0

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?

1 Answers1

0

Let's work $\pmod {7}$: in this case you have $a^6 \equiv 1$ for all $a \neq 0$. Then $$n^{31} \equiv (n^5)^{6} \cdot n \equiv 1 \cdot n \equiv n$$

If we work $\pmod {11}$, we have $a^{10} \equiv 1$ for all $a \neq 0$. Then $$n^{31} \equiv (n^3)^{10} \cdot n \equiv 1 \cdot n \equiv n$$

Crostul
  • 36,738
  • 4
  • 36
  • 72