Perfect powers in Arithmetic Progression

isid/ms/015/18 October 1, 015 http://www.isid.ac.in/ statmath/index.php?module=preprint Perfect powers in Arithmetic Progression Shanta Laishram and T. N. Shorey Indian Statistical Institute, Delhi Centre 7, SJSS Marg, New Delhi 110 016, India

PERFECT POWERS IN ARITHMETIC PROGRESSION SHANTA LAISHRAM AND T. N. SHOREY 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. 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 (1) n(n + d) (n + (k 1)d) = by l in variables n, d, k, b, y, l. If k =, we observe that () has infinitely many solutions. Therefore we always suppose that k 3. It has been conjectured (see [Tij88], [SaSh05]) that Conjecture 1.1. Equation (1) implies that (k, l) {(3, 3), (, ), (3, )}. It is known that (1) has infinitely many solutions when (k, l) {(3, ), (3, 3)(, )}. A weaker version of Conjecture 1.1 is the following conjecture due to Erdős. Conjecture 1.. Equation (1) implies that k is bounded by an absolute constant. For an account of results on (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 () a + b = c where a, b and c are coprime positive integers, then c c ɛ p p abc 1+ɛ. Key words and phrases. abc Conjecture, Arithmetic progressions, Erdős Conjecture. 1

SHANTA LAISHRAM AND T. N. SHOREY It is known as abc-conjecture; the name derives from the usage of letters a, b, c in (). 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 abc-conjecture is equivalent to the following: Conjecture 1.. 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 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. Theorem 1. The abc conjecture implies Conjecture (1.). The proof was first published in the Master s Thesis of first author [Lai0]. 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 due to Baker [Bak9] is the following: Conjecture 1.5. Explicit abc-conjecture: Let a, b and c be pairwise coprime positive integers satisfying (). Then where N = N(abc) and ω = ω(n). c < 6 (log N)ω N 5 ω! We observe that N = N(abc) whenever a, b, c satisfy (). 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. Assume Conjecture 1.5. Let a, b and c be pairwise coprime positive integers satisfying () and N = N(abc). Then we have (3) c < N 1+ 3. Further for 0 < ɛ 3, 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 PERFECT POWERS IN ARITHMETIC PROGRESSION 3 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. 3 ɛ 7 1 6 11 ω ɛ 1 9 7 17 175 58 660 N ɛ e 37.1101 e 0.75 e 335.71 e 679.585 e 100.763 e 389.57 e 6377 1 3 71 5 1 1 3 Thus c < N which was conjectured in Granville and Tucker [GrTu0]. As a consequence of Theorem, we prove Theorem 3. Assume Conjecture 1.5. Then the equation () n(n + d) (n + (k 1)d) = by l in integers n 1, d > 1, k, 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 3 is a considerable improvement of Saradha [Sar1] where it is shown that () 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) = p and θ(x) = log(θ(x)). p x,p d 1, π(x) = π 1 (x) = p x 1, Θ(x) = p x We write log i for log(log i). Here we understand that log 1 =. Lemma.1. We have (i) π(x) x ( 1 + 1.76 ) for x > 1. log x log x (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].

SHANTA LAISHRAM AND T. N. SHOREY 3. Proof of Theorem 3 Let n, d, k, b, y be positive integers with n 1, d > 1, k, b 1, y 1, gcd(n, d) = 1 and P (b) k. We consider the Diophantine equation (5) n(n + d) (n + (k 1)d) = by l. 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 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 = 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 (6) A i (k 1)! p ordp((k 1)!). A i S As a consequence, we have Lemma 3.1. Let α, β R with α 1, β < 1 and eβ < α. Let For (7) ( log( eα β ) + k log(αk) log k k max{ we have S 1 > βk. p d S 1 := S 1 (α) := {A i S 0 : A i αk}. 1 + 1.76 log k log(eα) + β log ( β eα ) log(αk) ), exp( 1 + 1.76 log k 1 β )} Proof. Let S = S 0, s 1 = S 1 and s = S S 1. Then s k π(k) s 1. We get from (6) that (8) k π(k) s 1 s 1! ([αk + i]) A i (k 1)! 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.

