19. Given the prime $p=71$ with 7 as one of its primitive roots, try to find all primitive roots of 71, and find a primitive root for $p^{2}$ and $2 p^{2}$.
19. Solution: From $\varphi(p-1)=\varphi(70)=24$, we know that $p=71$ has 24 primitive roots. The 24 natural numbers between 1 and 70 that are coprime with 70 are: 1, 3, 9, 11, 13, 17, 19, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 51, 53, 57, 59, 61, 67, 69.
$$\begin{array}{l}
7^{1}=7,7^{3}=343 \equiv 59,7^{9} \equiv 59^{3}=205379 \equiv 47,7^{11} \\
\equiv 7^{2} \cdot 47 \equiv 31,7^{13} \equiv 31 \cdot 7^{2} \equiv 28,7^{17} \equiv 7^{4} \cdot 28 \equiv \\
7 \cdot 59 \cdot 28 \equiv 62,7^{19} \equiv 7^{2} \cdot 62 \equiv 56,7^{23} \equiv 7^{3} \cdot 7^{3} \\
\cdot 56 \equiv 7 \cdot 59 \cdot 56 \equiv 53,7^{27} \equiv 7 \cdot 59 \cdot 53 \equiv 21, \\
7^{29} \equiv 7^{2} \cdot 21 \equiv 35,7^{31} \equiv 7^{2} \cdot 35 \equiv 11,7^{33} \equiv 7^{2} \\
\cdot 11 \equiv 42,7^{37} \equiv 7 \cdot 59 \cdot 42 \equiv 22,7^{39} \\
\equiv 7^{2} \cdot 22=1078 \equiv 13,7^{41} \equiv 7^{2} \cdot 13=637 \equiv 69,7^{43} \\
\equiv 7^{2} \cdot 69=3381 \equiv 44,7^{47} \equiv 7 \cdot 59 \cdot 44=18172 \\
\equiv 67,7^{51} \equiv 7 \cdot 59 \cdot 67=27671 \equiv 52,7^{53} \equiv 7^{2} \cdot 52 \\
=2548 \equiv 63,7^{57} \equiv 7^{2} \cdot 59 \cdot 63=26019 \equiv 33,7^{59} \equiv 7^{2} \\
\cdot 33=1617 \equiv 55,7^{61} \equiv 7^{2} \cdot 55=2695 \equiv 68,7^{67} \\
\equiv\left(7^{3}\right)^{2} \cdot 7^{61} \equiv(59)^{2} \cdot 68=236708 \equiv 65,7^{69} \equiv 7^{2} \\
\cdot 65=3185 \equiv 61 \neq(\bmod 71) .
\end{array}$$

Thus, the 24 primitive roots modulo 71 are
$61(\bmod 71)$.
Further calculations yield
$$\begin{array}{c}
7^{5}=16807 \equiv 1684\left(\bmod 71^{2}\right) \\
7^{10} \equiv(1684)^{2}=2835856 \equiv 2814\left(\bmod 71^{2}\right) \\
7^{20} \equiv(2814)^{2}=7918596 \equiv-815\left(\bmod 71^{2}\right) \\
7^{40} \equiv(-815)^{2}=664225 \equiv-1187\left(\bmod 71^{2}\right) \\
7^{70}=7^{10} \cdot 7^{20} \cdot 7^{40} \equiv(2814)(-815)(-1187)= \\
(2814)(967405) \equiv(2814)(-467)=-1314138 \\
\equiv 1563\left(\bmod 71^{2}\right)
\end{array}$$

Therefore, $7^{p-1} \not \equiv 1\left(\bmod p^{2}\right)$, and by Theorem 8 of this chapter, 7 is also a primitive root modulo $71^{2}=5041$. Furthermore, by Theorem 12 of this chapter, 7 is also a primitive root modulo $2 p^{2}=10082$.
$$\begin{array}{l}
7,59,47,31,28,62,56,53,21,35,11,42,22, \\
13,69,44,67,52,63,33,55,68,65, 
\end{array}$$