Every element of $Z (\sqrt{-5})$ is factorable into irreducible factors. I am trying to proof that each element of $R :=\mathbb{Z} (\sqrt{-5})$ is factorable into irreducible elements.
Let $x \in R, x = AB$ with $A = (a + b \sqrt{-5} ),$ $B = (c + d \sqrt{-5})$.
If $A = \pm 1$ or $B = \pm 1$ then $x$ irreducible and the proof is done. 
Otherwise, $x = AB, N(x) = N(A) N(B)$ with $N(A),N(B)  \neq 1$.
How can I proceed?
Thanks in advance!
 A: Your idea in the comment is right (see the third proof below).
An nonzero and noninvertible element $a$ of a domain $R$ fails to be a product of irreducible elements if, for every $b_1,b_2,\dots,b_n\in R$, with $b_i$ not invertible, at least one of the factors $b_i$ can be written as $b_i=cd$, where neither $c$ nor $d$ is invertible, that is, is not irreducible.
First proof (with a big gun)
Assume $a$ is not a product of irreducibles. In what follows all elements I write as $a_{ij}$ is assumed to be noninvertible. Since $a=a_{11}$ is not irreducible, $a_{11}=a_{21}a_{22}$ and it's not restrictive to assume $a_{22}$ not irreducible, so we can factor it and get
$$
a_{11}=a_{21}a_{22}=a_{31}a_{32}a_{33}
$$
where the elements are reordered so that $a_{33}$ is not irreducible. Continuing this way, we find, for every $n\ge1$,
$$
a=a_{n1}a_{n2}\dotsm a_{nn} \tag{*}
$$
This allows us to build the increasing chain of ideals
$$
(a)=(a_{11})\subsetneq(a_{21})\subsetneq\dots\subsetneq(a_{n1})
\subsetneq\dotsb
$$
and therefore $R$ is not Noetherian.
The ring $\mathbb{Z}[\sqrt{-5}]$ is a homomorphic image of $\mathbb{Z}[X]$ via the isomorphism $\mathbb{Z}[\sqrt{-5}]\cong\mathbb{Z}[X]/(X^2+5)$, so it is Noetherian, because $\mathbb{Z}[X]$ is Noetherian by Hilbert’s Basis theorem.
Second proof (with just a bludgeon)
However, the specific case is easier. From (*), applying the norm function
$$
N(x+y\sqrt{-5})=x^2+5y^2
$$
we get
$$
N(a)=N(a_{n1})N(a_{n2})\dotsm N(a_{nn})
$$
where $N(a_{nj})>1$, for $j=1,2,\dots,n$. Since $N$ has (positive) integer values, this is a contradiction, because it would follow that
$$
N(a)\ge2^n
$$
for every $n\ge1$.
Third proof (no big weapons)
Alternatively, do an induction on $N(a)$. If $N(a)=2$, the statement is (vacuously) true. Suppose it holds for every element $b$ with $2\le N(b)<N(a)$. Then either $a$ is irreducible and we are done, or $a=b_1b_2$, with $N(b_1)>1$ and $N(b_2)>1$; then $2\le N(b_1)<N(a)$ and $2\le N(b_2)<N(a)$ and the induction hypothesis applies.
Comment
Why showing the big gun proof? Because the same result in every Noetherian domain, not just a simple integral extension of the integers.
Note however that uniqueness of factorization need not hold. In particular it doesn't hold in $\mathbb{Z}[\sqrt{-5}]$, because
$$
6=2\cdot 3=(1+\sqrt{-5})(1+\sqrt{-5})
$$
where $2$ and $3$ are associate neither of $1+\sqrt{-5}$ nor of $1+\sqrt{-5}$.
