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Math Help - Prove that n is square or doubled square

  1. #1
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    Prove that n is square or doubled square

    Hey,
    N is positive integer number. Prove: if summation of all positive divisors of n is uneven, then n is square or doubled square of integer number.
    So far I have proved that in this case n is a square of integer number. How?

    First of all, every natural number can be shown as:
    Prove that n is  square or doubled square-3.jpg
    and we should know the equation for summation of all positive divisors:
    Prove that n is  square or doubled square-1.jpg
    which is equal to
    Prove that n is  square or doubled square-2.jpg
    As far as I know \alpha = \lambda
    Example: Divisors of number 15: 1,3,5,15 Sum=24
    From my equation: 15=3*5 Sum=(1+3)(1+5)=4*6=24
    I know that one example doesn't prove that this equation is true, nevermind.

    So if the sum of all postive divisors is uneven then every bracket should be uneven (if one was even then the product would be even).
    Prove that n is  square or doubled square-4.jpg
    Then we should have "pairs" inside all brackets, as shown on the picture, so that in every bracket sum+1 is uneven.
    From that we can deduce that \alphashould be even (there are only "pairs of elements"+1 in brackets)- hope you understand what I mean.
    If \alpha is even then it can be written as 2t (t is natural number) and then our initial equation would look like this:
    Prove that n is  square or doubled square-mimetex.gif (click to enlarge)
    That means n is square of natural number. But in my task there is also "doubled square". I found out that n would be doubled square if n is 2 to the uneven power.
    E.g.
    Prove that n is  square or doubled square-mimetex2.gif
    But how can I prove that?
    Thanks for your replies and sorry for mistakes. If something is unclear- simply ask
    Regards
    Lukasz
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  2. #2
    Super Member
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    Washington DC USA
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    Re: Prove that n is square or doubled square

    Your conclusion that \alpha had to be even assumed that the prime was an odd number. No such restriction applies when p=2. The factor will be odd no matter the \alpha when p=2.
    That's where the "or double a square" part will show up.
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