# Prove that the fundamental unit of the real quadratic field $\mathbb{Q}(\sqrt{m^2 - 1})$ is $m+\sqrt{m^2 - 1}$ if $m^2-1$ is square free

Prove that the fundamental unit of the real quadratic field $$\mathbb{Q}(\sqrt{m^2 - 1})$$ is $$m+\sqrt{m^2 - 1}$$ if $$m^2-1$$ is square free

I tried taking $$a + b\sqrt{m^2-1}$$ as an arbitrary unit and multiplied it with another unit $$c + d\sqrt{m^2-1}$$ to get some equations to work with, but it is going nowhere.

Any help will be appreciated.

• First of all considering the norm it's easy to see that it is a unit at all. Apr 29, 2021 at 14:32

Assume that $$m \geq 2$$.

If $$a + b\sqrt{m^2 - 1}$$ is a unit (with $$a, b\in \Bbb Z_{> 0}$$), then we have $$a^2 - (m^2 - 1)b^2 = \pm 1$$.

It is then easy to see that $$a \geq m$$, because $$a^2 = (m^2 - 1)b^2 \pm 1\geq m^2 - 2$$.

Therefore, $$(a, b) = (m, 1)$$ is the smallest solution to the Pell equation, and hence the fundamental unit of the order $$\Bbb Z[\sqrt{m^2 - 1}]$$ (which is equal to the ring of integers of $$\Bbb Q(\sqrt{m^2 - 1})$$ if $$m^2 - 1$$ is square free).

• It's not true that $b \in \mathbb Z$ in general. See my answer. Apr 29, 2021 at 14:43
• @principal-ideal-domain Of course the question only makes sense to ask for the fundamental unit of the (not necessarily maximal) order $\Bbb Z[\sqrt{m^2 - 1}]$. Apr 29, 2021 at 14:45

It's easy to see that $$\varepsilon = m +\sqrt{m^2-1}$$ is a unit at all since $$N (\varepsilon) = m^2-\sqrt{m^2-1}^2 = m^2-m^2+1=1.$$ But it is not always a fundamental unit, so the fundamental statement is false. See for example $$m=3$$. Here we have $$\mathbb Q(\sqrt{3^2-1}) = \mathbb Q(\sqrt{8}) = \mathbb Q(\sqrt{2})$$ but we have $$(1+\sqrt{2})^2 = 3+2\sqrt{2}=\varepsilon.$$

• I forgot to mention that $m^2-1$ is square free. Edited the question. Apr 29, 2021 at 14:46