Positive lower density for prime divisors of generic linear recurrences
PPOSITIVE LOWER DENSITY FOR PRIME DIVISORS OF GENERICLINEAR RECURRENCES
OLLI J ¨ARVINIEMI
Abstract.
Let d ≥ P ∈ Z [ x ] be a polynomial of degree d whoseGalois group is S d . Let ( a n ) be a linearly recuresive sequence of integers which has P asits characteristic polynomial. We prove, under the generalized Riemann hypothesis, thatthe lower density of the set of primes which divide at least one element of the sequence( a n ) is positive. Given a sequence of integers, it is natural to consider the set of primes which divide atleast one of its values (the prime divisors of the sequence). Here we consider the primedivisors of linearly recursive sequences. We assume that the elements a , a , . . . of thesequence and the coefficients c i in the defining recursion a n + d + c d − a n + d − + . . . + c a n = 0are integers. The order of ( a n ) is (the minimum possible value of) d .It is a well-known result often attributed to P´olya [16] that, excluding degenerate cases,a linearly recursive sequence has infinitely many prime divisors. One naturally wonders:how dense is the set of prime divisors of a linearly recursive sequence (with respect to theset of primes)?The case of second order sequences has been studied in a number of works under theassumption of the generalized Riemann hypothesis (GRH), under which it is known thatthe density of prime divisors exists, is positive (unless there are only finitely many primedivisors) and can, at least in principle, be computed explicitly (see [1], [5], [7], [9], [10,Section 8.4], [11], [12], [13], [20], [21]). Unconditional results are much more modest: in[14] it is proved that the number of primes p ≤ x which are prime divisors of such asequence is at least of magnitude log x .Higher order sequences have received considerably less attention. In [18] Roskam statesthat “Essentially nothing is known for sequences of order larger than 2”, and it is mentionedthat some very non-generic cases can be handled (see [1]). Roskam proceeds to provingthat, under a certain generalization of Artin’s conjecture on primitive roots, “generic”linear recurrences have a positive lower density of prime divisors. However, we note thata much stronger result under a significantly weaker assumption follows directly from thework of Niederreiter [15]. Nevertheless, while the regular version of Artin’s conjecture hasbeen proven by Hooley [6] under GRH (see [9] for a survey), even the weakened version ofthe generalization seems to be out of reach under standard conjectures. See Section 8 formore discussion.Here we prove, under GRH, that the set of prime divisors of a “generic” linear recur-rences has positive lower density. This seems to be the first such result which is applicableto “almost all” sequences and which assumes only standard conjectures. a r X i v : . [ m a t h . N T ] F e b Olli J¨arviniemi
Theorem 1.1.
Assume GRH. Let d ≥ be an integer and let P ∈ Z [ x ] be a polynomialwhose Galois group is the symmetric group S d and such that the quotient of no two roots of P is a root of unity. Let ( a n ) be a linearly recursive sequence of integers whose character-istic polynomial is P . The set of primes which divide at least one element of the sequence ( a n ) has a lower density of at least / ( d − . In particular, this lower density is strictlypositive. The assumption on the quotients of the roots of P (the non-degeneracy ) follows fromthe assumption on the Galois group for d ≥ d have Galois group isomorphic to S d , sothe result applies to “100%” of linear recurrences. The proof actually works for a slightlywider class of recurrences, and proves that almost all primes p such that P has suitablefactorization modulo p are prime divisors of the sequence. Theorem 1.2.
Assume GRH. Let P ∈ Z [ x ] be a polynomial which has the followingproperties. • d := deg( P ) ≥ . • P is irreducible. • There are infinitely many primes p such that P factorizes as the product of alinear polynomial and an irreducible polynomial of degree d − modulo p , that is,the Galois group of P contains an element whose cycle type is (1 , d − . • If | P (0) | > , the roots of P are multiplicatively independent, and if | P (0) | = 1 ,some (or, equivalently, any) d − roots of P are multiplicatively independent.Let ( a n ) be a linearly recursive sequence of integers whose characteristic polynomial is P .Then almost all primes p such that P factorizes as the product of irreducibles of degree and d − modulo p divide at least one element of the sequence ( a n ) . In particular, theset of primes which divide at least one element of the sequence ( a n ) has a strictly positivelower density. (There are non-trivial examples of polynomials P which do not satisfy the last condition[2].)The proof also adapts to reducible characteristic polynomials in certain special cases,but as such cases are rare, we do not discuss them in detail.The GRH used in the proofs is that the non-trivial zeros of the Dedekind zeta-functionof any number field lie on the line Re( s ) = 1 /
2. (For details, see Lemma 3.2 below and [8,Theorem 3.1].)We first provide a proof sketch, after which we give a detailed argument. We concludeby discussing challenges arising in the study of prime divisors of linear recurrences.
