Exercise 8.18

Proof. Recall that $\pi =-J(\chi ,\rho )\in \mathbb{Z}[i]$, so $\pi =a+\mathit{bi},a,b\in \mathbb{Z}$.

1)

We begin by proving that $\pi \equiv 1(\mathit{mod}2+2i)$ (see Chapter 11, Section 5).

For all $t\in {𝔽}_p^{\text{∗}}$, $\rho (t)=\pm 1$, so $\rho (t)-1\equiv 0(\mathit{mod}2)$.

Let’s verify that $\chi (t)-1\equiv 0(\mathit{mod}1+i)$. $\chi (t)\in \{1,-1,i,-i\}$, so $\chi (t)-1\in \{0,-2,i-1,-i-1\}$. As $2=(1-i)(1+i)$ and $i-1=i(1+i)$, we obtain

$$\forall <!--\; FUNCTION\; APPLICATION\; -->t\in {𝔽}_p^{\text{∗}},1+i\mid \chi (t)-1.$$

Thus

$$\forall <!--\; FUNCTION\; APPLICATION\; -->s\in {𝔽}_p^{\text{∗}},\forall <!--\; FUNCTION\; APPLICATION\; -->t\in {𝔽}_p^{\text{∗}},(\rho (s)-1)(\chi (t)-1)\equiv 0(\mathit{mod}2+2i).$$

Moreover, if $s=0,t=1$, then $\chi (t)=1$, and if $s=1,t=0$, then $\rho (s)=1$, therefore

$$\underset{s+t=1}{\sum }(\rho (s)-1)(\chi (t)-1)\equiv 0(mod2+2i).$$

This gives, when developing this expression, :

$$-\pi -\underset{t\in {𝔽}_p}{\sum }\chi (t)-\underset{s\in {𝔽}_p}{\sum }\rho (s)+p\equiv 0(mod2+2i).$$

As ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_t\chi (t)={\sum }_s\rho (s)=0$, we obtain

$$\pi \equiv p(\mathit{mod}2+2i).$$

Finally, $p\equiv 1(\mathit{mod}4)$, and $2+2i\mid 4$, since $4=(1-i)(2+2i)$, thus $p\equiv 1(\mathit{mod}2+2i)$, and

$$\pi \equiv 1(\mathit{mod}2+2i).$$

2)

By Corollary of Theorem 1, $N(\pi )=N(J(\chi ,\rho ))=p=a^2+b^2$.

We know that $p\equiv 1(\mathit{mod}4)$, $p=a^2+b^2$ and $a+\mathit{ib}\equiv 1(\mathit{mod}2+2i)$. Then we prove that $a$ is odd, $b$ is even, and $a\equiv 1(\mathit{mod}4)$ if $4\mid b$ and $a\equiv -1(\mathit{mod}4)$ if $4\nmid b$.

$a+\mathit{bi}\equiv 1(\mathit{mod}2+2i)$, thus $a+\mathit{bi}\equiv 1(\mathit{mod}2)$, so that $a$ is odd, and $b$ is even.

$\text{∙}$ If $4\mid b$, then $2+2i\mid b$.

Therefore $a\equiv 1(\mathit{mod}2+2i)$, and by complex conjugation, $a\equiv 1(\mathit{mod}2-2i)$, so $(2+2i)(2-2i)=8\mid {(a-1)}^2$, thus $4\mid a-1$.

$\text{∙}$ If $4\nmid b$, then $b=4k+2,k\in \mathbb{Z}$.

Therefore, $1\equiv a+\mathit{bi}\equiv a+2i(\mathit{mod}2+2i)$. As $2i\equiv -2(\mathit{mod}2+2i)$, $a\equiv 3\equiv -1(\mathit{mod}2+2i)$. By conjugation, $a\equiv -1(\mathit{mod}2-2i)$. Multiplying these congruences, we obtain $8\mid {(a+1)}^2$, thus $a\equiv -1(\mathit{mod}4)$.

3)

$\pi =-J(\chi ,\rho )=a+\mathit{bi}$ is such that $a^2+b^2=p$, $a$ odd, $b$ even and also $$(4\mid b\mathrm{and}a\equiv 1[4])\mathrm{or}(4\nmid b\mathrm{and}a\equiv -1[4]).$$

If $p=A^2+B^2$, $A$ odd and $B$ even, then also $p={(-A)}^2+B^2$, and $A\equiv 1(\mathit{mod}4)$ or $-A\equiv 1(\mathit{mod}4)$. So there exists a decomposition $p=A^2+B^2$ such that $A\equiv 1(\mathit{mod}4)$, and $A$ is uniquely determined (Ex. 12).

Let’s verify that $4\mid b$ if $p\equiv 1(\mathit{mod}8),4\nmid b$ if $p\equiv 5(\mathit{mod}8)$.

$p=a^2+b^2,a=2a^{\prime }+1,b=2b^{\prime }$, so $p=4{a{}^{\prime }}^2+4a^{\prime }+1+4{b{}^{\prime }}^2=8\frac{a^{\prime }(a^{\prime }+1)}{2}+1+4{b{}^{\prime }}^2$.

Hence $4\mid b⟺2\mid b^{\prime }⟺8\mid p-1$.

Therefore if $p\equiv 1(\mathit{mod}8)$, $\mathrm{Re}\pi =a=A$, and if $p\equiv 5(\mathit{mod}8)$, $\mathrm{Re}\pi =a=-A$.

In conclusion, by Exercise 8.17 :

if $p=A^2+B^2,A\equiv 1(\mathit{mod}4)$, and $N=N(x^4+y^4=1)$ in ${𝔽}_p$,

(a)

$N=p-3-6A$ if $p\equiv 1(\mathit{mod}8)$,

(b)

$N=p+1+2A$ if $p\equiv 5(\mathit{mod}8)$.

Note : if $p\equiv -1(\mathit{mod}4)$, then there is no character of order 4 on ${𝔽}_p^{\text{∗}}$, and $d=4\wedge (p-1)=4\wedge (4k+2)=2$. By Exercise 1, we obtain

Using Chapter 8, Section 3, we obtain

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