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

Printable View

- December 16th 2011, 07:47 AMehpocDivisibility 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 AMPlatoRe: Divisibility question
- December 16th 2011, 08:07 AMHallsofIvyRe: 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 AManonimnystefyRe: 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 AMAlso sprach ZarathustraRe: Divisibility question
- December 16th 2011, 08:32 AMehpocRe: Divisibility question
Thanks. Yes now I remember this in class.

I just never recognized it as five consecutive integers.