2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Why not prove it directly? Multiply it by its conjugate and see you get $1$. $$\begin {align} \left(n^2-1+n\sqrt d\right)\left(n^2-1-n\sqrt d\right)&=\left(n^2-1\right)^2-n^2d\\ &=\left(n^2-1\right)^2-n^2\left(n^2-2\right)\\&=1 \end {align}$$

$\endgroup$
  • $\begingroup$ Ah yes. I will accept your answer once seven minutes have lapsed. Thank you for your help. $\endgroup$ – Heinrich Wagner Feb 16 at 17:35
1
$\begingroup$

You showed $N(w) = ww' = 1$ for $w' \in \Bbb Z[\sqrt d]$ which implies $w$ is a unit. You need nothing more.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.