Applications of Prime factor method

Sum of two cubes   can be made to equal with another sum of two cubes, i.e., some numbers can be expressed as sum of two cubes in two different ways.After exposing the personality of 1729, these numbers are generally called as Ramanujan Numbers.
If the Ramanujan number is the product of three different  odd prime numbers, it will be odd.
a3 + b3 = c3 + d3  = R = Po1 Po2 Po3
(a+b) (a2 - ab - b2)  =(c+d) (c2 - cd + d2) = Po1 Po2 Po3
Since the prime factors cannot be factorized further,  (a+b) and (c+d) must be equal to  one of the three prime factors independently
If a+b= Po3
a2 - ab +b2  = Po1 Po2
Substituting the value of a or b  obtained from the first equation in the second equation,
3b2 - 3bPo1 + Po32 - Po1 Po2   = 0
Solving for b,
b = [3Po3 + √ 12Po1 Po2   - 3Po32 ]/6
a =  [3Po3 - √ 12Po1 Po2   - 3Po32 ]/6
a and b to be whole number, the term inside the square root must be a square number
If c + d= Po2
c2 - cd + d2  = Po1 Po3
It gives c = [3 Po2 +√ 12Po1 Po3   - 3Po22 ]/6 and d =[3 Po2 -√ 12Po1 Po3   - 3Po22 ]/6

  7  x   13 x   19    =        1729    =   13 + 123    =     93  +   103
 13 x   37 x   43    =      20683    = 103 + 273    =   193  +   243
 73 x 157 x 211    =  2418271    = 233 +1343   =   953  + 1163  
103x 241 x 307    =  7620661    = 453 +1963   =  1333 + 1743

When the odd Ramanujan number is expressed as the product of two prime numbers and one square of a prime number or a product of two prime numbers

19x 72 x 43   = 40033 = 163 + 333 = 93 + 344
132 x 67 x 163 = 1845649 = 423 + 1213 = 493 + 1203
(7x19) x 379 x 409 = 20616463 = 1113 + 2683 = 1513 + 2583

In the case of even Ramanujan numbers, 2n  will be one of the factors.If the other two facgtors are prime, then, (a+b) or (c+d) will be even . It can be expressed by the entire 2n  and it may include or not one of the prime factors.

 13 x 79 x 26 = 65728 = 123  + 403 = 313 + 333

Fetmalt's Last Theorem

RESOLVING FERMAT'S LAST THEOREM _PRIME FACTOR METHOD
                   

Introduction

 Fermat asserted in the early 17 th century that it is impossible to find two cubes or two higher powers that sum to another cube or a higher power respectively. For n > 2, there are no integer solutions to the equation  xn  + yn = zn. Fermat claimed to have a valid proof of his conjecture which appears as one of the marginal notes in his manuscripts. As the margin was too small to accommodate the proof, he left the space without any clue or description.Fermat made a number of similar marginal notes, all of which had been resolved satisfactorily except this one- hence the title of Fermat Last Theorem.

Since then many mathematicians world over attempted to resolve FLT .Euler7 in 1753 proved FLT for n = 4. Gauss2 corrected his attempt for n = 3 in 1825.Legendre6,in 1839 proved FLT for n = 5 Lame3, Pepin4 proved it for n = 7 in 18 th century.Germain added additional values of n to the proven list. D .Lehmer and E.Lehmer expanded  their work in 20 th century.Joe Butler and Richard Crandall5 brought Computer technology to FLT and prove it for all exponents n < 4,000,000. A proof1 of FLT was given by R .Taylor and A.Wiles on the basis of elliptic curve and modular theory.

 Nature loves symmetry but also to be always simple. In fact mathematical descriptions made by human beings to understand the nature are not exactly what the nature is. The nature will never follow our mathematics at all .It is only an approximate description of nature, it always requires some correction at later stage.Every natural problem must have its own natural proof under the vision of nature. It prognosticates that there must certainly be a simple and natural proof for FLT.A natural attempt is made to invistigate the existence of such proof.This paper makes a successful attempt to get a simple and marvellous proof  for FLT. 

