Show that $a^2 \equiv a \mod (1+i)$ for all $a \in \mathbb{Z}[i]$. This is a problem statement from a section on quotient rings in Abstract Algebra, so I assume it requires the use of the FIT for rings. When looking around for similar problems, I was only able to find examples with number theory, which isn't really what I'm looking for.
My main question here is: what does $\mathbb{Z}[i]$ mean? I'm familiar with the notation $\mathbb{Z}(i) = \{a+bi : a,b\in\mathbb{Z}\}$ and $\mathbb{Z}[X]$ (a polynomial ring), but never have I seen $\mathbb{Z}[i]$.
I think that I'll be able to figure this out knowing that, but I still would very much appreciate a hint on how to start. I assume you can just show that $a^2 - a \equiv a(a - 1) \equiv 0 \mod (1 + i)$, but I'm not really sure where to go from there.

I will be using the following result.
RESULT: Let $z\in\mathbb{Z}[i]$. Then $N(z)$ (read as norm of $z$) is even $\iff z\equiv 0\mod(1+i)$.
My answer to the question is as follows:
Let $a=m+in$. Then $$a^2-a=(m^2-n^2-m)+(2m-n)i.$$ We get
$$N(a^2-a)=(m^2-n^2-m)^2+(2m-1)^2.$$
Enumerating all possible cases for the integers $m,n$, we can show that $N(a^2-a)$ is even. Using the above result, the posted question holds.