PERFECT POWERS IN ARITHMETIC PROGRESSION 5 Suppose s 1 βk. If k π(k) s 1, then using Lemma.1 (i), we get (1 β) log k < 1 + 1.76 which is not possible by (7). Hence k π(k) > s log k 1. By using Lemma.1 (v), we obtain (αk) k π(k) (k 1)! ( π(k 1) k 1 ) k 1 1 < (αk) s 1 e e 1(k 1) < if s 1 = 0 ) s1 ( s 1! k 1 ) k 1 if s 1 > 0. ( k 1 s 1 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 ) s 1 s 1 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 ( ) βk ( ) k 1 ( ) βk ( ) k 1 (αk) k π(k) k 1 eα k 1 1 eα k < < βk β e β β e implying Using Lemma.1 (i), we obtain ( ) β log(eα) + β log < 1 eα k αke s 1 ( ) βk β (eα) k < eα (αk) π(k) 1. eα β eα log( ) + log(αk) ( 1 + 1.76 ) β log k log k e log(αk). k The right hand side of the above inequality is a decreasing function of k for k given log αk by (7). This can be verified by observing that = 1 + log α and differentiating log k log k 1.76+log α log(αk) with respect to k. This is a contradiction for k given by (7). log k k Corollary 3.. For k > 113, there exist 0 f < g < h < k with h f 8 such that max(a f, A g, A h ) k. Proof. By dividing [0, k 1] into subintervals of the form [9i, 9(i + 1)), it suffices to show S 1 () > ([ k] + 1) where S 9 1 is as defined in Lemma 3.1. Taking α =, β = 1, we obtain from Lemma 3.1 that for k 700, S 1 () > k > ([ k ] + 1). Thus we may 9 suppose k < 700 and S 1 () ([ k ] + 1). For each prime k with 113 < k < 700, 9 taking α = and βk = ([ k ] + 1) in Lemma 3.1, we get a contradiction from (8). 9 Therefore S 1 () > ([ k ] + 1) and the assertion follows. 9 (9) 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 h X l h. 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 w = u + v with gcd(u, v) = 1,

6 SHANTA LAISHRAM AND T. N. SHOREY (9) can be written as (10) Let G = gcd(wa g, ua f, va h ), (11) and we rewrite (10) as (1) Note that gcd(rx l f, sxl h ) = 1. wa g X l g = ua f X l f + va h X l h. r = ua f G, s = va h G, t = wa g G tx l g = rx l f + sx l h. 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 (18). By Theorem, we obtain (13) i.e., (1) tx l g < κ ɛ N(rstX f X g X h ) 1+ɛ Xg l N(rst) 1+ɛ (X f X g X h ) 1+ɛ < κ ɛ. t Here N ɛ = κ ɛ = 1 if ɛ > 3. For ɛ = 3/, 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 if N(rstX 8π fx g X h ) N 3. We will be taking ɛ = 3 5 for l > 7 and ɛ {, 1 } for l = 7. We have from (13) that 1 3 Putting X 3 = X f X g X h, we obtain rst(x f X g X h ) l < κ 3 ɛn(rst) 3(1+ɛ) (X f X g X h ) 3(1+ɛ). (15) X l 3(1+ɛ) < κ ɛ N(rst) 3 +ɛ = κ ɛ N( uvwa fa g A h G 3 ) 3 +ɛ. Again from (1), we have implying rs(x f X h ) l Therefore we have from (1) that (16) i.e., (17) ( rx l f + sx l h ) = t X l g ( ) 1 ( t l X f X h X g X 3 w A ) 1 l g rs g = Xg 3. uva f A h N(rst) 1+ɛ X Xg l g 3+3ɛ < κ ɛ t X l 3(1+ɛ) g ( t rs N(rst) (1+ɛ)(1 1 l ) < κ ɛ 1+ɛ l 3(1+ɛ) 1 t l ) 1+ɛ l N(rst) 1+ɛ Xg 3+3ɛ = κ ɛ (rst) 1+ɛ l 3(1+ɛ) 1 t l N(rs)(1+ɛ)(1 1 l ) N(t) ɛ+ (1+ɛ) l. 1+ɛ l

We also have from (17) that (18) PERFECT POWERS IN ARITHMETIC PROGRESSION 7 X l 3(1+ɛ) g N( uvaf Ah G < κ ɛ ) (1+ɛ)(1 1 l ) N( wag G 1+ɛ l (1+ɛ) )ɛ+ l 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 (10) and (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 (17) with ɛ = 3, 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. As a consequence, we have Corollary 3.. Let d be even and l 11. Then k 1. Proof. Let d be even and l 11. Then we get from (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 + 1.76 (1 + ) k log k 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 3.5. Let l 11. Then k < 00.

8 SHANTA LAISHRAM AND T. N. SHOREY Proof. Assume that k 00. By Corollary 3., we may suppose that d is odd. Further by Corollary 3., there exists f < g < h with h f 8 and max(a f, A g, A h ) k. 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, N ɛ = 1 in (13), we get 00 11 k 11 txg l < N 1+ 3 e 37.1(1+ 3 ) which is a contradiction. Hence we may suppose that N e 37.1 N 3. 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 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 (k) 3 since max(a f, A g, A h ) k. Therefore taking ɛ = 3 in (15), we obtain using l 11 and X > k that k 11 3(1+ 3 ) 6 < 5 8π 35 3 + 3 (k) 3( 3 + 3 ). This is a contradiction since k 00. Hence the assertion.. Proof of Theorem 3 for k < 00 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. i k Lemma.1. Let r k = [k + 1 π(k) log i ] and 15 log 10 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 (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 k + 1 π(k) This is a contradiction. Here are some values of (k, r k ). i k log i 15 log 10 < r i k k = [k + 1 π(k) log i 15 log 10 ].

