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Recall.

  • Definition $1$. $Ord_n(a)$ is the smallest number $m$ st. $\quad a^{m} \equiv 1(\bmod n)$
  • Definition $2$. We say $r$ is a primitive root modulo $n$ if $\operatorname{Ord}_{n} (r)=\phi(n)$
  • Note: $\left\{1,r, r^2,..., r^{\phi(n)-1}\right\}=U_n$

I want to find $r$ such that $ord_{18} (r)=\phi(18)=6$, that is we get by definition 1: $r^{m} \equiv 1(\bmod 18)=6$, so hence can I find $r$?

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    $\begingroup$ The expression in your final sentence makes no sense. Please clarify what you mean. $\endgroup$ – Bill Dubuque Apr 8 at 2:40
  • $\begingroup$ @BillDubuque I edited question. $\endgroup$ – James Ensor Apr 8 at 15:02
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    $\begingroup$ the question still does not make sense: $1(\bmod18)=6$?! $\endgroup$ – J. W. Tanner Apr 8 at 16:29
  • $\begingroup$ @J.W.Tanner, okey... if $ord_{18}(r)=6$ then how can i find r? $\endgroup$ – James Ensor Apr 8 at 22:29
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As $(r,18)=1,(r,6)=1$

As $\phi(18)=6,$ord$_{18}r$ must divide $6$

Now $5^2\equiv7\not\equiv1,5^3\equiv-1\not\equiv1\pmod{18}$

$\implies$ord$_{18}5=6$

Finally use

What integers have order $6 \pmod {31}$?

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  • $\begingroup$ How did you think $5^2=7, 5^3=-1$? $\endgroup$ – James Ensor Apr 7 at 22:36
  • $\begingroup$ @James, $$5^2=7+18\equiv7\pmod{18}$$ $\endgroup$ – lab bhattacharjee Apr 8 at 2:12

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