Example 8 Given $p=29$, try to find a primitive root of $p^{2}=841$.

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Let's verify that 14 is a primitive root of 29. Since
$$\varphi(p)=28=(4)(7)=(2)(14)$$

To verify that 14 is indeed a primitive root of \( p = 29 \), it is not necessary to check for all \( m (1 \leqslant m < 28) \) that
$$14^{m} \equiv 1 \pmod{29}$$

but only the following conditions need to be verified (see Theorem 1 in the previous section):
$$14^{2} \equiv 1, 14^{4} \equiv 1, 14^{7} \equiv 1, 14^{14} \equiv 1 \pmod{29}$$

Calculations yield:
$$\begin{array}{c}
14^{2} \equiv 22, \quad 14^{4} \equiv 22^{2} \equiv (-7)^{2} \equiv 22 \\
14^{7} \equiv (14)^{4}(14)^{2}(14) \equiv (20)(22)(14) \equiv 12 \\
14^{14} \equiv (12)^{2} \equiv -1 \pmod{29}
\end{array}$$

Thus, the conditions in (21) are indeed satisfied, and 14 must be a primitive root modulo 29.
Further calculations yield:
$$\begin{aligned}
(14)^{28} & = (196)^{14} = (38416)^{7} \equiv (571)^{7} = (571)(326041)^{3} \\
& \equiv (571)(574)^{3} = (327754)(329476) \equiv (605)(645) \\
& = 390225 \equiv 1 \pmod{29^2}
\end{aligned}$$

Therefore, 14 is not a primitive root modulo \( 29^2 \). To find a primitive root of \( 29^2 \) from 14, we consider \( 14 + 29 = 43 \), since
$$43 \equiv 14 \pmod{29}$$

43 is still a primitive root modulo 29. However,
$$\begin{aligned}
(43)^{28} & = (3418801)^{7} \equiv (136)^{7} = (136)(18496)^{3} \\
& \equiv (136)(-6)^{3} = -29376 \equiv 59 \equiv 1 \pmod{29^2}
\end{aligned}$$

Thus, by Theorem 8, 43 is a primitive root modulo \( 29^2 \).

The method used in the above example is generally applicable. That is, if \( g \) is a primitive root modulo \( p \geqslant 3 \) and
$$g^{p-1} \equiv 1 \pmod{p^2}$$

then \( g + p \) must be a primitive root modulo \( p^2 \).

Since it is known that a primitive root modulo a prime \( p \) always exists, the above result implies that a primitive root modulo \( p^2 \) also always exists for an odd prime \( p \geqslant 3 \).

Surprisingly, for any odd prime power \( p^s, s \geqslant 2 \), condition (17) is also a necessary and sufficient condition for a primitive root \( g \) of \( p \) to remain a primitive root of \( p^s \). When \( g \) does not satisfy (17), taking \( g + p \) will yield a primitive root of \( p^s \). This is the result stated below.