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Pythagorean Triples

Pythagorean Triples: 𝑥²+𝑦²=𝑧²
If 𝑧≠0, 𝑥𝑧²+𝑦𝑧²=1⇒a unit circle with rational solutions for the homogeneous polynomial.

Algebraic Approach

ℚ(𝑖)={𝑎+𝑏𝑖;𝑎,𝑏∊ℚ} is a field of number system that build the structure to the rational solutions of 𝑥²+𝑦²=1
⇒𝑥²+𝑦²=(𝑥+𝑦𝑖)(𝑥−𝑦𝑖)=1 ⇒(𝑥+𝑦𝑖)²(𝑥−𝑦𝑖)²=1 rasie to higher power ⇒((𝑥²−𝑦²)+(2𝑥𝑦)𝑖)((𝑥²−𝑦²)−(2𝑥𝑦)𝑖)=1 ⇒(𝑥²−𝑦²)²+(2𝑥𝑦)²=1 structure of the equation is still preserved ⇒𝑥⁴+𝑦⁴+2𝑥²𝑦²=1 Similarly, (𝑥²+𝑦²)²=1⇒𝑥⁴+𝑦⁴+2𝑥²𝑦²=1
Solutions to the equation are
35,45, −725,2425, −5276255,−336625, 164833390625,354144390625, ⋯
The solutions to the equation can be expanded by combining more than one set of solutions through binary operation. {𝑥²+𝑦²=(𝑥+𝑦𝑖)(𝑥−𝑦𝑖)=1𝑧²+𝑤²=(𝑧+𝑤𝑖)(𝑧−𝑤𝑖)=1⇒((𝑥𝑧−𝑦𝑤)+(𝑥𝑤+𝑦𝑧)𝑖)((𝑥𝑧−𝑦𝑤)−(𝑥𝑤+𝑦𝑧)𝑖)=1 ⇒(𝑥𝑧−𝑦𝑤)²+(𝑥𝑤+𝑦𝑧)²=1 For example, 35,45 and 1213,513⇒1665,6365 1665,6365 and 35,45⇒−204325,253325 That is 𝑍={(𝑥,𝑦)∊ℚ²:𝑥²+𝑦²=1} 𝑃(𝑥,𝑦),𝑄(𝑤,𝑧)∊𝑍⇒𝑃⊕𝑄∊𝑍, 𝑃⊕𝑄=(𝑥𝑧−𝑦𝑤),(𝑥𝑤+𝑦𝑧)) For binary operation:

• 𝑃⊕𝑄=𝑄⊕𝑃
• (𝑃⊕𝑄)⊕𝑅=𝑃⊕(𝑄⊕𝑅)
• 𝐸=(1,0)⇒𝐸⊕𝑃=𝑃

Higher Degree

Similar to second degree, Suppose (𝑥+𝑦∛2+𝑧∛22)(𝑥+𝑦∛2𝜁+𝑧∛22𝜁2)(𝑥+𝑦∛2𝜁2+𝑧∛22𝜁)=1 Let 𝜁3=1 and 𝜁2+𝜁+1=0, and 𝜁4= 𝜁3𝜁1=𝜁 ⇒𝑥3+2𝑦3−6𝑥𝑦𝑧+4𝑧3=1 ⇒{(𝑥,𝑦,𝑧)}∊ℚ3:𝑥3+2𝑦3−6𝑥𝑦𝑧+4𝑧3=1} For Higher Power (𝑥+𝑦∛2+𝑧∛22)n=2,3⇒𝑥'+𝑦'∛2+𝑧'∛22 (𝑥+𝑦∛2𝜁+𝑧∛22𝜁2)n=2,3⇒𝑥'+𝑦'∛2𝜁+𝑧'∛22𝜁2 By raise the equation to power of 2 (𝑥+𝑦∛2+𝑧∛2²)²(𝑥+𝑦∛2𝜁+𝑧∛2²𝜁²)²(𝑥+𝑦∛2𝜁²+𝑧∛2²𝜁)²=1 ⇒(𝑥'+𝑦'∛2+𝑧'∛2²)(𝑥'+𝑦'∛2𝜁+𝑧'∛2²𝜁²)(𝑥'+𝑦'∛2𝜁²+𝑧'∛2²𝜁)=1 ⇒𝑥'³+2𝑦'³−6𝑥'𝑦'𝑧'+4𝑧'³=1 where {𝑥'=𝑥²+4𝑦𝑧𝑦'=𝑦²+2𝑥𝑧𝑧'=2𝑥𝑦+2𝑧²

Source and Reference

https://www.youtube.com/watch?v=nS6YwdKIIKA
https://www.youtube.com/watch?v=ABr3QisSAWQ