SQUARES IN BLOCKS FROM AN ARITHMETIC PROGRESSION AND GALOIS GROUP OF LAGUERRE POLYNOMIALS

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1 SQUARES IN BLOCKS FROM AN ARITHMETIC PROGRESSION AND GALOIS GROUP OF LAGUERRE POLYNOMIALS N SARADHA AND T N SHOREY Abstract We investigate when a product of t 2 terms taken from a set of k successive terms in arithmetic progression is a perfect square Further we study the Galois group of Laguerre polynomials which is related to studying when a product with terms from two blocks of arithmetic progression is a square 1 Introduction 11 Squares in Arithmetic Progression Let k t 2, m 0, d, y 1 be integers with gcd(m, d) = 1 and d 1 < < d t are integers from [0, k 1] For any integer n > 1, let P (n) denote the greatest prime factor of n Put P (1) = 1 We consider the equation (1) = (m + d 1 d) (m + d t d) = by 2 with P (b) k Note that when t = k and d i = i, the product on the left hand side is the product of k consecutive positive integers Starting with the famous result of Erdős and Selfridge [5] that a product of two or more positive consecutive integers is never a perfect power, this equation has been studied by several mathematicians in recent years See for instance, [11],[12],[20] and the references therein In this paper, our main focus is on the parameter t Note that k t is the number of terms which are omitted from the k terms m, m + d,, m + (k 1)d It is desirable to know how large k t can be taken so that the equation (1) has only finitely may solutions in the various parameters involved Let b = d = 1 In [4], Erdős observed that there exists an absolute constant C such that if equation (1) with m > k 2 and k t C k log k holds then k is bounded by an effectively computable number 2000 Mathematics Subject Classification Primary-11D61; Secondary- Key words and phrases Squares in Arithmetic Progression, Greatest prime factor, Irreducibility of Orthogonal Polynomials, Galois group 1

2 2 N SARADHA AND TN SHOREY For any n > e e, we shall denote by log 2 n = log log n and log 3 n = log log log n We shall always understand that n > e e whenever we use log 2 n or log 3 n so that the values of these functions are defined Erdős also stated in [4] that he was unable to prove the above result with k t (1 ɛ)k log 2 k log k Using Brun s sieve and a result of Sprindzuk on hyper-elliptic equations [21], Shorey [19] proved the above statement of Erdős Further, in [2], Balasubramanian and Shorey improved slightly the value of k t as k t k ( log2 k log k log 3 k log k θ log k where θ is any real number They also showed that if k t is taken larger than the above bound, then there are m, k, d 1,, d t for which the product (m+d 1 ) (m+d t ) is a square In this paper we prove analogous result for d > 1 We denote by c 0, c 1, effectively computable numbers Theorem 11 Suppose (1) holds with d > 1 and P (y) > k Then there exist c 0 and c 1 depending only on d such that if ( log2 k (2) k t k log k log 3 k log k c ) 0, log k then k c 1 In the next theorem, we relax the condition that d is fixed in the above theorem Theorem 12 Let d > 1 and ɛ > 0 be arbitrarily small Suppose (1) holds with P (y) > k (i) Assume that ( ) log2 k (3) k t (1 ɛ)k log k Then there exists c 2 depending only on ɛ such that d (log 2 k) 1 ɛ implies that k c 2 (ii) Assume that (4) k t C k log k where C is some absolute constant Then there exist c 3 and c 4 depending only on C such that d < (log k) c 3 implies that k c 4 )

3 SQUARES IN AN ARITHMETIC PROGRESSION 3 The assumption P (y) > k is equivalent to the condition that the left hand side of (1) is divisible by a prime > k Next we prove an analogue of Theorems 11 and 12 without the assumption P (y) > k Corollary 11 Suppose (1) holds Assume that any of the following conditions hold (i) (2) holds, d fixed and m e3 c 0 k log k log 2 k (ii) (3) holds, d (log 2 k) 1 ɛ and m k(log k) 1 ɛ (iii) (4) holds, d < (log k) c 3 and m e C k Then k c 5 where c 5 depends on c 0, ɛ and C whenever (i), (ii) and (iii) hold, respectively On the other hand, we apply a combinatorial result of Erdős and Turk [6] to prove the following Theorem 13 For any integer m 1, d 1 there exist integers d 1,, d t in [0, k 1] with k t k ( ( ) ) m + kd log + 3 log k k and such that is a square Remark 1 Let (m + d 1 d) (m + d t d) m + kd < e3 c 0 k log k log 2 k Then there exist d 1,, d t in [0, k 1] satisfying (2) such that (m + d 1 d) (m + d t d) is a square This shows that Corollary 11 is best possible with respect to assumptions on m and t 12 Galois group of some orthogonal polynomials For a positive real number α, let L (α) m (x) denote the Laguerre Polynomial of degree m given by m L (α) (m + α)(m 1 + α) (j α) m (x) = ( x) j (m j)!j! We set j=0 (5) v m m!l (u/v) m = m j=0 ( x ) := g(x) = g(x, u, v) v ( m j ) (vm + u) (v(j + 1) + u)x j