For concreteness we consider the sequence a n defined by a n = 5 n + (3 + √ n + (3 − √ n , n = 1 , , . . . The characteristic polynomial ( x − x − (3 + √ x − (3 − √ p the period of the sequence ( a n )modulo p divides p −
1. We are unable to say anything about whether such primes areprime divisors of ( a n ) or not. rime divisors of linear recurrences p , however, turns out to be accessible.Write n = ( p + 1) k + r, k, r ∈ Z . We may view the numbers 3 ± √ F p ,and by norms (3 ± √ p +1 = 7. Hence a ( p +1) k + r ≡ k + r + 7 k (cid:16) (3 + √ r + (3 − √ r (cid:17) (mod p ) . The equation a ( p +1) k + r ≡ p ) may thus be written as (cid:18) (cid:19) k ≡ − (cid:32) √ (cid:33) r − (cid:32) − √ (cid:33) r (mod p ) . (2.1)Artin’s primitive root conjecture states that a given rational number a is a primitiveroot modulo p for infinitely many primes p as long as a is not − a = 2 thedensity is roughly 37%.It turns out that the order of a rational number a modulo primes is almost always almostmaximal assuming a (cid:54)∈ {− , , } (under GRH). More precisely, the density of primes p with ord p ( a ) ≥ ( p − /C goes to 1 as C → ∞ . (See [22, Section 5] or Lemma 3.2 below.)Note also that if ord p ( a ) = ( p − /h , then the function x → a x (mod p ) , x ∈ Z attainsall non-zero h th powers modulo p as its values.In this light, to prove that (2.1) is solvable for almost any prime p it suffices to showthat the equation x h ≡ − (cid:32) √ (cid:33) r − (cid:32) − √ (cid:33) r (mod p )(2.2)has a solution ( x, r ) with x (cid:54) = 0 for almost any prime p .Note the right hand side is a linear recurrence. Denote it by b r . We aim to prove thatthe sequence b , b , . . . almost always attains a h th power as its value modulo p .By using results from Galois theory and the Chebotarev density theorem, it is not veryhard to show that this is true, for example, if there is an infinite subsequence of ( b r ) whoseelements are distinct primes.Of course, we cannot guarantee that a linear recurrence has infinitely many prime values.However, there are only a very few cases where such an idea does not work. To name one,if b r is always two times a square, then if 2 is a quadratic nonresidue modulo p (whichhappens for a positive density of primes, and in this example case for all p we consider),the sequence b r may avoid all squares modulo p .In general, the only obstructions are that the values of the linear recurrence are almostperfect powers. By applying Zannier’s result on Pisot’s d th root conjecture [23] we reduceto considering whether a linearly recursive sequence arising in the proof is the power ofanother recurrence. From here only elementary observations are needed.As we already mentioned, the sequence a n considered here is not of the form Theorem1.1, and the general case is more complicated. There are two notable differences.To perform the “norm-trick” and to arrive to an equation of the form (2.1) we need tocontrol the norms of the roots of the characteristic polynomial in finite fields. To do so,in the situation of Theorem 1.1 we consider those primes p for which P factorizes as theproduct of irreducibles of degrees 1 and d − p . Olli J¨arviniemi
From here we are able to reduce to an equation similar to (2.2), though this time theright hand side is not necessarily a linear recurrence of integers but of algebraic numbers.By taking norms we reduce to the integer case, the same idea can be implemented andwe are able to show that the reduction of at least one term of the sequence of algebraicnumbers to F p is a h th power almost always.In Section 3 we state the GRH-conditonal result mentioned earlier. In Section 4 wereduce the problem to a polynomial equation in a similar manner as above. We notethat the equation is solvable for a set of primes of density 1 if certain field extensions arelinearly disjoint. We present the tool to handle such questions in Section 5. To apply thetool we have to check that our sequence is not roughly a perfect power, as done in Section6. We wrap up the proof in Section 7. The following lemma is used when transforming our problem into a polynomial equation.This lemma is the only part of the proof that relies on GRH.
Lemma 3.1.
