On the greatest and least prime factors of n! + 1, II


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1 Publ. Math. Debrecen Manuscript May 7, 004 On the greatest and least prime factors of n! + 1, II By C.L. Stewart In memory of Béla Brindza Abstract. Let ε be a positive real number. We prove that for infinitely many odd integers n the least prime factor of n! + 1 is at most 8 + ε n and that for infinitely many positive integers n the greatest prime factor of n!+1 exceeds 11 εn. 1. Introduction In 1856, Liouville [6] proved that p 1! + 1 is not a power of p for any prime p larger than 5. More than a century later Erdős and Graham [] asked if the equation p 1! + a p 1 = p k 1 has only finitely many solutions in positive integers a,k,p with p an odd prime. In 1991 Brindza and Erdős [1] resolved the question by proving that all solutions of 1 are smaller than an effectively computable number. A few years later Yu and Liu [10] and also Le [5] determined the complete list of solutions. Mathematics Subject Classification: 11D75,11J5. Key words and phrases: shifted factorials, prime factors. This research was supported in part by Grant A358 from the Natural Sciences and Engineering Research Council of Canada and by the Canada Research Chairs Program.
2 C.L. Stewart In 1976 we investigated with Erdős [3] the arithmetical character of integers of the form n! + 1 where n is a positive integer. For any integer m larger than 1 let Pm denote the greatest prime factor of m and let pm denote the least prime factor of m. By Wilson s theorem p divides p 1! + 1 whenever p is a prime. Since all prime factors of n! + 1 exceed n we see that pn! + 1 = n + 1 whenever n + 1 is a prime. We showed with Erdős [3] that if n + 1 is not a prime then Further, for almost all integers n, log n pn! + 1 > n + 1 o1 log log n. pn! + 1 > n + εnn 1, 3 where εn is any positive function that decreases to 0 as n. In [3] we indicated how to prove that for infinitely many integers n for which n + 1 is not a prime pn! + 1 is less than n. We observed, as a direct consequence of Wilson s theorem, that if p is a prime then p 1 i!i! 1 i+1 mod p, 0 i p 1. 4 Thus if p i!+1 for some odd integer i > 1 then, from 4, p p i 1!+1. For any positive real numbers θ and t with t > 1 θ we have and Note that 5 and 6 that p p i 1 = 1 p 1 p i p max p i, max θ, min θ, 1 t θ 1 t 5 1 t θ 1 t. 6 and so on taking θ = p i p p i 1 and t = p 1 p p p 1 = + p 1, we see from and p min i, p + p i 1 p 1,
3 On the greatest and least prime factors of n! + 1, II 3 for 0 < i < p 1. As i tends to infinity so also do p and p i. Thus for each ε > 0, pn! + 1 < + εn, 7 for infinitely many composite integers n + 1. Further Pn! + 1 > n, 8 for infinitely many positive integers n. We indicated in [3] that +ε in 7 could be replaced by δ for some positive number δ. Our first result will be of this character. Theorem 1. Let ε > 0. There are infinitely many odd integers n for which pn! + 1 < + ε n. 9 8 Observe that = With Erdős [3] we proved that holds with Pn! + 1 in place of pn! + 1 for all positive integers n. Of course this only is an improvement on for those integers n for which n + 1 is a prime. Recently Luca and Shparlinski [7] sharpened this result by proving that 1 Pn! + 1 > n o1 log n; 10 indeed they established 10 with Pn!+1 replaced by Pn!+fn where f is any nonzero polynomial with integer coefficients. In 00 Murty and Wong [8] showed that if the abc conjecture holds, then Pn! + 1 > 1 + o1n log n. In 1976 we improved on 8 with Erdős [3] by proving that there is a positive number δ such that Pn! + 1 > + δn, 11 for infinitely many integers n. Luca and Shparlinski [7] established 11 with +δn replaced by 5 + o1 n and showed that their result applies with n! + fn in place of n! + 1 where f is any nonzero polynomial with integer coefficients. Our next result gives a further improvement on 11.
