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1 International Journal of Number Theory c World Scientific Publishing Comany SOLVING n(n + d) (n + (k 1)d ) = by 2 WITH P (b) Ck M. Filaseta Deartment of Mathematics, University of South Carolina, Columbia, SC 29208, U.S.A. filaseta@mailbox.sc.edu S. Laishram Stat-Math Unit, Indian Statistical Institute, New Delhi, , India shanta@isid.ac.in N. Saradha School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha Road, Mumbai , India saradha@math.tifr.res.in Abstract: We examine the roblem of determining the solutions to the Diohantine equation n(n + d) (n + (k 1)d ) = by 2, where d is fixed and the largest rime divisor of b is no more than Ck. Here, C is fixed but arbitrary. Under some rather minor conditions, it is shown that there are finitely many solutions that can be effectively comuted. Some new, largely combinatorial, ideas are introduced into the general theory to handle the case of arbitrary C as considered here. 1. Introduction With n, d and k ositive integers, we set (n, k, d) = n(n + d) (n + (k 1)d ). Fix d as above and real numbers ε > 0 and C d. We are interested in establishing that the equation (n, k, d) = by 2 (1.1) Mathematics Subject Classification: 11D45, 11D72, 11B25 Keywords: Diohantine equations, arithmetic rogressions, large rime divisors 1

2 2 M. Filaseta, S. Laishram, N. Saradha has finitely many solutions in ositive integers n, k, b and y with gcd(n, d) = 1, k 3, n ( C d + εd ) k and P (b) Ck, (1.2) where P (b) denotes the largest rime dividing b. Solutions to equation (1.1) have been investigated extensively in the literature and even further with the exonent on y on the right relaced by an arbitrary integer 2. A articularly imortant result in this direction is the comlete solution to the case d = b = 1 by P. Erdős and J. L. Selfridge [6], which imlies that the roduct of two or more consecutive ositive integers is never a ower. A small samling of the literature on the subject includes the work of T. N. Shorey and R. Tijdeman [17], the exository work of T. N. Shorey [16] and the references there, and the more recent work by M. A. Bennett, N. Bruin, K. Győry and L. Hajdu [2], K. Győry, L. Hajdu, and Á. Pintér [8] and N. Saradha and T. N. Shorey [15]. As we shall describe at the beginning of the third section, traditional methods allow one to bound the number of solutions to (1.1) and (1.2) when C is small and, in articular, for C 2 (and a little beyond). But for large C, these methods fail. In this aer, we describe an imrovement on these methods which allows one to handle this roblem for larger C as well. Our main result is the following. Theorem 1.1. Fix a ositive integer d. Let ε (0, 1) and C d be arbitrary. There is a finite effectively comutable set S = S(d, ε, C) of 4-tules such that if (1.1) and (1.2) hold, then (n, k, b, y) S. The above result is formulated in the way that we will establish it. We note, however, that the condition gcd(n, d) = 1 can be droed and the exression εd aearing in the lower bound for n can be relaced by ε. Indeed, these are not actual imrovements on Theorem 1.1 but rather equivalent formulations of it. We also note that one cannot obtain an analogous result with ε = 0 in (1.2) since then (1.1) would have infinitely many solutions corresonding to k 3 arbitrary, n = (C d)k, b = (n, k, d) and y = 1. Acknowledgment: The authors exress their gratitude to the referee for some helful suggestions. 2. Preliminaries We suose as we may that b is squarefree. For m a ositive integer, we will also make use of the notation ν (m) = e where e m. Observe that it suffices to bound n and k as above for which (1.1) has a solution satisfying (1.2). In fact, for each fixed k 3, there are finitely many ossibilities for squarefree b satisfying the last condition in (1.2). For each such b and solution to (1.1), we can consider the roduct (n, 3, d) which will necessarily be a square times a ositive squarefree integer having all of its rime factors Ck. Thus, we obtain (n, 3, d) = b y 2,

