Friday, June 16, 2017

creative thoughts

More properties of Pythagorean triples
The  relation a2  + b= c2  has an inherent feature in which twice of cis equal to a  sum of two squares  where  the greater number is [c +(a – 1)] while the smaller number is [c + (a+1)].
2 x 52  = 12  +  72   ; (5 - 4)2  + (5 + 2)2
2 x 132  = 72  + 172 ; (13 – 6)2  + (13 + 4)2
2 x 252  = 172 + 312 ; (25 – 8)+ (25 + 6)2
2 x 412  = 312 + 492 ; (41 – 10)2  + (41 + 8)2
2 x 612  = 492  + 712 ; (61 – 12)2  + (61 + 10)2

2 x 852  = 712  + 972 ; (85 – 14)2  + (85 + 12)2






Friday, June 9, 2017

FLT and Beal conjecture

Resolving Fermat’s Last Theorem by Prime Factor Method and Its Proof in 5 steps
M.Meyyappan#
# Professor of Physics (Retd), Alagappa Government Arts College, Karaikudi - 630003,Tamilnadu, India)
Email: meydhanam@gmail.com
Abstract
In number theory Fermat’s last theorem states that no three positive integers a,b,c satisfy 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 .However if the exponents are not same then the three member multi-power relation becomes possible. An unique property associated with these relations is described by Beal conjecture. The Pythagoras theorem, FLT and Beal conjecture are different forms of one and the same multi-power relation under different condition. The inherent linkage between them helps to prove both the Fermat’s last theorem and Beal conjecture in five successive steps.
Key words
Fermat’s Last Theorem , equal power relations , multi-power relation, prime factors, Beal conjecture , number theory.
I.Introduction
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. Using the same method used to prove Fermat’s last theorem one can demonstrate the Beal conjecture which is close analogous to FLT7-10. The Pythagorean relation with all whole integers can be used to prove both FLT and Beal conjecture within five steps.
II.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 = P01nα P02nβ P03nγ = (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 cm is 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,
cn = P01nα P02nβ P03nγ [(k2 + 1)/2k]2
bn = P01nα P02nβ P03nγ [(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 = cn with a,b,c are all positive integers,, it is equal to say that FLT is disproved, otherwise it confirms. The above equation can be rearranged as [(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,
(γn/2 - βn/2} = 1 , No whole integral solution is obtained for all possible values of ,β and γ . It substantiates Fermat’s last theorem. However there are many solutions when n = 2, where the condition reduces to γ - β = 1
III.FLT and Beal Conjecture
The Pythagoras theorem, FLT and Beal conjecture are different form of one and the same multi-power relation under different condition. The most general form of three member multi-power relation is aα + bβ = cγ , where both the base numbers (a,b,c) and the exponents (α,β,γ) are all supposed to be whole integers. Under different conditions, it gives Pythagorean relation (α = β = γ = 2), Fermat’s last theorem (α = β = γ = n and n > 2) and Beal conjecture (α, β, γ are not equal, at least one is different and α = β = γ ≠ 2).
When a,b,c,α,β and γ are all positive integers with α,β,γ are greater than 2, then a,b,c will invariably have a common prime factor (Beal conjecture) ,when the exponent of one or two base numbers is 2, the common prime factor is not to be a necessary condition and when the exponents of all the base numbers are equal, the common prime factor present can be cancelled out and reduces ultimately to an irreducible form of the relation (Fermat’s last theorem). In this paper a simple proof for FLT and Beal conjecture is derived from the realistic Pythagoras theorem.
IV.FLT in Five Steps
Step.1,Consider any known Pythagorean triples (a,b,c) satisfying a2 + b2 = c2 , where a,b and c are all whole integers. Dividing by a2 throughout gives reduced Pythagorean relation as 1 + (b/a)2 = (c/a)2
Step,2 Let us suppose an unknown three member like power relation xn + yn = zn, where all possible values of x,y and z are allowed., which may be whole integers or fractions or irrational or even complex.
Step 3. Multiplying throughout the reduced Pythagorean relation by xp gives xp + xp (b/a)2 = xp (c/a)2
Step 4. Comparing term by term with the unknown relation yp = xp (b/a)2 + k = (xα)p and zp = xp (c/a)2 + k = (xβ)p where α and β are assumed to be whole integers and k will be zero when (b/a)2 = αp and (c/a)2 = βp if not k may have some positive or negative value. Since it is added in both sides, the equation will not be affected.
Step 5. Substituting the derived values for (b/a)2 and (c/a)2 in the reduced Pythagorean relation 1 + αp = βp . No integer or fractional values of α and β will satisfy this relation for all values of p except p = 2.
V.Beal conjecture in five steps
Step.1 Consider any known Pythagorean triples (a,b,c) satisfying a2 + b2 = c2 , where a,b and c are all whole integers. Dividing by a2 throughout gives reduced Pythagorean relation as 1 + (b/a)2 = (c/a)2
Step 2. Let us suppose an unknown three member unlike power relation xp + yq = zr where x,y,z,,p,q and r are considered to be whole integers.
Step 3. Multiplying throughout the reduced Pythagorean relation by xp gives xp + xα (b/a)2 = xp (c/a)2
Step 4.Comparing term by term with the unknown relation, yq = xp (b/a)2 = (xα)p and zr = xp (c/a)2 = xp+k where α and k are all assumed to be whole integers.
Step 5. Substituting the derived values for (b/a)2 and (c/a)2 in the reduced Pythagorean relation we arrive at a mathematically possible 1 + αp = xk. . Few typical solutions and the corresponding multi-power relations are 1 + 23 = 32 gives 33 + 63 = 35 , 1+ 7 = 23 gives 73 + 74 = 143 and few general solutions are 1 + (2n – 1} = 2n gives (2n - 1)n + ( 2n -1)n+1 = [2(2n -1)]n and 1 + nm = (nm + 1) gives ( nm + 1)m + [n(nm +1)]m = (nm + 1)m+1.
VI.Conclusion
The proof of FLT and Beal conjecture in five steps shows that three member equal power relation is not possible with all the base numbers are integers for all exponents greater than 2 since 1 + αn ≠ βn .However such relations are possible when irrational and complex numbers are allowed or when the exponents of the base numbers are allowed to be different since 1 + αn = βm ..When it is multiplied with a common multiplier (βn) which becomes
the common prime factor in Beal conjecture we get a multi-power relation βn + (αβ)n = βm+n, while the common factor αm gives αm + α m+n = (αβ)m .
References
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.
7. M.Meyyappan http://www.ijmttjournal.org/2017/volume-45number-1/IJMTT-V45P503
8. G.Gadzirayi Nyambuya “ A simple and General Proof of Beal Conjecture, Adv.in Pure.Math, 2014, 4, 518-521
9. R.C.Thiagarajan,” A proof to Beal’s Conjecture” Bull.Maths.Sci & appl, 2014,89-93
10. L.Torres di Gregorio, “Proof for the Beal conjecture and a new proof for Fermat’s last theorem” Pure & appl.Math.J, (2013) 2(5),149-155.

