QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS

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1 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS RICHARD GOTTESMAN AND KWOKFUNG TANG Abstract. For f(x) Z[x] and a Z, we let f n (x) be the nth iterate of f(x), P (f, a) = {p prime : p f n (a) for some n}, and D(P (f, a)) denote the natural density of P (f, a) within the set of primes. A conjecture of Jones [5] indicates that D(P (f, a)) = 0 for most quadratic f. In this paper, we find an exceptional family of (f,a) such that D(P (f, a)) > 0 by considering f t (x) = (x + t) 2 2 t and a t = f t (0) for t Z. We prove that if t is not of the form ±M 2 ±2 or ±2M 2 ±2, then D(P (f t, a t )) = 3. We also determine D(P (f t, a t )) in some cases when the density is not equal to 3. Our main technique involves the computation of the Galois group of the nth iterate of (x + t) 2 2 t. Our results suggest a connection between the arithmetic dynamics of the conjugates of x 2 and the conjugates of x Introduction Determining which prime numbers divide at least one term of a given sequence of integers has attracted a great deal of attention (see e.g. []). It is easy to prove that every prime divides some term in the Fibonacci sequence, but for other integer sequences, this problem can be tantalizingly difficult. Indeed, merely determining the density of primes dividing some term of a given integer sequence presents considerable complexity. We recall that if S is a set of primes, then the natural density of S within the set of primes is defined by #{p x : p S} D(S) := lim, provided this limit exists. x #{p x : p is prime} The tools of algebraic number theory have been successfully employed to investigate the density of primes which divide some integer sequences defined by a binary linear recurrence, i.e. a n = c a n + c 2 a n. Hasse [4] used the Chebotarev Density Theorem to determine, for a given non-zero integer a, the natural density of primes that divide some term in the sequence (a n + ) n=. In 985, Lagarias [6] used Kummer theory and the Chebotarev Density Theorem to show that the set of primes dividing some term in the Lucas sequence has density 2. Lagarias [6] has also shown, assuming 3 the Riemann Hypothesis for certain Kummer extensions, that the natural density of Date: September 20, Mathematics Subject Classification. Primary B37; Secondary 2F0.

2 2 RICHARD GOTTESMAN AND KWOKFUNG TANG primes dividing some term of an integer sequence defined by a non-degenerate binary linear recurrence relation exists and is positive. The focus of this paper is to study the natural density of primes that divide a different class of integer sequences: those obtained via iteration of a non-linear polynomial. Despite the large amount of research done on primes dividing linear recurrence sequences, relatively few results are known about primes dividing the integer sequences we consider. If f is a polynomial, we let f n denote the nth iterate of f, which is defined as the composition of f with itself n times. The Fermat numbers, F n = 2 2n +, give a famous example of a sequence obtained by iterating a quadratic polynomial. We note that if f(x) = (x ) 2 +, then f n (3) = F n. We obtain another nice example by observing that if f(x) = x 2 2 then f n (3) = L 2 n+, where L m denotes the mth Lucas number. If f(x) Z[x] and a Z, let P (f, a) denote the set of primes that divide at least one term in the sequence (f n (a)) n=. Recent work of Jones [5] establishes that the set P (f, a) is infinite provided f(x) Ax d and (f n (a)) n= is unbounded. We are interested in investigating the natural density of P (f, a). We remark that it is an open problem to show that the natural density of P (f, a) exists in general. Odoni [8] used the Hilbert Irreducibility Theorem to prove that for most polynomials f, the natural density of P (f, a) (if it exists) must be quite small. Odoni s result is illuminating but is not explicit. Indeed, it does not allow us to determine the natural density of P (f, a) for even a single polynomial f and a single integer a. A variant of Euclid s famous proof of the infinitude of the primes proceeds by showing that infinitely many primes divide the sequence (w n ) defined by w 0 = 2 and w n = + w w 2 w n. We note that w n = f(w n ) where f(x) = x 2 x +, and thus this Euclid-style proof is equivalent to the observation that P (x 2 x +, 2) is infinite. What is the natural density of primes that divide this sequence? Odoni [9] proved that this density equals 0 by carefully analyzing the Galois groups of the iterates of x 2 x+. Odoni s result indicates that although this Euclid-type iterative sequence has infinitely many prime divisors, this set of prime divisors is quite small since it has density 0. Can we ever do better than Euclid? That is, can we ever find a quadratic polynomial f and an integer a such that the natural density of P (f, a) is positive? Numerical evidence together with the recent work of Jones indicates that such (f, a) should be rare. Jones shows in [5] that D(P (f, a)) = 0 for two infinite families of quadratic f, and has made a plausible conjecture that D(P (f, a)) = 0 for essentially all quadratic f and all integers a. The main result of this paper is to prove that in some truly exceptional cases, we do obtain positive densities. Our main result is the following. Theorem.. Let t Z, f t (x) = (x + t) 2 2 t and D(t) = D(P (f t (x), f t (0))). Then for all integers t, D(t) exists. If t > 2, and if t is not of the form ±M 2 ± 2 or ±2M 2 ± 2, then D(t) =. Moreover, if t > 2, and if there does not exist an integer 3 s such that t 2 2s 2 = 4, then we have the following:

3 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 3 ) If t is of the form ±(M 2 + 2) or ±(2M 2 + 2), then D(t) =. 3 2) If t is of the form ±(2M 2 2), then D(t) = We now discuss what motivated us to study the family f t and the density D(t). We say that a quadratic polynomial f with critical point α is critically finite if the set {f i (α) : i } is finite. Jones conjecture [5, Conjecture 4.7] suggests that if D(P (f, a)) > 0 then f is a critically finite polynomial. A simple calculation shows that f(x) is a critically finite quadratic polynomial if and only if there exists an integer t such that f(x) = (x+t) 2 t+m, for m {0,, 2}. If m = then some computations suggest that for all integers a, D(P (f, a)) = 0 (where we disregard any terms equal to zero). In the case when m = 2 or m = 0, we believe that D(P (f, a)) > 0 if and only if there exists some integer n such that a = f n (0), and the sequence (f n (a)) n= is unbounded. The case when f(x) = (x + t) 2 t and a {f n (0) : n } is very well understood. In this case, we define D(t) := D(P ((x + t) 2 t, a)). It is relatively easy to show that D(t) = D({p : ord p t is odd}). It is in this context that D(t) has been extensively studied. It follows from the work of Hasse [4] that D(t) exists and is a rational number. It has been shown (see [2], [4]) that D(t) = provided there does not exist an integer 3 M such that t = ±M 2 or t = ±2M 2. The only remaining pairs (f, a) which (at least conjecturally) yield non-zero densities are given by f(x) = (x + t) 2 2 t = f t (x), a {f n (0) : n }. This is the subject of our paper. We now give an outline of the rest of the paper and a brief discussion of how we prove Theorem.. The Galois groups of the iterates of f play a vital role in our work as well as in the works of Jones and Odoni. Let K n denote the splitting field of f n over Q, G n = Gal(K n /Q), and λ n denote the proportion of elements of G n that fix at least one root of f n. In Section 2, we show that if for all integers n, G n is isomorphic to a subgroup of Aff(Z/2 n Z), the group of invertible affine linear transformations on Z/2 n Z, and if only finitely many primes ramify in n K n, then D(P (f, a)) exists and equals lim n λ n. This is the content of Theorem 2.4. We use Galois theory, the Chebotarev Density theorem, and some basic group theory to prove this result. We wish to apply Theorem 2.4 in our analysis of D(t). If f(x) = (x + t) 2 2 t then f(x) is critically finite and it follows from [5, Lemma 2.8] and [7, Corollary 2, p. 57] that only finitely many primes ramify in n K n. In Section 3., we prove that if f(x) = (x + t) 2 2 t, then for all integers n, G n is a subgroup of Aff(Z/2 n Z). It follows that D(t) exists and equals lim n λ n by Theorem 2.4. In Sections 3.2, 3.3, and 3.4, we determine the exact structure of G n in the cases dealt with in Theorem.. In Section 4, we use our results from Section 3 to give an exact formula for λ n and find the numerical value of D(t) = lim n λ n. Our analysis of D(t) helps to explain the remarkable similarities between the densities D(t) and D(t). It can be proven that if f(x) = (x + t) 2 t, then for all integers n, G n is a subgroup of Aff(Z/2 n Z). It follows that D(t) exists and equals lim n λ n by Theorem 2.4. It can also be proven that if there does not exist an integer M such

4 4 RICHARD GOTTESMAN AND KWOKFUNG TANG that t = ±M 2 or t = ±2M 2, then G n = Aff(Z/2 n Z) and thus D(t) = lim n λ n =. 3 We do not give proofs of these results in our paper since the density D(t) has been previously studied. We do prove that if f(x) = (x+t) 2 2 t and t is not of the form ±M 2 ± 2 or ±2M 2 ± 2 then G n = Aff(Z/2 n Z) and thus D(t) = lim n λ n =. It is 3 a wonderful fact that the Galois groups of iterates of (x + t) 2 t and (x + t) 2 2 t are so closely connected. We wonder if there exists some greater theory lurking in the background that would shine more light upon this mysterious connection. Acknowledgements. The authors are grateful to Rafe Jones and Jeremy Rouse for their generosity and guidance. We would also like to thank Ken Ono, Sharon Garthwaite, and Karl Mahlburg for making the 2006 REU at the University of Wisconsin, Madison a success. 2. Determining the Density via Galois Theory Our first task is to give a nice description of P (f, a) when a is an iterate of 0. Lemma 2.. Let f(x) Z[x] and let p be a prime such that there exists an integer m > p with the property that f m (x) has a root modulo p. Then p P (f, f(0)). Proof. Let a Z be such that f m (a) 0 (mod p). By the pigeonhole principle, there exist integers j and k with 0 j < k m, such that f j (a) f k (a) (mod p). Thus for all b 0, f j+b (a) f k+b (a)(mod p). Let c Z such that k + c = m and let n = j + c. We note that f n (a) f m (a) 0 (mod p) and n = j + c < k + c = m. Hence f m n (0) f m n (f n (a)) f m (a) 0 (mod p). We conclude that p f m n (0) and we have that p P (f, f(0)). Theorem 2.2. If f(x) Z[x] then P (f, f(0)) = {p : for all n, f n (x) has a root modulo p}. Proof. We first show the LHS is contained in the RHS. Let p be a prime such that p f m (0), for some m. Hence the orbit modulo p of 0 under f is a cycle, and so there exist infinitely many positive integers n such that p f n (0). It follows that f n (x) has a root modulo p for infinitely many positive integers n. Let k be a positive integer. Choose an integer i such that i k and f i (x) has a root modulo p. Let b Z be such that f i (b) 0 (mod p). We have that f k (x) must have a root modulo p since f k (f i k (b)) = f i (b) 0 (mod p). This proves that p is an element of the RHS. We now need to prove the reverse inclusion. Let p be a prime such that for all n, f n (x) has a root modulo p. Then there exists an integer m such that m > p and f m (x) has a root modulo p. We now apply Lemma 2. to conclude that p is an element of P (f, f(0)). We are interested in determining the natural density of P (f, f(0)). In order to proceed, we must establish a lemma on properties of lifts of certain elements contained

