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A Generalization of Fermat’s Last Theorem: The Beal Conjecture and Prize Problem R. Daniel Mauldin
ndrew Beal is a Dallas banker who The prize. Andrew Beal is very generously of- has a general interest in mathemat- fering a prize of $5,000 for the solution of this ics and its status within our culture. problem. The value of the prize will increase by He also has a personal interest in the $5,000 per year up to $50,000 until it is solved. The discipline. In fact, he has formulated prize committee consists of Charles Fefferman, a conjectureA in number theory on which he has Ron Graham, and R. Daniel Mauldin, who will act been working for several years. It is remarkable that as the chair of the committee. All proposed solu- occasionally someone working in isolation and tions and inquiries about the prize should be sent with no connections to the mathematical world for- to Mauldin. mulates a problem so close to current research ac- The abc conjecture. During the 1980s a con- tivity. jectured diophantine inequality, the “abc conjec- The Beal Conjecture ture”, with many applications was formulated by Masser, Oesterle, and Szpiro. A survey of this idea Let A, B, C, x, y, and z be positive integers with has been given by Lang [5] and an elementary dis- x y z x, y, z > 2. If A + B = C , then A, B, and C have cussion by Goldfeld [4]. This inequality can be a common factor. stated in very simple terms, and it can be applied Or, slightly restated: to Beal’s problem. To state the abc conjecture, let The equation Ax + By = Cz has no solution in us say that if a, b, and c are positive integers, then positive integers A, B, C, x, y, and z with x, y, and N(a,b,c) denotes the square free part of the prod- z at least 3 and A, B, and C coprime. uct abc. In other words, N(a,b,c) is the product of It turns out that very similar conjectures have the prime divisors of a, b, and c with each divisor been made over the years. In fact, Brun in his 1914 counted only once. The abc conjecture can be for- paper states several similar problems [1]. How- mulated as follows: ever, it is very timely that this problem be raised now, since Fermat’s Last Theorem has just recently For each > 0, there is a constant µ>1 such that been proved (or re-proved) by Wiles [6]. Some of if a and b are relatively prime (or coprime) and c the significant advances made on some problems = a+b, then closely related to the prize problem by Darmon and max( a , b , c ) µN(a, b, c)1+ . Granville [2] are indicated below. Darmon and | | | | | | ≤ Granville in their article also discuss some related Now let us show that if the abc conjecture holds, conjectures along this line and provide many rel- then there are no solutions to the prize problem evant references. when the exponents are large enough. Let k = log µ/log 2 + (3 + 3 ). Let min(x, y, z) > R. Daniel Mauldin is Regents Professor of mathematics at the University of North Texas, Denton, TX. His e-mail ad- k. Assume A, B, and C are positive integers with dresses are [email protected] and mauldin@ A and B relatively prime and such that dynamics.math.unt.edu. Ax + By = Cz . Setting a = Ax and b = By, we have
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z c = a + b = C . From the abc conjecture and the 657, 92623 + 153122832 = 1137 , 438 + 962223 = fact that N(Ax,By ,Cz) ABC, we have 300429072 , 338 + 15490342 = 156133 . The last ≤ max(Ax,By ,Cz) µ(ABC)1+ . five big solutions were found by Beukers and Zagier. ≤ Recently Darmon and Merel have shown that If max(A, B, C)=A, then we would have there are no coprime solutions with exponents x 3+3 (x, x, 3) with x 3 [3]. A µA ≥ ≤ Acknowledgment. Since I am not an expert in or log µ this field, I would like to thank Andrew Granville x +3+3 k, ≤ log A ≤ and Richard Guy for their expert help in prepar- ing this note. which is not the case. A similar argument for the other two possibilities for the maximum shows that References our original assumption is impossible. [1] V. Brun, Über hypothesenbildung, Arc. Math. Naturv- Next let us give an explicit version of the abc con- idenskab 34 (1914), 1–14. [2] H. Darmon and A. Granville, On the equations jecture: If a and b are coprime positive integers and m p q r c = a+b, then c (N(a, b, c))2. Let us see what this z = F(x, y) and Ax + By = cZ , Bull. London ≤ Math. Soc. 27 (1995), 513–543. implies for the prize problem. Suppose x y z x [3] H. Darmon and L. Merel, Winding quotients and A + B = C , with x y z. Again, since A and some variants of Fermat’s Last Theorem, preprint. y ≤ ≤ B are coprime, [4] D. Goldfeld, Beyond the Last Theorem, Math Hori- Cz (N(AxBy Cz))2 (ABC)2
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