Almost Perfect Powers in Products of Consecutive Integers

Monatsh. Math. 145, 19–33 (2005) DOI 10.1007/s00605-004-0278-8

Almost Perfect Powers in Products of Consecutive Integers By

´ . Pinte´r K. Gyo¨ry and A University of Debrecen, Hungary Communicated by W. M. Schmidt Received December 3, 2003; accepted in revised form July 7, 2004 Published online November 8, 2004 # Springer-Verlag 2004 Abstract. The classical equations (1) and (3) have a very extensive literature. The main purpose of recent investigations is to solve these equations with as large bound for the greatest prime factor PðbÞ of b as possible. General elementary methods have been developed for studying (1) and (3) which, however, cannot be applied if k is small. As a generalization of previous results obtained for small values of k, we completely solve Eq. (1) for k 4 5, under the assumption that PðbÞ 4 pk , the k-th prime (cf. Theorem 1). A similar result is established for Eq. (3) (cf. Theorem 2). In our proofs, several deep results and powerful techniques are combined from modern Diophantine analysis. 2000 Mathematics Subject Classification: 11D41 Key words: Exponential diophantine equations, binomial Thue equations, product of consecutive integers

1. Introduction By a celebrated theorem of Erd€os and Selfridge [8], the product of consecutive positive integers is never a perfect power. More generally, in [8] the equation mðm þ 1Þ ðm þ k 1Þ ¼ byn in integers m; b; y 5 1 and k; n 5 2

ð1Þ

was completely solved under the assumption PðbÞ < k. As above, PðbÞ denotes the greatest prime factor of b with the convention that Pð1Þ ¼ 1. In the case b ¼ k!, the complete solution of (1) was given by Erd€os [7] for k 5 4, and by Gy€ory [12] for k 4 3. A common generalization of the above results has been obtained by Saradha [21] (case k 5 4) and Gy€ory [13] (case k 4 3); they resolved Eq. (1) for PðbÞ 4 k. The proof of Saradha [21] depends on the ingenious elementary method of Erd€ os and Selfridge whereas Gy€ory [13] utilized in his proofs the theorems of Darmon and Merel [6] and Ribet [20] on generalized Fermat equations of the form xn þ yn ¼ 2 zn . . P.), the Research supported in part by the Hungarian Academy of Sciences (K. G. and A . P.), by grants T29330, Netherlands Organization for Scientific Research (NWO) (K. G. and A . P.), T38225 (K. G.), T34981 (A . P.) of the Hungarian National Foundation T42985 (K. G. and A . P.). for Scientific Research, and by FKFP grant 3272-13=066=2001 (A

. Pinter K. Gy€ ory and A

20

Denote by pðkÞ the least prime exceeding k. For k 5 9, Saradha [21] proved that under the assumptions m > kn and n 5 3 prime, Eq. (1) has no solution even with PðbÞ 4 pðkÞ . Recently, this has been extended by Hanrot, Saradha and Shorey [15] to the case k 2 f6; 7; 8g, by Bennett [1] to the case k 4 5; n 5 2, and by Saradha and Shorey [23] to the case k 5 3; n ¼ 2. In the proofs of [15] and [1], the main ingredients include results on ternary Diophantine equations coming from the theory of Galois representations and modular forms. Let pk denote the k-th prime. We note that pk 5 pðkÞ , and pk > pðkÞ if k 5 4. For k 4 5, we give a further generalization of the above-mentioned results. Theorem 1. For k 2 f2; 3; 4; 5g, Eq. (1) with PðbÞ 4 pk and with n > 2 if k ¼ 2 has only the solutions k ¼ 2;

14m43

and

m ¼ 8;

k ¼ 3;

14m44

and

m ¼ 8;

k ¼ 4;

1 4 m 4 7;

k ¼ 5;

1 4 m 4 8:

It should be observed that for each of these solutions ð2Þ

Pðmðm þ 1Þ ðm þ k 1ÞÞ 4 pk

holds. For k ¼ n ¼ 2, (1) has obviously infinitely many solutions. In fact, Saradha [21], Hanrot, Saradha and Shorey [15], Bennett [1] and Saradha and Shorey [23] obtained their results on Eq. (1) as a consequence of a theorem concerning the more general equation ðm þ d1 Þ ðm þ dk1 Þ ¼ byn in integers m; b; y 5 1; n 5 2; k 5 3

