- #1
saadsarfraz
- 86
- 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.
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.