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x**2 - y**2 = p**2

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  • Kermit Rose
    5. a general formula for the following Posted by: san_tan1 san_tan1@yahoo.co.in san_tan1 Date: Thu Apr 2, 2009 5:15 am ((PDT)) is there any general formula
    Message 1 of 1 , Apr 2, 2009
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      5. a general formula for the following
      Posted by: "san_tan1" san_tan1@... san_tan1
      Date: Thu Apr 2, 2009 5:15 am ((PDT))

      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. ....(1)?

      also can this be extended to finding an algorithm for generating duplets
      (a,b) (c,d) etc....such that given p we can find upto any number of
      desired duplets having property 1??

      that is 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)....upto any desired no.
      of duplets such that in each dupet(x,y) x,y are both odd and mutually
      prime..?



      Hello san_tan1.


      First we strive to make the formulas without regard to whether or
      not x and y are prime.

      if
      x = m1 * m2 + n1 * n2
      and
      y = m1 * n2 - m2 * n1

      then x**2 - y**2
      = (m1**2 m2**2 + n1**2 n2**2 - m1**2 n2**2 - m2**2 n1**2 )
      = m1**2 (m2**2 - n2**2) + n1**2 (n2**2 - m2**2)
      = m1**2 (m2**2 - n2**2) - n1**2 (m2**2 - n1**2)
      = (m1**2 - n1**2) (m2**2 - n2**2)

      = (m1 - n1)(m1+n1)(m2-n2)(m2+n2)

      p**2 = (m1 - n1)(m1+n1)(m2-n2)(m2+n2)


      m1 - n1 = (q1**2)*q2*q3*q4
      m1 + n1 = (q5**2)*q2*q6*q7
      m2 - n2 = (q8**2)*q3*q6*q9
      m2 + n2 = (q10**2)*q4*q7*q9

      m1 = ((q5**2)*q2*q6*q7 + (q1**2)*q2*q3*q4 )/2

      n1 = ((q5**2)*q2*q6*q7 - (q1**2)*q2*q3*q4 )/2

      m2 = ( (q10**2)*q4*q7*q9 + (q8**2)*q3*q6*q9 )/2

      n2 = ( (q10**2)*q4*q7*q9 - (q8**2)*q3*q6*q9 )/2


      x = m1 * m2 + n1 * n2

      y = m1 * n2 - m2 * n1


      Systematically assign values to q1,q2,. . . q10,

      and search for x and y being prime.

      There are probably ways to shortcut the search,
      but I don't know them.

      Kermit
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