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Pythagorean triples proof

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data
    prove the following proposition:
    Let P be the set of pythagorean triples; that is,
    P= {(a,b,c) : a, b, c [itex]\in[/itex] Z and a[itex]^{2}[/itex]+b[itex]^{2}[/itex]=c[itex]^{2}[/itex]}
    and let T be the set
    T= {(p,q,r) : p=x[itex]^{2}[/itex]-y[itex]^{2}[/itex], q= 2xy, and r = x[itex]^{2}[/itex]+y[itex]^{2}[/itex] where x,y[itex]\in[/itex] Z}

    show also that T ≠ P … that is that T is a “proper subset” of P … that is that there’s at least 1 member of P that is NOT in T.

    2. Relevant equations



    3. The attempt at a solution
    I understand what it's asking but i have no idea how to begin proving it. I thought at first to try and let x[itex]\in[/itex] P and go from there but then I didn't know how to work with that idea at all. Can someone point me the right direction on figuring this out?
     
  2. jcsd
  3. Feb 24, 2014 #2

    Dick

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    Use algebra to show your formula generates Pythagorean triples, i.e. show p^2+q^2=r^2. That's easy enough. To show it doesn't generate all of them takes a little more guesswork. You have to find a triple is doesn't generate. Here's a hint. Multiply some of the known triples by a constant.
     
  4. Feb 24, 2014 #3

    Dick

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    Science Advisor
    Homework Helper

    Use algebra to show your formula generates Pythagorean triples, i.e. show p^2+q^2=r^2. That should be easy enough. To show it doesn't generate all of them takes a little more guesswork. You have to find a triple is doesn't generate. Here's a hint. Multiply some of the known triples by a constant.
     
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