Date: Nov 28, 2012 1:49 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem

On Tuesday, November 27, 2012 9:58:32 PM UTC+2, quasi 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.
>
>
>
> quasi

If a > c and/or b > c it's obvious that a^p + b^p > c^p.
We ca choose z ,y ,z > p , x <= y < z and using modulus properties a ,b ,c, a <= b < c ,a < p that x = a + mp ,y = b + np, z = c + qp with m, n ,q naturals
(a + mp)^p?(a + mp)(modp)=a + mp + kp=a + p(m+k)...