Euclidean division

**lehder** · March 7th 2010, 06:21 AM

Hi everybody,

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

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

Do you have any idea?

**Prove It** · March 7th 2010, 07:15 AM

Originally Posted by lehder

Hi everybody,

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

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

Do you have any idea?

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

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

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

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

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

So the number is divisible by 4.

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

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

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

So the number is divisible by 4.

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

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