Using dynamical systems to prove that there are infinitely many primes

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1 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 1 Using dynamical systems to prove that there are infinitely many primes Abstract. In this article we discuss proofs that there are infinitely many primes. We begin with a guided survey of the many known proofs (including a few new minor variations). Then we see how one of these ideas has been pushed in surprising new directions, involving some beautiful themes from arithmetic dynamics. 1. A WIDE VARIETY OF PROOFS. There are many different proofs in the literature that there are infinitely many primes. Most of the proofs use the theorem that Every integer q > 1 has a prime factor. (This was first formally proved by Euclid on the way to establishing the Fundamental Theorem of Arithmetic.) The idea, given any finite list P of primes, is to find an integer q > 1 that is not divisible by any of the primes in P, so the prime factors of q are not on the list P, and therefore P could not have been the complete list of primes to begin with. This implies that no finite list contains all of the primes, and therefore there are infinitely many primes. Non-constructive proofs, developing Euclid s idea. One can determine such an integer q, by establishing that For any given integer N 0 one can find an integer q > 1 for which gcd(n, q) = 1. If N = N(P), the product of all of the primes in P, then any such integer q has at least one prime factor that does not belong to the list P. Euclid took N = N(P) and q = N + 1, though there are many other possibilities for N and q: For any integer N > 2 that is divisible by N(P), we could take q = N + 1 or q = N 1. For example, q = 2N(P) 1 or 2N(P) + 1, or perhaps N = n! where n is the largest prime in P. 1 Partition P into two non-empty sets P = M N. Let m be the product of the elements of M, and let n be the product of the elements of N, so that N = N(P) = mn. Let q = m + n. We claim that gcd(q, N) = 1, for if not some prime p divides both q and N, so that p has to divide either m or n (as their product equals N). Now if p divides m then p divides q m = n, and vice-versa, but m and n were selected so as to have no common prime factor, a contradiction. (In Euclid s proof, one of M and N is the empty set.) See also [22]. A famous example comes from the number of home runs hit by Hank Aaron when he overtook Babe Ruth s record: 715 = versus 714 = [16] beautifully reformulates this idea: Let N = N(P) and p be a prime divisor of 2N + 1. If P contains all of the primes, then p divides N, so that 2N/p is an even integer. Therefore 0 = sin( (1+2N)π ) = sin( π p p ), which is impossible as p > 1, so that 0 < π p < π and therefore sin( π p ) > 0. January 2014] DYNAMICAL SYSTEMS AND PRIMES 1

2 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 2 which together give the product of all the primes up to 17. Note that = 1429 is prime. Modifying the last construction, one might instead take q = m n or some other linear combination am + bn in which the prime factors of a divide m and the prime factors of b divides n. This works provided q 1 or 1. For each prime p P, we let N p = N(P \ {p}) = N(P)/p, and let a p be any integer whose prime factors all divide N p. Then take q = p P a pn p and again we need to be careful that q 1 or 1. There is an advantage to this construction: If P is the set of all primes x, and if 1 < q x 2 then q must itself be prime, and one can show (see [1]) that every prime between x and x 2 can be represented in this way. If f(x) is a polynomial with integer coefficients then p is a prime divisor of f if there exists an integer m for which f(m) is divisible by p. We will now show that any polynomial with integer coefficients, that is not of the form cx r, has infinitely many distinct prime divisors: Suppose that our polynomial is x r f(x) where f(x) = d j=0 a jx j is a polynomial with integer coefficients, for which a 0, a d 0 and d 1. Given a finite list P of primes for which N = N(P) > 2, we have d 1 f(a 0 Nx) = a 0 (1 + Nxg(x)) where g(x) := a k+1 (a 0 N) k x k. If m is an integer for which mg(m) 0 then 1 + Nmg(m) N 1 > 1 and so is divisible by some prime p, which evidently does not divide N. Therefore p is a prime that is not in the list P, and divides f(a 0 Nm). See [8] for much more about prime divisors of polynomials. We do not actually need to construct q to show that such an integer q exists: Euler defined the totient function k=0 φ(n) := #{1 m N : gcd(m, N) = 1}, and proved that φ(n) = p k N p k 1 (p 1) where the product is over all prime powers p k > 1 that divide N, such that p k+1 does not divide N. This formula should be used by the reader to show that if N 1, 2, 3, 4 or 6 then φ(n) > 2, which implies that there exists an integer q in the range 1 < q < N 1 for which gcd(n, q) = 1. This gives the key idea behind the most beautiful of all proofs that there are infinitely many primes, Furstenberg s 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, m) := {a + nm : n Z}, with m 0, which is a subset of S. Evidently each Z(a, m) is open, and it is also closed since Z(a, m) = Z \ Z(b, m). b: 0 b m 1, b a 2 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

3 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 3 If there are only finitely many primes p then A = p Z(0, p) is also closed, and so Z \ A is open. Now Z \ A is the set of integers without prime factors, and so Z \ A = { 1, 1}. However Z \ A = { 1, 1} is not open since { 1, 1} does not contain any arithmetic progression Z(1, m). Hence there cannot be finitely many primes, so there are infinitely many primes. These beautiful proofs only show that there cannot be finitely many primes and so do not hint at how to construct infinitely many primes (other than just testing whether each consecutive integer, 2, 3, 4,... is prime). In the next subsection we highlight proofs that have the added bonus that they point us to where to look for our primes. The construction of infinitely many primes. The idea is to Construct an infinite sequence of distinct, pairwise coprime, integers a 0, a 1,..., that is, a sequence for which gcd(a m, a n ) = 1 whenever m n. We then obtain infinitely many primes, prime factors of the a n, as proved in the following lemma. Proposition 1. Suppose that a 0, a 1,... is an infinite sequence of distinct, pairwise coprime, integers. Let p n be a prime divisor of a n whenever a n > 1. Then the p n form an infinite sequence of distinct primes. Proof. The p n are distinct for if not then p m = p n for some m n and so p m = gcd(p m, p n ) divides gcd(a m, a n ) = 1, a contradiction. As the a n are distinct, any given value (in particular, 1, 0 and 1) can be attained at most once, and therefore all but at most three of the a n have absolute value > 1, and so have a prime factor p n. Here a few ways to construct such sequences: Modification of Euclid s proof: Let a 0 = 2, a 1 = 3 and a n = a 0 a 1 a n for each n 1. If m < n then a m divides a 0 a 1... a n 1 = a n 1 and so gcd(a m, a n ) divides gcd(a n 1, a n ) = 1, which implies that gcd(a m, a n ) = 1. Therefore if p n is a prime divisor of a n for each n 0, then p 0, p 1,... is an infinite sequence of distinct primes. The recurrence for the a n can be re-wriiten as a n+1 = a 0 a 1 a n 1 a n + 1 = (a n 1)a n + 1 = f(a n ), where f(x) = x 2 x + 1. Fermat conjectured that the integers F n = 2 2n + 1 are primes for all n 0. His claim starts off correct: 3, 5, 17, 257, are all prime, but is false for F 5 = , as Euler famously noted. It is an open question as to whether there are infinitely many primes of the form F n. 2 Nonetheless we can prove that the F n are pairwise coprime, and so deduce that if p n is a prime divisor of F n, then p 0, p 1,... is an infinite sequence of distinct primes: Since the polynomial x + 1 divides x 2L 1 for any integer L 1, this also holds when we substitute in an integer in place of 2 The only Fermat numbers known to be primes, have n 4. We know that the F n are composite for 5 n 30 and for many other n besides. It is always a significant moment when a Fermat number is factored for the first time. It could be that all F n, n > 4 are composite, or they might all be prime from some sufficiently large n, onwards. Currently, we have no way of knowing what is the truth. January 2014] DYNAMICAL SYSTEMS AND PRIMES 3

