Question:
Prove that $n^{31} \equiv n$ mod $77$ for all integers $n$
I've read this question: How to prove $n^5 - n$ is divisible by $30$ without reduction But I don't know how to use the same method to prove my example.
Induction seems very long and unnecessary.
I tried to come up with something using:
- $n^7 \equiv n \mod{7}$ and
- $n^{11} \equiv n \mod{11}$
but I am not sure that's useful.
I am now assuming I have to show that $n^{31} - n$ is a multiple of 7 and 11 and thus a multiple of 77. How do I arrive in this conclusion?