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A nice problem...

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  • S. R. Sudarshan Iyengar
    Here goes a nice problem... IT would be nice if someone can give me links/references where I can find more details about this problem. Given a prime P its easy
    Message 1 of 3 , Dec 27, 2007
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      Here goes a nice problem... IT would be nice if someone can give me
      links/references where I can find more details about this problem.

      Given a prime P its easy to see that it always divides 1+ 2 + .. + (p-1)

      But it in fact divides 1^k + 2^k + ... + (p-1)^k for all k except 0
      and p-1.

      Its obvious if k is odd. But is there any way of proving it when k is
      even?

      Regards,
      Sudarshan
    • Payam Samidoost
      Happy new year to all ... Warning: p must be an odd prime! ... It appears as a simple induction: Let S(n) denote 1^n+ 2^n + ... + (p-1)^n Recalling the
      Message 2 of 3 , Dec 27, 2007
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        Happy new year to all

        > Given a prime p its easy to see that
        > it always divides 1+ 2 + .. + (p-1)
        Warning: p must be an odd prime!

        > But it in fact divides 1^k + 2^k + ... + (p-1)^k
        > for all k except 0 and p-1.

        > Its obvious if k is odd.
        > But is there any way of proving it when k is even?

        It appears as a simple induction:

        Let S(n) denote 1^n+ 2^n + ... + (p-1)^n

        Recalling the binomial expansion
        (a+1)^n = a^n + (n|1)a^(n-1)+ ... + (n|n)
        where (n|m) denotes the binomial coefficients n!/m!(n-m)!

        Summing it up for a=1,...,p-1 and after telescoping ^n terms from both sides
        of the equality we obtain a recursive relation for S(n-1).

        Replacing n with k+1 we have a recursive relation for S(k) as follows:

        p^(k+1) = 1 + (k+1|1)S(k) + (k+1|2)S(k-1) + ... + (k+1|k)S(1) + (p-1)

        Now the induction works independent of parity of k. Please fill the small
        details.

        Regards
        Payam


        [Non-text portions of this message have been removed]
      • Werner D. Sand
        ... (p-1) ... is ... This is not a property of the prime numbers but of the natural numbers: put n instead of p, or, if you want, 2n-1 (odd numbers). WDS
        Message 3 of 3 , Dec 31, 2007
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          --- In primenumbers@yahoogroups.com, "S. R. Sudarshan Iyengar"
          <sudarshanmysore@...> wrote:
          >
          > Here goes a nice problem... IT would be nice if someone can give me
          > links/references where I can find more details about this problem.
          >
          > Given a prime P its easy to see that it always divides 1+ 2 + .. +
          (p-1)
          >
          > But it in fact divides 1^k + 2^k + ... + (p-1)^k for all k except 0
          > and p-1.
          >
          > Its obvious if k is odd. But is there any way of proving it when k
          is
          > even?
          >
          > Regards,
          > Sudarshan
          >


          This is not a property of the prime numbers but of the natural
          numbers: put n instead of p, or, if you want, 2n-1 (odd numbers).

          WDS
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