On Wednesday, November 28, 2012 8:34:35 AM UTC-5, John Jens wrote: > Corrections was made. > > > > It's sufficient that only a < p.
The statement of the theorem is that
a^p + b^p = c^p has no solutions.
There is nothing about a < p.
At best your argument shows that there are no solutions with a < p, but then you'd also have to consider the case a >= p.
> > > > If a + b ? c?0 because 0<a?b<c implies b ? c < 0 ,0 ? a + b ? c < a < p > > If a + b ? c = 0 see (1) > > If a + b ? c ? 1 and because a + b ? c < a implies a ? 2 and because a < p implies > > p > 2.