4 4 C.L. Stewart For any set X we denote the cardinality of X by X. For any set A of positive integers and any positive integer n we put An = A {1,...,n}. The lower asymptotic density of A is lim inf An n. Theorem. Let ε > 0. The set of positive integers n for which 11 Pn! + 1 > ε n 1 has positive lower asymptotic density. As we remarked in [3] estimates, 3, 7, 8 and 11 hold with n! + 1 replaced by n! 1 and the same comment applies to the estimates 9 and 1. Further, the same techniques used to prove Theorems 1 and allow one to prove, for instance, that for each positive real number ε there exist infinitely many positive integers n for which Pn! + 1 > ε n and there exist infinitely many positive integers n for which Pn! + 1n! 1 > ε n.. Preliminary lemmas Let p 1,p,... denote the sequence of prime numbers and put d k = p k+1 p k for k = 1,,.... Our first lemma, due to HeathBrown, gives a bound on the frequency of large differences between consecutive prime numbers. Lemma 1. Let ε be a positive real number. There is a positive number c, which depends on ε, such that p k x d k < cx3 18 +ε. Proof. This is Theorem 1 of [4]. Our next result gives a bound for the size of the greatest common divisor of a collection of terms of the form k! + 1.
5 On the greatest and least prime factors of n! + 1, II 5 Lemma. Let n and t be positive integers with t and let i 1,...,i t be distinct positive integers from a subinterval of [1,n] of length l. Then gcdi 1! + 1,...,i t! + 1 < n l t Further, there exists a positive number c 1 such that if n exceeds c 1 and t 3 then gcdi 1! + 1,...,i t! + 1 < e n n l t Furthermore, let δ and ε be positive real numbers. There exists a positive number c, which depends on ε and δ, such that if n exceeds c, and then gcdi 1! + 1,...,i t! + 1 < exp 1 + εl log t t 3 t < n δ, 15 n l >, 16 log n 1 + log ne l + t log n max1,log log t t We remark that it is possible to replace the term max1,log log t on the right hand side of inequality 17 by ft where f is any real valued function to the real numbers of size at least 1 for which lim ft = t provided that c is modified to depend on f. Proof of Lemma. Let A be a positive integer and let k 1,k and k 3,k 4 be distinct pairs of integers with n k 1 > k 1 and n k 3 > k 4 1 for which A k i! + 1, 18 for i = 1,, 3, 4. Suppose, without loss of generality, that k 1 k k 3 k Note that {k 1,k,k 3,k 4 } may only contain 3 elements. Then, by 18, A k i! k i+1!, for i = 1,3,
6 6 C.L. Stewart and so A k i+1!k i k i 1 k i , for i = 1,3. But, by 18, gcda,k i+1! = 1 for i = 1,3 hence for i = 1, 3. Therefore Since A k i k i , 0 A k 1 k + 1 k 3 k k 1 k + 1 k 3 k = k 3 k 4! k1 k! k 3 k 4! gcda,k 3 k 4! = 1 and k 1 k k 3 k 4, we find that k1 k k3 k 4, A k 3 k 4! 1 k 1 k + 1 k 3 k Notice that k 1 k + 1 k 3 k < n k 1 k. Thus, provided that k 1 k + 1 k 3 k 4 + 1, 3 we deduce from 1, and the fact that m! m m e, when m is a positive integer, that A < n k 1 k k 3 k 4 ne k3 k 4. 4 k 3 k 4 Since gx = ne x x attains its maximum value at x = n we see that, when 3 holds, A < n k 1 k k 3 k 4 e n. 5 We now put A = gcdi 1! + 1,...,i t! + 1. We shall prove 13 first. Put µ = min{i a i b i a > i b, a,b {1,...,t}}
