AN ESTIMATE FOR THE LENGTH OF AN ARITHMETIC PROGRESSION THE PRODUCT OF WHOSE TERMS IS ALMOST SQUARE 1. INTRODUCTION

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1 AN ESTIMATE FOR THE LENGTH OF AN ARITHMETIC PROGRESSION THE PRODUCT OF WHOSE TERMS IS ALMOST SQUARE SHANTA LAISHRAM ABSTRACT. Erdős conjectured that ( n(n + d (n + ( d = y in positive integers n,, d >, y with gcd(n, d =, implies that is bounded by an absolute constant. Shorey and Tijdeman [6] showed that ( implies that is bounded by an effectively computable number depending only on ω(d, the number of distinct prime divisors of d. In this paper, an explicit bound for in terms of ω(d is presented.. INTRODUCTION For an integer x >, we denote by P(x and ω(x the greatest prime factor of x and the number of distinct prime divisors of x, respectively. Further we put P( = and ω( = 0. Let n, d,, b, y be positive integers such that b is square free, d,, P(b and gcd(n, d =. We consider the equation ( n(n + d (n + ( d = by in n, d,, b, y with P(b. For a survey of results on (, see [6], [4], [4] and [5]. Equation ( with d = has been solved completely in [] with P(b < and in [] with P(b =. Therefore we assume from now onwards that d >. Marszale [7] proved that ( implies is bounded by an effectively computable number 0 depending only on d. In fact the above assertion holds with 0 depending only on ω(d. This is due to Shorey and Tijdeman [6], who proved that ω(d > c where c is an effectively computable absolute constant. However the bound log 0 is very large. Further ( with ω(d = and /, 5} has been solved completely in [] and [8]. Therefore we shall always assume that ω(d. In this paper, we give an explicit bound for in terms of ω(d whenever ( holds. For ω(d, we define κ 0 = κ 0 (ω(d as in the Table below. ω(d κ 0 (d even κ 0 (d odd ω(d κ 0 (d even κ 0 (d odd Mathematics Subject Classification: Primary D6. Keywords: Arithmetic Progressions, Diophantine equations, squarefree integers, Congruences The research partially supported by TIFR Endowment Fund.

2 SHANTA LAISHRAM For ω(d, we define κ 0 = κ 0 (ω(d as.5ω(d4 ω(d if d is even κ 0 (ω(d = ω(d4 ω(d if d is odd. We prove Theorem. Equation ( implies that ( < κ 0. Theorem is a direct consequence of the following two propositions. Proposition. Let κ 0. Then ( implies that (4 (5 and hence (6 where d < 4c (, n < c ( n + ( d < 5c ( if d is odd 6 c = 8 if ord (d = if ord 4 (d. Proposition. Let κ 0. Then ( implies that (7 where n + ( d δ 5 6 δ = minord (d, }.. NOTATION AND PRELIMINARIES From (, we have (8 n + id = A i Xi for 0 i < with P(A i and (X i, p p =. Also we have (9 n + id = a i x i for 0 i < with a i squarefree. Since gcd(n, d =, we see that (0 (A i, d = (a i, d = (X i, d = (x i, d = for 0 i <. Let T = i 0 i <, X i = }, T = i 0 i <, X i }. Note that X i > for i T. For 0 i <, let ( ν(a i = j T, A j = A i }.

