Given is $\mathbb{Z}[\sqrt{n}]^* \neq \{\pm 1\}, n>0$. Show that there is a smallest element $x \in \mathbb{Z}[\sqrt{m}]^*$ with $x > 1$ and $\mathbb{Z}[\sqrt{m}]^* = \langle -1,x \rangle \cong (\mathbb{Z}/2\mathbb{Z}) \ \times \ \mathbb{Z} $.
I don't exactly know how to start. Can somebody please help me to understand how I can proof these two questions?
The elements of $\Bbb Z[\sqrt n]$ are $a+b\sqrt n$ with arbitrary $a,b\in\Bbb Z$.
Let $\overline{(a+b\sqrt n)}:=a-b\sqrt n$ be the conjugate of an element.
Let $N(x):=x\overline x$, i.e. $N(a+b\sqrt n)=a^2-nb^2$ be the norm function, which assigns an integer to every element, and preserves multiplication: $N(x\cdot y)=N(x)\cdot N(y)$.
Lemma: An element $u\in\Bbb Z[\sqrt n]$ is unit if and only if $N(u)=\pm 1$.
Proof: $\implies:\ N(u)N(u^{-1})=N(1)=1$ and as they are integers, $N(u)=N(u^{-1})=\pm1$.
$\ \ \qquad\Longleftarrow:\ 1=\pm N(u)=u\cdot \pm\bar u$, so $u$ has an inverse.
If $u=a+b\sqrt n$ is a unit, $a,b\ne0$, then so are $\bar u=a-b\sqrt n$ and $-u,\,-\bar u$, and thus exactly one of them has both coefficients positive.
So, assume $a,b>0$ in $u=a+b\sqrt n$.
Note that having $N(u)=-1$ means that $(b\sqrt n+a)(b\sqrt n-a)=1$.
As $a,b>0$ integers, clearly we have $a+b\sqrt n>1$. Consequently, $a-b\sqrt n$ and $b\sqrt n-a$ are both $<1\ $ (one of them is negative).
Ok, now consider the set $\{b\in\Bbb N: b>0, a+b\sqrt n$ is a unit$\}$. As its elements are natural numbers, it contains a smallest element $b_0$.
For $b_0$ choose the smallest positive $a_0$ such that $x:=a_0+b_0\sqrt n$ is a unit ($a_0^2-nb_0^2=\pm1$).
Final claim: $x$ is the smallest among the units $>1$, and every positive unit is a power of $x$.