Tuesday, May 30, 2017

creative thoughts

The Proof of FLT and Beal conjecture in 5 steps
M.Meyyappan#
# (Professor of Physics (Retd), Alagappa Government Arts College, Karaikudi-630002, India}
Abstract:
Both the Fermat’s  Last Theorem and Beal conjecture can be proved in five steps.
Key words:
FLT, Beal Conjecture, multi-power relations, number theory
Introduction
   In literature quite a large number of references are available for Fermat’s Last Theorem and Beal conjecture [1]{2}{3}. The Pythagoras theorem, FLT and Beal conjecture are different form of one and the same multi-power relation under different condition. The most general form of three member multi-power relation is xp  + y = zr , where both the base numbers (x,y,z) and the exponents (p,q,r) are all supposed to be whole integers. Under different conditions, it gives Pythagorean relation (p = q = r = 2), Fermat last theorem (p = q = r = n and n > 2) and Beal conjecture (p,q,r are not equal, at least  one is different and p = q = r ≠ 2).  
When x,y,z,p,q and r are positive integers with p,q,r are greater than 2, then x,y,z will invariably have a common prime factor, when the exponent of  one or two base numbers is 2, the common prime factor is not to be a necessary condition and when the exponents of all the base numbers are equal, the common prime factor present can be cancelled out and reduces ultimately to an irreducible form of the relation. In this paper a simple proof for FLT and Beal conjecture is derived from the realistic Pythagoras theorem.
1.FLT in Five Steps
Step.1,Consider any known Pythagorean triples (a,b,c) satisfying a2  + b2  = c2 ,  where a,b and c are all whole integers. Dividing by athroughout gives reduced Pythagorean relation as 1 + (b/a)= (c/a)2
Step,2 Let us suppose an unknown three member like power relation  xp + y= zp,  where x,y and z may or may not be whole integers,
Step 3. Multiplying throughout the reduced Pythagorean relation by xp  gives  xp  + xp (b/a)2  = xp (c/a)2
Step 4. Comparing term by term with the unknown relation  yp  = xp (b/a)2 and xp (b/a)2  + k =  (xα)p   and zp  = xp (c/a)2 and xp (c/a)2 + k  = (xβ)p   where α and β are assumed to be whole integers and k will be zero when (b/a)2  =  αp  and (c/a)2  = βp if not k  may have some positive or negative value
Step 5. Substituting the derived values for (b/a)and (c/a)2 in the reduced Pythagorean relation 1 + αp =    βp . No integer or fractional values of α and β will satisfy this relation for all values of p  except p = 2.
II. Beal conjecture in Five steps
                                                                      
