CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION

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1 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION L. HAJDU 1, SZ. TENGELY, R. TIJDEMAN 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 prove 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 (1) n(n + d)... (n + (k 1)d) = by l in positive integers n, d, k, b, y, l with l 2, k 2, 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. Here (beside the papers referred to later on in the introdution) we only 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], and the referenes given there. Now 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 2000 Mathematis Subjet Classifiation. 11D41, 11D25, 11B25. Key words and phrases. Perfet ubes, arithmeti progressions. 1) Researh supported in part by the Hungarian Aademy of Sienes and by the OTKA grants T48791 and K

2 2 L. HAJDU, SZ. TENGELY, R. TIJDEMAN 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 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. The reason for doing so is that in this ase one an use speial methods whih are not available for general exponents. 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 (2) n(n + d)... (n + (k 1)d) = by 3 in integers n, d, k, b, y where k 3, d > 0, gd(n, d) = 1, P (b) k, 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 (3) n + id = a i x 3 i (i = 0, 1,..., k 1) 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), ( 6, 5, 4, 6).

3 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 3 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 for k = 3 and for k 13. Then (n, d, k) belongs to the following list: (n, 1, k) with 30 n 4 or 1 n 5, (n, 2, k) with 29 n 3, ( 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). Note that the above theorem for k < 12 follows from Theorem 1.1 of Bennett, Bruin, Győry, Hajdu [1] under the ondition 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 giving that, 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 with some integer w where gd(m, w) = 1. We shall use this notation throughout the paper, without any further referene. Lemma 3.2. Let n, d be integers. Suppose that for five 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.

4 4 L. HAJDU, SZ. TENGELY, R. TIJDEMAN 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). 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]. We give the proof of the theorem for eah k > 4 separately. However, it turns out 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,

5 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 5 respetively. Hene in the sequel we deal only with the remaining values of k. We disuss the ases k = 5, 7 separately from the others, beause here the method of our proof is different from the one used for the larger values of k. Here already the brute fore method suffies to treat all the ases. Furthermore, here there are some ases left whih apparently annot be handled with any lassial tool. At these points we shall make use of 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 distribution of the primes 2, 3, 5 in the oeffiients a i (i = 0,..., 4) one-by-one. Cheking the progressions modulo 7 and 9 (using that x 3 is ongruent to ±1 or 0 both (mod 7) and (mod 9) for any integer x), 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) 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 that we are left with the single possibility Then we have (a 0, a 1, a 2, a 3, a 4 ) = (2, 9, 2, 5, 12). (4) x x 3 2 = 9x 3 1 and x 3 0 2x 3 2 = 6x 3 4. Fatorizing the first equation of (4), a simple onsideration yields that x 2 0 x 0 x 2 +x 2 2 = 3u 3 holds for some integer u. Put K = Q(α) with α = 3 2. Note that the ring OK 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 0 x 2 + α 2 x 2 2 an have only

6 6 L. HAJDU, SZ. TENGELY, R. TIJDEMAN 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 a similar algorithm as for k = 5, we get that the only possible exeptions for the theorem 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 (5) x x 3 6 = 9x 3 5 and x 3 6 3x 3 1 = 2x 3 0. Fatorizing the first equation of (5), just as in ase of k = 5, a simple onsideration gives that 4x 2 6 2x 1 x 6 +x 2 1 = 3u 3 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 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 2 6+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. Hene after dividing both sides of this equation by x 3 1, the theorem follows from Lemma 3.3 in this ase.

