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Coprimes in Fermat's Last Theorem

Date: 06/03/99 at 07:07:38
From: oliver
Subject: Coprime in Fermat's Theorem n = 2


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.


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   
Associated Topics:
High School Discrete Mathematics
High School Number Theory

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