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Re: [PrimeNumbers] Pythagoras and factoring

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  • David Litchfield
    ... But: 205 - 164 = 41 123 / 41 = 3 So we can get the factors straight out with this p. triangle. David
    Message 1 of 9 , Dec 1, 2001
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      > Then B+C=(U+V)^2, B-C=(U-V)^2 as in your examples. But it is *not* common
      > solution!
      > For instance, the least solution to A^2+B^2=C^2 with A=123 is
      >
      > 123^2+164^2=205^2
      >
      > Both your C-program and my pascal one find this solution. But,alas
      >
      > 205-164=41 is *not* square, neither is
      > 205+164=369=3^2*41.
      >

      But: 205 - 164 = 41
      123 / 41 = 3
      So we can get the factors straight out with this p. triangle.

      David
    • Mark Underwood
      Hello David, I don t even know what the Fermat method is, and I don t entirely comprehend what has been exchanged thus far, but here are my gleanings on the
      Message 2 of 9 , Dec 2, 2001
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        Hello David,

        I don't even know what the 'Fermat method' is, and I don't entirely
        comprehend what has been exchanged thus far, but here are my
        gleanings on the matter:


        let a^2 + b^2 = c^2, a,b,c relatively prime. a or b must be even, so
        let us assume b is odd.

        b^2 = c^2 - a^2 = (c-a)(c+a)

        It is clear that c-a and c+a are relatively prime, so each of c-a and
        c+a must be squares themselves. So let

        r^2 = c-a
        s^2 = c+a

        So then,

        c = (s^2 + r^2)/2
        a = (s^2 - r^2)/2
        b = rs

        If b is 21, the only combinations possible for r and s are
        (r,s) = (1,21) and (3,7). That would make for two pythagorean
        triplets easily calculated with the equations just above.

        If b is 35, the only combinations possible for r and s are
        (r,s) = (1,35) and (5,7). Two pythagorean triplets.

        If b is 105 = 3*5*7, the combinations possible for r and s are
        (r,s) = (3,35), (5,21), (7,15) and (1,105). Four pythagorean triplets.

        More generally, if b is composed of x different prime factors, then
        the number of pythagorean triples possible is 2^(x-1).

        If b is a prime p, there is only one possible combination, and that is
        (r,s) = (1,p).

        In other words, if b is a prime p, the only pythagorean triplet
        possible would be

        b = p
        a = (p^2 - 1)/2
        c = (p^2 + 1)/2


        Examples: (b,a,c), where a^2 + b^2 = c^2, b is prime:

        (3,4,5), (5,12,13), (7,24,25), (11,60,61).

        a little more mud in the water,
        Mark





        --- In primenumbers@y..., "David Litchfield" <Mnemonix@g...> wrote:
        > > With you now, cheers. Does this explain the other two p.
        triangles:
        >
        > Bad form answering your own emails (sorry) but I've just seen the
        > relationship:
        >
        > Using 35 as the example
        >
        > 35 = 5 * 7
        > p. triangle is 35 -> 120 -> 125
        >
        > 120 / 5 = 24
        > 125 / 5 = 25
        > 24 + 25 = 49
        > 49 sqrt = 7
        >
        > and
        >
        > p. triangle is 35 -> 84 -> 91
        > 84 / 7 = 12
        > 91 / 7 = 13
        > 12 + 13 = 25
        > 25 sqrt = 5
        >
        >
        > David
      • David Litchfield
        ... But: 205 - 164 = 41 123 / 41 = 3 So we can get the factors straight out with this p. triangle. David
        Message 3 of 9 , Dec 3, 2001
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          > Then B+C=(U+V)^2, B-C=(U-V)^2 as in your examples. But it is *not* common
          > solution!
          > For instance, the least solution to A^2+B^2=C^2 with A=123 is
          >
          > 123^2+164^2=205^2
          >
          > Both your C-program and my pascal one find this solution. But,alas
          >
          > 205-164=41 is *not* square, neither is
          > 205+164=369=3^2*41.
          >

          But: 205 - 164 = 41
          123 / 41 = 3
          So we can get the factors straight out with this p. triangle.

          David
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