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September 10th, 2012, 02:47 PM  #1 
Member Joined: Sep 2012 Posts: 69 Thanks: 0  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 (1m^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? 
September 10th, 2012, 02:57 PM  #2 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  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 .

September 10th, 2012, 06:13 PM  #3 
Member Joined: Sep 2012 Posts: 69 Thanks: 0  Re: Primitive Pythagorean Proof question.!
Thank you for the proof..it helped me a lot..


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