Date: Nov 28, 2012 2:00 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem

On Wednesday, November 28, 2012 12:26:12 AM UTC+2, Pubkeybreaker wrote:
> On Nov 27, 2:58 pm, quasi <qu...@null.set> wrote:
>

> > John Jens wrote:
>
> > >quasi wrote:
>
> > >>John Jens wrote:
>
> >
>
> > >>>http://primemath.wordpress.com/
>
> >
>
> > >>Copying part of the text from the link above (enough to
>
> > >>expose the error in Jens' reasoning) ...
>
> >
>
> > >>>Fermat?s little theorem states that if p is a prime number,
>
> > >>>then for any integer a, the number a^p is an integer multiple
>
> > >>>of p.
>
> >
>
> > >>>   a^p = a(mod p)
>
> >
>
> > >> Yes, but note that a^p = a (mod p) does not imply 0 <= a < p.
>
> >
>
> > >>>Assume that a,b,c naturals and p prime and
>
> >
>
> > >>>   0 < a <= b < c < p
>
> >
>
> > >>> ...
>
> >
>
> > >>>So we can?t find naturals 0 < a <= b < c < p with p prime to
>
> > >>>satisfy a^p + b^p = c^p.
>
> >
>
> > >> Sure, but that doesn't even come close to proving Fermat's
>
> > >> Last Theorem. All you've proved is the trivial result that if
>
> > >> a,b,c are positive integers with p prime such that
>
> > >> a^p + b^p = c^p then c >= p.
>
> >
>
> > >"Assume that a , b , c naturals and p prime and 0<a=b<c<p"
>
> >
>
> > Yes, you can assume anything you want, but then any conclusion
>
> > is conditional on that assumption.
>
> >
>
> > Without loss of generality, you can assume
>
> >
>
> >    0 < a <= b < c
>
> >
>
> > but how do you justify the inequality c < p?
>
> >
>
> > Of course you can take 2 cases:
>
> >
>
> >    (1) c < p
>
> >
>
> >    (2) C >= p
>
> >
>
> > The case you analyzed is the case c < p (the trivial case),
>
> > and you never even considered the other case. Thus, you
>
> > did not actually prove Fermat's Last Theorem.
>
>
>
>
>
> He might also want to check his proof when p = 2.....


From the condition 0 < a <= b < c < p , p must be bigger then 2 because don't exist minimum two numbers (a <= b, c) between 0 and 2.