4 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 4 x. Taking x = 2 2m with L = 2 n 1 m we deduce that F m divides F n 2 and so gcd(f m, F n ) = gcd(f m, 2) = 1, as the Fermat numbers are obviously all odd. 3 The F n -values can be determined by a simple recurrence, as follows: F n+1 = (2 2n + 1)(2 2n 1) + 2 = F n (F n 2) + 1 = f(f n ), where f(x) = x 2 2x + 2. The (a n ) n 0 and the (F n ) n 0 are both examples of sequences (x n ) n 0 for which x n+1 = f(x n ) for some polynomial f(x) Z[x]; the a n with the polynomial x 2 x + 1, and the F n with the polynomial x 2 2x + 2. The terms of the sequence are all given by the recursive formula, so that x n = f(f(... f(x 0 ))) = f n (x 0 ) where the notation f n means the polynomial obtained by composing f with itself n times. The key to the proof that the a n s are pairwise coprime is that a n 1 (mod a m ) whenever n > m 0; and the key to the proof that the F n s are pairwise coprime is that F n 2 (mod F m ) whenever n > m 0. These congruences are not difficult to prove using the polynomials that can define the sequences: Let f(x) = x 2 x + 1. Then f(0) = 1 and f(1) = 1, so we deduce that f(f(... f(1))) = 1; that is, f n (0) = 1 for all n 1. To use this information we require the following lemma: Lemma 1. Let g(x) be a polynomial with integer coefficients. If a, b and m are integers for which a b (mod m) then g(a) g(b) (mod m). Proof. We first prove by induction that a j b j (mod m) for all integers j 1. It is evidently true when j = 1, and then if it is true for j, we have a j+1 = a j a b j b = b j+1 (mod m) using the induction hypothesis. Therefore if g(x) = g 0 + d j=1 g jx j then as claimed. g(a) = g 0 + d g j a j g 0 + j=1 d g j b j = g(b) j=1 (mod m), Returning to iterations of the polynomial f(x) = x 2 x + 1: If n = m + k for some k 1 then a n = f k (f m (a 0 )) = f k (a m ) and so a n = f k (a m ) f k (0) 1 (mod a m ). This implies that gcd(a m, a n ) divides gcd(a m, 1) = 1, and so the a n are pairwise coprime. This new proof also works for the Fermat numbers: Let f(x) = x 2 2x + 2. Then f(0) = 2 and f(2) = 2 and so f n (0) = 2 for all n 1. Therefore if n = m + k for some k 1 then F n = f k (F m ) f k (0) 2 (mod F m ), 3 Goldbach was the first to use the prime divisors of the Fermat numbers to prove that there are infinitely many primes, in a letter to Euler in late July c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

5 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 5 which implies that gcd(f m, F n ) divides gcd(f m, 2) = 1, and so the F n are pairwise coprime. This reformulation of two of the best-known proofs of the infinitude of primes seems to open up a whole new vista of possibilities, which we will explore in the next section. For now, though, we return to our survey of proofs that there are infinitely many primes. Let a 0 be a non-zero integers. Given a 0,..., a n 1, we define a n by partitioning the set {0, 1,..., n 1} into two sets L M, and then letting a n = l L a l + m M a m. (1) We claim that the a j are pairwise coprime, for if n is the smallest integer for which there exists some k < n and some prime p which divides gcd(a n, a k ), say k L, then p divides both a n and l L a l. Therefore p divides their difference, m M a m, and so p divides some a m with m M, which contradicts the assumption of n s minimality, as p divides gcd(a m, a k ) where m k < n. In Somos and Haas [20], L M partitions the integers into the even and odd integers < n. Harris [12] begins with positive integers b 0, b 1, b 2 with gcd(b 0, b 2 ) = 1, and then lets a 0 = b 0, a 1 = b 0 b 1 + 1, a 2 = b 0 b 1 b 2 + b 0 + b 2. Thereafter, he uses (1) to define the a n, with M = {n 2}. This ties in with continued fractions, for if b n = a 0... a n 3 for all n 3 then the continued fraction [b 0, b 1,..., b n ] = a n /d n for some integer d n, for every n 0. Analytic proofs. The idea is to count the number of positive integers whose prime factors only come from a given set of primes P = {p 1 < p 2 <... < p k }, up to some large point x. These integers all take the form p e 1 1 p e 2 2 p e k k for some integers e j, each 0. (2) We are going to count the number of such integers up to x = 2 m 1, for an arbitrary integer m 1, by studying this formula: As each p j 2, the smallest prime, we have 2 e j p e j j p e 1 1 p e 2 2 p e k k 2 m 1, and so each e j m 1, which implies that there are at most m possibilities for the integer e j (namely the integers from 0 to m 1). Therefore the number of integers of the form (2), up to 2 m 1, equals #{p e 1 1 p e 2 2 p e k k 2 m 1} k #{integers e j : 0 e j m 1} = m k. j=1 Now if P is the set of all primes then every positive integer is of the form (2), and so the last equation implies that 2 m 1 m k for all integers m. We select m = 2 k, so that 2 k k 2, which is false as k 5 (since we know the primes 2, 3, 5, 7, 11). Therefore there cannot be only finitely many primes. We can modify the last proof to show that for every, not-too-fast-growing, infinite sequence of integers, 1 a 1 < a 2 <..., there are infinitely many primes p which each divide some a j (where j depends on p). By not too fast growing, we mean that, January 2014] DYNAMICAL SYSTEMS AND PRIMES 5