ð3Þ

with PðbÞ 4 k, where 0 ¼ d1 < < dk1 4 k 1 are given integers. Namely, under the assumption m > kn , (3) was solved in [21] for k 5 9; n 5 3, and in [15] for k 2 f6; 7; 8g; n 5 3. A complete solution of (3) was given in [22], [23] for b ¼ 1, in [1] for k 2 f3; 4; 5g, and in [23] for n ¼ 2. For recent surveys on further related results concerning Eqs. (1) and (3), we refer to [27], [14], [25] and [26]. In case k 2 f3; 4; 5g we prove an even stronger result which implies Theorem 1 as a consequence. Theorem 2. Let k 2 f3; 4; 5g. Then Eq. ð3Þ has only the following solutions in the positive integer m and integers 0 ¼ d1 < < dk1 4 k 1, where PðbÞ 4 pk1 ; n 5 2, and n > 2 if k ¼ 3: k

m

di

k

m

di

3 3 4 4 4

1,2,3,8 1,2,4,6,16 1,2,3,4,8 1,2,3,5,9,15,24 1,2,3,6

f0; 1g f0; 2g f0; 1; 2g f0; 1; 3g f0; 2; 3g

5 5 5 5

1,2,3,4,5,6,7 1,2,3,4,5,6,8,14 1,2,3,4,5,6,24 1,2,3,4,5,6,12

f0; 1; 2; 3g f0; 1; 2; 4g f0; 1; 3; 4g f0; 2; 3; 4g

Almost Perfect Powers in Products of Consecutive Integers

21

The case k ¼ 3 of this theorem is due to Bennett [1], since then pk1 ¼ k. However, for k 5 4 we have pk1 > k, hence, for k ¼ 4 and k ¼ 5, our Theorem 2 is a generalization of theorems of Bennett [1] and Saradha and Shorey [23] on Eq. (3). We note that each solution of Eq. (3) listed in Theorem 2 satisfies Pððm þ d1 Þ ðm þ dk1 ÞÞ 4 pk1 :

ð4Þ

Equations (1) and (3) have solutions for each k. For example, for given k 5 2, every m contained in f1; 2; . . . ; pkþ1 kg is a solution of (1) with PðbÞ 4 pk . We conjecture that under the assumptions of our theorems, (2) (resp. (4)) holds for each k 5 3 and for all solutions of (1) (resp. (3)). The following theorem will play a crucial rôle in the proof of Theorem 2. Let I denote the set of primes f19; 23; 29; 31; 37; 41; 43; 47; 73g. Theorem 3. Let n be an odd prime or n ¼ 4. All the solutions of the equation mðm þ 3Þ ¼ 2 3 yn in integers m; y 5 1 and ; 5 0

ð5Þ

are m ¼ 1; 3; 6; 9; 24, except possibly in the case n 2 I, when further solutions may exist with 3 -m; ¼ 1; ¼ 0 and y odd. A similar result has been proved recently by Bennett [1] with mðm þ 3Þ replaced by mðm þ 2t Þ; t 5 0 integer, which will also be utilized in the proof of our Theorem 3 (see Lemma 5 below). Theorem 3 will be deduced from the next theorem which seems to be of independent interest. For n 5 3, consider the equation xn 2yn ¼ 3 in integers x; y:

ð6Þ

It has the trivial solution ðx; yÞ ¼ ð1; 1Þ for n odd. Theorem 4. Let n be an odd prime or n ¼ 4. Apart from the possible exceptions n 2 I, the only non-trivial solutions to Eq. (6) are with n ¼ 3; ðx; yÞ ¼ ð5; 4Þ. Theorem 4 is probably true in the exceptional cases n 2 I as well. This would mean that for these n, Eq. (5) has no solution m > 1 with 3 -m. A complete proof of Theorem 4 for all n would yield some simplifications in the proof of our Theorem 2. It is interesting to note that, as in our recent joint paper [3] with Bennett, the proofs of our theorems require a combination of virtually almost all techniques from modern Diophantine analysis, including local methods, a reciprocity theorem in cyclomotic fields, applications of the theory of linear forms in logarithms of algebraic numbers, a computational method for finding the solutions of Thue equations, the hypergeometric method of Thue and Siegel and results concerning ternary equations based on Galois representations and modular forms. Theorems 3 and 4 will be proved in Section 2. The proofs of Theorems 1 and 2 are given in Section 3. The proof of Theorem 2 requires to distinguish several subcases. For n 5 3 and n 2 = I Eq. (3) is reduced in each case either to the case k 1 or to the equation mðm þ aÞ ¼ byn with a 2 f2t ; 3g; PðbÞ 4 3, to which

