Pythagorean Triples

Binary operation on 𝑥²+𝑑𝑦²=1⇒(𝑥,𝑦)⊕(𝑤,𝑧)=(𝑥𝑧−𝑑𝑦𝑤,𝑥𝑤+𝑦𝑧) Consider 𝑥²+𝑑𝑦²=1 over a finite field 𝔽_𝑝 Theorem: Wherther or not 𝑑 is a square in 𝔽_𝑝, there is a solution 𝑃 such that 𝑍 is generated by 𝑃, i.e. 𝑄=𝑃⊕⋯⊕𝑃 for all 𝑄∊𝑍
This single generator is similar to the primitive root mod 𝑃
Proof: Consider the field ext 𝐾=𝔽_𝑝(√𝑑). Denote: [𝑛]𝑃=𝑃⊕⋯⊕𝑃. Define 𝑓:𝑍→𝐾 by 𝑓(𝑄)=𝑥+𝑦√𝑑 with rational solution (𝑥,𝑦). Existence of a primitive root: 𝑔∊𝐾 s.t. {𝑔^𝑛: 𝑛=0, 1, ⋯}=𝐾^*. □ Fact: A subset of finitely many solutions will not generate 𝑍 under the operation. Key ingredients for parameterisation: 𝑥²+𝑑𝑦²=1

• A rational point (1,0)
• The degree being 2⇒𝑚⁻¹: 𝑄→𝑍₀

No easy way to generalize the operation for 𝑥³+𝑦³=1 ⇒(𝑥+𝑦)(𝑥+𝑦𝜁)(𝑥+𝑦𝜁²)=1, where 𝜁³=1 and 𝜁²+𝜁+1=0 ⇒(𝑥+𝑦)^𝑛(𝑥+𝑦𝜁)^𝑛(𝑥+𝑦𝜁²)^𝑛=1 Fact: Let 𝑍={(𝑥,𝑦):𝑥³+𝑦³=1} The existence of Φ:𝑄→𝑍 contradicts some genus formula especially some positive. That is the topological invariant to the complex solution of the cubic curve⇒no parameterization.
Idea: One rational point and a tangent line → 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 𝑥³+2𝑦³=3

𝑍={𝑥³+2𝑦³=3}. 𝑃=(1,1)∊𝑍. Let 𝐹(𝑥,𝑦)=𝑥³+2𝑦³−3 tangent line at (𝑎,𝑏)=∂𝐹∂𝑥(𝑥−𝑎)+∂𝐹∂𝑦(𝑦−𝑏)=3𝑎²(𝑥−𝑎)+6𝑏²(𝑦−𝑏)=0 So 3(𝑥−1)+6(𝑦−1)=0 and 𝑥³+2𝑦³=3 subst. 𝑧=𝑥−1, 𝑤=𝑦−1 3𝑧+6𝑤=0 and (𝑧+1)³+2(𝑤+1)³=3 6𝑤²(3−𝑤)=0⇒𝑤=3⇒𝑦=4, 𝑥=−5 Idea: repeat with (−5,4) Line: 25(𝑥+5)+32(𝑦−4)=0 and 𝑥³+2𝑦³=3 subst. 𝑧=𝑥+5, 𝑤=𝑦−4 𝑥=655/253, 𝑦=−488/253, 253=11*23 Therefore (1,1)→(−5,4)→(655/253,−488/253)
Idea: Two rational points and a secant line→a rational point
−741253(𝑥−1)−402253(𝑦−1)=0 and 𝑥³+2𝑦³=3 subst. 𝑧=𝑥−1, 𝑤=𝑦−1 741𝑧+402𝑤=0 and (𝑧+1)³+2(𝑤+1)³=3 27732342𝑤³15069223+419922𝑤²61009+1080𝑤247=0 Therefore 𝑤→0, −741253(𝑥−1)=0⇒𝑤𝑤+741253=0⇒a factor of cubic equation By long division ⇒𝑤𝑤+74125327732342𝑤15069223+9108061009=0 ⇒𝑥=2630918269, 𝑦=344918269⇒𝑥≈1.44, 𝑦≈0.18 Tangent line, 𝑃⊕𝑄 and Secant line, 𝑃 ⇒𝑅 Mordell-Weil Theorem: There are a finite subset of solutions that generate all solutions.

Source and Reference

https://www.youtube.com/watch?v=hrp0GdsqLEg
https://www.youtube.com/watch?v=PZlVYEihCh0