3 ALMOST SQUARES IN ARITHMETIC PROGRESSION We always suppose that there exist i 0 > i > > i ν(ai such that A i0 = A i = = A iν(ai. Similarly we define and ( Define ( R = a i 0 i < } ν(a i = j 0 j <, ai = a j }. ρ := ρ(d = if d if d. The letter p always denotes a prime number and p i the i th prime number. Let P < P < be odd prime divisors of d. Let r := r(d 0 be the unique integer such that (4 P P P r < (4c ( but P P P r+ (4c (. If r = 0, we understand that the product P P r =. Let d d and d = d be such that, d =. We write d gcd(d if ord d (d = d d, gcd(d, d = if ord (d and we always suppose that d is odd if ord (d =. We call such pairs (d, d as partitions of d. We observe that the number of partitions of d is ω(d θ where if ord θ := θ (d (d =, = 0 otherwise and we write θ for θ (d. In particular, by taing d = and d = d, the number of partitions of d is ω(d θ. (5 Suppose that A i = A j, i > j. Then from (8 and (0, we have (i jd = A i (X i X j = A i (X i X j (X i + X j such that gcd(d, X i X j, X i + X j = if d is odd and if d is even. Hence for any divisor d of d, we have a partition (d, d of d corresponding to A i = A j such that d (X i X j and d (X i + X j and it is the unique partition of d corresponding to the pair (i, j. Similarly, we have unique partition of d corresponding to every pair (i, j whenever a i = a j. As in Shorey and Tijdeman [6], the proof depends on comparing an upper bound and a lower bound for n + ( d. The upper bound of n + ( d given by Proposition is a consequence of Lemmas 5, 8,,, which are refinements of results in [6], [] and []. It is proved by counting the number of distinct a i s and looing at the number of partitions of d. The proof of Proposition is by counting the number of X i s greater than and calculating the maximal value of A i. Proposition is a consequence of Lemmas 4, 6, 7, 9, 0, 4. The new features of the paper are the refinement of the upper bound of

4 4 SHANTA LAISHRAM the multiplicities of A i with respect to partitions of d, counting the number of A i s with multiplicity greater than and the use of r to improve the lower bounds of the maximum of A i s. We shall follow the notation of this section throughout the paper. We use MATHEMAT- ICA for the computations in the paper. This is a part of my Master s thesis [6].. LEMMAS We begin with some estimates from Prime number theory. Lemma. We have (i π(ν ν ( +.5 for ν > log ν log ν (ii π(ν ν ( for ν 59 log ν log ν (iii p i i log i for i (iv log p <.00008ν for ν > 0 p ν (v ord p (! p log( for p <. p log p Proof. The estimates (i, (ii and (iii are due to Rosser and Schoenfeld [0]. For estimate (iv, see [, p.60] and [, Prop.7]. For a proof of (v, see [5, Lemma (i]. The next result is Stirling s formula, see [9]. Lemma. For a positive integer ν, we have πν e ν ν ν e ν+ < ν! < πν e ν ν ν e ν. Lemma. Let π d ( π(. Then (6 T > ( log ( log (n + ( d log π(. Proof. We use [, Lemma ] with t =, log p d p ordp((! 0 and π d ( π( ω(d +. Let n ( d. Then log n log(n + ( d log. This with [, (4.] and Lemma (i gives (6. For n < ( d, we have log( + log d > log(n + ( d log. This with [, (4.] and Lemma (i gives (6. Lemma 4. Let d = d d with gcd(d, d =. Let i 0 T be such that A i0 d. Then (7 ν(a i0 ω(d θ (d.

5 ALMOST SQUARES IN ARITHMETIC PROGRESSION 5 Proof. For simplicity, we write θ = θ (d. Assume that ν(a i0 > ω(d θ. Then there exists a sequence of indices i 0 > i > > i ω(d θ such that A i0 = A i = = A i ω(d θ. For each pair (i 0, i r, r =,, ω(d θ, we have a unique partition corresponding to the pair. But there are at most ω(d θ partitions of d. Since (i 0 i r d = A i0 (X i0 X ir (X i0 + X ir and A i0 d, we have > i 0 i r = A ( ( ( ( i 0 Xi0 X ir Xi0 + X ir Xi0 X ir Xi0 + X ir, d d where (d, d is the partition of d d d corresponding to pair (i 0, i r. This shows that we cannot have the partition ( d θ, θ corresponding to any pair. Hence there can be at most ω(d θ partitions of d with respect to ω(d θ pairs of (i 0, i r, r =, ω(d θ. Hence by Box Principle, there exist pairs (i 0, i r, (i 0, i s with r < s ω(d θ and a partition (d, d of d corresponding to these pairs. Thus d (X i0 X ir, d (X i0 + X ir and d (X i0 X is, d (X i0 + X is so that lcm(d, d (X ir X is. Since A ir = A is = A i0 and gcd(d, d, we have d a contradiction. > (i r i s d = (X i r X is A i0 lcm(d, d (X ir + X is gcd(d, d > (X i r + X is > =, By taing d = and d = d, the following result is immediate from Lemma 4 since θ (d = θ. Corollary. For i 0 T, we have ν(a i0 ω(d θ. Lemma 5. Let 7. Suppose n c ( or d 4c (. Then for 0 i 0 <, we have (8 ν(a i0 ω(d θ. Proof. Suppose that ν(a i0 > ω(d θ. We note that both x i + x j and x i x j are even when d is even. Continuing as in the proof of (7 with d = d, we see that there exists i, j with i > j and > a i 0 (x i + x j where d (xi x 0 if d is even and d (xi x 0 if d is odd. We have x i x j + d so that > a i 0 (x i + x j ( a j x j + d 4 n + d and hence 4 > + c ( if d 4c ( (c ( + if n c ( which is not true for 7.