22


Theorem 3 or Lemma 5 applies. The remaining cases n are treated by other (mostly classical) methods. 2. Proofs of Theorems 3 and 4 We first prove Theorem 4. In the proof we shall need some lemmas. The proofs of Lemmas 1 to 3 below are based on the theory of linear forms in logarithms. Lemma 1. Let 3 4 n 4 17 be an odd prime. All the solutions to (6) are ðx; yÞ ¼ ð1; 1Þ and, for n ¼ 3; ðx; yÞ ¼ ð5; 4Þ. Proof. This result is an immediate consequence of Lemmas 13 and 14 in [15]. Lemma 2. Let Fðx; yÞ ¼ axn ða þ 1Þyn , where a 5 1; n 5 3 and x; y are integers with jxj 6¼ jyj and xy 6¼ 0. Suppose that 2 0 < jFðx; yÞj < min ð2n 2Þa; n2 a3 : 3 n n Then x and y are necessarily of the same sign, whereby we may suppose x and y positive, whence x > y > 1; x 5 naðx yÞ; and n < 600: Proof. This is Theorem 3 in [17]. It has been proved by means of a variant of a sharp estimate of Laurent, Mignotte and Nesterenko [16] concerning linear forms in two logarithms. For any algebraic number of degree d over Q, whose minimal polynomial Q over Z is a di¼1 ðX ðiÞ Þ where the roots ðiÞ are complex numbers, we define the absolute logarithmic height of by d X 1 log jaj þ hðÞ ¼ log maxð1; jðiÞ jÞ : d i¼1 Lemma 3. Consider the linear form ¼ b2 log 2 b1 log 1 ; where b1 and b2 are positive integers and 1 ; 2 are non-zero algebraic numbers. Suppose that 1 and 2 are multiplicatively independent. Let D ¼ ½Qð1 ; 2 Þ : Q=½Rð1 ; 2 Þ : R: Let a1 ; a2 ; h; k be real positive numbers, and a real number > 1. Put ¼ log and suppose that b1 b2 h 5 D log þ þ log þ f ðKÞ þ 0:023; a2 a1 and ai 5 maxf1; j log i j log ji j þ 2Dhði Þg ði ¼ 1; 2Þ; a1 a2 5 2 ;


where f ðxÞ ¼ log

ð1 þ

23

pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi x x 1Þ x log x 3 3 log x1 þ þ log þ þ 6xðx 1Þ 2 4 x1 x1

and L ¼ 2 þ ½2h=;

K ¼ 1 þ ½kLa1 a2 :

Then we have the lower bound log jj 5 kL2 a1 a2 maxfðL 0:5Þ pffiffiffi þ log L3=2 þ L2 k maxfa1 ; a2 g þ L ; D log 2g; provided that k satisfies

pffiffiffi kU V k W 5 0

with U ¼ ðL 1Þ h;

V ¼ L=3;

1 W¼ 3

rffiffiffiffiffiffiffiffiffi 1 1 L : þ þ2 a1 a2 a1 a2

Proof. This is Theorem 1.5 in [18]. It is a variant of Theoreme 2 of Laurent, Mignotte and Nesterenko [16]. Combining several deep techniques, Bennett [2] obtained the following result.

Table 1 n

c1 ðnÞ

dðnÞ

n

c1 ðnÞ

dðnÞ

n

c1 ðnÞ

dðnÞ

19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103

9.4 13.03 17.39 17.92 21.92 25.83 26.62 30.46 34.78 39.18 39.96 44.76 48.36 52.83 58.27 62.70 67.56 73.71 78.29 79.16

15.46 17.66 29.95 30.55 32.51 36.08 33.95 40.16 35.37 48.34 55.93 43.56 54.8 48.11 54.65 49.64 60.29 62.14 50.36 60.85

107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211

83.55 84.18 89.22 100.47 105.34 111.44 112.15 122.53 123.41 129.07 134.8 139.95 146.07 151.4 152.2 163.78 164.81 170.17 170.8 183.12

50.84 58.97 77.93 72.61 71.51 79.94 77.27 85.82 89.04 81.61 93.64 82.87 87.71 83.92 91.69 84.40 91.51 104.53 110.41 124.02

223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317

195.74 201.15 202.11 207.5 213.74 214.95 226.83 233.75 240.15 246.54 247.72 254.62 260.46 261.67 274.23 289.0 294.7 296.38 302.73

112.93 116.91 100.61 102.49 105.66 95.14 115.64 113.23 119.49 124.75 134.21 119.17 116.79 118.21 129.73 124.89 130.14 130.18 134.63


24

Lemma 4. Let b > a be positive, relatively prime integers. Suppose that 19 4 n 4 317 is prime, and that c1 ðnÞ; dðnÞ are as in Table 1 and m ¼ nþ1 3 . If we have pffiffiffi pffiffiffim m b m a ec1 ðnÞ < 1 then, if p and q > 0 are integers, we may conclude that

