Cubes in products of terms in arithmetic progression

Transcription

1 Publ. Math. Debreen Proof-sheets for paper Ref. no.: 4453 (2009), 1 18 Cubes in produts of terms in arithmeti progression By LAJOS HAJDU (Debreen), SZABOLCS TENGELY (Debreen) and ROBERT TIJDEMAN (Leiden) Abstrat. Euler proved that the produt of four positive integers in arithmeti progression is not a square. Győry, using a result of Darmon and Merel, showed that the produt of three oprime positive integers in arithmeti progression annot be an l-th power for l 3. There is an extensive literature on longer arithmeti progressions suh that the produt of the terms is an (almost) power. In this paper we extend the range of k s suh that the produt of k oprime integers in arithmeti progression annot be a ube when 2 < k < 39. We prove a similar result for almost ubes. 1. Introdution In this paper we onsider the problem of almost ubes in arithmeti progressions. This problem is losely related to the Diophantine equation n(n + d)...(n + (k 1)d) = by l (1) in positive integers n, d, k, b, y, l with l 2, k 3, gd(n, d) = 1, P(b) k, where for u Z with u > 1, P(u) denotes the greatest prime fator of u, and P(±1) = 1. This equation has a long history, with an extensive literature. We refer to the researh and survey papers [3], [10], [11], [14], [16], [18], [19], [20], [23], [25], [26], [28], [29], [31], [32], [33], [34], [35], [36], [37], [38], [40], [41], the referenes given there, and the other papers mentioned in the introdution. Mathematis Subjet Classifiation: 11D41, 11D25, 11B25. Key words and phrases: perfet ubes, arithmeti progressions. L. Hajdu is supported in part by the Hungarian Aademy of Sienes and by the OTKA grants T48791 and K Sz. Tengely is supported in part by the Hungarian Aademy of Sienes, by the Magyary Zoltán Higher Eduational Publi Foundation, and by the OTKA grant T48791.

2 2 L. Hajdu, Sz. Tengely and R. Tijdeman In this paper we onentrate on results where all solutions of (1) have been determined, under some assumptions for the unknowns. We start with results onerning squares, so in this paragraph we assume that l = 2. Already Euler proved that in this ase equation (1) has no solutions with k = 4 and b = 1 (see [7] pp. 440 and 635). Obláth [21] extended this result to the ase k = 5. Erdős [8] and Rigge [22] independently proved that equation (1) has no solutions with b = d = 1. Saradha and Shorey [27] proved that (1) has no solutions with b = 1, k 4, provided that d is a power of a prime number. Later, Laishram and Shorey [19] extended this result to the ase where either d 10 10, or d has at most six prime divisors. Finally, most importantly from the viewpoint of the present paper, Hirata Kohno, Laishram, Shorey and Tijdeman [17] ompletely solved (1) with 3 k < 110 for b = 1. Combining their result with those of Tengely [39] all solutions of (1) with 3 k 100, P(b) < k are determined. Now assume for this paragraph that l 3. Erdős and Selfridge [9] proved the elebrated result that equation (1) has no solutions if b = d = 1. In the general ase P(b) k but still with d = 1, Saradha [24] for k 4 and Győry [12], using a result of Darmon and Merel [6], for k = 2, 3 proved that (1) has no solutions with P(y) > k. For general d, Győry [13] showed that equation (1) has no solutions with k = 3, provided that P(b) 2. Later, this result has been extended to the ase k < 12 under ertain assumptions on P(b), see Győry, Hajdu, Saradha [15] for k < 6 and Bennett, Bruin, Győry, Hajdu [1] for k < 12. In this paper we onsider the problem for ubes, that is equation (1) with l = 3. We solve equation (1) nearly up to k = 40. In the proofs of our results we ombine the approah of [17] with results of Selmer [30] and some new ideas. 2. Notation and results As we are interested in ubes in arithmeti progressions, we take l = 3 in (1). That is, we onsider the Diophantine equation n(n + d)...(n + (k 1)d) = by 3 (2) in integers n, d, k, b, y where k 3, d > 0, gd(n, d) = 1, P(b) k, n 0, y 0. (Note that similarly as e.g. in [1] we allow n < 0, as well.) In the standard way, by our assumptions we an write n + id = a i x 3 i (i = 0, 1,..., k 1) (3)

3 Cubes in produts of terms in arithmeti progression 3 with P(a i ) k, a i is ube-free. Note that (3) also means that in fat n + id (i = 0, 1,..., k 1) is an arithmeti progression of almost ubes. In ase of b = 1 we prove the following result. Theorem 2.1. Suppose that (n, d, k, y) is a solution to equation (2) with b = 1 and k < 39. Then we have (n, d, k, y) = ( 4, 3, 3, 2), ( 2, 3, 3, 2), ( 9, 5, 4, 6) or ( 6, 5, 4, 6). We shall dedue Theorem 2.1 from the following theorem. Theorem 2.2. Suppose that (n, d, k, b, y) is a solution to equation (2) with k < 32 and that P(b) < k if k = 3 or k 13. Then (n, d, k) belongs to the following list: ( 10, 3, 7), ( 8, 3, 7), ( 8, 3, 5), ( 4, 3, 5), ( 4, 3, 3), ( 2, 3,3), ( 9, 5, 4), ( 6, 5, 4), ( 16, 7, 5), ( 12, 7, 5), (n, 1, k) with 30 n 4 or 1 n 5, (n, 2, k) with 29 n 3. Note that the above statement follows from Theorem 1.1 of Bennett, Bruin, Győry, Hajdu [1] in ase k < 12 and P(b) P k with P 3 = 2, P 4 = P 5 = 3, P 6 = P 7 = P 8 = P 9 = P 10 = P 11 = Lemmas and auxiliary results We need some results of Selmer [30] on ubi equations. Lemma 3.1. The equations x 3 + y 3 = z 3, {1, 2, 4, 5, 10, 25, 45, 60, 100,150, 225, 300}, ax 3 + by 3 = z 3, (a, b) {(2, 9), (4, 9), (4, 25), (4, 45), (12, 25)} have no solution in non-zero integers x, y, z. As a lot of work will be done modulo 13, the following lemma will be very useful. Before stating it, we need to introdue a new notation. For u, v, m Z, m > 1 by u v (mod m) we mean that uw 3 v (mod m) holds for some integer w with gd(m, w) = 1. We shall use this notation throughout the paper, without any further referene.

