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

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

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