Asymptotically exact heuristics for prime divisors of the sequence {a k +b k } k=1

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1 arxiv:math/03483v [math.nt] 26 Nov 2003 Asymtotically exact heuristics for rime divisors of the sequence {a k +b k } k= Pieter Moree Abstract Let N (x count the number of rimes x with dividing a k + b k for some k. It is known that N (x c(x/logx for some rational number c( that deends in a rather intricate way on a and b. A simle heuristic formula for N (x is roosed and it is roved that it is asymtotically exact, i.e. has the same asymtotic behaviour as N (x. Connections with Ramanujan sums and character sums are discussed. Introduction Let bearime (indeed, throughoutthis notetheletter will beused to indicate rimes. Let g be a non-zero rational number. By ν (g we denote the exonent of in the canonical factorisation of g. If ν (g = 0, then by ord g ( we denote the smallest ositive integer k such that g k (mod. If k =, then g is said to be a rimitive root mod. If g is a rimitive root mod, then g j is a rimitive root mod iff gcd(j, =. There are thus ϕ( rimitive roots mod in (Z/Z, where ϕ denotes Euler s totient function. Let π(x denote the number of rimes x and π g (x the number of rimes x such that g is a rimitive root mod. Artin s celebrated rimitive root conjecture (927 states that if g is an integer with g > and g is not a square, then for some ositive rational number c g we have π g (x c g Aπ(x, as x tends to infinity. Here A denotes Artin s constant A = ( = ( Hooley [3], under assumtion of the Generalized Riemann Hyothesis (GRH, established Artin s conjecture and exlicitly evaluated c g. It is an old heuristic idea that the behaviour of π g (x should be mimicked by H (x = xϕ( /(, the idea being that the robability that g is a rimitive root mod equals ϕ( /( (since this is the density of rimitive Mathematics Subject Classification (200. N37, B83

2 roots in (Z/Z. Using the Siegel-Walfisz theorem (see Lemma below, it is not difficult to show, unconditionally, that H (x Aπ(x. Although true for many g and also on average, it is however not always true, under GRH, that π g (x H (x, i.e., the heuristic H (x is not always asymtotically exact. Nevertheless, Moree [5] found a quadratic modification, H 2 (x, of the above heuristic H (x involving the Legendre symbol that is always asymtotically exact (assuming GRH. A rime is said to divide a sequence S of integers, if it divides at least one term of the sequence S (see [] for a nice introduction to this toic. Several authors studied the roblem of characterising (rimedivisors of the sequence {a k +b k } k=. Hasse [2] seems to have been the first to consider the Dirichlet density of rime divisors of such sequences. Later authors, e.g., Odoni[9] and Wiertelak [] strengthened the analytic asects of his work. The best result to date, in the formulation of [4], seems to be as follows (recall that Li(x = x 2 dt/logt is the logarithmic integral: Theorem Let a and b be non-zero integers. Put r = a/b. Assume that r ±. Let λ be the largest integer such that r = u 2λ, with u a rational number. Let ε = sign(r and L = Q( u. We have ( x(loglogx 4 N (x = δ(rli(x+o log 3, x where the imlied constant may deend on a and b and δ(r, a rational number, is given in Table. Table : The value of δ(r L λ ε = + ε = L Q( 2 λ 0 2 λ /3 2 λ /3 L = Q( 2 λ = 0 7/24 7/24 L = Q( 2 λ = 5/2 2/3 L = Q( 2 λ 2 2 λ /3 2 λ /3 Starting oint in the roof of Theorem is the observation that 2ab divides the sequence {a k +b k } k= iff ord r( is even, where r = a/b. The condition that ord r ( be even is weaker than the condition that ord r ( = and now the analytic tools are strong enough to establish an unconditional result. Note that δ(r does not deend on ǫ in case λ = 0. For a generic choice of a and b, L will be different from Q( 2 and λ will be zero and hence δ(a/b = 2/3. It is not difficult to show [8] that the average density of elements of even order in a finite field of rime cardinality also equals 2/3. In this note analogs of H ( (x and H(2 (x of H (x and H 2 (x will be introduced and it will be shown that H (2 (x is always asymtotically exact. This leads to the following main result (where π(x;k,l denotes the number of rimes x satisfying l(mod k and ( / denotes the Legendre symbol: 2

