Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    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)
    Offline

    2
    ReputationRep:
    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
    Offline

    16
    ReputationRep:
    (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)
    Offline

    0
    ReputationRep:
    (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
    Offline

    16
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    (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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.