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Pythagorean Triples![]() Geometric Approach![]() π¦π₯β1. Since the rational solutions of π₯2+π¦2=1 always preserves the rationality of slope. In other words, the rational point (π₯,π¦) is mapped to a rational slope. π0={(π₯,π¦):π₯2+π¦2=1,π₯β 1}
π:π0ββ given by π(π)=
Similarly, the map π can be defined for more general curves. And the properties of π are
π¦-intercept![]() π¦π₯β1=π‘ βπ¦=π‘(π₯β1) and π₯2+π¦2=1 Let π§=π₯β1 βπ¦=π‘π§ and (π§+1)2+π¦2=1 β(π§+1)2+(π‘π§)2=1 βπ§2+2π§+π‘2π§2=0 β(1+π‘2)π§2+2π§=0 If π§β 0 β(1+π‘2)π§+π§=0 βπ§=β 21+π‘2βπ₯β1=β 21+π‘2βπ₯= π‘2β11+π‘2ββ βπ¦=π‘π§=π‘ 21+π‘2 2π‘1+π‘2βπβ1(π‘)=(π₯,π¦)= π‘2β11+π‘2,β 2π‘1+π‘2 Conic Section Example π₯2+2π¦2=1Similar to other conic section.![]() Let π(π₯,π¦),π(π€,π§) be the solution of the equation. β(π₯+π¦β2π)(π₯βπ¦β2π)=1 and (π€+π§β2π)(π€βπ§β2π)=1 β(π₯π§β2π¦π€)2+2(π₯π€+π¦π§)2=1 So πβπ=π where π =(π₯π§β2π¦π€,π₯π€+π¦π§) ![]() π¦π₯β1where π0={(π₯,π¦):π₯2+2π¦2=1, π₯β 1} πβ1(π‘)=(π₯,π¦), where π¦=π‘(π₯β1) Let π§=π₯β1β(π§+1)2+2π‘2π§2=1βπ§((1+2π‘2)π§+2)=0 βπ§=β2/(1+2π‘2) βπ₯= 2π‘2β11+2π‘2βπ¦= β2π‘21+2π‘2βπβ1(π‘)= 2π‘2β11+2π‘2, β2π‘21+2π‘2 Key Ingredients for Binary Operation: π₯2+ππ¦2=1π₯2+ππ¦2=1
Generalization to degree 3 with three variablesπ₯3+2π¦3β6π₯π¦π§+4π§3=1βmanipulate β2Generalization to twists: ππ₯2+ππ¦2=πβπ₯2+ππ¦2=πKey Ingredients for Parameterisation: π₯2+ππ¦2=1
Source and Referencehttps://www.youtube.com/watch?v=XiwGK8sdwpQhttps://www.youtube.com/watch?v=eLoQz1WRFu0 Β©sideway ID: 201100016 Last Updated: 16/11/2020 Revision: 0 Ref: ![]() References
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