Show $a+bi \equiv 0,1 \pmod{1+i}$ This is a problem (self-studier) from Ireland & Rosen (3.23):
Show $a+bi \equiv 0,1 \pmod{1+i}$ with $a,b \in\mathbb{Z}$.
There is an extensive hint/solution offered, some parts of which I would appreciate help understanding.
To begin, it says $i\equiv -1(1+i)\pmod{1+i}$ No problem here.
Then it says write $a+ib \equiv a-b \pmod{1+i}$. Here I think the $a$ remains intact since $(1+i) \nmid a$
Write $a-b$ as $2c+d$ where $d=0,1$. This makes sense as $a,b$ are integers and their difference will be either even or odd.
So here is where I am stuck.
Then $a+ib \equiv a-b \equiv 2c+d \equiv d\pmod{1+i}$ 
How is $2c \equiv 0 \pmod{1+i}$ when $a$  was not a few lines above, since they are both in $\mathbb{Z}$.
Thanks

I think the two equivalences $$a + bi \equiv a - b \pmod{1 + i}$$ and $$s \equiv  0, 1 \pmod 2 \qquad s \in \mathbb{Z}$$ along with Andre's comment  that $1 + i \mid 2$ gives a simple solution. 
