Is the Beal Conjecture trivial?

[Guest]

FYI: The Beal Conjecture makes the following mathematical statement.

"If [tex]A^{x}+B^{y}=C^{z}[/tex],

where A, B, C, x, y, and z are positive integers with. x, y, z > 2, then A, B, and C have a common prime factor." -- Wikipedia,
https://en.m.wikipedia.org/wiki/Beal_conjecture

[Guest]

The elegant Beal Conjecture is not a trivial problem since "The abc conjecture would imply that there are at most finitely many counterexamples to Beal's conjecture." -- Wikipedia

This conjecture demands an elegant proof. And in honor of the great mathematician, David Hilbert, we expect a clear and definitive proof of the Beal Conjecture within 48 weeks.

-- Dave,
https://www.researchgate.net/profile/David_Cole29/amp

[Guest]

Revised Proof of the Beal Conjecture:

The Beal Conjecture makes the following mathematical statement.

"If [tex]A^{x}+B^{y}=C^{z}[/tex],

where A, B, C, x, y, and z are positive integers with. x, y, z > 2, then A, B, and C have a common prime factor."

"The impossibility of the case A = 1 or B = 1 is implied by Catalan's conjecture, proven in 2002 by Preda Mihăilescu. (Notice C cannot be 1, or one of A and B must be 0, which is not permitted.)"

-- Wikipedia,
https://en.m.wikipedia.org/wiki/Beal_conjecture


Suppose [tex]A^{x}+B^{y}=C^{z}[/tex] with

[tex]gcd(A^{x}, B^{y}) =1[/tex] or [tex]gcd(A^{x}, C^{z}) = 1[/tex] or [tex]gcd(B^{y}, C^{z}) = 1[/tex] such that [tex]gcd(A^{x}, B^{y}, C^{z}) = 1[/tex] when x, y, and z are greater than two.

Therefore,

[tex]gcd(A^{2}, B^{2}) =1[/tex] or [tex]gcd(A^{2}, C^{2}) = 1[/tex] or [tex]gcd(B^{2}, C^{2}) = 1[/tex] such that [tex]gcd(A^{2}, B^{2}, C^{2}) = 1[/tex].

[tex]gcd(A^{x}, B^{y}) =1[/tex] which implies

[tex]i * A^{x} + j * B^{y}= 1[/tex] for some nonzero integers, i and j.


[tex]i * A^{x} + j * B^{y}= 1[/tex] implies

[tex](i * A^{x-2}) * A^{2} + (j* B^{y-2}) * B^{2}= 1[/tex] which implies

[tex](i * B^{2-y}) * A^{2} + (j* A^{2-x}) * B^{2}
= A^{2-x} * B^{2-y} = 1[/tex] iff x = y = 2. Contradiction!

Therefore, 1. [tex]gcd(A^{x}, B^{y}) > 1[/tex].

And similarly, we can show,

2. [tex]gcd(A^{x}, C^{z}) > 1[/tex] and

3. [tex]gcd(B^{y}, C^{z}) > 1[/tex].

[tex]gcd(A^{x}, B^{y}, C^{z}) = 1[/tex] implies

[tex]i * A^{x} + j * B^{y} + k * C^{z} = 1[/tex] for some nonzero integers, i, j, and k.

4. [tex]i * A^{x-2}* A^{2} + j * B^{y-2}* B^{2} + k * C^{z-2}* C^{2} = 1[/tex].

Next we divide equation four by [tex]A^{x-2}*B^{y-2}*C^{z-2}[/tex] which yields the following equation:

5. [tex](i * B^{2-y} ×C^{2-z}) * A^{2} + (j* A^{2-x} ×C^{2-z}) * B^{2} + (k * A^{2-x} ×B^{2-y}) * C^{2} = A^{2-x} * B^{2-y} * C^{2-z} = 1[/tex] iff x = y = z = Contradiction!

Hence, [tex]gcd(A^{x}, B^{y}, C^{z}) > 1[/tex],

and therefore, there exists a common prime factor, p, which divides A, B, and C.

Thus, the Beal Conjecture is true!

-- Dave,


https://www.researchgate.net/profile/David_Cole29

[Guest]

Revised Proof of the Beal Conjecture:

The Beal Conjecture makes the following mathematical statement.

"If [tex]A^{x}+B^{y}=C^{z}[/tex],

where A, B, C, x, y, and z are positive integers with. x, y, z > 2, then A, B, and C have a common prime factor."

"The impossibility of the case A = 1 or B = 1 is implied by Catalan's conjecture, proven in 2002 by Preda Mihăilescu. (Notice C cannot be 1, or one of A and B must be 0, which is not permitted.)"

-- Wikipedia,
https://en.m.wikipedia.org/wiki/Beal_conjecture


Suppose [tex]A^{x}+B^{y}=C^{z}[/tex] with

[tex]gcd(A^{x}, B^{y}) =1[/tex] or [tex]gcd(A^{x}, C^{z}) = 1[/tex] or [tex]gcd(B^{y}, C^{z}) = 1[/tex] such that [tex]gcd(A^{x}, B^{y}, C^{z}) = 1[/tex] when x, y, and z are greater than two.