7 On the greatest and least prime factors of n! + 1, II 7 and let k 1,k be elements of {i 1,...,i t } with k 1 k = µ. Since µ we see from 0 with i = 1 that A < n k 1 k n l t 1, l t 1 as required. We shall prove 14 next. There are t pairs of integers i,i with i > i which can be chosen from {i 1,...,i t }. Associated to each such pair is the difference i i and 0 < i i l. Therefore there are two such pairs, k 1,k and k 3,k 4 say, for which 0 k 1 k k 3 k 4 Thus, provided that 3 holds, by 5 and 6, A < n 1 t e n, hence, since t 3 and t 1 t 1, A < e n n l t 1. l l t 1. 6 It remains only to ensure that 3 holds. We may assume that A exceeds e n since otherwise the result is immediate. Note that if k 1 = k 3 then, since the pairs k 1,k and k 3,k 4 are distinct, k k 4 and so 3 holds. Thus we may assume that k 1 > k 3 ; a similar argument applies if k 3 < k 1. Further, we may assume, after renumbering k,k 3,k 4 if necessary, that k k 3 > k 4. Since, as in 0, A divides k 1 k we see that A n k 1 k. But A exceeds e n and so k 1 k n log n. By a version of the prime number theorem with an explicit error term, for n sufficiently large there is a prime p between k 1 and k. As a consequence p divides k 1 k + 1 and not k 3 k and so 3 holds and 14 follows. Finally, we shall prove 17. Let c 3,c 4,... denote positive numbers which are effectively computable in terms of ε and δ. Without loss of generality we may suppose that n i 1 > i > > i t 1.
8 8 C.L. Stewart Note that we may also suppose that since otherwise, by 16, and therefore, by 14, t 1 log n 1 8, 7 llog n t 1 llog 1 n3 4 nlog n 4 A < exp llog n 1 + ε t 1, for n larger than c 3, whence 17 holds. We consider the consecutive integers i j+1 +1,...,i j for j = 1,...,t 1. Notice that A divides i j!+1 and i j+1!+1 hence A divides i j i j Therefore A is at most n i j i j+1. If for some j, with 1 j t 1, n i j i j+1 <, tlog n 3 then n A exp, tlog n 1 and, by 16, 17 holds. Thus we may suppose that n i j i j+1, tlog n 3 for j = 1,...,t 1. Let m denote the number of the intervals [i j+1 +1,i j ] for j = 1,...,t 1 which do not contain a prime number. Let p 1,p,... denote the sequence of prime numbers and put d k = p k+1 p k for k = 1,,.... Then d k m n. tlog n 3 But, by Lemma 1, for n > c 4. In particular p k n p k n d k < n δ, m < t log n 3, n δ
9 and by 15, since t n, On the greatest and least prime factors of n! + 1, II 9 m < tn δ log n 3 < t 1 δ 3, 8 for n > c 5. Put t 1 = t 1 m, 9 and order the differences i j i j+1 with 1 j t 1 for which there is a prime in the interval [i j+1 + 1,i j ] according to size. Let us denote these differences by γ 1,...,γ t1 so that γ 1 γ γ t1. 30 Observe that γ γ t1 i 1 i t l. 31 For any real number x let [x] denote the largest integer of size at most x. Put t = [t 1 /log log t 1 1 ]. 3 Then, by 30 and 31, γ t t 1 t l. Thus, by 7, 8, 9, 30 and 3, for n > c 6, for h = 1,...,t. Next note that γ h < 1 + ε l t, 33 γ t γ t 1 + γ t 1 γ t + +γ 1 0 = γ t γ t 1 γ t + +γ 1 0 = γ t hence γ 1 0 = γ 1 γ t γ t 1 + γ t 1 γ t + + t γ 1 = γ t + + γ 1. 34
10 10 C.L. Stewart Put θ = minγ t γ t 1,...,γ γ 1,γ 1. Then, by 31 and 34, t t + 1 θ l. Therefore, by 7, 8, 9 and 3, for n > c 7, llog log t θ < 1 + ε t. 35 We have γ 1 = i r i r+1 for an integer r with 1 r t 1. Then A i r i r If θ = γ 1, we see that A < n θ, and by 35 our result follows. Thus we may suppose that θ = γ s γ s 1 for some integer s from {,...,t }. In particular, θ = i a i a+1 i b i b+1 with a and b distinct integers from {1,...,t 1}. Put k 1 = i a, k = i a+1, k 3 = i b and k 4 = i b+1. By construction there is a prime among the integers k + 1,...,k 1 and another prime among the integers k 4 + 1,...,k 3. Thus the larger of the two primes divides one of k 1 k +1 and k 3 k and not the other whence 3 holds. Note also that ne x x is an increasing function of x for x positive and less than n. Therefore, by 4, 7, 33 and 35, we find that log log t 1+εl net 1+ε l t A < n t, 1 + εl hence that A < exp 1 + εl log n log log t t + log t t log ne 1+εl + 36 t for n > c 8, as required.