4 4 N SARADHA AND TN SHOREY where u and v > 0 are integers with gcd(u, v) = 1 It is well known that if the Galois group of an irreducible polynomial f of degree m contains A m then it is either A m or S m according as whether the discriminant of f is a square or not In the case of g(x), Schur [18] showed that the discriminant is given by (6) D (u,v) m = m ( u ) j 1 j j v + j j=2 We write where D (u,v) m = by 2, Y Q (7) b = { 35 m(u+2v) (u+(m 1)v) v δ 1 if m 1, 3(mod 4) 35 (m 1)(u+2v) (u+mv) v δ 2 if m 0, 2(mod 4) with δ 1 = { 0 if m 1(mod 4) 1 if m 3(mod 4) and δ 2 = { 0 if m 0(mod 4) 1 if m 2(mod 4) Thus for any irreducible g(x) it is enough to find out when b, whose numerator is a product of two blocks of integers in arithmetic progression, is a square Let (u, v) = (1, 1) Schur [18] proved the irreducibility of g(x) Also he observed that b is a square if m is odd or if m is even and m + 1 is a square Thus in these cases the Galois group of g(x) is A m When m is even and m + 1 is not a square, then the Galois group is S m In the general case L (u/v) m,filaseta and Lam [7] showed that for all but finitely many positive integers m, the polynomial g(x) is irreducible over rationals unless u is a negative integer In [9], Hajir v provided a complement to the result of Filaseta and Lam by computing the Galois group of g(x) when m is large See also [10] In [1], Akhtari and Saradha gave explicit values m 0 depending on u and v such that for all m m 0, g(x) is irreducible Further they showed that there are at most n 0 number of polynomials g(x) which may not be irreducible thus making the result of [7] explicit for small values of v For results on more general polynomials see [23] In this paper, we make the result of Hajir explicit for certain values of (u, v) by using the expression for b as given in (7) The case v = 1 and 1 u 10 was treated by Banerjee, Filaseta, Finch and Leidy [3] We show

5 SQUARES IN AN ARITHMETIC PROGRESSION 5 Theorem 14 Let 182 if (u, v) {( 1, 2), (1, 2)} m 876 if (u, v) {( 1, 3), ( 2, 3), (1, 3), (2, 3)} 1325 if (u, v) {( 1, 4), ( 3, 4), (1, 4), (3, 4)} Then the Galois group of g(x) is S m For more values of (u, v), see Proposition 31 For all the values of (u, v) considered in Theorem 14, the irreducibility of g(x) is known When (u, v) {( 1, 2), (1, 2)} irreducibility of g(x) can be derived from the result of Schur [17] We refer to Finch & Saradha [8] for a more recent proof in the case (u, v) = (1, 2) Laishram and Shorey, in [13],[14],[15] showed that g(x) is irreducible for the rest of the values of (u, v) considered in Theorem 14 2 Theorems 11,12 and Lemmas for the proof of Theorems 11 and 12 We assume that a 0 exceeds a sufficiently large number In Theorem 11, we take a 0 depending on d while in Theorem 12, it is taken to be depending on ɛ or C according as (3) or (4) holds Further we denote by a 1, a 2,, effectively computable positive numbers depending on a 0 We assume in section 21 that (1) holds with P (y) > k and that k > a 0 We write m + d i d = A i X 2 i with P (A i ) k and gcd( p k p, X i ) = 1 for 1 i t Since P (y) > k, m + (k 1)d (k + 1) 2 Shorey and Tijdeman [22] (see Lemma 1) improved the above bound when t = k In this paper we deal with the case of t < k Since their result is applicable even in this general set up and is of independent interest, we state it as a lemma and give necessary steps of the proof For more details we refer to [22] Lemma 21 Let 0 < ɛ 0 < 1/2, 0 < δ < (1/2 ɛ 0 ) with δ fixed and t ( δ + ɛ 0)k Then for k a 0, there exists a 1 = a 1 (ɛ 0 ) such that Proof Let T 0 = {d 1,, d t } and Thus m + (k 1)d a 1 k 3 (log k) 2 T 0 = {d i T 0 : d i δk/2} ( ) 1 + δ T 0 + ɛ 0 k 2