1=n pffiffiffi pffiffiffim 1

b p

m

q ; bþ ma > 3:15 1024 ðm 1Þ2 nm1 ec1 ðnÞþdðnÞ

a q where

1 pffiffiffi pffiffiffim m c1 ðnÞþ1=20 m log bþ a e B pffiffiffi pffiffiffim C ¼ ðm 1Þ@1 A: m m c ðnÞ log b a e 1 0

Proof. See Theorem 7.1 in [2]. Proof of Theorem 4. It is easy to check that, for n ¼ 4, (6) is not solvable (mod 8). Hence it remains to consider the case when n is an odd prime. By Lemma 1, the theorem is proved for n 4 17. We now prove our theorem for the other primes n, except n 2 I. Suppose that (6) has a solution x; y with jxyj > 1. We may assume that x > 1 and y > 1. We first give a reasonable upper bound for the exponent n. Lemma 2 gives that x > y > 1, x=y 4 n=ðn 1Þ, and n < 600. In the sequel, we suppose that 300 < n < 600. By means of a continued fraction algorithm developed for determining the small solutions of Thue equations (see [19]) we deduce that Eq. (6) has no solution in positive integers x and y with maxðx; yÞ < 10500 : The CPU time is under one minute for an exponent. Set ¼ log 2 n log x=y. Then it follows from (6) that

n

y

6

ð7Þ jj 4 2 1 2 ¼ : x xn We apply now Lemma 3 to jj. In the notation of this paper we can choose 1 ¼ 2; b1 ¼ 1; 2 ¼ x=y; b2 ¼ n; ¼ 4:1; k ¼ 0:32 and set ¼ log 4:1 ¼ 1:4109 . . . : Set a1 ¼ 5:1 log 2 < 3:536 and x a2 ¼ 3:1 log þ 2 log x: y Then we deduce that 301 þ 2 log x 4 2:0001 log x: a2 4 3:1 log 300


25

We choose h as small as possible such that 1 600 þ þ f ð1 þ 0:32 L 3:536 2:0001 log xÞ þ log h 5 log 2 log x 3:536 þ 0:023: Since f is a strictly monotone decreasing function and x 5 10500 , we may take h ¼ 6:722; L ¼ 11; U ¼ 7:388; V ¼ 3:667 and sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 1 1 1 11 þ þ2 4 0:11: W< 3 3:536 2 log x 3:536 2 log x One can check that

pffiffiffi kU V k 5 W

and so Lemma 3 yields log jj 5 ð386:4 log x þ log log x þ 19:75Þ: On comparing this with (7), since x > 10500 we get n 4 387. Repeating this argument with the new bound for n, h can be chosen now so that 387 1 þ þ f ð1 þ 0:32 L 3:536 2:0001 log xÞ þ log þ 0:023: h 5 log 3:536 2 log x Then we may take h ¼ 6:283; L ¼ 10; U ¼ 6:416; V ¼ 3:334 and W as in the first step. The conditions of Lemma 3 are again fulfilled, hence Lemma 3 gives log jj 5 ð319:34 log x þ log log x þ 18:31Þ; which, on comparing with (7), implies that n 4 320. Thus, on noting that n is prime, we obtain that n 4 317. Next we apply Lemma 4 with b ¼ 2; a ¼ 1. It yields that

ffiffiffi

p

n 2 x > 1 ybðnÞ ;

y aðnÞ p ffiffiffi aðnÞ One can check that the condition m and bðnÞ are given in Table 2 below. where m and for the c1 ðnÞ occurring in 2 1 ec1 ðnÞ < 1 is fulfilled for m ¼ nþ1 3 Table 1. On the other hand, (6) implies that

ffiffiffi

p

n 2 x < 1 ;

y yn i.e. we have ynbðnÞ < aðnÞ: If n 6 2 I, then n bðnÞ > 0 and the continued fraction algorithm for small solutions completes the proof. In the proof of Theorem 3 some further lemmas will be needed. We note that in the proofs of Lemmas 5, 8 and 9 the theory of Galois representations and modular forms is involved.


26

Table 2 n

a(n)

b(n)

n

a(n)

b(n)

n

a(n)

b(n)

19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 97 101 103

:12 1046 :41 1052 :2 1064 :12 1065 :18 1072 :89 1079 :43 1079 :38 1088 :21 1093 :77 10105 :63 10109 :42 10111 :96 10122 :2 10122 :57 10132 :36 10137 :72 10158 :87 10160 :15 10166

26.83 28.47 33.86 35.99 39.21 42.39 44.6 47.41 51.49 55.64 57.25 61.98 64.37 73.39 78.83 81.99 92.46 95.82 97.4

107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 197 199 211

:93 10168 :12 10173 :61 10188 :88 10202 :11 10210 :12 10222 :32 10221 :12 10241 :14 10243 :17 10248 :65 10261 :11 10265 :53 10275 :24 10282 :24 10286 :11 10300 :39 10317 :5 10320 :15 10344

