Question 442481
If x^3 and x leave the same remainder upon dividing by 6, then x^3 ≡ x (mod 6) --> x^3 - x ≡ 0 (mod 6).


The left expression factors to


x(x^2 - 1) = x(x+1)(x-1). At least one of these numbers must be even, and exactly one of them is divisible by 3, so x(x+1)(x-1) ≡ 0 (mod 6), so our original claim is true.