# Prove that n^5 - 5n^3 + 4n is divisible by 120:

1. Jan 15, 2007

### murshid_islam

i need to prove that $n^5 - 5n^3 + 4n$ is divisible by 120 where n is a natural number. i can prove that using induction. but it got really long using induction. so i tried the following

by factoring i got,

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

now, to prove that it is divisible by 120, i have to prove that it is divisible by 3, 5 and 8. it is definitely divisible by 3 and 5 since it is a product of five consecutive natural numbers. also, at least one of the 5 factors is divisible by 4 (lets call it p) and at least two of the factors are divisible by 2 (one of them is p and the other one is p+2 or p-2). so p(p-2) or p(p+2) is divisible by 8. and now we can say that $n^5 - 5n^3 + 4n$ is divisible by 3, 5 and 8 and hence by 120.

what i want to know is whether there is anything wrong with my reasoning.

Last edited: Jan 15, 2007
2. Jan 15, 2007

### LeonhardEuler

Your reasoning looks fine with the minor exception of the statement "it is a product of five consecutive natural numbers" which is not true for n less than two. But regardless, if you just replace "natural numbers" with "integers", everything goes through.

3. Jan 15, 2007

### tehno

Math induction method easily solves such problems.

4. Jan 15, 2007

### Crosson

Another "hybrid" method would be to prove using induction the statement "the product of 5 consecutive natural numbers is divisible by 120".

5. Jan 16, 2007

### murshid_islam

thanks a lot.

6. Jan 16, 2007

### Gib Z

Heres a hint: Its the multiplication of 5 consecutive integers, and 5! =120 :)