6 6 SHANTA LAISHRAM Lemma 6. Equation ( implies that either or d 4c ( [ ] ω(d r. Proof. If r + [ ω(d ], then ω(d (r + giving d 4c ( by (4. Lemma 7. Let S A i 0 i < } and min A h U. Let t. Assume that A h S ( ( P Pt (9 S > Q t where Q t is an integer. Then (0 max A h δ Q t P P t + U. A h S Proof. For an odd p d, we have (Ah = p ( Ah Xh = p ( n p where ( is Legendre symbol, so that A h belongs to at most p distinct residue classes modulo p for each 0 h <. If d is even, then A h also belongs to a unique residue class modulo δ for each 0 h <. Hence by Chinese remainder theorem, A h belongs to at most ( P Pj ( distinct residue classes modulo δ P P j for each j, j t. Assume that (0 does not hold. Then Therefore contradicting (9. max A h (U δ Q t P P t. A h S S δ Q t P P t δ P P t ( P ( Pt Corollary. Let S and U be as in Lemma 7. Let S s > ( P, ( Pt, then ( max A h A h S 4 t+δ s + U. ( Pt Proof. Let (f ( P where Q t and f Pt ( Pt ( < s Q P t ( Pt ( f P is an integer. To see this, write s = Q ( P + R where 0 Q < Pt and 0 R < ( P ( Pt ( Pt +. If ( Pt Q ( P R > 0, then tae Q t = Q, f = Q ; if R = 0 and Q > 0, then tae Q t = Q, f = Q ; and if R = Q = 0, then tae Q t = Q and f = Pt. We arrange the elements of S in increasing order and let S S be the first (f ( P ( Pt + elements and

7 ALMOST SQUARES IN ARITHMETIC PROGRESSION 7 S consist of the remaining set. Then we see from Lemma 7 with t = t and Q t = f that max A h S A h δ (f P P P t + U = U. Now we apply Lemma 7 with U = U in S to derive max A h δ Q t P P P t + δ (f P P P t + U. A h S Hence to derive (, it is enough to prove Q t P P t + (f P P t 4 Q t(p (P t + f(p (P t }. By observing that it suffices to show that Q t (P (P t Q t P P t Q t P P t, f(p (P t fp P t fp P t, Q t + (Q t (f + P t + 6f P t P t 0 which is true since Q t and f Pt. Lemma 8. Let s i denote the i-th squarefree positive integer. Then ( and ( s i.6i for i 78 l s i (.6 l l! for l 86. Further let t i be i-th odd squarefree positive integer. Then (4 and (5 i= t i.4i for i 5 l t i (.4 l l! for l 00. i= Proof. The proof is similar to that of [, (6.9]. For ( and (4, we chec that s i.6i for 78 i 86 and t i.4i for 5 i, respectively. Further we observe that in a given set of 44 consecutive integers, there are at most 90 squarefree integers and at most 60 odd squarefree integers by deleting multiples of 4, 9, 5, 49, and, 9, 5, 49, respectively. Then we continue as in the proof of [, (6.9] to get ( and (4. Further we chec that ( holds at l = 86 and (5 holds at l = 00. Then we use ( and (4 to obtain ( and (5, respectively. Lemma 9. Let X > be a positive integer. Then (6 X i= ω(i η(xx log X