6 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 6 for any fixed ɛ > 0 there exists infinitely many N for which a N 2 Nɛ 1. If not then P := {p prime : p divides some a j } is finite. Taking m = [N ɛ ] for 0 < ɛ < 1 k, so that a N 2 Nɛ 1 < 2 m we have, by the above, N = #{n : a n a N } #{n : a n 2 m 1} m k N ɛk < N, a contradiction. Euler noted that if the positive integer n = p e 1 1 p e 2 2 p e k k, then we may write 1/n as 1/p e 1 1 1/p e 2 2 1/p e k k. Summing this over all integers n 1, we obtain n 1 n a positive integer 1 n = a 1,a 2,...,a k 0 a p e 1 1 p e p e k k ( ) ( ) 1 1 = p e 1 1 p e... 2 a ak 0 = k j=1 1 p e k k ( 1 1 p j ) 1. (3) The last equality holds because each sum in the second-to-last line is over a geometric progression. The right-hand side is a finite product of rational numbers, so is rational, whereas, as we will show, the left-hand side diverges to, a contradiction, and so there cannot be only finitely many primes. Now, to prove that the left-hand side diverges, notice that 2 k 1 n<2 k n a positive integer 1 n > 1 2 k 2 k 1 n<2 k n a positive integer 1 = 2k 1 2 k = 1 2 as 1/n > 1/2 k in this range for n. Therefore, partitioning the integers in [1, 2 K ) into intervals of the form [2 k 1, 2 k ), we have 1 n<2 K n a positive integer K 1 n = k=1 2 k 1 n<2 k n a positive integer 1 n > K 2, which diverges as 2 K grows. Even though there are infinitely many primes, a small modification of the proof of (3), leads to Euler s remarkable identity, n 1 n a positive integer 1 n = s p prime ( 1 1 p s ) 1. 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 > 1; that is, when the sum of the absolute values of the terms converges. In fact they are absolutely convergent even if s is a complex number so long as Re(s) > 1. This 6 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

7 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 7 function, discovered and developed by Euler, is known as the Riemann zeta-function, in honour of the remarkable use Riemann made of it in his plan to give a very good approximation for the number of primes up to any given large x. Arithmetic proofs. Fermat s little theorem implies that if p is an odd prime then 2 p 1 1 (mod p). If P = {p 1,..., p k } is a finite set of odd primes, let L = L(P) := (p 1 1) (p k 1). For each j we can write L = (p j 1)L j where L j is an integer, which can be given by the same product as L, excluding the factor p j 1. Then 2 L = (2 p j 1 ) L j 1 L j 1 (mod p j ). The number q := 2 L + 1 is odd, and q = 2 L = 2 (mod p) for every odd prime p P, and so q is not divisible by any of the primes in P. The Mersenne number q := 2 L+1 1 is odd, and q = 2 2 L = 1 (mod p) for every prime p P, and so q is not divisible by any of the primes in P. The number q := 2 L+2 3 is odd, and q = 4 2 L = 1 (mod p) for every odd prime p P, and so q is not divisible by any of the primes in P. For the next two proofs (that there are infinitely many primes) we use the following lemma: Lemma 2. If 2 m 1 (mod p) for odd prime p, then 2 g 1 (mod p) where g =gcd(m, p 1). Proof. Euclid s algorithm, applied to m and p 1, exhibits integers a and b for which am + b(p 1) = g. Therefore, by Fermat s little theorem, 2 g = 2 am+b(p 1) = (2 m ) a (2 p 1 ) b 1 a 1 b 1 (mod p). Suppose that there are only finitely many primes and let q be the largest prime. If p is a prime factor of the Mersenne number, 2 q 1, then 2 q 1 (mod p). Therefore 2 g 1 (mod p) where g = gcd(q, p 1) by Lemma 2. Now g divides q, so g must equal either 1 or q. However g cannot equal 1, else p divides 2 g 1 = 2 1 = 1. Therefore q = g which divides p 1. But then q p 1 < p, so p is a larger prime than q, contradicting its maximiality. Suppose that there are only finitely many primes p 1,..., p k and let 2 n be the highest power of 2 dividing some p j 1. The Fermat number q = 2 2n + 1 must be divisible by some odd prime p, and so 2 2n+1 1 = (2 2n ) 2 1 ( 1) (mod q), and therefore also mod p. By the lemma, 2 g 1 (mod p) where g = gcd(2 n+1, p 1). Now g divides 2 n+1 so g must be a power of 2, say g = 2 m. Moreover m n as g = 2 m divides p 1, by the definition of n. Therefore 0 q = 2 2n + 1 = (2 2m ) 2n m n m (mod p), so that p divides 2, which is impossible as p is an odd prime. This proof also yields that, for any integer N 1, there are infinitely many primes 1 (mod 2 N ), and suitable modifications even allow one to prove that for any integer m 2, there are infinitely many primes 1 (mod m). January 2014] DYNAMICAL SYSTEMS AND PRIMES 7