Assume GRH. Let P ∈ Z [ x ] be non-constant and irreducible. Assume P isnot a cyclotomic polynomial, i.e. no root of P is a root of unity, and that P is not theidentity. Let S denote the set of primes such that the equation P ( x ) ≡ p ) has atleast one solution f ( p ) . For C > let S C denote the set of primes p ∈ S for which theorder of f ( p ) modulo p is at least ( p − /C . The (lower) density of S C with respect to S approaches as C → ∞ . Note that we do not say anything about which root f ( p ) of P modulo p we choose ifthere are several of them. The result is true no matter how the choices are made.This result is equivalent to the following algebraic number theoretic formulation. Lemma 3.2.
Assume GRH. Let α be a non-zero algebraic number which is not a root ofunity. Let K = Q ( α ) and let O K denote the ring of integers of K . Let T denote the setof prime ideals of O K whose norm is a prime. For C > let T C denote the set of primes p of T such that the reduction of α in O K / p ∼ = F p has order at least ( p − /C , where p is the norm of p . The (lower) density of T C with respect to T approaches as C → ∞ (where ideals are ordered by norm).Proof. (Cf. [22, Section 5].) Note that almost all ideals of O K belong to T . For k ∈ Z + let T (cid:48) k = T k \ T k − . The results of Lenstra [8] imply that T (cid:48) k has a density for all k . Thisdensity is given by d ( T (cid:48) k ) = ∞ (cid:88) t =1 µ ( t )[ K ( ζ kt , α /kt ) : K ] , where the sum is absolutely convergent. Let f ( n ) = 1 / [ K ( ζ n , α /n ) : K ]. Now rearranginggives ∞ (cid:88) k =1 ∞ (cid:88) t =1 µ ( t ) f ( kt ) = ∞ (cid:88) K =1 f ( K ) (cid:88) d | K µ ( d ) = f (1) = 1 . (cid:3) rime divisors of linear recurrences We consider the setup of Theorem 1.1. The proof also works in the situation of Theorem1.2.Let P and ( a n ) be as in Theorem 1.1. Assume ( a n ) is never zero. Let S denote the setof primes p such that P factorizes as the product of polynomials of degree 1 and d − p . By the assumption and the Chebotarev density theorem the density of S is1 / ( d − S which are prime divisors of ( a n ) isone.We write a n = γ α n + . . . + γ d α nd , where α , . . . , α n are the roots of P and γ , . . . , γ d are constants, and (for further purposes)we define b n = b h,n := α hn (cid:18) − a n γ α n + 1 (cid:19) and c n = c h,n := N F/ Q ( b n ) = ( − d N hn d (cid:89) i =1 (cid:88) j (cid:54) = i γ j α nj γ i α ni , where F is the splitting field of P , N := N F/ Q ( α ) and h ∈ Z + is a parameter fixed later.For an algebraic number field K we let O K denote the ring of integers of K .Note that γ i ∈ Q ( α i ) \ { } (see e.g. [18, Section 2]).For each (unramified) prime p ∈ S there exists a homomorphism ϕ : O F → F p d − mapping one of the roots α i , say α , to an element of F p , and the other roots to elementsof F p d − whose degrees over F p are d −
1. Let K = Q ( α ).Note that for n = k ( p d − − / ( p −
1) + r one has ϕ ( α ) n = ϕ ( α ) kd + r ,ϕ ( α i ) n = N F pd − / F p ( ϕ ( α i )) k ϕ ( α i ) r , ≤ i ≤ d and N F pd − / F p ( α i ) = ϕ ( α ) · · · ϕ ( α d ) = Nϕ ( α ) , ≤ i ≤ d (Clearly ϕ ( α ) (cid:54) = 0 for all but finitely many p .) Hence ϕ ( a n ) = ϕ ( γ ) ϕ ( α ) kd + r + (cid:18) Nϕ ( α ) (cid:19) k ( ϕ ( γ ) ϕ ( α ) r + . . . + ϕ ( γ d ) ϕ ( α d ) r )so ϕ ( a n ) = 0 if and only if ϕ ( α d +11 /N ) k = − ϕ ( γ ) ϕ ( α ) r + . . . + ϕ ( γ d ) ϕ ( α d ) r ϕ ( γ ) ϕ ( α ) r . We then note that α d +11 /N is not a root of unity: Otherwise the conjugates α d +1 i /N areroots of unity as well. Hence the product (cid:89) ≤ i ≤ d α d +1 i /N = N is a root of unity. Hence N = ±
1, and thus the numbers α i are roots of unity. Thiscontradicts the non-degeneracy of P . Olli J¨arviniemi
We may hence apply Lemma 3.1 to α d +11 /N . It suffices to show that for any h ∈ Z + the equation x h = b r , x (cid:54) = q F p for almost all primes p ≡ h ). This may be reformulated as follows:almost all prime ideals of O K ( ζ h ) split in at least one of the fields K r := K ( ζ h , b /hr ) , r =1 , , . . . We use the following basic result from Galois theory [4].
