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Diophantine Pythagorean theorem CORRECT2

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  • Sebastian Martin
    Hello all Diophantine Pythagorean theorem.   Prove this Theorem:   “ If p is a ODD prime number then the following Diophantic Equation   p^2+(p+k)^2=q^2
    Message 1 of 3 , Aug 25, 2008
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      Hello all



      Diophantine Pythagorean theorem.
       
      Prove this Theorem:
       
      “ If p is a ODD prime number then the following Diophantic Equation
       
      p^2+(p+k)^2=q^2
       
      has only one solution. (p, k ,q) integers.
       
      k=((p-2)*p-1)/2
       
      q=(p^2+1)/2  “
       
       
       
       
      Countraexample for the composite case:
       
      p=9       9^2 +(9+3)^2=15^2           (9,3,15)
       
                    9^2+(9+31)^2=41^2         (9,31,41)
       
       
      Sebastián Martin Ruiz
       
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    • Sebastian Martin
      ... De: Chris Caldwell Para: Sebastian Martin Enviado: lunes, 25 de agosto, 2008 19:06:42 Asunto: RE: [PrimeNumbers]
      Message 2 of 3 , Aug 25, 2008
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        ----- Mensaje original ----
        De: Chris Caldwell <caldwell@...>
        Para: Sebastian Martin <sebi_sebi@...>
        Enviado: lunes, 25 de agosto, 2008 19:06:42
        Asunto: RE: [PrimeNumbers] Diophantine Pythagorean theorem CORRECT2

        Did you mean positive integers?  Otherwise let p = +/-q = -k.
        Yes, certainly. Sorry

        -----Original Message-----
        From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com] On Behalf Of Sebastian Martin
        Sent: Monday, August 25, 2008 9:54 AM
        To: Ignacio Larosa; Ignacio Larosa; Lista de Matemáticas; lista de primos; Phil Carmody; William Bouris; Yves Gallot
        Subject: [PrimeNumbers] Diophantine Pythagorean theorem CORRECT2







        Hello all



        Diophantine Pythagorean theorem.
         
        Prove this Theorem:
         
        " If p is a ODD prime number then the following Diophantic Equation
         
        p^2+(p+k)^2=q^2
         
        has only one solution. (p, k ,q) integers.
         
        k=((p-2)*p-1)/2
         
        q=(p^2+1)/2  "
         
         
         
         
        Countraexample for the composite case:
         
        p=9       9^2 +(9+3)^2=15^2           (9,3,15)
         
                      9^2+(9+31)^2=41^2         (9,31,41)
         
         
        Sebastián Martin Ruiz
         
        __________________________________________________


        __________________________________________________
        Correo Yahoo!
        Espacio para todos tus mensajes, antivirus y antispam ¡gratis!
        Regístrate ya - http://correo.yahoo.es

        [Non-text portions of this message have been removed]


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      • elevensmooth
        ... Pythagorean Triplets must be of the form p = a^2 - b^2 = (a+b)*(a-b) p+k = 2ab q = a^2 + b^2 For p to be prime, a-b=1. Simple substitution finishes the
        Message 3 of 3 , Sep 4, 2008
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          --- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
          wrote:

          > Prove this Theorem:
          >  
          > " If p is a ODD prime number then the following Diophantic Equation
          >  
          > p^2+(p+k)^2=q^2
          >  
          > has only one solution. (p, k ,q) integers.
          >  
          > k=((p-2)*p-1)/2
          >  
          > q=(p^2+1)/2  "

          Pythagorean Triplets must be of the form

          p = a^2 - b^2 = (a+b)*(a-b)
          p+k = 2ab
          q = a^2 + b^2

          For p to be prime, a-b=1. Simple substitution finishes the proof.

          William
          Poohbah of oddperfect.org
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