8 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 8 Euler exhibited the inspiring identity π 4 = The right-hand side can be re-written as the sum, over integers n 1, of χ(n)/n where 1 if n 1 (mod 4), χ(n) = 1 if n 1 (mod 4), 0 if n is even (One can write χ(n) = ( 4/n), a Jacobi symbol). It is not difficult to see that χ(.) is multiplicative, that is that if n = p e 1 1 p e 2 2 p e k k then χ(n) = χ(p 1 ) e 1 χ(p 2 ) e2 χ(p k ) e k. Imitating the proof of (3), but replacing each term of the form 1/m s by χ(m)/m s, we deduce that π 4 = n 1 n a positive integer χ(n) n k ( = j=1 1 χ(p j) p j ) 1. It is well-known that π (and so π/4) is irrational, but the right-hand side is a finite product of rational numbers, so is rational, a contradiction. Other types of proofs? There seem to be only a few that are genuinely different from those that we have outlined above. My favourite is a new proof by Levent Alpoge [2], based on van der Waerden s Theorem, a deep result in combinatorics. 2. DYNAMICAL SYSTEMS AND THE INFINITUDE OF PRIMES. One can model evolution by assuming that the future development of the subject of study follows from understanding its current state. 4 This gives rise to dynamical systems, a rich and bountiful area of study. One simple model is that the state of the object at time n is denoted by x n, and given an initial state x 0 one can find subsequent states via a map x n f(x n ) = x n+1 for some given function f(.). We call the set of points x 0, x 1,... the orbit of x 0 under the map x f(x). In the complex plane, orbits of linear polynomials are easy to understand, 5 but quadratic polynomials can give rise to evolution that is very far from what one might naively guess (the reader might look into the extraordinary Mandelbrot set). Here we are interested in the number theory of such orbits, when f is a polynomial with integer coefficients. It helps to have a little terminology to describe this evolutory process: Given an integer a 0 we generate the orbit, (a n ) n 0, by taking a n a n+1, where a n+1 = f(a n ). This orbit is said to be eventually periodic if there exist integers N 0 and m 1 such that a n+m = a n for all n N. If m is the minimal such integer then we 4 Not taking into account any random mutations. 5 One can verify: if f(t) = at + b then x n = x 0 + nb if a = 1, and x n = a n x 0 + b/(1 a) if a 1. 8 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

9 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 9 say that the period length is m. If N is the minimal such integer then the numbers a 0, a 1,..., a N 1 are pre-periodic points. For example if then the orbit of 9 under this map is the set of numbers 9, 2, 3, 5, 11, 13, 11, 13, 11, This is eventually periodic, with period of length two (which could also be written ), and pre-period of length four. Notice that once two of the a i s are the same number then the sequence must be periodic from that point onwards: For, if a N+m = a N then a n+m = a n for all n N, which follows from a suitable induction hypothesis as a n+m+1 = f(a n+m ) = f(a n ) = a n+1. Generating infinite sequences of coprime integers. In section 1.2, we constructed two infinite sequences of pairwise coprime integers which are each the orbit of an integer a 0 under a map x f(x), for some polynomial f(x). In both cases f n (0) = f(0) 0 for all n 1, which we can write as 0 a a... where a = f(0); so that 0 is a pre-periodic point for the map x f(x). The key idea in proving that the terms of the sequence a 0, a 1,... are pairwise coprime is encapsulated in the following lemma. Lemma 3. Let f(x) Z[x]. In any integer orbit, (a n ) n 0, under the map x f(x), we have that gcd(a m, a n ) divides f n m (0), whenever n > m 0. If 0 is a pre-periodic point for the map x f(x) then gcd(a m, a n ) divides the positive integer Proof. By Lemma 1 we have L(f) := lcm[f k (0) : k 1]. a n = f n m (a m ) f n m (0) (mod a m ), and so gcd(a m, a n ) divides gcd(a m, f n m (0)), which divides f n m (0). If 0 is pre-periodic then L(f) is the lcm of a finite number of non-zero integers, and so is itself a well-defined non-zero integer. The result follows since f n m (0) divides L(f). This shows that if 0 is pre-periodic then gcd(a m, a n ) belongs to a finite set of possibilities, and we might hope to prove that gcd(a n, L) = 1 for all n 1. If so then the a n are necessarily pairwise coprime, and therefore each a n with a n > 1, will contain its own private prime factor (We say that p n is a private prime factor of a n if p n divides a n but does not divide a m for every m n). We now run through various examples to help us formulate what we will aim to prove. January 2014] DYNAMICAL SYSTEMS AND PRIMES 9

10 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 10 In our first example, f(x) = x 2 x + 1, we have L(f) = 1, so the a n are pairwise coprime. Therefore if a n > 1, then a n has its own private prime factor p n. If the orbit of a 0 is not eventually periodic then the only possible exceptional a n (that is, for which a n 1), is 1, since 0 and 1 belong to the orbit with a period. However there are no rational solutions to f(x) = 1, so if 1 appears in an orbit, then it must be the first term of the orbit, that is a 0 = 1. Therefore if the orbit of a 0 is not eventually periodic then, for every integer n 1, the integer a n has its own private prime factor. In our second example, f(x) = x 2 2x + 2, we have L(f) = 2, so if a 0 is any odd integer whose orbit is not eventually periodic then the a n are pairwise coprime. Again 1 is the only possible exception (since 1 1), and if it appears in an orbit then a 0 = 1. Therefore if the orbit of a 0 is not eventually periodic then, for every integer n 1, the integer a n has its own private prime factor. If a 0 is even then each a n is even, as the map reduces to the map (mod 2), and so gcd(a m, a n ) = gcd(a m, 2) = 2. Moreover a n 2 (mod 4) for all n 1 as 0, 2 2. Therefore the a n /2 for n 1 are odd and pairwise coprime, and so a n has its own private prime factor provided a n > 2. Here 2 is the only possible exception (since 2 2), and if it appears in an orbit then a 0 = 2,. Therefore if the orbit of a 0 is not eventually periodic then, for every integer n 1, the integer a n has its own private, odd prime factor. We have now seen that, whether or not a 0 is even, every a n with n 1, has its own private prime factor which does not divide L(f). 6 If f(x) = x 2 6x + 6 then L(f) = 6. In this case gcd(a m, a n ) = gcd(a m, 6) = gcd(a 0, 6) as gcd(t 2 6t + 6, 6) = gcd(t 2, 6) = gcd(t, 6). In the example the a n /3 are pairwise coprime, but a 0 /3 and a 1 /3 do not have their own prime factors. In this example every a n with n 2 has its own private prime factor > 3, but not a 0 or a 1. Putting together what we have learned from these examples, we might hope for the following theorem: Theorem 1. Let f(x) Z[x] for which 0 is a pre-periodic point of the map x f(x). If the orbit of the integer a 0 is not eventually periodic then the integer a n has its own private prime factor, which does not divide L(f), provided n is sufficiently large. How big is sufficiently large? We will show that this holds provided n 2L(f) + 6, a bound which depends only on f and not on the choice of a 0 (as in the above examples). It may be that the Theorem holds for all n n 0 where n 0 is some number, perhaps even 2 or 3, which is independent of f as well as a 0, but we have no idea how to prove such a result. To prove Theorem 1 we will need to classify the possible orbits of 0 which are eventually periodic, and to better understand the orbit of a 0 reduced mod p k+1 where p k L(f). We will discuss and sketch the proofs here, but full details can be found in the arxiv version of this article. 3. POSSIBLE PERIODS, WITH 0 IN THE PRE-PERIOD We have already seen the example f(x) = x 2 ax + a for which 0 a a... 6 For this map, we always have the closed fomula a n = (a 0 1) 2n c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

