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

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Question 745014: if n > 2 ,then show that n^5-5n^3+4n is divisible by 120
Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
Clearly we assume that n is an integer...

First let's factor:


next we rearrange them is succession:

%0D%0A%28n-2%29%28n-1%29n%28n%2B1%29%28n%2B2%29%0D%0A
which shows that this is a product of 5 consecutive numbers starting with n-2
and
%0D%0A120=12%2A10=2%5E3%2A3%2A5=3%2A5%2A8%0D%0A



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.