Lemma 5.1.
Let h be a positive integer, let K be a number field containing a h th rootof unity, and let a , a , . . . , a k be a sequence of integers. Assume that for any integers ≤ e i < h , not all zero, one has a e /h · · · a e k /hk (cid:54)∈ K. Then [ K ( a /h , . . . , a /hk ) : K ] = h k . Lemma 5.2.
Let ( x n ) be a linearly recursive sequence of integers. Assume that there donot exist a number field K and integers A, B ≥ , D ≥ such that x An + B is always a D thpower of an element of K . Then there exist a subsequence x n , x n , . . . of ( x n ) and primes p , q , p , q , . . . such that • gcd( v p i ( x n i ) , v q i ( x n i )) = 1 for all i • the primes p , q , p , q , . . . are pairwise distinctProof. Note that the condition implies that ( x n ) has infinitely many prime divisors –otherwise choose A = 1 , B = 0, D = 2 and K to be Q ( √ p , √ p , . . . , √ p n ), where p , . . . , p n are the prime divisors of ( x n ). Note also that for all except finitely many primes p , sayfor all p not belonging to T , the sequence ( x n ) is periodic modulo p k for all k ∈ Z + .We will inductively choose the indices n i and the primes p i , q i , additionally requiringthat p i , q i (cid:54)∈ T . Assume we have already choosen some n , . . . , n k , p , . . . , p k , q , . . . , q k satisfying the conditions. We now pick n k +1 , p k +1 , q k +1 . Let S = { p , q , . . . , p k , q k } ∪ T .Pick some prime divisor p k +1 (cid:54)∈ S , let v p k +1 ( x n ) = t > n such that x n (cid:54) = 0.By periodicity modulo p t +1 k +1 , there exists an arithmetic progression An + B, n = 1 , , . . . such that v p k +1 ( x An + B ) = t for all n If there exist some prime q k +1 (cid:54)∈ S ∪{ p k +1 } and n ∈ Z + such that gcd( v q k +1 ( x An + B ) , t ) =1, we are done. Assume this is not the case.If for any prime q k +1 (cid:54)∈ S and any n we had gcd( v q k +1 ( x An + B ) , t ) = t , then we couldwrite | x An + B | = f ( n ) t (cid:89) s ∈ S s f s ( n ) , for some functions f, f s : Z + → Z ≥ . Then x An + B would always be a perfect t th power ina number field containing the t th roots of all primes of s and a 2 t th root of unity, contraryto the assumption.Hence there exist a prime q k +1 (cid:54)∈ S and n ∈ Z + such that t (cid:48) := gcd( v q k +1 ( x An + B ) , t ) < t .Repeat the above argument with q k +1 in place of p k +1 and t (cid:48) in place of t . The value of t decreases. It must happen that for some value of p k +1 and t we find a prime q k +1 withgcd( v q k +1 ( x An + B ) , t ) = 1. (cid:3) rime divisors of linear recurrences Lemma 5.3.
