This is a rather easy exercise - using either induction or direct proof (unless I am missing something in the question)
I have a back-door approach to this one . . .
As aman-cc pointed out: .
I recognized this is as part of the formula for the sum of consecutive squares:
Since the left side is an integer, the right side must be an integer.
. . That is, is a multiple of 6.
Therefore: . .for all integer values of
Consider , use the property of product of few consecutive integers .
This a solution I saw once:
. One of and is divisible by .
. Then at least one of , , and is divisible by .
Since and are relatively prime is divisible by .