6 6 N SARADHA AND TN SHOREY Further, for any d i T 0, (8) m + d i d m + kdδ/2 δ 2 (m + (k 1)d) δ 2 (k + 1)2 δk 2 /2 Write T 0 = T T 1 where T = {d i T 0 : X i = 1}; T 1 = {d i T 0 : X i 1} We use an argument of Erdős [4] to estimate T We shall refer to it as a combinatorial argument of Erdős Let U = {m + d i d, d i T } Then m + d i d = A i for any m + d i d U Since P (A i ) k, we have P (m + d i d) k if m + d i d U For every prime p k, choose an m + d i d U in which p appears to the maximum power By removing these elements from U, we get a subset U 1 U with U 1 T π(k) and m+d i d U 1 (m + d i d) k! By (8), we have, (δk 2 /2) T π(k) k k Thus T k log k log(δk 2 /2) + π(k) (1/2 + ɛ 0)k since δ is fixed and k is sufficiently large Hence T 1 = T 0 T δk/2 This is equation (312) of [22] with δ/2 in place of 1/8 From here, we follow the proof of [22] to get the assertion of the lemma By (2), (3) and (4) we get t 3k/4 Hence applying Lemma 21 with δ = 1/4 ɛ 0, a 1 = a 1 (ɛ 0 ), we have (9) m a 2 k 3 (log k) 2 if d < a 1 k 2 (log k) 2 /2 with a 2 = a 1 2 which we may assume Suppose there exist A i and A j such that A i = A j We may take X i > X j Then kd (d i d j )d = (m + d i d) (m + d j d) = A i X 2 i A j X 2 j = A i (X i X j )(X i + X j ) 2A i X i 2(A i Xi 2 ) 1/2 2m 1/2 2a 1/2 2 k 3/2 log k by (9) This implies that d a 3 k 1/2 log k with a 3 = 2a 1/2 2 Since we may assume that d < a 3 k 1/2 log k, we conclude that (10) A i A j for i j

7 SQUARES IN AN ARITHMETIC PROGRESSION 7 Hence while proving Theorems 11 and 12, we assume that all A i s are distinct Let S = {A 1,, A t } Corresponding to each prime p k, we remove an A i such that the prime appears to the maximum power in A i Let S 1 S be the remaining set of A i s Then S 1 t π(k) Further we have (11) A i k k A i S 1 The following lemma gives a subset of S 1 in which A i s are bounded in terms of k and the proof depends on a combinatorial argument of Erdős Lemma 22 Let 0 < η 1/2 Suppose h is a positive number such that h η log k/8 and S 1 k hk Then there exists a subset log k S 2 S 1 with at least ηk/2 elements satisfying A i 4e (1+η)h k for A i S 2 This lemma is by Shorey and Tijdeman (see Lemma 6 of [22]) As an application of this lemma we get Corollary 21 There exists a subset S 2 of S 1 with the following property (i) Suppose (2) holds with c 0 = a 0 Then S 2 a 0k and 2 log 2 k A i 4e 2 k log k log 2 k for A i S 2 (ii) Suppose (3) holds Then S 2 a 0k 2(log 2 k) ɛ and A i 4k(log k) 1 ɛ/2 for A i S 2 (iii) Suppose (4) holds Then S 2 k 4 and Further by (9), we have A i 4e 3C 2 +3 k for A i S 2 X i m/a i > a 4 k log k for any A i S 2 where a 4 = a 1/2 2 /(2e) in all the cases (i) (iii) Proof We give below the choice of the parameters h and η in Lemma 22 to get the assertion of the corollary (i) Take h = log 2 k log 3 k (a 0 2) and η = a 0 log 2 k (ii) Take h = (1 ɛ) log 2 k + 2 and η = a 0 (log 2 k) ɛ (iii)take h = C + 2 and η = 1 2