3 Theorem 2 Let a and b be non-negative natural numbers. Let N (x count the number of rimes x that divide some term a k +b k in the sequence {a k +b k } k=. Put r = a/b and ǫ = sgn(a/b. Assume that r ±. Let h be the largest integer such that r = r0 h for some r 0 Q and h. Put e = ν 2 (h. If ǫ =, then ( x(loglogx N (x = π(x;2 e+, 2 e+ 2 ν2( 4 +O, and if ǫ =, then N (x = π(x x, (r 0 /= ν 2 ( =e+ 2 e+ x, (r 0 /= ν 2 ( >e x, (r 0 /= ν 2 ( >e+ where the imlied constants deend at most on a and b. log 3 x ( x(loglogx 2 ν2( 4 +O log 3 x, 2 Preliminaries The roof of Theorem 2 requires a result from analytic number theory: the Siegel- Walfisz theorem, see e.g., [0, Satz 4.8.3]. For notational convenience we write ( instead of gcd(. Lemma Let C > 0 be arbitary. There exists c > 0 such that π(x;k,l = Li(x ϕ(k +O(xe c logx, uniformly for k log C x, (l,k =, where the imlied constant deends at most on C. Our two heuristics will be based on the following elementary observation in grou theory. Lemma 2 Let h and w 0 be integers. Let G be a cyclic grou of order n. Let G h = {g h : g G} and G h w = {gh : ν 2 (ord(g h = w}. We have #G h = n/(n,h and #G h 0 = 2 ν 2(n/(n,h n/(n,h. Furthermore, for w, we have #G h w = { 2 w ν 2 (n/(n,h n/(n,h if ν 2 (n/(n,h w; 0 otherwise. ( 2 If ν 2 (h ν 2 (n, then every element in G h has odd order. If ν 2 (h < ν 2 (n, then G h 0 G 2h. 3 We have { G h G h \G 2h if ν 2 (n = ν 2 (h+; G 2h if ν 2 (n > ν 2 (h+. If ν 2 (n ν 2 (h, then G h is emty. 3

4 Proof. Let g 0 be a generator of G. On noting that g m 0 = g m 2 0 iff m m 2 (mod n, the roof becomes a simle exercise in solving linear congruences. In this way one infers that G h = {g0 hk : k n/(n,h} and hence #G h = n/(n,h. Note that ord(g0 hk is the smallest ositive integer m such that n/(n,h divides mk. Thus ord(g0 hk will be odd iff ν 2(k ν 2 (n/(n,h. Using this observation we obtain that G h 0 = {g0 hk : k n (n,h, ν n 2(k ν 2 ( } (2 (n,h and hence #G h 0 = 2 ν 2(n/(n,h n/(n,h. Similarly G h w = {g0 hk : k n (n,h, ν n 2(k = ν 2 ( (n,h w} and hence we obtain (. 2 If ν 2 (h < ν 2 (h, then using (2 we infer that G h 0 {ghm 0 : m n (n,h, ν 2(m } = {g 2hk 0 : k n (n,2h } = G2h, where we have written m = 2k and used that (n,2h = 2(n,h. 3 Similar to that of art 2. Remark. Note that G h and G h 0 actually subgrous of G. with the induced grou oeration from G are 3 Two heuristic formulae for N (x In this section we roose two heuristics for N (x; one more refined than the other. Starting oint is the observation that if 2ab divides the sequence {a k +b k } k= if and only if ord r( is even, where r = a/b. Let h be the largest integer such that we can write r = r0 h with r 0 a rational number. Let ǫ = sgn(r. We will use Lemma 2 in the case G = G := (Z/Z = F. The first heuristic aroximation we consider is K ( (x = x, 2ab #G h,( ǫ/2 #G h where K ( (x is suosed to be an heuristic for the number of rimes x such thatord r (isodd. Fromourresultsbelowitwillfollowthatlim x K ( (x/π(x exists. Note that in case h =, this limit is the average density of elements of odd order (if ǫ =, resectively of order congruent to 2(mod 4 (if ǫ =. For a more detailed investigation of the average number of elements having order a(mod d vide [8]. Suose that 2ab. By assumtion r ǫg h. In the case ǫ =, the latter set has #G h,0 elements having odd order and so, in some sense, #Gh,0 /#Gh is the robability that ord r ( is odd. This motivates the definition of K ( (x in case 4,

