## Diophantine Pythagorean theorem CORRECT2

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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
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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• ... 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-----
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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• ... 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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