Prime Divisors of Palindromes

Transcription

1 Prime Divisors of Palindromes William D. Banks Department of Mathematics, University of Missouri Columbia, MO 6511 USA Igor E. Shparlinski Department of Computing, Macquarie University Sydney, NSW 109, Australia Abstract In this paper, we study some divisibility properties of palindromic numbers in a fixed base g. In particular, if P L denotes the set of palindromes with precisely L digits, we show that for any sufficiently large value of L there exists a palindrome n P L with at least log log n 1+o1 distinct prime divisors, and there exists a palindrome n P L with a prime factor of size at least log n +o1. 1 Introduction For a fixed integer base g, consider the base g representation of an arbitrary natural number n N: L 1 n = a k ng k, 1 k=0 where a k n {0, 1,..., g 1} for each k = 0, 1,..., L 1, and the leading digit a L 1 n is nonzero. The integer n is said to be a palindrome if its digits satisfy the symmetry condition: a k n = a L 1 k n, k = 0, 1,..., L 1. 1

2 It has recently been shown in [1] that almost all palindromes are composite. For any n N, the number L in 1 is called the length of n; let P L N denote the set of all palindromes of length L. In this paper, as in [1], we estimate exponential sums of the form S q L; c = n P L e q cn, where as usual e q x = expπix/q for all x R. Using these estimates, we show that for all sufficiently large values of L, there exists a palindrome n P L with at least log log n 1+o1 distinct prime divisors, and there exists a palindrome n P L with a prime factor of size at least log n +o1. Throughout the paper, all constants defined either explicitly or implicitly via the symbols O, Ω, and may depend on g but are absolute otherwise. We recall that, as usual, the following statements are equivalent: A = OB, B = ΩA, A B, and B A. We also write A B to indicate that B A B. Acknowledgements. The authors wish to thank Florian Luca for some useful conversations. Most of this work was done during a visit by the authors to the Universidad de Cantabria; the hospitality and support of this institution is gratefully acknowledged. W. B. was supported in part by NSF grant DMS , and I. S. was supported in part by ARC grant DP Preliminary Results For every natural number q with gcdq, g = 1, we denote by t q the order of g in the multiplicative group modulo q. For arbitrary integers a, b, K with K 1 we consider the exponential sums T q a, b = t q e q ag k + bg k and T q K; a, b = where the inversion g k is taken in the residue ring Z q. K e q ag k + bg k, Lemma 1. Let S be a set of primes coprime to g, with gcdt p1, t p = 1 for all distinct p 1, p S. Then for the integer q = p S p one has T q a, b = p S T p a, b.

3 Proof. Consider the Kloosterman sums K χ a, b; q = χc e q ac + bc 1 c q gcdc,q=1 as χ varies over the multiplicative characters of Z q. Denoting by X q the group of all such characters for which χg = 1, as in the proof of Lemma.1 of [1] one has T q a, b = t q ϕq χ X q K χ a, b; q. Because of the assumed property of the set S, we see that t q = p S t p, and therefore #X q = ϕq = ϕp = #X p. t q t p p S p S By duality theory, it follows that X q is the direct product of the groups {X p : p S}, hence every character χ X q has a unique decomposition χ = p S χ p where χ p X p for each p S. By the well known multiplicative property of Kloosterman sums, K χ a, b; q = p S K χp a, b; p, and therefore T q a, b = t q ϕq The result follows. χ X q p S K χp a, b; p = p S t p ϕp χ p X p K χp a, b; q. Lemma. Let S be a set of primes p such that p z, p 3 mod 4, gcdp, gg 1 = 1, and t p = Ωlog p for every p S. Suppose that gcdt p1, t p for all distinct p 1, p S. If z is sufficiently large, then for some absolute constant A > 0 and all a, b Z one has where q = p S p. Tq a, b tq p S gcda,b,p=1 1 3 A log plog log p 5,

4 Proof. If t q is odd, then gcdt p1, t p = 1 for all distinct p 1, p S, thus t q = p S t p. By Lemma 1, we also have T q a, b = p S T p a, b. Moreover, T p a, b = t p p 1 x Z p e p ax p 1/t p + bx p 1/tp for all p S. If gcda, b, p = 1, then since t p = Ωlog p, Theorem 1.1 of [] implies that the estimate e p ax p 1/t p + bx p 1/tp A p 1 1 log plog log p 5 x Z p holds for some absolute constant A > 0 provided that z is large enough. On the other hand, T p a, b = t p if gcda, b, p = p. This completes the proof in the case that t q is odd. If t q is even, then the multiplicative order of g modulo q is τ q = t q /, and for each p S the multiplicative order of g modulo p is τ p = t p / or τ p = t p according to whether t p is even or odd, respectively. Since each prime p S is congruent to 3 mod 4, it follows that τ p is odd, and we have τ q = p S τ p. We now write T q a, b = τ q e q af k + bf k + τ q e q agf k + bg 1 f k where f = g. Noting that τ p = Ωlog p for all p, we can apply the preceding argument to both of these sums with g replaced by g, and we derive the stated result in the case that t q is even. 4