5 ǫ =. In case ǫ = we use the observation that for is odd, r0 h has odd order iff r0 h has order congruent to 2(mod 4. Thus the elements in Gh of odd order are recisely the elements having order 2(mod 4 in G h and hence have cardinality #G h,. On using art of Lemma 2 we infer that K( (x = x, 2ab k( ( with { k ( (+ǫ/2 if ( = ν2 ( e; 2 e ν 2( (3 if ν 2 ( > e. An heuristic H ( (x for N (x is now obtained on merely setting H ( (x = π(x K ( (x. Put ω(n = n. On using (3 we then infer that H ( (x = π(x;2e+, 2 e x ν 2 ( >e +O(ω(ab. if ǫ = and H ( (x = π(x 2e x ν 2 ( >e +O(ω(ab. if ǫ =. In the context of (near rimitive rootsit is known that the analoga of H ( (x do not always, assuming GRH, exhibit the correct asymtotic behaviour, but that an aroriate quadratic heuristic, i.e. an heuristic taking into account Legendre symbols, always has the correct asymtotic behaviour [5, 6, 7] (in [7] the main result of [6] is roved in a different and much shorter way. With this in mind, we roose a second, more refined, heuristic: H (2 (x. If ν (r = 0 we can consider r = r0 h and r 0 as elements of G. We write (r 0 / = if r 0 is a square in G and (r 0 / = otherwise. First consider the case where ǫ := sgn(r =. If ν 2 ( e := ν 2 (h, then r has odd order by art 2 of Lemma 2. If ν 2 ( > ν 2 (h and (r 0 / =, then r G h, but r G 2h (by art 2 of Lemma 2 again. It then follows that r has even order. On the other hand, if (r 0 / = then r G 2h. This suggests to take K (2 (x = #G h,0 +, #G 2h x, ν 2 ( e x, (r 0 /= ν 2 ( >e where furthermore we require that 2ab. A similar argument, now using art 3 instead of art 2 of Lemma 2, leads to the choice K (2 (x = x, (r 0 /= ν 2 ( =e+ #G h, #G 2h + x, (r 0 /= ν 2 ( >e+ #G h,, #G 2h in case ǫ =, where again we furthermore require that 2ab. We obtain K (2 (x = x, 2ab k(2 (, with (+ǫ/2 if ν 2 ( e; k (2 ( = (+ǫ( r 0 /2 if ν 2 ( = e+; (4 (+( r 0 2 e ν 2( if ν 2 ( > e+. 5