                                                                                                                                                                                 2
Step.1 Consider any known Pythagorean triples (a,b,c) satisfying a2  + b2  = c2 ,  where a,b and c are all whole integers. Dividing by athroughout gives reduced Pythagorean relation  as 1 + (b/a)= (c/a)
Step 2. Let us suppose an unknown three member unlike power relation  xp + yq  = zwhere x,y,z,p,q and r are supposed to be whole integers.
Step 3. Multiplying throughout the reduced Pythagorean relation by xgives  xp  + xp (b/a)2  = xp (c/a)2
Step 4.Comparing term by term with the unknown relation, yq   = xp (b/a)2 = (xα)p  and zr  = xp (c/a)2 = xp+k  where α , β and k are all assumed to be whole integers.
Step 5. Substituting the derived values for (b/a)and (c/a)2 in the reduced Pythagorean relation 1 + αp  = xk. . Few possible solutions and the corresponding multi-power relations are 1 + 2 = 32 gives  33  + 63  = 35 ,  1+ 7 = 2gives  73  + 74 = 143  and few general solutions are  1 + (2n – 1} = 2n gives  (2n  - 1)n   + ( 2n -1)n+1  = [2(2n -1)]n  and 1 + nm  = (nm + 1)  gives ( nm  + 1)+ [n(nm +1)]m  = (n+ 1)m+1.
III. Conclusion:
The proof of FLT and Beal conjecture in five steps shows that three member equal power relation is not possible with all the base numbers are integers for all exponents greater than 2 since 1 + αn  ≠ βn .However such relations are possible when exponents of the base numbers are allowed to be different since 1 + αn  = βm.When it is multiplied with a common multiplier (βn) which becomes the common prime factor in Beal conjecture we get a multi-power relation βn  + (αβ)n  = βm+n, while the common factor αm  gives αm  + α m+n  = (αβ)m .
References
[1] M.Meyyappan http://www.ijmttjournal.org/2017/volume-45number-1/IJMTT-V45P503
[2]Gerd Falting,”The Proof of Fermat’s Last Theorem by R.Taylor and A,Wills” Notices AMS,          1995,42(7)743—746.