7 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 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. However, for larger values of k this approah would be extremely time onsuming, so more effiient proedure is needed. Note that it is not worth to use this proedure for k 7. Now we outline the main ideas of our approah. Note that in [17] a similar method is applied. In the main argument of our method we assume 3 d, 7 d and 13 d. We explain this ase 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. We may assume that the values r + i (i = 0,..., k 1) belong to the torus 31 r + i < 32 modulo 63. Sine we are interested only in the residues of every r + i modulo 7 and 9 up to ubes, we make a table ontaining these data. In fat we make a full table, but here we indiate only part of the torus, with 0 r + i < 11. r + i mod mod In the first row of the table we give the values of r + i, while in the seond and third rows the orresponding residues of r +i modulo 7 and modulo 9 up to ubes, respetively. Note that the lasses of the relation are represented by 0, 1, 2, 4 modulo 7, and by 0, 1, 2, 3, 4 modulo 9. 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)}. Based upon this observation, we introdue ertain filtering proedures. We start with explaining how these proedures work, then we illustrate the whole method on an example.

8 8 L. HAJDU, SZ. TENGELY, R. TIJDEMAN In what follows, we always assume k and r to be fixed. Further, in our method we apply these proedures in the order below. 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. (Here {l 1, l 2, l 3 } = {1, 2, 3}). In the opposite ase the pairs (u i, v i ) belonging to the a i with P (a i ) 2 may be ontained in E l3 and we need to exeute the following tests. 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. Suppose that E E l for some l {1, 2, 3} remains possible after the test Class over. 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 ondition does not hold, then the orresponding ase an be exluded. Test modulo 13. 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 and let U t = {i : a i = ±2 t, i {0, 1,..., k 1}} U 3 = {i : a i = ±5 γ, i {0, 1,..., k 1}, γ {0, 1, 2}}. Assume that 13 n + i 0 d with some i 0. Then as 13 d and 5 1 (mod 13), if i, j U t for any t {0, 1, 2, 3}, then i i 0 j i0 (mod 13) must hold. If this ondition is not valid, then the orresponding fixed hoie of the prime fators of the a i an be exluded. Further, if i i 0 j i0 (mod 13) holds for some i U t1, j U t2 with 0 t 1 < t 2 2, then we get the same onlusion. 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

9 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 9 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 has p 2, 3, 4, 5 (mod 7), then we an fix the exponent of p using the above information for d. 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. We remark that we used this proedure for 0 r k + 1, when in almost all ases it turned out that a i is even for r + i even. Further, we typially 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 subprogression, 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 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

10 10 L. HAJDU, SZ. TENGELY, R. TIJDEMAN 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. We onsider two possible distributions of the prime divisors. Note that here it is better to write p n + id rather than p a i. Indeed, if p divides n + id exatly on the third power say, then of ourse p n + id, however, p a i. Further, ertainly p n + id implies p n + jd whenever i j (mod p). 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 ase where 3 n + 2d, 5 n + d, 7 n + d, 11 n + 7d, 13 n + 8d, 17 n + 3d, 19 n, 23 n + 13d. Then Test modulo 13 does not work. However, using the strategy explained in Test modulo 7, we an fix several a i. Indeed, one 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 ε 7 with ε 7 {0, 1, 2}. Hene we get that ±11 ε 7 n + 7d d 1 (mod 7). This gives ε 7 = 0, thus a 7 = ±1. However, then P (a 4 a 7 a 10 ) 2, and the theorem follows by Indution in this ase. Note that in some ases it happens that already at Test modulo 7 we get a ontradition, implied by the fat that there is no valid hoie for an a i for some i.

11 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 11 The ase gd(3 7 13, d) 1. In this ase in our argument we shall use the fat that almost the 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. In the ase where gd(3 7, d) = 1, already the proedures Class over, Test modulo 7 and Indution suffie. Hene we may suppose that gd(3 7, d) > 1. Assume first that 7 d. Observe that then one of P (a 0 a 1... a 4 ) 5, P (a 5 a 6... a 9 ) 5 is always valid. Hene the statement follows from the part k = 5 in this ase. Suppose next that 3 d. Observe that if 11 a 4 a 5 a 6 then one of P (a 0 a 1... a 6 ) 7, P (a 4 a 5... a 10 ) 7 is valid. 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 then in ase of 5 n we have P (a 0 a 5 a 10 ) 2, whene using the identity n + (n + 10d) = 2(n + 5d) by Lemma 3.1 we get all solutions of the equation (2) in this ase. 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 always 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 again. 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 always holds, hene the theorem follows. 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. In ase of gd(3 7 13, d) = 1 by the help of the proedures desribed in the previous setion all ases an be exluded. 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

