Prove that $47|(2^{23} -1)$ using the fact that $47=3\cdot2^4-1$ and $3^5 = 2^3 \mod 47$.
This problem has not been an obvious/routine divisibility proof. I've been playing around with the two identities and using congruence and divisibility definitions to see how to prove this, but with no luck. Any hints/solutions are greatly appreciated.
$2^{23} = 3^{5} \times 2^{20} = (3 \times 2^{4})^{5} = 48^{5} = 1 \mod 47$. As such $2^{23} - 1 = 0 \mod 47$, which means that $47|2^{23} - 1$.
Notice that $2\equiv 7^2\mod 47$. Therefore, $2^{23}\equiv 7^{46}\equiv 1\mod 47$ by Fermat's little theorem. So, $47|2^{23}-1$
Though not using the facts you have proposed.
To use the fact you proposed, you could refer to the other answer.
$\begin{align}\bmod 47\!:\ \,{\rm by\ applying} &\,\ \ \ \ \,\overbrace{\color{#c00}{2^{\large 4}\ \equiv\ 3^{\large -1}}}^{\textstyle 3\cdot 2^{\large 4}\equiv 1}\ \ \, \text{as a}\textit{ rewrite rule}\\[.4em] \text{we obtain}\ \ \ 2^{\large\:\! 3+4\cdot 5} = \ &2^{\large 3}(\color{#c00}{2^{\large 4}})^{\large 5}\! \equiv \color{#0a0}{2^{\large 3}}(\color{#c00}3^{\large\color{#c00}{-5}})\equiv 1,\ {\rm by}\ \ \color{#0a0}{2^{\large 3}}\!\equiv\color{#c00}{3^{\large 5}} \end{align}$
