My Math Forum Primitive Pythagorean Proof question.!

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 September 10th, 2012, 01:47 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, 01:57 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 $u= dm; v = dn$ for some d greater than one. Then $u^2 - v^2, 2uv, u^2 + v^2$ is not a primitive pythagorean triple because all of the terms have a common factor of $d^2$.
 September 10th, 2012, 05:13 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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