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I am currently studying Number Theory at the undergraduate level. In my textbook, I have come across the following question:

Show that if p is an odd prime and n is a positive integer, then there is a primitive root of $p^n$. Hint: suppose g is a primitive root of $p^k$. Use the equation we proved in problem 4* to show that either g or g+p is a primitive root of pk+1.

*equation 4 is: (g+p)$\Phi$($p^k$) $\equiv$ g$\Phi$($p^k$) - $p^k$g$\Phi$($p^k$)-1 (mod pk+1).

I am completely lost on this problem, so any help would be very appreciated.

Liv
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    There is another piece of important information in this context: if $g$ is a primitive root modulo $p^k$, so that the order of $g$ modulo $p^k$ equals $p^{k-1}(p-1)$, then the order of $g$ modulo $p^{k+1}$ equals either $p^{k-1}(p-1)$ or $p^k(p-1)$. Do you see why this statement is true? Do you see how it can be used in conjunction with equation 4 to establish the statement you wrote? – Greg Martin Nov 18 '19 at 02:57
  • https://math.stackexchange.com/questions/227199/order-of-numbers-modulo-p2 – lab bhattacharjee Nov 18 '19 at 04:06

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