Proving the Impossibility of Odd Pythagorean Triples

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SUMMARY

The impossibility of both a and b being odd in a Pythagorean triple is established through modular arithmetic. Specifically, if both a and b are odd, then c² must equal 2 modulo 4, which contradicts the fact that 2 is not a quadratic residue modulo 4. Therefore, at least one of a or b must be even. This proof is valid for all odd integers, not just 1.

PREREQUISITES
  • Understanding of Pythagorean triples
  • Familiarity with modular arithmetic
  • Knowledge of quadratic residues
  • Basic algebraic manipulation skills
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  • Study the properties of quadratic residues in modular arithmetic
  • Explore the concept of Pythagorean triples in greater depth
  • Learn about proofs in number theory
  • Investigate other forms of Pythagorean triples, such as even-odd combinations
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Mathematicians, students studying number theory, educators teaching modular arithmetic, and anyone interested in the properties of Pythagorean triples.

saadsarfraz
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If we have a pythagorean triple a^2 + b^2 = c^2 and we need to show that a and b both cannot be odd. I found a proof from a website:

if a and b both odd, then we must have c[tex]^{2}[/tex][tex]\equiv[/tex]a[tex]^{2}[/tex]+b[tex]^{2}[/tex][tex]\equiv[/tex]1+1[tex]\equiv[/tex]2 (mod4), which is a contradiction, since 2 is not a square mod 4. Hence at least one of a and b must be even.

I didnt quite understand the proof as this is just when a and b are 1? what about other odd numbers.
 
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If a is odd then [tex]a^2=[1][/tex] in [tex]\mathbb{Z}_4[/tex] (The only odds in there are 1 and 3, both give 1 when squared).
 

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