PERFECT POWERS IN ARITHMETIC PROGRESSION 9 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. () 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, N ɛ = 1 so that κ ɛ = 1 in (13) to (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, we obtain from (1) and l 11 that p 37 l 7 π(k)+1 X 1+ 3 1 (19) 0 < N(uvwA f A g A h ) p. p max{a 0,b 0 1} 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: (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, we obtain from (17) and l 11 that p 3 l 7 π(k)+z1 1 X 1+ 3 3 (0) 0 < N(uvwA f A g A h ) p p max{a 1,b 1 1} 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.

10 SHANTA LAISHRAM AND T. N. SHOREY.1. k {, 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 (19) and (0) are contradicted. p k.. 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.1 with k = 11, we see that I 0 z 0 = 11 > 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 (19) and (0) are contradicted..3. 9 k 7 We take I 0 = {i [1, 17] : p (n + id) for 17 p k}. Then by r 17 = 10 and Lemma.1 with k = 17, we have I 0 z 0 = 17 (π(k) π(13)) = 3 π(k) 3 π(7) = 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 (19) and (0) are contradicted... k 53 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 < P (A j i ) mq 1. Therefore j 1 q ( I π(k) π(mq 1) + j π( mq 1 j 1 ) π(mq 1 ) ) j j=1 j= q 1 = 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) ]. We will choose q such that 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 [ T (k,m,q) q 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) ] < r q m, we get from Lemma.1 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.

PERFECT POWERS IN ARITHMETIC PROGRESSION 11 k 53 k < 89 89 k < 179 179 k < 39 39 k < 367 367 k < 33 (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) ] and we check that z q 0 3. The Subsection.3(9 k 7) is in fact obtained by considering m = 17, q = 1. Now we consider Cases I and II and try to get contradiction in both (19) and (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 0m]. 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 (0) is contradicted. Hence the assertion. 5. l = 7 Let l = 7. Assume that k exp(13006.). Taking α = 3, β = 1 3.1, we get 15 + 9 in Lemma S 1 (3) = {i [0, k 1] : A i 3k} > k( 1 15 + 9 ). 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 π(k)+[ k ] }. We have 15 J 0 > k( 1 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. Observe that uvw 3 70 since u + v = w 8, obtained at + 5 = 7. Taking ɛ = 1, we obtain from 3 (15) and max(a f, A g, A h ) 3k that p 3 π(k)+[ k 15 ] < 5 6 π 658 N(uvwA fa g A h ) 5 70 (3k)3 6 190π. Since π(k) > we have π(k) + [ k ] > k and hence p 15 15 π(k)+[ k 15 ] > k log k by 15 15 Lemma.1 (ii). Therefore ( log k ) ( 3 ( ) ) 1 350 (3 15)3 < 15 6 350 or k < 15 exp 5 190π 6 3 190π

1 SHANTA LAISHRAM AND T. N. SHOREY 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 in (13), we get k7 < Xg 7 < N 1+ 3 exp(3895 7) or k < exp( 3895 5 ) which is a contradiction. Now we take ɛ = 1 in (13) to get k 7 < Xg 7 < N 1+ 5 1 exp(666 17 ) or k < exp(13006.). Hence the assertion. 1 6. abc-conjecture implies Erdős conjecture: Proof of Theorem 1 Assume (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 i [ k ] i / I By Sylvester-Erdős inequality(see [ErSe75, Lemma ] for example), we get { ( ordp (k [ k ord p (Φ) ord p (i i p ) ( ] 1 (i p [ k]))!(i p [ k])!) if i p [ k], (k 1 ip ) ) ord p k [ (k [ k ])! otherwise. k ] Since ord p (r!s!) ord p ((r + s)!) and k [ k] = [ k+1 ], we see that p ordp(φ) p ordp((k 1 ip [ k+1 ] )) k+1 ordp([ p ]!). Using the fact that p ordp((x k)) x for any x k, we get Φ (k 1) π d(k) ([ k + 1 ])! k k e c 1 k by using Lemma.1 (i), (v). Let D be a fixed positive integer and let { k 1 J = 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 k 1 ] Dj + i k 1 for 1 i D and consequently J π(k). For each D D j J, let Φ j = A Dj+i. Then j J Φ j divides Φ implying i=1 Φ j Φ k k e c 1 k. j J

