<s>[INST] Showing $m+n\sqrt 2$ is a unit in set $R$ Let $R=\{m+n\sqrt 2|m,n\in\mathbb{Z}\}$. Show that $m+n\sqrt2$ is a unit in $R$ iff $m^2-2n^2=\pm 1$. Hint: Show that if $(m+n\sqrt2)(x+y\sqrt 2)=1$ then $(m-n\sqrt 2)(x-y\sqrt2)=1$ I needed help with the foward direction
This is what I have so far: ($\rightarrow$). Suppose that  $m+n\sqrt2$ is a unit. I want to show that $m^2-2n^2=(m+n\sqrt2)(m-n\sqrt2)=\pm 1$. Since $m+n\sqrt2$ is a unit there exists a $p\in R$ such that $p*(m+n\sqrt2)=1$ where $p=(x+y\sqrt 2)$. But from here i'm stuck showing $(m-n\sqrt 2)(x-y\sqrt2)=1$. Any hints or help would be great.Thanks 
Edit:
Using what I got as a hint below I got since $(m-n\sqrt 2)(x-y\sqrt2)=(m^2-2n^2)(x^2-2y^2)=1\implies (m^2-2n^2)=\frac{1}{x^2-2y^2}$. But since $m^2-2n^2$ is an integer it follows that $(x^2-2y^2)=\pm 1$ and $(m^2-2n^2)=\pm 1$ since $1*1=1$ and $-1*-1=1$. 
 [/INST] Let $(x,y)$ denote $x+y\sqrt 2$. Then, we have multiplication defined as $(x,y)(x',y')=(xx'+2yy',xy'+x'y)$. If we define $N(x,y)=x^2-2y^2$, we have that $$\begin{align}N(x,y)N(x',y')&=(x^2-2y^2)(x'^2-2y'^2)\\&=x^2x'^2-2(x'^2y^2+x^2y'^2)+4y'^2y^2\\&=(xx')^2+(2y'y)^2-2((x'y)^2+(xy')^2)\\&=(xx')^2+\color{red}{2(xx')(2yy')}+(2y'y)^2-2((x'y)^2+\color{red}{2(x'y)(xy')}+(xy')^2)\\&=(xx'+2yy')^2-2(xy'+x'y)^2\\&=N((x,y)(x',y'))\end{align}$$
Now, suppose $z=(x,y)$ and $w=(x',y')$ are such that $zw=1$. Then $N(wz)=N(w)N(z)=1$, so?
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