Date: Jan 7, 2013 1:22 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem

On Sunday, January 6, 2013 9:05:08 PM UTC+2, M_Klemm wrote:
> You should give a reason why you assume a < p.
>
> For p = 2 you have 3^2 + 4^2 = 5^2, and then by little Fermat 3 + 4
>
> congruent 5 (mod 2)
>
> but not 3 < 2.
>
>
>
> Regards
>
> Michael
>
>
>
> wrote in message
>
> news:696982e8-3228-48d0-b483-9cc3acb97341@googlegroups.com...
>

> > On Sunday, January 6, 2013 10:56:35 AM UTC+2, M_Klemm wrote:
>
> >> Hello,
>
> >>
>
> >>
>
> >>
>
> >> consider the case p =3, proved by Euler. Then you see that the assumption
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> >> a
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> >>
>
> >> < p in line 4 is not at all justified.
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> >>
>
> >> Regards
>
> >> Michael
>
> >
>
> > I'm sorry but I don't understand what are you trying to say.

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

Still a + b ? c>0 ,a + b - c must be at least 1 and with p=2 we can't find natural a between a + b - c and
p there's no Step 1 for p = 2

(If a + b ? c ? 0 we have a + b ? c

(a+b)^p?c^p and using binomial theorem

a^p+b^p<(a+b)^p=a^p+?+b^p?c^p)