3 Solving n(n + d) `n + (k 1)d = by 2 with P (b) Ck 3 where P (b ) Ck. There are finitely many ossibilities then for b, and we deduce (1.1) has finitely many integer solutions as solutions corresond to integer oints on the ellitic curve described by (n, 3, d) = b y 2. That these integer oints can be effectively comuted is a consequence of, for examle, Theorem 4.2 in [1] which asserts (in the case n = 3 there) the following: Let K be an algebraic number field, and let α 1, α 2 and α 3 be distinct algebraic integers in K. Then the equation Y 2 = (X α 1 )(X α 2 )(X α 3 ) (2.1) has only a finite number of solutions in algebraic integers X and Y in K and these can be effectively determined. As noted in [1], the result remains valid if a non-zero factor in K is included on the right side of (2.1) (in fact, this is easily seen to be equivalent to the assertion above). Thus, it suffices to show that (1.1) and (1.2) imly k is bounded. Beginning with (1.1), for 0 j < k, we can write n + jd = a j x 2 j, a j, x j Z, a j squarefree. We will want to know that P ( (n, k, d) ) is large, so we state this as a first result. Theorem 2.1. Fix a ositive integer d. Let ε (0, 1) and C d be arbitrary. There is a finite effectively comutable set S = S (d, ε, C) of 2-tules such that if n and k are ositive integers for which then k 2, gcd(n, d) = 1, n (C d + εd)k, (n, k) S, P ( (n, k, d) ) > Ck. Proof. An asymtotic form of Dirichlet s Theorem (such as that in [4],. 123, see (11) and the discussion after it) imlies that there is an exlicit k 0 such that if k k 0, then there is a rime in the set {n + (k εk + 1)d, n + (k εk + 2)d,..., n + (k 2)d, n + (k 1)d} for all n satisfying (C d + εd)k n e 3C k. We deduce that if P ( (n, k) ) Ck, then either k < k 0 or n > e 3C k. Suose that n > e 3C k. We show in this case that P ( (n, k) ) > Ck rovided k is sufficiently large. We use an idea of Erdős [5]. For each rime Ck, we consider n from the set T = {n, n + d,..., n + (k 1)d} (2.2)

4 4 M. Filaseta, S. Laishram, N. Saradha for which ν (n ) is maximal. If x is the number of integers < n in T and y the number > n, then x + y = k 1. We also have ( ) ( ) x y x + y ( ) ν m j + j j = ν (k 1)!. m T j=1 j=1 m n We deduce that Hence, e (n,k,d) log e (n,k,d) On the other hand, e (k 1)! n k k (n + kd) π(ck) k k (2n) π(ck). e k log k + π(ck) log(2n) k log k + 2Ck log k log(2n). log m T m > log n k k log n. Thus, rovided e (n,k,d) e < m T m log n log k + 2C log k log(2n). The latter holds rovided ( 1 2C ) 2C log 2 log n log k + log k log k. Since n > e 3C k, it suffices here for ( 1 2C ) ( ) 2C log 2 log k + 3C log k + log k log k, which is easily seen to hold for k k 0, say. We are left then with the task of considering finitely many k < max{k 0, k 0}. A result of G. Pólya [12] imlies that P (n(n+d)) tends to infinity with n. In articular, there is an n 0 such that if n > n 0, then for all k satisfying 2 k < max{k 0, k 0}, one has P ( (n, k, d) ) P ( n(n + d) ) C max{k 0, k 0} > Ck. What remains are airs (k, n) satisfying 2 k < max{k 0, k 0} and (C d+εd)k n n 0. There are finitely many such airs. The rime distribution results used

