1. ## Divisibility question

I am looking at a question that I have no clue how to solve.

It says....

show that n(n^2-1)(n^2-4) is divisible by 5! for all n>=3.

Any help appreciated. Thanks

2. ## Re: Divisibility question

Originally Posted by ehpoc
show that n(n^2-1)(n^2-4) is divisible by 5! for all n>=3.
This is clearly be done using induction.
You must at least show the base case, $\displaystyle n=3~.$

3. ## Re: Divisibility question

I would NOT use induction. I would observe that this can be factored as
(n-2)(n-1)n(n+1)(n+2)], five consecutive integers.

4. ## Re: Divisibility question

hi

this can be done easily by factoring the expression you got there:
n(n^2-1)(n^2-4)=n(n-1)(n+1)(n-2)(n+2)=(n-2)(n-1)n(n+1)(n+2)

and this is a product of five consecutive numbers so it is definitely divisible by 5,4,3,2 and 1. note that this is true only if the product above is positive or equal to zero,so it is true only for n>=3 which what you wanted to prove.

5. ## Re: Divisibility question

Originally Posted by ehpoc
I am looking at a question that I have no clue how to solve.

It says....

show that n(n^2-1)(n^2-4) is divisible by 5! for all n>=3.

Any help appreciated. Thanks

$\displaystyle \binom{n+2}{5}$

6. ## Re: Divisibility question

Thanks. Yes now I remember this in class.

I just never recognized it as five consecutive integers.