[3]Simon Singh, “Fermat’s Last Theorem” Fourth Estate Ltd., 1997

Thursday, April 13, 2017

creative thoughts

Curious Properties of Multi-Power relations
Abstract:-
Some of the interesting and inherent properties associated with the specific relation a2 + b2 + c2 = 2 d where a + b = c are investigated.
Introduction:-
In number theory there are many relations under equal sums of like powers and unlimited kinds of multi-power relations with different number terms in each side and with different exponents. When we impose some conditions, some of them are allowed and others are forbidden.
The sum of any number of squares can be shown to be equal to the sum of any number of squares. This property is not found in any other powers other than square where one or more restrictions disallow such formations. The Fermat’s Last Theorem is one such discrimination.  In this article, some of the interesting and inherent properties associated with the specific relation a2 + b2 + c2 = 2 d where a + b = c are investigated.
Property-1
An interesting property associated with the relation a2 + b2 + c2 = 2 d where a + b = c is  d+ ab = (a+b)2 = c2. One can prove it with the help of algebra, but it has more inherent properties. It is exemplified with few typical examples
12 + 122 + 132  = 2 x 157 ; 157 + 1 x 12 = 169 = 132
22 + 112 + 132 = 2 x 147 ;  147 + 2 x 11 = 169 = 132
32 + 102 + 132 = 2 x 139 ;  139 + 3 x 10 = 169 = 132
42 + 9+ 132 = 2 x  133 ;  133  + 4 x 9  = 169 = 132
52  + 8+ 132 = 2 x129  ; 129  + 5 x 8  = 169 = 132
6+ 72  + 13= 2 x127 ; 127  + 6 x 7 =  169 = 132
Property-2
If a2 + b2 + c2 = 2 d where a + b = c, then
a4  + b+ c4  = 2 d
                                                                     -2-
This is shown with some numerical examples.
12 + 72 +  82  =  2 x 57 and  1+ 74  + 8= 6498 = 2 x 572
22 + 6+ 82  =  2 x 52 and   24  + 64  + 84 = 5408 = 2 x 522   
32 + 5+ 8=  2 x 49 and   34  + 54  + 84  = 4802 = 2 x 492  
4+ 42 + 82  =  2 x 48 and   44  + 4+ 84 = 4608 = 2 x 482
If d happens to be a square number (d = e2) ,then the relation turns into a4  + b+ c4        = 2 e4. For example,
7+ 84 + 154     =  2 x 169=   2 x 134
54 + 164 + 214    =  2 x 3612 =  2 x 194
94 + 154 + 244    =  2 x 4412 =  2 x 214
114  + 244 + 354 = 2 x 9612  = 2 x 314
144 + 164 + 304 =  2 x 6762  = 2 x 264
Combining these two properties, we arrive yet another relation
a2 + b2 + c2 = 2 d
Squaring both sides, (a2 + b2 + c2)2 = (2d)2 = a4  + b+ c4  + 2(a2b2  + b2c2 + c2a2)
But  a4  + b+ c4 = 2d2  which demands that (a2b2  + b2c2 + c2a2) = d2
Taking the case 1+ 74  + 8=  2 x 572 , we have
12 72  + 72 82 + 82 12 =  72 + 56+ 82 =  572
Property-3
 If c = ab  and b = a+ 1 then a2  + b2  + c2  = d2 where  d is equal to c+1. For example,
1+ 2+ 22  = 32 ; 14 + 2+ 2= 33 = 3 x 11
2+ 32 + 62   = 72 ; 2+ 34  + 6= 1393 = 7 x199
32 + 4+ 122 = 132 ;34 + 4+ 124  = 21073 = 13 x 1621
42  + 52  + 202  = 212 ; 44 + 54 + 204 =160881 = 21 x 7661
It shows that a4  + b+ c4  invariably has d as a factor

                                                                  -3-
The readers may develop curiosity to disclose many other properties associated with this type of conditional relations.

                                                 .................

Tuesday, April 11, 2017

creative thoughts

Resolving Fermat’s Last Theorem by Prime Factor Method
Abstract:-
 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
Introduction:-
   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
                                                                      -2-
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.
                                                                      -3-
[(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
References:-
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.

Monday, April 10, 2017

creative thoughts- FLT a new proof

A versatile proof of Fermat’s last theorem
(Dr.M.Meyyappan,  Professor of Physics (Retd), Alagappa Government Arts College, Karaikudi- 630 001)
Abstract:-
   A simple and multi-purpose proof is proposed with well known arithmetic axioms to explain not only the Fermat’s assertion but also the non-existence of square triples (a,b,c) satisfying an  +  bn  = cn for all exponents including n = 2, reason for certain allowed and forbidden relations among equal sum of like powers and Beal conjecture (Mauldin/Tijdeman-Zagier conjecture) altogether successfully in a single frame.
Key words:-
Fermat’s Last Theorem - arithmetic axioms equal power relations – Multi-power  relation -  Beal conjecture.
Introduction:-
   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 Wills5,6 with the help of modern concepts of mathematics – elliptic curve and modular theory. In general there must be two or more ways to get the same result for a given mathematical problem. Since the law of nature is always as simple as possible, there must be more natural and naive proof for FLT. This paper describes the attempt undertaken to disclose the naive and accessible proof of FLT with the help of well known arithmetic axioms and number theory. The unequivocal representation reveals its versatility of the methodology.
Arithmetic axioms:-
1.In an  +  bn  = cn  , c is always greater than both a and b but  cm  is less than am  +  bm  when m < n, and greater when m > n. This is true even when m is a fraction
2.The conservation of odd-evenness in the three member equal sum of like powers necessitates that the allowed and irreducible set of triples (a,b,c) must be with (odd, even, odd), where as the apparently allowed triples with (odd, odd, even) is forbidden due to non-conservation of odd-evenness of the relation, as a doubly even component cannot
                                                                   