5 Solving n(n + d) `n + (k 1)d = by 2 with P (b) Ck 5 above are effective, so we deduce that the set S given in the theorem is effectively comutable, and the result follows. We note that [13] rovides some exlicit estimates that can be used for obtaining k 0 for small d, for examle d 72. The use of Pólya s result can be relaced by work in [11] for d {1, 2, 4} or a use of estimates on linear forms of logarithms or by use of algorithms for Thue equations as in [3] and [18]. The reader may also want to see [10] for related work on the largest rime factor of a roduct of consecutive numbers in an arithmetic rogression. Given Theorem 2.1, one can show effectively that, with the conditions on n in (1.2), the inequality P ( (n, k, d) ) > Ck holds for all but finitely many airs (n, k). This is more than enough to allow one to consider the case that the a j are distinct. This is accomlished as follows. Since some x j is divisible by a rime > Ck dk, we deduce that n + kd > a j x 2 j (kd + 1) 2 > k 2 d 2 + kd. Hence, n > k 2 d 2. Now, if a u = a v with u v, then (k 1)d a u x 2 u a v x 2 v = a u (x u + x v ) x u x v > a u x u a u x 2 u > k 2 d 2 = kd, which is imossible. Note that above we can also obtain that n + kd > C 2 k 2. In the oosite direction, the inequality n > k 2 d 2 easily imlies kd < n so that n + kd 2n. We also have as a consequence of (1.1) that P (a j ) Ck for each j. Before roceeding, we refer a stronger lower bound on n or, more recisely, on the numbers x j. We address that next. Lemma 2.2. If (1.1) and (1.2) hold and k is sufficiently large, then n > Ck 2.8. Proof. Assume n Ck 2.8. We saw above that n > k 2 d 2 k 2. The basic idea is to find a lower bound on the number of integers in the set T given in (2.2) that are squarefree. We show that at least 57% of the elements of T are squarefree by making use of the assumtion n Ck 2.8. We exlain first why this leads us to a contradiction. Let t = 0.57k. Suose there are t elements of T that are squarefree. Then log a j log (n + jd) > t log n 0.57k log(k 2 ) > 1.1k log k. 0 j<k 0 j<t On the other hand, each rime Ck divides at most k/ +1 of the numbers a j. Recall that the a j are squarefree and satisfy P (a j ) Ck. For k sufficiently large,

6 6 M. Filaseta, S. Laishram, N. Saradha we deduce that log 0 j<k a j ( ) k + 1 log k log + log 1.05k log k + 1.1Ck 1.1k log k. Thus, we have a contradiction. We finish the roof by showing that there are at least t squarefree numbers in T. We consider rimes in three different ranges. Let z = log k with k sufficiently large. We start with rimes z. Since gcd(n, d) = 1, the number of multiles of m 2 in T is 0 if m has a rime factor in common with d. Otherwise, the number of multiles of m 2 in T is k m 2 + R m = k m 2 + R m where R m {0, 1} and R m ( 1, 1]. Let P denote the roduct of the rimes z. Note that we are considering k sufficiently large. Then the sieve of Eratosthenes imlies that number of elements of T that are not divisible by 2 for every rime z is ( ) k µ(m) m 2 + R m = (1 1 ) 2 k + E, z where Since m P z E 2 π(z) 2 log k = k log k. (1 1 ) 2 (1 1 ) 2 = 6 π 2 > 0.6, we deduce that there are at least 0.59k elements of T that are not divisible by 2 for every rime z. Next, we observe that the number of elements of T divisible by 2 for some rime (z, kd] is bounded by z< kd ( k ) k m>z 1 m 2 + π(kd) k z 1 + 2kd log k < 0.01k. We deduce that there are at least 0.58k elements of T that are not divisible by 2 for every rime kd. Finally, we consider the rimes > kd for which 2 divides some element of T. Observe that necessarily < n + kd 2n. Recall that we are assuming n Ck 2.8. As a consequence k (n/c) 1/2.8 > n Thus, we are interested in rimes for which n 0.35 < kd < 2n.