Characteristics of Equal power relations

The numeral relation in the form xn + yn = zn is found to exist only when x,y,z called triple and n are all whole integers provided n =1 or 2. When n = 1 ; x+ y = z, then xn + yn < zn  for n  >  1; when n = 2 ; x2 + y2 = z2  then  x + y > z and xn + yn < zn when n > 2; when n = 3; x3 + y3 = z3 then xn + yn  > zn  when n < 3 and xn + yn  < zn  when n > 3 . In general xn + yn = zn  then  xn + yn  > zn when n = 1 to n-1 and xn + yn  < zn when n = n+1 and  above.
The choice of being odd and even is also restricted. In (x,y,z) all cannot be odd and any two cannot be even with third one being odd due to non-conservation of odd-evenness of the equation. Being (odd,odd,even) is forbidden for the same reason. All being even is not taken into consideration as it is reducible into another configuration. The only possibility is the members of the triple must be (odd ,even,odd).

Prime factor method - a Marvalous proof for FLT

For n > 2, there are no integer solutions to the equation  xn + yn = zn. Andrew-Wills  in the 1990 s, prove this conjecture on so called elliptic curves and modular forms. Without any complexity of mathematical description,  FLT can simply be proved with the prime factors associated with triples. At first let us establish this new method of proving FLT for the exponent  n= 3.All odd numbers will have only odd prime  factors denoted by Po ,  where as all even numbers will have invariably the even prime factor 2 .Besides that it may have one or more odd prime factors Po.
 In general an odd and even numbers can be represented in terms of its prime factors as
N(odd) = Po1n1 P02n2 Po3n3 ..................... Pomnm
N(even)  =  2n1 Po1n2 Po3n3 ..................... Pomnm
where n1, n2, n3 , .........nm may have any value from 0 to n depending upon the value of the number. n1 has a minimum value 1 in the case of even numbers. If all powers are zero, the number will reduce to 1 For a number greater than 1, atleast one or more exponents will be non-zero.
The most general representation of cube of an odd and even numbers is
N3(odd) = Po13n1 P023n2 Po33n3 ..................... Pom3nm
N3(even)  =  23n1 Po13n2 Po33n3 ..................... Pom3nm
If the number N is not prime, then it  will be divisible by a prime less than or equal to √N.

If a cube is divisible by a prime factor, it should be divisible by square and cube  of the prime factor if the number is singly that prime factor.If it is doubly that factor, then its cube will be divisible by its square cube,fourth power,fifth power and 6 th power of the prime factor and so on.
Making use of this fact one can prove that x3 + y3 = z3  cannot exist with x,y,z all whole integers.
x3 + y3 = z3 
(x+ y) (x2 - xy + y2) = z3
In this equal power relation, the prime factors representation must be same in both sides. Taking x=odd, y=even and z=odd then (x+y) = odd and (x2 - xy + y2) = odd.The conditions provided by the equal power of cubical relation for the values of its members are (i) x < z, (ii) y < z  (iii) x+y > z. and (iv) Since (x+y) is greater than z,  (x2 - xy + y2) must be less than z2.
(x+y) cannot be a single prime number. If so it will be a prime factor for z and z3 will be equal to some multiples of  (x+y)3 . It cannot be since (x+y) > z. The prime factors of z3, must be equal to the prime factors of the product (x+ y) (x2 - xy + y2) .What ever prime factors associated with z3, all of them must be shared with the two product terms as per the conditions stated above.
Let us represent z  in terms of two prime factors for simplicity.
z = P01 P02   and z3  = Po13 P023
One possible distribution of prime factors without violating the conditions is
(x+y) =   P012 P02  then (x2 - xy + y2)  =  P01 P022   , whatever may be the prime factors  associated with x and y .In fact the prime factors associated with x and y is  not at all required.
By solving these two equations, one can get values for x and y for a given value of z.
  y = P012 P02  - x
By substituting this value of y in the other relation we get,
3x2 -3x P012 P02  +  P014 P022  =   P01 P022
3x2 -3x P012 P02  +  P014 P022  -   P01 P022  = 0
Solving for x,we get,
x = 1/2 (P012 P02 ) + 1/6 √ (12P01 P022  - 3P014 P022) 
y =  1/2 (P012 P02 ) - 1/6 √ (12P01 P022  - 3P014 P022)  
The term inside the square root is negative and hence they must be complex in  nature as √ p is irrational for any prime p.
If z  = P01P02  = 3x7 = 21, then
x=[63 + i √ 1127]/2  and y = [63 - i √ 1127 ]/2
When the prime factors are interchanged, the value of z will  not change But it gives another permissible pair of value for x and y.
x =   [147 + i√ 7119]/2   and y = [147 - i√ 7119]/2
Instead of z,If x (odd) is taken as P01P02
x3 = z3 - y3  = (z - y) (z2 + zy + y2)
x = P01P02  and x3  = P013P023,
By distributing the prime factors as follows,
 z - y = P01 then z2 + zy + y2 =P012P023 since x+y > z , x > z - y
Substituting this value of z = P01 + y  in the other  equation
z2 + zy + y2 = 3y2 + 3y P01 +  P012 = P012P023
3y2 + 3y P01 +  P012 - P012P023 = 0
This is a quadratic equation in terms of y, solving for y we get,
y = [ -3P01 +  √  9 P012 -12 (P012 - P012P023)]  / 6
y = [ -3P01 + √ 12 P012P023 - 3P012 ] /6 and z = [ 3P01 + √ 12 P012P023 - 3P012 ]  /6
If x = P01 x P02  = 3x7 , then
y = [-3 + √  4113]  / 2 and z = [3 + √  4113] / 2
By interchaning the prime factors, we get yet another permissible value
y = [- 21 +√ 15729 ]/6  and z = [21 +√ 15729]/6