5 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 5 in affine groups, which we define in a moment. We begin our discussion with some preliminaries. Let R be a commutative ring. The set of invertible affine linear transformations of R, which we denote Aff(R), is a group under composition. We write Aff(R) = {(x ax + b) : a R and b R}. The group Aff(R) acts naturally on R. We say that an element g Aff(R) has a fixed point if there exists some x R such that g(x) = x. Let Z 2 denote the ring of 2-adic integers, A = Aff(Z 2 ), A n = Aff(Z 2 /2 n Z 2 ), and φ n : A A n denote the natural quotient map. We observe that A is canonically isomorphic to Aff(Z/2 n Z) and that A is the inverse limit of the A n. We note that {( ) } A n l k = : k Z/2 0 n Z and l (Z/2 n Z). We let v 2 denote the 2-adic valuation on Z 2 and note that for any positive integer m and for all x, y Z 2 such that x 0 (mod 2 m ) we have v 2 (x) = v 2 (x + 2 m y). Thus v 2 is well-defined on all of the nonzero elements of Z 2 /2 m Z 2. Lemma 2.3. Let g be an element of A such that there exists a positive integer n with the property that φ n (g) has a fixed point and φ n (g) is not equal to the identity element of A n. Then for all m n, φ m (g) has a fixed point. Proof. We write g : x ax + b where a and b are elements of Z 2. Let k be a positive integer and let a k and b k be the reductions modulo 2 k Z 2 of a and b. We give a necessary and sufficient condition for φ k (g) to have a fixed point. The element φ k (g) has a fixed point if and only if there exists some y Z 2 /2 k Z 2 such that a k y + b k = y. There are three cases to consider. If a k = and the equation a k y + b k = y has a solution then b k = 0 and thus φ k (g) is the identity element of A k. If b k = 0 then the equation a k y + b k = y always has a solution. Now let s consider the case when a k and b k 0. In this case, v 2 (a k ) and v 2 (b k ) are well-defined, and φ k (g) has a fixed point if and only if there exists some y Z 2 /2 k Z 2 such that (a k )y + b k = 0, which occurs if and only if v 2 (a k ) v 2 (b k ). We divide the rest of the proof of the lemma into two cases. Case : b n = 0. In this case, a n since φ n (g) is not equal to the identity element of A n. Thus for all m n, v 2 (a n ) = v 2 (a m ) < n. We note that b n = 0 implies that for all m n, b m 0 (mod 2 n Z 2 ). We now conclude that for all m n, there exists some y Z 2 /2 m Z 2 such that a m y + b m = y since b m 0 (mod 2 n Z 2 ) and a m 0 (mod 2 n Z 2 ). It follows that for all m n, φ m (g) has a fixed point. Case 2: b n 0. The assumption that φ n (g) has a fixed point implies that a n 0 and v 2 (a n ) v 2 (b n ). Moreover, for all m n, v 2 (a n ) = v 2 (a m ) and v 2 (b m ) = v 2 (b n ) since a n 0 and b n 0. Thus for all m n, v 2 (a m ) = v 2 (a n ) v 2 (b n ) = v 2 (b m ).

6 6 RICHARD GOTTESMAN AND KWOKFUNG TANG The fact that for all m n, v 2 (a m ) v 2 (b m ) implies φ m (g) has a fixed point for all m n. We can now use Lemma 2.3 to show that under an interesting hypothesis, the natural density of P (f, f(0)) exists and equals a limit involving Galois-theoretic quantities. Theorem 2.4. Let f(y) Z[y], K n denote the splitting field of f n (y) over Q, G n = Gal(K n /Q), and λ n denote the proportion of elements of G n which fix at least one root of f n (y). Assume that for all n, G n is isomorphic to a subgroup of Aff(Z/2 n Z) and that there are only finitely many primes which ramify in n K n. Then D(P (f, f(0)) exists and equals lim n λ n. Proof. We will first prove that lim n λ n exists by showing that (λ n ) n is a nonincreasing sequence of positive real numbers. Let n be a positive integer. We let S n denote the subset of G n consisting of elements that fix at least one root of f n (y), and we let χ n : G n+ G n be the natural quotient map. We claim that if σ S n+ then χ n (σ) S n. To see that this is the case, note that if σ S n+ then there exists some root α of f n+ (y) such that σ(α) = α. It follows that f(α) is a root of f n (y) and σ(f(α)) = f(σ(α)) = f(α). Hence χ n (σ) fixes f(α) and so χ n (σ) S n. Thus #S n #χ n (S n+ ) = #S n+ #Kerχ n, and λ n = #Sn #G n #S n+ #Kerχ n #G n = #S n+ #G n+ = λ n+. We have now established that lim n λ n exists. Let A x,n := {p x : f n (y) has a root modulo p but does not split completely modulo p} B x,n := {p x : f n (y) splits completely modulo p} R x := {p x : for all n, f n (y) has a root modulo p} T := {p : p ramifies in K n } n= #R Our characterization of P (f, f(0)) in Theorem 2.2 implies D(P (f, f(0)) = lim x x. π(x) We claim that A x,n T R x. Suppose that p A x,n T. Then p is unramified in K n and we let Frob p,n denote the Frobenius conjugacy class at the prime p in G n and Frob p denote the global Frobenius conjugacy class at the prime p in Gal( K n /Q). We remark that the number of fixed points of a permutation is invariant under conjugation and φ n (Frob p ) = Frob p,n. The fact that p A x,n T implies φ n (Frob p ) fixes some root of f n (y) but does not fix all the roots of this polynomial. This fact and the assumption that for all n, G n is isomorphic to a subgroup of Aff(Z/2 n Z), allows us to apply Lemma 2.3 to conclude, that for all m n, φ m (Frob p ) fixes some root of f m (y). Thus for all m n, f m (y) has a root modulo p and consequently p R x. We have now shown that A x,n T R x. It follows that #Ax,n T #Rx. The assumption π(x) π(x) that T is finite together with the Chebotarev Density Theorem imply λ n #A x,n T = lim #G n x π(x) lim inf x #R x π(x).

7 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 7 Hence lim n λ n lim inf x #R x π(x). We now give an upper bound for lim sup x #R x π(x) note that R x A x,n B x,n and the sets A x,n and B x,n are disjoint. Thus #R x #A x,n + #B x,n. It follows that lim sup x #R x π(x) lim sup x #A x,n π(x). To achieve this upper bound, + lim sup x #B x,n π(x) By the Chebotarev Density Theorem, the sequences of the right side of () have limits that exist and equal (λ n #G n ) and #R #G n, respectively. Thus lim sup x x π(x) (λ n #G n ) + #R #G n = λ n and so, lim sup x x lim π(x) n λ n. But we have already shown that lim inf x x lim π(x) n λ n. Thus D(P (f, f(0)) exists and equals #R lim n λ n. 3. Computing the Galois Groups 3.. Splitting Field and Affine Groups. We will prove Theorem. case by case. The cases t = 0, ±, ±2 are trivial and throughout we assume t is not one of these values. We first compute the splitting field of the n-th iterate of f(x) = (x+t) 2 (2+t). Fix t Z, and note that f n (x) = (((x + t) 2 2) 2 2) 2 (2 + t). (2) }{{} n squares Throughout the rest of this paper, we let n Z be such that n > 2, and we let γ C be a root of f n (x). Write γ + t = α +, and plug into (2). A simple induction α shows that 0 = f n (γ) = α 2n + t. α 2n It follows that α 2n + = t and f n (αζ i α 2n 2 + ) = 0 for i = 0,,..., n αζ2 i 2n, where ζ n 2 n is a primitive 2 n -th root of unity. Thus K n, the splitting field of f n (x) over Q, is Q({αζ2 i + n αζ2 i n }2n i=0 ). The identity (αζ2 k + )(ζ n αζ2 k 2 n + ) = (αζ2 k+ ζ + ) + (αζ n n 2 n αζ2 k+ 2 k + n n shows that αζ k+ 2 n + αζ k+ 2 n Q(αζ k 2 n + αζ k 2 n, αζk 2 + n αζ k 2 n, ζ 2 n + K n = Q(α + α, ζ 2 n +, αζ 2 n + ) ζ 2 n αζ 2 n ζ 2 n ) αζ2 k n () ). By induction, We observe that ζ 8, i Q(ζ 2 n, α) since n > 2. We will use this fact in our computation of Gal(K n /Q).

