James Constant

Beal's Conjecture is disproved for the same reasons Fermat's Last Theorem is proved.

For simple proofs see 

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

thus proving FLT

(3) . . . . . . . . . . . . x,y,z,m positive integers . . . . . . m>2

Disproof of Beal's Conjecture

When a=b=m, Beal's equation (1) becomes Fermat's equation (2). Clearly, Fermat's equation (2) is a special case of Beal's equation (1). The same procedure used in Fermat's equation (2) can be used to show that z cannot be an integer in Beal's equation (1). Start by rearranging Beal's equation (1)

(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^kk 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^kk integer, ka>=m-1 and ka>=b. See also Beal's Conjecture And The Problem of Multiplying Irrational Number's at Problem/bealprob.htm .

Mirabilis Disproof of Beal's Conjecture

Write  the  right  hand  side  of  equation (1)  as

(5)      x2a/2 + y2b/2                   x,y,a,b integers         a,b>2

which, by Pythagora's theorem, is a real number squared, say

(6)    v2=xa + yb          v2 >2 or = 2 when x=y=1     

in which  v2  is an integer and v is an integer or irrational number. Equality equation (1) is obtained by assuming

(7)   v2=zm                  m > 2  positive integer

in which zm is an integer and z is an integer or irrational number. Therefore, equality equation (1) cannot meet Beal's conditions 

(8)   x,y,z,m,a,b  positive integers    m,a,b>2 

because z is an integer or irrational number. QED

The Mirabilis proof is a partial, but nevertheless valid, disproof of Beal's equality equation (1). It simply says that Beal's conditions equation (8) are incorrect because z is an integer or irrational number. A full disproof of Beal's equality equation (1), which determines solely that z is an irrational number, is provided by the foregoing bilinear expansion of z equation (4).

An Alternative Full Proof 

Write v2/m as an exponential series with terms (ln v2/m)n/n!  and compare with terms of exponential series e with terms 1/n! which is known to converge. If  ln v2/m<1  the v2/m series necessarily converges when v2< em for all values of v,m and, therefore, from equation (7) v2=zm< em, z<e which can occur only when z= 2 when x=y=1.  Accordingly, the assumption that  equation  (7)  can be  written  with  z  as an integer >2 is refuted and, therefore, is an irrational number disproving Beal's equality equation (1). QED

1 The Beal Conjecture and Prize

2 James Constant, Proof of Fermat's Last Theorem

Copyright© 2003 James Constant

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