LTCC basic course: Analytic Number Theory. Andrew Granville

Transcription

1 LTCC basic course: Analytic Number Theory Andrew Granville

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3 3 Preface Riemann s seminal 860 memoir showed how questions on the distribution of prime numbers are more-or-less equivalent to questions on the distribution of zeros of the Riemann zeta function. This was the starting point for the beautiful theory which is at the heart of analytic number theory, which we introduce in this minicourse. Selberg showed how sieve bounds can be obtained by optimizing values over a wide class of combinatorial objects, making them a very flexible tool. We will also see how such methods can be used to help understand primes in short intervals. Key phrases: By a conjecture we mean a proposition that has not been proven, but which is favoured by some compelling evidence. An open question is a problem where the evidence is not especially convincing in either direction. The distinction between conjecture and open question is somewhat subjective and may change over time. We trust, though, that there will be no similar ambiguity concerning the theorems! The exercises: In order to really learn the subject the keen student should try to fully answer the exercises. We have marked several with if they are difficult, and occasionally if extremely difficult. Some exercises are embedded in the text and need to be completed to fully understand the text; there are many other exercises at the end of each chapter. Textbooks: There is a bibliography at the end of these notes of worthy books. Davenport s [Da] is, for us, the greatest introduction to the key ideas of the subject. Davenport keeps the focus narrow and gives a masterful, if terse, explanation of the proof of the prime number theorem and related issues. Similarly Bombieri s [Bo] is focussed on all there is to know about the large sieve and applications, and Crandall and Pomerance [CP] on computational issues. Titchmarsh [Ti] long ago write the definitive treatise on the Riemann zeta function, and Edwards [Edw] a wonderful historical development. Narkiewicz s historical treatise [Nar] on the distribution of prime numbers is a beautiful and informative read - the reader is asked to find more about who did what from that source. Ribenboim s [Rib] is source of joy to professional and amateur prime seekers alike. There are two recent tracts by the modern masters of the subject, [IK] and [MV], covering a wide cross-section of topics in analytic number theory; these are much broader than what we have attempted here and are recommended to someone who wishes to gain broad expertise in our subject. These notes have been cobbled together out of some of the course notes from two full length courses I have taught in the past, so please excuse any repetitions. For example, whether or not there are odd perfect numbers was an open question; however it is now known that if one exists it is > 0 300, which entitles us to conjecture that there are none.

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5 Contents Chapter. Proofs that there are infinitely many primes, without analysis.. Euclid and beyond.2. Various other non-analytic proofs 3.3. Primes in certain arithmetic progressions 4.4. Prime divisors of polynomials 7.5. A diversion: Dynamical systems proofs 9.6. Formulas for primes.7. Special types of primes 3 Chapter 2. Infinitely many primes, with analysis First Counting Proofs Euler s proof and the Riemann zeta-function Upper bound on the number of primes up to x An explicit lower bound on the sum of reciprocals of the primes Another explicit lower bound Binomial coefficients: First bounds Bertrand s postulate Big Oh and other notation How many prime factors does a typical integer have? How many primes are there up to x? The Gauss-Cramèr heuristic Smoothing out the prime counting function An attempt to prove the prime number theorem 33 Chapter 3. The prime number theorem Partial Summation Chebyshev s elementary estimates Multiplicative functions and Dirichlet series The average value of the divisor function and Dirichlet s hyperbola method The prime number theorem and the Möbius function: proof of Theorem Selberg s formula 43 Chapter 4. Introduction to Sieve Theory The sieve of Eratosthenes A first combinatorial Eratosthenes Integers without large prime factors: Smooth or friable numbers An upper bound on the number of smooth numbers Large gaps between primes 56 5

6 6 CONTENTS 4.6. Exercises 59 Chapter 5. Selberg s sieve applied to an arithmetic progression Selberg s sieve The Fundamental Lemma of Sieve Theory A stronger Brun-Titchmarsh Theorem Additional exercises 67 Chapter 6. Primer on analysis Inequalities Fourier series Poisson summation The order of a function Complex analysis Perron s formula and its variants Analytic continuation The Gamma function 80 Chapter 7. Riemann s plan for proving the prime number theorem A method to accurately estimate the number of primes Linking number theory and complex analysis The functional equation The zeros of the Riemann zeta-function Counting primes Riemann s revolutionary formula 88 Chapter 8. The fundamental properties of ζ(s) Representations of ζ(s) A functional equation A functional equation for the Riemann zeta function Properties of ξ(s) A zero-free region for ζ(s) Approximations to ζ (s)/ζ(s) On the number of zeros of ζ(s) 96 Chapter 9. The proof of the Prime Number Theorem The explicit formula Proving the prime number theorem Assuming the Riemann Hypothesis Primes in short intervals, via zeta functions 02 Chapter 0. The prime number theorem for arithmetic progressions Uniformity in arithmetic progressions Breaking the x-barrier Improvements to the Brun-Titchmarsh Theorem To be put later 07 Chapter. Gaps between primes, I 09.. The Gauss-Cramér heuristic predictions about primes in short intervals The prime k-tuplets conjecture 0

7 CONTENTS 7.3. A quantitative prime k-tuplets conjecture.4. Densely packed primes in short intervals 3 Chapter 2. Goldston-Pintz-Yıldırım s argument The set up Evaluating the sums over n Evaluating the sums using Perron s formula Evaluating the sums using Fourier analysis Evaluating the sums using Selberg s combinatorial approach, I Sums of multiplicative functions Selberg s combinatorial approach, II Finding a positive difference; the proof of Theorem Chapter 3. Goldston-Pintz-Yıldırım in higher dimensional analysis The set up The combinatorics Sums of multiplicative functions The combinatorics, II Finding a positive difference A special case Maynard s F s, and gaps between primes F as a product of one dimensional functions The optimal choice Tao s upper bound on ρ(f ) 32