be balanced with a singly even component, and it is found to be true for all exponents. This restriction is given up in multi-power relations.
3.If a is prime in a three member equal power relation, then c - b will invariably be equal to 1
4.The product of any two powers to give a single power follows some rules. If ax  x  by  = cz,  either the base numbers must be same or some of its power or the exponents must be same or some of its multiple. When b is equal to aα, where α may have any integer value,   then  z = x + α y   and c = a. When y is equal to αx, then z = x and c = abα .
5.In ax =  by, if a,b and the exponents are all whole integers,  then a = αy  and b = αx,  so that ax =  by = αxy
Proof of Fermat’s Last Theorem:-
In an  +  bn  = cn, an  is equal to a product of two factors (cn/2  -  bn/2) (cn/2 + bn/2) and an/2  > cn/2  - bn/2,, one can ingeniously separate these two factors as  (cn/2  -  bn/2) = an/2 /k where k  > 1 , not necessary to be an integer and may have all possible values (axiom 1), then  (cn/2 + bn/2) = k an/2 . Solving for bn  and cn  in terms of an  we get b = a[ (k2 – l) /2k]2 and  c= an [( k2 + 1) / 2k]2. In the same manner one can solve for an and cn in terms of bn also, which gives mathematically identical result.  an  +  bn  = cn  is then rearranged with its internal constituents as [(2k)2/na]n  + [(k2-1)2/n a]n  =  [ (k2 + 1)2/n a]n where ‘a’ is found to be a common prime factor of the triple  with an inherent condition  1 + [ (k2 – l) /2k]2   =  [( k2 + 1) / 2k]2 . Mathematically speaking such equal power relations have a common prime factor, since any two members can be expressed in terms of the other member. They are reducible due to this common prime factor. An irreducible set of triple for an exponent n will not have any common prime factor at all. Besides that at least two members will be either irrational or complex instead of being whole integers.  But the common prime factor remains in all the three member multi-power relations as stated in Beal conjecture, because it cannot be reduced as done in equal power relations. When reduced it becomes a generator of multi-power relation which will not have the common prime factor.
Whatever may be the way by which a member is factorized with respect to the remaining members, the internal constituents may differ, but in general, the structure of the basic relation will be same. If a = p, a prime, then the relation becomes, pn  + [(pn – 1)/2]2   =  [(pn + 1)/2]2, where the difference c - b = 1 (axiom-3), where no common prime factor exists.  If a = pq where p < q, then (cn/2  -  bn/2) = pn  and (cn/2 + bn/2) = qn ,it gives yet another form of the above expression (22/n pq)n  +  (qn – pn)2  =  (qn + pn)2, where the common prime factor may or may  not exist.

                                                                     
In fact all the three member equal or unequal power relations have the following general form an  + an [x2]  = an  [y2], , where [x2] and [y2] may have any integer or a quantum of value or fractional value but with same denominator satisfying the condition  1 +[x2]  = [y2]. For a given value of the exponent, one can change the base numbers only. Due to this limitation, the formation of  an + (βa)n = (γa)n is  practically impossible as both βn and γn  cannot be whole integer simultaneously. By choosing a quantum of value for [x2], one term can be converted into a modified power as required, but  it is not possible for the other with (1 +[ x2]).Even by adding any equal quantum of value in both side, the condition fails to support the relation. For example, by adding q both sides, the balance of the relation will not be affected. (anβ) + q = mn  and an ( 1 + β) + q = nx  which demands 1 = nx  - m x  and for which no integer solutions are available.
Any known relation with whole integers such as Pythagorean relation can even be considered for this purpose. It can be used as a key to solve problems related to equal sums of like and unlike powers with same or different number of terms in each side. The scope of this paper is restricted only to three member equal and unequal power relations.
General Proof of FLT:-
     Consider a basic relation (2k)2  + (k2-1)2  =   (k2 + 1)2,To convert it into an equal power relation with an exponent n, the requirements are k2 = an/4, (k2 -1)2 = bn and (k2 + 1)= cn , with a condition cn/2 -  bn/2  =  2.This is valid only when n = 2 for whole integral values of c and b. When an  + an [x2]  = an  [y2]  represents the equal power relation, the condition  is 1+ bn = cn  . It is found that the relation becomes distinct when n ≤ 2 and n ≥ 3. When n = 2, the terms within each bracket will be whole integral number and will not have any common prime factor in its irreducible form and the condition reduces to c – b = 2, which is practically possible. In the other case the condition is possible when b and c are fractions with equal denominator. When n is greater than 2, the condition becomes unrealistic and all the members cannot be expressed as whole integral number simultaneously.
   It is found that the terms inside each of the brackets can be made to be whole integer only when n = 2, which is the case of Pythagorean triples. When n ≥ 3 , not all the three members will be whole integers simultaneously and one or more members will invariably be irrational or complex. It explains the argumentative support put forwarded by Pierre de Fermat for cubical and higher power relations and stands as valid proof of FLT.                                                                 
Non existence of square triples
   All the members of a triple (a,b,c) corresponding to any exponent n cannot be squares. It is true even when the exponent n = 2.In fact it is a direct consequence of FLT. The non-existence of square triples can also be taken as an indication of FLT. If all the members of
                                                                        
