>If a^p= c^p- b^p is true for a , b , c ,naturals a < p , >is true for a rational , a < p and b , c naturals because >c^p- b^p is natural. > >We can divide a^p= c^p- b^p with k^p , k rational k > 1 >and note (a/k) = q , > >q^p = (c/k)^p - (b/k)^p with q rational q < 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
Sorry, I no longer have time for this.
There's no way I can get through to you.
Your logic is totally flawed, and that, together with your poor language skills, makes it impossible to have a worthwhile discussion with you.
Suffice it to say that your argument is total nonsense, with no redeeming value whatsoever. It's completely worthless.