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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
      > 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
    • David Litchfield
      ... ^ My method is not related to Fermat in the way you suggest (I qualify that with a big I think [and apology if I m thinking wrong]!) Your suggestion works
      Message 2 of 9 , Dec 2, 2001
        Marcel Martin wrote:
        >In your example, you split 21 using 21^2 = 29^2 - 20^2. With the Fermat
        >method, you would only need 21 = 5^2 - 2^2.
        >Notice that 29^2 - 20^2 = (5^2 + 2^2)^2 - (5^2 - 2^2)^2.
        ^
        My method is not related to Fermat in the way you suggest (I qualify that
        with a big I think [and apology if I'm thinking wrong]!)

        Your suggestion works for 21 but not, say, 35

        Adj = 35
        Hyp = 37
        Opp = 12

        37 + 12 = 49
        37 - 12 = 25
        49 sqrt = 7
        25 sqrt = 5
        5 * 7 = 35

        and so
        37^2 - 12^2 != (7^2 + 5^2)^2 - (7^2 - 5^2)^2

        I think it mere coincidence that it happens with 21 as most others I have
        tested do not follow A^2 - B^2 = (c^2 + d^2)^2 - (c^2- d^2) [29^2 - 20^2 =
        (5^2 + 2^2)^2 - (5^2 - 2^2)^2.]

        Will look deeper but perhaps the only relation to Fermat's method is that it
        uses squares.

        Cheers,
        David Litchfield
      • 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 3 of 9 , Dec 2, 2001
          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
          ... Cheers, David ... From: Marcel Martin Cc: Sent: Sunday, December 02, 2001 10:16 PM Subject: Re:
          Message 4 of 9 , Dec 3, 2001
            With you now, cheers. Does this explain the other two p. triangles:

            > 21 -> 28 -> 35
            > 21 -> 72 -> 75

            > 15 -> 20 -> 25
            > 15 -> 36 -> 39

            > 33 -> 44 -> 55
            > 33 -> 180 -> 183

            > 35 -> 84 -> 91
            > 35 -> 120 -> 125

            > These dimensions are for such p. triangles that when you subtract
            > the two dervied lengths from given number they produce the factors
            > e.g.
            > 77 -> 264 -> 275
            > 77 -> 420 -> 427

            > 275 - 264 = 11
            > 427 - 420 = 7

            > 7 * 11 = 77

            Cheers,
            David



            ----- Original Message -----
            From: "Marcel Martin" <znz@...>
            Cc: <primenumbers@yahoogroups.com>
            Sent: Sunday, December 02, 2001 10:16 PM
            Subject: Re: [PrimeNumbers] Pythagoras and factoring


            >
            > I sent an other post in which I pointed out the typo I made in the
            > first one. The right relation was
            > 29^2 - 21^2 = (5^2 + 2^2)^2 - (5^2 - 2^2)^2.
            >
            > The link with your method and the Fermat's one is
            >
            > Fermat: N = u^2 - v^2
            > Yours : N^2 = U^2 - V^2
            >
            > Assuming N = a*b, a and b odd, a > b.
            > With Fermat, it comes, u = (a+b)/2 and v = (a-b)/2. Then your
            > parameters are simply U = u^2 + v^2 and V = 2uv
            >
            > With 35 you get
            > u = 6, v = 1 thus U = 37 and V = 12.
            >
            > >and so
            > >37^2 - 12^2 != (7^2 + 5^2)^2 - (7^2 - 5^2)^2
            >
            > Of course. But
            >
            > 37^2 - 35^2 = (6^2 + 1^2)^2 - (6^2 - 1^2)^2
            > U^2 - N^2 = (u^2 + v^2)^2 - (u^2 - v^2)^2
            >
            > >Will look deeper but perhaps the only relation to Fermat's method is that
            it
            > >uses squares.
            >
            > No. It is just a complication of Fermat method. And, except maybe
            > some particular cases, it is much less efficient since you have to
            > work with numbers that are twice greater than the ones required in
            > the Fermat method.
            >
            > Marcel Martin
            >
            >
            > Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
            > The Prime Pages : http://www.primepages.org
            >
            >
            >
            > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
            >
            >
            >
          • David Litchfield
            ... 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
            Message 5 of 9 , Dec 3, 2001
              > 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 6 of 9 , Dec 3, 2001
                > 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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