PERFECT POWERS IN ARITHMETIC PROGRESSION

Perfect powers in Arithmetic Progression 1 PERFECT POWERS IN ARITHMETIC PROGRESSION Shanta Laishram Stat-Math Unit, Indian Statistical Institute New Delhi 110016, India Tarlok N. Shorey Department of Mathematics, Indian Institute of Technology Bombay Powai, Mumbai 400076, India Received: December 5, 013; Accepted: September 5, 015 Abstract The conjecture of Masser-Oesterlé, popularly known as abc-conjecture have many consequences. We use an explicit version due to Baker to solve the equation n(n + d) (n + (k 1)d) = by l in positive integer variables n, d, k, b, y, l such that b square free with the largest prime divisor of b at most k, k, l and gcd(n, d) = 1. 010 Mathematics Subject Classification: Primary 11P99; Secondary 11A41, 11E5. 1. Introduction Let n, d, k, b, y be positive integers such that b is square free with P (b) k, k, l and gcd(n, d) = 1. Here P (m) denotes the largest prime divisor of m with the convention P (1) = 1. We consider the equation n(n + d) (n + (k 1)d) = by l (1.1) in variables n, d, k, b, y, l. If k =, we observe that (1.4) has infinitely many solutions. Therefore we always suppose that k 3. It has been conjectured (see [Tij88], [SaSh05]) that Conjecture 1.1. Equation (1.1) implies that (k, l) {(3, 3), (4, ), (3, )}. E-mail address: shanta@isid.ac.in; Website: http://www.isid.ac.in/ shanta E-mail address: shorey@math.iitb.ac.in; Website: http://www.math.tifr.res.in/ shorey

S. Laishram and T. N. Shorey It is known that (1.1) has infinitely many solutions when (k, l) {(3, ), (3, 3)(4, )}. A weaker version of Conjecture 1.1 is the following conjecture due to Erdős. Conjecture 1.. Equation (1.1) implies that k is bounded by an absolute constant. For an account of results on (1.1), we refer to Shorey [Sho0b] and [Sho06]. The well known conjecture of Masser-Oesterle states that Conjecture 1.3. Oesterlé and Masser s abc-conjecture: For any given ɛ > 0 there exists a computable constant c ɛ depending only on ɛ such that if where a, b and c are coprime positive integers, then c c ɛ p a + b = c (1.) p abc It is known as abc-conjecture; the name derives from the usage of letters a, b, c in (1.). For any positive integer i > 1, let N = N(i) = p i p be the radical of i, P (i) be the greatest prime factor of i and ω(i) be the number of distinct prime factors of i and we put N(1) = 1, P (1) = 1 and ω(1) = 0. It has been shown in Elkies [Elk91] and Granville and Tucker [GrTu0, (13)] that abcconjecture is equivalent to the following: Conjecture 1.4. Let F (x, y) Z[x, y] be a homogenous polynomial. Assume that F has pairwise non-proportional linear factors in its factorisation over C. Given ɛ > 0, there exists a computable constant κ ɛ depending only on F and ɛ such that if m and n are coprime integers, then p κ ɛ (max{ m, n }) deg(f ) ɛ. p F (m,n) Shorey [Sho99] showed that abc-conjecture implies Conjecture 1. for l 4 using d k c 1 log log k. Granville (unpublished) gave a proof of the preceding result without using the inequality d k c 1 log log k. Furthermore his proof is also valid for l =, 3. 1+ɛ Theorem 1.1. The abc conjecture implies Conjecture (1.). The proof was first published in the Master s Thesis of first author[lai04]. We include the proof in this paper to have a published literature. This is given in Section 6. We would like to thank Professor A. Granville for allowing us to publish his proof. An explicit version of Conjecture 1. due to Baker [Bak94] is the following: Conjecture 1.5. Explicit abc-conjecture: Let a, b and c be pairwise coprime positive integers satisfying (1.). Then where N = N(abc) and ω = ω(n). c < 6 (log N)ω N 5 ω!.

