Prize offered for solving number conundrum.For more than 300 years, the Years, The the seven decades of Eleanor Pargiter’s life. [Br. Lit.: Benét, 1109] See : Time mathematical problem Mathematical problem may mean two slightly different things, both closely related to mathematical games:
Statement that there are no natural numbers x, y, and z such that xn + yn = zn, in which n is a natural number greater than 2. attracted a plethora of would-be conquerors. The theorem was finally proved just a few years ago by Andrew Wiles of Princeton University as part of a larger effort that illuminates links between number theory and geometry (SN: 11/5/94, p. 295). The complicated, lengthy proof that Wiles wile n. 1. A stratagem or trick intended to deceive or ensnare. 2. A disarming or seductive manner, device, or procedure: the wiles of a skilled negotiator. 3. Trickery; cunning. found hasn't satisfied everyone, however. Some still wonder whether a simpler proof could be found--one that Pierre de Fermat Noun 1. Pierre de Fermat - French mathematician who founded number theory; contributed (with Pascal) to the theory of probability (1601-1665) Fermat himself may have had in mind but failed to write down centuries ago. "The mystery remains: Is there an elementary proof?" asks Andrew Beal, a banker in Dallas and an amateur mathematician interested in number theory. In grappling with the problem, Beal has looked for solvable equations analogous to the one expressing Fermat's last theorem. "I would try to understand why the solutions of the new equation can't be converted to solutions of the Fermat equation," he says. As a result of those studies, Beal has formulated a conjecture involving equations of the form [A.sup.x] + [B.sup.y]= [C.sup.z]. The six letters represent whole numbers, with x, y, and z greater than 2. Fermat's last theorem involves the special case in which the exponents x, y, and z are the same. Beal noticed that when a solution of the general equation existed, A, B, and C had a common factor. For example, in the equation [3.sup.6] + [18.sup.3] = [3.sup.8], the numbers 3, 18, and 3 all have the factor 3. Using computers at his bank, Beal checked equations with exponents up to 100. "I couldn't come up with a solution that didn't involve a common factor," he says. The question is whether that is always true. Beal has now offered a prize of $50,000 to anyone who can prove the conjecture or $10,000 to anyone who can find a counterexample coun·ter·ex·am·ple n. An example that refutes or disproves a hypothesis, proposition, or theorem. Noun 1. counterexample - refutation by example . Interestingly, Beal's conjecture is closely related to questions of considerable concern to number theorists. "It is remarkable that occasionally someone working in isolation and with no connections to the mathematical [community] formulates a problem so close to current research activity," R. Daniel Mauldin of the University of North Texas in Denton comments in the December Notices of the American Mathematical Society Notices of the American Mathematical Society is a membership journal of the American Mathematical Society. It is published monthly except for the combined June/July issue. , where the Beal prize problem is announced. Mauldin heads the prize committee and will handle all inquiries and proposed proofs. Whether a proof is likely to materialize anytime soon isn't evident at the moment. "It's not clear how you would go about solving it or whether the methods of Wiles could be extended this far," says Andrew J. Granville of the University of Georgia Organization The President of the University of Georgia (as of 2007, Michael F. Adams) is the head administrator and is appointed and overseen by the Georgia Board of Regents. in Athens. Intriguingly, introducing a coefficient larger than 1, so that the equation reads, say, [A.sup.x] + 31[B.sup.y] = [C.sup.z], allows all sorts of possible solutions when x, y, and z are greater than 2. In the Beal conjecture, the coefficients are all 1, which appears to put the resulting equations and their solutions in a special category. The deeper question is one of finding a general method of identifying all solutions to the equation, whether there are just a few or an infinite number infinite number a number so large as to be uncountable. Represented by 8, frequently obtained by 'dividing' by zero. of them, Granville says. "That would be a wonderful advance." |
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