a triple are squares (a2,b2,c2)  for an exponent n, then  (a,b,c) would be a triple for the exponent 2n.If square triples exist for the exponent 2 than a fourth power must be equal to a sum of two fourth powers which is against the argument of the previous section. In  an  + (a[x2]1/n)n  = (a[y2]1/n)n,, if a is a square number say α, the other two members will be squares only if [x2] = α2n and [y2]  = β2n, which gives again an impossible condition       β2n - α2n = 1. It is not fulfilled even when n = 2.It predicts that all the members can be squares when the exponent is ½. The general form of the possible solution is given by 11/2 + (n2)1/2  = [(n+1)2]1/2. From (2k)2  + (k2-1)2  =   (k2 + 1)2  we can arrive at the same conclusion. The condition cn – bn  = 1 does not permit square triples for any exponent.
Forbidden power relations:-,
   In general equal sum of like powers is possible with equal number of terms in each side of the expression for all exponents. Let us restrict our interest to an  +  bn  = c+  dn .The basic relation to explain its characteristics is [x1]2 +[ y2] 2 =  [x2]2 + [y1]2 where 1 + [x1]2  =    [y1 ]2  and 1 + [x2]2  = [y2]2. In terms of independent variables k, k’, m and n it can be expressed as [ (k2 – m2) /2mk]2  +  [( k’2 + n2) /2nk’]2  =  [ (k’2 – n2) /2nk’]2  +  [( k2 + m2) /2mk]2. This can also be expressed as (a+b)2 + (c-d)2  = (a-b)2  + (c+d)2  with a condition ab = cd. For nth power relation each term must be nth power of an integer. If the two terms in one side of the relation are taken as an  + bn , then the terms in the other side will be (an +/- 1) and (bn -/+ 1). By adding and subtracting a suitable number without affecting the balance, both the term in the other side can be made to be nth power of different numbers respectively.  
   But there are some restrictions in the case of unequal number of like powers in each side of the relation. Fermat’s assertion, a sum of two nth powers greater than equal to 3 equal to nth power of whole numbers is an unique restriction. This can also be proved algebraically (Annexure). The restrictions are found to be different for different exponents.  For example, a sum three squares or three cubes can be shown to be a square or cube respectively
(4kk’)+ [2k’(k2 – 1)]2  + [(k2 + 1)(k’2  + 1)]2 = [(k2+ 1)(k’+ 1)]2
where 1 +  [(k2 -1) /2k]2    =   [(k2 + 1) /2k]2  and 1 +  [(k’2 -1) /2k]2    =   [(k’2 + 1) /2k]2
But no such expression exists for the exponents n = 4 and 5.However a sum of four fourth or fifth powers equalling a fourth or fifth power respectively exists.  It is the self adjustment that determines the possibility of formation of equal sum of like powers balanced with equal and unequal number of terms.
      Considering the case where a square is equal to a sum of three squares. [y2]2 =               (1 +[ x2])2 = 1 + [x2]2 + 2[x2].  To make  2[x2] to be a square, [x2] must be equal to 2 α2. For
                                                                    