Perfect powers in Arithmetic Progression 3 We observe that N = N(abc) whenever a, b, c satisfy (1.). We shall refer to Conjecture 1.3 as abc conjecture and Conjecture 1.5 as explicit abc conjecture. Conjecture 1.5 implies the following explicit version of Conjecture 1.3 proved in [LaSh1]. Theorem 1.. Assume Conjecture 1.5. Let a, b and c be pairwise coprime positive integers satisfying (1.) and N = N(abc). Then we have c < N 1+3/4 (1.3) Further for 0 < ɛ 3/4, there exists an integer ω ɛ depending only ɛ such that when N = N(abc) N ɛ = p p ωɛ p, we have c < κ ɛ N 1+ɛ where κ ɛ = 6 5 π max(ω, ω ɛ ) 6 5 πω ɛ with ω = ω(n). Here are some values of ɛ, ω ɛ and N ɛ. ɛ 3/4 7/1 6/11 1/ 34/71 5/1 1/3 ω ɛ 14 49 7 17 175 548 6460 N ɛ e 37.1101 e 04.75 e 335.71 e 679.585 e 1004.763 e 3894.57 e 6377 Thus c < N which was conjectured in Granville and Tucker [GrTu0]. As a consequence of Theorem 1., we prove Theorem 1.3. Assume Conjecture 1.5. Then the equation n(n + d) (n + (k 1)d) = by l (1.4) in integers n 1, d > 1, k 4, b 1, y 1, l > 1 with gcd(n, d) = 1 and P (b) k implies l 7. Further k < e 13006. when l = 7. We observe that e 13006. < e e9.5. Theorem 1.3 is a considerable improvement of Saradha [Sar1] where it is shown that (1.4) with k 8 implies that l 9 and further k 8, 3, 10, 10 7 and e e80 according as l = 9, l {3, 19}, l = 17, 13 and l {11, 7}, respectively.. Notation and Preliminaries For an integer i > 0, let p i denote the i th prime. We always write p for a prime number. For a real x > 0 and d Z, d 1, let π d (x) = 1, π(x) = π 1 (x) = 1, Θ(x) = p and θ(x) = log(θ(x)). p x,p d p x p x We write log i for log(log i). Here we understand that log 1 =.

4 S. Laishram and T. N. Shorey Lemma.1. We have (i) π(x) x log x ( 1 + 1.76 log x ) for x > 1. (ii) p i i(log i + log i 1) for i 1 (iii) θ(p i ) i(log i + log i 1.076869) for i 1 (iv) θ(x) < 1.000081x for x > 0 (v) πk( k e )k e 1 1k+1 k! πk( k e )k e 1 1k. The estimates (i) and (ii) are due to Dusart, see [Dus99b] and [Dus99a], respectively. The estimate (iii) is [Rob83, Theorem 6]. For estimate (iv), see [Dus99b]. The estimate (v) is [Rob55, Theorem 6]. 3. Proof of Theorem 1.3 Let n, d, k, b, y be positive integers with n 1, d > 1, k 4, b 1, y 1, gcd(n, d) = 1 and P (b) k. We consider the Diophantine equation n(n + d) (n + (k 1)d) = by l. (3.5) Observe that P (n(n + d) (n + (k 1)d)) > k by a result of Shorey and Tijdeman [ShTi90] and hence P (y) > k and also n + (k 1)d (k + 1) l. For every 0 i < k, we write n + id = A i X l i with P (A i ) k and (X i, p k p) = 1. Without loss of generality, we may assume that k = 4 or k 5 is a prime which we assume throughout in this section. We observe that (A i, d) = 1 for 0 i < k and (X i, X j ) = 1. Let S 0 = {A 0, A 1,..., A k 1 }. For every prime p k and p d, let i p be such that ord p (A i ) =ord p (n+id) ord p (n+i p d) for 0 i < k. For a S S 0, let S = S {A ip : p k, p d}. Then S S π d (k). By Sylvester-Erdős inequality(see [ErSe75, Lemma ] for example), we obtain A i (k 1)! p ordp((k 1)!). (3.6) A i S As a consequence, we have p d

