Proving you can write the Gaussian integers in a certain form Question: Show that any element of $\Bbb{Z}[i]$ can be written in the form $c+x$ where $c\in \{0,1,2,3,4\}$ and $x \in I$ where $I$ is the principal ideal $I= \langle 2-i\rangle$ generated by $2-i$
Answer:
I am given a hint to observe that $5=(2-i)(2+i) \in I$ but I don't see why I need it.
We have to prove that for any $a,b \in \Bbb Z$ 
$a+bi=c+x= c + (x+yi)\cdot(2-i)$
Which implies 
$a=c+2x+y$
$b=2y-x$
Doesn't this prove that I can write any member of $a+bi$ as a $c+x$?
 A: ${\rm mod}\,\ i\!-\!2\!:\,\  \color{#c00}{i\equiv 2}\,\Rightarrow\,0\equiv (\color{#c00}{2\!-\!i})(2\!+\!i)=\color{#0a0}{5},\ $ so $\,\ a +\color{#c00}i\,b\,\equiv\, (a+\color{#c00}2\,b) \bmod \color{#0a0}5\,\in\, \{0, 1, 2, 3, 4\}$
A: Elementary solution: it is clear that every gaussian integer is $\equiv n\mod 2+i$ where $n\in \mathbb{Z}$. Further any integer is congruent to $0,1,2,3,4$  modulo $5$, and thus modulo $2+i$.
Conceptual solution: The number of residue classes modulo an algebraic number is $N\alpha$ where $N$ is the norm. Some take this as the definition, in which case it is theorem that this is the product of the conjugates. Thus in this case, residue field has order $N(2+i)=5$ and there is only one such field. 
A: We will prove 
$$
\Bbb{Z}[i]/(2-i) \cong \Bbb{Z}/5\Bbb{Z}
$$
 First define
$$
\phi: \Bbb{Z} \rightarrow \Bbb{Z}[i]/(2-i),\quad\phi(z) = z + (2-i)\Bbb{Z}[i]
$$
Clearly $\phi$ is a homomophism and
$$
\ker \phi = (2-i)\Bbb{Z}[i] \cap \Bbb{Z}
$$
For any $z \in \ker \phi$
$$
z = (2-i)(a+bi)=2a+b+(2b-a)i
$$ 
where $a,b \in \Bbb{Z}$. But 
$$
2a+b+(2b-a)i \in \Bbb{Z}\iff 2b-a=0\iff a=2b
$$
Thus
\begin{align}
\ker \phi = (2-i)\Bbb{Z}[i] \cap \Bbb{Z} &= \{a,b \in \Bbb{Z}\mid (2-i)(a+bi)\}
\\
&= \{b \in \Bbb{Z}\mid 2\cdot 2b+b +0i \}
\\
&= \{b \in \Bbb{Z}\mid 5b \}
\\
&= 5\Bbb{Z}
\end{align}
Hence 
$$
\Bbb{Z}[i]/(2-i) \cong \Bbb{Z}/5\Bbb{Z}
$$