8 CHAPTER Proofs that there are infinitely many primes, without analysis Introduction The fundamental theorem of arithmetic states that every positive integer may be factored into a product of primes in a unique way. Moreover any finite product of prime numbers equals some positive integer. Therefore there is a -to- correspondence between positive integers and finite products of primes. This means that we can understand positive integers by decomposing them into their prime factors and studying these, just as we can understand molecules by studying the atoms of which they are composed. Once one begins to determine which integers are primes and which are not, one observes that there are many of them, though as we go further out, they seem to become sparser, a decreasing proportion of the positive integers. It is also tempting to look for patterns amongst the primes: Can we find a formula that describes all of the primes? Or at least some of them? Can we prove that there are infinitely many? And, if so, can we quickly determine how many there are up to a given point? Or at least give a good estimate? How are they distributed? Regularly, or unevenly? These questions motivate this mini-course. This first, preliminary chapter investigates what is known about the infinitude of prime numbers using only imaginative elementary arguments. This is not necessarily part of this mini-course but the student might enjoy playing with some of the concepts here... Euclid and beyond Ancient Greek mathematicians knew that there are infinitely many primes. Their beautiful proof by contradiction goes as follows: Suppose that there are only finitely many primes, say k of them, which we will denote by p = 2 < p 2 = 3 <... < p k. What are the prime factors of p p 2... p k +? Since this number is > it must have a prime factor (by the Fundamental Theorem of Arithmetic), and this must be some p j, for some j, j k (since, by assumption, p, p 2,..., p k is a list of all of the primes). But then p j divides both p p 2... p k and p p 2... p k +, and hence p j divides their difference,, which is impossible. 2 A positive integer n is prime if it is divisible only by and itself. Note that is not considered to be prime. An integer n is composite if it is divisible by two (not necessarily distinct) primes; note that this includes negative integers as well as 0. Neither word applies to or, or to p for any prime p. 2 Euclid gives this proof in Book 9 Proposition 20 of his Elements, assuming that there are just three primes. The reader is evidently meant to infer that the same proof works no matter how

9 2. PROOFS THAT THERE ARE INFINITELY MANY PRIMES, WITHOUT ANALYSIS This proof has the disadvantage that it does not exhibit infinitely many primes, but only shows that it is impossible that there are finitely many. It is possible to more-or-less correct this deficiency by defining the sequence a = 2, a 2 = 3 and then a n = a a 2... a n + for each n 2. Let p n be some prime divisor of a n, for each n. We claim that the p n are all distinct so we have an infinite sequence of distinct primes. We know that these primes are distinct since if m < n then (p m, p n ) divides (a m, a n ) = (a m, ) =, since a n (mod a m ) by our construction, which implies that (p m, p n ) = ; that is, p m and p n must be different. Fermat conjectured that the integers b n = 2 2n + are primes for all n 0. His claim starts off correct: 3, 5, 7, 257, are all prime, but is false for b 5 = , as Euler famously noted. Nonetheless we can prove that if p n is some prime divisor of b n for each n 0 then p 0, p,... is an infinite sequence of distinct primes, in this case because b n = b b 2... b n + 2 for each n, and so (b m, b n ) = (b m, 2) = for all m < n, since b n 2 (mod b m ). 3 A related proof involves letting q n be the smallest prime factor of n! +. We see that there cannot be a largest prime n since q n is always larger. Exercises Exercise... (Other proofs inspired by Euclid s proof) Suppose that we are given distinct primes p, p 2,..., p k, and let m denote their product, p p 2... p k. Show that each of the following integers N has a prime factor which is not equal to any prime in this list, and so deduce that there are infinitely many primes. a) For any r, r k, let n = p p 2... p r and N := n + m/n. b) (Reminiscent of proofs of the Chinese Remainder theorem): Let N := k i= m/p i. Exercise..2. In this question we give an algorithm that determines all of the primes. Let p < p 2 <... < p k be the distinct primes up to p k, and let m = p p 2... p k. a) Prove that if N = k i= N i, where the set of prime factors of N i is precisely {p, p 2,..., p k } \ {p i } for each i, then (N, m) =. b) Use the Chinese Remainder theorem to show that, for any given integer N with (N, m) =, there exist integers M i [, m] for which N k i= M i (mod m), where the prime factors of M i are exactly {p,..., p k } \ {p i }. c) Prove that if N k i= m/p i with (N, m) = then there exist integers N i as in part a, with each N i m. (Hint: Select N i = M i or M i m in part b). large a finite number of primes we assume there to be. The notation of those times was far less flexible than that of today, so that the astute reader necessarily had to deduce the full content of the statement of a theorem, or of a proof, from what was written, and could not necessarily learn all that was meant from what was actually written. Even Renaissance thinkers like Fermat and Descartes faced these notational limitations and deplored those who could not navigate these difficulties adroitly. 3 This proof appeared in a letter from Goldbach to Euler in July 730.

10 .2. VARIOUS OTHER NON-ANALYTIC PROOFS 3 d) The discussion in part c) gives us a way to determine, with proof, each prime N between x and x, given the primes up to x. 4 Find all the primes between 0 and 00 in this way, finding a representation of N as in part c). Exercise..3. Use the sequence of Fermat numbers to prove that, for each integer k, there are infinitely many primes (mod 2 k ). Exercise..4. Suppose that p = 2 < p 2 = 3 <... is the sequence of prime numbers. Use the fact that every Fermat number has a distinct prime divisor to prove that p n 2 2n +. What can one deduce about the number of primes up to x? Exercise..5. (Open questions). Are there infinitely many primes of the form a n? If p = 2 < p 2 = 3 <... is the sequence of prime numbers then are there infinitely many n for which p p 2... p n + is prime? For which p p 2... p n is prime? Let us determine an infinite sequence of primes by starting with prime q, and then letting q n be some prime divisor of q q 2... q n + for each n 2. Can the sequence q, q 2,... be a re-arrangement of the set of all primes? What if q n is the smallest prime divisor of q q 2... q n +?.2. Various other non-analytic proofs The Mersenne numbers take the form M n = 2 n. Suppose that p is prime and q is a prime dividing 2 p. The order of 2 mod q, must be divisible by p, and must divide q, hence p q. Thus there cannot be a largest prime p, since any prime factor q of M p is larger, and so there are infinitely many primes. Furstenberg gave an extraordinary proof using point set topology: Define a topology on the set of integers Z in which a set S is open if it is empty or if for every a S there is an arithmetic progression Z a,q := {a + nq : n Z} which is a subset of S, for some integer q 0. Evidently each Z a,q is open, and it is also closed since Z a,q = Z \ b: 0 b q, b a Z b,q. If there are only finitely many primes p then A = p Z 0,p is also closed, and so Z\A = {, } is open, but this is obviously false since A does not contain any arithmetic progression Z,q. Hence there are infinitely many primes. Exercises Exercise.2.. a) Prove that if f(t) Z[t] and r, s Z then r s divides f(r) f(s). b) Prove that if 2 n is prime then n is prime. c) Prove that if 2 n + is prime then n is either 0 or a power of 2. Exercise.2.2. Prove that if prime q divides 2 p, where p is prime, then 2 has order p mod q. Deduce that (2 p, 2 l ) = for all primes p and l. Exercise.2.3. (Open questions). Prove that there are infinitely many Mersenne primes, 2 p. (This is equivalent to asking whether there are infinitely many even perfect numbers, since n is an even perfect number if and only if it is of the form 4 The sieve of Eratosthenes also allows us to find the primes in ( x, x], given the primes up to x, but subsequently, the only way to prove to a skeptic that what you claim are the primes, are actually prime, would be for that person to redo the calculation. On the other hand the proof here provides easily verified proofs of primality.

