Divisibility properties of the Fibonacci entry point

For a prime $p$, let $Z(p)$ be the smallest positive integer $n$ so that $p$ divides $F_{n}$, the $n$th term in the Fibonacci sequence. Paul Bruckman and Peter Anderson conjectured a formula for $\zeta(m)$, the density of primes $p$ for which $m | Z(p)$ on the basis of numerical evidence. We prove Bruckman and Anderson's conjecture by studying the algebraic group $G : x^{2} - 5y^{2} = 1$ and relating $Z(p)$ to the order of $\alpha = (3/2,1/2) \in G(\F_{p})$. We are then able to use Galois theory and the Chebotarev density theorem to compute $\zeta(m)$.


Introduction and Statement of Results
Let F n denote the Fibonacci sequence defined as usual by F 0 = 0, F 1 = 1 and F n = F n−1 + F n−2 for n ≥ 2. If p is a prime number, the smallest positive integer m for which p|F m is called the Fibonacci entry of p, Z(p). For example, we have Z(11) = 10 since 11|F 10 = 55 is the smallest Fibonacci number that is a multiple of 11.
It is well-known that for every prime p, Z(p) ≤ p + 1 (in fact, a proof of this follows from Lemma 8 and Lemma 10 in Section 3). In 1913, Carmichael (see [3], Theorem XXI) proved that if m = 1, 2, 6, 12, then there is a prime number p so that Z(p) = m. It is not presently known if there are infinitely many primes p for which Z(p) = p + 1.
The main question we study is, given a positive integer m, how often does m divide Z(p)? A natural conjecture would be that Z(p) is "random" mod m and so the answer should be 1/m. However, Lagarias proved in 1985 (see [9] and [10]) that the density of primes p so that Z(p) is even is 2/3. More precisely, lim x→∞ #{p ≤ x : p is prime and Z(p) is even} where π(x) is the number of primes ≤ x. Motivated by this work, Bruckman and Anderson in [2] gathered numerical data and conjectured a formula for where M(m, x) = #{p ≤ x : p is prime and m|Z(p)}. Their conjecture is the following.
For an arbitrary positive integer m, we have where the product is over all prime powers occurring in the prime factorization of m, and 5 4 if m ≡ 10 (mod 20) 1 2 if 20|m.
The main result of the current paper is a proof of this conjecture.
Theorem 2. Conjecture 1 is true for every positive integer m.
One consequence of this result is that the numbers Z(p) show a bias toward being composite. For example, if q is prime, then ζ(q) = q q 2 −1 ≈ 1 q + 1 q 2 , and so the numbers Z(p) are divisible by q more often than entries in a random sequence.
Here L n is the nth Lucas number. These are defined by L 0 = 2, L 1 = 1 and L n = L n−1 + L n−2 for n ≥ 2. In Lemma 10, we use this to relate the Fibonacci entry point of p to the order of α ∈ G(F p ).
Next, we study the density of primes p for which m divides the order of α ∈ G(F p ). In the context of untwisted tori, questions of this nature are quite familiar. They were first considered by Hasse in [5] and [6], and very general results of this type are due to Ballot [1] and Moree [13]. Twisted tori of the type mentioned above are considered by Jones and the second author in [8], and in this paper criteria are given that will guarantee that if α ∈ G(Q) is fixed and ℓ is a prime number, then the density of primes p for which ℓ divides the order of α ∈ G(F p ) exists and equals ℓ ℓ 2 −1 . In the present paper, we extend these results to arbitrary positive integers m. To make the extension to prime powers, we consider preimages of α ∈ G(F p ) under multiplication by ℓ. We say that α has an ℓ n th preimage if there is a β ∈ G(F p ) so that ℓ n β = α. In Lemma 11, we relate the order of α with the number of preimages of α under the multiplication by ℓ. Questions of this type are common in arithmetic dynamics, and these have connections with the Diffie-Hellman key exchange protocol (see [4]). In this context, a generator α ∈ F × q is chosen and an integer x that encodes a message is also chosen. Computing y = α x given α and x is straightforward, but computing x given α and y is quite difficult (note that α is a preimage of y).
To study how often α has an ℓ n th preimage, we show that there are ℓ n elements P n,r of G(C) so that ℓ n P n,r = α. We let K ℓ k be the field obtained by adjoining all of the x and y-coordinates of the ℓ k th preimages to Q. Then, we essentially show that there is an nth preimage of α in G(F p ) if and only if the Artin symbol K ℓ k /Q p fixes an nth preimage for all prime ideals p above p. Finally, we compute Gal(K ℓ k /Q) and compute the relevant density. As a consequence, we are able to prove another conjecture of Anderson and Bruckman.