100.38 101.48 105.55 115.48 119.14 124.66 125.76 133.81 135.12 139.76 144.45 148.27 153.43 157.45 158.54 167.69 173 173.85 183.71

223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 313 317

:14 10357 :13 10367 :55 10360 :66 10369 :14 10380 :24 10376 :48 10402 :86 10410 :56 10422 :14 10434 :12 10439 :73 10441 :29 10449 :76 10450 :71 10473 :21 10491 :19 10503 :15 10514

193.79 197.63 198.85 202.63 207.46 208.99 217.78 223.34 228.23 233.08 234.52 239.96 244.06 245.54 254.73 266.48 272.31 276.86

Lemma 5. If m; t; ; ; y and n are nonnegative integers with n 5 3 and y 5 1, then the only solutions to the equation mðm þ 2t Þ ¼ 2 3 yn are with m 2 f2t ; 2t 1 ; 3 2t ; 2t 3 g: Proof. This is Theorem 1.1 in [1]. Lemma 6. The equations x3 þ 2 y3 ¼ 3z3 ;

¼ 0; 1; 2

have no solutions in non-zero integers x; y; z with gcdðx; y; zÞ ¼ 1 and jxyzj > 1. Proof. See Selmer [24]. Lemma 7. The equations x5 þ 2 y5 ¼ 3;

¼ 1; 2; 3

have no solutions in integers x; y with jxyj > 1. Proof. This is a special case of Lemma 13 in [15]. Lemma 8. If n > 3 is prime and 5 4 then the equation xn þ 2 yn ¼ 3zn in non-zero integers x; y; z with gcdðxn ; 2 yn ; 3zn Þ ¼ 1 has no solution.


27

Proof. This is a special case of Lemma 13 of [22]. Lemma 9. Suppose that n 5 7 is prime. If 5 2, then the equation xn þ 2 yn ¼ 3z2 has no solutions in non-zero coprime integers ðx; y; zÞ. Proof. This is a special case of Theorem 1.2 in [4]. Proof of Theorem 3. Let m; y 5 1, n 5 3, ; 5 0 be a solution of Eq. (5). We may assume that m > 1. We distinguish several cases. 1) If 3jm, then m ¼ 3t with t 2 N and (5) implies tðt þ 1Þ ¼ 2 32 yn :

ð8Þ

Now applying Lemma 5 to (8), we get that m 2 f3; 6; 9; 24g. It is clear that each of these m yields a solution of (5). 2) If 3 -m, then we deduce from (5) that ¼ 0 and m ¼ X n ; m þ 3 ¼ 2 Y n with positive integers X; Y, or conversely, which leads to the equation X n 2 Y n ¼ 3 in integers X; Y:

ð9Þ

This equation has the trivial solutions ðX; YÞ ¼ ð1; 1Þ, when ¼ 1 and ðX; YÞ ¼ ð1; 1Þ, when ¼ 2. These give in (5) m ¼ 1. For n ¼ 4, (9) is solvable (mod 3) only if 0 or 2 (mod 4). But it is easy to see that in this case (9) is solvable only if ¼ 2, when (9) gives X ¼ Y ¼ 1, which yields m ¼ 1. In what follows, we may and will assume that n 5 3 is a prime. If n ¼ 3, by Lemma 6 Eq. (9) has no non-trivial solution. For n 5 5 and ¼ 0, (9) does not hold. Further, for n 5 5 and 5 4, no nontrivial solution exists by Lemma 8. The same holds for n ¼ 5; ¼ 1; 2; 3, by Lemma 7. For n 5 7, Lemma 9 implies that (9) can have non-trivial solution only if ¼ 1 and 2 -Y. Hence it remains to consider the equation X n 2Y n ¼ 3;

ð10Þ

where n 5 7 prime, and X; Y 5 2 are unknown integers with Y odd. We can now apply Theorem 4 to Eq. (10), and the assertion of Theorem 3 follows. 3. Proofs of Theorems 1 and 2 We need again some lemmas. Lemma 10. If a 5 1 and n 5 3 are integers, then the equation ða þ 1Þxn ayn ¼ 1 in positive integers x; y has only the solution ðx; yÞ ¼ ð1; 1Þ. Proof. This is a consequence of Theorem 1.1 in [2]. Let ðcÞ denote Euler’s function, and Bm the m-th Bernoulli number. We recall that B2mþ1 ¼ 0 for m 51.


