Question 745014
Clearly we  assume that n is an integer...


First let's factor:


{{{
n^5-5n^3+4n=n(n^4-5n^2+4)=n(n^2-1)(n^2-4)=n(n-1)(n+1)(n-2)(n+2)
}}}

next we rearrange them is succession:


{{{
(n-2)(n-1)n(n+1)(n+2)
}}}

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

and
{{{
120=12*10=2^3*3*5=3*5*8
}}}




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.