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

## Main Question or Discussion Point

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.

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## Answers and Replies

LeonhardEuler
Gold Member
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.

Math induction method easily solves such problems.

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

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.
thanks a lot.

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