8 8 N SARADHA AND TN SHOREY Shorey[19] used the following three lemmas for proving a result similar to Theorem 11 in the case d = 1 The first lemma is a result of Sprindzuk [21] on hyper-elliptic equations Lemma 23 Suppose that f Z[X] has at least three simple roots Denote by H( 2) an upper bound for the height of f For integers a 0, x and y, the equation implies that f(x) = ay 2 log max( x, y ) ( a H) a 5 with a 5 depending only on the degree of f The next two lemmas are by Erdős [4] and they are based on combinatorial and sieve arguments Lemma 24 Let 0 < X < Y and b 1,, b s denote all the integers in the interval [X, Y ] such that all their proper divisors are less than X Then every integer in [X, Y ] is divisible by some b i Proof Observe that all primes in [X, Y ] are among the b i s Suppose an integer n [X, Y ] is not divisible by any of the b i s Then there is a proper divisor d 1 n and d 1 > X Since d 1 is also not divisible any b i, there is a proper divisor d 2 d 1 and d 2 > X Proceeding inductively, there exists a divisor d m of n such that d m > X and all the proper divisors of d m are X Thus d m = b i for some i with 1 i s Lemma 25 The number of integers n i x such that n i n j are distinct is bounded by 2x/ log x whenever x exceeds a sufficiently large absolute constant Lemma 26 Let d + 1/(2k) if (2) holds f(k) = (log 2 k) 1 ɛ + 1/(2k) if (3) holds 4e 3C /2 if (4) holds and If S 3 = {A i S 2 with A i kf(k)} (12) S 3 S 2 /2 then there exists a 6 such that log d a 6 log 2 k

9 SQUARES IN AN ARITHMETIC PROGRESSION 9 Proof The proof follows a combinatorial argument of Erdős and uses Brun s sieve Let e 1, e 2, denote absolute constants and we shall choose e 1 suitably at the end of the proof of this lemma Let b 1,, b s denote all integers between and kf(k) such that every proper k (log k) 2e 1 k divisor of b i is If b (log k) 2e 1 i > exceeds (log k) e 1 Hence by Brun s sieve, s k (log k) e 1 k (log k) e 1, then every prime divisor of b i + e 2kf(k) log 2 k e 3kf(k) log 2 k k Let S 4 = {A i S 3, A i } Then every element of S (log k) 2e 1 4 is divisible by at least one b i by Lemma 24 Further (13) S 4 S 3 k (log k) 2e 1 If every b i appears in at most two elements of S 4, then (14) S 4 2s 2e 3kf(k) log 2 k Thus by (12), (13) and (14) we get 2k S 2 2 S 3 2 S 4 + (log k) 2e 1 e 4kf(k) log 2 k This contradicts with the lower bound for S 2 in Corollary 21, since a 0 is sufficiently large Hence we may assume that there exists at least 3 elements,say A i1, A i2, A i3 in S 4 which are all divisible by a b ν Let B ij = b 1 ν A ij, j = 1, 2, 3, Y j = X ij, B j = B ij, d j = d ij for j = 1, 2, 3 R 2,1 = b 1 ν (d 2d d 1d), R 3,1 = b 1 ν (d 3d d 1d) We form the equation B 2B 3(Y 2 Y 3 ) 2 = (B 1Y R 2,1 )(B 1Y R 3,1 ) where B 1, B 2, B 3, R 2,1, R 3,1 have absolute values not exceeding d(log k) 2e 1+ɛ This is true since k is sufficiently large Finally we apply Lemma 25 to conclude that Thus we get log k log(max 1 i 3 ( Y i )) (d(log k) 3e 1 ) 6a 5 log d ( ) 1 3e 1 log 6a 2 k 5