Theorem 3 (Conjecture 2.1 of [2]). Let p be a prime and ǫ p = p 5 . Given a prime q and integers x, i and j with i ≥ j ≥ 0, let M(q, x, i, j) denote the number of primes p ≤ x such that q i (p − ǫ p ) and q j Z(p). Then To handle the case that m = r i=1 ℓ s i i is composite, we must show that the fields K ℓ s i i , 1 ≤ i ≤ r are (almost) linearly disjoint. The complication in the formula for ζ(m) arises from the fact that Q( An outline of the paper is as follows. We review the relevant algebraic number theory in Section 2. In Section 3 we connect the Fibonacci sequence with the arithmetic of G(F p ) and Z(p) with the order of α ∈ G(F p ), and in turn we connect that with the number of preimages. In Section 4 we define a Galois representation and prove that it is surjective (except in the case that 2 and 5 both divide m). Finally in Section 5 we compute the relevant densities and prove Theorem 3 and Theorem 2.

Background
We begin by reviewing some algebraic number theory. For an introduction to these ideas, see [12]. If L/K is a Galois extension of number fields and α ∈ L, define the norm of α to be N L/K (α) = σ∈Gal(L/K) σ(α).
If K/Q is a field extension, let O K be the ring of algebraic integers in K. We say that a prime number p ramifies in K if in the factorization we have e i > 1 for some i. For a fixed K, only finitely many primes p ramify, and those that ramify are precisely those that divide the discriminant ∆ K .
Suppose now that K/Q is a Galois extension and p is a prime ideal in O K . We say that p is a prime ideal above p if p ∩ Z = (p). This implies that O K /p is a finite extension of F p . If p is unramified in K/Q, then there is a unique element σ ∈ Gal(K/Q) for which for all α ∈ O K . This element is called the Artin symbol of p and is denoted K/Q This set is a conjugacy class in Gal(K/Q).
Let ζ n = e 2πi/n . It follows from the definition of the Artin symbol that if p is a prime and p is a prime ideal above p in O Q(ζn) , then K/Q p (ζ n ) = ζ p n . We are now ready to state the Chebotarev density theorem. This result is the key tool we will use to compute the densities mentioned in the introduction. To state it, let π(x) be the number of primes ≤ x. We will need a standard result in Galois theory. If K 1 and K 2 are two subfields of a field E, let K 1 , K 2 be the smallest subfield that contains both K 1 and K 2 . The following result describes Gal( K 1 , K 2 /F ) in terms of Gal(K 1 /F ), Gal(K 2 /F ) and Gal((K 1 ∩ K 2 )/F ).
Theorem 5 ( [11], Proposition 3.20). Let K 1 and K 2 be Galois extensions of a field F . Then K 1 ∩ K 2 and K 1 , K 2 are both Galois over F and In particular, Finally, given a group G, we will denote its order by |G|. If g ∈ G is an element, we will denote the order of g by |g|. If n is a non-zero integer and ℓ is a prime number we will denote by ord ℓ (n) the highest power of ℓ that divides n.

Connection between the Fibonacci sequence and G(F p )
In this section we will connect the order of the point α = (3/2, 1/2) ∈ G(F p ) with Z(p), the smallest positive integer n so that p|F n . Given a prime number ℓ we will also relate ord ℓ (|α|), ord ℓ (|G(F p )|) and the highest positive integer m so that there is The set G(K) becomes an abelian group with the group operation (x 1 , y 1 ) * (x 2 , y 2 ) = (x 1 x 2 + 5y 1 y 2 , x 1 y 2 + x 2 y 1 ).
The identity of G(K) is (1, 0), and the inverse of (x 1 , y 1 ) is (x 1 , −y 1 ). The group G is a twisted algebraic torus. It becomes isomorphic to G m over K( √ 5).
A simple consequence of this is the following.
Proof. We must prove this for two cases: p ≡ 1, 4 (mod 5) and p ≡ 2, 3 (mod 5). In the first case, 5 is a square in F p and Lemma 6 shows that G(F p ) ∼ = F × p which is cyclic. In the second case, 5 is a square in F p 2 and so G( This shows that G(F p ) is a subgroup of a cyclic group and is hence cyclic.
The following result gives a formula for the order of G(F p ).
