>Corrections was made. > >It's sufficient that only a < p.
But you never _proved_ the inequality a < p, so you don't get to use it.
Moreover, the equation
a^p + b^p = c^p
with the restrictions
a,b,c positive integers
does not imply min(a,b) < p.
To see this, just use p = 2 with a,b,c = 3,4,5.
You tried to argue that you can't have p=2 since the inequality min(a,b) < p would then force min(a,b) = 1, leading to an easy contradiction. But you can't use the inequality min(a,b) < p without proving it, and the example p = 2 with a,b,c = 3,4,5 makes it clear that you can't prove it.