Date: Nov 27, 2012 5:26 PM
Author: Pubkeybreaker
Subject: Re: From Fermat little theorem to Fermat Last Theorem
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.....