Loading ...
Sorry, an error occurred while loading the content.

Re: Diophantine Pythagorean theorem CORRECT2

Expand Messages
  • 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 1 of 3 , Sep 4, 2008
    • 0 Attachment
      --- 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
    Your message has been successfully submitted and would be delivered to recipients shortly.