Question

Prove: If n^3 ≡n (mod 6)

Answer 1

I remember this. this was annoying

Answer 2

Hint : \[n^3-n = n(n^2-1)=(n-1)n(n+1)\]

Answer 3

next, recall that any product of \(n\) consecutive integers is divisible by \(n!\)

not that annoying :)