Show that an element $ m + n \sqrt{2} $ of $ \mathbb{Z}[\sqrt{2}] $ is a unit if and only if $ m^{2} - 2 n^{2} \in \{ 1,-1 \} $. 
Show that an element $ m + n \sqrt{2} $ of $ \mathbb{Z}[\sqrt{2}] $ is a unit if and only if $ m^{2} - 2 n^{2} \in \{ 1,-1 \} $.

Okay, I have a pretty big hint as to how to do this problem, but I'm having a problem connecting the dots. Here's the hint: 
For the $\Rightarrow$ direction, suppose $m+n\sqrt2$ is a unit, so there exists $x+y\sqrt2\in\Bbb{Z}[\sqrt2]$ such that $(m+n\sqrt2)(x+y\sqrt2)=1$. Show that this implies $(m-n\sqrt2)(x-y\sqrt2)=1$ also, then multiply these equations.
For the $\Leftarrow$ direction, you need to suppose that $m^2-2n^2\in \{1,-1\}$, then use this assumption to define a multiplicative inverse for $m+n\sqrt2$.
I understand that there exists an element $x+y\sqrt2$ when multiplied by $m+n\sqrt2$ equals $1$. I don't understand how this implies the next statement and then I don't understand how to go from there to the fact that $(m+n\sqrt2)(m-n\sqrt2)=m^2+2n^2$.
Also, I know that $\Bbb{Z}[\sqrt2]$ is a subring of $\Bbb{R}$, so I can use those properties.
 A: If there exists $x,y\in\mathbf Z$ such that $(m+n\sqrt 2)(x+y\sqrt2)=1$ means the linear system 
$$\begin{cases}mx+2ny=1\\nx+my=0\end{cases}$$
has  a unique solution $(x,y)\in\mathbf Z^2$. It has a unique solution if and only if the matrix $\begin{pmatrix}m&2n\\n&m\end{pmatrix}$ is invertible, which is the case if and only if its determinant $m^2-2n^2$ is a unit in $\mathbf Z$.
A: I'm going to follow your hint. Given $a+b\sqrt{2}$, we define its "conjugate" as $\overline{a+b\sqrt{2}}=a-b\sqrt{2}$. This conjugate behaves like the usual conjugate for complex numbers, e.g., given $a+b\sqrt2, c+d\sqrt2\in \Bbb{Z}[\sqrt{2}]$, it holds $$\overline{a+b\sqrt{2}}\cdot \overline{c+d\sqrt{2}}=\overline{(a+b\sqrt{2})(c+d\sqrt{2})}\;.$$
Try to prove by yourself that this is indeed true. Now, if you have $(m+n\sqrt2)(x+y\sqrt2)=1 ...(\alpha)$, applying the conjugate to both sides of $(\alpha)$ we get $$\overline{(m+n\sqrt2)(x+y\sqrt2)}=\overline{1}=1$$ $$\implies \overline{(m+n\sqrt2)}\cdot \overline{(x+y\sqrt2)}=1$$ $$\implies (m-n\sqrt2)\cdot (x-y\sqrt2)=1 ... (\beta)$$
Then if we multiply $(\alpha)$ and $(\beta)$ we deduce that $$(m+n\sqrt2)(m-n\sqrt2)(x+y\sqrt2)(x-y\sqrt2)=1,$$ which leads to $(m^2-2n^2)(x^2-2y^2)=1$, so $m^2-2n^2\mid 1$, i.e., $m^2-2n^2\in \{\pm1\}$.
For the other implication, if $m^2-2n^2\in \{\pm 1\}$, then show that $\frac{m-n\sqrt{2}}{m^2-2n^2}$ is the inverse of $m+n\sqrt2$ in $\Bbb{Z}[\sqrt2] $.
A: First of all, notice that the irrationnality of $\sqrt{2}$ entails the uniqueness of the $(a,b) \in \mathbb{Z}^{2}$ in the $a+b\sqrt{2}$ notation. This allow you to define $N(a+b\sqrt{2})=a^{2}-2b^{2} \in \mathbb{Z}$. Show that $N(uv)=N(u)N(v)$ for all $x,y \in \mathbb{Z}[\sqrt{2}]$. So, if $u$ is invertible and $v$ is it's inverse, you get $N(u)N(v)=1$ and $N(u),N(v) \in \mathbb{Z}$, so $N(u) \in \{-1,1\}$.
A: In general :To determine invertible elements in $\mathbb Z[\sqrt n]$ where $n$ free of square::
Let be $u=a+b\sqrt n $ an invertible element in $\mathbb Z[\sqrt n]$, then exists $v=c+d\sqrt n \in \mathbb Z[\sqrt n] $ where $uv=1$ , then $N(uv)=N(u)N(v)=1\Rightarrow N(u)=1$. we  have three cases: 
$1)$ $n=-1 \Rightarrow N(u)=a^2+b^2=1\Rightarrow a=\pm 1, b=0$ or $a=0, b=\pm 1$, then $u=\pm 1$ or $u=\pm i$
$2)$ $n<-1 \Rightarrow N(u)=a^2-nb^2=1 \Rightarrow a=\pm 1, b=0$  ,then $u=\pm 1$
$3)$ $n>0 \Rightarrow N(u)=a^2-nb^2=1$ has infinitely many of solutions , then invertible element is $\{-1,+1 \}$ and infinite many of elements. 
for example for case $(3)$ in $\mathbb Z[\sqrt3]$ we have $u=2-\sqrt3$ is invertible element because $(2-\sqrt3)(2+\sqrt3)=1$
