Exercise 9.29

Proof. $a\equiv {(-1)}^{(p-1)∕4}[4]$ (Ex. 9.27(a)), $a{(-1)}^{(p-1)∕4}\equiv 1[4]$, thus $a{(-1)}^{(p-1)∕4}$ is primary (if $a\ne \pm 1$).

If $a=\pm 1$ is an unit, $a{(-1)}^{(p-1)∕4}=1$ and ${\chi }_{\pi }(a{(-1)}^{(p-1)∕4})=1={(-1)}^{(a^2-1)∕8}$, so we can suppose that $a$ is not an unit.

As $a{(-1)}^{(p-1)∕4}\equiv 1(\mathit{mod}4)$, the Law of Biquadratic Reciprocity (Prop. 9.9.8) gives

As $a\wedge b=1$ (since $p=a^2+b^2$), ${\chi }_a(b)=1$ (Prop. 9.8.5, with $a\ne 1$), so

If $a\equiv 1[4]$, Proposition 9.8.6 gives ${\chi }_a(i)={(-1)}^{(a-1)∕4}$. Write $a=4A+1,A\in \mathbb{Z}$. Then

If $a\equiv -1[4]$, then ${\chi }_a(i)={\chi }_{-a}(i)={(-1)}^{(-a-1)∕4}$ by the same proposition. Write $a=4A-1,A\in \mathbb{Z}$. Then

So, for each odd $a$, $a\ne \pm 1$,

Conclusion : if $\pi =a+\mathit{bi}$ is a primary irreducible such that $N(\pi )=p$, then

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