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

Question

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 $x^5\equiv 4\pmod 7$

Answer 1

Since $x^6\equiv1\bmod 7$, invert the exponent mod $6$: $\;5\times5\equiv1\bmod 6$,

so $x^5\equiv4\bmod7\implies x\equiv x^{5\times5}\equiv4^5=1024\equiv2\bmod7.$

Answer 2

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 $x\equiv 2\bmod 7$.