Given 2 is a primitive root mod 19, find all solutions to x^12 ≡ 7 (mod 19) (a) x^12 ≡ 6 (mod 19)

Question

I'm stuck on (another) problem under Number Theory. There are quite a few gaps on what was covered in my class so I'm having quite a bit of trouble. Could you please help me solve the following problem?

2 is a primitive root mod 19. Using this information, find all solutions to x^12 ≡ 7 (mod 19) and x^12 ≡ 6 (mod 19)

I think I would have to make use of the powers of 2 to solve this, but I can't get any further than that.

Any help would be much appreciated!

Answer 1

Hint:

Write $x=2^p$ (why ?) then you get $2^{12p}\equiv 2^y\pmod{19}$ where $y$ correspond to the representation for $7$ and $6$.

Then you get solve $12p\equiv y\pmod{18}$ (why ?)

Answer 2

Hint:

So you're given that $2$ has order $18\bmod 19$. In particular, $7\equiv 2^6\mod 19$.

Any solution can be written as $2^k$, where $k$ is unique module $18$. The equation can be rewritten as $$2^{12k}\equiv 2^6 \mod 19\iff 12k\equiv 6\mod 18\iff 2k\equiv 1\mod 3.$$

Can you continue?