4 4 L. Hajdu, Sz. Tengely and R. Tijdeman Lemma 3.2. Let n, d be integers. Suppose that for five values i {0, 1,..., 12} we have n + id 1 (mod 13). Then 13 d, and n + id 1 (mod 13) for all i = 0, 1,...,12. Proof. Suppose that 13 d. Then there is an integer r suh that n rd (mod 13). Consequently, n + id (r + i)d (mod 13). A simple alulation yields that the ubi residues of the numbers (r + i)d (i = 0, 1,..., 12) modulo 13 are given by a yli permutation of one of the sequenes Thus the statement follows. 0, 1, 2, 2, 4, 1, 4, 4, 1, 4, 2, 2,1, 0, 2, 4, 4, 1, 2, 1, 1, 2, 1, 4, 4,2, 0, 4, 1, 1, 2, 4, 2, 2, 4, 2, 1, 1,4. Lemma 3.3. Let α = 3 2 and β = 3 3. Put K = Q(α) and L = Q(β). Then the only solution of the equation C 1 : X 3 (α + 1)X 2 + (α + 1)X α = ( 3α + 6)Y 3 in X Q and Y K is (X, Y ) = (2, 1). Further, the equation C 2 : 4X 3 (4β + 2)X 2 + (2β + 1)X β = ( 3β + 3)Y 3 has the single solution (X, Y ) = (1, 1) in X Q and Y L. Proof. Using the point (2, 1) we an transform the genus 1 urve C 1 to Weierstrass form E 1 : y 2 + (α 2 + α)y = x 3 + (26α 2 5α 37). We have E 1 (K) Z as an Abelian group and (x, y) = ( α 2 α+3, α 2 3α+4) is a non-torsion point on this urve. Applying ellipti Chabauty (f. [4], [5]), in partiular the proedure Chabauty of MAGMA (see [2]) with p = 5, we obtain that the only point on C 1 with X Q is (2, 1). Now we turn to the seond equation C 2. We an transform this equation to an ellipti one using its point (1, 1). We get E 2 : y 2 = x 3 + β 2 x 2 + βx + (41β 2 58β 4). We find that E 2 (L) Z and (x, y) = (4β 2, 2β 2 + β + 12) is a non-torsion point on E 2. Applying ellipti Chabauty (as above) with p = 11, we get that the only point on C 2 with X Q is (1, 1).

5 Cubes in produts of terms in arithmeti progression 5 4. Proofs In this setion we provide the proofs of our results. As Theorem 2.1 follows from Theorem 2.2 by a simple indutive argument, first we give the proof of the latter result. Proof of Theorem 2.2. As we mentioned, for k = 3, 4 the statement follows from Theorem 1.1 of [1]. Observe that the statement for every k {6, 8, 9, 10, 12, 13, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 28, 29,31} is a simple onsequene of the result obtained for some smaller value of k. Indeed, for any suh k let p k denote the largest prime with p k < k. Observe that in ase of k 13 P(a 0 a 1... a pk 1) p k holds, and for k > 13 we have P(a 0 a 1... a pk ) < p k + 1. Hene, noting that we assume P(b) k for 3 < k 11 and P(b) < k otherwise, the theorem follows indutively from the ase of p k -term produts and p k + 1-term produts, respetively. Hene in the sequel we deal only with the remaining values of k. The ases k = 5, 7 are different from the others. In most ases a brute fore method suffies. In the remaining ases we apply the ellipti Chabauty method (see [4], [5]). The ase k = 5. In this ase a very simple algorithm works already. Note that in view of Theorem 1.1 of [1], by symmetry it is suffiient to assume that 5 a 2 a 3. We look at all the possible distributions of the prime fators 2, 3, 5 of the oeffiients a i (i = 0,...,4) one-by-one. Using that if x is an integer, then x 3 is ongruent to ±1 or 0 both (mod 7) and (mod 9), almost all possibilities an be exluded. For example, is impossible modulo 7, while (a 0, a 1, a 2, a 3, a 4 ) = (1, 1, 1, 10, 1) (a 0, a 1, a 2, a 3, a 4 ) = (1, 1, 15, 1, 1) is impossible modulo 9. (Note that the first hoie of the a i annot be exluded modulo 9, and the seond one annot be exluded modulo 7.) In ase of the remaining possibilities, taking the linear ombinations of three appropriately hosen terms of the arithmeti progression on the left hand side of (2) we get all solutions by Lemma 3.1. For example, (a 0, a 1, a 2, a 3, a 4 ) = (2, 3, 4, 5, 6)

