
Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 7, 2013 10:37 AM


On Monday, January 7, 2013 3:18:29 PM UTC+2, M_Klemm wrote: > "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
Yes it's only for step 1 when a,b,c naturals.