8 8 RICHARD GOTTESMAN AND KWOKFUNG TANG Lemma 3.. K n (α) = K n (ζ 2 n) = Q(ζ 2 n, α) Proof. It suffices to show that α K n (ζ 2 n) and ζ 2 n K n (α). These assertions follow from the identities ζ α = (α + ) (αζ 2 n α 2 n + ) αζ 2 n ζ 2 n ζ 2 n α ζ 2 n = (ζ 2 n + ) (αζ ζ 2 n 2 n + ) αζ 2 n α α Therefore, we have the diagram Q(ζ 2 n, α) = K n (α) = K n (ζ 2 n) K n Q We have that [Q(ζ 2 n, α) : K n ] 2 since ζ 2 n + K ζ 2 n n. The field Q(ζ 2 n, α) is Galois over Q because it is the splitting field of x 2n+ tx 2n + over Q, and any element σ Gal(Q(ζ 2 n, α))/q) is determined by its action on α and ζ 2 n. The possible actions are { α αζ i 2 n or (0 i < 2 n ) αζ σ : 2 i n ζ 2 n ζ j 2 (0 j < (3) n 2n, j odd). Thus [Q(ζ 2 n, α) : Q] = Gal(Q(ζ 2 n, α) : Q) 2 2n. Since K n is also Galois over Q, all elements in Gal(K n /Q) are inherited from Gal(Q(ζ 2 n, α)/q) via restriction. Given an element { α αζ2 k σ : (0 k < n 2n ) ζ 2 n ζ2 l (0 l < (4) n 2n, l odd). of Gal(Q(ζ 2 n, α)/q), the map { α αζ σ : 2 k n (5) ζ 2 n ζ2 l n restricts to the same automorphism in K n as σ, as one can easily verify by checking the actions of σ and σ on α +, ζ α 2 n + and αζ ζ 2 n 2 n + αζ 2. More succinctly, if we n denote σ := (k, l), and σ := (k, l), then (k, l) Kn = (k, l) Kn. This shows that [K n : Q] ( 2 2n ). The set {(k, l)} is a group under composition, and the map l k (k, l) is easily seen to give an isomorphism between the group {(k, l)} 0

9 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 9 and Aff(Z/2 n Z). If σ := (k, l) and αζ2 m + is a root of f n (x), then σ(αζ m n αζ2 m n 2 + ) = n αζ2 m n αζ2 k+lm +. This corresponds to the following matrix multiplication: n αζ k+lm 2 n ( ) ( ) ( ) l k m lm + k =. 0 This shows that Gal(K n /Q) is isomorphic to a subgroup of Aff(Z/2 n Z) for all values of t, because elements of the form (4) describe all the possible elements of the Galois group The generic case. We show that for a generic t, [Q(ζ 2 n, α) : Q] = 2 2n, which forces [K n : Q] = 2 2n since [Q(ζ 2 n, α) : K n ] 2 and [K n : Q] 2 2n, as shown in the previous subsection. We use the fact that [K n : Q] 2 2n to show that for a generic t, Gal(K n /Q) = Aff(Z/2 n Z). We first establish a lemma about quadratic subfields of Q(ζ 2 n). Lemma 3.2. For n 3, there are exactly 3 quadratic subfields of Q(ζ 2 n) over Q, namely Q( 2), Q(i), Q(i 2). Proof. It is easy to see that all 3 are quadratic subfields. Since Gal(Q(ζ 2 n)/q) = Z/2 n 2 Z Z/2Z is abelian, there are as many order 2 subgroups as index 2 subgroups, and it is straightforward to count that there are 3 order 2 subgroups. Thus there are exactly 3 quadratic subfields of Q(ζ 2 n) over Q, as claimed. Theorem 3.3. For any t Z, Gal(K n /Q) is isomorphic to a subgroup of Aff(Z/2 n Z). Moreover, if t cannot be written as ±M 2 ± 2 or ±2M 2 ± 2 where M Z, then Gal(K n /Q) = Aff(Z/2 n Z). Proof. We have already shown that Gal(K n /Q) is isomorphic to a subgroup of Aff(Z/2 n Z) for all t Z in the previous subsection. Let h(x) = x 2 tx + and let β be a root of h(x). Recall that α 2n is also a root of x 2 tx +. Without loss of generality, we take α 2n = β. We have the following diagram: Q(ζ 2 n, α) Q(ζ 2 n, β) Q(ζ 2 n) Q(β) Q