8 8 SHANTA LAISHRAM where (7 if X = X η := η(x = ω(i i= if < X < 48 X log X 0.75 if X 48. Proof. We chec that (6 holds for < X < 500. Thus we may assume X 500. Let s j be the largest squarefree integer X. Then by Lemma 8, we have.6j s j X so that j [ X.6]. We have ω(i = e i µ(e. Therefore X i= ω(i = X i= µ(e e i X e<x [ X e ] µ(e (X e<x µ(e e X We chec that there are 6990 squarefree integers upto 500. By using (, we have X 6990 ω(i X 6990 s i= i= i.6 i + [ X.6 ].6 i i= i= 6990 } implying (6. i= 4 X log X s i.6 i +.6 i=.658 log X Lemma 0. Let c > 0 be such that c ω(d >, µ and Then (8 C := µ ( + log X.6 }, C µ = A i ν(a i = µ, A i > ρδ }. cω(d µ(µ C µ c 8 η(cω(d ω(d ( ω(d θ (log c ω(d. [.6] X. s i i= Proof. Let i > i > i µ be such that A i = A i = = A iµ. These give rise to µ(µ pairs of (i, j, i > j with A i = A j. Therefore the total number of pairs (i, j with i > j and A i = A j is C. We now that there is a unique partition of d corresponding to each pair (i, j, i > j such C that A i = A j. Hence by Box Principle, there exists at least pairs of (i, j, i > j ω(d θ with A i = A j and a partition (d, d of d corresponding to these pairs. For every such pair (i, j, we write X i X j = d r ij, X i + X j = d s ij. Then gcd(x i X j, X i + X j = and 4 (X i Xj. Let r ij, s ij be such that r ij r ij, s ij s ij, gcd(r ij, s ij = and r ij s ij = 4 r ρ ijs ij. δ

9 Then ALMOST SQUARES IN ARITHMETIC PROGRESSION 9 r ij s ij = ρδ 4 r ijs ij = ρδ Xi X j 4 d = ρδ 4 i j A i < ρδ 4 A i < c ω(d c ω(d since A i > ρδ. There are at most ω(i possible pairs of (r c ij, s ij, and hence an ω(d equal number of possible pairs of (r ij, s ij. By Lemma 9, we estimate c ω(d i= i= ω(i η(c ω(d c ω(d (log c ω(d. Thus if we have C ω(d θ > η(cω(d c ω(d (log c ω(d, then there exist distinct pairs (i, j (g, h, i > j, g > h with A i = A j, A g = A h such that r ij = r gh, s ij = s gh giving X i X j = d r ij = X g X h and X i + X j = d s ij = X g + X h. Thus X i = X g, X j = X h implying (i, j = (g, h, a contradiction. Hence implying (8. C ω(d θ η(cω(d c ω(d (log c ω(d, The following Lemma is a refinement of [6, Lemma ]. Lemma. Let i > j, g > h, 0 i, j, g, h < be such that (9 and (0 a i = a j, a g = a h x i x j = d r, x i + x j = d r, x g x h = d s, x g + x h = d s where (d, d is a partition of d; r s (mod, r s (mod when d is even; and either r s (mod and a i a g (mod 4 or gcd(r, s when d is odd. Then we have either ( or (4 and (5 hold. a i = a g, r = s or a i = a g, r = s Proof. We follow the proof of [6, Lemma ]. Suppose that ( does not hold. Then ( a i r a g s 0, a i r a g s 0. We proceed as in [6, Lemma ] to conclude from d (a i x i a gx g that ( Thus we have d d = d 4 (ai r a gs d + (a ir a gs d + d(a ir r a g s s }. (a i r a gs d = a i(x i x j a g (x g x h 0