Perfect powers in Arithmetic Progression 5 Lemma 3.. Let α, β R with α 1, β < 1 and eβ < α. Let For ( log( eα β ) + k log(αk) log k k max we have S 1 > βk. S 1 := S 1 (α) := {A i S 0 : A i αk}. ) 1 + 1.76 log k log(αk) ), exp log(eα) + β log ( β eα ( ) 1 + 1.76 log k 1 β (3.7) Proof. Let S = S 0, s 1 = S 1 and s = S S 1. Then s k π(k) s 1. We get from (3.6) that k π(k) s 1 s 1! ( αk + i ) A i (k 1)! (3.8) i=1 A i S since elements of S S 1 are distinct and the product on the left side is taken to be 1 if k π(k) s 1. Suppose s 1 βk. If k π(k) s 1, then using Lemma.1 (i), we get (1 β) log k < 1 + 1.76/ log k which is not possible by (3.7). Hence k π(k) > s 1. By using Lemma.1 (v), we obtain ( π(k 1) k 1 ) k 1 1 e 1(k 1) if s 1 = 0 (αk) k π(k) < (k 1)! (αk) s 1 < s 1! k 1 s 1 ( ) e s1 αke ( k 1 s 1 e ) k 1 if s 1 > 0. We check that the expression for s 1 = 0 is less than that of s 1 = 1 since α 1. Observe that ( ) k 1 αke s1 s 1 is an increasing function of s 1 since s 1 βk and eβ < α. This can be verified by taking log of the above expression and differentiating it with respect to s 1. Therefore ( ) (αk) k π(k) k 1 eα βk ( ) k 1 k 1 ( ) 1 eα βk ( ) k k 1 < < βk β e β β e implying Using Lemma.1 (i), we obtain ( ) β log(eα) + β log < 1 eα k s 1 ( ) β βk (eα) k < eα (αk) π(k) 1. eα β eα log( ) + log(αk) ( 1 + 1.76 ) β log k log k log(αk). k The right hand side of the above inequality is a decreasing function of k for k given by (3.7). This can be verified by observing that log(αk)/ log k = 1+log α/ log k and differentiating (1.76 + log α)/ log k log(αk)/k with respect to k. This is a contradiction for k given by (3.7).

6 S. Laishram and T. N. Shorey Corollary 3.1. For k > 113, there exist 0 f < g < h < k with h f 8 such that max(a f, A g, A h ) 4k. Proof. By dividing [0, k 1] into subintervals of the form [9i, 9(i + 1)), it suffices to show S 1 (4) > ( k/9 + 1) where S 1 is as defined in Lemma 3.. Taking α = 4, β = 1/4, we obtain from Lemma 3. that for k 700, S 1 (4) > k/4 > ( k/9 + 1). Thus we may suppose k < 700 and S 1 (4) ( k/9 + 1). For each prime k with 113 < k < 700, taking α = 4 and βk = ( k/9 + 1) in Lemma 3., we get a contradiction from (3.8). Therefore S 1 (4) > ( k/9 + 1) and the assertion follows. Given 0 f < g < h k 1, we have (h f)a g X l g = (h g)a f X l f + (g f)a hx l h. (3.9) Let λ =gcd(h f, h g, g f) and write h f = λw, h g = λu, g f = λv. Rewriting h f = h g + g f as (3.9) can be written as Let G = gcd(wa g, ua f, va h ), w = u + v with gcd(u, v) = 1, wa g X l g = ua f X l f + va hx l h. (3.10) and we rewrite (3.10) as r = ua f G, s = va h G, t = wa g G (3.11) tx l g = rx l f + sxl h. (3.1) Note that gcd(rx l f, sxl h ) = 1. From now on, we assume explicit abc conjecture. Given ɛ > 0, let N(rstX f X g X h ) N ɛ which we assume from now on till the expression (3.18). By Theorem 1., we obtain i.e., tx l g < κ ɛ N(rstX f X g X h ) 1+ɛ (3.13) Xg l N(rst) 1+ɛ (X f X g X h ) 1+ɛ < κ ɛ. (3.14) t Here N ɛ = κ ɛ = 1 if ɛ > 3/4. For ɛ = 3/4, by abuse of notation, we will be taking either N ɛ = 1, κ ɛ = 1 or N ɛ = e 37.1101, κ ɛ 6/(5 8π) if N(rstX f X g X h ) N 3/4 and we will be using it without reference. We will be taking ɛ = 3/4 for l > 7 and ɛ {5/1, 1/3} for l = 7. We have from (3.13) that rst(x f X g X h ) l < κ 3 ɛn(rst) 3(1+ɛ) (X f X g X h ) 3(1+ɛ).

