Prove Proposition (6.19).
Hint: Use Lagrange’s Theorem:
Let $P(x) = c_m x^m + c_{m−1}x^{m−1} + \ldots + c_1x + c_0$ where $c_m \not \equiv 0 \pmod p$ and $p$ is prime. Then polynomial congruence $P (x) \equiv 0 \pmod p$ has at most m incongruent solutions.
Let's remind ourselves of Proposition (6.19).
Let $p$ be prime and $d \mid (p− 1)$. The congruence
$$ x^d \equiv 1 \pmod p $$
has exactly $d$ incongruent solutions.
Since $p$ is prime, we know it has a primitive root, which we can call $r$. Applying Propositions (6.15) and (6.16) to $x^d \equiv 1 \pmod p$ gives
$$ d \times \text{ind}_r(x) \equiv 0 \pmod {p-1} $$
This linear congruence has solutions for $\text{ind}_r(x)$ if $g=\gcd(p-1,d)=d$ divides 0, which it does. There are $g=d$ incongruent solutions modulo $p-1$.
This means there are $d$ incongruent solutions for $x$ modulo $p-1$.
We can extend this to say there are $d$ incongruent solutions for $x$ modulo $p$, because we know $x \not \mid p$.
Note: The author's solution uses Lagrange's Theorem because it doesn't assume that all primes have a primitive root.