# Can I find $r$, if $r^m\equiv 1 (\bmod 18)=6$

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 $$r$$?

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

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 $6 \pmod {31}$?

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