. . . . . . . . x,y,z,m,a,b positive integers . . . . . . m,a,b>2 BEAL'S CONJECTURE DISPROVED
James Constant
math@coolissues.com
Beal's Conjecture is disproved for the same reasons Fermat's Last Theorem is proved.
Beal's conjecture
A prize is offered for proof or disproof of Beal's conjecture1, stated as follows: If
(1). . . . . . . . . . . . x,y,z,m,a,b positive integers . . . . . . m,a,b>2
then x,y,z have a common factor >1.
Proof of Fermat's Last Theorem
A proof of Fermat's Last Theorem (FLT) is available using the binomial expansion2. In this proof it is shown that z cannot be an integer in the equality
(2) . . . . 
. . . . . . . . x,y,z,m positive integers
. . . . . . m>2
(3) . . . . 
. . . . . . . . x,y,z,m positive integers . . . . . . m>2
(4) . . . .
. . . . . . . . x,y,z,m positive integers . . . . . . m>2
and then expressing the parenthesis term as a binomial series, with results
1. Since m>2,
index 1/m is not an integer 
and the series
cannot terminate becoming
the binomial theorem.
2.
The series is absolutely
convergent for 
and divergent for 
When convergent, the parenthesis
term series converges to an irrational number and thus z
in equation (4) is an irrational number for any xa/m.
When divergent, the parenthesis term series and thus z in
equation (4) have no values.
3. For yb/xa=+1
the series converges absolutely since m>2 and thus 1/m>0.
The same argument applies. In this case 
, an irrational number except when x=2^k k integer, ka>=m-1 and
ka>=b.
In summary,
Beal's conjecture is disproved for the stated conditions x,y,z,m,a,b
positive integers and m,a,b>2. It is no longer a
conjecture but an observation that Beal's equality exists only
when xa=yb, x=2^k k integer, ka>=m-1 and
ka>=b. See also Beal's Conjecture And The Problem of Multiplying Irrational Number's at
1 The Beal Conjecture and Prize http://www.math.unt.edu/~mauldin/beal.html
2 James Constant, Proof of Fermat's Last Theorem http://fernat.coolissues.com/fermat.htm