Proof of the Beal Conjecture

Proof of the Beal Conjecture

Postby Guest » Fri Feb 22, 2019 6:25 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,
https://en.m.wikipedia.org/wiki/Beal_conjecture

Dave,
https://www.researchgate.net/profile/David_Cole29
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Feb 22, 2019 6:48 am

Proof of Beal Conjecture:

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

And suppose there is no common prime factor, p, for A, B, and C.

Hence, gcd(A, B) = 1 or gcd(A, C) = 1 or gcd(B, C) = 1 which imply,
[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].

...
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Feb 22, 2019 7:28 am

Proof of Beal Conjecture(con't):

Thus, we must consider four equations.

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

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

3. [tex]k * A^{x} + l * C^{z}= 1[/tex];

4. [tex]m * B^{y} + n * C^{z}= 1[/tex].

Note: i, j, k, l, m, and n are nonzero integers.

...
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Feb 22, 2019 8:04 am

Proof of Beal Conjecture(con't):

We add equations one and two, and we obtain the following statement.

5. [tex](i + 1) * A^{x} + (j + 1) * B^{y}= C^{z} + 1 > 1[/tex] which implies [tex]gcd(A^{x}, B^{y}) > 1[/tex].

Therefore, p | A and p | B. This is a contradiction!

...
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Feb 22, 2019 8:40 am

Proof of Beal Conjecture(con't):

We add equations one and three, and we obtain the following statement.

6. [tex]-(k + 1) * A^{x} + (l - 1) * C^{z}= B^{y} - 1 > 1[/tex] which implies [tex]gcd(A^{x}, C^{z}) > 1[/tex].

Therefore, p | A and p | C. This is a contradiction!

Similarly, by adding equations one and four, we obtain the final result, p | B and p | C. This is also contradiction!

Thus, the Beal Conjecture is true!

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


...
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Feb 22, 2019 9:08 am

Oops! We must also confirm that the critical statement,

gcd(A, B, C) > 1, before concluding p divides A, B, and C.

That challenge is not difficult... And yes, we can succeed! Amen!

The Beal Conjecture is still true!

Thank God! Amen!

Dave : - )

Go Blue!
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Feb 22, 2019 2:18 pm

Proof of Beal Conjecture(con't):

Thus, we must now consider equations one and eight.

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

8. [tex]a * A^{x} + b * B^{y} + c* C^{z}= 1[/tex].

Note: a, b, and c are nonzero integers.

We add equations one and eight, and we obtain the following equation.

9. [tex](a + 1) * A^{x} + (b + 1)* B^{y} + (c - 1) * C^{z}= 1[/tex].

However, we recall that [tex]gcd(A^{x}, B^{y}) > 1[/tex],
[tex]gcd(A^{x}, C^{z}) > 1[/tex], and
[tex]gcd(B^{y}, C^{z}) > 1[/tex] which imply

[tex](c - 1) * C^{z} < -1[/tex],
[tex](b + 1) * B^{y} < -1[/tex],
and [tex](a + 1) * A^{x} < -1[/tex], respectively.

Thus, equation nine is false. This is a contradiction!

Therefore, [tex]gcd(A^{x}, B^{y}, C^{z}) > 1[/tex] which implies p | A, p | B, and p | C.

And we confirm the Beal Conjecture is true!

-- Dave.

Go Blue!

Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Sat Feb 23, 2019 6:55 am

Oops!

If x = y = z = 2, then our previous proof fails exceptionally!

Hmm. We must reconsider our proof and provide a fix and an explanation too...
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Sun Feb 24, 2019 8:01 am

Revised Proof of the Beal Conjecture:

We recall our assumptions, [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].

Suppose
[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.

Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Sun Feb 24, 2019 9:19 am

Revised Proof of the Beal Conjecture (con't):

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

But statements, 1, 2, and 3 imply k = 0, j = 0, and i = 0, respectively. Three contradictions!

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

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

The Beal Conjecture is true!

-- Dave,


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


Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Sun Feb 24, 2019 11:46 am

Guest wrote:Revised Proof of the Beal Conjecture (con't):

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

Therefore,

[tex](i * B^{2-y} ×C^{2-z}) * A^{2} + (j* A^{2-x} ×C^{2-z}) * B^{2} + (j* 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
 

Re: Proof of the Beal Conjecture

Postby Guest » Sun Feb 24, 2019 12:21 pm

Revised Proof of the Beal Conjecture:

We recall our assumptions, [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].

Suppose
[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.

Therefore,

[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
 

Re: Proof of the Beal Conjecture

Postby Guest » Thu Feb 28, 2019 2:36 pm

Oops! The latest proof of Beal Conjecture is wrong!

-- Dave
Guest
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Mar 01, 2019 12:00 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,
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
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Mar 01, 2019 4:05 am

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
 

Re: Proof of the Beal Conjecture

Postby Guest » Fri Mar 01, 2019 5:08 am

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
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