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.

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.

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