6 6 L. Hajdu, Sz. Tengely and R. Tijdeman obviously survives the above tests modulo 7 and modulo 9. However, in this ase using the identity 4(n + d) 3n = n + 4d, Lemma 3.1 implies that the only orresponding solution is given by n = 2 and d = 1. After having exluded all quintuples whih do not pass the above tests we are left with the single possibility Here we have (a 0, a 1, a 2, a 3, a 4 ) = (2, 9, 2, 5, 12). x x3 2 = 9x3 1 and x 3 0 2x3 2 = 6x3 4. (4) Fatorizing the first equation of (4), a simple onsideration yields that x 2 0 x 0x 2 + x 2 2 = 3u3 holds for some integer u. Put K = Q(α) with α = 3 2. Note that the ring O K of integers of K is a unique fatorization domain, α 1 is a fundamental unit and 1, α, α 2 is an integral basis of K, and 3 = (α 1)(α+1) 3, where α+1 is a prime in O K. A simple alulation shows that x 0 αx 2 and x αx 0x 2 + α 2 x 2 2 an have only the prime divisors α and α + 1 in ommon. Hene heking the field norm of x 0 αx 2, by the seond equation of (4) we get that x 0 αx 2 = (α 1) ε (α 2 + α)y 3 with y O K and ε {0, 1, 2}. Expanding the right hand side, we dedue that ε = 0, 2 yields 3 x 0, whih is a ontradition. Thus we get that ε = 1, and we obtain the equation (x 0 αx 2 )(x 2 0 x 0 x 2 + x 2 2) = ( 3α + 6)z 3 for some z O K. Hene after dividing both sides of this equation by x 3 2, the theorem follows from Lemma 3.3 in this ase. The ase k = 7. In this ase by similar tests as for k = 5, we get that the only remaining possibilities are given by (a 0, a 1, a 2, a 3, a 4, a 5, a 6 ) = (4, 5, 6, 7, 1, 9, 10), (10, 9, 1, 7, 6, 5, 4). By symmetry it is suffiient to deal with the first ase. Then we have x x3 6 = 9x3 5 and x 3 6 3x3 1 = 2x3 0. (5) Fatorizing the first equation of (5), just as in ase of k = 5, a simple onsideration gives that 4x 2 6 2x 1x 6 + x 2 1 = 3u3 holds for some integer u. Let L = Q(β) with β = 3 3. As is well-known, the ring O L of integers of L is a unique fatorization

7 Cubes in produts of terms in arithmeti progression 7 domain, 2 β 2 is a fundamental unit and 1, β, β 2 is an integral basis of L. Further, 2 = (β 1)(β 2 + β + 1), where β 1 and β 2 + β + 1 are primes in O L, with field norms 2 and 4, respetively. A simple alulation yields that x 6 βx 1 and x βx 1 x 6 + β 2 x 2 1 are relatively prime in O L. Moreover, as gd(n, d) = 1 and x 4 is even, x 0 should be odd. Hene as the field norm of β 2 + β + 1 is 4, heking the field norm of x 6 βx 1, the seond equation of (5) yields x 6 βx 1 = (2 β 2 ) ε (1 β)y 3 for some y O L and ε {0, 1, 2}. Expanding the right hand side, a simple omputation shows that ε = 1, 2 yields 3 x 6, whih is a ontradition. Thus we get that ε = 0, and we obtain the equation (x 6 βx 1 )(4x 2 6 2x 1 x 6 + x 2 1) = ( 3β + 3)z 3 for some z O L. We divide both sides of this equation by x 3 1 and apply Lemma 3.3 to omplete the ase k = 7. Desription of the general method. So far we have onsidered all the possible distributions of the prime fators k among the oeffiients a i. For larger values of k we use a more effiient proedure similar to that in [17]. We first outline the main ideas. We explain the important ase that 3, 7, and 13 are oprime to d first. The ase gd(3 7 13, d) = 1. Suppose we have a solution to equation (2) with k 11 and gd(3 7, d) = 1. Then there exist integers r 7 and r 9 suh that n r 7 d (mod 7) and n r 9 d (mod 9). Further, we an hoose the integers r 7 and r 9 to be equal; put r := r 7 = r 9. Then n+id (r+i)d (mod q) holds for q {7, 9} and i = 0, 1..., k 1. In partiular, we have r + i a i s q (mod q), where q {7, 9} and s q is the inverse of d modulo q. Obviously, we may assume that r + i takes values only from the set { 31, 30,..., 31}. First we make a table for the residues of h modulo 7 and 9 up to ubes for h < 32, but here we present only the part with 0 h < 11. h h mod h mod In the first row of the table we give the values of h and in the seond and third rows the orresponding residues of h modulo 7 and modulo 9 up to ubes, respetively, where the lasses of the relation are represented by 0, 1, 2, 4 modulo 7, and by 0, 1, 2, 3, 4 modulo 9.

8 8 L. Hajdu, Sz. Tengely and R. Tijdeman Let a i1,..., a it be the oeffiients in (3) whih do not have prime divisors greater than 2. Put E = {(u ij, v ij ) : r + i j uij (mod 7), r + i j vij (mod 9), 1 j t} and observe that E is ontained in one of the sets E 1 := {(1, 1), (2, 2), (4, 4)}, E 2 := {(1, 2), (2, 4), (4, 1)}, E 3 := {(2, 1), (4, 2), (1, 4)}. We use this observation in the following tests whih we shall illustrate by some examples. In what follows we assume k and r to be fixed. In our method we apply the following tests in the given order. By eah test some ases are eliminated. Class over. Let u i r + i (mod 7) and vi r + i (mod 9) (i = 0, 1,...,k 1). For l = 1, 2, 3 put C l = {i : (u i, v i ) E l, i = 0, 1,...,k 1}. Chek whether the sets C 1 C 2, C 1 C 3, C 2 C 3 an be overed by the multiples of the primes p with p < k, p 2, 3, 7. If this is not possible for C l1 C l2, then we know that E E l3 is impossible and E l3 is exluded. Here {l 1, l 2, l 3 } = {1, 2, 3}. The forthoming proedures are applied separately for eah ase where E E l remains possible for some l. From this point on we also assume that the odd prime fators of the a i are fixed. Parity. Define the sets I e = {(u i, v i ) E l : r + i is even, P(a i ) 2}, I o = {(u i, v i ) E l : r + i is odd, P(a i ) 2}. As the only odd power of 2 is 1, min( I e, I o ) 1 must be valid. If this does not hold, the orresponding ase is exluded. Test modulo 13. Suppose that after the previous tests we an deide whether a i is even for the even values of i. Assume that E E l with fixed l {1, 2, 3}. Further, suppose that based upon the previous tests we an deide whether a i an be even for the even or the odd values of i. For t = 0, 1, 2 put U t = {i : a i = ±2 t, i {0, 1,...,k 1}}

