Looking at group $U{81}$ find $0<x<81$ such that:
$$x^{19}=8 \:(mod\:81)$$
I know that $o(8)=18$, is it true to say:
$$(x^{19})^{18}=1 \:(mod\:81)$$
and from here to use Euler's theorem which yields:
$$x^{18}=1 \:(mod\: 81) \ ?$$
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1 Answer
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$8$ isn't the only integer with order $18$, so solving $x^{18}\equiv 1\pmod{81}$ gives solutions which don't solve $x^{19}\equiv 8\pmod{81}$.
You can note that $(9-1)^{18}\equiv 1\pmod{81}$ to find the answer as $x=8$.
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$\begingroup$ Cam you elaborate more please? I didn't understand how to use what you suggested $\endgroup$– 3xhaustJun 1 at 11:57
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1$\begingroup$ @3xhaust; as $8^{18}\equiv1\pmod{81}$ then $8^{19}\equiv8\pmod{81}$ $\endgroup$– JMPJun 1 at 12:10
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$\begingroup$ But how can I find all the numbers that satisfy it? $\endgroup$– 3xhaustJun 1 at 13:32
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$\begingroup$ What's the explanation for that? Especially for the +1? $\endgroup$– 3xhaustJun 1 at 13:45
$x\equiv y\pmod{n}$for $x\equiv y\pmod{n}$. $\endgroup$