Why is $(3,1+\sqrt{-5})^2=(2-\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$ I want to show $(3,1+\sqrt{-5})^2=(2-\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. It's easy to see $(2-\sqrt{-5})\subset (3,1+\sqrt{-5})^2$ since $2-\sqrt{-5}=3-(1+\sqrt{-5})$ is in $(3,1+\sqrt{-5})^2$. But I have difficulty proving the converse inclusion. 
$(3,1+\sqrt{-5})^2=(3)(3)+(3)(1+\sqrt{-5})+(1+\sqrt{-5})^2=(3)(3,1+\sqrt{-5})+(2)(2-\sqrt{-5})$. How should I proceed?
 A: Adding the simplest proof I could come up with here:
The ideal $P=(3,1+\sqrt{-5})$ is one of the prime ideals lying above the rational prime $3$, and of index three in the ring of integers.
By multiplicativity of indices of ideals we have 
$$
[\Bbb{Z}[\sqrt{-5}]:P^2]=9.
$$
Combining this with the fact 
$$[\Bbb{Z}[\sqrt{-5}]:(2-\sqrt{-5})]=N(2-\sqrt{-5})=2^2+5\cdot(-1)^2=9,$$ your observation
that $(2-\sqrt{-5})\subseteq P^2$ and multiplicativity of index in a tower of abelian groups:
$$
9=[\Bbb{Z}[\sqrt{-5}]:(2-\sqrt{-5})]= [\Bbb{Z}[\sqrt{-5}]:P^2]\cdot =[P^2:(2-\sqrt{-5})]
$$
imply that $[P^2:(2-\sqrt{-5})]=1$, in other words $P^2=(2-\sqrt{-5})$.

Leaving my earlier arguments showing how the approach evolved.

From where you are in your last line you can continue as follows:
First verify that
$$3\cdot 3=9=(2+\sqrt{-5})(2-\sqrt{-5})\in(2-\sqrt{-5}).$$
Then check that
$$
3(1+\sqrt{-5})=3+3\sqrt{-5}=(2-\sqrt{-5})(-1+\sqrt{-5})\in(2-\sqrt{-5}).
$$
Together with your last line these two observation prove the reverse inclusion
$$
(3,1+\sqrt{-5})^2\subseteq(2-\sqrt{-5})
$$
so you are done.

The above (or the approach in Bill Dubuque's answer) is the kind of argument you can use, when first starting with this topic. A little bit later, after you have learned about class groups and such, you can take advantage of the fact the class number of this field is $h=2$. An immediate consequence is that the square of any ideal is principal.
 Therefore $P^2$ is of index $3^2=9$ in the ring $\Bbb{Z}[\sqrt{-5}]$. This means that - without any further calculations whatsoever - we know that
$$
P^2=(a+b\sqrt{-5}),
$$
where $a,b$ are rational integers and
$$
N(a+b\sqrt{-5})=a^2+5b^2=9.
$$
This leaves us the possibilities (up to sign) $a=3,b=0$ or $a=2,b=\pm1$. But you already had the key idea that $2-\sqrt{-5}\in P^2$, implying that $P^2\subseteq (2-\sqrt{-5})$,
so the claim follows.
A: Below I show how to derive it (vs. simply check it) using simple Euclidean-style reductions, i.e. by modding out some generators modulo another.  
Write $\ \overbrace{w = \sqrt{-5}}^{\large w^2\ =\ -5},\ $ so $\ I = (3,w\!+\!1)^2 = (3^2,\,\overbrace{(w\!+\!1)^2}^{\large \color{blue}{2w\,-\,4}\ \ },\,3(w\!+\!1))$ 
$ 9\in I\,\Rightarrow\, I = (9,\, I\ {\rm mod}\ 9),\,$ and $\,\ {\rm mod}\ 9\!:\ \color{blue}{2(w\!-\!2)}\equiv 0\!\! \overset{\large \times\,5}\iff w\!-\!2\equiv 0$
So $\, I = (\color{#c00}9, \color{#0a0}{w\!-\!2}, 3(w\!+\!1)) = (9,w\!-\!2,\color{#c00} 0)$ by $\, 3(w\!+\!1)\overset{\color{#0a0}{\large w\,\equiv\, 2}}\equiv 3^2\overset{\large\color{#c00}{9\,\equiv\,0}}\equiv\color{#c00} 0$
