- #1
kingwinner
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Definition: A primitive Pythagorean triple is a triple of natural numbers x,y,z s.t. [tex]x^2 + y^2 = z^2[/tex] and gcd(x,y,z)=1.
note: d|gcd(x,y) => d|x and d|y
=> [tex]d^2|x^2[/tex] and [tex]d^2|y^2[/tex]
Now [tex]z^2 = x^2 + y^2[/tex]
=> [tex]d^2|z^2[/tex]
=> d|z
Thus, it follows that for any Pythagorean triple x,y,z, we must have that
gcd(x,y)=gcd(x,z)=gcd(y,z)=gcd(x,y,z). Hence, we can replace gcd(x,y,z)=1 in the above definition by e.g. gcd(x,z)=1.
[quote from my textbook]
======================================
1) Why [tex]d^2|z^2[/tex] => d|z ? I tried writing down the meaning of divisibility and looked at the theorems about the basic properties of divisibility, but I still don't understand why it's true...
2) The above shows that d|gcd(x,y) => d|z, but then why does it FOLLOW from the above that gcd(x,y)=gcd(x,z)=gcd(y,z)=gcd(x,y,z)? I don't understand this part at all.
Any help is appreciated! :)
note: d|gcd(x,y) => d|x and d|y
=> [tex]d^2|x^2[/tex] and [tex]d^2|y^2[/tex]
Now [tex]z^2 = x^2 + y^2[/tex]
=> [tex]d^2|z^2[/tex]
=> d|z
Thus, it follows that for any Pythagorean triple x,y,z, we must have that
gcd(x,y)=gcd(x,z)=gcd(y,z)=gcd(x,y,z). Hence, we can replace gcd(x,y,z)=1 in the above definition by e.g. gcd(x,z)=1.
[quote from my textbook]
======================================
1) Why [tex]d^2|z^2[/tex] => d|z ? I tried writing down the meaning of divisibility and looked at the theorems about the basic properties of divisibility, but I still don't understand why it's true...
2) The above shows that d|gcd(x,y) => d|z, but then why does it FOLLOW from the above that gcd(x,y)=gcd(x,z)=gcd(y,z)=gcd(x,y,z)? I don't understand this part at all.
Any help is appreciated! :)