Loading ...
Sorry, an error occurred while loading the content.

p-tight integers

Expand Messages
  • Andrey Kulsha
    Given an odd prime p, let s call the positive integer p-tight if all its prime factors are 1 (mod 2p). Theorem: if p does not divide a-1, then (a^p-1)/(a-1)
    Message 1 of 6 , Dec 20, 2013
    • 0 Attachment
          Given an odd prime p, let's call the positive integer "p-tight" if all its prime factors are 1 (mod 2p).
       
          Theorem:
      if p does not divide a-1, then (a^p-1)/(a-1) is p-tight.
      (trivial proof follows from FLT)
       
          Conjecture:
      if p divides a-1, then (a^p-1)/(a-1)/p is p-tight.
       
          Any suggestions for a proof?
       
          Thanks a lot,
       
          Andrey
    • djbroadhurst
      ... This follows from Theorem 48 of Daniel Shanks delightful book: https://archive.org/details/SolvedAndUnsolvedProblemsInNumberTheory David
      Message 2 of 6 , Dec 20, 2013
      • 0 Attachment
        Andrey Kulsha wrote:

        > Given an odd prime p, let's call the positive integer
        > "p-tight" if all its prime factors are 1 (mod 2p).

        > Conjecture:
        > if p divides a-1, then (a^p-1)/(a-1)/p is p-tight.

        This follows from Theorem 48 of Daniel Shanks' delightful book:

        https://archive.org/details/SolvedAndUnsolvedProblemsInNumberTheory

        David
      • Andrey Kulsha
        ... This follows from Theorem 48 of Daniel Shanks delightful book: https://archive.org/details/SolvedAndUnsolvedProblemsInNumberTheory Thanks for the book.
        Message 3 of 6 , Dec 21, 2013
        • 0 Attachment
          
          > Conjecture:
          > if p divides a-1, then (a^p-1)/(a-1)/p is p-tight.

          This follows from Theorem 48 of Daniel Shanks' delightful book:

          https://archive.org/details/SolvedAndUnsolvedProblemsInNumberTheory
              Thanks for the book. Maybe I misunderstand something, but it's quite obvious that p divides (a^p-1)/(a-1) when a = 1 (mod p), because (a^p-1)/(a-1) = sum(a^k,{k,0,p-1}).

              The conjecture states that (a^p-1)/(a-1) isn't divisible by p^2. But I can't derive this from Theorem 48...
           
              Best regards,
           
              Andrey
        • djbroadhurst
          ... Ah, yes, Shanks says that all factors expect p are congruent to 1 mod p, but does not explicitly say that p occurs only once. Here is the extra step. Let a
          Message 4 of 6 , Dec 21, 2013
          • 0 Attachment

            Andrey Kulsha wrote:

            > The conjecture states that (a^p-1)/(a-1) isn't divisible by
            > p^2. But I can't derive this from Theorem 48...

            Ah, yes, Shanks says that all factors expect p
            are congruent to 1 mod p, but does not
            explicitly say that p occurs only once.

            Here is the extra step.

            Let a = 1 + r*p. Then
            S := (a^p-1)/(a-1)
            = sum(k=0,p-1,a^k)
            = sum(k=0,p-1,1+k*r*p) mod p^2
            = p mod p^2,
            since sum(k=0,p-1,k)=p*(p-1)/2 and p is odd.
            Hence S/p = 1 mod p.

            Happy now?

            David

          • djbroadhurst
            Summary: For every prime p 2, and for every integer a 1, it is proven that (a^p-1)/(a-1)/if(a%p==1,p,1) is an integer all of whose positive integral
            Message 5 of 6 , Dec 21, 2013
            • 0 Attachment

              Summary: For every prime p > 2, and for
              every integer a > 1, it is proven that
              (a^p-1)/(a-1)/if(a%p==1,p,1)
              is an integer all of whose positive integral
              divisors are congruent to 1 modulo 2*p.

              David

            • Andrey Kulsha
              Here is the extra step. Let a = 1 + r*p. Then S := (a^p-1)/(a-1) = sum(k=0,p-1,a^k) = sum(k=0,p-1,1+k*r*p) mod p^2 = p mod p^2, since sum(k=0,p-1,k)=p*(p-1)/2
              Message 6 of 6 , Dec 23, 2013
              • 0 Attachment
                
                Here is the extra step.

                Let a = 1 + r*p. Then
                S := (a^p-1)/(a-1)
                = sum(k=0,p-1,a^k)
                = sum(k=0,p-1,1+k*r*p) mod p^2
                = p mod p^2,
                since sum(k=0,p-1,k)=p*(p-1)/2 and p is odd.
                Hence S/p = 1 mod p.

                Happy now?
                    That was so simple.
                 
                    Thanks.
                 
                    Andrey :-)
              Your message has been successfully submitted and would be delivered to recipients shortly.