Exercise 9.34

Proof. Let $\pi =a+\mathit{bi}$ be a primary irreducible, with $a\wedge b=1$, so $b\ne 0$. We can apply the result of Exercise 9.29:

(a)

Suppose that $\pi \equiv 1[4]$.

Then $a\equiv 1[4],b\equiv 0[4],a=4A+1,b=4B,A,B\in \mathbb{Z}$.

As ${\chi }_{\pi }(-1)={(-1)}^{(a-1)∕2}$,

$${\chi }_{\pi }(a)={(-1)}^{\frac{a-1}{2}\frac{p-1}{4}}{(-1)}^{\frac{a^2-1}{8}},$$

where

$$p=\mathit{N\pi }=a^2+b^2,{(-1)}^{(p-1)∕4}={(-1)}^{\frac{a^2-1}{4}+\frac{b^2}{4}}={(-1)}^{4A^2+2A+4B^2}=1,$$

thus ${(-1)}^{\frac{a-1}{2}\frac{p-1}{4}}=1$.

$${\chi }_{\pi }(a)={(-1)}^{(a^2-1)∕8}={(-1)}^{2A^2+A}={(-1)}^A={(-1)}^{(a-1)∕4}=i^{(a-1)∕2}.$$

Conclusion: if $\pi \equiv 1[4]$, ${\chi }_{\pi }(a)=i^{(a-1)∕2}$.

(b)

Suppose that $\pi \equiv 3+2i[4]$.

Then $a\equiv 3[4],b\equiv 2[4],a=4A+3,b=4B+2,A,B\in \mathbb{Z}$. As in (a),

$${\chi }_{\pi }(a)={(-1)}^{\frac{a-1}{2}\frac{p-1}{4}}{(-1)}^{\frac{a^2-1}{8}},$$

where $a^2+b^2-1=16A^2+24A+16B^2+16B+12\equiv 4[8]$, so $\frac{a^2+b^2-1}{4}\equiv 1[2]$, thus ${(-1)}^{(p-1)∕4}={(-1)}^{(a^2+b^2-1)∕4}=-1$.

$${(-1)}^{\frac{a-1}{2}\frac{p-1}{4}}={(-1)}^{\frac{a-1}{2}}={(-1)}^{2A+1}=-1,$$

$$\frac{a^2-1}{8}=2A^2+3A+1,{(-1)}^{(a^2-1)∕8}={(-1)}^{3A+1}={(-1)}^{A+1}={(-1)}^{(a+1)∕4},$$

$${\chi }_{\pi }(a)=-{(-1)}^{(a+1)∕4}=-i^{(a+1)∕2}.$$

Moreover

$$\frac{a+1}{2}\equiv \frac{-a-1}{2}[4]⟺a+1\equiv -a-1[8]⟺2a\equiv -2[8]⟺a\equiv 3[4],$$

thus $i^{(a+1)∕2}=i^{(-a-1)∕2}$.

Conclusion : if $\pi \equiv 3+2i[4]$, ${\chi }_{\pi }(a)=-i^{(-a-1)∕2}$.