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`Pythagorean TriplesβExample π₯3+2π¦3=3βSource and Reference`

# Pythagorean Triples

Binary operation on π₯2+ππ¦2=1β(π₯,π¦)β(π€,π§)=(π₯π§βππ¦π€,π₯π€+π¦π§) Consider π₯2+ππ¦2=1 over a finite field π½π Theorem: Wherther or not π is a square in π½π, there is a solution π such that π is generated by π, i.e. π=πββ―βπ for all πβπ
This single generator is similar to the primitive root mod π
Proof: Consider the field ext πΎ=π½π(βπ). Denote: [π]π=πββ―βπ. Define π:πβπΎ by π(π)=π₯+π¦βπ with rational solution (π₯,π¦). Existence of a primitive root: πβπΎ s.t. {ππ: π=0, 1, β―}=πΎ*. β‘ Fact: A subset of finitely many solutions will not generate π under the operation. Key ingredients for parameterisation: π₯2+ππ¦2=1
• A rational point (1,0)
• The degree being 2βπβ1: πβπ0
No easy way to generalize the operation for π₯3+π¦3=1 β(π₯+π¦)(π₯+π¦π)(π₯+π¦π2)=1, where π3=1 and π2+π+1=0 β(π₯+π¦)π(π₯+π¦π)π(π₯+π¦π2)π=1 Fact: Let π={(π₯,π¦):π₯3+π¦3=1} The existence of Ξ¦:πβπ contradicts some genus formula especially some positive. That is the topological invariant to the complex solution of the cubic curveβno parameterization.
Idea: One rational point and a tangent line β given a rational point.
Intesection to degree 2 such that a cubic curve of degree 3 to degree 2 and the tangent line with

## Example π₯3+2π¦3=3

π={π₯3+2π¦3=3}. π=(1,1)βπ. Let πΉ(π₯,π¦)=π₯3+2π¦3β3 tangent line at (π,π)=βπΉβπ₯(π₯βπ)+βπΉβπ¦(π¦βπ)=3π2(π₯βπ)+6π2(π¦βπ)=0 So 3(π₯β1)+6(π¦β1)=0 and π₯3+2π¦3=3 subst. π§=π₯β1, π€=π¦β1 3π§+6π€=0 and (π§+1)3+2(π€+1)3=3 6π€2(3βπ€)=0βπ€=3βπ¦=4, π₯=β5 Idea: repeat with (β5,4) Line: 25(π₯+5)+32(π¦β4)=0 and π₯3+2π¦3=3 subst. π§=π₯+5, π€=π¦β4 π₯=655/253, π¦=β488/253, 253=11*23 Therefore (1,1)β(β5,4)β(655/253,β488/253)
Idea: Two rational points and a secant lineβa rational point
β741253(π₯β1)β402253(π¦β1)=0 and π₯3+2π¦3=3 subst. π§=π₯β1, π€=π¦β1 741π§+402π€=0 and (π§+1)3+2(π€+1)3=3 27732342π€315069223+419922π€261009+1080π€247=0 Therefore π€β0, β741253(π₯β1)=0βπ€π€+741253=0βa factor of cubic equation By long division βπ€π€+74125327732342π€15069223+9108061009=0 βπ₯=2630918269, π¦=344918269βπ₯β1.44, π¦β0.18 Tangent line, πβπ and Secant line, π βπ Mordell-Weil Theorem: There are a finite subset of solutions that generate all solutions.

## Source and Reference

ID: 201100017 Last Updated: 17/11/2020 Revision: 0 Ref:

References

1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science &amp; Engineering
2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions
3. Stroud, K.A., 2001, Engineering Mathematics
4. Coolidge, J.L., 1949, The Story of The Binomial Theorem

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