11 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 11 Now we exhibit several other examples. If f(x) = x 2 2 then f(0) = 2, f( 2) = 2 and f(2) = 2, that is Although the period has length 1, as in our previous example, here the pre-period has the longer length 2. We also deduce that gcd(a m, a n ) = 1 or 2 if m n. Next, for f(x) = x 2 x 1, we have The iterates of 0 are again periodic, but now with a period of length 2 rather than 1. We also deduce that gcd(a m, a n ) = 1 if m n. Finally for f(x) = 1 + x + x 2 x 3 the pre-period and period both have length 2, We deduce that gcd(a m, a n ) = 1 or 2 if m n. The main result of this section is to show that this more-or-less accounts for all of the possible types of periods with a pre-period involving 0. Theorem 2. If 0 is pre-periodic for the map x f(x) Z[x] then the orbit of 0 is 0 a a... for some integer a 0; or 0 a a a... for a = 2, 1, 1 or 2; or 0 1 a 1... or 0 1 a 1... for some integer a 0; or or All periods have length one or two. This surprising result follows from Lemma 1: Suppose that N is the smallest positive integer for which a N = a 0. We first prove that a N+j = a j for all j 0, by induction: This is true for j = 0 by the hypothesis; and if it is true for j then a N+j+1 = f(a N+j ) = f(a j ) = a j+1, as desired. Now assume that N > 1 so that a 1 a 0. Lemma 1 implies that a n+1 a n divides f(a n+1 ) f(a n ) = a n+2 a n+1 for all n 0. Therefore a 1 a 0 divides a 2 a 1, which divides a 3 a 2,..., which divides a N a N 1 = a 0 a N 1 ; and this divides a 1 a N = a 1 a 0, the non-zero number we started with. We deduce that a 1 a 0 a 2 a 1... a j+1 a j... a 1 a 0, and so these are all equal. Now there must be some j 1 for which a j+1 a j = (a j a j 1 ), else each a j+1 a j = a j a j 1 =... = a 1 a 0 in which case N 1 N 1 0 = a N a 0 = (a j+1 a j ) = (a 1 a 0 ) = N(a 1 a 0 ) 0. j=0 January 2014] DYNAMICAL SYSTEMS AND PRIMES 11 j=0

12 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 12 Therefore a j+1 = a j 1, and we deduce that a 2 = a N+2 = f N+1 j (a j+1 ) = f N+1 j (a j 1 ) = a N = a 0, as desired. Lemma 1 can be used to understand the length two periods of a given map x f(x): If there is any period of length two then all periodic points are roots of f(x) + x T where T is the sum of the two elements in a period; moreover if there is a periodic point of period one then it is unique and equals T/2. To prove this, let a b with a b, be a period of length two, and let c d be some other period. By Lemma 1, we have that a c divides f(a) f(c) = b d, which divides f(b) f(d) = a c, and so a c = ±(b d); and, similarly a d = ±(b c). Therefore a + b = c + d, for if not then a c = b d and a d = b c so that a = b which is false. So we can let T = a + b, and if c = d then T = 2c. Orbits and algebra. Before proving Theorem 2 we highlight linear transformations linking different orbits: If (a n ) n 0 is an orbit of the map x f(x) then (as a n+1 = f( ( a n ))); and ( a n ) n 0 is an orbit of the map x f( x) (a n d) n 0 is an orbit of the map x g d (x), where g d (x) := f(x + d) d (as g d (a n d) = f(a n ) d = a n+1 d). This leads to the following result which allows us to quickly identify eventually periodic orbits: Corollary 1. If the orbit of a 0 is eventually periodic then a 2 = a 4. Proof. If a 0 is in the period then a n+2 = a n for all n 0 as the period length is either one or two. If a 0 is pre-periodic then 0 is pre-periodic for the map x f(x + a 0 ) a 0, with orbit 0 a 1 a 0 a 2 a By inspecting all the cases of Theorem 2 we see that a 2 a 0 = a 4 a 0, and so the result follows. Does a given map come from a polynomial with integer coefficients? Suppose that f 0 (x) Z[x] is given, and f 0 (u j ) = v j for j = 1, 2,..., n, where the u j and v j are given integers. We now prove that there exists a polynomial f(x) Z[x] for which f(u j ) = v j for j = 1, 2,..., n, and f(u) = v for some given integers u and v, if and only if n j=1 (u u j) divides v f 0 (u): If f(u j ) = v j for j = 1, 2,..., n then the polynomial f(x) f 0 (x) has roots u 1,..., u n, which happens if and only if it is of the form m(x)g(x), for some g(x) Z[x] with m(x) := n j=1 (x u j). Therefore if f(u) = v then v f 0 (u) = m(u)g(u), and so m(u) divides v f 0 (u), as g(u) is an integer. On the other hand if m(u) divides v f 0 (u) then we can take f(x) = f 0 (x) + Cm(x) where C = (v f 0 (u))/m(u) Z. The polynomial f 0 (x) = 6 satisfies f 0 (0) = f 0 (6) = 6. Our result implies that there is a polynomial f for which f(0) = f(6) = 6 and f(3) = v, if and only if 9 = u(u 6) divides v f 0 (u) = v 6. On the other hand Lemma 1 implies that 3 = 3 0 divides f(u) f(0) = v 6, and 3 = 3 6 also divides f(u) f(6) = v 6, which is not as restrictive a requirement on v. This new result is therefore even more powerful than Lemma 1. We now proceed to a proof of Theorem c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

