# Number TheoryBeal's conjecture

#### sh911098

##### New member
I’m a number theory lover but not an expert in the area. Recently, motivated by the report of Peter Norvig, Director of Research at Google, I’m interested in searching for counterexamples of Beal’s conjecture.

Billionaire banker Andrew Beal formulated this conjecture in 1993. For a proof or counterexample published in a refereed journal, Beal initially offered a prize of US dollar 5,000 in 1997, but raised it to 1,000,000 this Jun.

Beal’s conjecture is that if

Ax + By = Cz.

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 conjecture is very understandable. Below are a few positive examples
33 + 63 = 35
where A = 3, B = 6, C = 3, x = 3, y = 3, z = 5; A, B, C have a common prime factor 3.
76 + 77 = 983
where common prime factor is 7. Actually we always have
[A(Am + Bm)]m + [B(Am + Bm)]m = (Am + Bm)m+1
where A, B > 0, and m > 2.

However, followings are negative examples:
2713+23×35×733 = 9193 = 776151559
34×293×893+73×113×1673=27×54×3593=3518958160000
because the conjecture requires A, B, and C each has their own exponent, the two equations do not fulfill the requirement.

Beal’s conjecture is a generation of Fermat’s last theorem, which has been proved by Andrew Willes in 1995. The theorem says

An + Bn = Cn

has no solution of positive integers when n > 2. Every one understands the theorem is just a special case of Beal’s conjecture with x=y=z.

It took more than three hundreds years to get complete solution of Fermat’s last theorem.

How many years will be needed to have a proof or a counterexample of Beal’s conjecture? Nobody knows.

Do you have courage and interest to face the challenge? To me it is too hard to prove it. But I found it is not difficult to develop a program for the job, and I did one much fast than that used by Peter Norvig.

The purpose of this thread is to discuss basic idea and programming details of the project, and to associate as more as possible people, search as wide as possible.

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