Therefore,

[tex]gcd(A^{2}, B^{2}) =1[/tex] or [tex]gcd(A^{2}, C^{2}) = 1[/tex] or [tex]gcd(B^{2}, C^{2}) = 1[/tex] such that [tex]gcd(A^{2}, B^{2}, C^{2}) = 1[/tex].

[tex]gcd(A^{x}, B^{y}) =1[/tex] which implies

[tex]i * A^{x} + j * B^{y}= 1[/tex] for some nonzero integers, i and j.


[tex]i * A^{x} + j * B^{y}= 1[/tex] implies

[tex](i * A^{x-2}) * A^{2} + (j* B^{y-2}) * B^{2}= 1[/tex] which implies

[tex](i * B^{2-y}) * A^{2} + (j* A^{2-x}) * B^{2}
= A^{2-x} * B^{2-y} = 1[/tex] iff x = y = 2. Contradiction!

Therefore, 1. [tex]gcd(A^{x}, B^{y}) > 1[/tex].

And similarly, we can show,

2. [tex]gcd(A^{x}, C^{z}) > 1[/tex] and

3. [tex]gcd(B^{y}, C^{z}) > 1[/tex].

[tex]gcd(A^{x}, B^{y}, C^{z}) = 1[/tex] implies

[tex]i * A^{x} + j * B^{y} + k * C^{z} = 1[/tex] for some nonzero integers, i, j, and k.

4. [tex]i * A^{x-2}* A^{2} + j * B^{y-2}* B^{2} + k * C^{z-2}* C^{2} = 1[/tex].

Next we divide equation four by [tex]A^{x-2}*B^{y-2}*C^{z-2}[/tex] which yields the following equation:

5. [tex](i * B^{2-y} ×C^{2-z}) * A^{2} + (j* A^{2-x} ×C^{2-z}) * B^{2} + (k * A^{2-x} ×B^{2-y}) * C^{2} = A^{2-x} * B^{2-y} * C^{2-z} = 1[/tex] iff x = y = z =2. Contradiction!

Hence, [tex]gcd(A^{x}, B^{y}, C^{z}) > 1[/tex],

and therefore, there exists a common prime factor, p, which divides A, B, and C.

Thus, the Beal Conjecture is true!

-- Dave,


https://www.researchgate.net/profile/David_Cole29

[Guest]

Oops! The latest proof of Beal Conjecture is wrong!

-- Dave

[Guest]

The Beal Conjecture makes the following mathematical statement.

"If [tex]A^{x}+B^{y}=C^{z}[/tex],

where A, B, C, x, y, and z are positive integers with. x, y, z > 2, then A, B, and C have a common prime factor."

"The impossibility of the case A = 1 or B = 1 is implied by Catalan's conjecture, proven in 2002 by Preda Mihăilescu. (Notice C cannot be 1, or one of A and B must be 0, which is not permitted.)"

-- Wikipedia,
https://en.m.wikipedia.org/wiki/Beal_conjecture

Short Proof of Beal Conjecture:

(Note: There exists a longer proof of Beal Conjecture. Hopefully, we will post it at a later date.

Relevant Reference Link:

https://math.stackexchange.com/questions/1993460/prove-a-statement-about-a-conditional-diophantine-equation )

[tex]gcd(A^{x}, B^{y}, C^{z}) =A^{x} > 1[/tex]

where [tex]A^{x} < B^{y} < C^{z} = A^{x} + B^{y}[/tex].

Thus, the Beal Conjecture is true!


-- Dave

[Guest]

Important Equations in our Proof of the Beal Conjecture:


Keywords: Fundamental Theorem of Arithmetic

For some real number, [tex]\nu > x[/tex], we had derived the following equations:

1. [tex]A^{x}=A^{x}[/tex].

2. [tex]B^{y}=((A^{\nu} - 1 )^{2 }/(4*A^{\nu} )) * A^{x}[/tex].

3. [tex]C^{z}=((A^{\nu} + 1 )^{2 }/(4*A^{\nu} )) * A^{x}[/tex].

Clearly, [tex]gcd(A^{x}, B^{y}, C^{z}) = A^{x} > 1[/tex].

-- Dave,

https://www.researchgate.net/profile/David_Cole29/amp

[Guest]

Note: We had obtained the previous three equations (see previous post) from the following important equations:

0. [tex]A^{x}=(A^{x/2 - \nu /2}) *(A^{x/2 + \nu /2}) = C^{z} - B^{y}
= (C^{z/2} - B^{y/2}) * (C^{z/2} + B^{y/2})[/tex]
for some real number, [tex]\nu >x[/tex].

00. [tex]A^{x/2 - \nu /2} = C^{z/2} - B^{y/2}[/tex].

000. [tex]A^{x/2 + \nu /2} = C^{z/2} + B^{y/2}[/tex].

-- Dave