9 and let Cubes in produts of terms in arithmeti progression 9 U 3 = {i : a i = ±5 γ, i {0, 1,..., k 1}, γ {0, 1, 2}}. Assume that 13 n + i 0 d for some i 0. Reall that 13 d and 5 1 (mod 13). If i, j U t for some t {0, 1, 2, 3}, then i i 0 j i0 (mod 13). If i U t1, j U t2 with 0 t 1 < t 2 2, then i i 0 j i 0 (mod 13). We exlude all the ases whih do not pass these tests. Test modulo 7. Assume again that E E l with fixed l {1, 2, 3}. Chek whether the atual distribution of the prime divisors of the a i yields that for some i with 7 n + id, both a i = ±t and r + i = t hold for some positive integer t with 7 t. Then t n + id (r + i)d td (mod 7) implies that d 1 (mod 7). Now onsider the atual distribution of the prime fators of the oeffiients a i (i = 0, 1,...,k 1). If in any a i we know the exponents of all primes with one exeption, and this exeptional prime p satisfies p 2, 3, 4, 5 (mod 7), then we an fix the exponent of p using the above information on n. As an example, assume that 7 n, and a 1 = ±5 γ with γ {0, 1, 2}. Then d 1 (mod 7) immediately implies γ = 0. Further, if 7 n and a 2 = ±13 γ with γ {0, 1, 2}, then d 1 (mod 7) gives a ontradition. We exlude all ases yielding a ontradition. Moreover, in the remaining ases we fix the exponents of the prime fators of the a i -s whenever it is possible. We remark that we used this proedure for 0 r k + 1. In almost all ases it turned out that a i is even for r+i even. Further, we ould prove that with r + i = 1 or 2 we have a i = ±1 or ±2, respetively, to onlude d 1 (mod 7). The test is typially effetive in ase when r is around k/2. The reason for this is that then in the sequene r, r+1,..., 1, 0, 1,..., k r 2, k r 1 several powers of 2 our. Indution. For fixed distribution of the prime divisors of the oeffiients a i, searh for arithmeti sub-progressions of length l with l {3, 5, 7} suh that for the produt Π of the terms of the sub-progression P(Π) L l holds, with L 3 = 2, L 5 = 5, L 7 = 7. If there is suh a sub-progression, then in view of Theorem 1.1 of [1], all suh solutions an be determined. An example. Now we illustrate how the above proedures work. For this purpose, take k = 24 and r = 8. Then, using the previous notation, we work with the following stripe (with i {0, 1,..., 23}): r + i mod mod

10 10 L. Hajdu, Sz. Tengely and R. Tijdeman In the proedure Class over we get the following lasses: C 1 = {0, 4, 6, 7, 9, 10, 12, 16}, C 2 = {3, 13, 18}, C 3 = {19, 21}. For p = 5, 11, 13, 17, 19, 23 put m p = {i : i C 1 C 2, p n + id}, respetively. Using the ondition gd(n, d) = 1, one an easily hek that m 5 3, m 11 2, m 13 2, m 17 1, m 19 1, m Hene, as C 1 C 2 = 11, we get that E E 3 annot be valid in this ase. By a similar (but more sophistiated) alulation one gets that E E 2 is also impossible. So after the proedure Class over only the ase E E 1 remains. From this point on, the odd prime divisors of the oeffiients a i are fixed, and we look at eah ase one-by-one. Observe that p n + id does not imply p a i. Further, p n + id implies p n + jd whenever i j (mod p). We onsider two subases. Suppose first that we have 3 n + 2d, 5 n + d, 7 n + d, 11 n + 7d, 13 n + 7d, 17 n + 3d, 19 n, 23 n + 13d. Then by a simple onsideration we get that in Test modulo 13 either or 4 U 1 and 10 U 2, 10 U 1 and 4 U 2. In the first ase, using 13 n + 7d we get 3d 2 (mod 13) and 3d 4 (mod 13), whih by 3d 3d (mod 13) yields a ontradition. In the seond ase we get a ontradition in a similar manner. Consider now the subase where 3 n + 2d, 5 n + d, 7 n + d, 11 n + 7d, 13 n + 8d, 17 n + 3d, 19 n, 23 n + 13d. This ase survives the Test modulo 13. However, using the strategy explained in Test modulo 7, we an easily hek that if a i is even then i is even, whih yields a 9 = ±1. This immediately gives d 1 (mod 7). Further, we have a 7 = ±11 ε7 with ε 7 {0, 1, 2}. Hene we get that ±11 ε7 n + 7d d 1 (mod 7). This gives ε 7 = 0, thus a 7 = ±1. Therefore P(a 4 a 7 a 10 ) 2. Now we apply the test Indution.