Putting X 3 = X f X g X h, we obtain Perfect powers in Arithmetic Progression 7 X l 3(1+ɛ) < κ ɛ N(rst) 3 +ɛ = κ ɛ N( uvwa f A g A h G 3 ) 3 +ɛ. (3.15) Again from (3.1), we have implying rs(x f X h ) l Therefore we have from (3.14) that ( rx l f + sx l h ) = t X l g 4 ( ) 1 ( ) t l X f X h X g X 3 w A 1 l g 4rs g = Xg 3. 4uvA f A h Xg l N(rst) 1+ɛ Xg 3+3ɛ < κ ɛ t ( t 4rs ) 1+ɛ l N(rst) 1+ɛ Xg 3+3ɛ = κ ɛ (4rst) 1+ɛ l 3(1+ɛ) 1 t l (3.16) i.e., X l 3(1+ɛ) g We also have from (3.17) that N(rst) (1+ɛ)(1 1 l ) < κ ɛ 4 1+ɛ 3(1+ɛ) 1 l t l X l 3(1+ɛ) g N( uva f A h G < κ ɛ N(rs)(1+ɛ)(1 1 l ) (1+ɛ) ɛ+ N(t) l. (3.17) 4 1+ɛ l ) (1+ɛ)(1 1 l ) N( wag G 4 1+ɛ l (1+ɛ) )ɛ+ l. (3.18) Lemma 3.3. Let l 11. Let S 0 = {A 0, A 1,..., A k 1 } = {B 0, B 1,..., B k 1 } with B 0 B 1... B k 1. Then In particular S 0 k 1. B 0 B 1 < B... < B k 1. Proof. Suppose there exists 0 f < g < h < k with {f, g, h} = {i 1, i, i 3 } and A i1 = A i = A and A i3 A. By (3.10) and (3.11), we see that max(a f, A g, A h ) G and therefore r u < k, s v < k and t w < k. Since X g > k, we get from the first inequality of (3.17) with ɛ = 3/4, N ɛ = κ ɛ = 1 that k l 3(1+ɛ) < (rs) (1+ɛ)(1 1 l ) (1+ɛ) ɛ+ t l < k +3ɛ implying l < 5 + 6ɛ = 5 + 9/. This is a contradiction since l 11. Therefore either A i s are distinct or if A i = A j = A, then A m > A for m / {i, j} implying the assertion.