11 4. PROOFS THAT THERE ARE INFINITELY MANY PRIMES, WITHOUT ANALYSIS 2 p (2 p ) with 2 p prime. 5 ) Prove that there are infinitely many Fermat primes, 2 2n +. Prove that there are integers n for which 2 2n + is composite. 6 Exercise.2.4. (Open). Prove that there are infinitely primes p for which 2 p is composite. (This is a conjecture, rather than an open question, because one can prove, and you should prove, that if p 3 (mod 4) and q = 2p + is also prime then q divides 2 p, so that 2 p is composite.) Exercise.2.5. We know that 2 2, 2 22, and are all prime. Is it true that the next term in the sequence is prime? How about all the terms of the sequence?.3. Primes in certain arithmetic progressions Any prime a (mod m) is divisible by (a, m), and so if (a, m) > there cannot be more than one prime a (mod m). Thus all but finitely many primes are distributed among the φ(m) arithmetic progressions a (mod m) with (a, m) =. We will eventually prove that all such arithmetic progressions contain infinitely many primes, and that the primes are roughly equally distributed amongst these φ(m) arithmetic progressions (mod m). For now we will prove results along these lines using only elementary methods. We begin by proving that there are infinitely many primes in each of the two feasible residue classes mod 3. There are infinitely many primes (mod 3), for if there are only finitely many, say p, p 2,..., p k, then N = 3p p 2... p k must have a prime factor q (mod 3), else N (mod 3), and so q divides both N and N + and hence their difference, which is impossible. A similar proof works for primes (mod 4), and indeed for much more general sets of primes see exercise.3.. How about the primes (mod 3)? Suppose that prime q divides a 2 + a + for some integer a. Then a 3 (mod q) (since a 2 + a + divides a 3 ) and so a has order or 3 mod q. In the first case a (mod q) in which case 0 a 2 + a + 3 (mod q) so that q = 3. Otherwise 3 divides q ; that is, q (mod 3). Therefore if there are only finitely many primes (mod 3), say p, p 2,..., p k, let a = 3p p 2... p k so that any prime divisor q of N = a 2 + a + is different from 3 and the p i s, and (mod 3), contradicting our assumption. To generalize this argument to primes (mod m), we replace the polynomial a 2 + a + by a polynomial that recognizes when a has order m. Evidently this must be a divisor of the polynomial a m, indeed a m divided through by all of the factors corresponding to orders which are proper divisors of m. So we define the cyclotomic polynomials φ n (t) Z[t], inductively, by the requirement t m = d m φ d(t) for all m, with each φ d (t) monic. The roots of t m are the distinct mth roots of unity, so our definition implies that the roots of φ m (t) are exactly the primitive mth roots of unity, that is those α C for which α m = 5 It is known that 2 p is prime for p = 2, 3, 5, 7, 3, 7, 9,..., , a total of 48 values as of late 205. There is a long history of the search for Mersenne primes, from the first serious computers to the first great distributed computing project, GIMPS (The Great Internet Mersenne Prime Search). 6 There are no primes known of the form 2 2 n + other than for n 4, and we know 2 2n + is composite for 5 n 30 and many other n besides. It is always a significant moment when a Fermat number is factored for the first time.

12 .3. PRIMES IN CERTAIN ARITHMETIC PROGRESSIONS 5 but α r for all r, r m. These can be written more explicitly as exp(2iπj/m) with (j, m) = so that φ m (t) has degree φ(m). A proof of the infinitude of primes (mod m), along these lines, is given by exercise.3.3. Exercises Exercise.3.. Let G be a proper subgroup of the multiplicative group of elements mod m (that is, the residue classes coprime to m). a) Show that if N is an integer with (N, m) = where N (mod m) is not an element of G, then N has a prime factor which is not an element of G. b) Given any finite set of primes p, p 2,..., p k which do not divide m, and a residue class b (mod m), show that there exists an integer N such that N b (mod m) and (N, p p 2... p k ) =. c) We will now prove that there are infinitely many primes whose residues mod m do not belong to the subgroup G. For if there are only finitely many, say p, p 2,..., p k, select b appropriately and then N as in part b), and finally use part a) to deduce the desired conclusion. d) Deduce that there are infinitely many primes in at least two of the arithmetic progressions 3 mod 8, 5 mod 8, and 7 mod 8. Exercise.3.2. a) Prove that the set of primitive mth roots of unity can be written in the form exp(2iπj/m) with (j, m) =. b) Prove that φ n (x) = d n (xd ) µ(n/d), where { ( ) k if n = p p 2... p k is squarefree; µ(n) := 0 if n is divisible by p 2 for some prime p. (Hint: Use the Mobius inversion formula). Exercise.3.3. The prime p is a primitive prime factor of a m if p divides a m but p does not divide a n for any n, n m. a) Show that every primitive prime factor of a m divides φ m (a). b) Prove that gcd(a m, a n ) = a (m,n). Deduce that if prime p divides φ m (a) but is not a primitive prime factor of a m then p divides φ d (a) for some proper divisor, d, of m. c) Prove that if d divides m and prime p divides gcd(φ m (a), φ d (a)) then m/d is a power of p. (Hint: Suppose that prime q divides m/d. Show that φ m (a) divides (a m )/(a m/q ), and φ d (a) divides a m/q. By considering (a m )/(a m/q ) (mod a m/q ), establish that gcd(φ m (a), φ d (a)) divides q.) d) Prove that φ m (0) = for all m >. Deduce that if p is a prime factor of φ m (ma) for any integer a then p (mod m). e) Use part d) to deduce that there are infinitely primes (mod m). In the next exercise we more precisely understand the primes that divide φ n (a) as n varies, for fixed a with a >. Exercise.3.4. a) Show that if d is the smallest integer for which p divides φ d (a) then d divides p.

