Prove that $p^2$ is the principal ideal $(2)$. 
Let $p$ be the ideal $\{2a+(1+\sqrt{-5})b\mid a,b\in\mathbb{Z}[\sqrt{-5}]\}$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $p^2$ is the principal ideal $(2)$.

I tried multiplying the ideal with itself and tried to simplify to see if I could relate it to the principal ideal $(2)$ but I can't seem to get it. This is what I have so far.
$(2a+(1+\sqrt{-5})b)^2=4a^2+4(1+\sqrt{-5})ab+(1+\sqrt{-5})^2b^2$
$=4a^2+(4+4\sqrt{-5})ab+(2\sqrt{-5}-4)b^2$
$=2(2a^2+(2+2\sqrt{-5})ab+(\sqrt{-5}-2)b^2)$
If $x\in p^2$, then $x$ is a multiple of $2$, so $x\in(2)$. Thus $p^2\subseteq (2)$.
I am unsure of how to show that $(2)\subseteq p^2$.
 A: Here’s another approach:
To multiply two ideals, say $I=(a_1,a_2,\cdots,a_m)$ and $J=(b_1,\cdots,b_n)$, where what’s in each pair of parentheses is a list of generators of the ideal, all you need to do is write down the products $a_ib_j$, all of them, and see what ideal they generate.
In the present case, we do this:
\begin{align}
\left(2,1+\sqrt{-5}\,\right)\left(2,1+\sqrt{-5}\,\right)&=\left(4,2+2\sqrt{-5},-4+2\sqrt{-5}\,\right)\\
&=\left(4,2+2\sqrt{-5},2\sqrt{-5}\,\right)\\
&=\left(4,2,2\sqrt{-5}\,\right)\\
&=\left(2,2\sqrt{-5}\,\right)=(2)
\end{align}
Notice that each simplification is reversible: you can go from left to right, as written, but also from right to left.
A: Let $\,w = 1\!+\!\sqrt{-5}.\,$ By $\,\rm 4NT =$ Euler's Four Number Theorem and $\,(2,3)\!=\!(1)\,$ we have
$$\,\underbrace{2\cdot 3 = \color{0a0}w\!\:\color{c00}{\bar w}\, \Rightarrow\, (2) = (2,\color{#c00}w)(\color{#90f}2,\color{#0a0}{\bar w})}_{\textstyle   a\,d\, =\, b\,c\ \underset{\small \rm 4NT}\Rightarrow\, (a) = (a,b)\,(a,c)} = (2,\color{#c00}w)^2\ \ {\rm by}\ \  \underbrace{\color{#0a0}{\bar w}\equiv -\color{#c00}w\!\!\!\pmod{\!\color{#90f}2}}_{\textstyle \bar w + w \,=\, 2}\qquad\qquad$$
As for integers,  $\rm 4NT$ constructs a nontrivial (ideal) factorization of $\,2\,$ from a factorization of any multiple $\,w\bar w\,$ of $\,2\,$ that is nontrivial (i.e. $\,2\nmid w,\bar w),\,$ i.e. a witness that $\,2\,$ is composite (not prime). To do so we simply take the ideal gcd (= ideal sum) of $\,2\,$ with each factor $\:\!w,\:\!\bar w\,$ (assuming any common factor of $\color{#0a0}{a,b,c,d}$ has been cancelled). The proof is the same as for gcds in the link, i.e.
$$ \begin{align}&(\color{#0a0}{a,b,c,d})\!=\!(1),\ \color{#c00}{ad = bc} \ \Rightarrow\  (a,b)\,(a,c) = (aa,ab,ac,\color{#c00}{bc}) = (\color{#c00}a)(\color{#0a0}{a,b,c,}\color{#c00}d) = (a)\\[.1em]
&\rm\small \color{#0a0}{A\!+\!B\!+\!C\!+\!D}\!=\!(1),\, \color{#c00}{AD\!=\!BC}\Rightarrow(A\!+\!B)(A\!+\!C)=A^2\!+\!AB\!+\!AC\!+\!\color{#c00}{BC}= \color{#c00}A(\color{#0a0}{A\!+\!B\!+\!C}\!+\!\color{#c00}D) = A\end{align}
\qquad$$

Alternatively, we can repeat the proof of $\rm 4NT$ in this special case as below, but doing that obfuscates the key arithmetical idea (factorization refinement) that $\rm 4NT$ conveys.
$w^2 =\, 2w\,-\,6\,$  for $\,w = 1+\sqrt{-5},\,$ so apply below with $\,a=2$
$\!\begin{align}w^2 = ab w + ac,\ \color{#c00}{(a,c)\!=\!(1)} \Rightarrow\ (a,w)^2\! &= (a^2,aw,w^2)\\
&=\, (a^2,aw,ac+abw)\\
&=\, a(a,\ \,\color{#0a0}w,\ \,c\:\!+\ \,b\color{#0a0}w)\\
&=\, a(\color{#c00}a,\ \ w,\ \,\color{#c00}c)\\
&=\, a(\color{#c00}1)
\end{align}$
The ideal arithmetic in Lubin's answer is a special case of the above. For an introduction to this simple ideal (and gcd) arithmetic see here and its links.
A: Let's call this ideal $\mathfrak{p}$. Note that $2 \in \mathfrak{p}^2$, since
$$
2 = (1+\sqrt{-5})(1 - \sqrt{-5}) \  - \ 2\cdot2.
$$
On the other hand, if $m \in \mathfrak{p}^2$, then $m$ is a sum of terms of the form
\begin{align}
(2a + (1 + \sqrt{-5})b)&(2a'+(1+\sqrt{-5})b') = \\
             & 4aa' + 2ab' + 2a'b + bb' - 5bb' + \bigg(2ab' + 2a'b + bb'+bb'\bigg)\sqrt{-5}.
\end{align}
and everything is divisible by $2$ when we group like terms.
This is a bit of a miracle proof; you can redo it conceptually by observing that $\mathfrak{p}$ is the ideal of all elements of the form $c + d\sqrt{-5}$ where $c$ and $d$ have the same parity.
