$d \in \mathbb{N}$ is square free with $d \equiv 2, 3 \mod 4$. We let $K=\mathbb{Q}( \sqrt{d})$ and let $\mathcal{O}_K$ be the ring of integers of $K$.
($1$) It is a consequence of Dirichlet’s unit theorem that there exists a unit $\epsilon \in \mathcal{O}_K^\times$ such that every unit $u$ can be written in a unique fashion as $u = \pm \epsilon^k$ with $k \in \mathbb{Z}$. Such a unit $\epsilon$ is called a fundamental unit of $K$ Show that if $\epsilon$ is a fundamental unit then the other fundamental units are precisely the numbers $\pm \epsilon, \pm \epsilon^{-1}$.
($2$)Show first that there is precisely one fundamental unit $\epsilon=u+v\sqrt{d}$, with $u,v \in \mathbb{N}$. Then show that $v \in \mathbb{N}$ is smallest possible such that such that either $dv^2+1$ or $dv^2−1$is a square.
($3$)Using the result from ($2$), give a description of all solutions in integers $x, y$ to the Diophantine equation $x^2 − 7y^2 = 1$.
My attempt:
$(1)$ If $u \in \mathcal{O}_K^\times$, then $u=\pm \epsilon^k$, for some $k \in \mathbb{Z}$. Then:
$u=\pm (-\epsilon)^k$ if $k$ is even and $u=\mp (-\epsilon)^k$ if $k$ is odd.
$u=\pm (\epsilon^{-1})^{-k}$
$u=\pm (-\epsilon^{-1})^{-k}$ if $k$ is even and $u=\mp (-\epsilon^{-1})^{-k}$ if $k$ is odd.
Therefore $- \epsilon, \epsilon^{-1}, -\epsilon^{-1}$ are fundamental units. Suppose now $\omega=\pm \epsilon^l$ is a fundamental unit: then $\epsilon=\pm \omega^k=\pm (\pm \epsilon^l)^k=\pm \epsilon^{lk}$ thus $lk=1$ and since $l, k \in \mathbb{Z}$ we must have $l= \pm 1$ and thus $\omega \in \{ \epsilon, -\epsilon, \epsilon^{-1}, -\epsilon^{-1}\}$.
$(2)$ If $\eta=a+b\sqrt{d}$ is a unit, $a,b \in \mathbb{Z}$, then $-\eta=-a-b\sqrt{d}$, $\eta^{-1}=\pm a \mp b\sqrt{d}$ and $-\eta^{-1}=\mp a \pm b\sqrt{d}$, depending whether $\eta$ has norm $1$ or $-1$. Therefore only one of the four possibilities has both coefficients in $\mathbb{N}$.
I don't know how to prove that $v \in \mathbb{N}$ is the smallest possible.
$(3)$ All the integers solution $x,y$ to the equation $x^2-7y^2=1$ are exactly the elements of norm $1$ in $\mathcal{O}_K^\times$. Moreover, if $\alpha=(x,y)$ is a solution then also $-\alpha=(-x,-y)$, $\alpha^{-1}=(x,-y)$, $-\alpha^{-1}=(-x,y)$ are solutions. Let $\epsilon=u+v\sqrt{7}$ be a fundamental unit of $\mathcal{O}_K$ with $u,v \in \mathbb{N}$. Then by part $2)$ either $7v^2+1$ or $7v^2-1$ is a square. If $7v^2+1$ is a square, then $\mathrm{N}(\epsilon)=1$ and therefore all the integer solutions to $x^2-7y^2=1$ are of the form $\pm \epsilon^{k}$, $k \in \mathbb{Z}$. If $7v^2-1$ is a square then $\mathrm{N}(\epsilon)=-1$ and therefore all the integer solutions to $x^2-7y^2=1$ are of the form $\pm \epsilon^{2k}$, $k \in \mathbb{Z}$.
EDIT:
$(3)$ the fundamental unit is $8+3\sqrt{7}$ and $\mathrm{N}(8+3\sqrt{7})=1$. Therefore all the solutions $(x_k,y_k)$ are such that $x_k+y_k \sqrt{7}=\pm(8 + 3 \sqrt{7})^k$ with $k \in \mathbb{Z}$.
For (2), show that $(u+v\sqrt{d})^k=u_k+v_k\sqrt{d}$ has $v\mid v_k$ for all $k\in\mathbb{Z}$. If $0<v'<v$ such that $v'd^2\pm 1$ is a square, say $u'^2$, then what is the norm of $u'+v'\sqrt{d}$?
For (3), you are supposed to determine the fundamental unit!