13 6. PROOFS THAT THERE ARE INFINITELY MANY PRIMES, WITHOUT ANALYSIS b) Use part a) and.3.3c) to show that if p divides φ m (a) but is not a primitive prime factor of a m then m = p r d for some r and some d < p. Deduce that there can be at most one such prime p. c) Prove that if odd prime p divides φ m/p (a) then (a m )/(a m/p ) p (mod p 2 ). Deduce, using part b), that if φ m (a) has no primitive prime factor then either φ m (a) = p, a prime dividing m, or m = p = 2 and a = ±2 k, for some k. d) Our goal now is to show that if m 3 and a 2 then φ m (a) > p where p is the largest prime factor of m, other than the case φ 3 ( 2) = φ 6 (2) = 3. Use.3.2b) to show that if a 2 then φ m (a) / a φ(m) ( 2 ) d. d A calculation reveals that this equals ; in particular it is > /4, and hence φ m (a) > 2 φ(m) 2. Next prove that 2 φ(m) 2 m for all m > 8; and then that 2 φ(m) 2 p, except when m =, 2, 3, 4, 6, 0. For m = 3, 4, 6, 0, show that φ m (a) > p for a 4. Verify the remaining cases to justify our claim. e) Deduce that φ m (a) has a primitive prime factor whenever a >, except the cases φ 3 ( 2) = φ 6 (2) = 3, φ 2 (±2 k ) and φ (a) with a 2. f) Deduce that there is a prime (mod m) which is 2 m. (Note that this cannot be improved when m = 3.) In the next exercise we see further applications and developments of the theory that we have developed in exercise.3.4. Exercise.3.5. a) Writing a d = + kp r where p k, prove that if p r 2 then a pd is divisible by p r+ but not by p r+2. Deduce that if p divides φ m (a) but is not a primitive prime factor of a m then p 2 does not divide φ m (a), except when m = 2 and a 3 (mod 4). b) Deduce that for each integer a with a 2, the integers (a m )/(a ) have a primitive (and thus distinct) prime factor for all integers m, 2, 3 or 6. {x n } n 0 is a Lucas sequence if x 0 = 0, x = and (.3.) x n+2 = bx n+ + cx n for all n 0, for given non-zero, coprime integers b, c; the integers x n = (a n )/(a ) form a Lucas sequence with b = a + and c = a for each n 0. In 93 Carmichael showed that if the discriminant := b 2 + 4c > 0 then x n has a primitive prime factor for each n, 2 or 6 except for F 2 = 44 where F n is the Fibonacci sequence (b = c = ), and for F 2 where F n = ( ) n F n (b =, c = ). It is much more difficult to prove that Lucas sequences with negative discriminant have primitive prime factors. Nonetheless, in 974 Schinzel succeeded in showing that x n has a primitive prime factor once n > n 0, for some sufficiently large n 0, if 0, other than in the periodic case b = ±, c =. Determining the smallest possible value of n 0 has required great efforts culminating in the beautiful work of Bilu, Hanrot and Voutier [BHV] who proved that n 0 = 30 is best possible (indeed if b =, c = 2 then x 5, x 8, x 2, x 3, x 8, x 30 have no primitive prime factors). Exercise.3.6. In exercise.3.3e we saw that there are infinitely many primes (mod 8). Now we prove that there are infinitely many primes in the other arithmetic progressions (mod 8).

14 .4. PRIME DIVISORS OF POLYNOMIALS 7 a) For b = 3, 5 or suppose that there are only finitely many primes b (mod 8), and let n b be their product. Establish a contradiction by considering the prime factors of n 2 b + b. (Hint: Use the law of quadratic reciprocity.) b) Generalize this argument to prime divisors of values of other quadratic polynomials..4. Prime divisors of polynomials In 837 Dirichlet proved that any linear polynomial mt+a Z[t] with (m, a) = takes on infinitely many prime values (which is equivalent to the fact that there are infinitely many primes a (mod m) if (a, m) = ). We wish to generalize this to any irreducible f(t) Z[t], with suitable restrictions. To formulate the restrictions there is one subtlety: the reason we need (m, a) = in the linear case is that (m, a) divides mn+a for every integer n, and in fact (m, a) = gcd{mn+a : n Z}. Hence let us define Content(f) to be the gcd of the integers f(n) as we vary over n Z. It is conjectured that for any irreducible f(t) Z[t] with Content(f) = there are infinitely many integers n for which f(n) is prime. (In fact that for any irreducible f(t) Z[t] there are infinitely many integers n for which f(n)/content(f) is prime.) The polynomial n 2 + n + 4 is prime for n = 0,,..., 39, though composite for n = 40. One can ask whether there are any polynomials f(t) Z[t] such that f(n) is prime for all integers n? The answer is no: We now show that if f(t) Z[t] has degree then f(n) is composite for infinitely many integers n. Since a non-zero polynomial has only finitely many roots, for example (f(t) )f(t)(f(t) + ), thus f(n) 2 for all but finitely many integers n. Select any such n and let m = f(n). Now f(n + km) f(n) 0 (mod m) for any integer k, by exercise.2.a, so that f(n + km) is composite all k except for the finitely many n + km which are roots of (f(t) m)f(t)(f(t) + m). One can use a minor variation to show that if f(t) Z[t] has degree d and there are more than 2d distinct integers n for which f(n) is prime then f(t) is irreducible. To prove this suppose f(t) = g(t)h(t); for each n where f(n) is prime we have that either g(n) = ± or h(n) = ±. Now there are no more than 2 deg(g) roots of (g(t) )(g(t) + ), and no more than 2 deg(h) roots of (h(t) )(h(t) + ), and therefore 2 deg(g) + 2 deg(h) = 2d distinct integers n for which f(n) prime. (The 2d in this result can be improved to d + 2, and this is probably best possible for all d 6. If we ask for f(n) to be prime then the correct bound is d.) We finish this section by proving that for any f(t) Z[t] of degree there are infinitely many distinct primes p for which p divides f(n) for some integer n. We may assume that f(n) 0 for all n Z else we are done. Now suppose that p,..., p k are the only primes which divide values of f and let m = p... p k. Then f(nmf(0)) f(0) (mod mf(0)) for every integer n, by exercise.2.a, so that f(nmf(0))/f(0) (mod m). Therefore f(nmf(0)) has prime divisors other than those dividing m for all n other than the finitely many n which are roots of (f(tmf(0)) f(0))(f(tmf(0)) + f(0)), a contradiction. Exercises on prime values of reducible polynomials Exercise.4.. a) Prove that the gcd of the coefficients of f divides Content(f). b) Give an example to show that Content(f) can be larger than the gcd of the coefficients of f.

