
Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Dec 29, 2012 6:00 AM


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.

