Exercise 9.38

Proof. Let $\pi =a+\mathit{bi}$ a primary irreducible. Then $a$ is odd, and $b$ is even, and $N(\pi )=a^2+b^2$. Then

By Lemma 6, section 7, $a\equiv 1[4],b\equiv 0[4]$, or $a\equiv 3[4],b\equiv 2[4]$.

$\text{∙}$

If $a\equiv 1[4],b\equiv 0[4]$, then ${(-1)}^{\frac{a+1}{2}}=-1,{(-1)}^{\frac{b^2}{4}}=+1$ , so $${\chi }_{\pi }(-1)={(-1)}^{\frac{a-1}{2}}.$$

$\text{∙}$

If $a\equiv 3[4],b\equiv 2[4]$, then ${(-1)}^{\frac{a+1}{2}}=1,{(-1)}^{\frac{b^2}{4}}=-1$, so $${\chi }_{\pi }(-1)=-1={(-1)}^{\frac{a-1}{2}}.$$

Conclusion: if $\pi$ is a primary irreducible in $\mathbb{Z}[i]$, then

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