1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook