# Fundamental unit of $\mathbb Q(\sqrt{n^2-1})$

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 $$\mathbb Q (\sqrt{n^2-1})$$ ?

• Can you explain what is fundamental unit? I don't know the definition. – S.D. Nov 8 at 13:43
• You might need $n^2-1$ to be squarefree. – 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 $$\mathbb{Z}[\sqrt{n^2-1}]=\mathbb{Z}[\varepsilon]$$.