Exercise 5.32

Proof. Let $p$ be an odd prime number, and $\zeta =e^{2\mathit{i\pi }∕p}$. Then ${\zeta }^p=1$.

Let

As ${\zeta }^j$ depends only of the class $\bar{j}\in {𝔽}_p$, this product can be written

since $f:{𝔽}_p\to {𝔽}_p,x\mapsto 2x$ is a bijection. So

Since ${\zeta }^0=1,\zeta ,\cdots ,{\zeta }^{p-1}$ are the roots of the polynomial $f(x)=x^p-1$, then $1+{\zeta }^0,\cdots ,1+{\zeta }^{p-1}$ are the roots of $g(x)={(x-1)}^p-1=f(x-1)$, so $g(x)={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{p-1}(x-(1+{\zeta }^k))$.

As $g(0)={(-1)}^p-1=-2=(-1-{\zeta }^0)\cdots (-1-{\zeta }^{p-1})=-{\prod <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^{p-1}({\zeta }^k+1)$, we obtain

so

As $\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi -\alpha )=\cos <!--\; FUNCTION\; APPLICATION\; -->(\alpha )$,

$\text{∙}$

Case 1: if $1\le j\le p∕4,0\le 2\mathit{\pi j}∕p<\pi ∕2$, thus $\cos <!--\; FUNCTION\; APPLICATION\; -->(2\mathit{\pi j}∕p)>0$.

$\text{∙}$

Case 2: if $p∕4<j\le (p-1)∕2,\pi ∕2<2\mathit{\pi j}∕p<\pi$, thus $\cos <!--\; FUNCTION\; APPLICATION\; -->(2\mathit{\pi j}∕p)<0$.

In the first case, $2\le 2j\le (p-1)∕2$ : the least residue of $2j$ is positive. In the second case $p∕2<2j\le p-1$ : the least residue of $2j$ is negative.

Let $\mu$ be the number of negative least residues of the integer $2j,1\le j\le (p-1)∕2$. We know from Gauss’ Lemma that $(2∕p)={(-1)}^{\mu }$. As $\mu$ is also the number of $j,1\le j\le (p-1)∕2$ such that $\cos <!--\; FUNCTION\; APPLICATION\; -->(2\mathit{\pi j}∕p)<0$,

If $p\equiv 1[8]$ , $p=8q+1,q\in \mathbb{N}$. For $1\le j\le (p-1)∕2$,

so $\mu =2q$ and $(2∕p)={(-1)}^{\mu }=1$.

If $p\equiv -1[8]$, $p=8q-1,q\in {\mathbb{N}}^{\text{∗}}$.

thus $\mu =2q$ and $(2∕p)={(-1)}^{\mu }=1$.

If $p\equiv 3[8]$, $p=8q+3,q\in \mathbb{N}$.

thus $\mu =2q+1$ and $(2∕p)={(-1)}^{\mu }=1$.

If $p\equiv -3[8]$, $p=8q-3,q\in {\mathbb{N}}^{\text{∗}}$,

thus $\mu =2q-1$ and $(2∕p)={(-1)}^{\mu }=1$. □