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
>
> Michael


Yes it's only for step 1 when a,b,c naturals.