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Re: [PrimeNumbers] Diophantine Pythagorean Conjecture

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  • Peter Kosinar
    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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