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.

Note that $ 2 = (1+i)(1-i) \equiv 0 \pmod{1+i} $ and any $ a + bi \in \mathbf Z[i] $ with $ a, b \in \mathbf Z $ is congruent to an integer modulo $ 1 + i $, thus any element of $ \mathbf Z[i] $ is congruent to either $ 0 $ or $ 1 $ modulo $ 1+i $.
