BEAi CONJECTroE Slobodan Stanojevic RETRACTED · 2- Also if we multiple with common factor Nnt...

BEAL CONJECTURE

Slobodan Stanojevic

12 Stayner Court, Chelsea VIC 3196 Melbourne, Victoria, Australia

[email protected]

Keywords and phrases: Beal Conjecture, Fermat´s Last Theorem, Analysis

Abstract

Beal Conjecture was formulated in 1997 and presented as a generalization of Fermat's Last

Theorem, within the field of number theory.

It states that, if:

-A, B, C, x, y, and z are positive integers, and x, y, and z > 2

-Then A, B and C must have a common prime factor.

This article presents solution and the proof for the Beal Conjecture.

Introduction

The aim of this paper is to provide proof of Beal Conjecture with the use of basic mathematics.

This article is divided into three parts:

The first part presents two theorems:

Theorem 1: Any two of exponents x, y, and z must be the same.

Theorem 2: The factors A, B and C must have a common prime factor.

The second part presents the solutions and proof of Beal Conjecture.

The third part includes analysis and discussion on topic of the solutions of Beal Conjecture

including some examples.

1. THEOREMS

Theorem 1

A, B, C, x, y and z are positive integers and x, y, z > 2

A, B and C must have a common prime factor.

Then any two of exponents x, y, or z must be the same.

Proof

The exponents x, y and z we can express as:

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The Bulletin of Society for Mathematical Services and Standards Online: 2015-06-01ISSN: 2277-8020, Vol. 14, pp 7-27doi:10.18052/www.scipress.com/BSMaSS.14.7CC BY 4.0. Published by SciPress Ltd, Switzerland, 2015

This paper is an open access paper published under the terms and conditions of the Creative Commons Attribution license (CC BY)(https://creativecommons.org/licenses/by/4.0)

https://doi.org/10.18052/www.scipress.com/BSMaSS.14.7

Case 1. For u = v = 0 ; x = y = z = n

We get the Fermat’s Last Theorem

An + B

n = C

n

It has no integer solutions for n > 2.

Case 2. For u ≠ v ; x ≠ y ≠ z ; 0 ≤ u,v < n, in case ( u or v or both ≥ 2 )

we get:

An + B

n+u = C

n+v

If A,B and C have a common prime factor N, then we can express them as:

A = KN

B = PN

C = SN

K, P, and S ≥ 1 and positive integers

N Common prime factor or Common factor, then we can write:

This is degree equation For K, P, and S ≥ 1 and n ≥ 3,

than in case 0 ≤ u,v < n, ( u or v or both ≥ 2 ; u ≠ v) it is not possible that common prime factor N

and A, B, and C are all integers simultaneously, except for the particular case where:

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Case 3: u ≠ v ; 0 ≤ u,v < 2

Then: u,v = {0,1}

a. u = 0 ; x = y = n

v = 1 ; z = n + 1

b. u = 1 ; x = z = n

v = 0 ; y = n + 1

If we substitute in equation:

We then have:

Based on the equations (a), (b) and Beal Conjecture:

A, B, C, x, y and z are positive integers and x, y, and z > 2.

A, B and C must have a common prime factor,

Then x, y, and z can never be all equal or different simultaneously. Any two of x, y, and z

must be the same.

Theorem 2

A, B, C, x, y and z are positive integers and x, y, z > 2.

Then A, B and C must have a common prime factor “Ni” or common factor N.

Proof

If we start from the opposite assumption for Beal Conjecture Ax+ B

y = C

z that:

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The Bulletin of Society for Mathematical Services and Standards Vol. 14 9

A, B and C have no common prime factor.

1. A, B and C are all different prime factors A ≠ B ≠ C. E.g. 2, 3, 5, 7, 11…

2. Or different common factors with no common prime factors.

E.g. A = 2x11, B = 3x7, and C = 5x13x17.

Then following Theorem 1

We have:

1. If A, B and C are all different prime factors A ≠ B ≠ C, quotient of two distinct prime numbers is

always irrational or rational (but not a positive integer), prime factor is divisible only by 1 and

themselves, then any one factor of A, B or C will be irrational or rational but can’t be positive

integer.

2. If any two of A, B, and C are coprime to each other and positive integers, then third one can’t be

a positive integer. In that case A, B and C can’t be positive integers simultaneously.

3. From equation (a), C will be positive integer if: are positive integer.

will be positive integers if

A and B is divisible by C with no remainder. From that it follows:

A= KC ; B = PC

Where K and P are positive integer ≥ 1 ;

If two of A, B and C are positive integers and have common prime factor, then third one must

have common prime factor too.

