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 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. Tags flt, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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