11 Cubes in produts of terms in arithmeti progression 11 The ase gd(3 7 13, d) 1. In this ase we shall use the fat that almost half of the oeffiients are odd. With a slight abuse of notation, when k >11 we shall assume that the oeffiients a 1, a 3,..., a k 1 are odd, and the other oeffiients are given either by a 0, a 2,..., a k 2 or by a 2, a 4,..., a k. Note that in view of gd(n, d) = 1 this an be done without loss of generality. We shall use this notation in the orresponding parts of our arguments without any further referene. Now we ontinue the proof, onsidering the remaining ases k 11. The ase k = 11. When gd(3 7, d) = 1, the proedures Class over, Test modulo 7 and Indution suffie. Hene we may suppose that gd(3 7, d) > 1. Assume that 7 d. Observe that P(a 0 a 1...a 4 ) 5 or P(a 5 a 6... a 9 ) 5. Hene the statement follows by indution. Suppose next that 3 d. Observe that if 11 a 4 a 5 a 6 then P(a 0 a 1...a 6 ) 7 or P(a 4 a 5... a 10 ) 7. Hene by indution and symmetry we may assume that 11 a 5 a 6. Assume first that 11 a 6. If 7 a 0 a 6 then we have P(a 1 a 2 a 3 a 4 a 5 ) 5. Further, in ase of 7 a 5 we have P(a 0 a 1 a 2 a 3 a 4 ) 5. Thus by indution we may suppose that 7 a 1 a 2 a 3 a 4. If 7 a 1 a 2 a 4 and 5 n, we have P(a 0 a 5 a 10 ) 2, whene by applying Lemma 3.1 to the identity n + (n + 10d) = 2(n + 5d) we get all the solutions of (2). Assume next that 7 a 1 a 2 a 4 and 5 n. Hene we dedue that one among P(a 2 a 3 a 4 ) 2, P(a 1 a 4 a 7 ) 2, P(a 1 a 2 a 3 ) 2 is valid, and the statement follows in eah ase in a similar manner as above. If 7 a 3, then a simple alulation yields that one among P(a 0 a 1 a 2 ) 2, P(a 0 a 4 a 8 ) 2, P(a 1 a 4 a 7 ) 2 is valid, and we are done. Finally, assume that 11 a 5. Then by symmetry we may suppose that 7 a 0 a 1 a 4 a 5. If 7 a 4 a 5 then P(a 6 a 7 a 8 a 9 a 10 ) 5, and the statement follows by indution. If 7 a 0 then we have P(a 2 a 4 a 6 a 8 a 10 ) 5, and we are done too. In ase of 7 a 1 one among P(a 0 a 2 a 4 ) 2, P(a 2 a 3 a 4 ) 2, P(a 0 a 3 a 6 ) 2 holds. This ompletes the ase k = 11. The ase k = 14. Note that without loss of generality we may assume that 13 a i with 3 i 10, otherwise the statement follows by indution from the ase k = 11. Then, in partiular we have 13 d. The tests desribed in the previous setion suffie to dispose of the ase gd(3 7 13, d) = 1. Assume now that gd(3 7 13, d) > 1 (but reall that 13 d). Suppose first that 7 d. Among the odd oeffiients a 1, a 3,..., a 13 there are at most three multiples of 3, two multiples of 5 and one multiple of 11. As 13 1 (mod 7), this shows that at least for one of these a i -s we have a i 1 (mod 7). Hene a i 1 (mod 7) for every i = 1, 3,...,13. Further, as none of

12 12 L. Hajdu, Sz. Tengely and R. Tijdeman 3, 5, 11 is a ube modulo 7, we dedue that if i is odd, then either gd(3 5 11, a i ) = 1 or a i must be divisible by at least two out of 3, 5, 11. Noting that 13 d, by Lemma 3.2 at most four numbers among a 1, a 3,..., a 13 an be equal to ±1. Moreover, gd(n, d) = 1 implies that 15 a i an be valid for at most one i {0, 1,..., k 1}. Hene among the oeffiients with odd indies there is exatly one multiple of 11, exatly one multiple of 15, and exatly one multiple of 13. Moreover, the multiple of 11 in question is also divisible either by 3 or by 5. In view of the proof of Lemma 3.2 a simple alulation yields that the ubi residues of a 1, a 3,..., a 13 modulo 13 must be given by 1, 1, 4, 0, 4, 1, 1, in this order. Looking at the spots where 4 ours in this sequene, we get that either 3 a 5, a 9 or 5 a 5, a 9 is valid. However, this ontradits the assumption gd(n, d) = 1. Assume now that 3 d, but 7 d. Then among the odd oeffiients a 1, a 3,..., a 13 there are at most two multiples of 5 and one multiple of 7, 11 and 13 eah. Lemma 3.2 together with 5 1 (mod 13) yields that there must be exatly four odd i-s with a i 1 (mod 13), and further, another odd i suh that ai is divisible by 13. Hene as above, the proof of Lemma 3.2 shows that the a i -s with odd indies are 1, 1, 4, 0, 4, 1, 1 (mod 13), in this order. As the prime 11 should divide an a i with odd i and a i 4 (mod 13), this yields that 11 a5 a 9. However, as above, this immediately yields that P(a 0 a 2... a 12 ) 7 (or P(a 2 a 4...a 14 ) 7), and the ase k = 14 follows by indution. The ase k = 18. Using the proedures desribed in the previous setion, the ase gd(3 7 13, d) = 1 an be exluded. So we may assume gd(3 7 13, d) > 1. Suppose first that 7 d. Among a 1, a 3,..., a 17 there are at most three multiples of 3, two multiples of 5 and one multiple of 11, 13 and 17 eah. Hene at least for one odd i we have a i = ±1. Thus all of a 1, a 3,..., a 17 are 1 (mod 7). Among the primes 3, 5, 11, 13, 17 only 13 is 1 (mod 7), so the other primes annot our alone. Hene we get that a i = ±1 for at least five out of a 1, a 3,...,a 17. However, by Lemma 3.2 this is possible only if 13 d. In that ase a i = ±1 holds for at least six oeffiients with i odd. Now a simple alulation shows that among them three are in arithmeti progression. This leads to an equation of the shape X 3 + Y 3 = 2Z 3, and Lemma 3.1 applies. Assume next that 13 d, but 7 d. Among the odd oeffiients a 1, a 3,...a 17 there are at most three multiples of 3, two multiples of 5 and 7 eah, and one multiple of 11 and 17 eah. Hene, by 5 1 (mod 13) there are at least two a i 1 (mod 13), whene all ai 1 (mod 13). As from this list only the prime 5 is a ube modulo 13, we get that at least four out of the above nine odd a i -s are

