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Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 7, 2013 9:01 AM
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On Jan 7, 8:18 am, "M_Klemm" <m_f_kl...@t-online.de> 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
I have already told him that regardless of the details of what he is
doing
that his proof CAN NOT work. I gave the reasons why.
But of course, like all cranks, he just ignores the reviews from
experts
and just prattles on.
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