10 0 SHANTA LAISHRAM and Since and we have (4 Also (5 and (6 (a i r a g s d = a i (x i + x j a g (x g + x h 0. n a j x j < a ix i x j < a i x i n + ( d n a h x h < a gx g x h < a g x g n + ( d, ai x i x j a g x g x h < ( d. a i x i a gx g = i g d ( d a j x j a h x h = j h d ( d n min 4 a i(x i + x j, } 4 a g(x g + x h. Hence we derive from (4, (5 and (6 that (7 (a i r a g s d < 4( d (8 n (a i r a g s d < 4 ( d and further considering the cases a i r > a g s, a i r > a g s }, a i r > a g s, a i r < a g s }, a i r < a g s, a i r > a g s } and a i r < a g s, a i r < a g s }, we derive (9 G(i, g = a i r a gs d + a ir a gs d < 4( d. Let d = d d be odd, gcd(d, d =. We have either r, s are even and hence r, r, s, s are even, or a i a g (mod 4 and r s (mod and hence r s (mod. Then reading modulo d and d separately in (, we have (40 Therefore (4 and (4 From (9, we have so that d 4 (a ir a g s and d 4 (a ir a g s. 4dd = 4d d a ir a gs d 4dd = 4d d a i r a g s d. 4d(d + d G(i, g < 4( d ( d + d d = d d < (. 4

11 ALMOST SQUARES IN ARITHMETIC PROGRESSION This gives (4. Again from (4 and (8, we see that 4ndd < 4 ( d, i.e, n < 6 ( d. From (4 and (7, we have 4dd < 4( d, i.e, d < (. Thus (5 is also valid. Let d = d d be even with ord (d = and d odd. Then the x i s are odd and therefore both r and s is even. We see from ( that (4 (ai 4d r a gs d and 4d (a i r a gs d. Since r s (mod, r s (mod, gcd(d, d = and d odd, we derive that Therefore d (ai r a g s, 4d (ai r a g s. dd = d d a i r a g s d, 4dd = 4d d a i r a g s d. Now we argue as above to conclude (4 and (5. Let d = d d be even with ord (d, gcd(d, d =. Then we see from ( that (4 holds. Since gcd(d, d =, r s (mod and r s (mod, we derive that (ai d r a gs, d (a i r a gs. Therefore dd = d d a i r a g s d, dd = d d a i r a g s d. Now we argue as above to conclude (4 and (5. Lemma. For a prime p <, let γ p = ord p ( a i R Let m > by any real number. Then ( (44 p γp γ p.5π(m p m a i z, γ p = + ord p((!. <p m p p p ( z <p m where (z, z = ( 4, if d is odd and (z, z = (4, if d is even. p p Proof. The proof is the refinement of inequality [, (6.4]. Let p h < p h+ where h is a positive integer. Then [ ] [ ] [ ] (45 γ p = p p p h Let p d. Then we see that g p is the number of terms in n, n + d,, n + ( d} divisible by p to an odd power. After removing a term to which p appears to a maximal power, the number of terms in the remaining set divisible by p to an odd power is at most [ ] ([ ] [ ] ([ ] + p p p p ( ɛ ([ p h ] + ( ɛ