13 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 13 When 0 is in the pre-period, and the period length is 1. We have the orbit 0 a a for any f(x) of the form f(x) = a + x(x a)g(x) for some g(x) Z[x]. If 0 b a a with b a is an orbit for x f(x) then b 0 a b a b for x g b (x) = f(x + b) b. Now 0 a b a b is an orbit of the map x a b and so such a polynomial g b exists (and therefore f(x) = g b (x b) + b also exists) if and only if ab = (0 ( b))((a b) ( b)) divides (a b) 0 = a b. We deduce that b divides a b and so divides a, and similarly a divides b; as b a, this implies that b = a. Therefore a 2 = ab divides 2a = a b, and so a = 2, 1, 1 or 2. One can verify that we have such an orbit for f(x) = ±(2x 2 1) if a = ±1, and for f(x) = ±(x 2 2) if a = ±2. There is no orbit 0 C B A A else there is an orbit c 0 a a a (for x g C (x)), by the previous paragraph: By Lemma 1 we then have c = c 0 divides f(c) f(0) = 0 b = a, but c a else c would equal one of a or a. Therefore a = 2 and c = 1 and we have c(c 2 4) = ±3 divides c 2 2 = 1, which is nonsense. When 0 is in the pre-period, and the period length is 2. The period b a b is an orbit of the map x a + b x, and so we can extend this to 0 b a b if and only if ab = (0 a)(0 b) divides (a + b 0) b = a, and so b = ±1. If there is an orbit 0 C B A B then there is an orbit C 0 b a b with b = ±1, and we may assume b = 1 by replacing g(x) by g( x) if b = 1. Now 0 1 a 1 is an orbit for the polynomial 1 + ax x 2, so can be extended by C 0 if and only if C(C + a)(c + 1) divides C 2 + ac 1. We see that C divides 1, so must equal 1 (as C b), and so 2(a + 1) divides a, and therefore a = 2. Hence we have the orbit (as well as ). This is an orbit for the polynomial 1 + x + x 2 x 3. This cannot be extended by r 0 else r = 0 ( r) divides 1 0 = 1, so equals ±1, which is impossible, as 1, 1 2 not 0. This completes the proof of Theorem PRIVATE PRIME FACTORS. PROOF SKETCH OF THEOREM 1. When 0 a a under the map x f(x), we now sketch a proof that for any integer a 0, there exists a divisor D = D(a 0 ) of a such that the integers a n /D are pairwise coprime, and coprime with a, for all n n 1 (f), which is the maximum of p 1 + k over all prime powers p k dividing a: By Lemma 3 we know that if m < n then gcd(a m, a n ) =gcd(a m, a), so we need only worry about the prime divisors p of a. One begins by showing that if m is the smallest integer 0 for which p a m then m p 1 by the pigeonhole principle. Suppose that p k a and p j a n. As a n (a n L) divides a n+1 L, we deduce that p 2j a n+1 if 2j k, and p k a n+1 if 2j > k. This implies that p k a n whenever n n 1 (f), and the claim follows. Define a sequence (b j ) j 0 of integers by b j := a n1 +j/d for all j 0, so that b j+1 = g(b j ) for all j 0, where g(x) = f(dx)/d. Here g(x) Z[x] since g(0) = a/d, so that 0 g(0) g(0). By the same methods as above one can show that for any map x g(x) in which the orbit of 0 looks like 0 b b, if 1 or 1 appears in an orbit, then it must be one of the first three terms. Therefore a n1 +j/d = b j 2 if j 3. We deduce that a n has its own private prime factor, not dividing a, for each n n 1 (f) + 3. If 0 a a a with a = ±1 or ±2, then by Lemma 3, either the a n are pairwise coprime or, when a = ±2, the a n are even for all n 1 and the a n /2 are January 2014] DYNAMICAL SYSTEMS AND PRIMES 13

14 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 14 pairwise coprime and odd for all n 2. These numbers are then all > 1 in absolute value for n 5, and so each a n contains its own private prime factor once n 5. In every other case in Theorem 2, the orbit of 0 eventually has period 2. The orbit of 0 under the map x f 2 (x) is of the form 0 a a, of period 1. Moreover the orbit of a 0 under x f(x) is partitioned into the two orbits a 0, a 2, a 4,... and a 1, a 3, a 5,... under the map x f 2 (x), and we have gcd(a m, a n ) = 1 whenever m and n have different parity (that is, belong to different orbits), since f k (0) = 1 or 1 whenever k is odd. We apply the results of the first paragraph to the two orbits under f 2 (.), which are coprime to each other, and so deduce that each a n has its own private prime factor, which does not divide L(f) = a, once n 2n 1 (f) + 6. Finally observe that p + k 1 p k L(f) whenever p k L(f), so that n 1 (f) L(f), and the result follows. 5. OTHER DYNAMICAL SYSTEMS TO FIND AN INFINITY OF PRIMES. In Theorem 1 we saw that if 0 is a pre-periodic point for the map x f(x) then any given integer orbit is either eventually periodic, or every a n has its own private prime factor once n is sufficiently large. Do we expect something like this be true for other polynomials f(x) Z[x]? If f(x) has degree d > 1 then one can show that if (a n ) n 0 is not eventually periodic then there exist real numbers 1 α > 0 and β, which depend only on f, and τ > 1 which depends on f and a 0, such that If n is sufficiently large then a n is the integer nearest to ατ dn + β. (4) The a n grow so fast that the product of the gcd(a m, a n ) with m < n is far smaller than a n, so it seems very likely that a n has its own new prime factors. However this argument cannot be made into a proof since some of the primes that divide the a m with m < n, might divide a n to a very high power. We will discuss this possibility further in section 5. Before this more general discussion, we highlight another class of polynomials f that have been used to construct infinitely long sequences of coprime integers, in an elementary way. Another iteration to describe Euclid s construction. Let A 0 = 2, and A n+1 = A n (A n + 1) for all n 0, so that A 1 = 2 3, A 2 = 2 3 7, A 3 = , etc. Taking a 0 = A 0 with a n = A n /A n 1 for all n 1, we again obtain Euclid s sequence (see, e.g. [21, 18])! Whereas before we described the iteration as a n+1 = f(a n ) with f(x) = x 2 x + 1, now we have A n+1 = F (A n ) with F (x) = x(x + 1). For the proof with the Fermat numbers we may similarly let F (x) = x(x + 2). Does this generalize? Do we obtain all the same sequences as before? The idea is to take G(x) Z[x] with G(0) 0, and let F (x) = xg(x). Then, for given integer A 0 = a 0, we let We deduce, by induction, that A n+1 = F (A n ) = A n G(A n ), and then a n+1 = A n+1 /A n = G(A n ) for all integers n 0. A n = a 0 a 1 a n. 14 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