10 0 RICHARD GOTTESMAN AND KWOKFUNG TANG Since Q Q(β) Q(ζ 2 n) Q(β), and [Q(β) : Q] 2, we have Q(β) Q(ζ 2 n) = Q or Q(β). We show that the latter case cannot hold under the conditions of Theorem 3.3. Since Q(β) = Q( t 2 4), Lemma 3.2 gives that Q(β) is a subfield of Q(ζ 2 n) for the following values of t: t 2 4 = s 2, which implies t = ±2; t 2 4 = s 2, which implies t = ±2; t 2 4 = 2s 2 (6) t 2 4 = s 2, which implies t = ±2; We will show that if (t, s) is a solution to (6) then t = ±2M 2 ± 2 or t = ±M 2 ± 2. It is easy to see that both t and s have to be even, so we let t = 2t and s = 2s. Plugging into (6) gives (t ) 2 = 2(s ) 2. This implies that t is odd, so let t = 2k +. Evaluating and expanding gives 2(k 2 + k) = (s ) 2, which implies that s is even. Let s = 2u. Substituting and simplifying gives k(k+) = u 2. Since k and k + are relatively prime, either ±k is a square or ± k is a square. Since t = 4k+2, t must be of the 2 2 form ±2M 2 ± 2 or ±M 2 ± 2. These ideas along with a simple calculation show that t satisfies (6) if and only if t can be simultaneously expressed as ±2M 2 +2 and ±N 2 2 or as ±2M 2 2 and ±N This observation will be used in the proof of Lemma 3.8. We have thus shown, that for the values of t considered in Theorem 3.3, Q(ζ 2 n) and Q(β) are linearly disjoint over Q and [Q(β) : Q] = 2. We will now determine when [Q(ζ 2 n, α) : Q(ζ 2 n, β)] = 2 n, which will allow us to show [Q(ζ 2 n, α) : Q] = 2 2n. We first prove the following proposition. Proposition 3.4. Let b = [Q(ζ 2 n, α) : Q(ζ 2 n, β)]. Then b is a power of 2 and b 2 n. Moreover, if b > then b = 2 c if and only if α 2c Q(ζ 2 n, β) and α 2c / Q(ζ 2 n, β). Proof. Let m(x) be the minimal polynomial of α over Q(ζ 2 n, β). Then m(x) (x 2n β), and m(x) = b i= (x αζa i 2 n) where b = [Q(ζ 2 n, α) : Q(ζ 2 n, β)] and the a i s are distinct, with 0 a i < 2 n. But since m(x) Q(ζ 2 n, β)[x], we have that m(0) Q(ζ 2 n, β), and so α b Q(ζ 2 n, β). Now α 2n = β Q(ζ 2 n, β) and thus α gcd(b,2n) Q(ζ 2 n, β), because gcd(b, 2 n ) can be written as a Z-linear combination of b and 2 n. The polynomial x gcd(b,2n) α gcd(b,2n) Q(ζ 2 n, β)[x] has α as a root and thus the degree of this polynomial is greater than or equal to deg m(x) = b. Thus gcd(b, 2 n ) = b and so b = 2 c for some c n. The polynomial x 2c α 2c has α as a root and has degree less than deg m(x). Thus this polynomial is not contained in Q(ζ 2 n, β)[x] and so α 2c / Q(ζ 2 n, β). Now, suppose that k Z such that α 2k Q(ζ 2 n, β) and α 2k Q(ζ 2 n, β). These two conditions determine a unique k and thus k = b = [Q(ζ 2 n, α) : Q(ζ 2 n, β)].

11 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS We wish to show the hypotheses of Theorem 3.3 exclude the case c < n. Assume that c < n, and note this implies ±α 2n = β Q(ζ 2 n, β). By a standard theorem in Galois theory [3], the linear disjointness of Q(β) and Q(ζ 2 n) implies that Gal(Q(ζ 2 n, β)/q(β)) = Gal(Q(ζ 2 n)/q) = Z/2Z Z/2 n 2 Z. Thus Q(ζ 2 n, β) also has 3 quadratic subfields over Q(β), namely Q( 2, β), Q(i, β) and Q(i 2, β). We will determine when Q( β) is equal to Q(β) or one of these 3 quadratic subfields. We will show that under the hypotheses of Theorem 3.3, Q( β) is not equal to Q(β) or one of these 3 quadratic subfields. This will be a contradiction and will prove that c = n. Case I: If Q( β) = Q(β), then β = a + bβ (a, b Q) (7) β = a 2 + 2abβ + b 2 β 2 β 2 + 2ab β + a2 b 2 b = 0 2 Since β 2 tβ + = 0 and β / Q, comparing coefficients we have a 2 = b 2, which implies that a = ±b, and t = 2ab = M 2 ± 2, where M is an integer. b 2 Case II: If β Q( 2, β), then write β = a + b 2 + cβ + dβ 2 (a, b, c, d Q). (8) The nontrivial element in Gal(Q( β)/q(β)) maps β to β, and 2 to 2. Applying it to (8) gives β = a b 2 + cβ dβ 2 (9) Adding (8) and (9) gives a + cβ = 0, which forces a = c = 0 since β / Q. Thus β = 2(b + dβ), which implies β 2 + 4bd β + b2 2d 2 d = 0 2 if d 0. Then b 2 = d 2, so t has the form 2M 2 ± 2, where M is an integer. If d = 0, then β = 2b 2, which means β is rational, and implies that t = ±2. Note that in the case d 0, we have β = c2 2( ± β) (c2 Q). (0) This relation will be useful later. Almost identical calculations show that Case III: If β Q(i, β), then t = M 2 ± 2 and β = c 3 i( ± β) (c 3 Q). Case IV: If β Q(i 2, β), then t = 2M 2 ± 2 and β = c 4 i 2( ± β) (c 4 Q).