Proof. In the case that p ≡ 1, 4 (mod 5) the previous lemma gives |G(F p )| = |F × p | = p − 1. However the following proof handles both cases. A line through the point (1, 0) intersects the curve x 2 − 5y 2 = 1 in another rational point if and only if the slope is rational; the same argument applies in F p . Such a line has the form y ≡ m(x − 1) (mod p). Every point in G(F p ) other than (1, 0) lies on one such a line. Computing the intersection of this line with The argument above shows that this map is surjective, and a straightforward calculation shows that f is injective. It follows that |G(F p ) = |S|. We have that |S| = p − 1 when p ≡ 1, 4 (mod 5) and |S| = p + 1 otherwise. This completes the proof.
The next result shows that the Fibonacci and Lucas sequences occur as the coordinates of multiples of α = (3/2, 1/2) ∈ G(Q).
The next result connects the order of α ∈ G(F p ) with Z(p).
Lemma 11. Suppose that ℓ and p are prime numbers and α ∈ G(F p ). If ℓ ∤ |α|, then there are infinitely many preimages of α under multiplication by ℓ. Suppose that ℓ divides |α| and there is an ℓ m -th preimage of α in G(F p ), but no ℓ m+1 -th preimage. Then Proof. Let α ∈ G(F p ), H = α and φ : H → H be given by φ(x) = ℓx. We have that φ is an automorphism of H if and only if gcd(ℓ, |H|) = 1 and this implies that α has an ℓ m th preimage for all m if and only if ord ℓ (|α|) = 0.
On the other hand, suppose that ℓ divides |α| and write α = sγ, where γ is a generator of the cyclic group G(F p ). Write s = gcd(s, |G(F p )|)s ′ and note that s ′ γ is also a generator of G(F p ). By replacing γ with s ′ γ, we may assume that s divides |G(F p )|.
Let m = ord ℓ (s). Then (s/ℓ m )γ is an ℓ m th preimage of α. It is easy to see, however that if an ℓ m+1 st preimage β m+1 of α existed, then its order would be ℓ m+1 |α|, and this does not divide |G(F p )|. Finally, the order of sγ is |α| = |G(Fp)| s and this gives as desired.

Galois theory
For a positive integer m, let γ m = m 3+ The isomorphism in Lemma 6 shows that if m is a positive integer, there are precisely m elements P m,r ∈ G(C) with mP m,r = α. These are given by Note that m(P m,r − P m,s ) = α − α = 0 and so P m,r − P m,s is a point in G(C) with order dividing m.
To study the frequency with which m divides |α| in G(F p ), we need to know how often the points P m,r "live in G(F p )." More precisely, this means that if K m is the field obtained by adjoining the x and y-coordinates of all the P m,r to Q, we need to determine how often the rational prime p is contained in a prime ideal p ⊂ O Km so that K m /Q p (P m,r ) = P m,r .
If the above equation is true, this implies that the point P m,r (which for all but finitely many primes p can be thought of as an element of the finite field O Km /p) is fixed by a generator of the Galois group of O Km /p over F p , i.e., there is an element β ∈ G(F p ) so that mβ = α. We must first understand Gal(K m /Q) and its action on the points {P m,r }. For simplicity, let Let I(m) = {ax + b : a ∈ (Z/mZ) × , b ∈ (Z/mZ)} denote the affine group over Z/mZ. The group law on I(m) is composition.
Proof. The field K m lies in the Galois extension Q(γ m , ζ m , √ 5) over Q. Then K m is separable as it is an intermediate field of a separable extension, it is a splitting field as we are adjoining all conjugates of the coordinates P m and Q m . We conclude that the extension K m /Q is Galois. For σ ∈ Gal(K m /Q), σ(mP m,r ) = mσ(P m,r ) = α. Therefore Gal(K m /Q) must take a preimage of α to another preimage of α. Hence σ(P m,0 ) = P m,bσ = P m,0 + b σ Q m . Additionally σ(mQ m ) = mσ(Q m ) = 0. So Gal(K m /Q) must take an element of order m to another element of order m. Hence σ(Q m ) = a σ Q m . This yields σ(P m + rQ m ) = σ(P m ) + rσ(Q m ) = P m + (a σ r + b σ )Q m . From the definition of ρ, it is easy to see that ρ is a homomorphism. The kernel of ρ is the set {σ ∈ Gal(K m /Q) : ρ(σ) = x}. If σ ∈ ker ρ, then σ(P m + rQ m ) = P m + rQ m for all r. This implies that P m and Q m are both fixed by σ. Since the coordinates of P m and Q m generate K m over Q, it follows that σ fixes K m and so ker ρ = 1.
