MHB Quotient remainder theorem problem.

[tmt1]

For any int $$n $$ , prove that $$ 4 | n (n^2 - 1) (n + 2)$$.

I know I have to use the quotient remainder theorem, but I'm wondering how to go about this problem.

I'm not sure how to phrase this problem in English.

 

[Ackbach]

Couple of hints.

1. $n(n^2-1)(n+2)=(n-1)\, n \,(n+1)(n+2)$.
2. By the Quotient Remainder Theorem, there exist integers $q$ and $r$ such that $n=4q+r$, where $0\le r<4$.

Does this suggest anything to you?

 

[tmt1]

Couple of hints.

1. $n(n^2-1)(n+2)=(n-1)\, n \,(n+1)(n+2)$.
2. By the Quotient Remainder Theorem, there exist integers $q$ and $r$ such that $n=4q+r$, where $0\le r<4$.

Does this suggest anything to you?

Right, so $(n-1)\, n \,(n+1)(n+2)$ is 4 consecutive integers. I get that if you take any arbitrary integer, if it is not divisible by 4, you can increment it by some int $c$ such that $0< c < 4$ and get an int that is divisible by 4.

Therefore, if you have 4 consecutive integers, one of those integers will be divisible by 4, and as a result the product of those 4 integers will be divisible by 4. I just don't know how to state or prove this formally.

 

[Ackbach]

Right, so $(n-1)\, n \,(n+1)(n+2)$ is 4 consecutive integers. I get that if you take any arbitrary integer, if it is not divisible by 4, you can increment it by some int $c$ such that $0< c < 4$ and get an int that is divisible by 4.

Therefore, if you have 4 consecutive integers, one of those integers will be divisible by 4, and as a result the product of those 4 integers will be divisible by 4. I just don't know how to state or prove this formally.

Why not plug in $4q+r$ in for $n$, and see what comes out in the wash?