12 2 RICHARD GOTTESMAN AND KWOKFUNG TANG Thus [Q(ζ 2 n, α) : Q(ζ 2 n, β)] = 2 n for all t that cannot be written as ±M 2 ± 2 or ±2M 2 ± 2. The fact that Q(ζ 2 n) and Q(β) are linearly disjoint implies [Q(ζ 2 n, β) : Q] = [Q(ζ 2 n) : Q][Q(β) : Q] = 2 n 2 = 2 n. Thus [Q(ζ 2 n, α) : Q] = [Q(ζ 2 n, α) : Q(ζ 2 n, β)][q(ζ 2 n, β) : Q] = 2 n 2n = 2 2n, if t is not of the form ±M 2 ±2 or ±2M 2 ±2. In this case, Gal(Q(ζ 2 n, α)/q) consists of all the elements of the form (3), and Gal(K n /Q) is the restriction of these elements to K n. From here it is easy to see that Gal(K n /Q) is isomorphic to Aff(Z/2 n Z) by the discussion on Page 8. Remark. Direct calculations show that t is of the form 2M 2 ± 2 if and only if β = c2 2( β) for some c2 Q, and t is of the form M 2 ± 2 if and only if β = c ( β) for some c Q. We also have that t is of the form M 2 ± 2 if and only if β = c 3 i( β) for some c 3 Q and t is of the form 2M 2 ± 2 if and only if β = c 4 i 2( β) for some c 4 Q. Note also that the ± sign involving the expression for t and that of β are exactly opposite Some Exceptional Cases. Now we move on to a few exceptional cases, namely those t which do not satisfy (6) and are of the form ±2M 2 ± 2 or ±(M 2 + 2). We will prove the following. Theorem 3.5. If t does not satisfy (6) and t is of the form ±2M 2 ± 2 or ±(M 2 + 2), then Gal(Q(ζ 2 n, α)/q) = 2 2n. We will show that for these values of t, no root of x 4 β is contained in Q(ζ 2 n, β), which implies α 2n 2 / Q(ζ 2 n, β). We also have that α 2n = ± β Q(ζ 2 n, β) (see Cases II-IV on Page ). Proposition 3.4 and the fact that α 2n Q(ζ 2 n, β) and α 2n 2 / Q(ζ 2 n, β) imply [Q(ζ 2 n, α) : Q(ζ 2 n, β)] = 2 n. The assumption that t does not satisfy (6) implies Q(ζ 2 n) and Q(β) are linearly disjoint and thus [Q(ζ 2 n, β) : Q] = [Q(ζ 2 n) : Q][Q(β) : Q] = 2 n 2 = 2 n. Putting these ideas together gives that if no root of x 4 β is contained in Q(ζ 2 n, β) then [Q(ζ 2 n, α) : Q] = 2 2n. Let 4 β denote a root of x 4 4 β. We note that if β / Q(ζ 2 n, β) then no root of x 4 β is contained in Q(ζ 2 n, β) since i Q(ζ 2 n, β) and the roots of x 4 β are ± 4 β and ±i 4 β. We wish to prove that 4 β / Q(ζ 2 n, β). We first prove a lemma. Lemma 3.6. There are 3 subfields of Q(ζ 2 n, β) which are of degree 4 over Q(β), namely Q( 2, i, β), Q( 2 + 2, β) and Q(i 2 + 2, β). Proof. Since Q(ζ 2 n) and Q(β) are linearly disjoint, Gal(Q(ζ 2 n, β)/q(β)) = Gal(Q(ζ 2 n)/q) = Z/2Z Z/2 n 2 Z. Since the Galois group is abelian, the number of order 4 subgroups is the same as the number of index 4 subgroups. It is easy to see that there are 3 order 4 subgroups, 2 of them isomorphic to Z/4Z and the remaining one isomorphic to Z/2Z Z/2Z. Since Q( 2, i, β), Q( 2 + 2, β) and Q(i 2 + 2, β) are distinct degree 4 extensions of Q(β) contained in Q(ζ 2 n, β), we are done.

13 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 3 Proposition 3.7. Assume the hypotheses of Theorem 3.5. Let 4 β denote a root of x 4 β. If [Q( 4 β) : Q(β)] = 4, then no root of x 4 β is contained in Q(ζ 2 n, β). Proof. Suppose that some root of x 4 β is contained in Q(ζ 2 n, β). It then follows that all the roots of x 4 β are contained in Q(ζ 2 n, β), since i Q(ζ 2 n, β). The fact that Q(β) and Q(ζ 2 n) are linearly disjoint implies Gal(Q(ζ 2 n, β)/q(β)) is abelian and thus Q( 4 β) is a Galois extension of Q(β). If [Q( 4 β) : Q(β)] = 4 then x 4 β is irreducible over Q(β). We have that x 4 β splits completely in Q( 4 β) since Q( 4 β) is a Galois extension of Q(β) and x 4 β is irreducible over Q(β) and has a root in Q( 4 β). Thus 4 β, i 4 β Q( 4 β) and so i Q( 4 β). We have that Q( 4 β) = Q(i, 2, β) because by Lemma 3.6, the field Q(i, 2, β) is the only degree 4 extension of Q(β) contained in Q(ζ 2 n, β) that contains i. If β > 0 we already obtain a contradiction because we can take 4 β R and then i / Q( 4 β). Thus we assume that β < 0 from now on. Note that Gal(Q(i, 2, β)/q(β)) = Z/2Z Z/2Z. Since the roots of x 4 β are ± 4 β and ±i 4 β, and every element of Gal(Q( 4 β)/q(β)) is completely determined by its action on 4 4 β, each element can described by σ j : β i j 4 β, where j {0,, 2, 3}. Note that every non-trivial element of the Galois group has order 2. Let σ j = τ Gal(Q( 4 β)/q) be such that τ(i) = i and τ( 2) = 2. Then 4 β = τ(τ( 4 β)) = τ(i j 4 β) = i 2j 4 β = 4 β, and so 4 2j which implies that j = 2, and thus τ = σ 2. It follows that σ and σ 3 map i to i. One of them fixes 2, while the other maps 2 to 2. Let 4 β = a + bi + c 2 + di 2 (a, b, c, d Q(β)) Applying σ 2 gives 4 β = a + bi c 2 di 2 Adding gives 2a + 2bi = 0. Thus a = b = 0, because i / Q(β) (recall that β R). We now have 4 β = 2(c+di), and we apply σ to it to get ±i 4 β = 2(c di) (the ± sign is there because we do not know explicitly what σ does to 2). Multiplying the first equation by ±i and subtracting the second equation, we see that ±(ci d) = c di, 2(±i) 2 is a which implies that c = ±d. Thus 4 β = 2c( ± i) = 2cζ 8 ±, since primitive 8-th root of unity. Squaring gives β = ±4c 2 i Q(β, i), which implies that Q( β) = Q(β, i). This is only possible if t = M 2 ± 2 by Case III on page. We only consider the case when t = (M 2 + 2) because the case t = M is not covered in Theorem 3.5. Note that in this case, β = c3 i( + β) (c 3 Q). () Since β < 0, we can take 4 β to be real. We have the equations { 4 β = ±2c = s + s 2 β (s, s 2 Q) β = c ( + β) (c Q). (2)

