Date: Dec 31, 2012 3:51 AM
Author: quasi
Subject: Re: From Fermat little theorem to Fermat Last Theorem

John Jens wrote:

>I intentionally choose a < p ,there's no crime to choose
>an a natural, a < p.


Right.

So let's assume you've proved that with the conditions

p prime, p > 2
a,b,c positive integers with a < p

the equation a^p + b^p = c^p cannot hold.

>If for a < p ,a,b,c, naturals , we can multiply the
>inequality a^p + b^p != c^p with any rational number,
>it will remain a inequality.


But that doesn't help you prove, subject to the conditions

p prime, p > 2
a,b,c positive rationals with a < p

that the equation a^p + b^p = c^p cannot hold.

You are making a rather blatant logical error.

Assume that there exist a,b,c,p with the conditions

p prime, p > 2
a,b,c positive rationals with a < p

such that a^p + b^p = c^p.

You can scale a,b,c up to get positive integers A,B,C
such that A^p + B^p = C^p, but in scaling up, you can no
longer guarantee A < p, so there's no contradiction.

quasi