6 Now we ut H (2 (x = π(x K(2 (x as before. On invoking Lemma 2, H(2 (x can then be more exlicitly written as if ǫ = and H (2 (x = π(x;2e+, 2 e+ H (2 (x = π(x if ǫ =. x, (r 0 /= ν 2 ( =e+ 2 e+ x, (r 0 /= ν 2 ( >e x, (r 0 /= ν 2 ( >e+ +O(ω(ab, (5 +O(ω(ab, (6 4 Asymtotic analysis of the heuristic formulae Inthissectionwedeterminetheasymtoticbehaviour ofh ( (xandh(2 (x. We adot the notation from Theorem 2 and in addition write D for the discriminant of Q( r 0. Note that D > 0. Theorem 3 Let A > 0 be arbitrary. The imlied constants below deend at most on A. We have where H ( (x = δ (rli(x+o(xlog A x+o(ω(ab, δ (r = { 2 e /3 if ǫ = +; 2 e /3 if ǫ =. In articular, if L Q( 2, then H ( (x is an asymtotically exact heuristic for N (x. 2 We have H (2 (x = δ(rli(x+o(d2 xlog A x+o(ω(ab. In articular, H (2 (x is an asymtotically exact heuristic for N (x. The roof of art 2 requires a few facts from algebraic number theory, the roof of art does not even require that and is an easier variant of the roof of art 2 (and is left to the interested reader. The roof of art 2 rests on a few lemmas. Lemma 3 Let n be a non-zero integer and K = Q( n a quadratic number field of discriminant. Let A > and C > 0 be ositive real numbers. Then ( ( = Li(x [K(ζ 2 k : Q] x +O, [K(ζ 2 k+ : Q] log A x x, (n/= ν 2 ( =k uniformly in k with k satisfying 2 k+3 log C x, where the imlied constant deends at most on A and C. 6

7 Proof. By quadratic recirocity a rime satisfies (n/ = iff is in a certain set of congruences classes modulo 4. Thus the rimes we are counting in our sum are recisely the rimes that belong to certain congruences classes modulo 2 k+2, but do not belong to certain congruence classes of modulus 2 k+3. The total number of congruence classes involved is less than 8. Now aly Lemma. This yields the result but with an, as yet, unknown density. On the other hand, the rimes that are counted are recisely the rimes x that slit comletely in the normal number field K(ζ 2 k, but do not slit comletely in the normal number field K(ζ 2 k+. If M is any normal extension then it is a consequence of Chebotarev s density theorem that the set of rimes that slit comletely in M has density /[M : Q]. On using this, the roof is comleted. Lemma 4 Let m be fixed. With the notation as in the revious lemma we have ( ( 2 ν2( = Li(x 2 k [K(ζ 2 k : Q] 2 x +O [K(ζ 2 k+ : Q] log A x x, (n/= ν 2 ( m k=m where the imlied constant deends at most on A. Proof. We have x, (n/= ν 2 ( m = m k=m x, (n/= ν 2 ( =k 2 k +O( x 4 m, where we used the trivial bound x, ν 2 ( m 2 ν2( = O(x/4 m. Choose m to be the largest integer such that 2 m+3 log C x. Aly Lemma 3 with any C > A/2. It follows that x, (n/= ν 2 ( m = Li(x = Li(x m k=m k=m ( 2 k [K(ζ 2 k : Q] [K(ζ 2 k+ : Q] ( 2 k [K(ζ 2 k : Q] [K(ζ 2 k+ : Q], +O( x 4 m ; +O( x 4 m, where we used that ϕ(2 k [K(ζ 2 k : Q] 2ϕ(2 k. On noting that O(x/4 m = O( 2 xlog A x, the result follows. Lemma 5 We have H (2 (x = δ 2(rLi(x+O(D 2 xlog A x+o(ω(ab, where δ 2 (r = ( 2 e 2e+ 2 k [L(ζ 2 k : Q] (7 [L(ζ 2 k+ : Q] if ǫ = and if ǫ =. k=e+ δ 2 (r = 2 e+ + [L(ζ 2 e+ : Q] 2 e+ k=e+2 2 k ( [L(ζ 2 e+2 : Q] [L(ζ 2 k : Q], (8 [L(ζ 2 k+ : Q] 7

