Date: Jan 7, 2013 8:18 AM
Author: Michael Klemm
Subject: Re: From Fermat little theorem to Fermat Last Theorem

"John Jens" wrote

> The reason is to prove FLT .

> Let's split in three steps :

> Step 1--> prove a^p + b^p != c^p with a < p ,a,b,c, naturals

> Step 2--> extend to rationals , still a < p

> Step 3--> pick A >= p, assume A^p + b^p = c^p and scaling down to A/k < p

> ,k rational -->contradiction to step 2

> If a + b ? c>0 because 0<a?b<c implies b ? c < 0 ,

> 0 ? a + b ? c < a < p

> then a + b ? c ? 1 and because a + b ? c < a implies a ? 2 and because a <

> p implies p > 2 ...

> .... and using binomial theorem

Is this intended to be a proof of step 1?

If yes, it is essentially correct, because a^p + b^p = c^p together with a <

p implies

a^p + b^p <= a + b^p < p + b^p < (b+1)^p <= c^p, a contradiction.

The inequality (b+1)^p <= c^p is however not necessaryly true for rational b

and c with b < c.

Regards

Michael