15 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 15 Theorem 3. Let F (x) = xg(x) with G(x) = G(0) + x h H(x) where H(x) Z[x], and G(0), H(0) 0. If (a n ) n 0 is not eventually periodic then the integers a n, with n max{ G(0) + 4, H(0) + 2}, each have their own private prime factor. Proof sketch. If the (a n ) n 0 are eventually periodic then the A m all belong to the finite set {l Z : G(l) = a n for some n 1}. Therefore A m+d = A m for some m + d > m, so either some A m = 0, or 1 = A m+d /A m = a m+d a m+d 1... a m+1, so either a m+1 = 1 or a m+2 = 1 or a m+1 = a m+2 = 1. In the first case, if it is the smallest such m then a m = 0 and each subsequent a n = G(0). If a n = 1 then A n = A n 1 and therefore a n+1 = G(A n ) = G(A n 1 ) = a n, so a n = 1 for all n m + 2. Finally if a m+1 = a m+2 = 1 then the A n have period two by an analogous induction argument and a n = 1 for all n m + 1. In this proof we saw that if a N a N+1 1 then a n is periodic for all n N + 2. Hence if (a n ) n 0 is not eventually periodic then a N a N+1 2 for all N. Therefore if N 2k(F ) + 2 then A N 1 > max{2 G(0), H(0) }, where k(f ) is the largest integer k for which 2 k max{2 G(0), H(0) }. We now deduce if a n > 1 for all n 2k(F ) + 1 then (a n ) n 0 is not eventually periodic. For if (a n ) n 0 is eventually periodic then a N = 1, 0 or 1 for some N 2k(F ) + 2. Then G(A N 1 ) = a N, and so A N 1 = A N 1 0 divides G(A N 1 ) G(0) = a N G(0). Now a N = G(0) else A N 1 G(0) G(0), a contradiction. Therefore A N 1 is a root of G(x) a N = x h H(x), and so is a root of H(x) as A N 1 0. Therefore A N 1 H(0) and so A N 1 H(0), another contradiction. Now, if m < n then a m A m A n 1 and so a n = G(A n 1 ) G(0) (mod a m ), which implies that gcd(a m, a n ) = gcd(a m, G(0)) which divides gcd(a n, G(0)). Therefore if prime p G(0) then p can divide at most one of the a n. Suppose that p k G(0). As before (for those f for which 0 is pre-periodic), if m is the smallest integer for which p divides A m then m p 1. Then p a m, and p a m A n for all n m, and therefore a n+1 = G(A n ) G(0) 0 (mod p). which implies that p j+1 a m a m+1 a m+j A m+j for all j 0. Therefore if n m + k + 1( p + k) then a n = G(A n 1 ) G(0) (mod p k+1 ) and so p k a n. Therefore, if D is the product of those prime powers p k G(0) for which p divides some a n, then D a n and the a n /D are pairwise coprime, and coprime with G(0), when n G(0) + 1. These a n /D will have their own private prime factor unless G(A n 1 ) = a n = D or D. That is A n 1 is a root of G(x) d for some positive or negative divisor d of G(0). If d G(0) then A n 1 divides G(0) d, so that A n 1 G(0) + d 2 G(0) ; if d = G(0) then A n 1 is a root of H(x), and so divides H(0), so that A n 1 H(0). Therefore n 2k(F ) + 1, and the result follows. The taking the pth power example. Let F (x) = (x + 1) p 1 and A 0 = a 1 so that A n = a pn 1 for all n 0 and a n := ap n 1 a pn 1 1. Now G(0) = p and, by Fermat s Little Theorem, F (b 1) = b p 1 b 1 (mod p) and so A n a 1 = A 0 (mod p) by induction. Therefore if a 1 (mod p) then gcd(a n, p) divides (A n, p) = (A 0, p) = 1. This implies that if a > 1 and a 1 (mod p) then the integers (a n ) n 0 are pairwise coprime, and so have their own private prime factor if n 1. January 2014] DYNAMICAL SYSTEMS AND PRIMES 15

16 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 16 Now if prime q n divides a n then a pn 1 (mod q n ), but a pn 1 1 (mod q n ) as q n a n 1 (as a n and a n 1 are coprime). The order of a (mod q n ) is the least integer r > 0 for which a r 1 (mod q n ). Lemma 2 implies that r =gcd(r, p n ), so r divides p n, but r does not divide p n 1 and therefore r = p n. Fermat s Little Theorem, together with Lemma 2, then implies that r = p n divides q n 1; that is, q n 1 (mod p n ) for all n 1. In particular q n 1 (mod p m ) whenever n m; that is, we obtain an infinite sequence of distinct primes 1 (mod p m ). If a 1 (mod p) then a pn 1 (mod p) for all n 0 and so p 1 p 1 a n = (a pn ) j 1 j = p 0 (mod p). j=0 j=0 Therefore gcd(a m, a n ) = gcd(a m, p) = p; and so the integers (a n /p) n 0 are pairwise coprime. This implies that if a > 1 then the integers (a n ) n 0 each have their own private prime factor, not equal to p, for all n 1. If p = 2 then a n = a 2n and so a n+1 = f(a n ) where f(x) = x 2 2x + 2, which we recognize from earlier. However the sequence a n is new when p > 2; in fact, it could not have been generated by a map a n f(a n ), where f(x) Z[x]. For if it is then, as a n+1 /a p n 1 as n, we deduce that f(x) = x p + terms of lower degree, and the limit of (f(a n ) a p n)/a p 1 n x p 1. However (f(a n ) a p n)/an p 1 cannot equal f(a n ) for some fixed f(x) Z[x]., as n, will yield the coefficient of = (a n+1 a p n)/a p 1 n as n, so a n+1 Constructing F from f. Suppose that 0 a a under the map x f(x). Let F (x) = f(x + a) a = x(x + a)h(x) for some H(x) Z[x]. Now, given a 0 and then a n+1 = f(a n ) for all n 0, let A 0 = a 0 a and define A n+1 = F (A n ) for all n 0. We claim that A n = a n a for all n 0: This is true for n = 0 and then A n+1 = F (A n ) = f(a n + a) a = f(a n ) a = a n+1 a. Now A n+1 = a n A n H(A n ) so all of the prime factors of the a n divide the A n. On the other hand, if prime p H(A n ) but p ah(0)h( a) then p A n+1 and p A n (as p H(0)) so p a 0 a 1 a n 1, and p a n = A n + a (as p H( a)), and therefore a m a (mod p) as p A m so p a m for all m n + 1. (The same argument works here if F (x) = x r (x + a) s H(x).) Other polynomials? What of the sequences (a n ) n 0 generated by x f(x) where 0 is neither pre-periodic, nor a root of f(x)? At the start of this section we noted that, if the a n are not eventually periodic, and If n is sufficiently large then a n is the integer nearest to ατ dn + β. for some τ = τ(a 0 ) > 1, and 0 < α 1. Now if n = m + k then gcd(a m, a n ) = gcd(a m, f k (0)) and so is bounded by min{ a m, f k (0) }, which implies that m n 1 gcd(a m, a n ) 0 m n/2 a m 0 k n/2 f k (0) C dn/2 for some constant C = C(a 0 ) (τ(a 0 )τ(0)) 2. This is evidently far, far smaller than a n for n large, so only a small part of a n comes from its gcds with the a m for m < n, 16 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

