This page intentionhally left blank A PRIMER OF ANALYTIC NUMBER THEORY This undergraduate introduction to analytic number theory develops analytic skills in the course of a study of ancient questions on polygonal numbers, perfect numbers, and amicable pairs The question of how the primes are distributed among all integers is central in analytic number theory This distribution is determined by the Riemann zeta function, and Riemann’s work shows how it is connected to the zeros of his function and the significance of the Riemann Hypothesis Starting from a traditional calculus course and assuming no complex analysis, the author develops the basic ideas of elementary number theory The text is supplemented by a series of exercises to further develop the concepts and includes brief sketches of more advanced ideas, to present contemporary research problems at a level suitable for undergraduates In addition to proofs, both rigorous and heuristic, the book includes extensive graphics and tables to make analytic concepts as concrete as possible Jeffrey Stopple is Professor of Mathematics at the University of California, Santa Barbara A PRIMER OF ANALYTIC NUMBER THEORY From Pythagoras to Riemann JEFFREY STOPPLE University of California, Santa Barbara    Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge  , United Kingdom Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521813099 © Jeffrey Stopple 2003 This book is in copyright Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press First published in print format 2003 - isbn-13 978-0-511-07316-8 eBook (EBL) -  eBook (EBL) isbn-10 0-511-07316-X - isbn-13 978-0-521-81309-9 hardback - isbn-10 0-521-81309-3 hardback - isbn-13 978-0-521-01253-9 paperback -  paperback isbn-10 0-521-01253-8 Cambridge University Press has no responsibility for the persistence or accuracy of s for external or third-party internet websites referred to in this book, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate This book is dedicated to all the former students who let me practice on them Contents Preface Chapter Chapter Chapter Chapter Interlude Chapter Interlude Chapter Chapter Interlude Chapter Chapter Chapter 10 Interlude Chapter 11 Chapter 12 Chapter 13 page ix Sums and Differences Products and Divisibility Order of Magnitude Averages Calculus Primes Series Basel Problem Euler’s Product Complex Numbers The Riemann Zeta Function Symmetry Explicit Formula Modular Arithmetic Pell’s Equation Elliptic Curves Analytic Theory of Algebraic Numbers Solutions Bibliography Index 24 43 64 83 96 111 146 159 187 193 216 229 254 260 274 295 327 375 379 vii Solutions 369 Now, · > 31/2 So, we choose the representative a = −31 + 20 = −11 ≡ 20 mod 31 The theorem says that the number of points is 31 − 11 − = 19 (12.4.1) For small primes p ≤ 13, you can define a function nsmall, which counts the number of points on the curve y = x − 432d modulo p by brute force and ignorance, testing all possible x and y values nsmall[p_, d_] := (pointcount = 1; Do[ Do[ If[Mod[x^3 - 432*d^2, p] == Mod[y^2, p], pointcount = pointcount + 1], {x, 0, p - 1}], {y, 0, p - 1}]; pointcount) The function rep takes a prime p ≡ mod as input and returns the unique L ≡ mod such that p = L + 27M rep[p_] := If[Mod[p, 3] == 1, Do[M = Sqrt[(4*p - L^2)/27]; If[IntegerQ[M], If[Mod[L, 3] == 1, Return[L], Return[-L]] ], {L, 1, Sqrt[4*p]}] ] For primes p > 13, the function nbig, which uses (12.4) and (12.5), will be more efficient nbig[p_, d_] := If[Mod[p, 3] == 2, p + 1, L = rep[p];a = Mod[d^((p - 1)/3)*L, p]; If[a > p/2, a = a - p]; p + a + ] The function bsd takes x and d as input and returns the product over primes p < x of the ratio N p / p for the curve y = x − 432d It is very inefficient if you want many different values of x, because it recomputes previously used values of N p Try to think of another 370 Solutions way to this bsd[x_, d_] := Product[p = Prime[j]; If[p a, Return[{a, b - 2Sign[b]a, a + c - Abs[b]}], If[Abs[b] > c, Return[{a + c - Abs[b], b - 2Sign[b]c, c}], If[a > c, Return[{c, -b, a}], Return[{a, b, c}]]]]) To carry out the complete reduction on a list {a,b,c}, use Mathematica’s FixedPoint function: FixedPoint[red,{a,b,c}] (13.1.8) See Table 13.1 (13.1.9) You might something like classno[d_] := (h = 0; Do[ If[Mod[b, 2] == Mod[d, 2], Do[c = (b^2 - d)/(4a); If[IntegerQ[c], If[b == || b == a || a == c, h = h + 1, 372 Solutions Table S.1 Examples of Representation Numbers 11 13 17 19 23 29 31 37 −159 −163 −164 −424 −427 −431 2 2 0 0 0 0 0 0 0 0 2 2 0 0 2 2 2 0 0 0 0 2 2 0 2 0 h = h + ] ], {a, Max[b, 1], Floor[Sqrt[(b^2 - d)/4]]}] ], {b, 0, Floor[Sqrt[Abs[d]/3]]}]; h) (13.1.10) Modulo · = 20, the only squares are 0, 1, 4, 5, 9, and 16 −23 ≡ −3 ≡ 17 mod 20 is not a square Modulo 12, −23 ≡ is a square And modulo 24, − 23 ≡ ≡ 12 ≡ 52 ≡ 72 ≡ 112 So, has representations (13.1.11) The representation numbers are in Table S.1 Observe that the class numbers of the discriminants are, respectively, 10, 1, and 6, 2, 21 (taken from Table 13.1) The point here is that discriminants with relatively smaller class numbers tend to represent fewer small primes than those with larger class numbers (13.1.12) You might something like repno[n_, d_] := (count = 0; Do[ If[Mod[m^2, 4n] == Mod[d, 4n], count = count + 1], {m, 0, 2n - 1}]; count) Solutions 373 (13.2.1) Take ∞ x 3k+2 x 3k+1 − 3k + 3k + f −3 (x) = k=0 thus f −3 (1) = L(1, ␹ −3 ) Then, ∞ ∞ f −3 (x) = x 3k − x 3k+1 = (1 − x) k=0 x 3k k=0 1−x = = 1−x + x + x2 Completing the square, we can write f −3 (x) = (x + 1/2)2 + 3/4 Then, via u-substitution (or looking in a table of integrals), √ f −3 (x) = √ arctan((2x + 1)/ 3) + C expansion, Because we require that f −3 (0) = from the series √ the usual choice of C = won’t Instead, arctan(1/ 3) = ␲/6 √ forces us to pick C = −␲/(3 3) Then, √ √ f −3 (x) = √ arctan((2x + 1)/ 3) − ␲/(3 3) √ has f −3 (0) = 0, and f −3 (1) = ␲/(3 3) = L(1, ␹ −3 ) (13.3.1) Consider, first, the terms in the sum with n < |d| With the hypothesis about ␴, we still have that n −␴ < e/n, as before Then, we can estimate ␹ d (n)n −␴ < n