# Proving that $2+\sqrt{2}$ is irreducible in $\mathbb{Z}[\sqrt{2}]$.

I'm asked to show that $$x=2+\sqrt{2}$$ is irreducible in $$\mathbb{Z}[\sqrt{2}]$$ by using the norm map $$N:\mathbb{Z}[\sqrt{2}]\rightarrow \mathbb{Z}^+:a+\sqrt{2}b\mapsto |a^2-2b^2|$$

Now, if $$x=yz$$, then $$2=N(x)=N(y)N(z)$$ forcing wlog $$N(y)=1$$. I'm now stuck trying to show that $$y$$ must be a unit and would appreciate any help.

Use the definition of the norm. If $$y=c+d\sqrt{2}$$ then $$N(y)=(c+d\sqrt{2})(c-d\sqrt{2})=1$$.
Write $$y=u+\sqrt2 v$$, $$N(y)=u^2-2v^2=(u+\sqrt 2 v)(u-\sqrt2 v)=1$$ or $$-1$$ implies that $$u-\sqrt 2v$$ or $$-(u-\sqrt2 v)$$ is the inverse of $$y$$.