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Math Help - Divisibility by prime

  1. #1
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    Divisibility by prime

    Let  \frac{m}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \mbox{...} + \frac{1}{p-1} .  p is a prime number > 2 . Prove that  m is divisible by  p . If  p > 3 , Prove that  m is divisible by  p^2 .

    I've done the first bit, I can't seem to get ideas for the 2nd part of the problem. Please help!
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  2. #2
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    Quote Originally Posted by sashikanth View Post
    Let  \frac{m}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \mbox{...} + \frac{1}{p-1} .  p is a prime number > 2 . Prove that  m is divisible by  p . If  p > 3 , Prove that  m is divisible by  p^2 .

    I've done the first bit, I can't seem to get ideas for the 2nd part of the problem. Please help!

    Check the following:

    http://people.brandeis.edu/~gessel/h...pers/wolst.pdf

    http://math.uci.edu/~tchoi/notes/wolstenholme.pdf

    Wolstenholme's theorem - Wikipedia, the free encyclopedia

    This is a very interesting, non-trivial, result.

    Tonio
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  3. #3
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    Thanks a lot! I was trying to solve the 2nd bit in a manner I solved the first bit, which is a bit different from what was given on the website and I made no headway. Thank you, this is a beautiful solution.
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  4. #4
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    Quote Originally Posted by sashikanth View Post
    Thanks a lot! I was trying to solve the 2nd bit in a manner I solved the first bit, which is a bit different from what was given on the website and I made no headway. Thank you, this is a beautiful solution.

    Would you guys mind sharing the way to do the first bit of the problem?
    The way I did was - gouped the ith and (p-i)th term and when I summed it p came out from all the grouping.

    Is there a better way to do it?
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  5. #5
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    Quote Originally Posted by aman_cc View Post
    Would you guys mind sharing the way to do the first bit of the problem?
    The way I did was - gouped the ith and (p-i)th term and when I summed it p came out from all the grouping.

    Is there a better way to do it?
    Let m be our numerator.

    Then, m\equiv \sum_{i=1}^{p-1} (p-1)! i^{-1} \bmod p, where i^{-1} is inverse modulo p.

    But \sum_{i=1}^{p-1} (p-1)! i^{-1} \equiv (p-1)! \sum_{i=1}^{p-1}i\bmod{p} since i^{-1} is a permuation of \{1,2,...,p-1\}.

    Now note \sum_{i=1}^{p-1} i = \frac{p(p-1)}{2}.
    Therefore,  m is divisible by p.
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  6. #6
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    The above post outlines my method for the 1st bit.
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