 Generalization

Let us consider the most general case xn + yn = zn. The exponent n may be either even or odd. This method is unique for both odd and even exponents.

The the general form can be rewritten as
xn = zn - yn = (zn/2 - yn/2) ( zn/2 + yn/2)
x is assumed as x =Po1n1 P02n2 Po3n3 ..................... Pomnm,  then
xn =   Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm
Distributing the prime factors associated with xn between (zn/2 - yn/2) and ( zn/2 + yn/2) without violating the conditions (i) xk+yk > z k when k < n , xk +yk = z k when k = n and xk+yk < z k when k >n (ii) x< z, y< z  .
The advantage of this method is it requires the prime factors associated with one member only  and not for the remaining two other members since the power of a member xn is represented in terms of product of sum and difference of two terms.
According to any one of the permissible distributions of prime factors,
Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm  = (zn/2 - yn/2) ( zn/2 + yn/2)
( zn/2 + yn/2) = Po1nn1- m1 P02nn2-m2 Po3nn3-m3 ..................... Pomnnm-mm
and
(zn/2 - yn/2)  =   Po1 m1 P02m2 Po3m3 ..................... Pommm
where nn s and mns may take any value depending upon the prime factors  associated with members of the triple.
Adding  we get
2zn/2 = Po1nn1- m1 P02nn2-m2 Po3nn3-m3 ..................... Pomnnm-mm +  Po1 m1 P02m2 Po3m3 ..................... Pommm
Squaring both sides
4zn = Po12(nn1- m1) P022(nn2-m2) Po32(nn3-m3) ..................... Pom2(nnm-mm) +  Po1 2m1 P022m2 Po32m3 ..................... Pom2mm + 2Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm

zn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po12nn1-4 m1 P022nn2- 4m2 Po32nn3-  4m3 ..................... Pom2nnm- 4mm +  1 + 2Po1nn1-2m1 P02nn2-2m2 Po3nn3 -2m3..................... Pomnnm-2mm)

zn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po1nn1-2 m1 P02nn2- 2m2 Po3nn3- m3 ..................... Pomnnm- 2mm +  1)2 .
Similarly by substracting we get,
2yn/2 = Po1nn1- m1 P02nn2-m2 Po3nn3-m3 ..................... Pomnnm-mm -  Po1 m1 P02m2 Po3m3 ..................... Pommm
Squaring both sides
4yn = Po12(nn1- m1) P022(nn2-m2) Po32(nn3-m3) ..................... Pom2(nnm-mm) +  Po1 2m1 P022m2 Po32m3 ..................... Pom2mm - 2Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm

yn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po12nn1-4 m1 P022nn2- 4m2 Po32nn3-  4m3 ..................... Pom2nnm- 4mm +  1 - 2Po1nn1-2m1 P02nn2-2m2 Po3nn3 -2m3..................... Pomnnm-2mm)
yn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po1nn1-2 m1 P02nn2- 2m2 Po3nn3- 2m3 ..................... Pomnnm- 2mm - 1)2

The roots of both z and y will be either complex or irrational. The complex and irrational features are exposed by its constituent prime factors.

