Consider the ideal $\langle 1+i \rangle$ in $\mathbb{Z}[i]$. Consider the ideal $\langle 1+i \rangle$ in $\mathbb{Z}[i]$. 
$(a)$ Make use of the given description of this ideal,
$\hspace{75pt}$ $\langle 1+i \rangle = \{a+bi:a+b \text{ is even}\}=\{\alpha\in \mathbb{Z}[i]:N(\alpha) \text{ is even}\}$
$\hspace{15pt}$ to show that for all $a+bi\in \mathbb{Z}[i]$, there exists $c\in \mathbb{Z}$, such that
$\hspace{100pt} $ $\langle 1+i \rangle+(a+bi)=\langle 1+i \rangle +c$.
$(b)$ Use part $a$ to prove that $\mathbb{Z}[i]/\langle 1+i \rangle$ has only two elements.

I have no idea how to go about showing part $a$. I know that since $\langle 1+i \rangle$ is the ideal, it is equivalent to $0$, but I'm stuck after that. Any hints?
 A: There is an idea in your comment that can be used in the proof, but you should try to be as precise as possible when getting used to making arguments in quotient spaces. For example, here $1 \neq -i$, but what you probably mean is $1 + \langle 1+i\rangle = -i + \langle 1+i\rangle$, which is true.
Here's some scratch work/hints regarding how to proceed. Given $a+bi \in \mathbb{Z}[i]$, you're trying to show that there exists $c \in \mathbb{Z}$ such that $(a+bi)-c = (a-c) + bi \in \langle 1+i\rangle$. The statement $(a-c)+bi \in\langle1+i\rangle$ is true if and only if
$$
    (a-c) + bi = d(1+i) = d + di
$$
for some integer $d$. So what should $c$ be equal to? (Hint: Figure out what $d$ should be equal to first from this last equation.)
Edit: I forgot that you're supposed to make use of the rather unnecessary description of the ideal $\langle 1+i\rangle$ that they've provided. Prahlad provided an answer that uses this, or if you formulate a proof based on my suggestions above the 'right' way, you could just note that $2d$ is even.
A: For (a), if $m+ni$ has even norm, then $m\equiv n\pmod{2}$, so just take
$$
u = \frac{n-m}{2}, \text{ and } v = \frac{n+m}{2}
$$
then $u, v\in \mathbb{Z}$ and $m+ni = (1+i)(u+vi)$
For the second half, use Euclidean division.
A: Just to get you started:
$\langle 1+i \rangle$ is a principal ideal in $\mathbb{Z}[i]$, which consists of all multiples of $1+i$ in $\mathbb{Z}[i]$.
Thus, any element of $\langle 1+i \rangle$ looks like $(1+i)(m+ni)$ where $m,n \in \mathbb{Z}$.
So, an element of $\langle 1+i \rangle$ is of the form $(m-n)+(m+n)i$.
Comparing this with $a+ib$, we see that $a = m-n$ and $b = m+n$, do you now see that $a+b$ must be even?
Can you make any progress from here?
A: For part $(a)$, recall
$$ (a+bi) + \langle 1 + i \rangle = c + \langle 1 + i \rangle $$
if and only if
$$ a+bi-c \in \langle 1 + i \rangle. $$
An example might be helpful here.  Let's say we start with $a+bi = 3 + 4i$.  Then
$$ (3+4i) + \langle 1 + i \rangle = (-1) + \langle 1 + i \rangle $$
because
$$ (3+4i)-(-1) = 4(1+i) \in \langle 1 + i \rangle. $$
In other words, in this example, you can subtract off $4$ of $1 + i$'s because each of the $1+i$'s are in the ideal $\langle 1 + i \rangle$.  Now what might you do for an arbitrary $a + bi$?
