Irreducibility and factoring in $\mathbb Z[i], \mathbb Z[\sqrt{-3}]$ 
In $\mathbb Z[i]$, prove that $5$ is not irreducible.
  In $\mathbb Z[\sqrt{-3}]$, factor $4$ into irreducibles in two distinct ways.

I am completely stumped on how to do this. I really need all the help I can get and a possible walkthrough.
 A: For the first assignment you correctly found
$$5 = (1+2i)(1-2i)\ .$$
This already shows that $5$ is reducible in $\mathbb Z[i]$, as claimed.
To provide a little intuition on how to find these factors, consider the third binomial rule for the product
$$(a+bi)(a-bi) = a^2 - (bi)^2 = a^2+b^2$$
So any sum of perfect squares is reducible with such factors. Now it should be easy to see that $5$ fulfills this criteria: $5=4+1 = 2^2 + 1^2$. This immediately gives rise to two factorisations of $5$ in $\mathbb Z[i]$, one for each choice of $(a,b)$:
$$5 = (1+2i)(1-2i) = (2+i)(2-i)$$

Here is a starter on the second assignment:
$$(a+b\sqrt{-3}) \cdot (c+d\sqrt{-3}) = ac - 3bd + (ad+bc)\sqrt{-3}$$
So you need to find two distinct integer solutions to the equations
$$\begin{align*}
ac - 3bd & = 4 \\
ad+bc & = 0
\end{align*}$$
One solution is $(a,b,c,d) = (1,-1,1,1)$ corresponding to
$$4 = (1-\sqrt{-3})(1+\sqrt{-3})$$
Now you must find another solution to get the second factorisation.
A: The concept of being irreducible in $\mathbb{Z}[i]$ is similar to the idea of being prime in regards to $\mathbb{R}$. In this case, we must check if 5 can be written as $(a+b i)(c+d i)$ where $a,b,c,d \in \mathbb{Z}$. As you found, it can be written as $5 = 1 + 4 = 1 - 4 i^2 = (1 + 2 i)(1-2 i)$ and thus is not irreducible.
We apply the same concept to show that 4 is reducible in $\mathbb{Z}[\sqrt{-3}]$. We look for $a,b,c,d \in \mathbb{Z}$ such that $4 = (a + b \sqrt{-3})(c + d \sqrt{-3})$. Fairly obviously, we see that $4 = 1 + 3 = 1 - (\sqrt{-3})^2 = (1+\sqrt{-3})(1-\sqrt{-3})$.
