by 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!
[b][size=150]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!
[/size][/b]