cubical relation [y2]3 = (1 +[ x2])3 = 1 + [x2]3 + 3[x2][y2], and 3[x2][y2] to be a cube, it must be equal to 27α3 .  The roots of the quadratic equation x2 (1+ x2) = 9 α3  is given by x2  = [-1 ± √ (1+ 36 α3 )]/2, when α = 2, we get x2 = 8 or – 9 which gives a numeral relation  93 = 13 + 63  + 83. Non-existence of equal sum of like powers in the form a4 = b4 + c4 + d conjectured by Euler can be explained successfully with the same fundamental expression- (c2)2 = ( a2 + b2)2 = a4 + b4 + 2a2b2 ,where 2a2b2 cannot be expressed as a fourth power of a whole number with all possible values of a and b. 
Beal conjecture
    Even though the conditional equal power relation with three members is unachievable for all exponents n ≥ 3, it is found that multi-power relations are obtainable.  According to Beal conjecture7,8,9 when the base numbers a,b and c and the exponents x,y and z greater than 2 in a+ by  = cz are all positive, then a,b,c will invariably have a common prime factor. The secret behind the multi-power relations is making up a power of an integer from the product of two terms - a power of an integer am and a number of any kind [x2] and  [y2] respectively obeying the condition 1 + [x2] = [y2] .This can be accomplished under certain ways. The product am . [x2] becomes am + n  when [x2] = an , where the base number is unchanged but the exponent gets changed. The product is equal to (aα)m when [x2] = αm where the exponent remains same but the base number is changed. It is equal to (aα)m + n  when [x2] = an αn + m  where both the base number and exponent are changed. It is (aα)n when [x2] =   an - m α where n > m  and it is equal to αn / (am – n)  where m > n . In all the above cases,  the base number of the product is some multiples of the base number in the multiplicand, i.e., the base of the multiplicand preserves the base of the product as its generic character. The product can be made to be equal to βn which may miss the generic character when [x2] = βn/am. In such cases both the base and exponent get changed. The acceptable multi-power relations are completely determined by the simultaneous fulfilment of the requirement by both the products am [x2] and am [y2] satisfying the condition  1 + [x2] = [y2].  
   Even though am + b= c is not possible when a,b,c are all whole integers and m ≥ 3 , multi-power relation in the form  ax + by = cis possible for certain permissible values of x,y,z.  Like equal power relations, all the three member multi-power relations have the same basic structure  a + ax[(k2 -1)/2k]2  =  ax[(k2 +1)/2k]2, where a and k are two independent variables.  This can be converted into a multi-power relation by properly choosing the values of [x2] and [y2] and converting the products to a power of an integer. For example  ax  + ax + m  =  ax + n  is not  possible  with same base number since 1 + am  ≠ an  for all possible values of a,m and n. But ax  + ax + m = (aα)is possible due to the soluble condition 1 + am = αx. When a = 2,α = 3, m = 3 and x = 2, we get 22 + 25 = 62. Similarly ax                                                     
(aα)x = ax + m is allowed due to attainable condition 1 + αx = am  which gives a multi-power relation 33 + 63 = 35.
      From the general form of the three member equal power relation, one can derive one or more conditions for any particular requirement. Its numeral form can be used as a generator for the development of such multi-power relations. In fact three member multi-power relation is not possible for all possible set of exponents due to insoluble condition. Table.1 shows some of the allowed and forbidden 3 member multi-power relations with a typical numerical example.
Table.1 Allowed and forbidden three member multi-power relations
ax + by = cz                               conditions                               numerical example ___________________________________________________________________________
an + (aα)n  = (aβ)n                            1 + αn = βn                                               not possible (FLT)
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a+ (aα)3  = (aβ)4                       1 + α3 = aβ4                                       93 + 183 = 94
(aα)3 + (aβ)3 = a4                                 α+ β3 = a                                       703+ 1053 = 354
a3 + (aα)4    = (aβ)3                          1 +  a α4 = β3                                     73 + 74 = 143
(aα)3 + a4    = (aβ)3                                     α3 + a   = β3                                                                 383  + 194 = 573
a3 + (aα)3    = (aβ)5                             1 + α3 = a2β5                                                   33  + 63  = 35
(aα)3 + (aβ)3 = a5                                α3 + β3 = a2                                                        83  + 83 = 45
a+ (aα)3  = (aβ)3                              a+ α3 = β3                                                    135 + 913 = 1043
a+ α3  = β3                                         a5 = (β – α)(β2+ βα +α2)                       not possible       
a3 + (aα)5  = (aβ)3                               1 + a2 α5 =  β3                                               913 + 135 = 1043
a3 + (aα)4  = (aβ)5                              1  = a(a β5 – α4)                                             not possible
In fact multi-power relations can be obtained directly from the generator- any simple numeral expression, by multiplying each term with a common multiplier, which provides a common prime factor to the relation. For example, the condition 1 + 1 = 2 when multiplied by 2m gives 2m + 2m  = 2m + 1 ,and the common multiplier 22m gives  4m + 4m = 22m + 1  where as the condition 1 + 7 = 2gives 73 + 74 = 143  with a common multiplier 73. an + bn + 1 = cis the
                                                                       -7-
most common type of multi-power relation. The general form of this type of relation is given by (2n -1)n + (2n -1)n + 1 = [2(2n – 1)]n
     A generator can provide many multi-power relations each with different multipliers. It is exemplified with few examples                                          
1 + by  = cz        →        1 + 23 = 32                                                                   
Only certain common multiplier alone can make each term into a power of an integer. 1 is very convenient member as it is the only multiplicand to make product with multiplier of any form. 
 In the above case either bz  or cy or αzy will be the  convenient multipliers.
[1 + by  = cz] x bz  = bz  + by + z  = (bc)z    →   22  + 25 = 62   
[1 + by  = cz] x cy = cy  + (cb)y  = cy + z      →  3+ 6= 35
[1 + by  = cz] x αzy = αzy + (αzb)y = (αyc)z → 76 + (98)3 = (1029)2
Similarly, ax  + b= c →  13  + 33  = 28, multiplying with the common multiplier cx
[ax  + b= c] x cx  = (ac)x  + (bc)x  = cx+1  → 283  + 743  = 28with common multiplier 283
                                                                                     →7843 + 23523 = 287 with common multiplier 286
If ax  + by  = citself is taken as generator, then one of the common multipliers will be αxyz
(ax  + by  = cz ) x αxyz = (aαyz)x  + (bαxz)y = (cαxy)z
The generators usually do not have common prime factor, but will have in its reducible form. Hence the generators may or may not have a common prime factor depending upon its reducibility. In fact the generators are irreducible form of multi-power relations. When multiplied with a common multiplier to get a multi-power relation, it acquires a common prime factor.
One can show that the common prime factor is compulsory for all the three member multi-power relations but vanishes in all the three member equal power relations .The common prime factor is not a compulsory requirement in multi-power relations with at least one member is a square or more than three members. The non-existence of common factor in equal power relations for all exponents n ≥ 3 can also be taken as an indirect proof of FLT.
Exempted multi-power relations

