Coprimes in Fermat's Last TheoremDate: 06/03/99 at 07:07:38 From: oliver Subject: Coprime in Fermat's Theorem n = 2 Hi! I was wondering, in Fermat's Conjecture of n = 2 x^2 + y^2 = z^2 which can be rearranged to: (y/2)^2 = (z-x)/2 * (z+x)/2 (z-x)/2 and (z+x)/2 are coprime? I have done some examples to convince myself, but I can't find a book anywhere on this. I also can't find what topic it would come under. Is this very easy and am I being stupid? PLease help and point me in the right direction. Thanks, Oliver Date: 06/03/99 at 12:12:17 From: Doctor Rob Subject: Re: Coprime in Fermat's Theorem n = 2 (z-x)/2 and (z+x)/2 do not have to be coprime. For example, x = 300, y = 400, and z = 500 gives a solution to the equation, and (z-x)/2 = 100 and (z+x)/2 = 400 have a common factor of 100. What is true is that if x, y, and z have no factor in common, and y is even, then (z-x)/2 and (z+x)/2 also have no factor in common. Proof: Suppose that d divided both (z-x)/2 and (z+x)/2. Then d would divide their sum, z, and their difference, x. If d divided both z and x, then d^2 would divide z^2 - x^2 = y^2, so d would divide y, and d would be a factor in common among x, y, and z. Thus d = 1. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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