13 Cubes in produts of terms in arithmeti progression 13 equal to ±1. Reall that 7 d and observe that the ubi residues modulo 7 of a seven-term arithmeti progression with ommon differene not divisible by 7 is a yli permutation of one of the sequenes 0, 1, 2, 4, 4, 2, 1, 0, 2, 4, 1, 1, 4, 2, 0, 4, 1, 2, 2, 1, 4. Hene remembering that for four odd i we have a i = ±1, we get that the ubi residues of a 1, a 3,...,a 17 modulo 7 are given by 1, 1, 4, 2, 0, 2, 4, 1, 1, in this order. In partiular, we have exatly one multiple of 7 among them. Further, looking at the spots where 0, 2 and 4 our, we dedue that at most two of the a i -s with odd indies an be multiples of 3. Swithing bak to modulo 13, this yields that a i = ±1 for at least five a i -s. However, this ontradits Lemma 3.2. Finally, assume that 3 d. In view of what we have proved already, we may further suppose that gd(7 13, d) = 1. Among the odd oeffiients a 1, a 3,...,a 17 there are at most two multiples of 5 and 7 eah, and one multiple of 11, 13 and 17 eah. Hene as 7 d and 13 1 (mod 7), we get that the ubi residues modulo 7 of the oeffiients a i with odd i are given by one of the sequenes 1, 0, 1, 2, 4, 4, 2, 1, 0, 0, 1, 2, 4, 4, 2, 1, 0, 1, 1, 1, 2, 4, 0, 4, 2, 1, 1. In view of the plaes of the values 2 and 4, we see that it is not possible to distribute the prime divisors 5, 7, 11 over the a i -s with odd indies. This finishes the ase k = 18. The ase k = 20. By the help of the proedures desribed in the previous setion, in ase of gd(3 7 13, d) = 1 all solutions to equation (2) an be determined. Assume now that gd(3 7 13, d) > 1. We start with the ase 7 d. Then among the odd oeffiients a 1, a 3,...,a 19 there are at most four multiples of 3, two multiples of 5, and one multiple of 11, 13, 17 and 19 eah. As 13 1 (mod 7), this yields that a i 1 (mod 7) for all i. Hene the primes 3, 5, 11, 17, 19 must our at least in pairs in the a i -s with odd indies, whih yields that at least five suh oeffiients are equal to ±1. Thus Lemma 3.2 gives 13 d, whene a i 1 (mod 13) for all i. Hene we dedue that the prime 5 may be only a third prime divisor of the a i -s with odd indies, and so at least seven out of a 1, a 3,..., a 19 equal ±1. However, then there are three suh oeffiients whih belong to an arithmeti progression. Thus by Lemma 3.1 we get all solutions in this ase. Assume next that 13 d. Without loss of generality we may further suppose that 7 d. Then among the odd oeffiients a 1, a 3,..., a 19 there are at most four multiples of 3, two multiples of 5 and 7 eah, and one multiple of 11, 17 and

14 14 L. Hajdu, Sz. Tengely and R. Tijdeman 19 eah. As 5 1 (mod 13) this implies a i 1 (mod 13) for all i, whene the primes 3, 7, 11, 17, 19 should our at least in pairs in the a i -s with odd i. Hene at least four of these oeffiients are equal to ±1. By a similar argument as in ase of k = 18, we get that the ubi residues of a 1, a 3,...,a 19 modulo 7 are given by one of the sequenes 1, 0, 1, 2, 4, 4, 2, 1, 0, 1, 1, 1, 4, 2, 0, 2, 4, 1, 1, 4, 4, 1, 1, 4, 2, 0, 2, 4, 1, 1. In view of the positions of the 0, 2 and 4 values, we get that at most two orresponding terms an be divisible by 3 in the first ase, whih modulo 13 yields that the number of odd i-s with a i = ±1 is at least five. This is a ontradition modulo 7. Further, in the last two ases at most three terms an be divisible by 3, and exatly one term is a multiple of 7. This yields modulo 13 that the number of odd i-s with a i = ±1 is at least five, whih is a ontradition modulo 7 again. Finally, suppose that 3 d. We may assume that gd(7 13, d) = 1. Then among the odd oeffiients a 1, a 3,..., a 19 there are at most two multiples of 5 and 7 eah, and one multiple of 11, 13, 17 and 19 eah. Hene Lemma 3.2 yields that exatly four of these oeffiients should be 1 (mod 13), and exatly one of them must be a multiple of 13. Further, exatly two other a i -s with odd indies are multiples of 7, and these a i -s are divisible by none of 11, 13, 17, 19. So in view of the proof of Lemma 3.2 a simple alulation gives that the ubi residues of a 1, a 3,..., a 19 modulo 13 are given by one of the sequenes 0, 2, 4, 4, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 4, 4, 2, 0, 2, 4, 2, 1, 1, 4, 0, 4, 1, 1, 1, 1, 4, 0, 4, 1, 1, 2, 4, 2. In the upper ases we get that 7 divides two terms with a i 2 (mod 13), whene the power of 7 should be 2 in both ases. However, this implies d, hene 7 d, a ontradition. As the lower ases are symmetri, we may assume that the very last possibility ours. In that ase we have 7 a 5 and 7 a 19. We may assume that 11 a 17, otherwise P(a 6 a 8...a 18 ) 7 and the statement follows by indution. Further, we also have 13 a 7, and 17 a 9 and 19 a 15 or vie versa. Hene either P(a 3 a 8 a 13 ) 2 or P(a 4 a 10 a 16 ) 2, and indution suffies to omplete the ase k = 20. The ase k = 24. The proedures desribed in the previous setion suffie to ompletely treat the ase gd(3 7 13, d) = 1. So we may assume that gd(3 7 13, d) > 1 is valid. Suppose first that 7 d. Among the odd oeffiients a 1, a 3,..., a 23 there are at most four multiples of 3, three multiples of 5, two multiples of 11, and one multiple of 13, 17, 19 and 23 eah. We know that all a i belong to the same ubi

