Exercise 4.21

Proof. Let $x={\bar{g}}^{(p-1)∕d}\in {𝔽}_p^{\text{∗}}$, where $g$ is a primitive root modulo $p$. For all $k\in \mathbb{Z}$,

So the order of ${\bar{g}}^{(p-1)∕d}$ is $d$.

$\text{∙}$ If $\bar{a}={\bar{g}}^{\mathit{kd}}$, then $\bar{a}=x^d$, where $x={\bar{g}}^k$, so $\bar{a}$ is a $d$th power.

$\text{∙}$ If $\bar{a}\ne \bar{0}$ is a $d$th power, $\bar{a}=x^d,x\in {𝔽}_p^{\text{∗}}$. As $x\in ⟨\bar{g}⟩$, $x={\bar{g}}^k$, so $\bar{a}={\bar{g}}^{\mathit{kd}}$.

So, if $a\not\equiv 0(\mathit{mod}p)$, $a$ is a $d$th power iff $a\equiv g^{\mathit{kd}}(\mathit{mod}p)$ for some $k$.

By example (Ex. 4.20), $2$ is a primitive root modulo $19$, so the 6th powers modulo 19 are $2^0=1,2^6=7,2^{12}=11$. □