Hi I am given this problem and I am supposed to use fermat's theorem. Here is it is: Prove that $$24^{31} \equiv 23^{32} \pmod{19}$$
We are supposed to solve it by setting up the congruence like this:
$$24^{31} \equiv x \pmod{19}$$
$$23^{32} \equiv x \pmod{19}$$
I am just confused how to break it down. I was hoping someone can guide me through this question
mod $\,19\!:\,\ \color{#c00}{24}\equiv \overbrace{ 5 \equiv \color{#c00}{4^{-1}}}^{\Large\! 5\,\cdot\, 4\,\equiv \,1 }\,$ so $\ 23^{32}\color{#c00}{24}^{-31}\!\equiv \overbrace{4^{32}(\color{#c00}{4^{-1}})^{-31}}^{\Large 4^{63}\,\equiv\,\ \color{#0a0}2^{\,\color{#0a0}{2\cdot 9}\,\cdot 7}}\!\equiv \color{#0a0}{(2^{18})}^7\equiv\color{#0a0}{(1)}^7\!\equiv 1\,$ by $\,\rm\color{#0a0}{Fermat}$