15 8. PROOFS THAT THERE ARE INFINITELY MANY PRIMES, WITHOUT ANALYSIS c) If a polynomial f(t) of degree d only takes on integer values then show that there exist integers b 0, b,..., b d for which f(t) = d j=0 b j( t j). d) Prove that Content(f) = gcd 0 j d b j = gcd 0 n d f(n). Exercise.4.2. Use the proof above to show that if f has degree d then there exists an integer n, with n d( + max m d f(m) ), for which f(n) is composite. Can you significantly improve this? Exercise.4.3. Suppose that m, m 2,..., m k are a set of pairwise coprime integers such that m j divides f(a j ) for some integer a j, for each j. Prove that there are infinitely many integers n for which m m 2... m k divides f(n). Exercise.4.4. a) Suppose that g(t) Z[t], where m,..., m k are integers that satisfy g(m i ) =, and n,..., n l are integers that satisfy g(n j ) =. Prove that k i= (m i n j ) divides 2 for each j, and l j= (m i n j ) divides 2 for each i. b) Deduce that if k + l 4 with k, l then k + l = 4 and the four integers are consecutive. In fact g(x) = ±G(±x + a) for some choice of sign ± and some integer a where either G(x) x(x ) or x(x+)(x 2)+ mod (x 2)(x )x(x+). In particular show that if k + l > deg(g) then g(x) = ±G(±x + a) where G(x) is either x(x + )(x 2) + with k = 3, l =, or x(x ) with k = l = 2, or 2x 2 with k = 2, l =, or 2x with k = l =, or x with k = l =. c) Suppose that f(t) Z[t] is reducible of degree d, for which there are more than d integers n with f(n) is prime. Deduce that f(t) has a proper factor g(t), as in part b). d) Suppose that f(t) Z[t] is reducible of degree d, for which there are d + 4 integers n for which f(n) is prime. Deduce that there exist integers a and b such that f(t) = g(t + b) where g(t) = ((t a)(t a ) )(t(t ) ) so that the prime values are g(2) = g(a ) = a 2 3a+, g() = g(a) = a 2 a, g(0) = g(a + ) = a 2 + a and g( ) = g(a + 2) = a 2 + 3a +. (These are prime for a = 4, 5, 0, 55 and 550 and, we conjecture, for infinitely many values of a. This is the same as asking that t 2 t is prime for t = a, a, a + and a + 2.) e) Show that if d > 4 then for any reducible polynomial f(t) Z[t] of degree d, the values of f(n) can be prime no more than d + 2 times. f) We conjecture that r 2 i r i is prime for some infinite sequence of integers r, r 2,.... For any given integer d 2, let d 2 f(t) = (t 2 + t )( + a (t + r i )), a degree d, reducible polynomial, with f( r i ) prime for i d 2. We would like to find four more prime values, for values of n where n 2 + n =. The prime k-tuplets conjecture states that if a t + b,..., a k t + b k Z[t] and Content( k j= (a jt + b j )) = then there are infinitely many integers n for which a n + b,..., a k n + b k are all prime. Use the prime k-tuplets conjecture to deduce that there exist an integer a such that f(t) takes on prime values at d + 2 different integer values for t. If one asks for prime values of f(n)/content(f) then the answer is surprisingly different: The number of prime values that can be taken by f(n)/content(f) where i=

16 .5. A DIVERSION: DYNAMICAL SYSTEMS PROOFS 9 f(t) Z[t] is a reducible polynomial of degree d is something like 2 cd/ log d, for some constant c > 0. 7 See [ChRu] for details. The final exercise in this section leads to the wonderful theorem that a polynomial f(t) with non-negative integer coefficients, can be proved to be irreducible simply by finding one integer n for which f(n) is prime. Exercise.4.5. a) For a given polynomial f(t) = d i=0 a it i Z[t] define H := max 0 i d a i /a d. Show that if f(α) = 0 then α < H +. b) Show that if f is reducible and n is an integer for which f(n) is prime then there exists a root α of f(α) = 0 for which n α. c) Deduce that if n is an integer, with n H + 2, for which f(n) is prime then f(t) is irreducible. d) Show that these results cannot be much improved by considering the example f(t) = t(t + p) for p prime. e) Show the following result. If p = a 0 +a a d 0 d is a prime written in base 0, then the polynomial a 0 + a t a d t d is irreducible. (Hint: For α as in part b), prove that f(α)/α d a d 9 j 2 / α j, as Re(/α) > 0.) Generalize this argument to any base b 3 in place of A diversion: Dynamical systems proofs The prime divisors of a sequence of integers > form an infinite sequence of primes if the integers in the sequence are pairwise coprime. To generalize the constructions from section., we simplify the description of the sequences a n and b n. a n+ = a a 2... a n = (a a 2... a n )a n = (a n )a n, so that a n+ = f(a n ) where f(t) := t 2 t +. (Similarly b n+ = g(b n ) where g(t) := t 2 2t + 2.) How do we explain the fact that a n (mod a m ) for all m < n? Well and, thereafter, a m+ = f(a m ) f(0) = (mod a m ) a n+ = f(a n ) f() = (mod a m ) by induction on n m +. (Similarly b m+ = g(b m ) g(0) = 2 (mod b m ) and b n+ = g(b n ) g(2) = 2 (mod b m ) by induction.) So the only requirements on f seem to be that f(0) = f() =, and f = + t(t ) is the simplest such polynomial, though any polynomial + h(t)t(t ), where h(t) Z[t] has positive leading coefficient, will work. In this section we will determine all polynomials f(t) Z[t] that can be used into this framework. We introduce a little terminology from dynamical systems: A number a is said to be preperiodic for f, if the sequence a, f(a), f(f(a)),... is eventually periodic. 7 It is standard practice in this subject to let log be in base e, whereas the logarithm in base b is denoted log b.