14 4 RICHARD GOTTESMAN AND KWOKFUNG TANG The second equation comes from multiplying () by i. Now Q( 4 β) = Q(β) is Galois over Q, so let σ Gal(Q(β)/Q) map β to β. Then σ( β) = c ( + β ) = c (+β) β = β by (2), and thus σ( 4 β) = ±i 4 β. This gives s + s 2 β = ±i 4 β by (2), a contradiction because the right side is complex while the left side is real. Thus, to show that 4 β / Q(ζ 2 n, β), it suffices to show that [Q( 4 β) : Q(β)] = 4. Lemma 3.8. If t is of the form ±2M 2 ± 2 and t does not satisfy (6), then [Q( 4 β) : Q(β)] = 4. Proof. For these values of t, since Q( β) Q(β) (see Case I on page ), we know that [Q( 4 β) : Q(β)] equals 2 or 4. If [Q( 4 β) : Q(β)] = 2, the minimal polynomial 4 of β over Q(β) is of degree 2. The minimal polynomial divides x 4 β and thus the possible values of the constant term of the minimal polynomial (which must be an element of Q(β)) are ± β and ±i β. The former is impossible because Q( β) Q(β) for these values of t. If i β Q(β), then Q( β) = Q(β) = Q( β). But β = t t 2 4 is a root of x 2 ( t)x+, and so Q( β) = Q( β) implies that 2 t = M 2 ± 2 by Case I on page, or equivalently t = M 2 2. Our assumption that t is of the form ±2M 2 ± 2 and t does not satisfy (6) implies t is not of the form M 2 2. This follows from the remarks made shortly after (6). Our proof of Lemma 3.8 is now complete. We have thus proven that if t is of the form ±2M 2 ± 2 and t does not satisfy (6) then Gal(Q(ζ 2 n, α)/q) = 2 2n by combining Proposition 3.7 and Lemma 3.8. We will now show that if t is of the form ±(M 2 + 2) and t does not satisfy (6) then Gal(Q(ζ 2 n, α)/q) = 2 2n by proving the following lemma. Lemma 3.9. If t is of the form ±(M 2 + 2) and t does not satisfy (6) then no root of x 4 β is contained in Q(ζ 2 n, β). Proof. Assume that some root of x 4 β is contained in Q(ζ 2 n, β). We then have that all roots of x 4 β are contained in Q(ζ 2 n, β) since i Q(ζ 2 n, β). We first consider the case when t is of the form M and t does not satisfy (6). In this case, t > 0 and so the polynomial x 4 β has a real root, which we denote by 4 β. In this case, we also have that Q( β) = Q(β) and β = c( β) for some c Q. Thus [Q( 4 β) : Q(β)] 2. This fact combined with Lemma 3.2 and the fact that Q( 4 β) R imply Q( 4 β) = Q(β) or Q( 4 β) = Q(β, 2). Hence Q( 4 β) is Galois over Q. Let σ Gal(Q( 4 β)/q) such that σ(β) =. Thus σ( β) = c( /β) = c( β) = β β β and so σ( 4 β) = ±i 4 β / R. But σ( 4 β) must be real since Q( 4 β) R and Q( 4 β) is Galois over Q. This is a contradiction. We now consider the case when t is of the form M 2 2 and t does not satisfy (6). In this case, β is a root of x 2 (M 2 + 2)x + and so we apply our argument in the previous case when t is of the form M to conclude that 4 β / Q(ζ 2 n, β) =

15 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 5 Q(ζ 2 n, β). Since n > 2 we have ζ 8 Q(ζ 2 n, β) and thus 4 β / Q(ζ 2 n, β) if and only 4 if β / Q(ζ 2 n, β). Thus 4 β / Q(ζ 2 n, β) since 4 β / Q(ζ 2 n, β). The proof of this lemma is now complete Galois Groups. Using Theorem 3.5, we can compute Gal(Q(ζ 2 n, α)/q) explicitly if t is of the form ±2M 2 ± 2 or ±(M 2 + 2) and t does not satisfy (6). Proposition 3.0. Assume that t does not satisfy (6). If t = ±(2M 2 + 2) or t = ±(M 2 + 2), then Gal(K n /Q) = Aff(Z/2 n Z). If t = ±(2M 2 2), then Gal(K n /Q) = 2 2n 2, and Gal(K n /Q) is isomorphic to the following subgroups of Aff(Z/2 n Z) respectively: { ( ) } j i either j ± (mod 8) and i even if t = 2M 0 or j ±3 (mod 8) and i odd 2 2 { ( j i 0 ) either j or 3 (mod 8) and i even or j or 3 (mod 8) and i odd } if t = (2M 2 2) Proof. We first look at the case when Q( β) = Q( 2, β), which occurs if t = 2M 2 ±2 (see Case II on page ). From (0) we have β = c 2( ± β) (c Q). (3) Also recall that the sign in t = 2M 2 ± 2 and in (3) are opposite (see Remark ). Let σ Gal(Q(ζ 2 n, α)/q). We know that α 2n = β, so α 2n = ± β. Without loss of generality let α 2n = β. We now have two cases, depending on whether σ fixes β or maps it to. β I) σ(β) = β: Then σ has the form { α αζ2 i σ : (0 i < n 2n ) ζ 2 n ζ j 2 (0 j < (4) n 2n, j odd). By (3), we see that σ( β) = β if and only if σ( 2) = 2. Note also that σ( β) = σ(α 2n ) = (αζ i 2 n)2n = β( ) i (5) If σ( 2) = 2, then σ( β) = β, and thus i has to be even by (5). We note that 2 = ζ8 + ζ 8, so σ( 2) = 2 if and only if ζ j 8 + = ζ ζ j 8 + ζ 8 8, which happens exactly when j ± (mod 8). If σ( 2) = 2, then σ( β) = β, and i is odd. We also have that σ( 2) = 2 if and only if j ±3 (mod 8).

16 6 RICHARD GOTTESMAN AND KWOKFUNG TANG II) σ(β) = β : Then σ has the form σ : { α αζ2 i n (0 i < 2 n ) ζ 2 n ζ j 2 n (0 j < 2n, j odd), (6) and σ( β) = cσ( 2)( ± ) by (3). β Suppose σ( 2) = 2 (by the same argument, this corresponds to the case j ± (mod 8)). Then σ( β) = { β β if the sign in (3) is +, which implies that i is even if the sign in (3) is, which implies that i is odd Suppose σ( 2) = 2 (by the previous argument, this corresponds to the case j ±3 (mod 8)). Then σ( β) = { β β if the sign in (3) is +, which implies that i is odd if the sign in (3) is, which implies that i is even Thus we have the following classification of possible automorphisms: { { σ(β) = β : σ(β) = : β (i, j) (i, j) ± : either j ± (mod 8) and i even or j ±3 (mod 8) and i odd + : : either j ± (mod 8) and i even or j ±3 (mod 8) and i odd either j ± (mod 8) and i odd or j ±3 (mod 8) and i even (7) where the signs in the chart are that in (3), while (i, j) and (i, j) are defined after (5). Thus, in both the + case and the case, the total number of (i, j) and (i, j) s is 2 2n. Since we know that [Q(ζ 2 n, α) : Q] = 2 2n, these are all the elements of Gal(Q(ζ 2 n, α)/q). The key fact which we will now use throughout the proof of this Proposition is that (i, j) Kn = (i, j) Kn. In the + case, the elements on the right coincide with the elements on the left when restricted to K n, so Gal(K n /Q) can be described just by the restriction of the elements on the left to K n.

