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Larson 3.3.19b

  1. Jan 4, 2008 #1
    [SOLVED] larson 3.3.19b

    1. The problem statement, all variables and given/known data
    What is the largest number N for which you can say that n^5-5n^3+4n is divisible by N for every positive integer N.

    EDIT: change the last N to n

    2. Relevant equations



    3. The attempt at a solution
    I have just been plugging in things for n and seeing what happens. If n=N-2,N-1,N,N
    +1,N+2, then n^5-5n^3+4n is divisible by N because -2,-1,0,1,2 are the roots of that equation. If n=N+3, we get that 120 = -120 must equal 0 mod N. So, N=3 is a lower bound. So N must be a factor of 120. Should I just keep keep plugging in numbers for n and setting them equal to 0 mod N? It seems like that will give me a solution but that won't prove that this particular N works for all values of n.
     
    Last edited: Jan 4, 2008
  2. jcsd
  3. Jan 4, 2008 #2

    morphism

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    I suppose you meant to say "What is the largest number N for which you can say that n^5-5n^3+4n is divisible by N for every positive integer n."

    Now, n^5-5n^3+4n=(n-2)(n-1)n(n+1)(n+2) is the product of 5 consecutive integers hence divisible by 5!=120. I guess this is what you said, but I didn't really follow your reasoning. Next note that for n=3, the expression is 120.
     
  4. Jan 4, 2008 #3
    Very nice! I was just beating around that solution...I just could quite hit it on the head.
     
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