Show that if $\beta \mid \alpha$ in $\mathbb Z[i]$ then $N(\beta) \mid N(\alpha)$ Show that if $\beta \mid \alpha$ in $\mathbb Z[i]$  then $N(\beta) \hspace{1mm}| \hspace{1mm} N(\alpha)$ where $\alpha$ is a prime in $\mathbb Z[i]$ and $N(a + bi) = a^2 + b^2$.
So we are working with Gaussian integers and primes there. How can this be true? Why is it related to the divisibility of the norm?
 A: Hint:
$$N(\alpha\beta)=N(\alpha)N(\beta)$$
If $\beta\mid\alpha$, you can write $\alpha=\beta\gamma$. Now, what does $N(\alpha)$ look like?
A: Because the norm function is chosen for the express purpose of enabling comparison of numbers, especially if one number is a divisor of another.
Algebraically, $N(\alpha \beta)$ should be such that $|N(\alpha)| < |N(\alpha \beta)|$ and $|N(\beta)| < |N(\alpha \beta)|$, and also $N(\alpha) \mid N(\alpha \beta)$ and $N(\beta) \mid N(\alpha \beta)$.
Consider for example $\gcd(3, -21)$ in $\mathbb Z$. Define $N(n) = n^2$. Then, although $-21 < 3$, $9 < 441$, which tells us that 3 is closer to 0 than $-21$. So next we see if 3 is a divisor of $-21$. Indeed $-21 = -3 \times -7 + 0$.
Of course you can also use the absolute value function, but I chose to use squaring instead to hopefully make the similarity to the Gaussian integers clearer. Norms of Gaussian integers are never negative, so we don't need to worry about taking the absolute value of a norm.
So if $\alpha = a + bi$, then $N(\alpha) = a^2 + b^2$. For example, $N(-2 + i) = 4 + 1 = 5$. And $N(5 + 0i) = 25 + 0 = 25$. Since 5 is a divisor of 25, this suggests that $-2 + i$ is a divisor of 5. We see that $$(-2 - i)(-2 + i) = 4 - 2i + 2i - (-1) = 4 + 1 = 5.$$

Norms are also useful when Gaussian integers are coprime. For example, $\gcd(-2 + i, 6) = 1$ and we readily see that $6 = (-2 - i)(-2 + i) + 1$.
A: It suffices to prove that if $\beta\mid\alpha$ then $\overline{\beta}\mid\overline{\alpha}$, because
$$N(a+bi)=a^2+b^2=(a+bi)(a-bi)=(a+bi)(\overline{a+bi}).$$
