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

$a + ib \equiv a + (-1)b$ because we have just shown that $i \equiv -1$.
But $i \equiv -1$ implies $1 = -1 \cdot -1 \equiv i \cdot i = -1$. Add 1 to both sides to get $2 \equiv 0$. All equivalences modulo $(1+i)$.
