Dave's Proof of the Beal Conjecture

Dave's Proof of the Beal Conjecture

Postby Guest » Fri Mar 01, 2019 6:03 am

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 on the 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.

[/size][/b]

Important Equations in our Proof of the Beal Conjecture:


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.




[b]Note: We had obtained the previous three equations (see equations one, two, and three above) 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,
https://www.researchgate.net/profile/David_Cole29/amp
Guest
 

Re: Dave's Proof of the Beal Conjecture

Postby Guest » Sat Mar 02, 2019 8:38 am

Guest wrote: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 on the 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.

Important Equations in our Proof of the Beal Conjecture:


For some real number, [tex]0 < \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.




Note: We had obtained the previous three equations (see equations one, two, and three above) 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]0 < \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,
https://www.researchgate.net/profile/David_Cole29/amp


Note: [tex]0 < \nu < x[/tex].
Guest
 

Re: Dave's Proof of the Beal Conjecture

Postby Guest » Sat Mar 02, 2019 9:04 am

Hmm. [tex]0 < \nu < x[/tex]?
or [tex]\nu > x[/tex]? The choice depends on data.

[tex]\nu > 0[/tex] is probably best overall.

Guest
 

Re: Dave's Proof of the Beal Conjecture

Postby Guest » Sat Mar 02, 2019 10:17 am

Observations:

[tex](A^{\nu} + 1 )^{2 }/(4*A^{\nu} )[/tex] and

[tex](A^{\nu} - 1 )^{2 }/(4*A^{\nu} )[/tex] are positive integers!

[tex](A^{\nu} + 1 )^{2 }/(4*A^{\nu} ) - (A^{\nu} - 1 )^{2 }/(4*A^{\nu} ) = 1[/tex].
Guest
 


Re: Dave's Proof of the Beal Conjecture

Postby Guest » Mon Mar 04, 2019 1:46 pm

Some Empirical Data on the Beal Conjecture:

[tex]34^{5}+ 51^{4} = 85^{4} = 52,200,625[/tex]

= [tex]17^{4} * (2^{5} * 17 + 3^{4})= 17^{4}*5^{4} = 52,200,625[/tex]

= [tex]17^{4} * (625) = 17^{4}*625= 52,200,625[/tex]

= [tex]17^{4} * (1 + 624) = 17^{4}*625 = 52,200,625[/tex].

Therefore, [tex]A^{x} = 17^{4}[/tex];

[tex]B^{y} = 17^{4} * 624[/tex];

[tex]C^{z} = 17^{4} * 625[/tex];

And gcd([tex]A^{x}, B^{y}, C^{z}[/tex]) = [tex]A^{x} = 17^{4}[/tex].

Relevant Reference Link:

http://www.norvig.com/beal.html
Guest
 

Re: Dave's Proof of the Beal Conjecture

Postby Guest » Mon Mar 04, 2019 2:24 pm

Guest wrote:
Guest wrote: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 on the 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} \le B^{y} < C^{z} = A^{x} + B^{y}[/tex].

Thus, the Beal Conjecture is true!


-- Dave.

Important Equations in our Proof of the Beal Conjecture:


For some real number, [tex]\nu > 0[/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.




Note: We had obtained the previous three equations (see equations one, two, and three above) 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 > 0[/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,
https://www.researchgate.net/profile/David_Cole29/amp


Note: [tex]\nu > 0[/tex].


Note: [tex]A^{x} \le B^{y} < C^{z} = A^{x} + B^{y}[/tex].
Guest
 

Re: Dave's Proof of the Beal Conjecture

Postby Guest » Wed Jul 31, 2019 8:07 pm

Note: The proof of the ABC Conjecture implies at most a finite exceptions/counterexamples to the Beal Conjecture...

Dave's Proof of the Beal Conjecture claims there are no exceptions/counterexamples to the Beal Conjecture.

Relevant Reference Links:

'Dave's Proof of the ABC Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=8385;

'Beal Conjecture',

https://en.wikipedia.org/wiki/Beal_conjecture.
Guest
 

Re: Dave's Proof of the Beal Conjecture

Postby Guest » Sun Sep 06, 2020 10:21 pm

https://docs.google.com/presentation/d/ ... p=drivesdk

The a, b &c are lines in a triangle. The cosine law can be manipuated to c^z and rearranged showing Beal's conjecture is correct.
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