17 0. PROOFS THAT THERE ARE INFINITELY MANY PRIMES, WITHOUT ANALYSIS Proposition.5.. Let f(t) Z[t] have degree >, positive leading coefficient, and f(0) 0. Suppose that 0 is a preperiodic point for f but that 0 is not part of the period, and let l be the least common multiple of the integers in the sequence f(0), f(f(0)), f(f(f(0))),.... If a 0 Z with a n+ = f(a n ) for all n 0, and (a n, l) = for all n 0, then we obtain an infinite sequence of distinct primes by selecting one prime factor from each a n. Proof. Let w 0 = 0 and w n+ = f(w n ) for all n 0, so that a m+ = f(a m ) f(0) = w (mod a m ) and, thereafter, a m+j+ = f (a m+j ) f(w j ) = w j+ (mod a m ) by induction on j. Therefore if m < n then (a m, a n ) = (a m, w n m ) which divides (a m, l), which equals by the hypothesis. The rest of the proof follows as above. To be able to use Proposition.5. we need a good idea of when 0 is a preperiodic point, which turns out to be simpler than one might guess: Proposition.5.2. Suppose that f(t) Z[t] and that the sequence {u n : n 0}, with u 0 Z and u n+ = f(u n ) for all n 0, is periodic. The shortest period has length or 2. Proof. Suppose that u n+p = u n for all n 0. One knows that x y divides f(x) f(y) for any integers x, y; in particular that u n+ u n divides f(u n+ ) f(u n ) = u n+2 u n+. Therefore u u 0 divides u 2 u, which divides u 3 u 2,..., which divides u p u p, which divides u p+ u p = u u 0. That is, we have a sequence of integers that all divide one another and so must all be equal in absolute value. If they are all 0 then p =. If not then they cannot all be equal, say to d 0, else 0 = (u u 0 ) + (u 2 u ) + (u 3 u 2 ) + + (u p u p ) = pd. Therefore there must be two consecutive terms that have opposite signs, yet have the same absolute value, so that u n+2 u n+ = (u n+ u n ) and therefore u n+2 = u n. But then applying f, p n times to both sides, we deduce that u 2 = u 0 and therefore p = 2. This allows us to classify all such polynomials f: Theorem.5.3. Suppose that f(t) Z[t] and that 0 is a preperiodic point for f but is not in the period. The basic possibilities are: a) The period has length, and either f(t) = u with 0 u u..., or, for u = or 2, f(t) = (2/u)t 2 u with 0 u u u... ; b) The period has length 2, and either f(t) = + ut t 2 with 0 u..., or f(t) = + t + t 2 t 3 with Other examples arise by replacing f(t) by f( t), or by adding a polynomial multiple of k i= (t a i) where the a i are the distinct integers in the orbit of 0. Proof by Exercises Exercise.5.. Let f(t) Z[t], and assume that f has a period of length, say f(u) = u. Prove that a) f must be of the form f(t) = u + (t u)g(t) for some g(t) Z[t]. b) If f(v) = u with v u then f(t) = u + (t u)(t v)g(t) for some g(t) Z[t].

18 .6. FORMULAS FOR PRIMES c) If f(w) = v then v w = w u = ± or ±2, = δ say and g(t) = 2/δ + (t w)h(t) for some h(t) Z[t]. d) If f(x) = w then (x u)(x v) divides (w u), which is impossible. Exercise.5.2. Assume f(t) Z[t], and f has a period of length 2, say f(u) = v and f(v) = u. Prove that a) f must be of the form f(t) = v + u t + (t u)(t v)g(t) for some g(t) Z[t]. b) If f(w) = v then w v = ±, so that g(t) = w v + (t w)h(t) for some h(t) Z[t]. c) If f(x) = w then x u = ±. If x u = w v = δ then 2 = (x v)(w v + (x w)h(x)); this implies that x v = δ, 2δ, δ or 2δ each of which can be ruled out. If x u = (w v) then u, x, w, v are consecutive integers (in this order), and h(t) = + (t x)j(t) for some j(t) Z[t]. d) Show that if f(y) = x then y u divides x v = 2, and y v divides x u =, which is impossible. Exercise.5.3. a) Deduce the cases of the theorem by setting x = 0, then w = 0 and then v = 0. Exercise.5.4. The second case in Theorem.5.3a) with u = 2, gives f(t) = t 2 2, with The sequence starting with x 0 = 4 and then let x n+ = f(x n ) for all n 0 is known as the Lucas-Lehmer sequence, who proved that the Mersenne number 2 n is prime if and only if it divides x n 2. Prove that gcd(x m, x n ) = 2 for all m n and deduce that there is an infinite sequence of distinct odd primes {p n : n } where each p n is a prime divisor of x n. Exercise.5.5. Going back to the proof of Proposition.5.2, now suppose that f(x) A[x] where A is the ring of integers of some number field, and that u 0 u... u p u 0 where p is prime, and u 0 A. Prove that the quotient (u i+m u i )/(u m u 0 ) is a unit in A for all i and all m, m p. We have considered iterations of the map n f(n) where f(t) Z[t]. If one allows f(t) Q[t] then it is an open question as to the possible period lengths in the integers. Even the simplest imaginable case, f(x) = x 2 + c, with c Q, is not only open but leads to the magnificent world of dynamical systems (see []). It would be interesting to know what primes divide the numerators of x n, when x n = x 2 n Formulas for primes Let p = 2 < p 2 = 3 <... be the sequence of primes and define α = m p m /0 m2 = Then p m = [0 m2 α] 0 2m [0 (m )2 α]. Is such a magical number α truly interesting? If one could easily describe α (other than by the definition that we gave) then it might provide an easy way to determine the primes. But with its artificial definition it does not seem like it can be used in any practical way. At first one might suppose that, useful or not, α is quite unique; however the following exercise shows that this is not so.

