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Math Help - Construction of rational points and pythagorean triples

  1. #1
    Member courteous's Avatar
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    Question Construction of rational points and pythagorean triples

    Intro:
    Construction of rational points and pythagorean triples-unit_circle.jpg

    Above is a unit circle, x^2+y^2=1.

    1. First (trivial) solution is x=1, \text{ }y=0.
    2. For second (non-trivial) solution, QR has a rational slope y=t(x+1) and we plug this \displaystyle{ x^2+y^2=x^2+t^2(x+1)^2=x^2(1+t^2)+x(2t^2)+(t^2-1)=1 }.
      This gives (non-trivial) solution of x=\frac{1-t^2}{1+t^2}, \text{ } y=\frac{2t}{1+t^2}.


    Simple enough. Now the exercise.
    The parameter  t in the pair \left( \frac{1-t^2}{1+t^2}, \text{ } \frac{2t}{1+t^2} \right) runs through all rational numbers if t=\frac{q}{p} and p, \text{ } q run through all pairs of integers.

    Deduce that if \text{ }(a,b,c) is any Pythagorean triple then
    \displaystyle{ \frac{a}{c}=\frac{p^2-q^2}{p^2+q^2}, \text{ } \frac{b}{c}=\frac{2pq}{p^2+q^2} }
    for some integers  p and  q.
    I'm having problem with a minus in \frac{a}{c} part:
    \displaystyle{ \frac{a^2}{c^2}=x^2\Rightarrow \frac{a}{c}=x=\frac{1-t^2}{1+t^2}=\frac{1-(\frac{p}{q})^2}{1+(\frac{p}{q})^2}=\frac{q^2-p^2}{q^2+p^2}\neq \frac{p^2-q^2}{q^2+p^2} }

    Use previous exercise to prove Euclid's formula for Pythagorean triples.
    How would I use the previous exercise to prove that Euclid's generates all triples?
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  2. #2
    Super Member
    earboth's Avatar
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    Quote Originally Posted by courteous View Post
    ...

    Simple enough. Now the exercise.
    I'm having problem with a minus in \frac{a}{c} part:
    \displaystyle{ \frac{a^2}{c^2}=x^2\Rightarrow \frac{a}{c}=x=\frac{1-t^2}{1+t^2}=\frac{1-(\frac{p}{q})^2}{1+(\frac{p}{q})^2}=\frac{q^2-p^2}{q^2+p^2}\neq \frac{p^2-q^2}{q^2+p^2} }

    ...
    In the quoted text of the exercise t is given as t = \frac qp.

    You used t = \frac pq.
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  3. #3
    Member courteous's Avatar
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    OK, first one is done.
    What about the second one?
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