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.

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 $4 = 1 + 3 = 1 - (\sqrt{-3})^2 = (1+\sqrt{-3})(1-\sqrt{-3})$.