The Chinese remainder theorem allows one to see that if the prime factorization of m = r i=1 ℓ s i i , then We will therefore begin by studying the images of Gal(K m /Q) → I(m) in the case that m is a prime power. This case was studied by Jones and the second author in [8].
In particular, Proposition 4.1 and Theorem 4.2 (with the special case d = 5) handle the case that m = ℓ k is a prime. Specifically, if F is a number field, the map is surjective for all k if and only if Using this result, we deduce the following result concerning the surjective of the map ρ : Gal(K m /Q) → I(m).
Lemma 14. If ℓ > 2 is prime, then every minimal subfield of K ℓ k /Q is either Q (P ℓ ) (or some conjugate), or is contained in Q(ζ ℓ 2 ). If ℓ = 2, then every minimal subfield of .
Proof. Lemmas 12 and 13 translate this into a group theory problem. Let N = {ax + b ∈ I(ℓ k ) : a ≡ 1 (mod ℓ)}. This is a normal subgroup of I(ℓ k ). If G is a group, let Φ(G) denote the Frattini subgroup of G, the intersection of the maximal subgroups of G. Since N G, Φ(N) ⊆ Φ(G). Since the order of N is a power of ℓ, every maximal subgroup has index ℓ and so if n ∈ N, then n ℓ ∈ Φ(N). This implies that and in particular, that the kernel of the map from I(ℓ k ) → I(ℓ 2 ) is contained in Φ(G). Since I(ℓ 2 ) is solvable, every maximal subgroup has prime power index. Using the surjective homomorphism φ : I(ℓ 2 ) → (Z/ℓ 2 Z) × , we can identify the maximal subgroups of index less than ℓ and also show there are none of index > ℓ. If M ⊆ I(ℓ 2 ) is a maximal subgroup of index ℓ then either x + 1 ∈ M, in which case for some e with 0 ≤ e ≤ ℓ 2 − 1. All of these subgroups are conjugates of {ax : a ∈ (Z/ℓ 2 Z) × }. Translating back to fields, we obtain the desired result when ℓ > 2.
Lemma 15. Suppose that ℓ is prime. Then K ℓ k /Q is ramified only at 5 and ℓ. If ℓ = 2, then every minimal subfield of K ℓ k /Q is ramified at ℓ.
Proof. From the formulas for the P ℓ k ,r it is clear that K ℓ k is contained in the splitting field of x ℓ k − 3+ is a power of ℓ times a unit, it follows that ∆ L/Q is a power of 5 times a power of ℓ. If L 1 and L 2 are two extensions of Q ramified only at primes in a set S, then L 1 , L 2 /Q is ramified only at primes in S (see Theorem 4.67 of [12]). It follows from this that the splitting field of x ℓ k − 3+ √ 5 2 is ramified only at 5 and ℓ and hence K ℓ k /Q is too.
To prove the second claim it is enough, by Lemma 14, to prove that Q(P ℓ ) is ramified at ℓ. From the surjectivity of ρ proven in Lemma 13, it follows that Gal(K ℓ /Q) acts transitively on the P ℓ,r . Therefore, the Galois closure of Q(P ℓ ) over Q is K ℓ . Since Q(ζ ℓ + ζ −1 ℓ ) is ramified at ℓ and is contained in K ℓ , it follows that K ℓ /Q is ramified at ℓ and this implies that Q(P ℓ ) is ramified at ℓ as well.
Lemma 16. Let m be an arbitrary positive integer. The map ρ : Proof. First note that if gcd(m 1 , m 2 ) = 1, then K m 1 , K m 2 = K m 1 m 2 . It is clear that K m 1 , K m 2 ⊆ K m 1 m 2 . For the reverse direction note that if m 1 x + m 2 y = 1 then m 1 m 2 (yP m 1 + xP m 2 ) = m 2 yα + m 1 xα = α.
Thus yP m 1 + xP m 2 is preimage of α under multiplication by m 1 m 2 . Further, since G( K m 1 , K m 2 ) contains elements of order m 1 and m 2 , it must contain Q m 1 m 2 and so K m 1 m 2 = K m 1 , K m 2 .