12 SHANTA LAISHRAM where ɛ = or 0 according as h is even or odd, respectively. We note that the above expression is always positive. Combining this with (45 and [ ] + =, p i p i p i p i we have [ ] [ ]} γ p γ p h + ɛ p p h +ɛ p + + p h + ɛ } + h + ɛ h +ɛ = p ( ( ph +ɛ +.5(h + ɛ. p Since p h p γ p γ p < p log and h <, we get log p +.5 log log p + p ɛ p +.5ɛ.5 p When d is even, we have γ γ = ord ( < + log log Now (44 follows immediately. +.5 log log p + p p. + by Lemma (v. Lemma. Suppose that n c ( or d 4c ( or both. Let ϱ ω(d θ be the greatest integer such that R ϱ = a i ν(a i = ϱ} φ. For κ 0, we have r = (i, j ai = a j, i > j} 4ϱ( ω(d if d is odd g(ϱ := ϱ( ω(d θ if d is even. Proof. We have = ϱ µr µ and R = µ= where r µ = R µ = a i ν(a i = µ}. Each R µ gives rise to µ(µ r µ pairs of i, j with i > j such that a i = a j. Then r = ϱ µ= µ(µ r µ = R + ϱ µ= ϱ µ= r µ (µ (µ r µ. Suppose that the assertion of the Lemma does not hold. Then g(ϱ > R + ϱ (µ (µ µ= r µ. We have ϱ (µ (µ (ϱ (ϱ g(ϱ r µ g(ϱ := g 0 (ϱ. µ= We see that g 0 (ϱ is an increasing function of ϱ. Since ϱ ω(d θ, we find that R < g 0 ( ω(d θ = ( ω(d θ (z ω(d θ + := g where z = 7 if d is odd and if d is even. Thus R > g. Since the a i s are squarefree, we have by Lemma 8 that a i z g 4 ( g! a i R

13 ALMOST SQUARES IN ARITHMETIC PROGRESSION where z 4 =.6 if d is odd and.4 if d is even. Also, we have p a i R a i R a i (! ( p< <p m p m p γp γ p where γ p, γ p and m are as in Lemma. This with (44 and Lemma (iv gives ( ( a i <!.5π(m z p p z p p p..705 Comparing the lower and upper bounds, we have ( z g 4! (46 ( g! >.5π(m+ z <p m p p p ( z z <p m <p m p p. By Lemma, we have z g 4! ( g + ( g! < zg 4 e g g e. g ( g e + Since κ 0, we find that g < z 5 for ω(d where z 5 = 7, 8 for d odd and d even, respectively. Thus ( g ( + z g 4! z4 ( g ( g! < e g if ω(d ( + g if ω(d. Hence we derive from (46 that ( (47 for ω(d and (48 ( g > log for ω(d. g > z 5 z z 4 z log z 5 z 5 z z 4 p p.705 <p m ( (z 4 (z 5 z 5 e + ( + log( g log( g + log z 4 ( (.5π(m log + log z <p m p p <p m log( g + log z 4 p p p (.5π(m log log ( z5 z 5 z log + log z 4 (z 5 log z 5 <p m p p p Let ω(d. Taing m = min(000, κ 0 in (47, we observe that the right hand side of (47 is an increasing function of and the inequality does not hold at = κ 0. Hence (47 is not valid for all κ 0. For instance, when ω(d = 4, d odd, we have κ 0 = 5700 and g = 855. With these values, we see that the right hand side of (47 exceeds 855 at = 500, a contradiction. Hence (47 is not valid for all 500.

14 4 SHANTA LAISHRAM Let ω(d. Taing m = 000 in (48, we derive that if d is odd log For d odd, we see that g >.8 log log κ 0 logκ 0 = > 7 4ω(d > g, if d is even. a contradiction. Similarly, we get a contradiction for d even ω(d4 ω(d ω(d log4 + log + log ω(d Lemma 4. Let κ 0 = κ 0 (ω(d. Assume that d < 4c (. Let T = 0 i < X i > } defined in section be such that C + + C if ω(d = T > C := C + + C + ω(d+ if ω(d =, 4, 5 C if ω(d 6 9 where C and C = 9, 4, 95, 806 for ω(d =,, 4, 5, respectively. Then maxa i δ if ω(d = (49 C 0 where C 0 = C 0 (ω(d = ω(d i T C [ ] if ω(d. 4 Proof. We see that for ω(d 6, 0 ( 4c ω(d ( ω(d > d ω(d where c is given by Proposition. Hence there exists a partition d = d d of d with Therefore (50 by Lemma 4. Let d < 0 ω(d with ω(d and ω(d ω(d. ν(a i ω(d ω(d for A i 0 ω(d T = i T Ai > δ ρ (5 c }, T ω(d = T T, where c = 6 if ω(d =, c = 4 if ω(d =, 4, 5 and c = if ω(d 6. Further let (5 S = A i i T }, S = A i i T } and S = s. Then considering residue classes modulo δ ρ, we derive that δ ρ maxa c ω(d i δ ρ(s + A i S

