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Pythagorean Triples

Diophantine equation → Rational solutions. Pythagorean theorem 𝑎²+𝑏²=𝑐² 3 variables with integer solutions of Pythagorean triples.

Arithmetric Approach

Example 3²+4²=5²

From Pythagorean theorem 𝑥²+𝑦²=𝑧² 32+42=52⇒𝑥2+𝑦2=𝑧2=(𝑦+1)2
⇒𝑥2+𝑦2=𝑦2+2𝑦+1
⇒𝑥2=2𝑦+1 linear form Since right hand side is odd, therefore 𝑥 must be odd. Let 𝑥=2𝑘+1⇒𝑥2=(2𝑘+1)2=4𝑘2+4𝑘+1=2𝑦+1⇒2𝑘(𝑘+1)=𝑦
⇒𝑥=2𝑘+1⇒𝑦=2𝑘2+2𝑘⇒𝑧=2𝑘2+𝑘+1
Pythagorean triples {[3,4,5],[5,12,13],[7,24,25],[9,40,41],⋯}

Example 6²+8²=10²

From Pythagorean theorem 𝑥²+𝑦²=𝑧² 62+82=102⇒𝑥2+𝑦2=𝑧2=(𝑦+2)2
⇒𝑥2+𝑦2=𝑦2+4𝑦+4
⇒𝑥2=4𝑦+4=4(𝑦+1) linear form Since right hand side is even, therefore 𝑥 must be even. Let 𝑥=2𝑙⇒𝑥2=(2𝑙)2=4𝑙2=4(𝑦+1)⇒𝑙2−1=𝑦
let 𝑙=𝑘,
⇒𝑥=2𝑘⇒𝑦=𝑘2−1⇒𝑧=𝑘2+1
Pythagorean triples {[4,3,5],[6,8,10],[8,15,17],[10,24,26],⋯}

General Case: 𝑐=𝑏+𝑠

Goal: {𝑥²+𝑦²=𝑧²}={𝑥=𝑥(param),𝑦=𝑦(param),𝑧=𝑧(param)}
cases and reduction: primitive solutions if gcd(𝑥,𝑦,𝑧)=1 Let 𝑃(𝑥0,𝑦0,𝑧0) be a primitive solution.
Let 𝑧0=𝑦0+𝑠, 𝑠∊ℤ 𝑥20+𝑦20=𝑧20 ⇒𝑥20+𝑦20=(𝑦0+𝑠)2 ⇒𝑥20+𝑦20=𝑦20+2𝑠𝑦0+𝑠2 ⇒𝑥20=2𝑠𝑦0+𝑠2 Let 𝑥0=𝑡 ⇒𝑡2=2𝑠𝑦0+𝑠2 ⇒𝑦0=𝑡2−𝑠22𝑠 ⇒𝑧0=𝑦0+𝑠=𝑡2+𝑠22𝑠 ⇒𝑥20+𝑦20=𝑧20⇒(𝑡)2+𝑡2−𝑠22𝑠2=𝑡2+𝑠22𝑠2 ⇒(2𝑠𝑡)2+(𝑡2−𝑠2)2=(𝑡2+𝑠2)2 ⇒𝑃1(𝑥1,𝑦1,𝑧1)=(2𝑠𝑡,𝑡2−𝑠2,𝑡2+𝑠2) be solution with common factor between 𝑠 and 𝑡 Let 𝑑 be common factor of 𝑠 and 𝑡, and 𝑠=𝑑𝑚, 𝑡=𝑑𝑛 ⇒𝑥1=2𝑠𝑡=2(𝑑𝑚)(𝑑𝑛)=𝑑2(2𝑚𝑛) ⇒𝑦1=𝑡2−𝑠2=(𝑑𝑛)2−(𝑑𝑚)2=𝑑2(𝑛2−𝑚2)=𝑑2(𝑚2−𝑛2) ⇒𝑧1=𝑡2+𝑠2=(𝑑𝑛)2+(𝑑𝑚)2=𝑑2(𝑛2+𝑚2)=𝑑2(𝑚2+𝑛2) ⇒𝑥21+𝑦21=𝑧21⇒𝑑2(2𝑚𝑛)+𝑑2(𝑚2−𝑛2)=𝑑2(𝑚2+𝑛2)⇒(2𝑚𝑛)+(𝑚2−𝑛2)=(𝑚2+𝑛2) ⇒𝑃2(𝑥2,𝑦2,𝑧2)=(2𝑚𝑛,𝑚2−𝑛2,𝑚2+𝑛2) be solution without common factor between 𝑚 and 𝑛. Check

• Is ther an odd common prime factor 𝑝? Let 𝑝|𝑚 and 𝑝|𝑦₂⇒𝑝|𝑛⇒contradict to gcd(𝑚,𝑛)=1
• test an even factor by Mod 2 𝑚20011 𝑛20101 𝑚2−𝑛20110 𝑚2+𝑛20110 Case both 𝑚² and 𝑛² are even⇒contradict to gcd(𝑚,𝑛)=1
  Case either 𝑚² or 𝑛² are even⇒no factor of 2 and 𝑚≢𝑛 Mod 2⇒(𝑥₂,𝑦₂,𝑧₂) is primitive.
  Case both 𝑚² and 𝑛² are odd, 𝑚≡𝑛≡1 Mod 2.⇒𝑥₂=2𝑚𝑛,𝑦₂=𝑚²−𝑛²,𝑧₂=𝑚²+𝑛² are even.
  Let 𝑚+𝑛=2𝐴; 𝑚−𝑛=2𝐵⇒𝑚=(𝐴+𝐵); 𝑛=(𝐴−𝐵)
  ⇒𝑥₂=2𝑚𝑛=2(𝐴+𝐵)(𝐴−𝐵)=2(𝐴²−𝐵²)
  ⇒𝑦₂=𝑚²−𝑛²=(𝑚+𝑛)(𝑚−𝑛)=(2𝐴)(2𝐵)=4𝐴𝐵
  ⇒𝑧₂=𝑚²+𝑛²=(𝐴+𝐵)²+(𝐴−𝐵)²=(𝐴²+2𝐴𝐵+𝐵²)+(𝐴²−2𝐴𝐵+𝐵²)=2(𝐴²+𝐵²)
  2 is the common factor
  ⇒𝑥₃=𝐴²−𝐵²
  ⇒𝑦₃=2𝐴𝐵
  ⇒𝑧₂=𝐴²+𝐵²
  If both 𝐴 and 𝐵 are odd, then the conversion can be repeated since 𝑃(𝑥,𝑦,𝑧) are always of the same form with the two terms on the left hand side are changed alternatively when 2 is the common factor of 𝑃(𝑥,𝑦,𝑧). When there is no more even common factor for 𝑃(𝑥,𝑦,𝑧), gcd(𝐴,𝐵)=1 or (𝑥,𝑦,𝑧) is primitive, and 𝐴 and 𝐵 can only be in opposite parity.

Therefore the primitive solution 𝑃₀(𝑥₀,𝑦₀,𝑧₀) is 𝑥0=2𝑢𝑣
𝑦0=𝑢2−𝑣2
𝑧0=𝑢2+𝑣2
where gcd(𝑢,𝑣)=1 and 𝑢≢𝑣 Mod 2

Source and Reference

https://www.youtube.com/watch?v=bTenb0VPa3A
https://www.youtube.com/watch?v=4D9ttfBNIJI
https://www.youtube.com/watch?v=eRXWpWgP0dQ