I was checking the following Fermat's little theorem exercise:
Show that $n^{23}+6n^{13}+4n^{3}$ is a multiple of $11$
I've started by stating each congruence individually suposing that each $n,6n$ and $4n$ are primes with $11$, for the first one I have:
$$n^{10} \equiv 1 \mod {11}$$ $$ \equiv n^{3} \mod {11}$$
I've stated the second one this way
$$6n^{10} \equiv 1 \mod {11}$$
But honestly I don't know how to go ahead as long as I don't have a number to evaluate with $11$.
Also I'm considering that I have:
$$n + 6n + 4n = 11n$$
This may have some relation with the proof but could be affected because of the powers of each one. Any help will be really appreciated.