15 Cubes in produts of terms in arithmeti progression 15 lass modulo 7. As 3 4 (mod 7), 5 2 (mod 7) and among the oeffiients a 1, a 3,..., a 23 there are at most two multiples of 3 2 and at most one multiple of 5 2, we get that these oeffiients are all 1 (mod 7). This yields that the primes 3, 5, 11, 17, 19, 23 may our only at least in pairs in the oeffiients with odd indies. Thus we get that at least five out of a 1, a 3,...,a 23 are 1 (mod 13). Hene, by Lemma 3.2 we get that 13 d and onsequently a i 1 (mod 13) for all i. This also shows that the 5-s an be at most third prime divisors of the a i -s with odd indies. So we dedue that at least eight out of the odd oeffiients a 1, a 3,..., a 23 are equal to ±1. However, a simple alulation shows that from the eight orresponding terms we an always hoose three forming an arithmeti progression. Hene this ase follows from Lemma 3.1. Assume next that 13 d, but 7 d. Among the oeffiients with odd indies there are at most four multiples of 3, three multiples of 5, two multiples of 7 and 11 eah, and one multiple of 17, 19 and 23 eah. Hene, by 5 1 (mod 13) we dedue a i 1 (mod 13) for all i. As before, a simple alulation yields that at least for four of these odd oeffiients a i = ±1 hold. Hene looking at the possible ases modulo 7, one an easily see that we annot have four multiples of 3 at the plaes where 0, 2 and 4 our as ubi residues modulo 7. Hene in view of Lemma 3.2 we need to use two 11-s, whih yields that 11 a 1 and 11 a 23. Thus the only possibility for the ubi residues of a 1, a 3,...,a 23 modulo 7 is given by the sequene 2, 1, 0, 1, 2, 4, 4, 2, 1, 0, 1, 2. However, the positions of the 2-s and 4-s allow to have at most two a i -s with odd indies whih are divisible by 3 but not by 7. Hene swithing bak to modulo 13, we get that there are at least five a i -s whih are ±1, a ontradition by Lemma 3.2. Finally, assume that 3 d, and gd(7 13, d) = 1. Then among a 1, a 3,...,a 23 there are at most three multiples of 5, two multiples of 7 and 11 eah, and one multiple of 13, 17, 19 and 23 eah. Hene by Lemma 3.2 we get that exatly four of the oeffiients a 1, a 3,..., a 23 are 1 (mod 13), and another is a multiple of 13. Further, all the mentioned prime fators (exept the 5-s) divide distint a i -s with odd indies. Using that at most these oeffiients an be divisible by 7 2 and 11 2, in view of the proof of Lemma 3.2 we get that the only possibilities for the ubi residues of these oeffiients modulo 13 are given by one of the sequenes 2, 2, 4, 2, 1, 1, 4, 0, 4, 1, 1, 2, 2, 1, 1, 4, 0, 4, 1, 1, 2, 4, 2, 2. By symmetry we may assume the first possibility. Then we have 7 a 3, 11 a 1, 13 a 15, and 17, 19, 23 divide a 5, a 7, a 13 in some order. Hene P(a 4 a 9 a 14 ) 2, or 5 n + 4d whene P(a 16 a 18 a 20 ) 2. In both ases we apply indution.

16 16 L. Hajdu, Sz. Tengely and R. Tijdeman The ase k = 30. By the help of the proedures desribed in the previous setion, the ase gd(3 7 13, d) = 1 an be exluded. Assume now that gd(3 7 13, d) > 1. We start with the ase 7 d. Then among the odd oeffiients a 1, a 3,...,a 29 there are at most five multiples of 3, three multiples of 5, two multiples of 11 and 13 eah, and one multiple of 17, 19, 23 and 29 eah. As (mod 7), this yields that a i 1 (mod 7) for all i. Hene the other primes must our at least in pairs in the a i -s with odd indies, whih yields that at least six suh oeffiients are equal to ±1. Further, we get that the number of suh oeffiients 0, 1 (mod 13) is at least eight. However, by Lemma 3.2 this is possible only if 13 d, whene a i 1 (mod 13) for all i. Then 5 and 29 an be at most third prime divisors of the oeffiients a i -s with odd i-s. So a simple alulation gives that at least ten out of the odd oeffiients a 1, a 3,..., a 29 are equal to ±1. Hene there are three suh oeffiients in arithmeti progression, and the statement follows from Lemma 3.1. Assume next that 13 d, but 7 d. Then among the odd oeffiients a 1, a 3,..., a 29 there are at most five multiples of 3, three multiples of 5 and 7 eah, two multiples of 11, and one multiple of 17, 19, 23 and 29 eah. From this we get that a i 1 (mod 13) for all i. Hene the primes different from 5 should our at least in pairs. We get that at least five out of the oeffiients a 1, a 3,..., a 29 are equal to ±1. Thus modulo 7 we get that it is impossible to have three terms divisible by 7. Then it follows modulo 13 that at least six a i -s with odd indies are equal to ±1. However, this is possible only if 7 d, whih is a ontradition. Finally, assume that 3 d, but gd(7 13, d) = 1. Then among the odd oeffiients a 1, a 3,..., a 29 there are at most three multiples of 5 and 7 eah, two multiples of 11 and 13 eah, and one multiple of 17, 19, 23 and 29 eah. Further, modulo 7 we get that all primes 5, 11, 17, 19, 23 divide distint a i -s with odd indies, and the number of odd i-s with a i 0, 1 (mod 7) is seven. However, heking all possibilities modulo 7, we get a ontradition. This ompletes the proof of Theorem 2.2. Proof of Theorem 2.1. Obviously, for k < 32 the statement is an immediate onsequene of Theorem 2.2. Further, observe that b = 1 implies that for any k with 31 < k < 39, one an find j with 0 j k 30 suh that P(a j a j+1... a j+29 ) 29. Hene the statement follows from Theorem 2.2. Aknowledgement. The authors are grateful to the referee for the useful remarks.