ACKNOWLEDGEMENT

The author is thankful to Fermat who gave an open  problem to think over in recreational mathematics.This proof is dedicated to Mahatma Gandhi in India and Fermat in France.

Fermat's Last Theorem- applications of Prime factors method

Fermat's Last Theorem- Applications
If x is given a whole integer. Then
x3 = z3 - y3  = (z - y) (z2 + zy + y2)
x(odd) = P01P02  and x3  = P013P023
Since x+y > z , x > z - y
 The distribution of the prime factors of x3 among its two product factors should satisfy this condition.
z - y = P01  then z2 + zy + y2 =P012P023
z= P01 + y substituting this value of z in the other  equation
z2 + zy + y2 = 3y2 + 3y P01 +  P012 = P012P023
3y2 + 3y P01 +  P012 - P012P023 = 0
This is a quadratic equation in terms of y, solving for y we get,
y = [ -3P01 +  √  9 P012 -12 (P012 - P012P023)]  / 6
y = [ -3P01 + √  12 P012P023 - 3P012 ]  /6
If x = P01P02  = 3x7 , then
y = [-3 + √ 4113]  / 2 and c = [3 +  √ 4113] / 2

The conservation of odd-evenness of the cubical relations demands that z - y = 23n where n is a number.
when n=1 z - y= 8 and y is singly even,when n = 2 x - y = 64 and y is doubly even and so on.
y 3  = 23n P013 ........... (any number of odd prime factors)
y3 = z3 - x3 = (z - x) (z2 +zx + x2)
where z - x is even where as z2 +zx + x2 is odd, hence z - x = 23n or one or more prime factors may be attached with it.. All the remaining prime factors multiplied together  must be equal to z2 +zx + x2 . For example
 y = 2P01  then    y3 = 23P013 = z3 - x3  = (z - x) (z2 +zx + x2)
z - x  = 8 or z = x + 8. Substituting this value of z in z2 +zx + x2 = P013
3a2 +24a +64 - P013  = 0
or a =[  -12 + √  3P013  - 48] / 3
If P01  = 7 , x = - 4 +  √109  and  z = 4 + √  109
If P01  = 11, x = [ -12 + √3945]/3 and z = x+ 8  = [ 12 + √ 3945]/3

Fermat Last Theorem- A simple and valid proof

A simple and natural  proof for Fermat's Last Theorem - 3

3. conservation of number digit in equal power relations (14-12-2015)

In all numeral relations, the number digit must be conserved. Number digit of a numbern is a single digit number by adding all the digits of the given number. The process of adding the digits of the number is repeated until a signle digit number called number digit is obtained.
In Pythagorus relation the sum of number digit of x2 and y2 must be equal to the number digit of z2. The number digit of the square of a number is equal to the square of the number nigit of that number.
It is noted that the number digit of a square can have only 1,4,7 and 9.
12 =   1 =    1        112  = 121  =   4
22  =  4  =   4        122  = 144  =   9
32  =  9  =   9        132  = 169  =   7
42  =  16  = 7        142  = 196  =   7
52  =  25 =  7        152  = 225  =   9
62  =  36 =  9        162  = 256  =   4
72 =   49 = 4         172  =  289 =   1
82 =   64 =  1        182  = 324  =   9
92 =   81 =  9        192 =  361  =   1
102 = 100 = 1       202  = 400  =   4

         Number digit                       Number digit of z2
         _______________________________________________
               x    /  y        1           4            7            9
         _______________________________________________
                      1                        2           5           8             1
         _______________________________________________
                      4                        5           8           2             4
         _______________________________________________
                      7                        8           2           5             7
         _______________________________________________
                      9                        1            4           7            9
         _______________________________________________

The conservation of number digit demands that the number digit of one of the two squares added together must be 9.All the numbers that are some multiples of 3 will have number digit 9. Thus one of the Pythagorean triples in the sum of two squares must be a multiple of 3.
Even though the conservation of number digit allows the possibility of having number digit of x,y,z as 9, it is not taken into consideration as the numeral relation becomes reducible.
In the case of cubical relation the conservation of number digits the number digit of a cube must be equal to the number digit of sum of the two cubes.The number digit of cubes can be either 1or 8 or 9 and they vary cyclically in the natural series.