Suppose first that 10 ∤ m. To prove that ρ is surjective, we will prove (by induction on the number of distinct prime factors of m) that |Gal(K m /Q)| = |I(m)|. The base case (n = 1) is handled by Lemma 13. Suppose that m = n i=1 ℓ s i i is the prime factorization of m. Since K m/ℓ sn n = K ℓ s 1 1 , K ℓ s 2 2 , . . . , K ℓ s n−1 n−1 we have that K m/ℓ sn n is ramified only at 5 and ℓ 1 , . . . , ℓ n−1 , while K ℓ sn n is ramified only at 5 and ℓ n . Moreover, by Lemma 15 every minimal subfield of K ℓ sn n is ramified at ℓ n . It follows therefore that K ℓ sn n ∩ K m/ℓ sn n is unramified everywhere, and since Q has no unramified extensions we have that K ℓ sn n ∩ K m/ℓ sn n = Q. From Theorem 5 we obtain that | K ℓ sn n , K m/ℓ sn n : Q| = |K ℓ sn n : Q| · |K m/ℓ sn n : Q| = |I(ℓ sn n )| · |I(m/ℓ sn n )| = |I(m)|, which proves the desired claim.
To determine |K 2 s 1 5 s 2 : Q| we will show that K 2 s 1 ∩ K 5 s 2 = Q( √ 5) by determining the subfields of K 2 s 1 that are ramified only at 5. We have that Q( √ 5) ⊆ K 2 s 1 and the subgroup of I(2 s 1 ) corresponding to Q( √ 5) is The field corresponding to this subgroup is L = Q Remark. The method from the previous sections is very general and can be used to establish the surjectivity of Galois representations attached to arbitrary one-dimensional tori.

Density computations
In this section, we will translate conditions on when preimages of α exist in G(F p ) into statements about the Frobenius conjugacy class Km/Q p . We will then count the sizes of these classes and use this to prove Theorem 2. We will start by focusing on the prime power case.
As in the previous section, let ℓ be a prime number, k ≥ 1 and K ℓ k be the field obtained by adjoining all the x and y-coordinates of P ℓ k ,r to Q. Let ρ : Gal(K ℓ k /Q) → I(ℓ k ) be the homomorphism defined in Lemma 12. We define C k,n,ℓ = {σ ∈ Gal(K ℓ k /Q) : σ(ℓ k−n P ℓ k ,r ) = ℓ k−n P ℓ k ,r for some r with 0 ≤ r ≤ ℓ k − 1}.
We will make the convention that if n < 0, then C k,n,ℓ is empty. Since P ℓ k ,r is a ℓ k th preimage of α, ℓ k−n P ℓ k ,r is an ℓ n th preimage of α.
Lemma 17. Let p = 2, 5, ℓ be a prime number. Then, there is an nth preimage of α in G(F p ) if and only if Proof. First suppose that K ℓ k /Q p ⊆ C k,n,ℓ and let p be a prime ideal above p in O K ℓ k .
Let σ = K ℓ k /Q p fix ℓ k−n P ℓ k ,r . We may consider ℓ k−n P ℓ k ,r as an element of G(O K ℓ k /p) (note that 2 and 5 are the only primes that divide the denominators of coordinates of preimages of α). Since σ fixes ℓ k−n P ℓ k ,r and σ acts as the Frobenius automorphism on O K ℓ k /p, it follows that ℓ k−n P ℓ k ,r ∈ G(F p ), as desired.
To show the reverse implication we will first show that for any prime p above p, the reduction mod p map on ℓ n th preimages of α is injective. Any nth preimage of α has the form P ℓ n ,r = P ℓ n + rQ ℓ n . If P ℓ n + r 1 Q ℓ n ≡ P ℓ n + r 2 Q ℓ n (mod p) with r 1 ≡ r 2 (mod ℓ n ), then (r 1 − r 2 )Q ℓ n is congruent to the identity mod p. Every element of order ℓ is a multiple of (r 1 − r 2 )Q ℓ n and this implies that the y-coordinate of Q ℓ is ≡ 0 (mod p) and this implies that This is a contradiction because N Q(ζ ℓ )/Q (ζ ℓ − ζ −1 ℓ ) = ℓ. Hence the reduction mod p map is injective on ℓ n th preimages.
Finally, suppose there is an ℓ n th preimage of α in G(F p ). For a prime ideal p above p, we consider F p ⊆ O K ℓ k /p. Since Gal(K ℓ k /Q) acts on the ℓ n th preimages of α in K ℓ k , and the reduction map is injective, this implies that K ℓ k /Q p ∈ C k,n,ℓ , as desired.
