Use Fermat's Little Theorem to solve the equation $x^{13} \equiv 2\;(\mod\ 23)$ I have to use Fermat's Little Theorem to solve the equation
$$x^{13} \equiv 2\;(\mod\ 23)$$
But I have no idea how to do so.
I understand that $2^{23} \equiv 2\;(\mod\ 23)$.
But I don't know how to solve what the question is asking.
Fermat's little theorem says that if $x$ is relatively prime to $23$, then $x^{22}\equiv1\pmod{23}$.

Moreover, $$13\times17=221=22\times10+1.$$
So if
$$x^{13}\equiv2\pmod{23},$$
then
$$(x^{13})^{17}\equiv x^{22\times10+1}\equiv x^{1}\equiv2^{17}\pmod{23}.$$
Now $$\begin{align}2^2&amp;\equiv4\\2^4&amp;\equiv16\\2^8&amp;\equiv256\equiv3\\
2^{16}&amp;\equiv9\end{align}.$$
So $2^{17}\equiv18\pmod{23}$. As a consequence, we find that $18$ is a solution to $x^{13}\equiv2\pmod{23}$.



Hope this helps.
Let us find $a,b$ such that $13a+22b=1,$

By Observation $b=3, a=-5$

Using Fermat's Little Theorem,

$$x=x^{22(3)-5(13)}=(x^{22})^3(x^{13})^{-5}\equiv1^3\cdot2^{-5}\pmod{23}$$

Let us find $c,d$ such that $-32c+23d=1\implies c\equiv32^{-1}\pmod{23}$

Use my method from  Solving a Linear Congruence

Or using Fermat's Little Theorem, $2^{-5}\equiv2^{17}\pmod{23}$