Number digit                       Number digit of z3                             _________________________________
               x    /  y        1           8           9
         ___________________________________
                      1                        2           9           1          
         ___________________________________
                      8                        9           7           8            
         ___________________________________
                      9                        1           8           9           
         ____________________________________

A cube cannot have number digit 2 and 7 .This table predicts the possibility of formation of cubical relations  only when the number digits of x,y,z as (1,8,9;1,9,1; 8,9,8). ,where (9,9,9) is ruled out as it gives reducible triples.It clearly shows that one of the cubes must have a number digit 9, that is one of the base numbers must be a multiple of 3. where as the other numbers must give remainder 1 or 2 and 2 or 1  respectively on division by 3.,if both of them are in the same side if not the two members must give same remainder on division by 3. Accordlingly the various possibilities of the triple are (3na + 1,3nb + 2, 3nc),or (3na+2 ,3nb+1 ,3nc),(3na + 1 , 3nb, 3nc + 1) (3na +2 , 3nb , 3nc + 2).


 simple and natural  proof for Fermat's Last Theorem

5.Prime factor method for FLT (Marvalous proof)

FLT can be simply proved with the prime factors associated with triples. At first let us make an attempt for n= 3.All odd numbers will have only odd prime  factors (Po),  where as all even number will have invariably the even prime factor 2 .Besides that it may have one or more odd prime factors (Po)

Odd  number-prime factors             even number-prime factors

                         2                         2
3                        3                                             4                         22
5                        5                                             6                         2,3
7                        7                                             8                         23
9                        32                                           10                        2,5
11                     11                                            12                        22,3
13                     13                                            14                        2,7
15                     3,5                                           16                        24
17                     17                                            18                        2,32
19                     19                                            20                        22,5
21                     3,7                                           22                        2,11
23                     23                                            24                        23,3
25                     52                                             26                       2,13
27                     33                                                28                        22,7
29                     29                                            30                        2,3,5
31                     31                                            32                        25
33                     3,11                                         34                        2,17 

