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