June 24th, 2014, 03:27 AM |
#1 |

Banned Camp Joined: Aug 2010 Posts: 170 Thanks: 4 | FLT Proof
Fermat’s Last Theorem This theorem basically states that X^n + Y^n ≠ Z^n, n > 2 if X,Y,Z and n are all positive integers. This inequation can be rewritten as Z^n – Y^n ≠ X^n. Andrew Wiles, a British mathematician, produced a lengthy proof of over 100 pages in the mid 1990s. Proof Let it be initially assumed that Fermat’s inequation is true i.e. it is an equation. If X, Y and Z have a highest common factor other than 1, P say, P^n can be cancelled out to give a new Fermat equation. Therefore right from the onset it will be assumed that X, Y and Z do not have any common factors other than 1 i.e. only ‘primitive’ equations/inequations need be considered here. Z^n-Y^n can be rewritten as [Z-Y][Z^(n-1) + Z^(n-2) Y+ Z^(n-3)Y^2 + ....ZY^(n-2) + Y^(n-1)] Let [C-B] be any of the prime factors of A^n = C^n-B^n where C=Z[K^(1/n-1)], B=Y[K^(1/n-1)], A, B, C and K are positive integer constants. This means that [C-B] should also be a factor of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] since [C-B][C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)]= A^n. More importantly this also means that [C-B]^(n-1) is a factor of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)]. Algebraic long division of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] by [C-B] always results in an algebraic remainder of nB^(n-1). This means that [C-B] needs to go into nB^(n-1) exactly K{[C-B]^(n - 2)} times for the algebraic long division process to be complete, where K is some positive integer constant i.e. nB^(n-1) = K{[C-B]^(n - 1)}. Therefore [C-B]^(n - 1) must be a factor of B^(n-1) unless [C-B] > B. This is a contradiction as C-B cannot be a factor of B since C and B do not share any common factors other than 1. Fermat's Last Theorem is therefore proved. |

June 24th, 2014, 03:31 AM |
#2 |

Banned Camp Joined: Aug 2010 Posts: 170 Thanks: 4 |
Fermat’s Last Theorem This theorem basically states that X^n + Y^n ≠ Z^n, n > 2 if X,Y,Z and n are all positive integers. This inequation can be rewritten as Z^n – Y^n ≠ X^n. Andrew Wiles, a British mathematician, produced a lengthy proof of over 100 pages in the mid 1990s. Proof Let it be initially assumed that Fermat’s inequation is true i.e. it is an equation. If X, Y and Z have a highest common factor other than 1, P say, P^n can be cancelled out to give a new Fermat equation. Therefore right from the onset it will be assumed that X, Y and Z do not have any common factors other than 1 i.e. only ‘primitive’ equations/inequations need be considered here. Z^n-Y^n can be rewritten as [Z-Y][Z^(n-1) + Z^(n-2) Y+ Z^(n-3)Y^2 + ....ZY(n-2) + Y^(n-1)] Let [C-B] be any of the prime factors of A^n = C^n-B^n where C=Z[K^(1/n-1)], B=Y[K^(1/n-1)], A, B, C and K are positive integer constants. This means that [C-B] should also be a factor of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] since [C-B][C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)]= A^n. More importantly this also means that [C-B]^(n-1) is a factor of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)]. Algebraic long division of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] by [C-B] always results in an algebraic remainder of nB^(n-1). This means that [C-B] needs to go into nB^(n-1) exactly L{[C-B]^(n - 2)} times for the algebraic long division process to be complete, where L is some positive integer constant i.e. nB^(n-1) = L{[C-B]^(n - 1)}. Therefore [C-B]^(n - 1) must be a factor of B^(n-1) unless [C-B] > B. This is a contradiction as C-B cannot be a factor of B since C and B do not share any common factors other than 1. Fermat's Last Theorem is therefore proved. |