Show that the Gauss ring $\mathbb{Z}[i]$ is Euclidean ring Show that the Gauss ring 
$$ (\mathbb{Z}[i]=\{a+bi \mid a,b \in \mathbb{Z}\},+,\cdot ) $$ is Euclidean ring with a norm $d(a+bi)=a^{2}+b^{2}$
How to prove that theorem?
I started checking the first condition of norm which is $\forall a,b \in \mathbb{Z}[i] d(a)<d(a\cdot b)$
I started $a^2+b^2\leqslant (ac-bc)^2+(ad+cb)^2$
and i stucked on $a^2+b^2\leqslant a^2(c^2+d^2)+2b^2 c^2-2abc^2+2abcd$
What is next and how to prove the second condition?
 A: Here is an Elegant Proof
It is well known that $(\mathbb{Z}[i]=\{a+bi \mid a,b \in \mathbb{Z}\},+,\cdot )$ in integral domain. Consider, $N:\Bbb Z[i] \to \Bbb N$ defined by
$$\color{blue}{N(z) = z\bar{z}=|z|^2 =a^2+b^2~~~ \text{for}~~z= a+ib.}$$
We want to show that $N$ define an integral function for our Ring $\Bbb Z[i].$

*

*$N(0) = 0$ and $N(z) \gt 0$ for $z\neq  0.$

*$z,w,q\in \Bbb Z[i]\setminus\{0\} $ such that $z=wq$ i.e $w|z$ we have $N(q) \gt 0 \implies N(q) \ge 1$ since $N(q) \in \Bbb N$
Then, $$N(w) \le N(w)N(q) = |w|^2|q|^2 = |wq|^2 =N(wq) =N(z)$$
So, if  $w|z$ then  $N(w) \le N(z)$.

*

*Now we want to show the Euclidean division property. we use the following


Lemma: for every $x\in \Bbb R$ there exists a unique $u\in \Bbb Z$ such that $\color{blue}{ |x-u|\le \frac{1}{2}}$
Proof: Let denote by  $\lfloor \ell \rfloor$ is the floor of $\ell$. Then We know that
$$\lfloor x+\frac{1}{2} \rfloor\le x+\frac{1}{2}  \lt \lfloor x+\frac{1}{2}\rfloor +1\implies-\frac{1}{2}\le x -\lfloor x+\frac{1}{2} \rfloor\lt \frac{1}{2} $$
Taking $ u= \lfloor x+\frac{1}{2}$
The unicity  follows from the unicity follows from the unicity of the floor.

Now let  $z,w\in \Bbb Z[i]\setminus\{0\} $ then $\frac{z}{w}$ can be written as
$$\color{red}{ \frac{z}{w}= x+iy :=\frac{z\bar{w}}{|w|^2}~~~~~ \text{with }~~~x,y\in \Bbb Q.}$$
From the Lemma there exist $u,v\in \Bbb Z$ such that
$\color{blue}{ |x-u|\le \frac{1}{2}~~~~\text{and}~~~|y-v|\le \frac{1}{2}}.$
Then,we can write
$$\color{red}{ \frac{z}{w}= x+iy = q +t~~~~~ \text{with }~~~q\in \Bbb Z, t\in \Bbb Q.}$$
Where,  $ \color{blue}{q= u+iv ~~\text{and}~~~t =x-u+i(y-v)}$. Then we have, $$\color{blue}{ z= qw +tw \implies r:= tw = z-qw \in \Bbb Z.}$$
Hence $\color{red}{z= qw +r}$ with $q,r \in \Bbb Z$ with $r=tw$ where we have,
$$\color{blue}{ t =x-u+i(y-v),~~~|x-u|\le \frac{1}{2}~~~~\text{and}~~~|y-v|\le \frac{1}{2}}$$
Which means that, $$N(t) =|x-u|^2+|y-v|^2\le \frac{1}{2}$$
Therefore,
$$ N(r) =N(tw) =N(t)N(w) \le \frac12N(w) \lt N(w)$$
That is $$ \color{red}{N(r)  \lt N(w).}$$

conclusion N is divivion for the Ring $\Bbb Z[i].$

A: Hint: 
For first condition $|z_1|,|z_2| \geq 1$ or zero
About second one: $\mathbb{Z}$ satisfies Euclidean algorithm  
