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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 (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? 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.. Tags primitive, proof, pythagorean, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ducnhuandoan Number Theory 11 January 19th, 2014 10:54 PM James1973 Abstract Algebra 1 November 19th, 2013 09:47 PM rhymin Applied Math 4 March 24th, 2013 12:20 PM billymac00 Number Theory 19 March 19th, 2013 08:38 PM nonlinear Number Theory 10 October 3rd, 2011 11:03 PM

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