12 12 L. HAJDU, SZ. TENGELY, R. TIJDEMAN 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 ai 1 (mod 7) for every i = 1, 3,..., 13. Further, as none of 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 there an be at most four a i = ±1 among a 1, a 3,..., a 13. 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, respetively. 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 a i 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 theorem follows by indution again. The ase k = 18. In ase of gd(3 7 13, d) = 1 by the help of the proedures desribed in the previous setion all ases 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 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 at least for 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 by Lemma 3.1 the theorem is proved in this ase.

13 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 13 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 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 2 and 4 values, from this we see that it is not possible to appropriately distribute the prime divisors 5, 7, 11 for the a i -s with odd indies. Hene the theorem follows also in this ase. The ase k = 20. In ase of gd(3 7 13, d) = 1 by the help of the proedures desribed in the previous setion all ases 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 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,

14 14 L. HAJDU, SZ. TENGELY, R. TIJDEMAN 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 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 of the 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 first two 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 and 7 d, a ontradition. As the latter two 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 the theorem follows by indution also in any ase.

15 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 15 The ase k = 24. In ase of gd(3 7 13, d) = 1 by the help of the proedures desribed in the previous setion all ases an be exluded. 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 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 the theorem follows from Lemma 3.1 in this ase. 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 3-s 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 should be valid. 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, then the positions of the 2-s and 4-s allow to use at most two primes 3 at plaes whih are not divisible by 7. Hene swithing bak to modulo 13, we get that there are at least five a i = ±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 more is a multiple of 13. Further, all

16 16 L. HAJDU, SZ. TENGELY, R. TIJDEMAN the listed primes (exept the 5-s) should our separately. Using that at most one of 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 that the we have 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 we may further assume that 5 n + 4d, otherwise P (a 4 a 9 a 14 ) 2, and by indution we are done. However, then P (a 16 a 18 a 20 ) 2, and the statement follows by indution. The ase k = 30. In ase of gd(3 7 13, d) = 1 by the help of the proedures desribed in the previous setion all ases 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. Further, 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 in this ase. 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

17 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 17 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 must our separately, 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. Hene the theorem follows also in this ase. 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 always find an appropriate j with 0 j k 30, suh that P (a j a j+1... a j+29 ) 29 is always valid. Hene the statement follows from Theorem 2.2 also in this ase. Referenes [1] M. Bennett, N. Bruin, K. Győry, 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, 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, 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 pp. [8] P. Erdős, Note on produts of onseutive integers (II), J. London Math. So. 14 (1939), [9] P. Erdős, J. L. Selfridge, The produt of onseutive integers is never a power, Illinois J. Math. 19 (1975), [10] P. Filakovszky, 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, pp [14] K. Győry, Perfet powers in produts with onseutive terms from arithmeti progressions, More Sets, Graphs and Numbers (E. Győri, G. O. H Katona, L. Lovász, eds.), Bolyai Soiety Mathematial Studies 15 (2006), pp

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19 CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION 19 [38] T. N. Shorey, R. Tijdeman, Some methods of Erdős applied to finite arithmeti progressions integers and related equations, The Mathematis of Paul Erdős I, Springer 1997, pp [39] Sz. Tengely, Note on the paper An extension of a theorem of Euler by Hirata- Kohno et. al, Ata Arith. (submitted). [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, pp Lajos Hajdu Institute of Mathematis University of Debreen and the Number Theory Researh Group of the Hungarian Aademy of Sienes P.O.Box Debreen Hungary address: hajdul@math.klte.hu Szabols Tengely Institute of Mathematis University of Debreen P.O.Box Debreen Hungary address: tengely@math.klte.hu Robert Tijdeman Mathematial Institute Leiden University P.O.Box RA Leiden The Netherlands address: tijdeman@math.leidenuniv.nl