Consider x, y, z, k and n positive integers (n

, n > 2).

Then x

^{n}, y

^{n}, z

^{n}, and x

^{n-k} y

^{k} are also positive integers.

Then x + y = z is a positive integer, and (x + y)

^{n} = z

^{n }is a positive integer as well.

By the

Binomial expansion,

(x + y)

^{n} = z

^{n }= x

^{n} + y

^{n} + f(x, y, n, k),

where f(x, y, n, k) represents the remaining terms in the Binomial expansion; each term is a function of powers n,k of positive integers (x,y)

multiplied by the

binomial coefficient
(from Pascal's triangle) for that term - each such term is a nonzero positive integer.

f(x, y, n, k) is the sum of these terms, which are also non zero positive integers, so f(x, y, n, k) is a positive integer as well.

For Fermat's theorem to be false, z

^{n }= x

^{n} + y

^{n} , so that f(x, y, n, k) would have to be equal to zero, which is a contradiction, since f(x, y, n, k)

0 by construction/observation.

Fermat's theorem is therefore proved. (Q.E.D.)

I would like my Abel prize in small unmarked currency bills, in paper sacks.....

Note: I spent

** weeks **on this project as an undergraduate at UCSB in the 60's ...

-------------------------------------------------------------

**Easy version for Grapes:**

Let x, y, z, n, be positive integers,

**n > 2**
For some z, let (x + y) = z

Then (x + y)

^{n} = z

^{n}
Then z

^{n} =(x + y)

^{n }= x

^{n} + y

^{n} + f(positive integers) ( from Binomial Theorem if not by inspection)

z

^{n} = x

^{n} + y

^{n} iff f(positive integers) = 0

f(positive integers)

0 ( Binomial Theorem, or inspection)

Therefore, z

^{n} x

^{n} + y

^{n} for n > 2 (Fermat's Theorem)

QED

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