```Date: Dec 31, 2012 3:51 AM
Author: quasi
Subject: Re: From Fermat little theorem to Fermat Last Theorem

John Jens wrote:>I intentionally choose a < p ,there's no crime to choose>an a natural, a < p.Right.So let's assume you've proved that with the conditions   p prime, p > 2   a,b,c positive integers with a < pthe equation a^p + b^p = c^p cannot hold.>If for a < p ,a,b,c, naturals , we can multiply the >inequality a^p + b^p != c^p with any rational number,>it will remain a inequality.But that doesn't help you prove, subject to the conditions   p prime, p > 2   a,b,c positive rationals with a < pthat the equation a^p + b^p = c^p cannot hold.You are making a rather blatant logical error.Assume that there exist a,b,c,p with the conditions   p prime, p > 2   a,b,c positive rationals with a < psuch that a^p + b^p = c^p.You can scale a,b,c up to get positive integers A,B,Csuch that A^p + B^p = C^p, but in scaling up, you can nolonger guarantee A < p, so there's no contradiction.quasi
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