Let $p$ be prime and $\gcd (a, p) = 1$. Prove that $x^m \equiv a \pmod p$ has a solution if and only if a $a^{\frac{p-1}{g}} \equiv 1 \pmod p$ where $g = \gcd (m, p− 1)$.
We'll use Proposition (6.17) which states that if $n$ has a primitive root and $a$ and $n$ are relatively prime, the congruence $x^m \equiv a \pmod n$ has a solution if and only if $a^{\phi(n)/g} \equiv 1 \pmod n$ where $g = \gcd (m, \phi (n))$. Additionally, there are exactly $g$ incongruent solutions.
Since $p$ is prime, it has a primitive root (stated on page 255, but proven later in the book). We're given $a$ and $p$ are relatively prime. And so the premises of Proposition (6,17) are satisfied. This means the conclusion that $x^m \equiv a \pmod p$ has a solution if and only if
$$ a^{\frac{\phi(p)}{g}} \equiv a^{\frac{p-1}{g}} \equiv 1 \pmod p $$
where $g=\gcd(m, \phi(p)) = \gcd(m, p-1)$.
The statement to be proven is simply Proposition (6.17) with $n$ as prime $p$.