 In general an odd and even numbers can be represented in terms of its prime factors as
n(odd) = Po1n1 P02n2 Po3n3 ..................... Pomnm
n(even)  =  2n1 Po1n2 Po3n3 ..................... Pomnm
where n1, n2, n3 , .........nm may have any value from 0 to n depending upon the value of the number. n1 has a minimum value 1 in the case of even numbers. If all powers are zero, the number will reduce to 1, hence atleast any one of the power will be greater than one.
The most general representation of cube of an odd and even numbers is
n3(odd) = Po13n1 P023n2 Po33n3 ..................... Pom3nm
n3(even)  =  23n1 Po13n2 Po33n3 ..................... Pom3nm
If a cube is divisible by a prime factor, it should be divisible by square and cube  of the prime factor if the number is singly that prime factor.If it is doubly that factor, then its cube will be divisible by its square cube,fourth power,fifth power and 6 th power of the prime factor and so on.
Making use of this fact one can prove that x3 + y3 = z3  cannot exist with x,y,z all whole integers.
x3 + y3 = z3 
(x+ y) (x2 - xy + y2) = z3
In this equal power relation, the prime factors representation must be same in both sides. Taking x=odd, y=even and z=odd then (x+y) = odd and (x2 - xy + y2) = odd.The conditions provided by the equal power of cubical relation for the values of its members are (i) x < z, (ii) y < z  (iii) x+y > z. and (iv) Since (x+y) is greater than z,  (x2 - xy + y2) must be less than z2 .
(x+y) cannot be a single prime number. If so it will be a prime factor for z and z3 will be equal to some multiples of  (x+y)3 . It cannot be since (x+y) > z. The prime factors of z3, must be equal to the prime factors of the product (x+ y) (x2 - xy + y2) .What ever prime factors associated with z3, all of them must be shared with the two product terms as per the conditions stated above.
Let us represent z  in terms of two prime factors for simplicity.
z = P01 P02   and z3  = Po13 P023
One possible distribution of prime factors without violating the conditions is
(x+y) =   P012 P02  then (x2 - xy + y2)  =  P01 P022   , whatever may be the prime factors  associated with x and y .In fact the prime factors associated with x and y is  not at all required.
(x+y)3 = x3 + y3 + 3xy(x+y) = z3 + 3xy(x+y)
substituting the corresponding prime factors, we get,
(P016 P023) = P013 P023 + 3xy(P012 P02)
(P014 P022) = P01 P022 + 3xy
3xy = P014 P022 -  P01 P022
4xy = (4/3)(P014 P022 -  P01 P022 )
(x-y)2 = (x+y)2 - 4xy
          = P014 P022 - (4/3)(P014 P022 -  P01 P022 )
         = [3 P014 P022 - 4 (P014 P022 -  P01 P022 )]/3
        = [4 P01 P022 - P014 P022]/3
x-y =  √[ (4/3)(P01 P022) - (1/3)P014 P022 ]
x+ y =  P012 P02 
By adding and subtracting we can get x and y,
2x = P012 P02  + √[4/3)(P01 P022) - (1/3)(P014 P022) ] 
2y = P012 P02   - √[4/3)(P01 P022) - (1/3)(P014 P022) ] 
Both x and y will not be whole integer as the odd prime factors cannot be divided perfectly by 2 and √P01  cannot be rational.
When P01=3 and Po2 = 7, we get,
z = 21 ,which gives x = [63 + i √1127]/2 , y = [63 - i √ 1127]/2
The very same result can be obtained by solving the two equations
(x+y) =   P012 P02 
 (x2 - xy + y2)  =  P01 P022
  y = P012 P02  - x
By substituting this value of y in the other relation we get,
3x2 -3x P012 P02  +  P014 P022  =   P01 P022
3x2 -3x P012 P02  +  P014 P022  -   P01 P022  = 0
Solving for x,we get,
x = 1/2 (P012 P02 ) + 1/6 √ (12P01 P022  - 3P014 P022) 
y =  1/2 (P012 P02 ) - 1/6 √ (12P01 P022  - 3P014 P022)  
The term inside the square root is negative and hence it must be complex in  nature.It gives the same value for x and y as determined before.
 Simple and natural  proof for Fermat's Last Theorem

6. Generalization of Prime factor method for FLT (Marvalous proof)
 This proof is dedicated to Mahatma Gandhi in India and Fermat in France.
Let us consider the most general case xn + yn = zn
Case(i) When n is even

The the general form can be rewritten as
xn = zn - yn = (zn/2 - yn/2) ( zn/2 + yn/2)
x is assumed as x =Po1n1 P02n2 Po3n3 ..................... Pomnm,  then
xn =   Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm
Distributing the prime factors associate with xn between (zn/2 - yn/2) and ( zn/2 + yn/2) without violating the conditions (i) xk+yk > z k when k < n , +yk = z k when k = n and xk+yk < z k when k >n (ii) x< z, y< z  .
This method does not require the prime factors associated with the remaining two other members since xn is represented in terms of product of sum and difference of two terms.
According to any one of the permissible distributions of prime factors,
Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm  = (zn/2 - yn/2) ( zn/2 + yn/2)
( zn/2 + yn/2) = Po1nn1- m1 P02nn2-m2 Po3nn3-m3 ..................... Pomnnm-mm
and
(zn/2 - yn/2)  =   Po1 m1 P02m2 Po3m3 ..................... Pommm
where nn s and mns may take any value depending upon the prime factors  associated with members of the triple.
Adding  we get
2zn/2 = Po1nn1- m1 P02nn2-m2 Po3nn3-m3 ..................... Pomnnm-mm +  Po1 m1 P02m2 Po3m3 ..................... Pommm
Squaring both sides
4zn = Po12(nn1- m1) P022(nn2-m2) Po32(nn3-m3) ..................... Pom2(nnm-mm) +  Po1 2m1 P022m2 Po32m3 ..................... Pom2mm + 2Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm

zn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po12nn1-4 m1 P022nn2- 4m2 Po32nn3-  4m3 ..................... Pom2nnm- 4mm +  1 + 2Po1nn1-2m1 P02nn2-2m2 Po3nn3 -2m3..................... Pomnnm-2mm)

zn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po1nn1-2 m1 P02nn2- 2m2 Po3nn3- m3 ..................... Pomnnm- 2mm +  1)2


