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Any hints or strategies would be greatly appreciated:

If $m$ is an integer and $b^k$ is a primitive root mod $m$, then $b$ is a primitive root mod $m$.

I am reviewing material from my elementary number theory course. I had a true/false question that I think is true (by process of elimination of how many "trues" the problem set was supposed to have). But I and am absolutely stuck on trying to prove it.

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  • $\begingroup$ Welcome to the site, and please use Latex to format your equations. (Put them between a pair of $'s). $\endgroup$ – user111187 Jun 14 '14 at 21:42
  • $\begingroup$ Almost always start with the question, not your life story. $\endgroup$ – Thomas Andrews Jun 14 '14 at 21:42
  • $\begingroup$ How does your material define "primitive root"? $\endgroup$ – Thomas Andrews Jun 14 '14 at 21:45
  • $\begingroup$ Have you tried contraposition? Instead of "$b^k$ primitive $\implies$ $b $ primitive", try proving "$b $ not primitve $ \implies $ $b^k$ not primitive". $\endgroup$ – zo0x Jun 14 '14 at 21:48
  • $\begingroup$ Text is Rosen. Definition is: If $r$ and $n$ are relatively prime integers with $n > 0$ and if $ord_n r = phi(n)$, then $r$ is called a primitive root mod $n$. $\endgroup$ – Meghan K. Jun 14 '14 at 22:07
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If powers of $b^k$ generate every coprime residue class mod m, then powers of $b$, which powers are a superset of the first set of powers, also generate every coprime residue class mod m. So $b$ is also a primitive root. (A coprime residue class mod m is an invertible one, i.e., $\{a|(a,m)=1\}$.)

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