17 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 7 In the case, the elements of Gal(Q(ζ 2 n, α)/q), restricted to K n, have the same effects as the following set acting on K n : { α αζ2 k σ = (k, l) : (0 k < n 2n ) ζ 2 n ζ2 l (0 l < n 2n, l odd). Thus in the case, Gal(K n /Q) = Aff(Z/2 n Z) by the arguments at the very end of page 8. For Q( β) = Q(i 2, β), which occurs if t = 2M 2 ± 2 (see Case IV on page ), we recall that β = ci 2( ± β) (c Q), (8) An almost identical argument shows that in this case, we have the following classification of the Galois group: { { σ(β) = β : σ(β) = : β (i, j) (i, j) ± : either j or 3 (mod 8) and i even or j or 3 (mod 8) and i odd + : : either j or 3 (mod 8) and i even or j or 3 (mod 8) and i odd either j or 3 (mod 8) and i odd or j or 3 (mod 8) and i even This time, in the + case, Gal(K n /Q) is the full affine group, while in the case Gal(K n /Q) is given by restrictions of the left side of the chart. By the sign relations in Remark, the proposition follows. Thus we have shown that if t = ±(2M 2 + 2) and t does not satisfy (6), then Gal(K n /Q) is the full affine group, and in this case K n actually contains ζ 2 n. We will now prove that if t = ±(M 2 + 2) and t does not satisfy (6), then Gal(K n /Q) = Aff(Z/2 n Z). In this case, Gal(Q(ζ 2 n, α)/q) = 2 2n, and so we can determine Gal(Q(ζ 2 n, α)/q) explicitly using a method that is analogous to the method used in the beginning of the proof of this Proposition. When t = M 2 + 2, Gal(Q(ζ 2 n, α)/q) is as follows: σ(β) = β : (i, j) j odd and i even σ(β) = : β (i, j) j odd and i odd

18 8 RICHARD GOTTESMAN AND KWOKFUNG TANG When t = M 2 2, σ(β) = β : (i, j) either j (mod 4) and i even or j 3 (mod 4) and i odd σ(β) = : β (i, j) either j (mod 4) and i even or j 3 (mod 4) and i odd In both cases, the restriction of Gal(Q(ζ 2 n, α)/q) to K n is equivalent to the restriction of (4) to K n, and thus Gal(K n /Q) = Aff(Z/2 n Z), as claimed. 4. Density Results Theorem 4.. Assume that t does not satisfy (6). If Gal(K n /Q) = Aff(Z/2 n Z), then D(t) = 3. If t = ±(2M 2 2), then D(t) = Proof. By Theorem 2.4, to find the density, it suffices to know the proportion of elements in Gal(K n /Q) that have a fixed( point. ) a b If G n = Aff(Z/2 n Z), an element g = G 0 n (where a (Z/2 n Z) and b Z/2 n Z) has a fixed point if and only if there exists some m Z/2 n Z such that m = am + b, which is equivalent to b a, where h is the subgroup generated by h in Z/2 n Z. If a =, g has a fixed point if and only if b = 0. Given a = 2 i for i > 0, there are 2 n i possible values of b (they are just multiples of 2 i ), and given an i > 0, there are 2 n i 2 n i possible values of a such that a = 2 i. Thus the number of elements in G n that have a fixed point is n ( ) n + 2 n i (2 n i ) = + 22n 3 [ ] 4 i= and thus the proportion of g G n having a fixed point is + [ ( 2 2n 3 4 )n ], which has limit as n. This proves the first claim of the theorem. 3 For t = 2M 2 2, Gal(K n /Q) is isomorphic to the following subgroup of Aff(Z/2 n Z) by Proposition 3.0: { ( ) } j i j ± (mod 8), i even. 0 j ±3 (mod 8), i odd ( ) j i An element g = has a fixed point if and only if i j. This is 0 impossible when i is odd, so we only have to consider the case when j ± (mod 8), and i is even. If j =, g has a fixed point if and only if i = 0, so we consider j. If j (mod 8), then j 8, while if j (mod 8), j = 2. Using

19 QUADRATIC RECURRENCES WITH A POSITIVE DENSITY OF PRIME DIVISORS 9 the reasoning above, we see that the number of elements in Gal(K n /Q) that have a fixed point is n + 2 n k (2 n k ) + 2 n (2 n 3 ). }{{} k=3 }{{} from j (mod 8) contributions from j (mod 8) Since Gal(K n /Q) = 2 2n 2 in this case, the proportion of elements having a fixed point converges to + 2 ( ) = 7, as claimed. The case for t = (2M 2 2) is identical. Combining Theorem 3.3, Proposition 3.0 and Theorem 4., we have now proven Theorem.. References [] Christian Ballot. Density of prime divisors of linear recurrences. Mem. Amer. Math. Soc., 5(55):viii+02, 995. [2] Koji Chinen and Leo Murata. On a distribution property of the residual order of a (mod p). I. J. Number Theory, 05():60 8, [3] David S. Dummit and Richard M. Foote. Abstract algebra. John Wiley and Sons, [4] Helmut Hasse. Über die Dichte der Primzahlen p, für die eine vorgegebene ganzrationale Zahl a 0 von gerader bzw.ungerader Ordnung mod. p ist. Math. Ann., 66:9 23, 966. [5] Rafe Jones. On the density of prime divisors of quadratic recurrences. preprint. [6] J. C. Lagarias. The set of primes dividing the Lucas numbers has density 2/3. Pacific J. Math., 8(2):449 46, 985. [7] W ladys law Narkiewicz. Elementary and analytic theory of algebraic numbers. Springer Monographs in Mathematics. Springer-Verlag, Berlin, third edition, [8] R. W. K. Odoni. The Galois theory of iterates and composites of polynomials. Proc. London Math. Soc. (3), 5(3):385 44, 985. [9] R. W. K. Odoni. On the prime divisors of the sequence w n+ = + w w n. J. London Math. Soc. (2), 32():, 985. Department of Mathematics, Brown University, Providence, RI 0292 Current address: 38 Arbor Road, Roslyn Hts.,NY 577 address: Richard Gottesman@brown.edu Department of Mathematics, University of Chicago, Chicago, IL Current address: Department of Mathematics, University of Chicago, Chicago, IL address: tang.409@gmail.edu