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Proof of pythagorean triple

  1. Jan 25, 2009 #1
    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.
     
  2. jcsd
  3. Jan 26, 2009 #2
    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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