## Devlin's Angle |

Dedicated followers of mathematical gossip will know that Fermat's Last Theorem said that for exponents n greater than 2, the equation

x^{n} + y^{n} = z^{n}

has no whole number solutions for x, y, and z (apart from trivial answers where one of the unknowns has the value 0).

First proposed in the seventeenth century by the great French mathematician Pierre De Fermat, the "theorem" resisted numerous attempts at solution until British mathematician Andrew Wiles of Princeton University found a proof in 1994. Wiles's achievement was portrayed in a BBC television Horizon documentary "The Proof" last year and described in a small rash of popular books.

Much of the general interest in Fermat's challenge can be attributed to the offer of a cash prize, the Wolfskehl Prize, to the person who first proved the theorem. Established in 1908 by Paul Wolfskehl, a German physician and amateur mathematician, the prize lost much of its value in the German inflation of the 1930s, but was still worth about $50,000 when Wiles collected the award earlier this year.

Beal's Problem is like Fermat's, but instead of focusing on equations with one exponent, n, there are three: m, n, and r. Thus, Beal's equation looks like this:

x^{m} + y^{n} = z^{r}

The idea is to look for whole number solutions to this equation where the solution values for x, y, and z have no common factor (i.e., there is no whole number greater than 1 that divides each of x, y, and z). Beal has conjectured that if the exponents m, n, and r are all greater than 2, then his equation has no such solution for x, y, and z.

Just as Fermat's equation can be solved if the exponent n is equal to 1 or 2 (the best known solution for n=2 is the "Pythagorean triple" x=3, y=4, z=5), so too you can find solutions to Beal's equation if you allow one of the exponents, m, n, r to be equal to 1 or 2. for example:

1^{1} + 2^{3} = 3^{2}

2^{5} + 7^{2} = 3^{4}

Also, just as it was known in the early 1980s that, for any exponent n, the corresponding Fermat equation could only have a finite number of solutions, so too it is known that for any three exponents m, n, and r greater than 2, the corresponding Beal equation has only a finite number of solutions. In the Fermat case, Wiles showed that the finite number for any exponent is in fact zero. Beal's conjecture is that it is zero for each of his equations as well.

Fermat's last theorem is the special case of Beal's conjecture you get when the exponents m, n, and r are equal. So the new problem is a generalization of the problem Wiles solved three years ago.

So, is Beal's conjecture likely to achieve the fame of Fermat's last theorem? Like Fermat's problem, Beal's question is easy to state, involving nothing more complicated than simple equations to be solved for unknown whole numbers. And like Fermat's last theorem, Beal's conjecture postulates that there are no solutions of the specified kind.

Now throw a cash prize into the mix, and you have all the ingredients for a new mathematical saga.

This year, Beal's prize stands at $5,000. Thereafter, it will increase by $5,000 each year for every year the problem remains unsolved, up to a limit of $50,000, the same amount as the Wolfskehl Prize when it was finally awarded.

Much is known of Fermat. But who is Beal? He is forty-four years of age, the father of five children, and the founder, owner, and chairman of Beal Bank, the largest locally owned bank in Dallas. He recently founded Beal Aerospace, which is designing and building a next-generation rocket for launching satellites into earth orbit. And, of course, he is a keen amateur mathematician, who has spent many hours pondering Fermat's last theorem.

Beal has appointed a committee of three distinguished professional mathematicians to screen potential prize-winning proofs of his conjecture. Anyone who thinks they have a solution should send it to Professor Daniel Mauldin at the University of North Texas in Denton, Texas, USA, who chairs the prize committee. But beware. There are indications that, as with Fermat's last theorem, solving Beal's problem might require a large dose of heavy mathematical machinery. But then again . . .

** - Keith Devlin **

Devlin's Angle is updated at the beginning of each month.