V.3 No 1

53

On the Fermat equation A^x + B^y = C^z

On general technique to seek the solution

of the Fermat equation A^x + B^y = C^z

Sergey B. Karavashkin and Olga N. Karavashkina

Special Laboratory for Fundamental Elaboration SELF

187 apt., 38 bldg., Prospect Gagarina, 38, Kharkov 61140, Ukraine

Phone: +38 (0572) 276624; e-mail: selftrans@yandex.ru, selflab@mail.ru

Abstract

This notice is our respond to a challenge for mathematicians presented on www.sciencenews.org : to find and prove the general solution for the Fermat last theorem A^x + B^y = C^z (where A, B, C and x, y, z are whole numbers). By the very approach to the search of solution which we can trace on this web site, we see that the hosts represent their problem rather as a school conundrum than as a serious study in the number theory. This is why as a first step towards the solution, we suggested them to consider the very possibility of general solution, since the problem multiply branches, as it is typical for the number theory. We reflected it in the below notice. However it looks like it is much easier to challenge than to be responsible in mathematics. As we see, the authors of challenge decided the best simply "to be not delivered" our proof than to lose so dear illusion that if they shut their eyes, it will be dark for all us around. So we decided to publish this notice in our journal, thinking that our prompt of approach will enable those not many scientists, for whom the knowledge development means more than their arrogance, to progress in their search of solution for the power-type equation A^x + B^y = C^z. Finally, this progress in knowledge is actually the responsibility of everyone involved in scientific research.

Classification by MSC 2000: 11D41; 11D61; 11D72

Keywords: number theory; power-type functions; Fermat last theorem.

On the web site Science News Online, www.sciencenews.org/sn , 15/11/97, there was presented the problem, how to find the general solution of the Fermat equation

with the arbitrary integer numbers A, B, C and x, y, z . It was shown there that in some particular cases this equation is true. As an example, there was mentioned an equation

Before seeking the technique to solve, it would be helpful first to prove that for different numbers A, B, C and x, y, z  the sought general technique branches.

To prove it, consider the simplest case

With it (1) takes the following form:

To find the general solution of (4), introduce an additional condition

where  a  is some integer number which we specify. We see that (5) does not limit the solution (4), as  a  can be taken arbitrarily.

Given (5), the equation (4) takes the following shape:

As B  is an integer, it follows from (7) that there appear two versions:

1. a is the divisor of B;
2. a is the divisor of B ² .

In the first case B can be presented as

Knowing the value  A  through the given parameter  a, we can easily find the rest numbers. The solution for this case will be

We see from (10) that with a = 1  and arbitrary  k we yield the spectrum of solutions at the condition (8) and necessarily odd  k .