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Show that if $r$ is a primitive root modulo the positive integer $m$, then $ {\overline r }$ is also a primitive root modulo m if $ {\overline r }$ is an inverse of $r$ modulo $m$.

My TA did not go over any proofs in my self study class. We just learned the topic and did some examples. I was wondering what this proof might look like. It seems it might help because it looks similar to the problems with examples.

Can someone please show me?

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    $\begingroup$ What is the definition of "primitive root" that you're using? $\endgroup$ – vadim123 Apr 29 '13 at 17:04
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A primitive root of $m$ is a number $g$ such that $g,g^2,g^3, \dots, g^{\varphi(m)}$ are all incongruent modulo $m$. So $g$ is a primitive root of $m$ precisely if $g$ has order $\varphi(m)$ modulo $m$.

So it is enough to show that for any $x$, if $xy\equiv 1\pmod{m}$, then $x^k\equiv 1\pmod m$ if and only if $y^k\equiv 1\pmod{m}$. This is immediate, since $(xy)^k=x^ky^k$.

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Since $r$ is a primitive root modulo $m$, we know it's invertible, so $$ r^n\equiv 1\text{ (mod }m\text{)}\Longleftrightarrow (r^{-1})^n\equiv 1\text{ (mod }m\text{)} $$

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    $\begingroup$ Oh, I apparently can't read today. My apologies. $\endgroup$ – Potato Apr 29 '13 at 17:58
  • $\begingroup$ @Potato Haha, no problem. :) $\endgroup$ – Warren Moore Apr 29 '13 at 17:58

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