Exercise 9.35

Proof. We give a generalization of Exercises 9.32 and 9.33 : if $n\equiv 1[4],n\ne 1$, then ${\chi }_n(1+i)=i^{(n-1)∕4}$.

By Exercises 9.32 and 9.33, we know that if $p\equiv 1[4]$ is a rational prime, then

and if $q\equiv 3[4]$, in other words $-q\equiv 1[4]$, where $q$ is a rational prime, then

Let $n\in \mathbb{Z},n\equiv 1[4],n\ne 1$.

If $n>0$, $n=q_1{q}_2\cdots q_k{p}_1{p}_2\cdots p_l$, where $q_i\equiv -1[4],p_i\equiv 1[4]$, thus $k$ is even.

If $n<0$, $n=-q_1{q}_2\cdots q_k{p}_1{p}_2\cdots p_l$, with $k$ odd. In both cases,

so we can write

where $s_i\equiv 1[4],1\le i\le N$.

the last equality resulting of Exercise 9.44.

Conclusion : if $n\in \mathbb{Z},n\equiv 1[4],n\ne 1$, then ${\chi }_n(1+i)=i^{(n-1)∕4}$.

Let $\pi =a+\mathit{bi},a\wedge b=1$ a primary irreducible. As $a(1+i)=a+b+i(a+\mathit{bi})$, $a(1+i)\equiv a+b[\pi ]$, so

As $\pi =a+\mathit{bi}$ is primary, $a+b\equiv 1[4]$.

If $a+b=1$, then ${\chi }_{\pi }(a){\chi }_{\pi }(1+i)={\chi }_{\pi }(a+b)=1=i^{3(a+b-1)∕4}$. If not, the Law of Biquadratic Reciprocity (Proposition 9.9.8) gives

Now $b\equiv -a(\mathit{mod}a+b)$, so $a+\mathit{bi}\equiv a(1-i)\equiv -\mathit{ia}(1+i)(\mathit{mod}a+b)$. Therefore

Since $n\equiv 1[4]$, ${\chi }_n(i)={(-1)}^{(n-1)∕4}$ (Prop.9.8.6), thus

Consequently, since $a+b\equiv 1[4]$, ${\chi }_{a+b}(-1)=1$ .

As $a\wedge b=1,(a+b)\wedge a=1$, thus ${\chi }_{a+b}(a)=1$ (Prop 9.8.5).

$a+b\equiv 1[4]$, thus ${\chi }_{a+b}(i)={(-1)}^{(a+b-1)∕4}$ (Prop. 9.8.6).

From the first part of this proof, ${\chi }_{a+b}(1+i)=i^{(a+b-1)∕4}$, so

Conclusion : if $\pi =a+\mathit{bi}$ is a primary irreducible, such that $a\wedge b=1$, then

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