28

Lemma 11. Let n > 3 be a prime, and c a rational integer with ððcÞ; nÞ ¼ 1 which is not a perfect n-th power. Suppose that (i) the second factor of the class number of the n-th cyclomatic field is not divisible by n, and that (ii) n3 does not divide the numerator of any of the numbers B2tn ; t ¼ 1; 2; . . . ; ðn 3Þ=2. Then equation xn þ yn ¼ czn

ð11Þ

has no solution in relatively prime non-zero integers x; y; z with njz. Proof. This is the main result of Gandhi [9]. We note that the condition (i) is satisfied for all odd primes n < 5500; see e.g. [5]. Lemma 12. Let n > 3 and c be as in Lemma 11, and suppose that cn1 6 2 ðmod n2 Þ. Then Eq. (11) has a solution in pairwise relatively prime integers x; y; z with n -z only if n1

rn1 1 ðmod n2 Þ for every divisor r of c. Proof. See Gy€ory [11, Satz 1]. The proof depends on Eiseinstein’s reciprocity theorem in cyclomotic fields. We first prove Theorem 2. Proof of Theorem 2. The case k ¼ 3 is proved in Corollary 1.5 of [1]. It remains to prove the theorem for k ¼ 4, 5, when pk1 > k. We deduce from (3) that n þ di ¼ ai xni

ð12Þ

with positive integers ai ; xi such that Pðai Þ 4 maxfPðbÞ; k 1g; 1 4 i 4 k 1: Further, by assumption PðbÞ 4 pk1 , hence it follows that Pðai Þ < pk1 for all but at most one i. We distinguish two cases according as n 5 3 or n ¼ 2. First consider the case n 5 3. Then we may assume without loss of generality that either n is a prime or n ¼ 4. First let k ¼ 4. If in (3) d1 ; d2 ; d3 are consecutive integers, then (3) can be written in the form mðm þ 1Þðm þ 2Þ ¼ 2 3 5 yn with unknown integers ; ; 5 0. We omit on the left side one factor such that the remaining product is not divisible by 5. Then we arrive in each possible case at an equation to which Lemma 5 applies, and we get the solutions m ¼ 1; 2; 3; 4; 8 of (3). If d1 ; d2 ; d3 are not consecutive, then (3) takes the form mðm þ 1Þðm þ 3Þ ¼ 2 3 5 yn

ð13Þ

mðm þ 2Þðm þ 3Þ ¼ 2 3 5 yn :

ð14Þ

or


29

First consider (13), and assume that 5 -mðm þ 1Þ or 5 -ðm þ 1Þðm þ 3Þ. Then omitting on the left side of (13) the factor m þ 3 or m, respectively, we arrive again at an equation to which Lemma 5 applies, and we obtain the solutions m ¼ 1; 2; 3; 5; 15. If, in (14), 5 -mðm þ 2Þ or 5 -ðm þ 2Þðm þ 3Þ, we get in a similar way the solutions m ¼ 1; 2; 6. There remain the cases when in (13) 5jm þ 1, or in (14) 5jm þ 2. It is easy to check that in these cases m ¼ 9; 24 and m ¼ 3 are solutions of (13) and (14), respectively. We now deduce from (13) and (14) that 0

0

mðm þ 3Þ ¼ 2 3 yn1 ; 0

ð15Þ

0

with integer ; 5 0; y1 5 1. On applying Theorem 3 to (15), we infer that if n is not contained in I, Eqs. (13) and (14) have no more solution. In the following, we may assume that n 2 I. Continuing the proof of Theorem 2 for k ¼ 4, consider Eqs. (13) and (14) in the case when 5jm þ 1 and 5jm þ 2, respectively. Then, by Theorem 3, 3 -mðm þ 3Þ and 2jjmðm þ 3Þ. First assume that, in (13), m þ 1 is even and, in (14), m þ 2 is even. Then we get two independent systems of equations, namely m ¼ Z n ; m þ 1 ¼ 21 3 5 Y n ; m þ 3 ¼ 2X n and m ¼ 2X n ; m þ 2 ¼ 21 3 5 Y n ; m þ 3 ¼ Z n ; where X; Y; Z are coprime, positive integers. We infer in both cases that X n 1 ¼ cY n

ð16Þ

Z n 1 ¼ 2cY n ;

ð17Þ

and where c ¼ 2

2

3 5 . It is easy to check that for n 2 I and for r 2 f2; 3; 5g r n1 6 1 ðmod n2 Þ

holds. This implies that at least one of the congruences cn1 2n1 ðmod n2 Þ

ð18Þ

ð2cÞn1 2n1 ðmod n2 Þ

ð19Þ

and

cannot hold. We can now apply Lemma 12 to (16) or (17) according as (18) or (19) does not hold, and we get that n divides Y. But it is easy to verify that the condition (ii) of Lemma 11 is satisfied for each prime n 2 I. Hence Lemma 11 proves that for these primes n, Eqs. (16) and (17) have no non-trivial solutions with njY. There remain the cases when, in (13), m þ 1 is odd and, in (14), m þ 2 is odd. Then we deduce from (13) and (14) that m ¼ 2X n ; m þ 1 ¼ 3 5 Y n ; m þ 3 ¼ Z n

