Hi everybody,

I must show that the number $\displaystyle a=n^2(n^2-1)$ is divisible by 12.

I know just that (n-1).n.(n+1) is divisible by 6

Do you have any idea?

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- Mar 7th 2010, 06:21 AMlehderEuclidean division
Hi everybody,

I must show that the number $\displaystyle a=n^2(n^2-1)$ is divisible by 12.

I know just that (n-1).n.(n+1) is divisible by 6

Do you have any idea? - Mar 7th 2010, 07:15 AMProve It
If the number is divisible by 6, you know it must be divisible by 3.

We can show that the number is also divisible by 4.

If it is divisible by 4 and divisible by 3, it must be divisible by 12.

Case 1: $\displaystyle n$ is odd. Therefore $\displaystyle n = 2m + 1$.

So $\displaystyle n^2(n^2 - 1) = (2m + 1)^2[(2m + 1)^2 - 1]$

$\displaystyle = (4m^2 + 4m + 1)(4m^2 + 4m + 1 - 1)$

$\displaystyle = (4m^2 + 4m + 1)(4m^2 + 4m)$

$\displaystyle = 4(4m^2 + 4m + 1)(m^2 + m)$.

So the number is divisible by 4.

Case 2: $\displaystyle n$ is even. Therefore $\displaystyle n = 2p$.

So $\displaystyle n^2(n^2 - 1) = (2p)^2[(2p)^2 - 1]$

$\displaystyle = 4p^2[4p^2 - 1]$.

So the number is divisible by 4.

Since the number is divisible by 3 and by 4, the number must be divisible by 12.