Here is the real solution to Fermat's Theorem (see my ignore list)

The Binomial Expansion is of the form (a + b)^{n}= a^{n}+ b^{n}+rem(a,b)where rem (a,b) all powers of a and b and coefficients are positive integers. Fermat's Theorem is:

a^{n}+ b^{n}c^{n}for n > 2, a and b positive integers

In order for Fermat's Theorem to befalse(i.e., a^{n}+ b^{n}= c^{n)}for some n > 2, a and b positive integers,rem(a,b) must be equal to zero, which it is not, by inspection (sincerem(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...

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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

c^{2}- (2ab) = a^{2}+ b^{2}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.)

QED

Try calling the Abel committee again... they may be finished their conversation with John....

(humbly, I await your applause.....

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