Similarly by substracting we get,
2yn/2 = Po1nn1- m1 P02nn2-m2 Po3nn3-m3 ..................... Pomnnm-mm -  Po1 m1 P02m2 Po3m3 ..................... Pommm
Squaring both sides
4yn = Po12(nn1- m1) P022(nn2-m2) Po32(nn3-m3) ..................... Pom2(nnm-mm) +  Po1 2m1 P022m2 Po32m3 ..................... Pom2mm - 2Po1nn1 P02nn2 Po3nn3 ..................... Pomnnm

yn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po12nn1-4 m1 P022nn2- 4m2 Po32nn3-  4m3 ..................... Pom2nnm- 4mm +  1 - 2Po1nn1-2m1 P02nn2-2m2 Po3nn3 -2m3..................... Pomnnm-2mm)
yn = (1/4) [Po1 2m1 P022m2 Po32m3 ..................... Pom2mm ( Po1nn1-2 m1 P02nn2- 2m2 Po3nn3- 2m3 ..................... Pomnnm- 2mm - 1)2

The roots of both z and y will be either complex or irrational.This is true even when n is odd.
Applying these relations for a cubical relation,
n=3, n1= 1 , n2  = 1, n3, n4.....  nm = 0 and m1, m2 = 1,

x3= P013 P023

z3 = (1/4) [ Po12  P022 (Po1 P02 + 1)2 ]
y3 = (1/4)  [ Po12  P022 (Po1 P02 - 1)2 ]

If P01 =  3 and  P02 = 7, then x = 21  and x 3 = 9261
y3  = (1/4) [441 x400] = 44100
z3 =(1/4)[441 x 484 ] = 53361

Fermat Last Theorem FLT

A simple and natural  proof for Fermat's Last Theorem - 2

2.Conservation of odd-evenness in  equal odd power relations

Before to understand the simple,natural and marvellous proof  (may be the one what Fermat missed to add in the margin), let us be conversant with the inherent mathematical properties associated with equal power relations xn  + yn = zn. When n is odd the conservation of odd-evenness does not permit the integer solutions.

This can be exemplified with the general form of cubical relation itself.
x3 + y3 = z3
(x+y) (x2 - xy + y2) = 8nz3
If x and y are both odd, then x+y will be even, while x2 - xy + y2 will be odd. This relation shows that the evenness of (x+y) is exactly same as theat of the evenness of z3.If z is singly even its value will be 8nz3, if it is doubly even, it will be 64nz3 and so on so fourth. If z is signly even z3 = 8 nz3 where nz is odd.Hence x+ y = 8 or 8nz [ x + y > z (=2nz) ] and (x2 - xy + y2) =nz3 or nz2 [(x2 - xy + y2)  < z2(= 4nz2) ]

If x+y = 8, then y = 8- x ,on substitution  in the other relation we get
3x2 - 24 x + 64 - nz3 = 0
It gives x =  [12 - √3 nz3 -48 ] / 3 and y = [12 + √3nz3 -48] / 3 , both x and y will be irrational for all real and integer values of nz.
If x+y = 8nz, then y = 8nz - x
3x2 -24 nz x + 64 nz 2 - nz 2 = 0
3x2 -24 nz x + 63 nz 2  = 0
It gives x =[ 8nz   +i nz √  20 ] / 2 and y = [8nz   - i nz √  20 ] / 2  
Both x and y become complex.
For any even integer values for z, both x and y are either irrational or complex. This is true even for odd integer values for z.
The concept of conservation of odd-evenness proves the FLT for n = 3.

If z is doubly even then z3 = 64 nz3 so that x+y= 64 or 64 nz and (x2 - xy + y2) =  nz3 or nz2
It gives the value of x and y in terms of nz
x = [96 + i √ 3nz3 - 3072]/ 3 and y = 96 + i √3nz3 - 3072]/ 3

Fermat Last Theorem -FLT

A simple and natural  proof for Fermat's Last Theorem - 1

Introduction:
 Fermat asserted in the early 17 th century that it is impossible to find two cubes or two higher powers that sum to another cube or a higher power respectively. For n > 2, there are no integer solutions to the equation  xn  + yn = zn. Fermat claimed to have a valid proof of his conjecture which appears as one of the marginal notes in his manuscripts. As the margin was too small to accommodate the proof, he left the space without any description.Fermat made a number of similar marginal notes, all of which had been resolved satisfactorily except this one- hence the title of Fermat Last Theorem.

