Fermat Modus Tollens

FERMAT, MODUS TOLLENS Manuel Oliveros Martínez, Arquitecto 2009-09-06 (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

c  a  b

n

a b

say

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. Making

c  x  b

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

n

n

( x  b )  a  b

n

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

k0 n 1

n k k n  combin(n k)x b   a

k0

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. Restricting ourselves to the integer field, for we are investigating if c can be an integer, every and all natural numbers c can be expressed as x+b as above, what exacts its nth potence to admit the binomial expansion in terms of Newton´s binomy. Now if c needs to range the entire integer field it may -even if b remains fixed- by just varying x, and so if it needs to do so it can approach as much as feasible for any integer to meet any solution for it can range the entire field. This means that for x+b be an integer and so also (x+b)^n, it follows that if we think any number be an entire with an entire nth root, the nth potence of the root, that is, the number must be able to take the form n k k  combin(n k)x b  n

k0

When we examine the left side of the equation we see that whatever the x we may be ranging in the integer field, so, for every x we are using, the left term of the equation never admits the binomial decomposition of Newton's binomy, for always lacks the b^n term that we have effectively truncated at both sides of the equation in our simplification in order to get proof. So, modus tollens, since it can't take the required form for x+b to be an integer when x and b are, x+b it is not an integer. b being an integer, x is not, and c is not an integer. We have proven Fermat's Theorem.