How can I solve $x^{5}\equiv 13 \pmod{18}$.

Question

I need to solve the congruence of $ x^{5}\equiv 13 \pmod{18}$ and find whether it is solvable.

Actually I can solve $x^{k}\equiv \pmod{q}$ when $q$ is odd. But now it is composite, and I do not know how to apply the theorem. Can you help me?

By the dupe: $!\bmod \phi(18)=6!:\ \dfrac{1}{\color{#c00}5}\equiv \dfrac{1}{-1}\equiv -1,$ so raising $,x^{\large \color{#c00}5}\equiv 13,$ to power $, \dfrac{1}{\color{#c00}5}\equiv-1,$ yields $,x\equiv \dfrac{1}{13}\equiv \dfrac{-35}{-5}\equiv 7,$ by inverse reciprocity $\ \ $ — Bill Dubuque, Apr 30 '20 at 00:29

lhf · Answer 1 · 2020-04-29T23:58:21.830

0

Hint: By Euler–Fermat, $x^{6}\equiv 1 \bmod{18}$.

edited Apr 29 '20 at 23:58

answered Apr 29 '20 at 23:43

lhf

If $x$ and 18 are relatively prime. – hamam_Abdallah Apr 29 '20 at 23:48
@hamam_Abdallah, they must be if $x^{5}\equiv 13 \bmod{18}$. – lhf Apr 29 '20 at 23:52
But the theorem is available when q is odd prime. 18 is $3^{2} *2$. Gcd(17,12)=1. – sankar Apr 29 '20 at 23:56
Thank you. I think I got the point. – sankar Apr 30 '20 at 00:00

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