10 10 N SARADHA AND TN SHOREY ( ) 1 Choosing e 1 = 36a 5 ɛ, we obtain where (15) a 6 = log d a 6 log 2 k ( ) 1 ɛ 12a 5 22 Proofs of Theorems 11 and 12 We assume that k a 0 where a 0 is taken sufficiently large as in section 21 Suppose (4) is valid By Corollary 21 and Lemma 26, we have S 3 = S 2 and hence d (log k) a 6 The assertion of Theorem 12(ii) follows by taking c 4 = a 6 Note that by (15) and ɛ arbitrarily small we have a 6 < 1 ɛ Suppose (2) or (3) is valid The by assumption on d, we have (16) log d < a 6 log 2 k Further by (16) and Lemma 26, S 3 < S 2 2 Let S 3 be the complement of S 3 in S 2 Then A i > kf(k) for A i S 3 and by Corollary (21), in either case (17) S 3 S 2 2 a 0k 4(log 2 k) Let A i, A j, A µ, A ν be elements of S 3 with A i A j = A µ A ν and X µ X ν > X i X j which we may assume We consider Suppose T 0 Then T = (m + d i d)(m + d j d) (m + d µ d)(m + d ν d) T = m(d i + d j d µ d ν )d + (d i d j d µ d ν )d 2 2kdm + k 2 d 2 On the other hand, T = A i A j (X i X j ) 2 A µ A ν (X µ X ν ) 2 2A i A j X i X j 2(A i A j ) 1/2 (A i X 2 i ) 1/2 (A j X 2 j ) 1/2 2kf(k)m By (9) and the assumptions on d we have k 2 d 2 < m Therefore 2kdm > m(2kf(k) 1) implying that d > f(k) 1/(2k) which contradicts our assumption on d

11 SQUARES IN AN ARITHMETIC PROGRESSION 11 Thus S 3 has the property that products of any two elements of S 3 are distinct Since S 3 S 2, by Corollary 21, all these elements do not exceed 4e2 k log k if (2) holds and 4k(log k) 1 ɛ/2 if (3) holds In either log 2 k case, we may apply Lemma 25 with x = 4e2 k to get log 2 k S 3 8e2 k log 2 k This contradicts (17) since k a 0 is sufficiently large 23 Proof of Corollary 11 We may suppose that k > max{c 1, c 2, c 5 } Then P (y) < k by Theorems 11 and 12 Corresponding to every prime p k, we delete a term in the product in which the prime appears to the maximum power Then the product of the remaining terms does not exceed k! and there are at least t π(k) terms in the product Thus m t π(k) k k Suppose condition (i) in Corollary 11 holds Then ( log m < log k 1 + log 2 k log k log 3 k log k c 0 log k + 3 ) log k which contradicts the assumption on m The cases (ii) and (iii) are similar 24 Proof of Theorem 13 The proof depends on the combinatorial lemma due to Erdős and Turk [6]: Lemma 27 Let G be a set of positive integers and let ω(g) denote the number of prime divisors of g G g There exists a subset G G with G G ω(g) such that the product of all elements of G is a square Let T 1 = {m, m + d,, m + (k 1)d} Suppose the greatest prime factor of every element of T 1 does not exceed k Then ω(t 1 ) π(k) By Lemma 27 with G = T 1 we get a subset T 2 T 1 satisfying such that T 2 k π(k) m+d i d T 2 (m + d i d) is a square Next we assume that some element of T 1 is divisible by a prime > k Let T 3 be the set of elements of T 1 with P (m+d i d) k and T 4 be the complement of T 3 in T 1 By our assumption T 4 Since

12 12 N SARADHA AND TN SHOREY every element of T 4 is divisible by a prime > k and product of all the elements of T 1 is divisible by k! we get which, by k! k k e k, implies Thus T 4 k log k k T 4 (m + kd) k /k! ( log T 3 > k k ( log log k ( ) m + kd k ( ) m + kd k ) + 1 ) + 1 Taking G = T 3, ω(g) = ω(t 3 ) π(k) < k, we get a subset T 5 of T 3 with T 5 k k ( ( ) ) m + kd log + 3 log k k such that (m + d i d) m+d i d T 5 is a square 3 Theorem Lemmas for the proof of Theorem 14 We begin with a lemma which depends on Theorems 1 and 2 of Ramaré and Rumley [16] on explicit results for Prime Number Theorem for Arithmetic Progression which are known for small values of the common difference Lemma 31 Let u, v be integers with v > 1 and gcd(u, v) = 1 Let ε be given by Table 1 of [16] (Here we identify v with k and u with l in Theorems 1 and 2 of [16]) (i) Suppose x Then for any y > 2εx, there is a prime p in 1 ε the interval (x, x + y] with p u(mod v) (ii) Suppose x 0 x < with x 0 > (µϕ(v)) 2 Then there is a prime p in the interval (x, x + y] with p u(mod v) provided ( 2µϕ(v)x y > max x0 µϕ(v), (ε ) x 0 + µϕ(v))x (1 ε) := Y (x) x 0 where µ is the value given in Table 2 of [16] under the column for θ Proof Let θ(x, v, u) = p x p u(mod v) log p