5 Lemma 3. If y is sufficiently large, there is a set S [ylogy, y] of primes p with p 3 mod 4 and gcdp, gg 1 = 1, of cardinality at least #S = Ωy 1/4 log y, such that gcdt p1, t p for any distinct p 1, p S, and t p p 1/4 for all p S. Proof. According to Lemma 1 of [3] taking k = 1, u = 3 and v = 16 in that lemma, for every sufficiently large value of y there are at least Ωy/ log y primes p y with p 3 mod 16 such that either p = r 1 r + 1 where r 1, r y 1/4 are primes, or p = r where r 0 is a prime. Clearly, the interval [ylog y, y] also contains a set L of Ωy/ log y such primes. Note that for y large enough, we have that p gg 1 for each p L. Take the smallest such prime p 1 L and put it into the set S. Next, remove all primes p L for which gcdp 1, p 1 1 > ; since this condition implies that gcdp 1, p 1 1 y 1/4, we remove at most Oy 3/4 such primes at this step. Now take the smallest remaining prime p L and add it to S, then remove the Oy 3/4 primes p L for which gcdp 1, p 1 >. Continuing in this manner, we eventually put Ω#Ly 3/4 = Ωy 1/4 log y primes into the set S. Noting that each t p > and t p p 1, it follows that t p y 1/4 p 1/4 for every p S. We also need the following bound for incomplete sums: Lemma 4. For any prime p with gcdp, g = 1 and any natural number K t p, the following bound holds: max Tp K; a, b p 1/ log p. gcda,b,p=1 Proof. It is easy to see that for any h = 0,...,t p, t p e p ag k + bg k e tp hk = t p e p ax p 1/t p + bx p 1/tp χx p 1 x F p where χx is a certain multiplicative character on F p. Applying the Weil bound to the last sum see Example 1 in Appendix 5 of [6]; also Theorem 3 of Chapter 6 in [4], and Theorem 5.41 and the comments to Chapter 5 in [5], we derive that t p e p ag k + bg k e tp hk p 1/. Now using the standard reduction from complete sums to incomplete ones, we obtain the desired result. 5

6 A relation between the sums S q L; c and T q K; a, b has been found in [1] which we now present in a slightly modified form. Lemma 5. Let K = L/. Then Sq L; c g g K Tq K; c, cg L 1 K/. Proof. As in the proof of Lemma 3.1 of [1] we have Sq L; c K g 1 g e q ac g k + g L 1 k. a=0 Then, using the arithmetic-geometric mean inequality, we derive that Sq L; c g 1 K = g 1 K K g 1 e q ac g k + g L 1 k g 1 a,b=0 a,b=0 K/ K/ K e q ca b g k + g L 1 k. Estimating each inner sum trivially as K for all a and b except for a = 1, b = 0, we obtain the desired statement. 3 Exponential Sums with Palindromes Theorem 6. There exists a constant B > 0 such that for all sufficiently large values of L and any prime p L / log 4 L such that gcdp, gg 1 = 1, the following bound holds: max Sp L; c #PL exp L/ log plog log p B. gcdc,p=1 Proof. Taking K = L/, we have by Lemma 5: Sp L; c g g K Tp K; c, cg L 1 K/. 6

7 Suppose that gcdc, p = 1. Let us write K = Qt p + R where Q 0 and 0 R < t p. Let us first consider the case K t p. Since p g tp 1, it is clear that t p = Ωlog p; using Theorem 1.1 of [] as in the proof of Lemma, it follows that for all sufficiently large primes p, Tp c, cg L 1 tp 1 1 log plog log p C 0 for some constant C 0 > 0. Moreover, for any prime p coprime to gg 1, it is clear that t p 1 and that Tp c, cg L 1 < tp. Therefore, adjusting the value of C 0 if necessary, we see that the bound 3 holds for every prime p such that gcdp, gg 1 = 1. Thus, in the case that K t p we have Tp K; c, cg L 1 = Q Tp c, cg L 1 + Tp R; c, cg L 1 1 Qt p 1 + R = K log plog log p C 0 Qt p log plog log p C 0 K 1 When K < t p we apply Lemma 4 to deduce that Tp K; c, cg L 1 p 1/ log p Klog p 1, 1 log plog log p C 0 since K L p 1/ log p. Thus, in this case, we have a much stronger bound. Therefore, for sufficiently large p, g Tp K; c, cg L 1 K g 1 log plog log p C 0 g 1 exp g log plog log p C 0 Using this estimate in together with the obvious relation #P L g L/, we derive the stated result

