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

  1. Jan 15, 2007 #1
    i need to prove that [itex]n^5 - 5n^3 + 4n[/itex] 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,

    [tex]n^5 - 5n^3 + 4n = (n-2)(n-1)n(n+1)(n+2)[/tex]

    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 [itex]n^5 - 5n^3 + 4n[/itex] 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.

    thanks in advance.
    Last edited: Jan 15, 2007
  2. jcsd
  3. Jan 15, 2007 #2


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    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.
  4. Jan 15, 2007 #3
    Math induction method easily solves such problems.
  5. Jan 15, 2007 #4
    Another "hybrid" method would be to prove using induction the statement "the product of 5 consecutive natural numbers is divisible by 120".
  6. Jan 16, 2007 #5
    thanks a lot.
  7. Jan 16, 2007 #6

    Gib Z

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    Homework Helper

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