Single Page Fermat Theorem Proof

SINGLE PAGE PROOF OF THE THEOREM OF FERMAT Manuel Oliveros Martínez, Arquitecto 2009-09-02 (REVISION 4) (C) CERTIFIED TO THE PUBLIC DOMAIN a b c natural integers positive bigger than or equal to 1, n exponent equal or bigger than 3, proof it is not possible n

n

n

say c  a  b ab and so there's not a natural c that satisfies the preceding equations in such conditions. To say that there's no natural c such that c^n=a^n+b^n when a and b naturals and n  3 is to say that the set of the nth power of naturals and the sum of of 2 of such powers is disjoint, so no c^n can become a^n+b^n whatever the a and b. For the sake of demonstration, assume this assertion is false and that exists such c^n. c  x  b

Making

n

n

( x  b )  a  b

n

n k k n n  combin(n k)x b   a  b n

k0

eliminating the common b^n term at left and right n 1

n k k n  combin(n k)x b   a

k0 or establishing a polynomial on x of the nth power at the left n 1

n k k n  combin(n k)x b   a  0

k0

This is a polynomial with positive coefficients on all x^n, x^(n-1) ... x and then -a^n. To negate Fermat's theorem we need to find some integer solution to x, for it might produce a natural c satisfying the equality.  

If any solution to it can´t be integer, Fermat theorem is proven. If contrarily, we meet an integer solution to it (x natural and lesser than a, we have proven in the shorthand article), Fermat's theorem, is false.

Now I will proceed to explain what makes that the above equation impossible to meet for an integer, waiting for better symbolical or whatever more formal proof in what I am also working at. a

n

n 1

entirely pinpoints (as a itself) at the integer field

 combin(n k)x

k0

n k

b

k



this term, even if integer, lacking the b^n term that would reconciliate it with some sum x+b of integers to be elevated to the nth potence, causes the pair x+b originating the summation be ejected from the integer field (not pinpoint on it), and since b integer, x can never be

As a corollary, the single real number x+b in the positive field that meets the equation, is irrational, for it belongs to the class of the root of a natural number rhe root of which is not an integer.