Then equation (a) An + B

n = C

n C for: A = KC and B = PC becomes:

(a) (KC) n

+ (PC) n = C

n+1

Where C = N is a common factor

K and P are positive integers ≥ 1

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We can show the same for equation (b) An + B

n B = C

n for A = KB and C = SB

(b) (KB) n + B

n+1 = (SB)

n

Where B = N is common factor

K and S are positive integers ≥ 1

Then equation (a) and (b) becomes:

(1) (KN) n + (PN) n =N n+1

(2) (KN) n + N n+1 = (SN) n

From the above we can conclude that any one of A, B and C must be the common factor itself.

Based on the equations (1) and (2) and Beal Conjecture if:

A, B, C, x, y and z are positive integers, and x, y, z > 2 we have proved that the factors of A, B

and C must have a common prime factor Ni or common factor N for every value n ≥ 3, and K,

P, S positive integers ≥ 1

2. PROOF THE BEAL CONJECTURE

Based on the theorems 1 and 2 we have:

From equation (1):

And from equation ( 2):

These are solutions and proof of the Beal Conjecture for any value of:

1- Substituted in equation (1), (1.1), (2), (2.1) any value of K,P,S ≥ 1 and n ≥ 3 we obtain infinite

number of solutions of the Beal Conjecture.

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2- Also if we multiple with common factor Nnt

equations (1) and (2), t = {1,2,3,..},or multiple or

divide with common prime factor Nin, where

N = N1xN2x…, and N1 = N2 = ..,

E.g N = 8 = 2x2x2 = N1N2N3 ; N1 = N2 =N3, or

N = 18 = 3 x 3 x 2 = N1 N2 N3

We get a different form of the same equations.

3- If we substitute the value of any factor of A, B or C in a different form

E.g: 39 = 27

3 or 3

6 = 9

3 or 2

8= 4

4 or 2

9= 8

3

Then we get a different form of the same equations.

1. Following above and equations (1), (1.1), (2), and (2.1), for K, P, and S ≥ 1 and n ≥ 3, we

obtain infinite number of solutions of the Beal Conjecture.

2. If we multiple equations (1) and (2), with common factor Ntn

or multiply or divide with

common prime factor Ni tn

, as shown, t = {1,2,3,…}, we get a different form of the same

equations.

3. If we substitute the value of any factor of A, B or C in different form, as shown, Eg. (29 =

83); we get a different form of the same equations. This is a solution and proof of Beal

Conjecture.

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3. ANALYSIS AND DISCUSSION

A. ANALYSIS WHEN “N” IS COMON FACTOR N = N1 x N2 x…

If Ax + By = Cz

Where A, B and C are positive integers and have a common factor N, we can write:

A = KN N = N1 x N2 x…

B = PN Ni= N1 or N2 or….

C = SN

Where Ni is:

1. Common prime factor such as:

Ni = {2,3,5,7,11,13,17,19,…}

Where N is:

2. Common factor which contains common prime factor such as:

N = 65 =5x13 =N1xN2 or,

N = 175 = 5x5x7 = N1 N2 N3

3. K, P, S positive integers ≥ 1

K, P, S ={ 1,2,3,4,…….}

Then we can write:

(KN)X + (PN)Y = (SN)Z

Where N = N1 N2 is common factor

and N1 and N2 are common prime factors

We then have:

If the exponents x, y and z are positive integers, all greater than 2, we can express them as:

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If we substitute the expression for A, B, C, x, y and z in Beal conjecture we obtain the following

form of the Beal conjecture.

(KiNi)n + (PiNi)

n+u = (SiNi)

n+v

From this equation we can calculate a common prime factor Ni for every value of the indicated

conditions.

We get:

We get:

N = 28 is common factor

N = 2 x 2 x 7 = N1 x N2 x N3

N1 = 2, N2 = 2, N3 = 7 is common prime factor. And we get the same expression if we write:

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14 BSMaSS Volume 14

B. ANALYSIS FOR THE CASE WHEN

0 ≤ u,v < n

Kn N

n +

Pn+u N

n+u = S

n+v N

n+v

Case 1. u ≠ v 0 ≤ u,v < 2

u,v ={ 0, 1}

a) u = 0 b) u =1

v = 1 v = 0

Case 2. u ≠ v 0 ≤ u,v < n

(u or v or both ≥ 2)

Case 3. u = v = m

0≤ m ≤ n-1

ANALYSIS CASE 1

a) u = 0

v = 1

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N will be positive integer if

K and P positive integer ≥ 1

b) u = 1

v = 0

N will be positive integer if S > K

and S and K positive integer ≥ 1

This is the solution and proof of Bealʼs Conjecture.

ANALYSIS CASE 2

If x, y, and z are all different in case 0 ≤ u,v < n, ( u or v or both ≥ 2 ; u ≠ v) then it is not possible

that common prime factor N and A, B, and C are all integers simultaneously, except for the

particular case where:

Then equation (1) takes the form

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If the equation 2

3 N

n + N

n = N

n+2 is multiplied by N

tn, ( t = 1, 2, 3, 4, ..), we obtain different form of

same equation .