8 Proof. This easily follows on combining the revious lemma with equation (5, resectively (6. Remark. From (7 and (8 we infer that δ 2 ( r δ 2 ( r = 3 2 e+ + 2 [L(ζ 2 e+ : Q] 2 [L(ζ 2 e+2 : Q]. The number δ 2 (r can be readily evaluated on using the following simle fact from algebraic number theory: Lemma 6 Let K be a real quadratic field. Let k. Then { 2 k if k 2 or K Q( 2; [K(ζ 2 k : Q] = 2 k if k 3 and K = Q( 2. Proof. If K is a quadratic field other than Q( 2 then there is an odd rime that ramifies in it. This rime, however, does not ramify in Q(ζ 2 n, so in this case K and Q( 2 are linearly disjoint. Note that ζ 8 + ζ8 = 2 and hence Q( 2 Q(ζ 8. Using the well-known result that [Q(ζ n : Q] = ϕ(n, the result is then easily comleted. The result of this evaluation is stated below. Lemma 7 We have δ 2 (r = δ(r. After all this reliminary work, it is straightforward to rove the two main results of this note: Proof of Theorem 3. Left to the reader. 2 Combine the latter lemma with Lemma 5. Comarison with Theorem shows that H (2 (x N (x as x and thus H (2 (x is an asymtotically exact aroximation of N (x. Proof of Theorem 2. Combine art 2 of Theorem 3 (with any A > 3, Theorem and equations (5 and (6. 5 Two alternative formulations 5. An alternative formulation using Ramanujan sums Recall that the Ramanujan sum c n (m is defined as k n, (k,n= e2πikm/n. It is well-known that c n (m Z and, more in articular, that c n (m = ϕ(n µ(n/(n,m ϕ(n/(n,m. This is known as Hölder s identity. It imlies that c n (m = c n ((n,m. For our uroses the following weak version of Hölder s identity will suffice: 0 if ν 2 (t < v; c 2 v(t = ϕ(2 v if ν 2 (t = v ; (9 ϕ(2 v if ν 2 (t v. 8

9 Another elementary roerty of Ramanujan sums we need is that for arbitrary natural numbers n and m { c d (m = if n m; (0 n 0 otherwise. d n Suose that ν (r = 0, then ord r ([F : r ] =. Note that ord r( is off iff 2 ν2( [F : r ]. Using identity (0 it then follows that N (x = π(x c 2 v([f : r ]+O(ω(ab. ( x, 2ab v ν 2 ( Corollary below shows that if in the latter double sum the summation is restricted to those v satisfying in addition v e, resectively v e +, then K ( (x, resectively K(2 (x is obtained. This in combination with Theorems and 3 leads to the following theorem: Theorem 4 We have, in the notation of Theorem 2, N (x = π(x ( x(loglogx c 2 ν([f 4 : r ]+O log 3, x x, 2ab 2 v (,2h and x, 2ab e+2 v ν 2 ( c 2 ν([f : r ] = O ( x(loglogx 4 log 3 x, where the imlied constant deends at most on a and b. Remark. Note that v min(ν 2 (,e+ is equivalent with 2 v (,2h. Lemma 8 Let,ǫ and e be as in Theorem 2 and let 2ab. We have c 2 ν([f : r ] = k ( (. 2 We have v min(ν 2 (,e v min(ν 2 (,e+ Corollary For j 2 we have x, 2ab c 2 ν([f v min(ν 2 (,e+j : r ] = k(2 (. c 2 ν([f : r ] = K(j (x. Proof of Lemma 8. Let us consider the case ǫ = and ν 2 ( > e (the remaining cases are similar and left to the reader. Since ( ( /2e (mod we see that and hence r is a 2 e th-ower mod and thus ν 2 ([F : r ] e. Hence the sum in the statement of the lemma reduces to 2 ν 2( v e ϕ(2v = 2 e ν2( = k ( (, where (9, (3 and the identity d nϕ(d = n are used. 9