7 Solving n(n + d) `n + (k 1)d = by 2 with P (b) Ck 7 Observe that, for each such, there is at most one multile of 2 in T. Furthermore, if a 2 is such a multile, then a is a ositive integer satisfying a n + kd 2 2n n = 2n0.3 < 2k 0.3/0.35 < k 0.9. On the other hand, if we also have a second rime q (kd, 2n] for which aq 2 T, then kd > a 2 aq 2 = a + q q a + q > kd, an imossibility. Thus, the rimes > kd for which there is an element of T divisible by 2 corresond to distinct ositive integer multiliers a < k 0.9. In articular, we deduce that there are < k 0.9 such rimes. Hence, there are also < k 0.9 < 0.01k elements of T divisible by the square of a rime exceeding kd. We obtain then that there are at least 0.57k elements of T that are not divisible by the square of a rime, and the result follows. The main idea behind the roof of Lemma 2.2 comes from the study of gas between squarefree numbers. Using [7], the lower bound can easily be sharened further to obtain that n k 5 ε for any ε > 0. For our uroses, we only need the above weaker version of the lemma. In fact, something considerably weaker would also do. Our interest is in the following result. Corollary 2.3. Let α be such that 0 < α k 0.8, and let T be as in (2.2). For k sufficiently large, the numbers x j, with 0 j < k, for which a j αk are distinct. Proof. By Lemma 2.2, we have n > Ck 2.8. If a j a j x 2 j n that αk, then we obtain from x 2 j n αk > Ck2.8 αk = Ck1.8 α dk. We deduce that there can be at most one multile of x 2 j follows. in T, and the corollary 3. The Main Lemma For the moment, consider the case d = 2. Let t j denote the jth odd squarefree number. The rior aroach to obtaining the solutions to (1.1) given (1.2) is to combine a lower bound and an uer bound on 0 j<k log a j. The lower bound is obtained from log a j log t j+1 (3.1) 0 j<k 0 j<k and a fairly recise estimate for this last sum. The uer bound is obtained using an aroach of Erdős already used in the roof of Theorem 2.1. Here, the aroach can be described roughly as follows. For each rime Ck, one can bound the

8 8 M. Filaseta, S. Laishram, N. Saradha number of a j divisible by by estimating the number of j {0, 1,..., k 1} for which ν (n + jd) is odd. If the number of such j is s(), then log a j s() log. (3.2) 0 j<k We deduce by combining the above estimates that s() log log t j+1. 0 j<k Ideally, we want aroriate estimates for each side of this inequality to lead to a contradiction when k is large. Observe that the right side is indeendent of C. As a consequence, this aroach seemingly is bound to fail when C is large. We modify the above idea. To understand the modification, it hels to examine s() more closely. The condition gcd(n, d) = 1 in (1.2) imlies that s() = 0 if d. We observe that otherwise we have 1 if k Ck 2 if k/2 < k s() 3 if k/3 < k/2.. These bounds on s() are in some sense best ossible. Although we cannot hoe to do better, what we will show is that, as varies, if s() takes many values near the uer bound indicated above, then tyically a j is considerably larger than t j+1. In other words, if the values of s() are near the uer bounds suggested above, at least on average, then the lower bound for 0 j<k log a j given by (3.1) can be imroved. We elaborate on the details of this idea next. We no longer restrict d to being 2. Our main imrovement is based on the following lemma. Lemma 3.1. Let k be a sufficiently large integer. Fix ositive real numbers α and, ossibly deending on k, with < 1. Let Then there are at most A = [1, αk] {a 0, a 1,..., a k 1 } and J = [k, k). α 2 2 2α + 4dα + 2 different rimes in J with the roerty that a for two or more different a A. Proof. We only consider α < k 1/2 since the result is trivial for larger (and somewhat smaller) values of α. Observe that if J and a for some a A, then 2 1 a αk k = α/.

9 Solving n(n + d) `n + (k 1)d = by 2 with P (b) Ck 9 As a/ is also an integer, there are α/ ossibilities for a/. Suose that and q are rimes in J such that a i = a u a j and q = a v q, (3.3) where a i, a j, a u, a v A, a i, a j, q a u, q a v. Since a i x 2 i n and a i αk, we have that x i n/(αk). On the other hand, a i x 2 i n + (k 1)d 2n and a i so that x i 2n/ 2n/(k). Similar arguments hold for bounding x j, x u and x v. Hence, n/(αk) xi, x j, x u, x v 2n/(k). (3.4) Observe that (ai /)x 2 i (a j /)x 2 j (k 1)d/ d/ (au /q)x 2 u (a v /q)x 2 v (k 1)d/q d/. Setting we see that X = x 2 v( (ai /)x 2 i (a j /)x 2 j) x 2 j ( (au /q)x 2 u (a v /q)x 2 v), X d(x2 j + x2 v) 4dn 2 k. From (3.3), we also have X = a i x 2 i x 2 v x 2 jx 2 u x i x v + x j x u x i x v x j x u 2n x i x v x j x u. αk Therefore, x i x v x j x u 2dα 2. (3.5) For the moment, view, x i and x j as fixed. We bound the number of distinct airs (x u, x v ) satisfying (3.4) and (3.5). Observe that if δ = gcd(x i, x j ), then a i x 2 i and a j x 2 j are both divisible by δ2. Two multiles of δ 2 in the arithmetic rogression n + jd with difference d must differ by at least δ 2 d kδ 2 d. On the other hand, a i x 2 i and a jx 2 j differ by at most (k 1)d. Hence, δ 1/. For a fixed integer t [ 2dα/ 2, 2dα/ 2 ], if x i x v x j x u = t, then the integer airs (x, y) satisfying x i x x j y = t are given by x = x v + x js δ, y = x u + x is, where s Z. δ Due to (3.4), we are interested in the case that x j n/(αk) n δ 1/ = and αk x v + x js 2n δ. k