17 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 17 and so we believe that a n is likely to have a prime divisor that the a m, with m < n, do not. However it is feasible that some of the primes dividing the a m might divide a n to extraordinarily high powers. We know of no unconditional way to show that this is impossible, but we can recover such a result if we assume the abc-conjecture, 7 so called because it limits how big is the contribution of prime powers to any solution of the equation a + b = c in pairwise coprime integers. Applied to this equation with b = 1, so that c = a + 1, one gets the following consequence of the abc-conjecture: For any fixed ɛ > 0 there exists a constant κ ɛ > 0 such that for any integer a we have p κ ɛ a 1 ɛ p prime p a(a+1) The product here is over all primes p which divide the polynomial f(x) = x 2 + x evaluated at x = a. There is a far-reaching generalization which is a consequence of the abc-conjecture. It applies to the prime divisors of any given polynomial (see [15, 6, 9]): Consequence of the abc-conjecture: Suppose that f(x) Z[x] has at least two distinct roots (in other words, f(x) is not of the form a(x + c) d ). For any fixed ɛ > 0 there exists a constant κ ɛ,f > 0 such that for any integer a we have p κ ɛ,f a 1 ɛ p prime p f(a) We say that p n is a primitive prime factor of a n if p n divides a n but does not divide a m for any m < n. It is possible that a primitive prime factor, p n of a n, divides a M for some M > n, so p n is not as unique as a private prime factor. We now show, assuming the abc-conjecture, that if n is sufficiently large then a n has a primitive prime factor: If not then every prime factor of a n must divide a m for some m < n, and so p = p prime p a n p prime p n 1 m=0 gcd(am,an) p n 1 m=0 gcd(a m, a n ) C dn/2, as we proved above. On the other hand, applying (4), and then taking ɛ = 1 in the 3 abc-conjecture (so that κ = κ 1/3,f > 0), we deduce that, for sufficiently large n, τ dn 1 /2 κ a n 1 2/3 p prime p a n=f(a n 1 ) p C dn/2, which is impossible, as d n 1 goes to infinity, as n grows, much faster than d n/2. Therefore we have proved: 7 This has recently been claimed to have been proved by Mochizuki, but the experts have as yet struggled to agree, definitively, that Mochizuki s extraordinary ideas are all correct. January 2014] DYNAMICAL SYSTEMS AND PRIMES 17

18 Mathematical Assoc. of America American Mathematical Monthly 121:1 September 23, :44 p.m. MonthlyArticleCorrectSubmittedVersion.tex page 18 Theorem 4. Assume the abc-conjecture. Suppose that f(x) Z[x] has at least two distinct roots. If (a n ) n 0 is generated by x f(x) and is not eventually periodic, then a n has a primitive prime factor for all sufficiently large n. Primitive prime factors of rational functions. Now suppose that φ(t) = f(t)/g(t), with f(t), g(t) Z[t]. Beginning with a rational number r 0, we define r n+1 = φ(r n ) for all n 0, and write r n = a n /b n where gcd(a n, b n ) = 1 and b n 1. Suppose that f(0) = 0 but f(t) is not of the form at d. Ingram and Silverman [13] showed that if the orbit of r 0 is not eventually periodic then a n has a primitive prime factor for all sufficiently large n. It is believed that some result of this sort should hold whether or not f(0) = 0, and indeed this has been confirmed in [11] under the assumption of the abc-conjecture. Faber and Granville [7] showed that for any rational function φ of degree 2, if the orbit of r 0 is not eventually periodic then either there is a primitive prime factor in the numerator of a n+1 a n for all sufficiently large n, or there is a primitive prime factor in the denominator of a n+1 a n for all sufficiently large n. All these cases give rise to new proofs that there are infinitely many primes! 6. PRIMITIVE PRIME FACTORS, FOR DEGREE ONE F, AND BEYOND. The Mersenne numbers, M n = 2 n 1 can be generated by taking M 0 = 0 and then M n = f(m n 1 ) for all n 1, with f(x) = 2x + 1. The growth of the numbers involved, now that f is a linear polynomial, is simply exponential (rather than doubly exponential in n, as we saw in (4) when the degree of f is 2), provided that the leading coefficient of f has absolute value > 1. 8 If n r (mod m) then we can write n = qm + r to obtain M n = 2 qm+r 1 = (2 m ) q 2 r 1 1 q 2 r 1 = 2 r 1 (mod M m ) and so gcd(m n, M m ) = M r. We deduce, by induction, that gcd(m n, M m ) = M gcd(m,n). In this case the gcd s grow large, but we can still hope to find a primitive prime factor of M n when n is sufficiently large. For example, for n = 2, 3, 4, 5, the M n have primitive prime factors 3, 7, 5, 31 then M 6 = 63 does not, and subsequently we have 127, 17, 73, 11,...; indeed M n has a primitive prime divisor for all n > 6. More than that, the Bang-Zsimondy theorem states that if x n = an b n for all a b n 0, with a > b 1 then x n has a primitive prime divisor for all n > 1, other than the example x n = M 6 above. These provide yet more infinite sequences of distinct prime numbers. A Lucas sequence (a n ) n 1 begins a 0 = 0, a 1 = 1 and then satisfies the secondorder linear recurrence a n = pa n 1 + qa n 2 for all n 2, for arbitrary integers p and q, in which the discriminant := p 2 + 4q is non-zero. 9 In 1913, Carmichael showed that if > 0 then a n has a primitive prime factor for each n > 12, the most exceptional case being the Fibonacci sequence F n = F n 1 + F n 2 in which F 2 = 1, F 6 = F3 3 and F 12 = F3 4 F4 2 have no primitive prime factors. In 2001, Bilu, Hanrot and Voutier [4] proved, following up on groundbreaking work of Schinzel [19], that if < 0 then a n has a primitive prime factor for each n > 30, the 8 If f(x) = x + b then f n(a) = a + nb. If f = x + b then f n(a) = a or b a as n is even or odd. 9 The sequence (x n) n 0 with x n = an b n, is the special case with p = a + b and q = ab. a b 18 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121