You can do it without a table by factoring x (x^2-1) (x^2+1) = x (x-1) (x+1) (x^2 - 4 + 5) = x (x-1) (x+1) ((x-2)(x+2) + 5) Now one of the five consecutive

You can do it without a table by factoring x (x^2-1) (x^2+1) = x (x-1) (x+1) (x^2 - 4 + 5) = x (x-1) (x+1) ((x-2)(x+2) + 5) Now one of the five consecutive numbers (x-2), (x-1), x, (x+1), (x+2) must be divisible by 5, so the whole is as well. I agree with you about induction.