8 S. Laishram and T. N. Shorey As a consequence, we have Corollary 3.. Let d be even and l 11. Then k 14. Proof. Let d be even and l 11. Then we get from (3.6) with S = S 0 that A i (k 1)! ord((k 1)!) = (i + 1). A i S i+1 k 1 Observe that the right hand side of the above inequality is the product of all positive odd numbers less than k. On the other hand, since gcd(n, d) = 1, we see that all A i s are odd and S S 0 π(k) k 1 π(k) by Lemma 3.3. Hence A i S A i k 1 π(k) i=1 (i 1). Observe here again that the right hand side of the above inequality is the product of first positive k 1 π(k) odd numbers. Hence we get a contradiction if (k 1 π(k)) 1 > k 1. Assume k + π(k). By Lemma.1 (i), we get 1 /k + (/ log k)(1 + 1.76/ log k) which is not possible for k 30. By using exact values of π(k), we check that k + π(k) is not possible for 15 k < 30. Hence the assertion. Lemma 3.4. Let l 11. Then k < 400. Proof. Assume that k 400. By Corollary 3., we may suppose that d is odd. Further by Corollary 3.1, there exists f < g < h with h f 8 and max(a f, A g, A h ) 4k. Since n + (k 1)d > k l, we observe that X f > k, X g > k, X h > k implying X > k. First assume that N = N(rstX f X g X h ) < e 37.1. Then taking ɛ = 3/4, N ɛ = 1 in (3.13), we get 400 11 k 11 txg l < N 1+3/4 e 37.1(1+3/4) which is a contradiction. Hence we may suppose that N e 37.1 N 3/4. Note that we have u + v = w h f 8. We observe that uvw is even. If A f A g A h is odd, then h f, g f, h g are all even implying 1 u, v, w 4 or N(uvw) 6 giving N(uvwA f A g A h ) 6A f A g A h. Again if A f A g A h is even, then N(uvwA f A g A h ) N((uvw) )A f A g A h 35A f A g A h where (uvw) is the odd part of uvw and N((uvw) ) 35. Observe that N((uvw) ) is obtained when w = 7, u =, v = 5 or w = 7, u = 5, v =. Thus we always have N(uvwA f A g A h ) 35A f A g A h 35 (4k) 3 since max(a f, A g, A h ) 4k. Therefore taking ɛ = 3/4 in (3.15), we obtain using l 11 and X > k that k 11 3(1+ 3 4 ) 6 < 5 8π 35 3 + 3 4 (4k) 3( 3 + 3 4 ). This is a contradiction since k 400. Hence the assertion. 4. Proof of Theorem 1.3 for 4 k < 400 We assume that l 11. It follows from the result of Saradha and Shorey [SaSh05, Theorem 1] that d > 10 15. Hence we may suppose that d > 10 15 in this section.

Perfect powers in Arithmetic Progression 9 Lemma 4.5. Let r k = k + 1 π(k) ( i k log i)/(15 log 10) and Then I(k) r k. I(k) = {i [1, k] : P (n + id) > k}. Proof. Suppose not. Then I(k) r k 1. Let I (k) = {i [1, k] : P (n + id) k} = {i [1, k] : n + id = A i }. We have A i = n + id (n + d) for i I (k). Let S = {A i : i I (k)}. Then S k + 1 r k. From (3.6), we get (k 1)! A i S A i (n + d) S > d k+1 rk π(k). Since d > 10 15, we get i k k + 1 π(k) log i 15 log 10 < r k = i k k + 1 π(k) log i. 15 log 10 This is a contradiction. Here are some values of (k, r k ). k 7 11 13 17 18 8 30 36 r k 3 6 7 10 10 18 18 3 We give the strategy here. Let I k = [0, k 1] Z and a 0, b 0, z 0 be given. Let obtain a subset I 0 I k with the following properties: 1. I 0 z 0 3.. P (A i ) a 0 for i I 0. 3. I 0 [j 0, j 0 + b 0 1] for some j 0. 4. X 0 = max i I0 {X i } > k and let i 0 I 0 be such that X 0 = X i0. For any i, j I 0, taking {f, g, h} = {i, j, i 0 }, let N = N(rstX f X g X h ). Observe that X 0 p π(k)+1 and further for any f, g, h I 0, we have N(uvw) p b 0 1 p and N(A f A g A h ) p a 0 p. We will always take ɛ = 3/4, N ɛ = 1 so that κ ɛ = 1 in (3.13) to (3.18). Case I: Suppose there exists i, j I 0 such that X i = X j = 1. Taking {f, g, h} = {i, j, i 0 } and ɛ = 3/4, we obtain from (3.14) and l 11 that p 37 7 l 1+ 3 1 4 π(k)+1 X0 < N(uvwA f A g A h ) p max{a 0,b 0 1} p. (4.19) Case II: There is at most one i I 0 such that X i = 1. Then {i I 0 : X i > k} z 0 1. We take a 1, b 1, z 1 and find a subset U 0 I 0 with the following properties:

10 S. Laishram and T. N. Shorey 1. U 0 z 1 3, z 0 / z 1 z 0.. P (A i ) a 1 for i U 0. 3. U 0 [i, i + b 1 1] for some i. Let X 1 = max i U0 {X i } p π(k)+z1 1 and i 1 be such that X i1 = X 1. Taking {f, g, h} = {i, j, i 1 } for some i, j U 0 and ɛ = 3/4, we obtain from (3.17) and l 11 that p 3 7 π(k)+z1 1 X l 1+ 3 3 4 0 < N(uvwA f A g A h ) p max{a 1,b 1 1} p (4.0) since l 11. One choice is (U 0, a 1, b 1, z 1 ) = (I 0, a 0, b 0, z 0 ). We state the other choice. Let b = max(a 0, b 0 1). For each b 0 / 1 < p b 1, we remove those i I 0 such that p (n + id). There are at most (π(b 1) π(b 0 / 1)) such i. Let I 0 be obtained from I 0 after deleting those i s. Then I 0 z 0 (π(b 1) π(b 0 / 1)). Let U 1 = I 0 [j 0, j 0 + b 0 1] and U 1 = I 0 [j 0 + b 0, j 0 + b 0 1]. Let U 0 {U 1, U } for which U 0 = max( U 1, U ) and choose one of them if U 1 = U. Then U 0 z 0 / π(b 1)+π(b 0 / 1) = z 1. Further P (A i ) b 0 / 1 = a 1 and b 1 = b 0 /. Further X 1 = max i U0 {X i } p π(k)+z1 1. Our choice of z 0, a 0, b 0 will imply that z 1 3. 4.1. k {4, 5, 7, 11} We take I 0 = U 0 = I k, a i = b i = z i = k for i {0, 1} and hence N(uvwA f A g A h ) p. And the assertion follows since both (4.19) and (4.0) are contradicted. p k 4.. k {13, 17, 19, 3} We take I 0 = {i [1, 11] : p (n + id) for 13 p 3}. Then by r 11 = 6 and Lemma 4.5 with k = 11, we see that I 0 z 0 = 11 4 > 11 r 11 11 I(11). Therefore there exist an i I 0 I 11 and hence X i > 3. We take U 0 = I 0, a i = b i = 11, z 1 = z 0 for i {0, 1} and hence N(uvwA f A g A h ) p 11 p. And the assertion follows since both (4.19) and (4.0) are contradicted. 4.3. 9 k 47 We take I 0 = {i [1, 17] : p (n + id) for 17 p k}. Then by r 17 = 10 and Lemma 4.5 with k = 17, we have I 0 z 0 = 17 (π(k) π(13)) = 3 π(k) 3 π(47) = 8 > 17 r 17 17 I(17) implying that there exists i I 0 with X i > k. We take a i = 13, b i = 17, z i = 3 π(k) for i {0, 1} and hence N(uvwA f A g A h ) p 13 p. And the assertion follows since both (4.19) and (4.0) are contradicted.