17 Cubes in produts of terms in arithmeti progression 17 Referenes [1] M. Bennett, N. Bruin, K. Győry and L. Hajdu, Powers from produts of onseutive terms in arithmeti progression, Pro. London Math. So. 92 (2006), [2] W. Bosma, J. Cannon and C. Playoust, The Magma algebra system, I. The user language, J. Symboli Comput. 24 (1997), [3] B. Brindza, L. Hajdu and I. Z. Ruzsa, On the equation x(x + d)...(x + (k 1)d) = by 2, Glasgow Math. J. 42 (2000), [4] N. Bruin, Chabauty methods and overing tehniques applied to generalized Fermat equations, CWI Trat, Vol. 133, Stihting Mathematish Centrum, Centrum voor Wiskunde en Informatia, Amsterdam, [5] N. Bruin, Chabauty methods using ellipti urves, J. Reine Angew. Math. 562 (2003), [6] H. Darmon and L. Merel, Winding quotients and some variants of Fermat s Last Theorem, J. Reine Angew. Math. 490 (1997), [7] L. E. Dikson, History of the Theory of Numbers, Vol. II: Diophantine analysis, Chelsea Publishing Co., New York, 1966, xxv+803. [8] P. Erdős, Note on produts of onseutive integers (II), J. London Math. So. 14 (1939), [9] P. Erdős and J. L. Selfridge, The produt of onseutive integers is never a power, Illinois J. Math. 19 (1975), [10] P. Filakovszky and L. Hajdu, The resolution of the equation x(x + d)...(x + (k 1)d) = by 2 for fixed d, Ata Arith. 98 (2001), [11] K. Győry, On the diophantine equation n k = x l, Ata Arith. 80 (1997), [12] K. Győry, On the diophantine equation n(n + 1)... (n + k 1) = bx l, Ata Arith. 83 (1998), [13] K. Győry, Power values of produts of onseutive integers and binomial oeffiients, Number Theory and Its Appliations, (S. Kanemitsu and K. Győry, eds.), Kluwer Aad. Publ., 1999, [14] K. Győry, Perfet powers in produts with onseutive terms from arithmeti progressions, More Sets, Graphs and Numbers, Vol. 15, (E. Győri, G. O. H Katona, L. Lovász, eds.), Bolyai Soiety Mathematial Studies, 2006, [15] K. Győry, L. Hajdu and N. Saradha, On the Diophantine equation n(n + d)...(n + (k 1)d) = by l, Canad. Math. Bull. 47 (2004), , 48 (2005), 636. [16] G. Hanrot, N. Saradha and T. N. Shorey, Almost perfet powers in onseutive integers, Ata Arith. 99 (2001), [17] N. Hirata-Kohno, S. Laishram, T. N. Shorey and R. Tijdeman, An extension of a theorem of Euler, Ata Arith. 129 (2007), [18] S. Laishram, An estimate for the length of an arithmeti progression the produt of whose terms is almost square, Publ. Math. Debreen 68 (2006), [19] S. Laishram and T. N. Shorey, The equation n(n + d)...(n + (k 1)d) = by 2 with ω(d) 6 or d 10 10, Ata Arith. 129 (2007), [20] R. Marszalek, On the produt of onseutive terms of an arithmeti progression, Monatsh. Math. 100 (1985), [21] R. Obláth, Über das Produkt fünf aufeinander folgender Zahlen in einer arithmetisher Reihe, Publ. Math. Debreen 1 (1950), [22] O. Rigge, Über ein diophantishes Problem, 9th Congress Math. Sand., Helsingfors 1938, Merator 1939, [23] J. W. Sander, Rational points on a lass of superellipti urves, J. London Math. So. 59 (1999),

18 18 L. Hajdu, Sz. Tengely and R. Tijdeman : Cubes in produts of terms... [24] N. Saradha, On perfet powers in produts with terms from arithmeti progressions, Ata Arith. 82 (1997), [25] N. Saradha, Squares in produts with terms in an arithmeti progression, Ata Arith. 86 (1998), [26] N. Saradha and T. N. Shorey, Almost perfet powers in arithmeti progression, Ata Arith. 99 (2001), [27] N. Saradha and T. N Shorey, Almost squares in arithmeti progression, Compositio Math. 138 (2003), [28] N. Saradha and T. N. Shorey, Almost squares and fatorisations in onseutive integers, Compositio Math. 138 (2003), [29] N. Saradha and T. N. Shorey, Contributions towards a onjeture of Erdős on perfet powers in arithmeti progressions, Compositio Math. 141 (2005), [30] E. Selmer, The diophantine equation ax 3 +by 3 +z 3 = 0, Ata Math. 85 (1951), [31] T. N. Shorey, Perfet powers in produts of arithmetial progressions with fixed initial term, Indag. Math. N.S. 7 (1996), [32] T. N. Shorey, Powers in arithmeti progression, A Panorama in Number Theory, (G. Wüstholzed, ed.), Cambridge University Press, 2002, [33] T. N. Shorey, Powers in arithmeti progression (II), New Aspets of Analyti Number Theory, Kyoto, 2002, [34] T. N. Shorey, Diophantine approximations, Diophantine equations, transendene and appliations, Indian J. Pure Appl. Math. 37 (2006), [35] T. N. Shorey and R. Tijdeman, Perfet powers in produts of terms in an arithmeti progression, Compositio Math. 75 (1990), [36] T. N. Shorey and R. Tijdeman, Perfet powers in produts of terms in an arithmeti progression II, Compositio Math. 82 (1992), [37] T. N. Shorey and R. Tijdeman, Perfet powers in produts of terms in an arithmeti progression III, Ata Arith. 61 (1992), [38] T. N. Shorey and R. Tijdeman, Some methods of Erdős applied to finite arithmeti progressions integers and related equations, The Mathematis of Paul Erdős I, Springer, 1997, [39] Sz. Tengely, Note on the paper An extension of a theorem of Euler by Hirata-Kohno et. al, Ata Arith. 134 (2008), [40] R. Tijdeman, Diophantine equations and diophantine approximations, Number Theory and Appliations, Kluwer Aad. Press, 1989, [41] R. Tijdeman, Exponential diophantine equations , Number Theory: Diophantine, Computational and Algebrai Aspets, (K. Győry, A. Pethő and V. T. Sós, eds.), Walter de Gruyter, Berlin New York, 1998, LAJOS HAJDU AND SZABOLCS TENGELY INSTITUTE OF MATHEMATICS UNIVERSITY OF DEBRECEN AND THE NUMBER THEORY RESEARCH GROUP OF THE HUNGARIAN ACADEMY OF SCIENCES 4010 DEBRECEN, P.O. BOX 12 HUNGARY hajdul@math.klte.hu tengely@math.klte.hu ROBERT TIJDEMAN MATHEMATICAL INSTITUTE LEIDEN UNIVERSITY 2300 RA LEIDEN, P.O. BOX 9512 THE NETHERLANDS tijdeman@math.leidenuniv.nl (Reeived Otober 30, 2008; revised November 18, 2008)