ANALYSIS CASE 3

For the common factor N to be a positive integer, the following conditions must be satisfied:

Sn must be divisible by K

n+m + P

n+m without a remainder, in that case we can write:

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Kn+m

+ Pn+m

= S

The following condition must be fulfilled:

- must be positive integers ≥ 1

As we can see equation:

35275

5+ 2

5275

5= 275

6

becomes equation (1)

KnN

n +P

n N

n= N

n+1

For K = 3 n = 5

P = 2 v =1

S = 1 u = 0

For the common factor N to be a positive integer, the following conditions must be satisfied:

Kn must be divisible by S

n+m - P

n+m without a remainder, in that case we can write:

The following condition must be fulfilled:

- must be positive integer ≥ 1

E.g 1 n = 3

m = 1

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S = 3

P = 2

As we can see equation

24

654 +

+ 655 = 3

4 65

4

1304 + 65

5= 195

4


For: K = 2 n = 4

S = 3 u = 1

P = 1 v = 0

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As we can see equation:


We get the equation for FERMATʼS CONJECTURE

Or Fermat’s Last Theorem

Following the Theorem 1: x, y, and z can never be all equal or different simultaneously, for x,

y, and z > 2

The equation An + B

n = C

n for n > 2 has no integer solutions other than trivial solution in

which at least one of the variables is zero.

For n = 2 we get Pythagorean theorem.

C. EXAMPLES AND DISCUSSION

EXAMPLE: 1

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If N is common factor:

And we get the same if we write:

If multiply

Where is:

N1 = 5 common prime factor

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Or: Ni = f( Ki, Pi, Si)-common prime factor Ki, Pi, Si, ≥ 1- positive integers

Multiple equation with Nnt

or Nint

t = {1,2,3,….},

we obtain infinite number of solutions, in different form.

EXAMPLE: 2

For N is common factor:

N = 15 is common factor

N = 5 x 3 = N1 x N2

N1 = 5 and N2 = 3 are common prime factors. And we get the same if we write:

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N = 31 is a common prime factor

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N2 = 17 is a common prime factor

For: n ≥ 3 ; u = 1 ; v = 0 ; K, P, S ≥ 1

Or Ni = f(Ki, Pi, Si) is a common prime factor

Ki, Pi, Si are positive integers ≥ 1

Multiple or divide equations with Nnt or Nint t = {1,2,3…}, we get a different form of equation and

obtain an infinite number of solutions.

From the analysis above we can conclude that every solution is a part off and belongs to the

solution set of equations (1), (1.1), (2) and (2.1), which are solution of the Beal Conjecture.

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D. MORE EXAMPLES

of applications of equations (1), (1.1), (2) and (2.1) as solutions of Beal Conjecture

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CONCLUSION

Starting from the Beal Conjecture that if

Ax + B

y = C

z where

A, B, C, x, y, and z are positive integers and x, y, and z >2 than A, B, and C must have a

common prime factor.

Than equations (1), (1.1), (2), (2.1), are solutions of the Beal Conjecture for every value of K,

P and S ≥ 1 ; n ≥ 3; and (u,v) = {0, 1} including particular case where (u = 0 ; v = 2), as shown.

This is the solution and proof of Beal Conjecture.

REFERENCES:

[1] Don Blazys, Proof of Bealʼs Conjecture By: Don Blazys

[2] Leonardo Torres Di Gregorio,“Proof for the Beal Conjecture and new proof for Fermatʼs Last

Theorem” In Pure and Applied Mathematics Journal,

[3] pp 149-155, Published Online September 20, 2013.

[4] Prof. dr. K. Rama Gandhi and Reuven Tint, Proof of Bealʼs Conjecture Bulletin of

Mathematics Sciences & Applications pp. 61-64 (2013)

[5] Jamel Ghanouchi, A proof of Bealʼs Conjecturehal-00710017 version 4 – 22 Jun 2012

[6] Byomkes Chanora Ghosh, The proof the Bealʼs Conjecture , Calcuta Mathematical Society,

EA-374 Sector-1 Salt lake City

[7] Charles William Johnson, A proof and Counterexamples Earth/matrix Editions 4-22 Aug.

2002

[8] Mauldin, R.D,A Generalization of Fermatʼs Last Theorem: The Beal Conjecture and prize

problem, AMS Notices 44, No 11, Dec. 1997, 1436-1437

[9] Raj C Thiagarajan, PhD A Proof to Bealʼs Conjecture 29. Aug. 2013. www.atoa.com, Rev 5,

14. Jan., page 1

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