n^3-n is a product of 3 consecutive integers....
i.e. It can be expressed as n(n-1)(n+1).
For divisibility by 4,
A product of consecutive integers of the form n^3-n i.e. n(n-1)(n+1) is always divisible by 4,if n is ODD.
Its given that n=2k+1 which will always be odd.Hence sufficient.
Also,using Kaplan's strategy of picking numbers,
If n=2k+1,then by picking simple numbers,it can be easily observed that the product will always be a factor of 4.
eg: K=1 implies n=3 , so the expression is 3*2*4...a factor of 4.
k=2 implies n =5, so the expression is 5*4*6...a factor of 4.
To conclude in general,
A product of consecutive integers of the form n^3-n i.e. n(n-1)(n+1) is always divisible by 4,if n is ODD.
A product of consecutive integers of the form n(n+1)(n+2) or n^3+3n^2+2n is always divisible by 4 if n is EVEN!
I hope this helps...
