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• ## Diophantine Pythagorean Conjecture

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• Hello all Diophantine Pythagorean theorem. Prove this Conjecture: “ If p is a ODD prime number then the following Diophantic Equation p^2+(p+k)^2=q^2 has
Message 1 of 2 , Sep 7, 2008
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Hello all

Diophantine Pythagorean theorem.

Prove this Conjecture:

 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) positive 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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• Hello Sebastian, ... Proof: Rewrite the equation as p^2 = (q+(p+k)) * (q-(p+k)). We know that q+p+k p and that p is a prime, so q+p+k = p^2 and q-(p+k) = 1.
Message 1 of 2 , Sep 7, 2008
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Hello Sebastian,

> ? 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) positive integers.
>
> k=((p-2)*p-1)/2
> q=(p^2+1)/2 ?

Proof:
Rewrite the equation as p^2 = (q+(p+k)) * (q-(p+k)). We know that
q+p+k > p and that p is a prime, so q+p+k = p^2 and q-(p+k) = 1.

Peter

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
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