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

Question

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.

Answer 1

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).

Answer 2

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 $$(1+\sqrt{2})^2 = 3+2\sqrt{2}=\varepsilon.$$