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Diophantine Pythagorean Conjecture

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  • Sebastian Martin
    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 1:31 PM
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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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    • 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 2 of 2 , Sep 7 7:11 PM
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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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