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Re: a general formula for the following

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  • David Broadhurst
    ... Yes: (4*n^2+1)^2 - (4*n^2-1)^2 = (4*n)^2 ... No: for p = 0 mod 4, the number of odd coprime pairs [x,y] with p = x^2 - y^2 is 2^k where k is the number of
    Message 1 of 3 , Apr 2, 2009
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      --- In primenumbers@yahoogroups.com,
      "san_tan1" <san_tan1@...> wrote:

      > is there any general formula for triples (a,b,c) such that
      > a,b are mutually prime and both odd and also
      > a^2-b^2=c^2

      Yes: (4*n^2+1)^2 - (4*n^2-1)^2 = (4*n)^2

      > given p,
      > p^2= a^2-b^2=c^2-d^2=e^2-f^2.......
      > is there any way to generate (a,b) (c,d) ,(e,f)....up
      > to any desired no. of duplets such that in each duplet(x,y)
      > x,y are both odd and mutually prime

      No: for p = 0 mod 4, the number of odd coprime pairs [x,y]
      with p = x^2 - y^2 is 2^k where k is the number of distinct
      odd prime divisors of p. This procedure will print them all:

      {pairs(p)=local(x,y,c,d);if(p%4==0,fordiv(p/2,d,
      c=p/2/d;if(c>d,x=c^2+d^2;y=c^2-d^2;if(gcd(x,y)==1,
      print([x,y])))));}

      pairs(2^3*3^2*5*7*11);

      [192099601, 192099599]
      [12006241, 12006209]
      [7684009, 7683959]
      [3920449, 3920351]
      [2371681, 2371519]
      [1587721, 1587479]
      [480649, 479849]
      [245809, 244241]
      [158041, 155591]
      [149521, 146929]
      [101161, 97289]
      [96889, 92839]
      [66529, 60479]
      [52369, 44431]
      [38329, 26471]
      [29401, 9799]

      David
    • David Broadhurst
      PS: I left out a square sign, here restored: for p = 0 mod 4, the number of odd coprime pairs [x,y] with p^2 = x^2 - y^2 is 2^k where k is the number of
      Message 2 of 3 , Apr 2, 2009
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        PS: I left out a square sign, here restored:

        for p = 0 mod 4, the number of odd coprime pairs [x,y]
        with p^2 = x^2 - y^2 is 2^k where k is the number of distinct
        odd prime divisors of p.

        David^2
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