```Date: Jan 7, 2013 10:37 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem

On Monday, January 7, 2013 3:18:29 PM UTC+2, M_Klemm wrote:> "John Jens" wrote> > > > > The reason is to prove FLT .> > > Let's split in three steps :> > > Step 1--> prove a^p + b^p != c^p with a < p ,a,b,c,    naturals> > > Step 2--> extend to rationals , still a < p> > > Step 3--> pick A >= p, assume  A^p + b^p = c^p and scaling down to A/k < p > > > ,k rational -->contradiction to step 2> > > > > If a + b ? c>0 because 0<a?b<c implies b ? c < 0 ,> > > 0 ? a + b ? c < a < p> > > then a + b ? c ? 1 and because a + b ? c < a implies a ? 2 and because a < > > > p implies p > 2 ...> > > .... and using binomial theorem> > > > Is this intended to be a proof of step 1?> > If yes, it is essentially correct, because a^p + b^p = c^p together with a < > > p implies> > a^p + b^p <= a + b^p < p + b^p < (b+1)^p <= c^p, a contradiction.> > > > The inequality (b+1)^p <= c^p is however not necessaryly true for rational b > > and c with b < c.> > > > Regards> > MichaelYes it's only for step 1 when a,b,c naturals.
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