Exercise 9.27

Proof. Let $\pi =a+\mathit{bi}$ be a primary prime in $\mathbb{Z}[i]$, $b\ne 0$, such that $p=N(\pi )$. Then

By Lemma 6, Section 7, $a$ is odd, $b$ even, and

(a)

$\text{∙}$

Case 1: $a\equiv 1[4],b\equiv 0[4]$. Then $a=4A+1,b=4B,A,B\in \mathbb{Z}$, so $(a^2+b^2-1)∕4=4A^2+4B^2+2A$ is even :

${(-1)}^{(p-1)∕4}={(-1)}^{(a^2+b^2-1)∕4}=1$, and $a\equiv 1[4]$, thus $a\equiv {(-1)}^{(p-1)∕4}[4]$.

$\text{∙}$

Case 2: $a\equiv 3[4],b\equiv 2(\mathit{mod}4)$.

$a=4A+3,b=4B+2,a^2+b^2-1=16A^2+24A+9+16B^2+16B+4-1\equiv 4[8]$, so $(a^2+b^2-1)∕4\equiv 1[2],{(-1)}^{(p-1)∕4}={(-1)}^{(a^2+b^2-1)∕4}=-1$, and $a\equiv -1[4]$, thus $a\equiv {(-1)}^{(p-1)∕4}[4]$.

In both cases,

$$a\equiv {(-1)}^{(p-1)∕4}[4].$$

(b)

In every case, $b\equiv a-1[4]$, thus $$b\equiv {(-1)}^{(p-1)∕4}-1[4].$$

In other words, for all primary primes $\pi =a+\mathit{bi}$ such that $N(\pi )=p$,