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Induction Division

  1. Oct 11, 2011 #1
    show (n^2)! is divisible by (n!)^n+1

    base case 1 divides 1

    so I did the induction and got

    (n^2+2n)!=((n+1)^n * (n!) *(n!)^n+1

    (n^2+2n)*.......*(n^2)!=((n+1)^n * (n!) *(n!)^n+1

    can we conclude that (n^2)! is divisible by (n!)^n+1 for all n from there because we have the bold terms on both sides?
  2. jcsd
  3. Oct 11, 2011 #2


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    The inductive step:

    Assume that for some k, (k2)! is divisible by (k!)k+1.

    Now prove that ((k+1)2)! is divisible by ((k+1)!)k+2.
  4. Oct 11, 2011 #3
    I have done all the work its just a lot to type out so I put the last line down. I havent taken number theory and im wondering if the point I got it to can I concluded that the it is divisible because of our induction hypothesis
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