30


and m ¼ Z n ; m þ 2 ¼ 3 5 Y n ; m þ 3 ¼ 2X n ; whence we get two (independent) system of equations: 2X n þ 1 ¼ 3 5 Y n ¼ Z n 2

ð20Þ

2X n 1 ¼ 3 5 Y n ¼ Z n þ 2:

ð21Þ

and We prove that for n 2 I, (20) and (21) have no non-trivial solution. For each fixed n, we find primes of the form p ¼ 2kn þ 1 for which the first equation of (20) is impossible modulo p, apart possibly from some exceptional pairs ð; Þ. For each of the remaining pairs ð; Þ we find an appropriate prime p1 ¼ 2k1 n þ 1 for which (20) has no solution mod p1 . To illustrate our algorithm we give an example. Set n ¼ 37. In the first step we consider the first equation of (20) modulo 149, 223, 593, 1259, 1481 or 1999 and we get that apart from the pairs ð; Þ ¼ ð1; 0Þ and ð31; 24Þ, the equation under consideration has no solution. Then for the remaining pairs it is easy to check the impossibility of (20) modulo 149. We shall refer to this approach as ‘the local method’ (cf. e.g. [3]). By means of this method one can prove that (20) and (21) have no solutions in ðX; Y; ZÞ. This completes the proof of Theorem 2 for n 5 3 and k ¼ 4. Let now k ¼ 5. 1) If in (3) d1 ; d2 ; d3 ; d4 are consecutive integers, then we may write (3) in the form mðm þ 1Þðm þ 2Þðm þ 3Þ ¼ 2 3 5 7 yn where ; ; ; 5 0 are unknowns. We omit on the left side one factor such that the remaining product is not divisible by 7. Then we arrive in each possible case at an equation to which Theorem 2 with k ¼ 4 applies, and we get the solutions m ¼ 1; 2; 3; 4; 5; 6 and 7. 2) Next suppose that in (3) d1 ; . . . ; d4 are not consecutive. Then d4 ¼ 4. If 7jm or m þ 4, then omitting the factor divisible by 7 we get to the case k ¼ 4. Then we obtain that m ¼ 3, 14, or m ¼ 3, 24, or m ¼ 3, according as fd2 ; d3 ; d4 g ¼ f1; 2; 4g or fd2 ; d3 ; d4 g ¼ f1; 3; 4g, or fd2 ; d3 ; d4 g ¼ f2; 3; 4g. We assume now that 7 -mðm þ 4Þ. At most one of m þ 1; m þ 2; m þ 3 can be divisible by 7. Hence we deduce from (3) that one of the following equations hold: 0

0

0

0

0

0

0

0

0

mðm þ 2Þðm þ 4Þ ¼ 2 3 5 yn1 ; mðm þ 1Þðm þ 4Þ ¼ 2 3 5 yn1 ; mðm þ 3Þðm þ 4Þ ¼ 2 3 5 yn1 ; where 0 ; 0 ; 0 are non-negative integers, and y1 is a positive integer.

ð22Þ ð23Þ ð24Þ


31

a) First consider Eq. (22). Omitting one factor on the left side so that the remaining product is not divisible by 5, we can apply Lemma 5 to each possible remaining equation. We get the solutions m ¼ 1; 2; 4; 6; 8 or m ¼ 1; 2; 4; 6 of (3), according as fd2 ; d3 ; d4 g ¼ f1; 2; 4g or f2; 3; 4g. b) Next consider Eq. (23). If 5 -m, then omitting the factor m þ 1 or m þ 4 on the left hand side, we can apply Lemma 5 and we get m ¼ 1; 2; 4 as further solutions of (3) if fd2 ; d3 ; d4 g ¼ f1; 3; 4g. If 5jm, we infer from (23) that 00

00

ðm þ 1Þðm þ 4Þ ¼ 2 3 yn2 ; with integers 00 ; 00 5 0; y2 5 1. Then, by Theorem 3, we get the solution m ¼ 5 of (3) for fd2 ; d3 ; d4 g ¼ f1; 2; 4g. Further, we have still to consider the cases when n 2 I; 3 -m þ 1; 00 ¼ 1 and y2 odd. We distinguish two subcases: If 2jm, then m þ 4 is also even. But then 2km þ 4 and so 2km. Hence we deduce from (23) that 0

