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Rubel's Problem: from Hayman's List to the Chabauty Method

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Rubel's Problem: from Hayman's List to the Chabauty Method i “NLMS_492” — 2020/12/21 — 10:40 — page 24 — #24 i i i 24 FEATURES Rubel’s Problem: from Hayman’s List to the Chabauty Method EDWARD CRANE AND GENE S. KOPP Walter Hayman’s list Research Problems in Function Theory has been updated for its 50th anniversary. Problem 4.27, posed by Lee Rubel, was related to the famous Prouhet-Tarry-Escott problem. Some modern tools of computational number theory can be used to nd explicit counterexamples. Walter Hayman’s list a problem in function theory but rather a problem in number theory. It was posed by the American Research Problems in Function Theory [9] is a analyst Lee Rubel (1927–1995), whose work spanned collection of problems collated by Walter Hayman in dierential equations, approximation theory, and the 1967 and fondly known as Hayman’s list. It gave a big theory of analog computing. Among many other impetus to the development of geometric function surprising results, Rubel showed that there is a single theory, the part of complex analysis concerned with entire function whose derivatives are dense in the the geometric properties of analytic functions. space of entire functions with respect to the topology of locally uniform convergence. Here is problem 4.27, Walter loved to as it appears in [9]: encourage younger mathematicians. He and his wife Margaret 4.27 (Lee Rubel). Let f ¹xº be a real polynomial founded the British of degree n in the real variable x such that Mathematical Olympiad, f ¹xº = 0 has n distinct (real) rational roots. and the mathematics Does there necessarily exist a (real) non-zero genealogy website lists number t such that f ¹xº − t = 0 has n distinct his 20 PhD students (real) rational roots? and 133 mathematical (I can prove this for n = 1,2,3.) descendants. His own research was You might like to try the case n = 3 for yourself. groundbreaking, particul- Walter Hayman. arly his work on the Photo credit: MFO The phrasing suggests Rubel hoped for an armative Bieberbach conjecture answer, extrapolating from the low-degree cases. and in the area of Nevanlinna theory, which relates the growth and covering properties of meromorphic In a straw poll of our local number theorists at the functions. He was awarded the LMS De Morgan medal University of Bristol, all had the opposite intuition to in 1995. Fifty years after the publication of the list, Rubel’s. We’ll give a calculation-free proof below that ≥ Walter Hayman teamed up with Eleanor Lingham the answer is no, at least for even degrees n 8. (the editor of this newsletter) to gather the state What is more interesting is to try to exhibit explicit ≥ of the art on each of the problems. The resulting counterexamples. For each degree n 4 we pose Fiftieth Anniversary Edition [10] was published in 2019. the following challenge: We were sorry to learn that Walter Hayman passed away on January 1st 2020. Problem Rubel¹nº: Exhibit f 2 Q»x¼ of degree n such that f − t has n distinct rational roots if and only if t = 0. Rubel’s problem If f is a polynomial of degree n, t < 0, and both The fourth chapter of Hayman’s list is about f and f − t have n distinct rational roots, then we polynomials. Among the problems there, number can rescale the roots of f and f − t to obtain an 4.27 stands out for the strange reason that it is not ideal solution of the Prouhet-Tarry-Escott problem (see i i i i i “NLMS_492” — 2020/12/21 — 10:40 — page 25 — #25 i i i FEATURES 25 inset). Those are hard to nd, but this doesn’t mean rst case f − t cannot have two distinct real roots, it is easy to prove that any particular polynomial f and in the second it cannot have two distinct 2-adic is a solution of Rubel¹nº. roots. Ruling out repeated roots stops Rubel’s problem from Prouhet–Tarry–Escott problem being trivial. For example f = x ¹x − 1º2 ¹x ¸ 1º2 has a local minimum at 1 and a local maximum at −1, so Given k,n 2 N, nd two distinct sets of f − t has at most three real roots if t < 0. integers A and B, both of size n, such that The constraint that f has n distinct rational roots Õ i Õ i makes the problem a lot more fun. It implies that a = b for i = 1,...,k. for K = R