Tuesday, April 11, 2017

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Resolving Fermat’s Last Theorem by Prime Factor Method
 In number theory Fermat’s last theorem states that no three positive integers a,b,c  satisfying the equation an  +  bn  = cn for all exponents including n ≥ 3.  This can be proved simply by considering the prime factors associated with one of the members a or b in the equation. It is found that when the exponent is greater than 2 , one or two numbers are invariably either complex or irrational. The non-existence of the relation with all are whole integers proves FLT for all exponents n ≥ 3.
Key words:-
Fermat’s Last Theorem - equal power relations – prime factors
   For all exponents greater than 2 , the numerical relation in the form  an  +  bn  = cn  cannot exist when a,b and c are restricted to have whole integers which is called Fermat’s Last Theorem (FLT) , but is true when irrational and complex numbers are allowed for at least  one or two members. After the Fermat’s assertion, many people 1-3 attempted to prove FLT for different values of exponent and verified with the help of number theory. After the introduction of computer technology people attempted to support the proof of FLT for a wide range of exponents 3 - 4,000,000 4. A proof of FLT was given by Taylor and Wills 5,6 with the help of modern concepts of mathematics – elliptic curve and modular theory. In general there must be two or more ways, one simple and another complicated, to get the same result for a given mathematical problem. Nature loves symmetry and likes always to be very simple. In fact it is our description that shows the nature to be illusory, which implies that there must be more natural and naive proof for FLT. This article describes the simple and natural proof of FLT.
Prime factor method – A simple proof of FLT:-                                                                                                                                                                                                            
Without any complexity in mathematical description, FLT can be proved with the knowledge of the prime factors associated with one of the members of the triple (a,b,c). Any number can be represented in terms of its prime factors. An odd number will have
only odd prime factors denoted by Po where as all even numbers will invariably have 2 as one of the prime factors and besides that it may have one or more odd prime factors. Accordingly an od and even number can be represented in terms of its prime factors as,
                    N(odd) = P01α P02β P03γ......................... Pomω
                          N(even) = 2m P01α P02β ........................... P0mω
Where α,β,γ,...........ω may have any value from 0 depending upon the value of the base numbers a,b,c. m has a minimum value of 1 in the case of singly even and its value increases depending upon the degree of evenness. If all the powers are zero, the base number becomes 1 and for a number greater than unity, at least one or more exponents will be non-zero.
     In an  +  bn  = cn, an  is equal to a product of two factors (cn/2  -  bn/2) (cn/2 + bn/2) and if the prime factors associated with a (or b) is known , then they can be distributed  for these two factors. By solving the two equations, one can determine the unknown base numbers in terms of the known. For simplicity let us assume a = P01α P02β P03γ
                               an  = P01P02P03nγ  = (cn/2  -  bn/2) (cn/2 + bn/2)
In any three member equal power relation an + bn  = cn, c is always greater than both a and b, but cis  less than am + bm when m < n  and greater when m > n . This is true even when m is a fraction. Since (cn/2  -  bn/2) < an/2,  
                     cn/2 + bn/2  = k[P01nα/2 P02nβ/2 P03nγ/2]
                            cn/2 -  bn/2  = [P01nα/2 P02nβ/2 P03nγ/2]/k
where k >1 , need not be an integer and may have all possible values fraction, complex,  and irrational.
Solving these two equations, we get,
                     c= P01P02P03 [(k2 + 1)/2k]2
                    bn = P01P02P03 [(k2 – 1)/2k]2
and substituting these values , we get 
an + an [(k2 - 1)/2k]2  = an [(k2 + 1)/2k]2
 If one is able to convert it into an  + bn  = cwith a,b,c are all positive integers,, it is equal to say that FLT is disproved, otherwise it confirms.
[(2k)2/na]n + [a(k2 - 1)2/n]n = [a(k2 + 1)]2/n]n
The terms inside each bracket will be whole number only when n = 2. For n ≥ 3 one or more numbers will be either irrational or complex.
When 2k = αn/2   then (2k)2/n  = α
(k2 - 1) = βn/2  then (k2 - 1)2/n   = β, and
(k2 + 1)  = γn/2  then (k2 + 1)2/n = γ
Subtracting one from the other,
n/2  -  βn/2}   = 1 , No whole integral solution is obtained for all possible values of ,β andγ . It substantiates Fermat’s last theorem. However there are many solutions when n = 2, where the condition reduces to γ -  β = 1
1.Simon Singh, ‘Fermat’s Last Theorem’ (1997) Fourth Estate Ltd.
2.Jackson,A. ‘Fermat’s Enigma ‘ (1997) Review, AMS.
3.Edward,H.M.,”Fermat’s Last Theorem – a genetic introduction to the Number Theory’ (1997),Springer.
4.Wagstaff,S., AMS Notices No.167,(1976)p A-53,244                                                                   
5.Andrew Wills, ‘Modular elliptic curve and Fermat’s last theorem’ (1995) Annals of Maths., 141, 443-551.

6.Gerd Faltings.,  “ The proof of Fermat’s Last Theorem by R.Taylor and A.Wills” AMS,1995,42 (7),743-746.

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