15 so that S = s + c ω(d ρ (5 that ( T + 0 ω(dω(d if ω(d 6 and Therefore ALMOST SQUARES IN ARITHMETIC PROGRESSION ( 6 0 T +. We see from Corollary, (50, (5 and c ω(d 6 ω(d 0 ω(d + ω(d + ω(d ( + 48 ω(d ω(d = + 8 if ω(d = 48 ( + ω(d ω(d = + ω(d+ if ω(d =, 4, 5. T > C T C 4 := C + C if ω(d =,, 4, 5 if ω(d 6. C + 4 Let C, C µ be as in Lemma 0 with c = 6 if ω(d =, c = 4 if ω(d =, 4, 5 and c = if ω(d 6. Then C 4 < T = S + µ (µ C µ. Now we apply Lemma 0 and use κ 0 η( ω(d (log ω(d ω(d ( ω(d θ for ω(d 6 to get Thus C 4 < S + C if ω(d 5 S + if ω(d 6. S > C. Let ω(d =. Then considering the A i s modulo δ, we see that maxa i δ [ ] + δ A i S C 48 4 δ C which gives (49. Now we tae ω(d. Since d < 4c (, we have r [ ω(d ] by r ( Lemma 6. By (4, we have C > (4c r ( Pj >. We now apply Corollary with s = [ C + ] and U = to get j= which yields (49. maxa i A i S 4 r+δ [ + ] ω(d C 4 [ ]+δ C 4. PROOF OF PROPOSITION We assume that either n c ( or d 4c (. Then ν(a i0 ω(d θ for 0 i 0 < by Lemma 5. Let ϱ be as defined in the statement of Lemma. Then ν(a i0 ϱ. By Lemma, there are at least zϱ( ω(d distinct pairs (i, j with i > j and a i = a j, where z = 4 if d is odd and if d is even. Since there can be at most ω(d θ

16 6 SHANTA LAISHRAM possible partitions of d, by Box principle, there exists a partition (d, d of d and at least zϱ pairs of (i, j with a i = a j, i > j corresponding to this partition. We write x i x j = d r (i, j and x i + x j = d r (i, j. Let d be odd. Suppose there are at least ϱ distinct pairs (i, j,, (i ϱ, j ϱ, with the corresponding r (i, j even. Then i,, i ϱ, j,, j ϱ } > ϱ. Hence we can find l, m ϱ with (i l, j l (i m, j m, a il = a jl, a im = a im and a il a im. Now the result follows by Lemma. Thus we may assume that there are at most ϱ pairs (i, j with r (i, j even. Then there are at least ϱ + distinct pairs (i, j with r (i, j odd. Since a i,, ( mod 4, we can find at least ϱ pairs with a i a g ( mod 4 for any two such pairs (i, j, (g, h. Then there exist two distinct pairs (i, j, (g, h with a i = a j, a g = a h and a i a g from these pairs. Also r (i, j r (g, h( mod. This gives (4 and (5 by Lemma which is a contradiction. Let d be even. We observe that 8 (x i x j and gcd(x i x j, x i + x j =. We claim that there are at least ϱ pairs with r (i, j r (g, h(mod and r (i, j r (g, h( mod for any two such distinct pairs (i, j and (g, h. If the claim is true, then there are two pairs (i, j (g, h with i > j, g > h, a i = a j, a g = a h and a i a g since ν(a i ϱ. This implies (4 and (5 by Lemma, contradicting our assumption. Let ord (d =. Then d is odd, implying that r (i, j is even. We can choose at least ϱ pairs whose r s are of the same parity. Thus the claim is true in this case. Let ord (d. Then we have either ord (d = implying that all the r s are odd, or ord (d = implying that all the r s are odd. Thus the claim follows. Finally, let ord (d =. Then d and d so that r and r are of the opposite parity for any pair and hence the claim holds. 5. PROOF OF PROPOSITION In this section, we assume that κ 0 = κ 0 (ω(d. In view of Proposition, we may assume that d < 4c (. We may also assume that X i is a prime for each i T in the proof of Proposition. Otherwise n + ( d ( + 4, which implies the assertion. Since d < 4c (, d has at least one prime divisor otherwise d > ω(d, giving a contradiction. Thus π d ( π(. Let n + ( d L for some L > 0. By Lemma and Lemma (i, we have (5 T > ( log( log L log ( +.5. log log We see from [5] that n(n + d (n + ( d is divisible by at least π( π d ( π( π(+ primes exceeding. Hence we have n+( d 4. Thus by taing L = 4 in (5, we get T > ( log( log( ( +.5. log log