10 10 M. Filaseta, S. Laishram, N. Saradha We deduce that there can be at most 2 2n αk + 1 = ( 2 2α/ ) + 1 k n different values of s. Hence, we have ( 2 2α/ ) + 1 as an uer bound on the number of ossibilities for x u and x v satisfying (3.4) and x i x v x j x u = t for a fixed t [ 2dα/ 2, 2dα/ 2 ]. Letting t vary, we get the uer bound 2 2α + 4dα + 2 B = 2 on the total number of distinct airs (x u, x v ) that can satisfy (3.4) and (3.5). This includes the solution x u = x i and x v = x j. Set 4dα + 2 N = α 2 2 2α Observe that if we consider N airs (u, v) of ositive integers with each of u and v being α/, then the igeon-hole rincile imlies that there must be some air that occurs > B times. Assume that there are N different rimes in J with the roerty that a i and a j for distinct a i, a j A. Then some air (a i /, a j /) occurs for > B rimes. Let P be such a set of rimes in J so that, in articular, P > B. (3.6) For P, let x i and x j be as before so that a i x 2 i and a j x 2 j are among the numbers n, n + d,..., n + (k 1)d. Thus, if q is in P, then we have that a i = a i q q, a j = a j q q, x i x jq x j x iq 2dα 2 and, furthermore, that there are B distinct ossibilities for the air (x iq, x jq ). From (3.6), we deduce that some air (x iq, x jq ) is reeated. Recalling that α < k 1/2, we obtain a contradiction to Corollary 2.3. Hence, the roof is comlete. 4. Proof of Theorem 1.1 We will make use of the following result. Lemma 4.1. Let s j denote the jth squarefree ositive integer. There is an m 0 such that if m is an integer m 0, then m s j (1.6) m m!. (4.1) j=1 We note that the above result is an easy consequence of the fact that the squarefree integers have asymtotic density 6/π 2. The reader can consult [9] for details. For the aroach below, we can also manage with the weaker and trivial estimate

11 Solving n(n + d) `n + (k 1)d = by 2 with P (b) Ck 11 s j j instead of Lemma 4.1. Presumably, Lemma 4.1 will, however, hel in obtaining effective results for secific C. Recall that we need only consider k sufficiently large. For such a k, we set α = k 0.12 and = e 33C in Lemma 3.1. Let U be the set of j {0, 1,..., k 1} for which a j αk, and let W be the set of j {0, 1,..., k 1} for which a j > αk. In articular, we have U + W = k. It suffices to consider k large and, in articular, k 2m 0. We set 30Ck m = k log k in Lemma 4.1. We use that either (i) U > m or (ii) U m. In the case of (i), we have m a j s j (1.6) m m!. j U j=1 We use the simle inequality m! m m /e m which follows by observing e m = j=0 mj /j! m m /m!. Thus, still in the case of (i), we deduce m log a j log s j m log(k/2) + m ( log(1.6) 1 ) j U j=1 m log k + m ( log(1.6) 1 log 2 ) k log k (30C )k. Observe that rimes k can divide at most one a j. Hence, Lemma 3.1 imlies that there is a constant C deending on C and d such that for all but C α 3.5 C k 0.5 rimes [k, Ck], there is at most one j U such that a j. For each of the C k 0.5 rimes [k, Ck] for which there is more than one j U such that a j, we use that there are at most k + 1 k k + 1 such j. Observe also that such are necessarily k so that log log k for such. For each < k, we simly use the uer bound k/ + 1 on the number of j U for which a j. We obtain log a j log + C k 0.5 log k + j U k k log. It is not difficult to estimate these sums, but we note that one can aeal to Theorem 4 and Theorem 6 of [14] which give log < x + x for all x > 1 2 log x x

