I do prove the equality false in Fermat's last theorem via the Binomial Expansion for n >2.

(also for n=2, if one excludes Pythagorean triples; but not necessary for Fermat's proof.)

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iff

The point is that a and b must be independent (a,b); i.e., not on the same number line, but on orthogonal number lines. If they are on the same number line, then c =(a+b) = d, so

; that is, the identity of a and b is not preserved in the Binomial expansion because the lhs and rhs of c=(a +b) refer to the same single integer.

By the Binomial Expansion, true for all positive numbers, not just integers:

That is, (once again)

since rem(a,b,n) > 0

, n > 2 (Fermat's Theorem, my proof)

QED

(Try really, really hard to understand Fermat's Theorem AND the Binomial Expansion)

rem(a,b,n) is simply everything that is not

or

in the Binomial Expansion.

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c < (a+b) merely says

so

which is a true expression, but is a metric expression, not a proof of Fermat's Theorem, so I don't have to prove it, trivial as it is. It is irrelevant to the proof. (In particular, it is not true for all integers a, b, c, in particular for c > (a + b) - c cannot satisfy both equations.

c=(a+b) sets the foundation for the proof (by contradiction)

My proof is true for all integers (and numbers) in Fermat's equation. And I don't really need all the analysis in terms of vectors, the Binomial Theorem is just fine as it is. (I had been trying to justify the Binomial Theorem in terms of vectors by including Pythagorean triples for the case n = 2, but Fermat specifies n>2, so that analysis is also irrelevant to the actual proof).

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