How to find all divisors of an element in $\mathbb{Z}[\sqrt{-5}]$? 
How do I find all divisors of $6$ in $\mathbb{Z}[\sqrt{-5}]$?

If I set 
$$
6=(a+b\sqrt{-5})(c+d\sqrt{-5}),
$$
where $a,b,c,d\in\mathbb{Z}$, I get
$$\tag{*}
\begin{cases}
6=ac-5bd,\\
0=ad+bc.
\end{cases}
$$
(*) is a system of equations in $\mathbb{Z}$, is there a systematic way to solve it? (With systematic I mean without need of guessing.)
I know the following divisors of $6$:
$$
2,3,(1-\sqrt{-5}),(1+\sqrt{-5})
$$
are they all?
The question if these divisors are irreducible has been answered more times: here and here. But I have not found how to find out all divisors (actually here the question has been asked, but non answered).
 A: Starting from your system of equations
$$\tag{*}
\begin{cases}
6=ac-5bd,\\
0=ad+bc,
\end{cases}$$
the second shows that $ad=-bc$. If $ac=0$ then $6=-5bd$, a contradiction, so $ac\neq0$ and hence 
$$d=-\frac{bc}{a}\qquad\text{ and }\qquad b=-\frac{ad}{c}.$$
Let $g:=\gcd(a,c)$ and let $a',c'\in\Bbb{Z}$ be such that $a=a'g$ and $c=c'g$. The above shows that $a'\mid b$ and $c'\mid d$. Let $e,f\in\Bbb{Z}$ be such that $b=a'e$ and $d=c'f$. Then
$$0=ad+bc=a'gc'f+a'ec'g=a'c'g(e+f),$$
which shows that $f=-e$. It follows that
$$6=ac-5bd=a'gc'g-5a'ec'f=a'c'(g^2+5e^2),$$
This clearly implies $e\in\{-1,0,1\}$, and the corresponding values of $a$, $b$, $c$ and $d$ are easily found:
For $e=0$ we have $b=d=0$ and $6=ac$, corresponding to factorizations of $6$ in $\Bbb{Z}$.
For $e=\pm1$ we get $g=1$ and $a'c'=1$, so $a'=c'=\pm1$, corresponding to the factorizations
\begin{eqnarray*}
6&=&(1+\sqrt{-5})(1-\sqrt{-5}),\\
6&=&(1-\sqrt{-5})(1+\sqrt{-5}),\\
6&=&(-1-\sqrt{-5})(-1+\sqrt{-5}),\\
6&=&(-1+\sqrt{-5})(-1-\sqrt{-5}),
\end{eqnarray*}
which are all the same factorization, up to rearrangement of the factors and multiplication by units.
