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    sorry if the answer is obvious.

    the standard thing where you do:
    a^2=b^2+c^2

    a=k(x^2+y^2)
    b=k(2xy)
    c=k(x^2-y^2)
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    http://www.cut-the-knot.org/pythagoras/pythTriple.shtml

    http://planetmath.org/encyclopedia/P...anTriples.html

    there is a selection of verifications on this page but it boils down to:

    (x^2+y^2)^2 = (2xy)^2 + (x^2 - y^2)^2
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    (Original post by The Muon)
    there is a selection of verifications on this page but it boils down to:

    (x^2+y^2)^2 = (2xy)^2 + (x^2 - y^2)^2
    But (x^2+1)^2 = (2x)^2+(x^2-1)^2... so it doesn't quite boil down to that.

    One way of looking at it is to say primitive Pythagorean triples correspond bijectively with rational points on the unit circle, and that we can parameterize these using half tangents... (and we all know the half tangent identities)
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    (Original post by SimonM)
    But (x^2+1)^2 = (2x)^2+(x^2-1)^2... so it doesn't quite boil down to that.
    Surely this is fine? y = 1
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    (Original post by Mathletics)
    Surely this is fine? y = 1
    My point was that just because there is an identity that satisfies it, doesn't mean they all have to be of that form. My identity satisfies the equation, does that mean I've found all solutions... no
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    (Original post by SimonM)
    My point was that just because there is an identity that satisfies it, doesn't mean they all have to be of that form. My identity satisfies the equation, does that mean I've found all solutions... no
    I see.
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    (Original post by SimonM)
    But (x^2+1)^2 = (2x)^2+(x^2-1)^2... so it doesn't quite boil down to that.

    One way of looking at it is to say primitive Pythagorean triples correspond bijectively with rational points on the unit circle, and that we can parameterize these using half tangents... (and we all know the half tangent identities)
    I wasn't saying that was why it worked. I should have explained better. That was merely a verification that it would work. The links were the crucial part.
 
 
 

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