On the Norm of Eisenstein integer Let us start with the set $\mathbb{Z[w]}=\{a+bw: a,b\in\mathbb{Z}\}$ where $w=(-1+i\sqrt{3})/2$. This set lies inside the set of complex numbers. Let us define the function
$$N:\mathbb{Z[w]}\to\mathbb{Z}^+\cup\{0\}$$ by $N(a+bw)=a^2+b^2-ab$. It is easy to check that for any $\alpha,\beta\in\mathbb{Z}[w]$ we have $$N(\alpha\cdot\beta)=N(\alpha)\cdot N(\beta).$$
Now, let $\alpha=a+bw$ and $\beta=c+dw$ be two elements of $\mathbb{Z}[w]$ with $\beta\neq 0$. Then
$$\frac{\alpha}{\beta}=t+sw$$ for some $t,s\in \mathbb{Q}$. Let us choose two integers $p$ and $q$ such that $|t-p|\leq\frac{1}{2}$ and $|s-q|\leq\frac{1}{2}$. Write
$$\phi=(t-p)+(s-q)w.$$
In here (please see Proposition 1.1), he wrote that $N(\phi)=(t-p)^2+(s-q)^2-(t-p)(s-q)$. I got confused with this because $t-p$ and $s-q$ may not be integers. Is there any reason how he come up with $N(\phi)$? 
 A: Presumably Prof. Valmir Buçaj is still alive, so you could ask him. My guess is that he likes to avoid fractions as much as possible, and though it greatly obscures things for specific examples, it perhaps simplifies the process of working out the proof of the theorem.
Declare $$\alpha = a + b \omega = \frac{2a - b}{2} + \frac{b \sqrt{-3}}{2} = \frac{v}{2} + \frac{w \sqrt{-3}}{2}$$ and $$\beta = c + d \omega = \frac{2c - d}{2} + \frac{d \sqrt{-3}}{2} = \frac{x}{2} + \frac{y \sqrt{-3}}{2}.$$ I think I did that correctly... someone will loudly point it out whatever mistake I've made in there. (Also, there is a slight inefficiency in defining $b = w$ and $d = y$, but I wanted to make the point that $\omega \neq w$).
It is assumed that $N(\alpha) \geq N(\beta)$, but if not, the Euclidean algorithm ought to switch them so that they satisfy that loose inequality. And $N(\beta)$ is easily calculated thus: $$N(\beta) = \frac{x^2}{4} + \frac{3y^2}{4}.$$ It's definitely possible that neither of those summands are integers from $\mathbb Z$. But if $x, y \in \mathbb Z$, $N(\beta)$ will work out to an integer just the same. e.g., $$N\left(\frac{5}{2} + \frac{\sqrt{-3}}{2}\right) = \frac{5^2}{4} + \frac{3 \times 1^2}{4} = \frac{25}{4} + \frac{3}{4} = \frac{28}{4} = 7.$$ This is, in my opinion, easier than calculating $N(3 + \omega)$. It also makes it much easier to locate $\beta$ in the complex plane.
Now we wish to compute $$\frac{\alpha}{\beta}.$$ Suppose $$\alpha = \frac{3}{2} + \frac{5 \sqrt{-3}}{2},$$ and $\beta$ is the same as in the previous example. Then $$\frac{\alpha}{\beta} = \frac{\frac{3}{2} + \frac{5 \sqrt{-3}}{2}}{\frac{5}{2} + \frac{\sqrt{-3}}{2}} = \ldots$$ I'm sorry, I don't feel like working out that computation, I'd rather go with $$\frac{\alpha}{\beta} = \frac{4 + 5 \omega}{3 + \omega} = \ldots$$ Yeah, that looks a bit more manageable. I'll worry about locating them in the complex plane on another occasion.