The previous lemma allows us to determine, based on K/Q p , when preimages exist. The next allows us to determine the group order mod ℓ k .
Proof. Let p be a prime above p in O K ℓ k and σ = K ℓ k /Q p . Note that σ(ζ ℓ k ) = ζ p ℓ k and define ǫ ∈ {±1} by σ( Comparing the x-coordinates, we obtain that a ≡ ±p (mod ℓ k ). Using this fact and comparing the y-coordinates gives that a ≡ ǫp (mod ℓ k ). Also, ǫ = 1 if and only if √ 5 ∈ O K ℓ k /p and so ǫ = p 5 . Finally, we use Lemma 8 to conclude that This yields the desired result.
Here a σ denotes the coefficient of x in ρ(σ) = a σ x+b σ ∈ I(ℓ k ). Combining Lemma 11, Lemma 17, and Lemma 18, we see that if p is a prime with p ≡ ±1 (mod ℓ k ), then ℓ t divides |α| if and only if K ℓ k /Q p ⊆ D k,t,ℓ .
We are now able to prove Theorem 3 using similar reasoning.
Proof of Theorem 3. Fix a prime q = 2 and integers i and j with i ≥ j ≥ 0. By Lemma 10, we have ord q (Z(p)) = ord q (|α|) for α = (3/2, 1/2) ∈ G(F p ). Let k = i + 1 and p = 2, 5, q be a prime. As in the proof of Lemma 19, we have ord q (p − ǫ p ) = i and ord q (Z(p)) = j if and only if when σ ∈ K q k /Q p , then ord q (a σ − 1) = i and ord q (b σ ) = i − j, (with the exception that when j = 0, there is a q n th preimage of α for all n and so ord q (b σ ) ≥ i). When i and j are both positive, there are (q − 1) 2 q 2k−2i+j−2 pairs of (a σ , b σ ). When i = 0, there are (q −2)q 2k−2 pairs, and when i ≥ 1 and j = 0, there are (q − 1)q 2k−2i−1 pairs. Applying the Chebotarev density theorem proves the desired result. Now, we turn to the composite case. Let m = r i=1 ℓ s i i . Take M = r i=1 ℓ S i i to be a multiple of m. The discussion following the proof of Lemma 18 now implies the following statement. If p ≡ ±1 (mod ℓ S i i ) for all i, 1 ≤ i ≤ r, then m divides the order of α in G(F p ) if and only if for In the case that m is coprime to 10, it follows from Lemma 16 that Gal( In the case that m is a multiple of 10, Gal(K M /Q) is an index 2 subgroup of the direct product and counting is more tricky. For this reason, we now define for t 1 ≥ 1 and t 2 ≥ 1 D k,t 1 ,t 2 ,10 = σ ∈ Gal(K 10 k /Q) : a σ ≡ 1 (mod 2 k ) and a σ ≡ 1 (mod 5 k ) and if n 1 = ord 2 (a σ − 1), n 2 = ord 5 (a σ − 1), then σ ∈ C k,n 1 −t 1 +1,2 and σ ∈ C k,n 2 −t 2 +1,5 } .
This sum of four geometric series is easily evaluated to give the stated answer. In the case that t 1 = 0, there are again 2 k−n 1 −1 · 4 · 5 k−n 2 −1 choices for a, and 2 k−n 1 · (5 k − 5 k−n 2 +t 2 −1 ) choices for b. Summing yields the stated result.
We are now ready to prove the main result.
Proof of Theorem 2. Let m be a positive integer with gcd(m, 10) = 1 and fix an ǫ > 0. Let m = r i=1 ℓ s i i be the prime factorization of m and note that lim k→∞ |D k,t,ℓ | |Gal(K ℓ k /Q)| = ζ(ℓ t ) by Lemma 19. Choose a positive real number η small enough so that 1 − ǫ 2 ≤ (1 − η) r and (1 + η) r ≤ 1 + ǫ 2 . Now, let M = r i=1 ℓ S i i be chosen with S i is sufficiently large that Combining the Chebotarev density theorem with the observation that ℓ s i i divides the order of α ∈ G(F p ) if and only if K ℓ S i /Q p ⊆ D S i ,s i ,ℓ i (provided p ≡ ±1 (mod ℓ S i i )) we obtain that the number of primes p ≤ x for which m divides |α| satisfies (ζ(ℓ s i i ) + η) ≤ ζ(m)(1 + ǫ/2).