Prove that if $x \equiv x_{1}(\bmod m)$ is a solution to (1), then by $(a, m)=1$ we know $\left(x_{1}, m\right)=1$. Therefore, by Theorem 5 in §2, there must be a $y_{1}$ such that
$$x_{1} \equiv g^{y_{1}}(\bmod m)$$

Thus, we have
$$g^{n y_{1}} \equiv a(\bmod m)$$

Furthermore, by Property I in §3, we have
$$n y_{1} \equiv \gamma(a)(\bmod \varphi(m))$$

This shows that $y \equiv y_{1}(\bmod \varphi(m))$ is a solution to the linear congruence
$$n y \equiv \gamma(a)(\bmod \varphi(m))$$

Conversely, if $y \equiv y_{1}(\bmod \varphi(m))$ is a solution to (5), then by Property I in §3, we can also deduce that equation (4) holds. Therefore, when $x_{1}$ is given by equation (3), $x \equiv x_{1}(\bmod m)$ must be a solution to (1). This proves that the congruence equation (1) $((a, m)=1)$ and the congruence equation (5) have solutions simultaneously or not. Moreover, for any
$$x_{2} \equiv g^{y_{2}}(\bmod m)$$

By Property IV in §1 (taking $a=g$), we know that $x_{1} \equiv x_{2}(\bmod m)$ if and only if
$$y_{1} \equiv y_{2}(\bmod \varphi(m))$$

Therefore, when the congruence equation (1) $((a, m)=1)$ has solutions, it has the same number of solutions as the congruence equation (5). By Theorem 2 in Chapter 4, §2, the necessary and sufficient condition for (5) to have solutions is that equation (2) holds. When it has solutions, there are exactly $(n, \varphi(m))$ solutions. This, together with the previous discussion, proves the theorem.

Theorem 1 provides a method for solving equation (1) $((a, m)=1)$ when the modulus $m$ has a primitive root: (i) use the index table to find the index $\gamma(a)$ of $a$; (ii) solve the congruence equation (5); (iii) if (5) has solutions, then for each solution $y_{1}(\bmod \varphi(m))$, use the index table to find $x_{1}$ satisfying equation (3). The resulting $x_{1}(\bmod m)$ are all the solutions to (1).