                                                                       
Beal assertion is not a necessary condition in all multi-power relations with more than three members or one of the members is a square. e.g., ( with more than three members 94 = 84 + 74  + 4; 35 = 43 + 23 + 23 + 13 ;  54 = 34 + 25 + 83) and ( with one member is a square 83 = 132 + 73; 1053 = 1812 + 1043 ; 174 = 1612 + 2402). When one of the members is a square, then the three member multi-power relation becomes
a2  + a2 [(k2 – 1)/2k]2  = a2 [(k2 + 1)/2k]2                                                                          
a2  + ay [1/ay – 2]  [(k2 – 1)/2k]2  = az [1/az – 2] [(k2 + 1)/2k]2
 [1/ay – 2]  [(k2 – 1)/2k]2 = αy  and [1/az – 2] [(k2 + 1)/2k]2 = βz  from which the required condition can be derived as
1  = (az – 2) βz - (ay – 2) α= ay – 2[ az – y βz – αy ] which has no integral solution.
When it is converted into a form  a2 + b3  = c3 , the common factor is either lost or changed by the denominator of [x2] and [y2] .  When the common multiplier is some multiples of the common denominator of the fraction representing [x2] and [y2] , the multi-power relation may lose its common prime factor. That is what is found in multi-power relation with one member is a square.
 Like three member equal power relations, in multi-power relations also all the members cannot be squares or any equal higher powers .
a2x + a2x [x2] = a2x [y2]  .Assuming x > y > z, one can construct a multi-power relation as
a2x + a2y(a2x -2y [x2]) = a2z (a2x -2 z [y2])  →  ax + (aβ)2y  = (aγ)2z
with a condition 1 + β2y/(a2x –2y) = γ2z /(a2x –2z).Since the denominators are different βy must be divisible by (ax –y) and γz  by (ax –z) which gives a modified condition as 1 + m2 = n2 , which has no practical solution for any integer values of m and n. Even by self adjustment also, one cannot make it to be true.
Acknowledgement
      The author is grateful to Pierre de Fermat who gave an open problem to think over at least if not to try to solve by people of all age group. This proof is dedicated to Mahatma Gandhi, a man I love most.                                                                 
Reference:-
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.
7.Daniel Mauldin,R., ‘A generalization of Fermat’s Last theorem: The Beal conjecture and prize problem’ (1997) , American Mathematical Society, 44,1436-39.
8, Golden G.Nyambuya, ‘A simple and general proof of Beal conjecture’ (2014) Adv.in Pure. Maths., 4,518-521.
9.wimhessellink.nl/fermat/WilesEnglish.htm
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Annexure. 
 The algebraic method also confirms FLT. In the case of cubical expression,
a3 + b3  = (a+b)(a2 - ab + b2) = c3
Since a+b > c, a+b= kc where k > 1 and a2  - ab + b2  = c2 /k, Solving for a and b
a =  (kc/2) – (ck/6)[ (12/k3) – 3]1/2  and b =  (kc/2) + (ck/6)[ (12/k3) – 3]1/2
when k =1 , the second term becomes c/2 and a = 0, b = c. For all values k > 1, the second term becomes irrational or complex.

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