13 SQUARES IN AN ARITHMETIC PROGRESSION 13 There exists a prime u(mod v) in (x, x + y] provided θ(x + y, v, u) θ(x, v, u) > 0 (i) Suppose x By Theorem 1 and Table 1 of [16], we have θ(x + y, v, u) θ(x, v, u) x + y (1 ε) x (1 + ε) ϕ(v) ϕ(v) = 1 (y (2x + y)ε) ϕ(v) which is positive if y > 2εx 1 ε (ii) Suppose x 0 x < First let x + y < Then by Theorem 2 and Table 2 of [16] we get θ(x + y, v, u) θ(x, v, u) x + y µ(x + y) x ϕ(v) x0 ϕ(v) µx x0 = y ϕ(v) µ (2x + y) x0 The right hand side of the above inequality is > 0 provided y > 2µϕ(v)x x0 µϕ(v) Next, let x + y Then θ(x + y, v, u) θ(x, v, u) x + y (1 ε) x ϕ(v) ϕ(v) µx x0 = y (1 ε) xε ϕ(v) ϕ(v) µx > 0 x0 provided y > x(ε x 0 + µϕ(v)) (1 ε) x 0 Now we turn to some lemmas required for computing Galois groups of irreducible Laguerre polynomials Hajir [9] gave a criterion for an irreducible polynomial to have a large Galois group For this purpose he used Newton Polygons We refer to [9] for more details Here we need Lemma 31 in [9] Lemma 32 Let f(x) = ( m m j=0 j polynomial of degree m Suppose there exists a prime p satisfying (i) m/2 < p < m 2 (ii) ord p (c j ) 0 for 0 j m ) c j X j Q[X] be an irreducible

14 14 N SARADHA AND TN SHOREY (iii) ord p (c j ) = 1 for 0 j m p (iv) ord p (c p ) = 0 Then the Galois group of f over Q contains A m We apply Lemma 32 to the polynomial g(x) given by (5) to get the following lemma Lemma 33 Let g(x), given by (5) be an irreducible Laguerre polynomial of degree m Assume that u < v Suppose there exists a term vl + u for some l with 1 l m such that vl + u is a prime p and mv + v + u (18) p m 3, v + 1 then the Galois group of g(x) contains A m Proof We apply Lemma 32 with c j = (mv + u) ((j + 1)v + u) Since mv+v+u v+1 > m 2 for v > 0, condition (i) holds As c j s are integers (ii) also holds Consider for any integer r > 0, rp = rvl + ru ru u(mod v) if and only if v (r 1) since gcd(u, v) = 1 Thus the least multiple of p exceeding p occurring in the arithmetic progression u+v, u+2v,, u+ mv, is (v + 1)p By our assumption (18) on p (19) (v + 1)p > mv + u Hence in c j s the prime p occurs at most once Thus ord p (c j ) 1 Further (20) ord p (c j ) = 1 for 0 j l 1 and ord p (c j ) = 0 for l j m In particular, ord p c p = 0 giving (iv) Also from (19) we get mv + u < vp + vl + u implying l > m p This together with (20), shows that (iii) is satisfied Now the lemma follows As a consequence of Lemmas 31 and 33 we get Corollary 31 Let g(x) given by (5) be irreducible and u < v Put M = mv+v+u Suppose any of the following two cases hold v+1 (i) Let M and { v < (1 ε)/2ε (21) m > (4 2ε)v+(1+ε)u+3(1 ε) := m 1 ε 2εv 1

