Prove that for every positive integer, this polynomial is divisible by 8

Question

prove that: $$8\mid (n-1)n(n+1)(n+2)$$

I tried to simplify this expression but had no luck.

Answer 1

Hint:

prove that of $4$ consecutive numbers at least one is a multiple of $4$, and that one of the others is a multiple of $2$.

Answer 2

With four consecutive numbers, you can write them like $4m, 4m+1, 4m+2, 4m+3$ (or possibly a shifted version of this like $4m-1, 4m, 4m+1, 4m+2$). In any case, taking four consecutive natural numbers, one of them is divisible by $4$ and a separate one is divisible by $2$. Thus the product is divisible by $8$.

Answer 3

For an integer $n>1$ the number $${n+2\choose 4}$$ is an integer, we recognize that $$3{n+2\choose 4}=3\cdot\frac{(n+2)(n+1)n(n-1)}{4!}=\frac{(n+2)(n+1)n(n-1)}8$$ Hence $(n+2)(n+1)n(n-1)$ is a multiple of $8$.