Here is the real solution to Fermat's Theorem (see my ignore list)
The Binomial Expansion is of the form (a + b)n = an + bn + rem(a,b) where rem (a,b) all powers of a and b and coefficients are positive integers. Fermat's Theorem is:
an + bn cn for n > 2, a and b positive integers
In order for Fermat's Theorem to be false (i.e., an + bn = cn) for some n > 2, a and b positive integers, rem(a,b) must be equal to zero, which it is not, by inspection (since rem(a,b) > 0, with a and b and the coefficients are positive integers for all powers of a and b and all n > 2.
(Edit: I had a senior moment where I wrote "true" a short while ago) I have written it correctly numerous times in other threads. If this is what Grapes is referring to, I can only say "my bad"; sorry, Grapes if you couldn't figure that out from the context.) .. (There is a whole other thread where I say it correctly, but WAY over engineer the problem - I simply couldn't believe it was that easy. I'm starting to believe the whole thing is a huge joke, like believing Trump can be president (just ask Vince McMahon if it is just another WWE passion play). Well, ok, if it weren't so serious...
For the case n = 2 (note: Fermat's Theorem is not relevant here, so this is actually a fool's errand w.r.t. to the hypothesis),
we must have
c2 - (2ab) = a2 + b2 is a positive integer
Therefore for Fermat's theorem to be true for this case, we must have a = 0 or b = 0, so the equation is not satisfied.
(there must be an (integer) metric for the integers as a ring without division, so 0 is the midpoint of all possible integer "distances" (lengths) defined in relation to 0 (a - a) = (b - b) = 0 for all a and b.)
Try calling the Abel committee again... they may be finished their conversation with John....
(humbly, I await your applause.....