Friday, May 18, 2018

FLT - Proof from Beal's relation


Proof of Fermat’s Last Theorem from Beal’s relation
Abstract:  Multi-power relations with reference to Beal are discussed with  numerical examples. Some additional properties of Beal’s conjecture are also identified to prove FLT
Keywords: Fermat Last Theorem, Beal Conjecture, Number theory
Introduction
Fermat’s last theorem says that integral solutions for ax  +  by   = cz  are not possible when all the exponents are same and greater than or equal to 3 . However it is possible if atleast one of the exponents is different. In such multi-power relations, according to Beal [1-3] the members when  all the  three members  a,b,c and the exponents x,y,z  are all positive integers such that (x,y,z) > 2, invariably have one or more common factors. Conversely the equation  ax  +  by   = cz has no solutions for any positive integers a,b,c,x,y and z with (x,y,z)  > 2  for any set of co-prime (a,b,c).The three members like or unlike power relations described by  the Fermat’s assertion and Beal conjecture respectively  have some similarities and differences.
The observed similarities are (i)  All the members can be even, then the relation will be reducible. All the members cannot be odd due to non-conservation of oddness (ii) One or two members but not all the three may be square or higher power but all the members cannot be square or with the same higher power, (iii) all the three members cannot be simultaneously primes (iv) FLT and Beal conjecture are mathematically true with fractional and rational or complex numbers, and (v) In irreducible form, both the FLT and Beal conjecture do not have a common prime factor among the members

Among the differences the most significant are (i) In ax  + by  = cz , FLT requires that a,b,c cannot be positive integers when the exponents x,y,z are same and greater than 2, but they can be in Beal’s expression  when the exponents are different (ii) In FLt, a + b will always be greater than c. but it can be greater, lesser or even be equal depending upon the value of the exponents

In this paper Beal’s expression in the form of an + an  = 2m is studied to investigate more properties of Beal’s conjecture and by using such properties FLT is proved.  

 Multi-Power Relation of type an + an  = 2m

We know that an + an  =  2 an, To make 2an  to be a power of a number the only possibility is “a” itself must be  equal to 2x . Under this condition it gives  2nx + 2nx = 2nx+1 , where n and x have all positive whole integral values. For different values of x we have

2n  +   2n    =  2n + 1
4n    +  4n    =  22n + 1
8n    +  8n   =  23n + 1
16n  +  16n  = 24n + 1
32n   + 32n   =  25n + 1

All of them have common prime factor 2 with which they can be reduced to an irreducible form 1 + 1 = 2, which is a fundamental requirement for this type of relations. Any power of 2 can be a common multiplier to generate multi-power relations with bigger numbers. For example  2n  +   2n    =  2n + 1 gives  22n+1  + 22n+1  = 4n+1 when all the members are multiplied by 2n+1 .

This method of getting multi-power relations, gives an ingenuous explanation for Fermat’s Last Theorem. Sum of two identical cubes can be expressed as a square or fourth or fifth power of a number, but never as a single cube or with any exponents of multiples of three.
(2x2)3 + (2x2)3 =(22x3)2      (2x4)3  + (2x4)3  = (2x3)4                (23x5)3 + (23 x5)3  = (22 x3)5
23 + 23 =  42                                       323 + 323   = 164                                  83  +  83  =  45
83 + 83  =  322                                  1623 + 1623 = 544                             2563  + 2563  = 325
183  + 183  = 1082                        5123  + 5123 = 1284                         19443 + 19443  = 1085

When the exponent of the sum is in the form of  3n+2, the Beal, expression of this type is
[22n+1 xn ]3  + [22n+1 xn ]3 = (22 x3)3n +2, and it is (2n xn)3 + (2n xn)3  = (2x3) 3n+1  when the exponent is in the form of 3n + 1. It is noted that the exponent of the sum will never be in the form of 3n.. To be true, it should satisfy the relation  (2n xn)3   + (2n xn)3   = (2m  x)3n which necessitates  the condition 23n + 1  = 23mn
No integer values for n and m satisfy this condition, which substantiates the Fermat’s last Theorem without any further argument.

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