PERFECT POWERS IN ARITHMETIC PROGRESSION 13 Thus there exists j 0 J such that ( ) Φ j0 k k 1 e c 1 k J ) ( k k e c 1 k ) 1 k 1 D π(k) c D k D. Let H := D (n + (Dj 0 + i)d). i=1 Since A Dj0 +ixdj l 0 +i n + (k 1)d, we have X Dj 0 +i ( n+(k 1)d A Dj0 ) 1 l. Thus +i p = p H p>k D X Dj0 +i (n + (k 1)d) D l (Φj0 ) 1 l i=1 Therefore p = p p Φ j 0 (n + (k 1)d) D l (Φj0 ) 1 D(1 l c 1 l ) k D(1 1 l ) (n + (k 1)d) D l. p H p H p k p H p>k On the other hand, we have H = F (n + Dj 0 d, d) where F (x, y) = D (x + iy) is a binary form in x and y of degree D such that F has distinct linear factors. From Conjecture 1., we have p c 3 (n + Dj 0 d) D ɛ. p H i=1 Comparing the lower and upper bounds of p H get +ɛ 1 D(1 k > c (n + (k 1)d) 1 l ). We now use n + (k 1)d > k l to derive that +ɛ l(1 D(1 c 5 > k 1 ) 1 l ). p and using n + Dj 0 d > n+(k 1)d, we Taking ɛ = 1 and putting D = 0, we get l c 6 > k l 1 1 8(l 1) k since l. This is a contradiction since k k 0 and k 0 is sufficiently large.

1 SHANTA LAISHRAM AND T. N. SHOREY Acknowledgments The second author would like to thank ISI, New Delhi where this work was initiated during his visit in July 011. References [Bak9] A. Baker, Experiments on the abc-conjecture, Publ. Math. Debrecen 65(00), 53 60. [Dus99a] P. Dusart, The k th prime is greater than k(ln k + ln ln k 1) for k, Math. Comp. 68 (1999), 11 15 [Dus99b] P. Dusart, Inégalitiés explicites pour ψ(x), θ(x), π(x) et les nombres premiers, C. R. Math. Rep. Acad. Sci. Canada 1(1)(1999), 53-59. [Elk91] N. Elkies, ABC implies Mordell, Int. Math. Res. Not. 7 (1991), 99 109. [ErSe75] P. Erdős and J. L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math. 19 (1975), 9-301. [GrTu0] A. Granville and T. J. Tucker, It s as easy as abc, Notices of the AMS, 9(00), 1-31. [Lai0] S. Laishram, Topics in Diophantine equations, M.Sc. Thesis, TIFR/Mumbai University, 00, online at http: //www.isid.ac.in/ shanta/mscthesis.pdf. [LaSh1] S. Laishram and T. N. Shorey, Baker s Explicit abc-conjecture and applications, Acta Arith., 155 (01), 19 9. [Rob55] H. Robbins, A remark on Stirling s formula, Amer. Math. Monthly 6 (1955), 6-9. [Rob83] G. Robin, Estimation de la fonction de Tchebychef θ sur le k-ieme nombre premier et grandes valeurs de la fonction ω(n) nombre de diviseurs premiers de n, Acta Arith. (1983), 367 389. [Sar1] N. Saradha, Applications of Explicit abc-conjecture on two Diophantine Equatiosn, Acta Arith., 151 (01), 01 19. [SaSh05] N. Saradha and T. N. Shorey, Contributions towards a conjecture of Erdos on perfect powers in arithmetic progressions, Compositio Math., 11 (005), 51-560. [Sho06] T. N. Shorey, Diophantine approximations, Diophantine equations, Transcendence and Applications, Indian Jour. of Pure and Applied Math, 37 (006), 9 39. [Sho0b] T.N. Shorey, Powers in arithmetic progression (II), Analytic Number Theory, RIMS Kokyuroku (00), Kyoto University. [Sho99] T. N. Shorey, Exponential diophantine equations involving products of consecutive integers and related equations, Number Theory ed. R.P. Bambah, V.C. Dumir and R.J. Hans-Gill, Hindustan Book Agency (1999), 63-95. [ShTi90] T. N. Shorey and R. Tijdeman, On the greatest prime factor of an arithmetical progression, A tribute to Paul Erdős, ed. by A. Baker, B. Bollobás and A. Hajnal, Cambridge University Press (1990), 385 389. [Tij88] R. Tijdeman, Diophantine equations and diophantine approximations, Number Theory and Applications, Kluwer (1988), 15 3. E-mail address: shanta@isid.ac.in Stat Math Unit, Indian Statistical Institute, 7 SJS Sansanwal Marg, New Delhi 110016, India E-mail address: shorey@math.iitb.ac.in Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 00076, India