Question: > 
> 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 \} $.
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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.

Answer: 
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 $\mathbf Z$.
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 $N(u) \in \{-1,1\}$.