Euler in 1753 proved FLT for n =4. Gauss corrected his attempt for n = 3 in 1825.Legendre,in 1839 proved FLT for n =5 Lame proved it for n =7 in 18 th century.Germain added additional values of n to the proven list.D .Lehmer and E.Lehmer expanded  their work in 20 th century.Joe Butler and Richard Crandall brought Computer technology to FLT and prove it for all exponents n < 4,000,000. A proof of FLT was given by R .Taylor and A.Wiles on the basis of elliptic curve and modular theory.

 Nature loves symmetry but also to be always simple. In fact mathematical descriptions made by human beings to understand the nature are not exactly what the nature is. There must be a simple and natural proof for FLT with no complexity in mathematical descrption.
1.Conservation of odd-eveness in equal even power relations


The numeral relation in the form an + bn = cn is found to exist only when a,b,c and n are all whole integers provided n =1 or 2.
When n = 1 ; a+ b = c, then an + bn < cn  for n  >  1

when n = 2 ; a2 + b2 = c2  then  a + b > c and an + bn < cn when n > 2

when n = 3; a3 + b3 = c3 then an + bn  > cn  when n < 3 and an + bn  < cn  when n > 3
In general an + bn = cn  then  an + bn  > cn when n = 1 to n-1 and an + bn  < cn when n = n+1 and all higher powers.

When n - 2; a2 + b2 = c2 and are called Pythagorus relations, where the sum of two squares is equal to a square. The set of values (a,b,c) is called Pythagorean triple.
The Pythagorean triples cannot have all possible values as they have to satisfy the Pythagorus relation.
Condition.1
 a             b             c
even      even       even
odd         odd         odd
odd        even        even                                                                                     
even       odd         odd
odd        odd          even

Out of various individual possibilities, the first one will be redicible where 2 will be a common factor for all the triple.Ultimately it will reduce to the permissible form.
The second one is forbidden as the sum of two odd numbers cannot be an odd number.This is true in  the case of third one .
The conservation of odd-evenness necessitates that only fourth and fifth cases are allowed.
Consider the fifth case (odd, odd, even)
(2na + 1)2 + (2nb + 1)2 = (2nc)2

where na, nb and nc may be odd or even. On expansion we have,
4(na2 + nb2 ) + 4(na + nb) + 2 = 4 nc2

2 (na2 + nb2 ) + 2(na + nb) + 1 = 2 nc2
It is found that LHS is odd while RHS is even, what ever may be the values of nx  ny  and nz.The non-conservation of odd-eveness demands that this type of relation is forbidden. Hence the only possibility  of (a,b,c) is (odd ,even, odd).

(2na + 1)2 + (2nb)2  = (2nc + 1)2

4(na2 + nb2 ) + 4na  + 1 = 4 nc2 + 4nc + 1

(na2 + nb2 ) + na  =   nc2 + nc 

nb2  = (nc - na) (nc + na + 1)
If nb  is odd  ,both the factos must be odd. If nc - na is odd nc + na will also be odd or nc + na + 1 will be even. .Hence nb must necessarily be even.

We can prove that  in all the equal pwer relations what ever may be the value of exponent n  ,the former form is forbidden while the later one is permitted.. It is very easy to establish this fact when the exponent is even.
When n = 4
(2nx + 1)4 + (2ny  + 1)4 = (2nz)4

16(nx4 + ny4 ) + 32 (nx3 + ny3) + 24(nx2 + ny2 ) + 8(nx+ ny) + 2 = 16 nc4
It is reduced to 
8(nx4 + ny4 ) + 16 (nx3 + ny3) + 12(nx2 + ny2 ) + 4(nx+ ny) + 1 = 8 nc4
 It is impossible  due to non-conservation of odd-evenness.It shows that  the sum of two odd numbers raised to any even power is always singly even,when divided by 2 gives  an odd number. It implies that the only possibility of formation of triplet with even exponent is (odd,even,odd).