11 On the greatest and least prime factors of n! + 1, II 11 For any prime p let tp denote the number of positive integers k for which p k! + 1. In [9, Theorem 7.5] we noted that tp < m + 1m + where m = 3p To see this observe that if n and s are positive integers and p divides both n!+1 and n+s!+1 then p divides n+s! n! hence n+s n+1 1 mod p. In particular, n is a solution of the polynomial congruence x + s x mod p, and by Lagrange s theorem the number of such solutions is at most s. Let I be an interval of length l 1 and let n 1 < n < < n k denote all the solutions of x! mod p in I. Plainly k 1 n i+1 n i l, 38 i=1 and by our earlier observation at most s of the terms in brackets in the above sum are equal to s. Therefore k 1 n i+1 n i i=1 where u is defined by the inequalities u s, 39 s=1 u u+1 s k 1 < s. s=1 s=1 Thus k is at most u+1u+ and by 38 and 39 l uu + 1u > u Since all integers n for which p n! + 1 lie in the interval [1,p 1], 37 follows from 40 with l = p. Further, from 40 we obtain Lemma 3 below, a result which is the special case of Lemma of Luca and Shparlinski [7] with fx equal to 1. Lemma 3. There exists a positive number c such that if p is a prime number and I is an interval of the positive real numbers of length l with l 1 then the number of integers k in I for which p divides k! + 1 is at most cl 3.
12 1 C.L. Stewart For any nonzero integer m and any prime p we denote by ord p m the exponent of the largest power of p which divides m. As usual m p is the padic absolute value of m normalized so that m p = p ordpm. Lemma 4. There exists a positive number c 1 such that if p is a prime number, n a positive integer and I a subinterval of [1,n] of length l then log pord p i Ii! + 1 < 3 llog llog n + c 1llog n + n log n. 41 Further, for each pair of positive real numbers ε and ε 1 there exist positive numbers c and c 3 such that if l exceeds ε 1 n and n exceeds c 3 then log pord p + 1 < 1 + ε i Ii! 9 llog l + c n log n. 4 Proof. Let i 1,...,i u be the integers i in I for which p divides i! + 1. By Lemma 3 there is a positive number c such that u cl Put h t = ord p i t! + 1 for t = 1,...,u. We may suppose that Then, by 13 of Lemma, h 1 h u. p ht < n l t 1, 44 for t =,...,u. In particular by 43 and 44, log ph + + h u < llog n 1 + cl 3 1 t 1 dt < 3 llog llog n + c 4llog n, 45
13 On the greatest and least prime factors of n! + 1, II 13 where c 4 is a positive number. Since h 1 log p n log n 46 and ord p i Ii! + 1 = h h u, follows from 45 and 46. Let c 5,c 6,... denote positive numbers which depend on ε and ε 1 and suppose that l exceeds ε 1 n. Then by 43, u cn 3 < n , for n > c 5. Thus for n > c 6, 15 of Lemma holds with δ = 1 36 and, since l > ε 1 n, 16 of Lemma also holds. Therefore by 17 of Lemma, for n > c 7, log ph t < 1 + εl log t log e ε 1 log n max1,log log t + + t t t 1 48 for t = 3,...,u. Since the expression on the right hand side of 48 is a decreasing function of t for t > e, we see that and so log ph h u cl3 < 1 + εl 3 log t t + log e ε 1 t + log n max1,log log t t 1 dt log ph h u < 1 + ε 9 llog l + c 8 n log n. 49 Since h 1 + h + h 3 log p < 3n log n, 50 4 now follows from 47, 49 and 50. Lemma 5. Let ε be a positive real number. There exists a positive number c, which depends on ε, such that if p is a prime number, n an integer with n and I a subinterval of [1,n] of length l then log p ord p i Ii! + 1 < 9 llog n + εnlog n + cn log n. 51