4.4. k 53 Perfect powers in Arithmetic Progression 11 Given m and q such that mq < k, we consider the q intervals I j = [(j 1)m + 1, jm] Z for 1 j q and let I = q j=1 I j and I = {i I : m P (A i ) k}. There is at most one i I such that mq 1 < P (A i ) k and for each j q, there are at most j number of i I such that (mq 1)/j < P (A i ) (mq 1)/(j 1). Therefore I π(k) π(mq 1) + q 1 j=1 q j= ( j π( mq 1 j 1 ) π(mq 1 j = π(k) + π( mq 1 ) qπ(m 1) =: T (k, m, q). j Hence there is at least one j such that I j I T (k, m, q)/q. We will choose q such that T (k, m, q)/q < r m. Let I 0 = I j \ I and let j 0 be such that I 0 [(j 0 1)m + 1, j 0 m]. Then p (n + id) imply p < m or p > k whenever i I 0. Further I 0 z 0 = m T (k, m, q)/q. Since T (k, m, q)/q < r m, we get from Lemma 4.5 with k = m and n = (j 0 1)m that there is an i I 0 with X i > k. Further P (A i ) < m for all i I 0. Here are the choices of m and q. k [53, 89) [89, 179) [179, 39) [39, 367) [367, 433) (m, q) (17, 3) (8, 3) (36, 5) (36, 6) (36, 10) We have a 0 = m 1, b 0 = m and z 0 = m T (k, m, q)/q and we check that z 0 3. The Subsection 4.3(9 k 47) is in fact obtained by considering m = 17, q = 1. Now we consider Cases I and II and try to get contradiction in both (4.19) and (4.0). For these choices of (m, q), we find that the Cases I are contradicted. Further taking U 0 = I 0, a 1 = a 0 = m 1, b 1 = b 0 = m, z 1 = z 0, we find that Case II is also contradicted for 53 k < 89. Thus the assertion follows in the case 53 k < 89. So, we consider k 89 and try to contradict Cases II. Recall that we have X i > k for all but at most one i I 0. Write I 0 = U 1 U where U 1 = I 0 [(j 0 1)m + 1, (j 0 1)m + m/] and U = I 0 [(j 0 1)m +m/ + 1, j 0 m]. Let U 0 = U 1 or U 0 = U according as U 1 z 0 / or U z 0 /, respectively. Let U 0 = {i U 0 : p A i for m/ p < m}. Then U 0 z 1 := z 0 / (π(m 1) π(m/)) = (m T (k, m, q)/q )/ (π(m 1) π(m/)) 3. Further p (n + id) with i U 0 imply p < m/ or p > k. Now we have Case II with a 1 = m/ 1, b 1 = m/ and find that (4.0) is contradicted. Hence the assertion. ) ) 5. l = 7 Let l = 7. Assume that k exp(13006.). Taking α = 3, β = 1/15 + /9 in Lemma 3., we get ( 1 S 1 (3) = {i [0, k 1] : A i 3k} > k 15 + ). 9

1 S. Laishram and T. N. Shorey For i s such that A i S 1 (3), we have X i > k and we arrange these X i s in increasing order as X i1 < X i < <. Then X ij p π(k)+j. Consider the set J 0 = {i : X i p }. We have π(k)+ k/15 ( 1 J 0 > k 15 + ) 9 k ( ) 15 + k 1 + 1. 9 Hence there are f, g, h J 0, f < g < h such that h f 8. Also A i 3k and X = (X f X g X h ) 1/3 p π(k)+ k/15. First assume that N = N(rstX f X g X g ) exp(6377) N 1/3. Observe that uvw 70 since u + v = w 8, obtained at + 5 = 7. Taking ɛ = 1/3, we obtain from (3.15) and max(a f, A g, A h ) 3k that p 3 π(k)+ < 5 k 15 6 π 6458 N(uvwA 5 70 (3k)3 f A g A h ) 6 190π. Since π(k) > we have π(k) + k/15 > k 15 and hence p π(k)+ k/15 > (k/15) log(k/15) by Lemma.1 (ii). Therefore ( log k ) ( 3 ( 350 (3 15)3 < 15 6 or k < 15 exp 45 190π 350 6 190π ) 1 ) 3 which is a contradiction since k exp(13006.). Therefore we have N = N(rstX f X g X h ) < exp(6377). We may also assume that N > exp(3895) otherwise taking ɛ = 3/4 in (3.13), we get k 7 < X 7 g < N 1+3/4 exp(3895 7/4) or k < exp(3895/4) which is a contradiction. Now we take ɛ = 5/1 in (3.13) to get k 7 < X 7 g < N 1+5/1 exp(6466 17/1) or k < exp(13006.). Hence the assertion. 6. abc-conjecture implies Erdős conjecture: Proof of Theorem 1.1 Assume (3.5). We show that k is bounded by a computable absolute constant. Let k k 0 where k 0 is a sufficiently large computable absolute constant. Let ɛ > 0. Let c 1, c, be positive computable constants depending only on ɛ. Let I = {i p : p k and p d} where i p be as defined in beginning of Section 3. Let S = {A i : k/ i < k or i I} and Φ = A i. i k i / I

