Exercise 5.1

Proof. $\text{∙}$ $a=5,p=7$.

The array of values of the least residues modulo $p=7$, for $1\le k\le (p-1)∕2$.

So the number of negative least residues is $\mu =1$, and $\left(\frac{5}{7}\right)={(-1)}^{\mu }=-1$.

$\text{∙}$ $a=3,p=11$.

So $\mu =2$, $\left(\frac{3}{11}\right)={(-1)}^{\mu }=1$.

$\text{∙}$ $a=6,p=13$.

So $\mu =3$, $\left(\frac{6}{13}\right)={(-1)}^{\mu }=-1$.

$\text{∙}$ If $a=-1$, and $p$ is an odd prime, the values of the least residues of $-k$ modulo $p$ for $k=1,2,\dots ,(p-1)∕2$ are $-k$, all negative. So the number of negative least residues is $\mu =(p-1)∕2$, and $\left(\frac{-1}{p}\right)={(-1)}^{(p-1)∕2}$. □