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 Pythagorean Triples
 Arithmetric Approach
  Example 32+42=52
  Example 62+82=102
  General Case: 𝑐=𝑏+𝑠
 Source and Reference

Pythagorean Triples

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

Arithmetric Approach

Example 32+42=52

image 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

image 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

image 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

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ID: 201100014 Last Updated: 14/11/2020 Revision: 0 Ref:

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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science & 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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