February 22nd, 2014, 10:54 AM |
#1 |

Banned Camp Joined: Aug 2010 Posts: 170 Thanks: 4 | FLT Proof
Fermat's Last Theorem says that A^x + B^x cannot be equal to C^x where A,B,C and x are positive integers and x>2. One can rewrite this as C^x - B^x cannot be equal to A^x. If A,B and C have a highest common factor other than 1, it can be cancelled out to get a new equation. Therefore right from the onset it will be assumed that A,B and C do not have any common factors other than 1. Let A = C - K, where K is some positive integer. If A and C don't have any common factors other than 1 then A,C and K and therefore A,C,K and B don't have any common factors other than 1. C^x - B^x = (C- K)^x can be rearranged after binomial expansion to get B^x + K^x = KC f(C,K) when x is even B^x - K^x = KC g(C,K) when x is odd where f(C,K) and g(C,K) are functions of C and K. It can clearly be seen that the left hand side of each equation is the left hand side of Fermat's equation in the 2 different forms. If C is a positive integer herein lies the contradiction i.e. the right hand side of the equations imply that K must be a factor of the left hand side of the equations and therefore K must be a factor of B which is a contradiction. If C is a positive decimal number and not a positive integer then it will now be shown that the right hand side of the equations will always work out to be positive decimal numbers while the left hand side of the equations will always be positive integers since B and K are positive integers. One can rewrite the above equations to read B^x + K^x = (C-K)^x - C^x - K^x. x is even B^x - K^x = (C-K)^x - C^x + K^x. x is odd Concentrating on the 3 terms on the right hand side of both equations it should be clear that a decimal number to a certain number of decimal places takeaway a decimal number to a smaller number of decimal places plus or minus a positive integer will always give a positive decimal number. Therefore the left hand side of each equation cannot be equal to the right hand side because one side will be a positive integer and the other will be a positive decimal number. For the case of x=2 i.e. Pythagoras theorem this does not necessarily need to be the case i.e. the right hand side does not always have to be a decimal number. |

February 26th, 2014, 08:44 AM |
#2 | |

Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 | Re: FLT Proof Quote:
To make it simple here's example x=5. B^5-K^5=5CK(C-K)(C^2-CK+K^2); Nothing prevents to assume K=k^5 and B=bk. Therefore no declared contradiction obtained. The proof can be restricted to prime exponents i.e. to only odd x. The cases of composite exponents (including even) are corollaries. | |