# Divisibility question

• December 16th 2011, 07:47 AM
ehpoc
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
• December 16th 2011, 07:52 AM
Plato
Re: Divisibility question
Quote:

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, $n=3~.$
• December 16th 2011, 08:07 AM
HallsofIvy
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.
• December 16th 2011, 08:08 AM
anonimnystefy
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.
• December 16th 2011, 08:12 AM
Also sprach Zarathustra
Re: Divisibility question
Quote:

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

$\binom{n+2}{5}$
• December 16th 2011, 08:32 AM
ehpoc
Re: Divisibility question
Thanks. Yes now I remember this in class.

I just never recognized it as five consecutive integers.