or K = Q , every suciently small a 2A b 2B p perturbation f˜ of f still factors completely over K . A simple appplication of the pigeonhole This is because each root of f can be perturbed to principle shows that solutions exist whenever a nearby root of f˜ in K . This is shown in the p-adic n ¡ k ¹k ¸1º/2. A solution is called ideal when case by Hensel’s lemma. So we cannot hope to solve n = k ¸ 1. In that case we have Rubel¹nº by nding a local obstruction. Ö Ö ¹x − aº − ¹x − bº = c a 2A b 2B for some non-zero constant c. The largest k A non-constructive solution of Rubel’s problem for which an ideal solution is known is 11. This solution was found in 1999 by Nuutti Kuosa, We will answer Rubel’s problem by showing that for Jean-Charles Meyrignac and Chen Shuwen: each n ≥ 4 , there exists a solution to Rubel¹2nº. A = f±35,±47,±94,±121,±146,±148g, Choose rationals 0 < a1 < ··· < an such that the polynomial f = În ¹x −a2º has n−1 distinct critical B = f±22,±61,±86,±127,±140,±151g . i=1 i values. This condition holds generically in An ¹Qº. For this example, we have Case 1: f is a solution of Rubel¹nº. 2 c = 67440294559676054016000 Then F := f ¹x º is a solution of Rubel¹2nº, since it has the 2n distinct roots ±a ,...,±a . However, if = 212.39.53.72.112.132.17.19.23.29.31 . 1 n F −t has 2n distinct rational roots for some nonzero − If you can prove that there is no ideal t, then f t must have n distinct rational roots, solution of Prouhet-Tarry-Escott with k ≥ 12, contrary to the hypothesis. then you will also have proved that every Case 2: f is not a solution of Rubel¹nº. ≥ polynomial of degree n 13 with n distinct Dene G ¹x,wº = ¹f ¹xº − f ¹wºº/¹x − wº. Thinking Rubel¹ º rational roots is a solution of n . See of G as a polynomial in x, its coecients are Borwein [5] for more information about the polynomials in w, so its discriminant h belongs to Prouhet-Tarry-Escott problem. Q»w¼. The roots of h in C are the n − 2 non-critical preimages under f of each of the n − 1 critical values of f . Each of these is a simple root of h. So h has degree d = ¹n − 1º¹n − 2º and has no No local obstructions repeated complex root. In particular, d ≥ 6, and the discriminant of h is non-zero. It is easy to write down an f 2 Q»x¼ of degree n ≥ 3 for which there is no t 2 Q such that f − t If f − t has rational roots q1 < ··· < qn, we have 2n has n distinct roots. For instance, this is true for rational points on the hyperelliptic curve Y 2 = h¹X º, f = x3 ¸ x, because it maps R to R injectively. given by For another example, f = x3 − 2x is an injective mapping from Q to Q. If x < y, but f ¹xº = f ¹yº, then x2 ¸ xy ¸ y 2 = 2, but for x,y 2 Q, not both © Ö ª ­q , ± ¹q − q º® , k = 1,...,n . zero, the valuation of x2 ¸ xy ¸ y 2 at 2 is always even. ­ k i j ® ­ i < j ® These are both examples of local obstructions. In the « i,j <k ¬ i i i i i “NLMS_492” — 2020/12/21 — 10:40 — page 26 — #26 i i i 26 FEATURES Since disc¹hº < 0, this is a smooth ane curve over Q of genus g = d¹d − 2º/2e = ¹n − 1º¹n − 2º/2 − 1. Faltings’ theorem Although its closure in the projective plane is not smooth, it can be embedded in a smooth curve In 1922 Louis Mordell proved the seminal result in projective 3-space, by a map that takes rational that the abelian group of rational points on points to rational points. Faltings’ theorem says that any elliptic curve dened over Q is nitely any smooth projective curve of genus at least 2 has generated. At the end of same paper, he only nitely many rational points (see inset). It follows conjectured a restricted version of what that the curve Y 2 = h¹X º has only nitely many became known as the Mordell conjecture, that rational points. any smooth algebraic curve dened over Q with genus at least 2 has only nitely many We deduce that there are only nitely many rational rational points. − values of t for which f t has n distinct rational The Mordell conjecture was proved by Gerd < ··· < roots. Let t1 tm be the complete sorted list Faltings in 1983, and he won a Fields Medal of these rational values. Since f is not a solution of in 1986 for this work. A few years later, Rubel¹ º ≥ n , m 2. Consider the degree 2n polynomial Paul Vojta gave a very dierent proof using Diophantine approximation and Arakelov F := ¹f − t1º¹f − t2º .
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