15 SQUARES IN AN ARITHMETIC PROGRESSION 15 (ii) Let x 0 < M < with Also let x0 > max{µϕ(v)(2v + 1), (22) m > m 2 where m 2 is the maximum of µvϕ(v) 1 ε vε } µϕ(v)(u 2v 3) + x 0 (4v + u + 3) x0 µϕ(v)(2v + 1) and µϕ(v)(v + u) + x 0 (4v + u + 3 3ɛv 3ε) (1 ε vε) x 0 µvϕ(v) Then the Galois group of g(x) contains A m Here ε and µ are as in Lemma 31 Proof In Lemma 31, we take x = mv+v+u and x + y = m 3 Suppose v+1 x > and by (i) of Lemma 31, there exists a prime u(mod v) in (x, x + y] if y = m 3 mv + v + u > 2ε mv + v + u v ε v + 1 On simplifying we get the condition on m as in the statement of the corollary Now we apply Lemma 33 to conclude that Galois group of g(x) contains A m Next we suppose M < Then by (ii) of Lemma 31 we require m 3 mv + v + u ( ) mv + v + u > Y v + 1 v + 1 On simplifying we get the assertion of the corollary In the next lemma we determine when the Galois group is exactly A m or S m Lemma 34 Let g(x) given by (5) be irreducible with u < v and m 9 If M 10 10, assume that 1 + ε (23) 1 ε < v < 1 ε { } v(3 + ε) and m > max m 1, 2ε v(1 ε) 1 ε If M < 10 10, assume that there exists x 0 < M satisfying x0 > max{µϕ(v)(2v + 1), µvϕ(v) } 1 ε vε (24) v( x 0 µϕ(v)) > x 0 + µϕ(v) v(1 ε) x 0 > x 0 + µϕ(v)

16 16 N SARADHA AND TN SHOREY and (25) m > max { m 2, v(3 x 0 + µϕ(v)) + µuϕ(v) v( x 0 µϕ(v)) x 0 µϕ(v), v(3 x0 + 2µϕ(v)) v(1 ε) x 0 x 0 µϕ(v) Then the Galois group of g(x) is S m Proof By Corollary 31, under the given conditions, the Galois group of g(x) contains A m Thus we need only to determine when b given by (7) is not a square Case 1 Let u be odd Let h be the least integer such that Then (26) Let u + (2h 2)v m < u + 2hv m u < 2h m u + 2 v v { m m 1 if m 1, 3(mod 4) = m if m 0, 2(mod 4) From the expression for b in (7), we see that the terms u + 2hv, u + (2h + 2)v,, u + (m )v are not repeated in b For m 9, we observe that 2h m u + 2 m v We apply Lemma 31 with x = u + 2hv, x + y = u + m v to conclude that there exists a prime p = u + 2rv with h r m /2 provided (i) y = (m 2h)v > 2ε (u + 2hv) if u + 2hv ε or (ii) y = (m 2h)v > Y (u + 2hv) if x 0 u + 2hv < The first inequality (i) simplifies to m v > 2hv + 2ε (u + 2hv) 1 ε ( ) 1 + ε 2hv + 2εu 1 ε 1 ε From the upper bound for 2h in (26), we see that the above inequality is satisfied if ( ) 1 + ε m v > (m u + 2v) + 2εu 1 ε 1 ε }

17 SQUARES IN AN ARITHMETIC PROGRESSION 17 On simplifying, we get (23) Arguing similarly the second inequality (ii) simplifies to (24) and (25) Further any multiple of p say sp = su + 2rsv su u(mod 2v) if and only if 2v (s 1) since u is odd and gcd(u, v) = 1 Thus the least multiple of p which is u(mod 2v) is (1 + 2v)p It can be easily seen that (2v + 1)p > (2v + 1)m > mv + u Hence p divides b to the first power and hence b is not a square Case 2 Let u be even, say, 2u 0 Then (7) simplifies to { 2 (m 1)/2 35 m(u 0 +v)(u 0 +2v) (u 0 +(m 1)v/2) if m 1, 3(mod 4) b = v δ 1 2 m/2 35 (m 1)(u 0 +v)(u 0 +2v) (u 0 +mv/2) if m 0, 2(mod 4) v δ 2 Let h be the least integer such that u 0 + (h 1)v m < u 0 + hv As in Case 1, we find a prime p = u 0 + h 1 v with h h 1 m /2 which occurs only to the first power in b under the conditions (23), (24) and (25) The result follows We summarize our discussions from Lemma as follows Proposition 31 Let v take the value of k given by Tables 1 and 2 of [16] Also let ε and µ be the values corresponding to each k under the columns and θ in Tables 1 and 2, respectively Suppose the following conditions hold (i) u is any integer with gcd(u, v) = 1 and u < v (ii) g(x) given by (5) is irreducible (iii) Either (23) holds or (24) and (25) hold Then the Galois group of g(x) is S m Proof Of Theorem 14 Let m It suffices to show that (23) holds Let (u, v) {( 1, 2), (1, 2)} Then ε = and (23) holds For other values of (u, v) we have ε = and we find that (23) holds Let m < We show that (24) and (25) hold to complete our proof Let (u, v) {( 1, 2), (1, 2)} Then µ = Take x 0 = 119 and 122, respectively so that all conditions in (24) and (25) are satisfied for m 168 and 175 Since M = mv+v+u > x v+1 0, we find that m 178 and 182, respectively Thus the Galois group of g(x) is S m for m 182 For other values of (u, v) we give the following data:

18 18 N SARADHA AND TN SHOREY (u, v) µ x 0 m ( 1, 3) ( 2, 3) (1, 3) (2, 3) ( 1, 4) ( 3, 4) (1, 4) (3, 4) Table References [1] SAkhtari and NSaradha, Irreducibility of some orthogonal polynomials, Math Indag 21 (2011), [2] R Balasubramanian and TN Shorey, Squares in products from a block of consecutive integers, Acta Arith 65 (1994), [3] P Banerjee, M filaseta, C E Finch and J R Leidy, On classifying Laguerre polynomials which have Galois group the alternating group, to appear in J de Théorie des Nombres de Bordeaux [4] P Erdős, On the product of consecutive integers III, Indag Math 17 (1955), [5] P Erdős and JL Selfridge, The product of consecutive integers is never a power, Illinois J Math 19 (1975), [6] P Erdős and J Turk, Products of integers in short intervals, Acta Arith 44 (1984), [7] M Filaseta and TY Lam, On the irreducibility of the generalized Laguerre polynomials, Acta Arith 105 (2), (2002), [8] C Finch and N Saradha, On the irreducibility of certain polynomials with coeeficients as products of terms in an arithmetic progression, Acta Arith, 143 (2010), [9] F Hajir, On the Galois group of generalized Laguerre polynomials, J Théor Nombres Bordeaux 17 (2005), no2, [10] F Hajir, Algebraic properties of a family of generalized Laguerre polynomials, CanadJ Math 61 (2009), no3, [11] S Laishram, Topics in Diophantine Equations, MSc thesis, 2004, Tata Institute of Fundamental Research [12] S Laishram, Some topics in Number Theory, PhD thesis, 2007, Tata Institute of Fundamental Research [13] S Laishram and TN Shorey, Irreducibility of generalized Hermite- Laguerre polynomials II, Math Indag, NS, 20(3) (2009), [14] S Laishram and TN Shorey, Irreducibility of generalized Hermite- Laguerre polynomials, Approx Commentarii Mathematici, 2012

19 SQUARES IN AN ARITHMETIC PROGRESSION 19 [15] S Laishram and TN Shorey, Irreducibility of generalized Hermite- Laguerre polynomials III, in preparation [16] ORamaré and R Rumley, Primes in Arithmetic progression, Math Comp 65 (1996), [17] I Schur, Einige Sätze über Primzahlen mit Anwendeungen auf Irreduzibilit ätsfragen, II, Sitzungsber Preuss Akad Wiss Berlin Phys- Math Kl, 14 (1929), [18] I Schur, Affektlose Gleichungen in der Theorie der Laguerreschen und Hermiteshen Polynome, J Reine Angew Math 165 (1931), [19] T N Shorey, Perfect powers in values of certain polynomials at integer points, Math proc Cambridge Philos Soc 99 (1986), [20] TN Shorey, Diophantine Approximations, Diophantine Equations, Transcendence and Applications, Indian J Pure ApplMath, 37(1): (2006), 9-39 [21] V G Sprindzuk, Hyperelliptic diophantine equations and class numbers of ideals [Russian], Acta Arith 30 (1976), [22] TN Shorey and R Tijdeman, Perfect powers in products of terms in an arithmetical progression, Composito Math 75 (1990), [23] TN Shorey and R Tijdeman, Generalizations of some irreducibility results by Schur,, Acta Arith 145 (2010), no 4, Addresses of the authors : School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha Road, Mumbai , INDIA Department Of Mathematics, Indian Institute of Technology, Powai, Mumbai , INDIA s: saradha@mathtifrresin shorey@mathiitbacin