19 2. PROOFS THAT THERE ARE INFINITELY MANY PRIMES, WITHOUT ANALYSIS Exercise.6.. Show that there are uncountably many numbers α (0, ) such that one can read the prime numbers out from the decimal digits, in-between the 0 s. An even less practical formula for primes is derived as follows: Wilson s theorem tells us that n is a prime if and only if n divides (n )! +. We deduce that [ ( ( ))] { (n )! + if n is prime; cos 2π = n 0 if n is composite. Summing this over all integers up to x we have an exact formula for the number of primes up to x. This formula is surely useless, because we have no idea how to sum up such terms. - - In 97 Matijasevič indicated how to construct a polynomial f in many variables, such that the positive values taken by f when each variable is an integer, is precisely the set of primes. 8 One can find many different such polynomials; we will give one below with 04 variables of degree 25, and it is known that one can cut the degree to as low as 5 though for an astronomical cost in terms of the number of variables. No one knows the minimum possible degree, or the minimum possible number of variables. Our polynomials below is in 26 variables, only allowed to take positive integer values, since we can write each variable as a sum of four squares plus. 9 Our polynomial, which is reducible, is k + 2 times (n + l + v y) 2 (2n + p + q + z e) 2 (wz + h + j q) 2 (ai + k + l i) 2 ((gk + 2g + k + )(h + j) + h z) 2 (z + pl(a p) + t(2ap p 2 ) pm) 2 (p + l(a n ) + b(2an + 2a n 2 2n 2) m) 2 ((a 2 )l 2 + m 2 ) 2 (q + y(a p ) + s(2ap + 2a p 2 2p 2) x) 2 ((a 2 )y 2 + x 2 ) 2 (6(k + ) 3 (k + 2)(n + ) 2 + f 2 ) 2 (e 3 (e + 2)(a + ) 2 + o 2 ) 2 (6r 2 y 4 (a 2 ) + u 2 ) 2 (((a + u 2 (u 2 a)) 2 )(n + 4dy) 2 + (x + cu) 2 ) 2. For this to take a positive value, the displayed quantity must equal, and k + 2 be a prime. The displayed quantity equals minus a sum of squares, so each of these squares must equal 0. Evidently there are some ideas hidden in the algebraic expressions given for the squares, and the only way to appreciate this polynomial is to understand its derivation see [JSW]. In the current state of knowledge it seems that this absolutely extraordinary and beautiful polynomial is entirely useless in helping us better understand the distribution of primes! 8 One can also construct such polynomials for the set of Fibonacci numbers, for the set of Fermat primes, for the set of Mersenne primes and the set of even perfect numbers, and indeed any diophantine set. 9 Lagrange proved that n is a non-negative integer if and only if it can be written as the sum of four squares of integers.

20 .7. SPECIAL TYPES OF PRIMES 3.7. Special types of primes In section.2 we asked whether there are infinitely many primes of the form 2 n +, or of the form 2 n, both of which are open questions. One can generalize this by asking whether there are infinitely many primes of the form k 2 n ± or of the form k ± 2 n for given integer k. At first sight this seems like a much more difficult question but Erdős showed, ingeniously, how these questions can be resolved for certain integers k: Let b n = 2 2n + be the Fermat numbers (remember that b 0, b, b 2, b 3, b 4 are prime and b 5 = ), and let k be any positive integer such that k (mod 64b 0 b b 2 b 3 b 4 ) and k (mod ). Now if n (mod 2) then k 2 n = b 0 0 (mod b 0 ); if n 2 (mod 4) then k 2 n = b 0 (mod b ); if n 4 (mod 8) then k 2 n = b 2 0 (mod b 2 ); if n 8 (mod 6) then k 2 n = b 3 0 (mod b 3 ); if n 6 (mod 32) then k 2 n = b 4 0 (mod b 4 ); if n 32 (mod 64) then k 2 n = b 5 0 (mod 64); and if n 0 (mod 64) then k 2 n = 0 (mod ). Every integer n belongs to one of these arithmetic progressions (these are called a covering system of congruences), and so we have exhibited a prime factor of k 2 n + for every integer n. This proof works for all k in some arithmetic progression mod 2 64, and therefore we have shown that for a positive proportion of integers k, there is no prime p such that (p )/k is a power of 2. Exercise.7.. Select k as above. a) Prove that k 2 n + is composite for every integer n 0.. b) Prove that 2 n + k is composite for every integer n 0. (That is, there is no prime p for which p k is a power of 2.) c) Let l be any positive integer for which l k (mod b 6 2). Prove that l 2 n and 2 n l is composite for every integer n 0. Deduce that a positive proportion of odd integers m cannot be written in the form p + 2 n with p prime. Exercise.7.2. (Open question) John Selfridge found the smallest k known for which k 2 n + is always composite: At least one of the primes 3, 5, 7, 3, 9, 37, and 73 divides n +. Is this the smallest such k? Exercise.7.3. Let l be any integer for which l (mod 64b 0 b 2 b 3 b 4 ) and 2 23 (mod ), so that k = l 4 (mod 64b 0 b 2 b 3 b 4 ) and is 2 25 (mod ). If n 2 (mod 4) then, writing n = 4m + 2 we have k 2 n + = 4(l2 m ) 4 + and the polynomial 4t 4 + = (2t 2 + 2t + )(2t 2 2t + ). Prove that k 2 n + is composite for every integer n 0. This last exercise shows that we can have k 2 n + composite for all n for reasons other than having a covering system. Exercise.7.4. Prove that b n 2 = b 0 b... b n cannot be written in the form p + 2 k + 2 l where p is prime and k > l 0. (Hint: Consider divisibility by b r where 2 r is the highest power of 2 dividing k l.)