Let h ∈ Z + and let F be a number field containing a h th root of unity. Let ( x n ) be a linearly recursive sequence of integers satisfying the assumption of Lemma 5.2.Then there exists a subsequence x n , x n , . . . of ( x n ) such that the extensions F ( x /hn ) /F, F ( x /hn ) /F, . . . are linearly disjoint and of degree h .Proof. Note first that there exists a constant c (depending on F ) such that if x is an integerwith x /h ∈ F , then all prime divisors of x which are greater than c have multiplicitydivisible by h . (If p is a prime with h (cid:45) v p ( x ), then p is ramified in the Q ( x /h ). If Q ( x /h ) ⊂ F , then p is ramified in F , and only finitely many primes ramify in F .)Let ( x n i ) be a subsequence constructed in Lemma 5.2. We may assume that the corre-sponding primes p i , q i are larger than c . We prove that this subsequence works by showingthat [ F ( x /hn , x /hn , . . . , x /hn k ) : F ] = h k for all k . Apply Lemma 5.1. Assume that x e /hn · · · x e k /hn k ∈ F, ≤ e i < h. (5.1)By the choice of c , (5.1) implies that the prime divisors of x e n · · · x e k n k which are larger than c have multiplicity divisible by h . By the choice of the primes p i , q i this implies h | e i forall i . (cid:3) In this section we show that Lemma 5.3 may be applied to the sequence ( c n ). Assumenot, so c An + B is always a D th power of an element in a fixed number field.By a result of Zannier [23], the only case when a linear recurrence is always a D th poweris when it is the D th power of a linear recurrence. We may hence write c An + B = d Dn , where d is a linearly recursive sequence (whose elements are not necessarily integers).Write d as an exponential polynomial d n = t (cid:88) m =0 (cid:32) n t u m (cid:88) k =1 e m,k σ nm,k (cid:33) , where the coefficients e m,k are non-zero, σ m, , σ m, , . . . , σ m,u m are non-zero and pairwisedistinct, and i m > m ≤ t .We first show that t = 0, i.e. that the characteristic polynomial of ( d n ) has no repeatedroots. Assume not. Now one sees that d Dn may be written as n Dt (cid:32) u t (cid:88) k =1 e t,k σ nt,k (cid:33) D + e n , where e n is an exponential polynomial whose polynomial terms have degree less than Dt .Since the representation of a linear recurrence as an exponential polynomial is unique, onesees that c An + B = d Dn cannot hold for all n ∈ Z .We may thus write d n = e σ n + . . . + e u σ nu . Olli J¨arviniemi
Each of σ i can be written in the form α x i, α x i, · · · α x i,d d , where x i,j are rational numbers[17]. Let M ∈ Z + be such that x i,j M ∈ Z for all i, j . Write the equation c AMn + B = d DMn as ( − d N h ( AMn + B ) d (cid:89) i =1 (cid:88) j (cid:54) = i γ j α Bj α AMnj γ i α Bi α AMni = u (cid:88) i =1 e i d (cid:89) j =1 α x i,j Mn D Note then that if (cid:81) di =1 α f i i ∈ Q for some integers f i , then f i = f j for all i, j . Indeed:Since the Galois group of P is S d , we have (cid:81) di =1 α f (cid:48) i i = (cid:81) di =1 α f i i for any permutation f (cid:48) i of f i . Hence α f i i α f j j = α f j i α f i j , so ( α i /α j ) f i − f j = 1, which by non-degeneracy of P implies f i = f j .Hence, if N (cid:54) = ±
1, one has (cid:81) di =1 α f i i = 1 only if f i = 0 for all i . By basic results onlinear recurrences, this implies that for any Q ∈ C [ X ± , . . . , X ± d ] in d variables we have Q ( α n , α n , . . . , α nd ) = 0for all integers n if and only if Q is identically zero. Hence( − d N hB ( X · · · X d ) hAM d (cid:89) i =1 (cid:88) j (cid:54) = i γ j α Bj X AMj γ i α Bi X AMi = u (cid:88) i =1 e i d (cid:89) j =1 X x i,j Mj D identically as elements of C [ X ± , . . . , X ± d ].In particular, the left hand side is a perfect D th power in C [ X ± , . . . , X ± d ]. Performa suitable transformation of the form X i → c i X i , clear out constants and simplify. Oneobtains that ( X · · · X d ) ( h − A (cid:48) d (cid:89) i =1 ( X A (cid:48) + . . . + X A (cid:48) d − X A (cid:48) i )is a D th power of a polynomial, where A (cid:48) := AM . This is clearly impossible if D (cid:45) ( h − A (cid:48) .Otherwise we may drop the term ( X · · · X d ) ( h − A (cid:48) . One sees that the polynomials X A (cid:48) + . . . + X A (cid:48) d − X A (cid:48) i are pairwise coprime, and hence each of them must be a D th power. This is not the case,as can be seen, for example, by considering the partial derivative of X A (cid:48) + . . . + X A (cid:48) d withrespect to X at ( X , X , . . . , X d ) = ( x , , . . . , x A (cid:48) + ( d −
