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September 10th, 2012, 02:47 PM   #1
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Primitive Pythagorean Proof question.!

For Pythagorean triples a^2 + b^2 = c^2, if we divide c^2 on both side of equation, we get (a^2/c^2) + (b^2/c^2) = 1
The last question is same as x^2 + y^2 = 1 which is circle.
if we consider a line through point (-1,0) , there must be another rational point on the circle.
The x and y i found in terms of slope m is (1-m^2/1+m^2), (2m/1+m^2)
if we replace m with some rational number v/u, we get (u^2 - v^2)/(u^2+v^2) and (2uv)/(u^2 + v^2)
so the Pythagorean triple (a,b,c) = ((u^2 - v^2)/(u^2+v^2), (2uv)/(u^2 + v^2), 1)
if we rearrange this we get ((u^2 - v^2), (2uv), (u^2 + v^2))
The question is to prove this will be primitive Pythagorean triple only if gcd(u,v) = 1 and one of the u or v has to be odd number.
Anyone can do this?
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September 10th, 2012, 02:57 PM   #2
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Re: Primitive Pythagorean Proof question.!

Well, gcd(u,v) = 1 implies that at least one of u and v is odd, since if they are both even, then gcd(u,v) is at least 2. If gcd(u,v) is not equal to one, then u and v have a common factor greater than one and we may write for some d greater than one. Then is not a primitive pythagorean triple because all of the terms have a common factor of .
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September 10th, 2012, 06:13 PM   #3
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Re: Primitive Pythagorean Proof question.!

Thank you for the proof..it helped me a lot..
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