Perfect powers in Arithmetic Progression 13 By Sylvester-Erdős inequality(see [ErSe75, Lemma ] for example), we get ord p (Φ) ord p (i i p ) i k i / I ( ord p (k k 1 (ip k ( (k 1 ip ord p ) (k ) k )! otherwise. k k ))!(ip ) k )! if ip k, Since ord p (r!s!) ord p ((r + s)!) and k k/ = (k + 1)/, we see that p ordp(φ) p ord p(( k 1 ip k+1 )) p ordp( k+1 Using the fact that p ordp((x k)) x for any x k, we get by using Lemma.1 (i), (v). Let D be a fixed positive integer and let J =!). Φ (k 1) πd(k) ( k + 1 k )! k e c 1 k { k 1 D j k 1 } 1 : {Dj + 1, Dj +,, Dj + D} I = φ. D We shall choose D = 0. Let j, j J be such that j j. Then Dj + i Dj + i for 1 i, i D otherwise D(j j ) = (i i ) and i i < D. Further we also see that k/ Dj + i k 1 for 1 i D and consequently J (k 1)/D π(k). For each j J, let Φ j = D A Dj+i. Then j J Φ j divides Φ implying i=1 Thus there exists j 0 J such that Let Φ j0 Φ j Φ k k e c 1 k. j J ( ) k k 1 e c 1 k J ) H := (k k e c 1 k ) 1 k 1 D π(k) c D k D. D (n + (Dj 0 + i)d). i=1

14 S. Laishram and T. N. Shorey Since A Dj0 +ix l Dj 0 +i n + (k 1)d, we have X Dj 0 +i ((n + (k 1)d)/A Dj0 +i) 1/l. Thus D p = X Dj0 +i (n + (k 1)d) D l (Φj0 ) 1 l p H p>k i=1 Therefore p = p p Φ j 0 (n + (k 1)d) D l (Φj0 ) 1 l p H p H p k p H p>k c D(1 1 l ) k D(1 1 l ) (n + (k 1)d) D l. On the other hand, we have H = F (n + Dj 0 d, d) where F (x, y) = D (x + iy) i=1 is a binary form in x and y of degree D such that F has distinct linear factors. From Conjecture 1.4, we have p c 3 (n + Dj 0 d) D ɛ. p H Comparing the lower and upper bounds of p H p and using n+dj 0d > (n+(k 1)d)/, we get We now use n + (k 1)d > k l to derive that Taking ɛ = 1/ and putting D = 0, we get +ɛ 1 D(1 k > c 4 (n + (k 1)d) 1 l ). +ɛ l(1 D(1 c 5 > k 1 ) 1 l ). l c 6 > k l 1 8(l 1) k 1 since l. This is a contradiction since k k 0 and k 0 is sufficiently large. Acknowledgments The second author would like to thank ISI, New Delhi where this work was initiated during his visit in July 011. The authors also like to thank the referee for remarks in an earlier draft of this paper.

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