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Math Help - Divisibility of (n^5 - n) by 30

  1. #1
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    Divisibility of (n^5 - n) by 30

    hey guys am having major problems with this question:
    show that n^5 - n is always divisible by 30
    i tried mathematical induction, which would have worked if it needed to be divisible by 10, but not 30!
    please help!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by wik_chick88 View Post
    hey guys am having major problems with this question:
    show that n^5 - n is always divisible by 30
    i tried mathematical induction, which would have worked if it needed to be divisible by 10, but not 30!
    please help!
    Is...

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

    Now observing (1) we deduce that either n, or n-1 or n+1 contains 2 or 3 so that 6 devides (1). Let suppose now that neither n, nor n-1 nor n+1 contains 5 so that is n=5 k \pm 2 or n= 5 k \pm 3. Is...

    n= 5 k \pm 2 \implies n^{2}+1 = 25 k^{2} \pm 20 k +4 +1

    n= 5 k \pm 3 \implies n^{2}+1 = 25 k^{2} \pm 30 k +9 +1

    ... and in both cases n^{2}+1 contains 5...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by wik_chick88 View Post
    hey guys am having major problems with this question:
    show that n^5 - n is always divisible by 30
    i tried mathematical induction, which would have worked if it needed to be divisible by 10, but not 30!
    please help!
    Or you could note it suffices to check for n=0,1,2,-2,-1 which is trivial.
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  4. #4
    Senior Member Shanks's Avatar
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    hint:it suffices to prove that n^5-n is congruent to 0 mod 2 and 3 and 5. Apply the Fermat's little theorem.
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  5. #5
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    you don't need induction, you can prove it directly.

    suppose n>1.

    n^5 - n = n(n-1)(n+1)(n^2+1)

    one of n-1,n, or n+1 is divisible by 3.

    one of n,n+1 is divisible by 2. hence n^5 - n is divisible by 6.

    now if n = 5k, 5|n. if n = 5k+1, 5|n-1, and if n = 5k+4 5|n+1. so in those 3 cases, it is obvious, 5|n^5 - n.

    but if n = 5k+2, n^2+1 = (5k+2)^2 + 1 = 25k^2 + 20k + 5 = 5(5k^2 + 4k + 1).

    and if n = 5k+3, n^2+1 = (5k+3)^2 + 1 = 25k^2 + 30k + 10 = 5(5k^2 + 6k + 2).

    so in all cases, n^5 - n is divisivble by 2,3 and 5, and since these are all co-prime, their product 30 divides n^5 - n.

    my apologies to chisigma, i wrote this up before dinner, and didn't post it until after you replied
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