14 14 C.L. Stewart Proof. Let c 1,c,... denote positive numbers which depend on ε. By 4 of Lemma 4, if l exceeds εn and n exceeds c 1, log p ord p + 1 < i Ii! 9 llog n + 9 εnlog n + c n log n. 5 On the other hand if l εn then by 41 of Lemma 4, log p ord p + 1 < i Ii! 3 εnlog n + c 3 n log n; 53 plainly 53 holds if l. Therefore from 5 and 53, we obtain 51 with c 4 in place of c for n > c 1, hence 51 holds for n and our result follows. 3. Proof of Theorem 1 Let δ be a positive real number with δ < Put δ = 10δ, λ = + δ, 54 8 and λ 1 = λ + δ. Note that λ < 3. Let c 1,c,... denote positive numbers which depend on δ. We may suppose that there exist only finitely many odd positive integers n for which pn! + 1 λ 1 n, since otherwise the theorem holds. Thus there exists a positive integer c 1 such that for each odd integer n with n > c 1, pn! + 1 > λ 1 n. 55 We shall show that this leads to a contradiction and the theorem then follows. Since 55 holds, we also have Pn! + 1 < λ n, 56 λ 1
15 On the greatest and least prime factors of n! + 1, II 15 for all odd integers n with n > c. To see this, observe that if q = Pn!+1 with n > 1 and q λ n, 57 λ 1 then q is odd and, by 4, q q n 1! + 1. But then 1 pq n 1! + 1 q = 1 n+1 q n 1. q We have q > λ and, by 57, n q λ 1 λ hence 1 1 n+1 q q = qλ λ 1 q λ. λ But qλ q λ < λ 1 for n > c 3 since q > n. Thus pq n 1! + 1 λ 1 q n 1. Furthermore, q n 1 > c 1 for n > c 4 by 57 and this contradicts 55. Therefore 56 holds. The proof now proceeds by a comparison of estimates for Z = N n! + 1. n=1 n odd, n>c Put R = {n Z n odd, c < n N}. Observe that if p n! + 1 with n in R then, by 55, p > λ 1 n and, by 56, p < λ λ 1n. In particular, Put Since n! n e n, I p = λ 1 λ p < n < 1 λ 1 p. λ 1 λ p,min N, 1 p. λ 1 Z > exp n log n n n R
16 16 C.L. Stewart so log Z > 1 δ N log N, 58 4 provided that N exceeds c 5. On the other hand Z = Z 1 p 1 n! + 1. p p Put n I p R Ap = log pord p n I p R p n! + 1. Then Z exp Ap. p< λ λ 1 N Thus by 51 of Lemma 5, with ε = δ, log Z log N 9 p< λ λ 1 N lp + δnlog N + c 6 N log Nπ λ λ 1 N, 59 where lp, the length of I p, is given by lp = 1 λ 1 p when p λ 1 N λ 1 λ and by λ 1 lp = N p when p λ 1 N. λ By 54, 59 and the prime number theorem, log Z log N 9 p< λ λ 1 N lp + 4δN log N + c 7 N. 60