2) = 0.The case N = ± Q ∈ C [ X ± , . . . , X ± d − ] wehave Q ( α n , α n , . . . , α nd − ) = 0for all integers n only if Q is zero. Proceeding as before, we have that d (cid:89) i =1 ( X A (cid:48) + . . . + X A (cid:48) d − X A (cid:48) i )is a D th power of a polynomial in the variables X , . . . , X d − (with possibly negativeexponents in the monomials), where X d is shorthand for 1 /X · · · X d − . Note that theterm X A (cid:48) + . . . + X A (cid:48) d − rime divisors of linear recurrences D th powerof a polynomial. We aim to prove that almost all primes of K ( ζ h ) split in at least one of the fields K ( ζ h , b /hn ). By the Chebotarev density theorem it suffices to construct a subsequence b n , b n , . . . of ( b n ) such that the fields K ( ζ h , b /hn ) /K ( ζ h ) , K ( ζ h , b /hn ) /K ( ζ h ) , . . . are linearly disjointIn Section 6 we checked that we may apply Lemma 5.3 to the norm sequence ( c n ). Let c n , c n , . . . denote a subsequence given by the lemma with the base field F ( ζ h ), so F ( ζ h , c /hn i ) /F ( ζ h ) , i = 1 , , . . . are of degree h and linearly disjoint. We claim that this implies that F ( ζ h , b /hn i ) /F ( ζ h ) , i = 1 , , . . . are of degree h and linearly disjoint, too.By Lemma 5.1 it suffices to show that b e /hn · · · b e k /hn k ∈ F ( ζ h ) , ≤ e i < h implies e i = 0 for all i . But if b en · · · b e k n k is an h th power in F ( ζ h ), the norm c en · · · c e k n k isa h th power in F ( ζ h ), too. By the choice of c n i this happens only if e i = 0 for all i .Finally, note that linear disjointness over F ( ζ h ) implies linear disjointness over K ( ζ h ),so K ( ζ h , b /hn ) /K ( ζ h ) , K ( ζ h , b /hn ) /K ( ζ h ) , . . . are linearly disjoint, as desired. The presented proof considers the primes p such that P has factorization type (1 , d − p . Naturally one wonders whether other factorization types could be handled aswell.Unfortunately, this is the only case our argument is able to handle (for irreducible P ):The proof relies crucially on transforming the problem of zeros of linear recurrences toperfect power values of linear recurrences modulo primes. To do so, we first extract normsof the roots of P via the substitution n = k ( p d − − / ( p −
1) + r . One has much lesscontrol over the norms in finite fields if P has more than two irreducible factors modulo p .We then apply Lemma 3.1, which requires that P has a root in F p . These two limitationsresult in having to consider the factorization type (1 , d − P remains irreducible modulo p , assuming an analogue of Lemma 3.1 holdsfor roots of P of degree d over F p (namely that the multiplicative orders of the rootsare of magnitude p d almost always). Note that given a linear recurrence ( a n ) with asquarefree characteristic polynomial P , the period of ( a n ) modulo p is, for large enoughprimes p , equal to the least common multiple of the multiplicative orders of the roots of P in extensions of F p . Hence Roskam’s assumption is equivalent to the period of ( a n )modulo p being of order p d for almost all primes p for which P is irreducible modulo p .A theorem of Niederreiter [15, Theorem 4.1] gives a stronger result under a weakerassumption: as long as the period of the sequence modulo p is at least of magnitude p d/ ,0 Olli J¨arviniemi the sequence does not only attain the value 0 (mod p ), but the sequence is approximatelyequidistributed modulo p . (While we have managed to avoid the consideration of theperiod of a n (mod p ), our approach does not yield equidistribution results or even non-trivial lower bounds for the number of values attained by a n modulo p .)We hence see that analogies of Lemma 3.1 to roots of P in extensions of F p are centralto understanding the behavior of linear recurrences modulo primes. While it seems likelythat such variants of Lemma 3.1 hold (one can present a similar heuristic as for Artin’sconjecture), our understanding is very limited. Unconditionally, we only know that givenan integer a, | a | >
1, the order of a modulo p is almost always > p / [3]. Under GRHone has Lemma 3.1, and an involved variant of Hooley’s classical (conditional) solution ofArtin’s conjecture yields an analogue of Lemma 3.1 in the case where P is of degree 2 andremains irreducible modulo p [19]. It seems that all other cases are open, and as explainedin [19], Hooley’s argument does not adapt to higher degrees without new ideas.We conclude by remarking that the case where P splits into d linear factors modulo p seems to be the most difficult to analyze. In these cases the period of the linear recurrencemodulo p divides p −
1, and heuristically there is a positive density of split primes which arenot prime divisors of the sequence (see [18]). For example, we are not able to say essentiallyanything about the prime divisors of 3 n + 2 n + 1 other than that there are infinitely manyof them. In contrast, heuristics suggest that if P is irreducible and deg( P ) ≥
3, thenalmost all non-split primes are prime divisors of the corresponding sequence (excludingdegenerate cases). This suggests that the lower bound 1 / ( d −
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