8 4 Congruences with Palindromes Let us denote P L q = { n P L : n 0 mod q }. Proposition 4. of [1] asserts that if gcdq, gg 1 = 1, then for L 10 + q log q the following asymptotic formula holds: #P L q = #P L #PL + O exp L. q q q Here we obtain a nontrivial bound on #P L q without any restrictions on the size or the arithmetic structure of q. Theorem 7. For all positive integers L and q, the following bound holds: #P L q #P L q 1/. Proof. Let r be the largest integer for which r L mod and g r q. Clearly, g r q. We observe that every palindrome n P L can be expressed in the form n = g L+r/ k 1 + g L r/ m + k where k 1, k < g L r/, g L r/ k 1 + k is a palindrome of length L r, and m < g r. Note that for each choice of k, the value of k 1 is uniquely determined by the palindromy condition. Let d = gcdq, g. If n P L is divisible by q, then d k ; since k 0 there are at most g L r/ /d choices for k. Since g r q, it follows that for each choice of k there are at most d values of m < g r such that the congruence g L+r/ k 1 + g L r/ m + k 0 mod q holds. Therefore, #P L q g L r/ #P L q 1/. 5 Prime Divisors of Palindromes Let ωn denote the number of distinct prime divisors of an integer n. Theorem 8. For all sufficiently large L, there is a palindrome n whose length is L and for which log log n ωn = Ω. log log log n 8

9 Proof. Define y by the equation C 1 y 1/4 log y 1 = log L, where C 1 is the constant implied by the Ω-symbol in Lemma 3, and let S be a set of primes of cardinality #S = C 1 y 1/4 log y with the properties stated in that lemma. Putting q = p S p, by Lemma we see that max Tq a, b C tq 1 gcda,b,q<q log ylog log y 5 for some constant C > 0 provided that L is large enough. In particular, supposing that gcdc, q = 1, we obtain the estimate Tq c, cg L 1 C tq 1 4 log ylog log y 5 since gcdg, q = 1 for sufficiently large L. Taking K = L/, we have by Lemma 5: Sq L; c g g Tq K; c, cg L 1 K/. 5 K As in the proof of Theorem 6, we now write K = Qt q +R with integers Q 0 and 0 R < t q. Because K = L/ t q 1/ t q we have Q 1. Thus, provided that L is large enough, using 4 we derive T q K; c, c = Q T q c, c + T q R; c, c C Qt q 1 + R log ylog log y 5 C Qt q = K log ylog log y K C, 5 log ylog log y 5 since Qt q Q > R. Applying this to 5, it follows that Sq L; c #PL exp C 4 L/ log ylog log y 5 9

10 for some constant C 4 > 0, provided that gcdc, q < q and L is sufficiently large. Now let us denote P L q, a = { n P L : n a mod q }. 6 By the same arguments given in the proof of Proposition 4. of [1], it is easily shown that the preceding estimate implies #P L q, a = #P L q + O #P L exp C 4 L/ log yloglog y 5. In particular P L q, 0 > 0 for sufficiently large L. Taking any n P L q, 0 we obtain ωn ωq #S = Ωy 1/4 log y, and since L log n the result follows. Theorem 9. There is a constant C > 0 such that for all sufficiently large L p p L log L C gcdp,gg 1=1 n P L n Proof. Repeating the same arguments as in the proof of Proposition 4.1 of [1], we derive from Theorem 6 that #P L p, a = #P L p + O #P L exp L/ log plog log p B where, B is defined in Theorem 6 and as before, P L p, a is defined by 6. In particular, #P L p, 0 > 0 provided that L is large enough. Theorem 9 immediately implies that L ω = Ω. log L C n P L n We now use Theorem 7 to derive a more precise result. Theorem 10. For all sufficiently large L, L ω = Ω. log L n P L n 10

11 Proof. Let W = n P L n, s = ωw. For each prime p, we denote by r p the exact power of p dividing W; then and this implies that n P L n = r p = p rp, p W #P L p α. α=1 By Theorem 7 we have the estimate thus, #P L p W r p #P L α=1 p α/ #P L p 1/ ; log p p r 1/ p log p = log W Lg L. p W Denoting by p j the j-th prime number, we obtain L p W which finishes the proof. log p p 1/ s j=1 log p j p 1/ j s log s 1/ 6 Remarks It is an open question posed in [1] as to whether there exist infinitely many prime palindromes in a given base g, and the solution appears to be quite difficult. Indeed, since the collection of palindromes in any base forms a set as thin as that of the square numbers, this question is likely to be as difficult as that of showing the existence of infinitely many primes of the form p = n + 1. At the present time, however, even the question as to whether there exist infinitely squarefree palindromes remains open. 11

12 References [1] W. Banks, D. Hart and M. Sakata, Almost all palindromes are composite, to appear in Math. Res. Lett. [] T. Cochrane, C. Pinner and J. Rosenhouse, Bounds on exponential sums and the polynomial Waring problem mod p, J. London Math. Soc no., [3] D. R. Heath-Brown, Artin s conjecture for primitive roots, Quart. J. Math., , [4] W.-C. W. Li, Number theory with applications, World Scientific, Singapore, [5] R. Lidl and H. Niederreiter, Finite fields, Cambridge University Press, Cambridge, [6] A. Weil, Basic number theory, Springer-Verlag, New York,