10 2 The case ν 2 ( e. The quantity under consideration agrees with that of art and by (4 we obtain that k ( ( = (+ǫ/2 = k(2 (. The case ν 2 ( = e+. Now ( 2 e+ (mod,(r0 h 2 e+ ( r 0 (mod and hence r 2 e+ ǫ( r 0 (mod. It follows that ν 2 ([F : r e+ if ǫ( r 0 = andν 2 ([F : r ] = e if ǫ( r 0 =. Using (9 the quantity under consideration is seen to reduce to ( ϕ(2 v +ǫ( r 0 2e = +ǫ(r 0. 2 v e v e By (4 this equals k (2 (. The case ν 2 ( > e+. Now r ( /2+e ( r 0 mod. Proceeding as before the quantity under consideration reduces to ( ϕ(2 v +( r 0 2e = 2 e ν2( (+( r An alternative formulation involving character sums Let G be a cyclic grou of order n and g G. It is not difficult to show that, for any d n, ord(χ=d χ(g = c d([g : g ]. Using this and noting that χ(r = χ(ǫχ h (r 0, equation ( can be rewritten as N (x = π(x χ(ǫχ h (r 0 +O(ω(ab, (2 x, 2ab ord(χ 2 ν 2 ( where the sum is over all characters of F having order dividing 2 ν2(. Note that if χ is of order 2 v, then χ h is the trivial character if v e and a quadratic character if v = e+. If in the main term of (2 only those characters of order dividing h are retained, i.e. those for which χ h is the trivial character, then (x is obtained (this is a reformulation of art of Lemma 8 and hence, by art of Theorem 3 the naïve heuristic. If in (2 only those characters of order dividing 2h are retained, i.e. those for which χ h is the trivial or a quadratic character, then the asymtotically exact heuristic is obtained. The error term assertion in Theorem 4 can be reformulated as: H ( Proosition We have ( x(loglogx χ(ǫχ h 4 (r 0 = O log 3, x x, 2ab 2 e+2 ord(χ 2 ν 2 ( where the imlied constant deends at most on a and b. In the setting of near rimitive roots it is already known that for the main term of the counting function of (near rimitive roots only the contributions coming from characters that are either trivial or quadratic need to be included [6]. 0

11 6 Conclusion There is a naïve heuristic for N (x that in many, but not all, cases is asymtotically exact. There is a quadratic modification of this heuristic involving the Legendre symboll that is always asymtotically exact. The same henomenon is observed (assuming GRH in the setting of Artin s rimitive root conjecture. References [] C. Ballot, Density of rime divisors of linear recurrences, Mem. Amer. Math. Soc. 5 (995, no. 55. [2] H. Hasse, Über diedichte der Primzahlen, fürdieeinevorgegebene ganzrationale Zahl a 0 von gerader bzw. ungerader Ordnung mod. ist, Math. Ann. 66 (966, [3] C. Hooley, Artin s conjecture for rimitive roots, J. Reine Angew. Math. 225 (967, [4] P. Moree, On the divisors of a k +b k, Acta Arith. 80 (997, [5] P. Moree, On rimes in arithmetic rogression having a rescribed rimitive root, J. Number Theory 78 (999, [6] P. Moree, Asymtotically exact heuristics for(near rimitive roots, J. Number Theory 83 (2000, [7] P. Moree, Asymtotically exact heuristics for (near rimitive roots. II, Jaan. J. Math. 29 (2003, [8] P. Moree, On the average number of elements in a finite field with order or index in a rescribed residue class, Finite Fields Al., to aear. [9] R.W.K. Odoni, A conjecture of Krishnamurthy on decimal eriods and some allied roblems, J. Number Theory 3 (98, [0] K. Prachar, Primzahlverteilung, Sringer, New York, 957. [] K. Wiertelak, On the density of some sets of rimes, for which n ord a, Funct. Arox. Comment. Math. 28 (2000, Korteweg-de Vries Institute, University of Amsterdam, Plantage Muidergracht 24, 08 TV Amsterdam, The Netherlands. moree@science.uva.nl