21 4. PROOFS THAT THERE ARE INFINITELY MANY PRIMES, WITHOUT ANALYSIS Exercise.7.5. Combine the ideas above to show that there are infinitely many integers m which cannot be written in the form p + 2 k + 2 l where p is prime and k l 0.

22 CHAPTER 2 Infinitely many primes, with analysis In this chapter the reader is introduced to some basic counting concepts while proving that there are infinitely many primes, giving upper and lower bounds for the number of primes, or certain functions of the number of primes, and culminating in a discussion of the inter-relation between counting functions. 2.. First Counting Proofs It is instructive to see several proofs involving simple counting arguments, as these will give a flavor of things to come. Suppose that there are only finitely many primes, say p < p 2 <... < p k. Let m = p p 2... p k. If d is squarefree then d divides m, and so the number of integers up to m that are divisible by d is m/d. Therefore, by the inclusion-exclusion principle, the number of positive integers up to m that are not divisible by any prime is m k i= m p i + i<j k m p i p j = m k ) ( pi = i= k (p i ). Now 3 is a prime so this quantity is 3 = 2. However is the only positive integer that is not divisible by any prime, and so we obtain a contradiction. The astute reader will observe that this counting argument holds with or without the assumption: We have simply evaluated Euler s function φ(m), the number of positive integers m that are coprime with m. Another version of this proof is to count more accurately the number of integers up to x: By the fundamental theorem of arithmetic we wish to count the number, N, of k-tuples of non-negative integers (e, e 2,..., e k ) such that p e pe pe k k x. In other words we wish to count the number or lattice points (e, e 2,..., e k ) Z k inside the k-dimensional tetrahedron T (x) defined by e,..., e k 0 with (2..) e log p + e 2 log p e k log p k log x. If one adjoins to each lattice point a unit cube in the positive direction, one obtains a shape S whose volume is exactly the number of lattice points, and which is roughly the same shape as the original tetrahedron itself. The tetrahedron T (x) has volume (2..2) k! k i= log x log p i, so we might expect this to be a good approximation to the number of such lattice points; actually it is a lower bound on the number of lattice points since T (x) S. On the other hand S is contained in the tetrahedron given by adding in each 5 i=

23 6 2. INFINITELY MANY PRIMES, WITH ANALYSIS direction to the diagonal boundary of the tetrahedron, that is T (xp... p k ). Therefore if p < p 2 <... < p k are the only primes then x = N = S T (xp... p k ) for all x; and so if m = p p 2... p k then (2..3) x k! k i= log(mx) log p i, which is certainly false if x is large enough (see exercise 2..). One can easily obtain good enough upper bounds on the number of solutions to (2..) with less work: In any solution to (2..) we must have e j log p j is no more than the left side of (2..) and thus than log x, so that 0 e j (log x)/(log p j ), and therefore N k i= (log xp j)/(log p j ). Exercises Exercise 2... Prove that (2..3) is false for x sufficiently large. constant C such that (2..3) is false for x = (Ck log m) k. Find a It is useful to be able to count the number of integers up to x divisible by a given integer d. These are the set of integers of the form dn where n and dn x, in other words these are in -to- correspondence with the set of integers n in the range n x/d. There are [x/d] such integers (where [t], the integer part of t, denotes the largest integer t), and [x/d] x/d. What about the number of positive integers up to x which are a (mod d)? Exercise Give a precise expression for the number of positive integers up to x which are a (mod d) in terms of the least positive residue of a (mod d)? Give an approximation, involving a smooth function involving only the variables x and d, that is out from the correct count by at most? 2.2. Euler s proof and the Riemann zeta-function In the eighteenth century Euler gave a different proof that there are infinitely many primes, one which would prove highly influential in what was to come later. Suppose again that the list of primes is p < p 2 < < p k. Euler observed that the fundamental theorem of arithmetic implies that there is a -to- correspondence between the sets {n : n is a positive integer} and {p a pa pa k k : a, a 2,..., a k 0}. Thus a sum involving the elements of the first set should equal the analogous sum involving the elements of the second set: n n a positive integer n s = a,a 2,...,a k 0 (p a pa pa k k )s = (p a a 0 )s a2 0 k ( = ), p s j j= (p a2 2 )s... a k 0 (p a k k )s for any real number s > 0. The last equality holds because each sum in the secondto-last line is over a geometric progression. Euler then noted that if we take s =

24 2.2. EULER S PROOF AND THE RIEMANN ZETA-FUNCTION 7 then the right side equals some rational number (since each p j > ) whereas the left side equals, a contradiction (and thus there cannot be finitely many primes). We prove that n /n diverges in exercise 2.2. below What is wonderful about Euler s formula is that something like it holds without assumption, involving the infinity of primes; that is (2.2.) ( ) p s. n n a positive integer n s = p prime One does need to be a little careful about convergence issues. It is safe to write down such a formula when both sides are absolutely convergent, which takes place when s >. In fact they are absolutely convergent even if s is a complex number so long as Re(s) >. We have just seen that (2.2.) makes sense when s is to the right of the horizontal line in the complex plane going through the point. Like Euler, we want to be able to interpret what happens to (2.2.) when s =. To not fall afoul of convergence issues we need to take the limit of both sides as s +, since (2.2.) only obviously holds for values of s larger than (2.2.). To do this it is convenient to note that the left side of (2.2.) is well approximated by diverges as s +. We deduce that (2.2.2) p prime which, upon taking logarithms, implies that (2.2.3) p prime ( ) = 0 p p =. dt t s = s, and thus So how numerous are the primes? One way to get an idea is to determine the behaviour of the sum analogous to (2.2.3) for other sequences of integers. For instance n n converges, so the primes are, in this sense, more numerous than 2 the squares. We can do better than this from our observation, just above, that n n s s is convergent for any s > (see exercise below). In fact, since n n(log n) converges, we see that the primes are in the same sense more 2 numerous than the numbers {n(log n) 2 : n }, and hence there are infinitely many integers x for which there are more than x/(log x) 2 primes x. There is another derivation of (2.2.) that is worth seeing. One begins with n n, the sum of /n s over all integers n. Now suppose that we wish to remove s the even integers from this sum. Their contribution to this sum is n n even writing even n as 2m, and hence n (n,2)= n s = n n s = m n s n n even (2m) s = 2 s m m s ( n s = ) 2 s n n s.