\begin {align} \left(n^2-1+n\sqrt d\right)\left(n^2-1-n\sqrt d\right)&=\left(n^2-1\right)^2-n^2d\\ &=\left(n^2-1\right)^2-n^2\left(n^2-2\right)\\&=1 \end {align}$$