Showing that $1 + \sqrt{5}$ is irreducible in $\mathbb{Z}[\sqrt{5}]$

Question

Consider the ring $\mathbb{Z}[\sqrt{5}]$. How can we show that the element $1 + \sqrt{5}$ is irreducible in this ring?

Answer 1

$\textbf{Hint:}$ Suppose $1 + \sqrt{5} = xy$. What can be norms of $x$ and $y$ be?

Answer 2

First, since $N(1+\sqrt{5})=1^2-1^2\cdot 5=-4$, we have that $1+\sqrt{5}$ is not a unit. Thus $1+\sqrt{5}=xy$ for $x,y\in\mathbb{Z}[\sqrt{5}]$, and suppose that neither are units. What does this tell us about, say, $N(x)$? Since $N(xy)=N(x)N(y)=-4$, it follows $N(x)=\pm 2$. If $x=a+b\sqrt{5}$ for $a,b\in\mathbb{Z}$, then $a^2-5b^2=\pm 2$, hence $a^2\pmod{5}\in\{2,3\}$. This is a contradiction (can you see why?) Hence one of $x,y$ is a unit and $1+\sqrt{5}$ is indeed irreducible.