I'm trying to solve $x^{18} \equiv 64 \pmod {13^2}$ and while trying I'm losing my mind.
First question was to prove that $ 2$ is a primitive root for $13^n$ for all natural $n$, and then I had to find all solutions for the above equation.
First I wrote $=x^{12}x^{6}=x^{18} \equiv -1 \pmod {13}$ so I know that all primitive roots solve this equation but not only, so I divide the original equation to $(x^3-2)(x^3+2)^2(x^9+8)\equiv 0\pmod {13^2}$.
I hand checked all three of those expression from 1 to 12 and figured out that there is only a solution for $(x^9+8)\equiv 0\pmod {13}$ with $2,5,6$, so I used Hensel theorem to find all solutions, and to make it easy I had to prove that 5 is a primitive root modulo 13 on the way, and still didn't get all the solutions. Is there any shorter way? It takes a lot of time and it is just one section in one of seven question in a 2 hours test.
Thank you!