if n > 2 ,then show that n^5-5n^3+4n is divisible by 120

Clearly we assume that n is an integer...

First let's factor:


next we rearrange them is succession:


which shows that this is a product of 5 consecutive numbers starting with n-2
and




Now notice that if n>2 the each product is
1*2*3*4*5
2*3*4*5*6
3*4*5*6*7
and so on
each of which is a multiple of 3 (every third digit), 5 (every fifth digit), and 8 (there will always be a multiple of 2 and a multiple of 4).
Therefore they will always divide 120.