Give a proof by cases that shows that $n(n ^2 − 1)(n + 2) $ is a multiple of $4$, for all integers $n$

Question

Give a proof by cases that shows that $n(n ^2 − 1)(n + 2) $ is a multiple of $4$, for all integers $n$

I have already done case 1 where n is even. I am doing case 2 where n is odd and I'm a bit confuse how to finish off the problem..Uploaded a picture of my work.. Is this work sufficient enough to prove that it is a multiple of 4? The 6k would not have any affect of the result in general?

I feel it is wrong because of the 6k. Right now my thoughts are if (statement) is a multiple of 4, then (statement) is a multiple of 2 as well. And work from there.

My work

Answer 1

Hint: Write your term in the form $$(n-1)n(n+1)(n+2)$$

Answer 2

Notice that $n^2-1=(n-1)(n+1)$
So if $n$ is odd, then both of the factors above are even.
From this you can show that 4 must divide $n(n^2-1)(n+2)$

Answer 3

You went wrong at this stage: $$n(n^2-1)(n+2)=(2k+1)((2k+1)-1)(2k+3)$$ Can you see where?

Answer 4

Okay, I think I got it.

n(n+1)(n-1)(n+2) =(2k+1)(2k)(2k+2)(2k+3) =($4k^2$+2k)(2k+2)(2k+3) =($8k^3$+$12k^2$+4k)(2k+3)

=4($2k^3$+$3k^2$+k)(2k+3)

edit 1:lemme know if there's any objections

Answer 5

There are two cases :

If $n$ is odd, then $n-1$ and $n+1$ both even . the product $(n-1)(n+1)$ is a multiple of $4$, so is $(n-1)n(n+1)(n+2)$.

If $n$ is even the $n+2$ is also even. The product $n(n+2)$ is a multiple of $4$ , so is $(n-1)n(n+1)(n+2)$.

Answer 6

$[(n^2-1)][n(n+2)].$

1) Let $n$ be odd.

$[(n^2-1)] =(n+1)(n-1);$

$n+1, n-1$ are even,

$2| (n+1)$ and $2|(n-1)$, hence

$4|(n-1)(n+1).$

2) Let $n$ be even.

$[n(n+2)];$

$2|n$ and $2|(n+2)$, hence

$4|n(n+2).$