Solve $x^5\equiv 4\pmod 7$ We know about calculating  $x^2\equiv 2\pmod 7$ using quadratic residue properties in order to find out whether a solution exists or not.
I wonder is there any way to determine that $x^n\equiv k\pmod v$, where $v\ge 2$, $k\in\Bbb Z$, and $n\ge 3$?  As I asked in title:

Solve $x^5\equiv 4\pmod 7$


Here is another way to determine this. Over the field $\Bbb F_7$ the Berlekamp algorithm gives the factorisation
$$
x^5-4=(x^4 + 2x^3 + 4x^2 + x + 2)(x - 2).
$$
Hence $x=2$ is the solution of the equation $x^5=4$. We may rewrite this as $x\equiv 2\bmod 7$.
