My first post, so I hope it is not totally idiotic to ask...but: if $n^2-1$ is not a square, why is then $n+\sqrt{n^2-1}$ a fundamental unit of $\mathbb Q (\sqrt{n^2-1})$ ?

  • $\begingroup$ Can you explain what is fundamental unit? I don't know the definition. $\endgroup$ – S.D. Nov 8 at 13:43
  • $\begingroup$ You might need $n^2-1$ to be squarefree. $\endgroup$ – GreginGre Nov 8 at 13:54

I will assume that $n^2-1$ is squarefree. Note that $\varepsilon=n+\sqrt{n^2-1}$ is a unit $>1$. Note that $n^2-1$ is not congruent to $1$ modulo $4$, so the ring of integers of you quadratic extension is $\mathbb{Z}[\sqrt{n^2-1}]$ (this is true because $n^2-1$ is square free). Note that for all $a,b\in\mathbb{Z}$ , we have $a+b\sqrt{n^2-1}=(a-nb)+b\varepsilon$ ,so $\mathbb{Z}[\sqrt{n^2-1}]=\mathbb{Z}[\varepsilon]$.

Now apply Theorem 1 of the following paper: http://iml.univ-mrs.fr/editions/biblio/files/louboutin-JRamanujanMathSoc23(2008).pdf

There is probably a direct argument, though, but I don't have time to think about it right now.


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