Date: Dec 29, 2012 6:00 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem
If a > p

...

So we can?t find naturals 0<a?b<c with a<p with p odd

prime to satisfy a^p+b^p=c^p.

We can extend this to a , b , c rational numbers ,

0<a?b<c and a<p .

Let?s pick d positive integer , p < d , d?b < c and

assume that d^p+b^p=c^p .

We can find k rational number such d/k < p and we have

(d/k)^p + (b/k)^p = (c/k)^p which is

false of course because d/k < p ,which implies that

we can?t find positive integers to satisfy

a^p+b^p=c^p with p > 2.