17 Further p< λ λ 1 N lp = On the greatest and least prime factors of n! + 1, II 17 p λ 1 N = 1 λ 1 1 λ 1 p + λ 1 λ p λ 1 N p + N λ 1 N<p< λ λ 1 N λ 1 N<p< λ λ 1 N 1 N λ 1 λ p λ 1 λ p< λ λ 1 N p. Thus by the prime number theorem and Abel summation, for N > c 8, λ1 λ lp < 1 + δ + λ 1 λ λ N 1 λ 1 log N p< λ λ 1 N λ < 1 + δ λ 1 λ 1 N log N, and so by 60, and the fact that λ 1 exceeds λ, 1 + δ λ log Z < 9 λ 1 λ N log N + 4δN log N + c 7 N. 61 Comparing 58 and 61 we find that for N > c 9, 1 δ 4 < 1 + δ 9 λ λ λ 1 + 4δ + c 8 log N. By 54 we obtain a contradiction for N sufficiently large and the result now follows. 4. Proof of Theorem We may suppose that 0 < ε < Put γ = the set of positive integers for which 18ε and let Bγ be Pn! + 1 γn. 6 We shall show that for n sufficiently large the set Bγ {1,...,n} has cardinality at least ε 3n and hence the result follows. Accordingly suppose that N is a positive integer for which Bγ {1,...,N} ε N. 63 3
18 18 C.L. Stewart Let c 1,c,... denote positive numbers which depend on ε. Our proof proceeds by a comparison of estimates for Z = N n=1 n Bγ n! + 1. Since n! n n e, N Z > exp n log n n Bγ {1,...,N} N log N whence, by 63, n= log Z > 1 ε N log N, 64 provided that N > c 1. Notice that if p n! + 1 and n Bγ then n < p < γn hence 1 γ p < n < p. Put I p = and Then 1 γ p,min{n,p}, Ap = log pord p n Ipn! + 1. Z exp Ap. p<γn Thus, by 51 of Lemma 5 with ε 6 in place of ε, log Z < ε log N lp Nlog N + c N log N πγn, 65 p<γn where lp, the length of I p, is given by lp = 1 1 p for p N γ and lp = N 1 p for p > N. γ
19 Since γ < 11 On the greatest and least prime factors of n! + 1, II 19 it follows from 65 and the prime number theorem that log Z < log N lp + εn log N + c 3 N p<γn Further p<γn lp = p N = p N 1 1 p + γ p + N<p<γN N<p<γN N 1 γ N 1γ p p<γn p. Thus, by the prime number theorem and Abel summation, for N > c 4, N γ 1N lp < 1 + ε + log N log N γ N log N p<γn and so by 66, log Z < 1 + ε γ 1N log N + εn log N + c 3 N Comparing 64 and 67 we find that for N > c 5, 1 ε γ 1 < 1 + ε + ε + c 3 9 log N. But γ < 11 and so for N sufficiently large we obtain a contradiction. Thus 63 does not hold for N sufficiently large and the result now follows. References [1] B. Brindza and P. Erdős, On some Diophantine problems involving powers and factorials, J. Austral. Math. Soc. Ser. A , 1 7. [] P. Erdős and R. Graham, Old and new problems and results in combinatorial number theory, Monographie 8 de L Enseignement Mathématique, Genève, [3] P. Erdős and C.L. Stewart, On the greatest and least prime factors of n! + 1, J. London Math. Soc 13, no. 1976, [4] D.R. HeathBrown, The difference between consecutive primes, III, J. London Math. Soc. 0, no. 1979,
20 0 C.L. Stewart [5] M. Le, On the Diophantine equation x p 1 + p 1! = p n, Publ. Math. Debrecen , [6] J. Liouville, Sur l équation 1 3 p = p m, J. Math. Pure Appl , [7] F. Luca and I.E. Shparlinski, Prime divisors of shifted factorials to appear. [8] M.R. Murty and S. Wong, The ABC conjecture and prime divisors of the Lucas and Lehmer sequences, Number Theory for the Millenium, III, Urbana, IL, 000, A.K. Peters, Natick, MA, 00, [9] C.L. Stewart, On divisor properties of arithmetical sequences, Ph.D. thesis, University of Cambridge, [10] K. Yu and D. Liu, A complete resolution of a problem of Erdös and Graham, Rocky Mountain J. Math , DEPARTMENT OF PURE MATHEMATICS UNIVERSITY OF WATERLOO WATERLOO, ONTARIO CANADA NL 3G1 URL:
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