Show that if $103x^5 \equiv 1$ has a solution x, then $x \equiv n \mod 23$ The full question is this:

find an integer $0<n<23$ such that, if $x∈Z$ is a solution to 
$103x^5 \equiv 1 \mod 23$,
then 
$x \equiv n \mod 23$.

I'm not sure how to approach it, though I think Fermat's Little Theorem and/or Euclid's algorithm will help.
I thought maybe the fact that $103 (-2) \equiv 1 \mod 23$ could be helpful too, but I'm not sure.
 A: $103x^5\equiv1\bmod23\iff   103^9x^{45}\equiv1\bmod23$
$\overset{Fermat}\iff 103^9x\equiv1\bmod23\iff x\equiv(-2)^9\bmod23.$
Can you take it from here?
A: Here is a variation of J. W. Tanner's answer, probably more suitable for hand computation.
$103x^5\equiv1\bmod23$
$\iff 11x^5\equiv1\bmod23$
$\iff x^5 \equiv -2 \bmod23$
$\iff x \equiv x^{45}\equiv (-2)^9 \equiv 17\bmod23$
As you have noticed, $103 (-2) \equiv 1 \bmod 23$ is definitely helpful. It let's you skip the first reduction.
A: Hint $ $ It reduces to solving $\,x^{\large 5}\equiv -2\pmod{\!23}\,$ by easy congruence arithmetic. Now apply
Key Idea $ $ To take $k$'th roots when $k$ is coprime to period $n,\,$   raise to power $\,1/k\pmod{\!n},\,$ i.e.
$$\ \ \ \ \ \ \,\bbox[8px,border:2px solid #c00]{ 
\text{if $\:\!\  x^{\large n} =\, 1\, =\, a^{\large n} \ $ then }\ x^{\large k} = a \iff x =  a^{\large (1/k)_n}}$$ 
so $\bmod 23\!:\ x,a\not\equiv 0\,\Rightarrow\, x^{\large 22} \equiv 1\equiv a^{\large 22} $ hence $\, x^{\large 5}\equiv a\iff x\equiv a^{\large (1/5)_{22}}\equiv a^{\large 9}\ $ by 
$$\bmod 22\!:\,\ \dfrac{1}5\equiv \dfrac{3(\color{#c00}{-7})}5\equiv \dfrac{3}1\,\dfrac{\color{#c00}{15}}5\equiv 9\qquad$$
Alternatively by inverse reciprocity $$\bmod 22\!:\,\ \dfrac{1}5\equiv \dfrac{1+22\color{#c00}{(2)}}{5}\equiv 9\qquad\qquad$$
by $\bmod 5\!:\ 0\equiv 1\!+\!22\color{#c00}k\equiv 1\!+\!2k\iff 2k\equiv 4\iff \color{#c00}{k\equiv 2}$
