$2$ is irreducible in $\mathbb{Z}[\sqrt 5]$

Question

• This is not a duplication because other questions on this subject refers to $\mathbb{Z}[\sqrt{\textbf{-}5}]$ and in addition I have a question about a specific part in the proof.

Show that $2$ is irreducible in $\mathbb{Z}[\sqrt5]$.

Suppose $$2=(a+b\sqrt5)(c+\sqrt5d) \\\Rightarrow 2=(a-b\sqrt5)(c-\sqrt5d) \\\Rightarrow 4=(a^2-5b^2)(c^2-5d^2) $$

I don't understand why does the last transition hold?

Answer 1

$4 = 2 \cdot 2 = (a+b\sqrt5)(c+\sqrt5d) \cdot (a-b\sqrt5)(c-\sqrt5d) = (a^2-5b^2)(c^2-5d^2)$

Answer 2

You have two expressions for $2$. Multiply them and simplify using $$a^2-b^2=(a-b)(a+b)$$