12 12 M. Filaseta, S. Laishram, N. Saradha and x log < log x + E log x for all x 319, where E = is a constant. Since k is sufficiently large, we easily deduce that log a j k log k + (1.5C + log )k = k log k 31.5Ck. j U This contradicts the lower bound we had for the sum; hence, we are done in the case of (i). Suose now that (ii) holds. Then we must have W 30Ck/ log k. Since all of the rime divisors of each a j are Ck and the a j are squarefree, we deduce k 1 log a j j=0 ( ) k + 1 log k log + log k log k + 2Ck, k where again we can aeal to the estimates indicated above from Theorem 4 and Theorem 6 of [14]. On the other hand, k 1 log a j = log a j + log a j log s j + W log(αk). j=0 j U j W j=1 Since U k, in the last sum each s j is easily 2k. On the other hand, αk = k 1.12 > 2k. As U + W = k, we can find a lower bound for the right-hand exression above by setting U = k U 30Ck = m and W = log k 30Ck. log k We deduce k 1 m log a j log s j + j=0 j=1 30Ck log ( k 1.12). log k Aealing to the earlier estimate for this last sum, we deduce k 1 log a j k log k (30C )k Ck 1.12 log k k log k + 2.3Ck. j=0 Thus, in this case, we also obtain a contradiction. Summarizing, we deduce that if k is sufficiently large, then there are no solutions to (1.1) and (1.2). As we have seen, this imlies the result stated in the theorem.

13 References Solving n(n + d) `n + (k 1)d = by 2 with P (b) Ck 13 [1] A. Baker, Transcendental Number Theory, Cambridge Univ. Press, Cambridge, [2] M. A. Bennett, N. Bruin, K. Győry and L. Hajdu, Powers from roducts of consecutive terms in arithmetic rogression, Proc. London Math. Soc. 92 (2006), [3] Y. Bilu and G. Hanrot, Solving Thue equations of high degree, J. Number Theory 60 (1996), no. 2, [4] H. Davenort, Multilicative Number Theory, 2nd edition (revised by H. L. Montgomery), Sringer-Verlag, New York, [5] P. Erdős, On consecutive integers, Nieuw Arch. Wisk. (3) 3 (1955), [6] P. Erdős and J. L. Selfridge, The roduct of consecutive integers is never a ower, Illinois J. Math. 19 (1975), [7] M. Filaseta and O. Trifonov, On gas between squarefree numbers II, J. London Math. Soc. 45 (1992), [8] K. Győry, L. Hajdu, and Á. Pintér, Perfect owers from roducts of consecutive terms in arithmetic rogression, Comos. Math. 145 (2009), [9] S. Laishram, An estimate for the length of an arithmetic rogression the roduct of whose terms is a square, Pub. Math. Debr. 68 (2006), [10] S. Laishram and T. N. Shorey, The greatest rime divisor of a roduct of terms in an arithmetic rogression, Indag. Math. 17 (2006), [11] D. H. Lehmer, On a roblem of Störmer, Illinois J. Math. 8 (1964), [12] G. Pólya, Zur arithmetischen Untersuchung der Polynome, Math. Z. 1 (1918), [13] O. Ramar and R. Rumely, Primes in arithmetic rogressions, Math. Com. 65 (1996), [14] J. B. Rosser and L. Schoenfeld, Aroximate formulas for some functions of rime numbers, Illinois J. Math. 6 (1962), [15] N. Saradha and T. N. Shorey, Contributions towards a conjecture of Erdős on erfect owers in arithmetic rogression, Comos. Math. 141 (2005), [16] T. N. Shorey, Powers in arithmetic rogression, A anorama of number theory or the view from Baker s garden (Zrich, 1999), Cambridge Univ. Press, Cambridge, 2002, [17] T. N. Shorey and R. Tijdeman, Perfect owers in roducts of terms in an arithmetical rogression, III, Acta Arith. 61 (1992), [18] N. Tzanakis and B. M. M. de Weger, On the ractical solution of the Thue equation, J. Number Theory 31 (1989),