17 ALMOST SQUARES IN ARITHMETIC PROGRESSION 7 The right hand side of the above inequality is an increasing function of and + + C if ω(d = T > + + C if ω(d = ( C + ω(d+ if ω(d = 4, if ω(d Now we see from Lemma 4 that (49 holds with 5 if ω(d = 6 if ω(d = C = 4 if ω(d = 4, 5 5 if ω(d This gives n+( d C 0 C. Hence (7 is valid for ω(d 4. Now we tae ω(d =,. Putting L = C 0 5 in (5, we derive that T > C ω(d+ if ω(d = C 4 + ω(d+ if ω(d =. We apply Lemma 4 again to get maxa i δ 5 so that n + ( d i T 6 δ 5 6, which implies (7. This completes the proof. REFERENCES [] B. Brindza, L. Hajdu and I. Z. Ruzsa, On the equation x(x + d (x + ( d = by, Glasgow. Math. J. (000, [] P. Dusart, Autour de la fonction qui compte le nombre de nombres premiers, Ph.D thesis, Université de Limoges, 998. [] P. Erdős and J.L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math. 9 (975, 9-0. [4] K. Győry, Power values of products of consecutive integers and binomial coefficients, Number Theory and its Applications (Kluwer Acad. Publ., (999, [5] S. Laishram and T. N. Shorey, Number of prime divisors in a product of terms of an arithmetic progression, Indag Math., 5(4 (004, in press. [6] S. Laishram, Topics in Diophantine Equations, M.Sc. Thesis, University of Mumbai, 004. [7] R. Marszale, On the product of consecutive elements of an arithmetic progression, Monatsh. für. Math. 00 (985, 5-. [8] A. Muhopadhyay and T. N. Shorey, Almost squares in arithmetic progression (II, Acta Arith., 0 (00, -4. [9] H. Robbins, A remar on Stirling s formula, Amer. Math. Monthly 6, (955, 6-9. [0] J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois Jour. Math 6(96, [] N. Saradha, On perfect powers in products with terms from arithmetic progressions, Acta Arith., 8 (997, [] N. Saradha and T. N. Shorey, Almost squares in arithmetic progression, Compositio Math. 8 (00, 7-. [] L. Schoenfeld, Sharper bounds for the Chebyshev functions θ(x and ψ(x, II, Math. Comp. 0 (976, 7-60.

18 8 SHANTA LAISHRAM [4] T. N. Shorey, Exponential diophantine equations involving products of consecutive integers and related equations, Number Theory edited by R. P. Bambah, V. C. Dumir and R. J. Hans-Gill, Hindustan Boo Agency (999, [5] T.N.Shorey, Powers in arithmetic progression, A Panorama in Number Theory or The View from Baer s Garden, ed. by G.Wüstholz, Cambridge University Press (00, 5-6. [6] T. N. Shorey and R. Tijdeman, Perfect powers in products of terms in an arithmetical progression, Compositio Math. 75 (990, SCHOOL OF MATHEMATICS, TATA INSTITUTE OF FUNDAMENTAL RESEARCH, HOMI BHABHA ROAD, MUMBAI , INDIA address: shanta@math.tifr.res.in