0

m ¼ 2 3 5 X n ; m þ 1 ¼ Y n ; m þ 4 ¼ 2Z n or 0

0

m ¼ 3 5 X n ; m þ 1 ¼ 2Z n ; m þ 4 ¼ Y n ; according as 2jm or 2 -m, where X; Y; Z are coprime positive integers. Hence we infer that 0

0

2 3 5 X n þ 1 ¼ Y n ¼ 2Z n 3

ð25Þ

or 0

0

3 5 X n þ 4 ¼ Y n ¼ 2Z n þ 3:

ð26Þ

Later, we shall solve these systems of equations for n 2 I. c) Consider now Eq. (24). If 5jm or m þ 3, then omitting the factor divisible by 5 we can again apply Lemma 5 and we get the new solutions m ¼ 5 or m ¼ 5; 12 of (3), according as fd2 ; d3 ; d4 g ¼ f1; 3; 4g or f2; 3; 4g. If 5 -mðm þ 3Þ, then we omit the factor m þ 4 and we deduce that 00

00

mðm þ 3Þ ¼ 2 3 yn2 :

ð27Þ

with integers 00 ; 00 5 0; y2 5 1. Then in view of Theorem 3, we get m ¼ 6 as a new solution of (3) for fd2 ; d3 ; d4 g ¼ f1; 3; 4g. Further, we have still to deal with the cases when n 2 I and 3 -m; 00 ¼ 1; y2 odd. If 2jm, then we infer that 2km and 2km þ 4. Thus (24) gives that 0

0

m ¼ 2Z n ; m þ 3 ¼ Y n ; m þ 4 ¼ 2 3 5 X n or 0

0

m ¼ Y n ; m þ 3 ¼ 2Z n ; m þ 4 ¼ 3 5 X n ; according as 2jm or 2 -m, where X; Y; Z are again coprime positive integers. Therefore, it follows that 0

0

2 3 5 X n 1 ¼ Y n ¼ 2Z n þ 3

ð28Þ


32

or 0

0

3 5 X n 4 ¼ Y n ¼ 2Z n 3:

ð29Þ

Now we solve the systems of Eqs. (25), (26), (28) and (29) for n 2 I. We may assume that in each case 0 4 0 ; 0 4 n 1. By the local method, Eqs. (25) and (28) have no solutions, and similarly, if ð0 ; 0 Þ 6¼ ð1; 0Þ then (26) and (29) possess no solutions. (For example, for n ¼ 37 we can work with the same primes as above). If ð0 ; 0 Þ ¼ ð1; 0Þ then (26) and (29) imply 3X n þ 4 ¼ Y n ¼ 2Z n þ 3 and 3X n 4 ¼ Y n ¼ 2Z n 3, respectively. From Lemma 10 one can deduce that all the solutions are ðX; Y; ZÞ ¼ ð1; 1; 1Þ and ð1; 1; 1Þ, respectively, which do not provide any solution for (3). This proves Theorem 2 for n 5 3 and k ¼ 5. Assume now that in (3) n ¼ 2. Then, for k ¼ 4, 5, and for each given tuple 0 ¼ d1 < < dk1 4 k 1 we obtain the equations ðm þ d1 Þðm þ d2 Þðm þ d3 Þ ¼ 2 3 5 y2 ; and ðm þ d1 Þðm þ d2 Þðm þ d3 Þðm þ d4 Þ ¼ 2 3 5 7 y2 ; where ; ; ; 2 f0; 1g. Employing the computational package MAGMA (for the theoretical background see [10] and [28]) one can solve all these elliptic equations and we find only the positive solutions m listed already in the statement of Theorem 2. Proof of Theorem 1. For k ¼ 2; 3 and 4, Theorem 2 implies immediately Theorem 1 with the choice di ¼ i 1 for i ¼ 1; . . . ; k. Next consider the case k ¼ 5, when pk ¼ 11. On the left hand side of (1) at most one factor is divisible by 11. If, in (1), 11 -b, then omitting the factor m þ 4 and using (12), we get to the case k ¼ 4 of Theorem 2 and the assertion easily follows. In the case 11jb we use again (12) and omit the factor on the left side which is divisible by 11 to arrive at an equation of the form (3) with k ¼ 5. Now Theorem 2 applies and completes the proof. Acknowledgements. The authors thank the referees for their useful and helpful remarks.

References [1] Bennett MA (2004) Products of consecutive integers. Bull London Math Soc (to appear) [2] Bennett MA (2001) Rational approximation to algebraic numbers of small height: the Diophantine equation jaxn byn j ¼ 1: J Reine Angew Math 535: 1–49 (2004) On the diophantine equation 1k þ 2k þ þ xk ¼ yn . [3] Bennett MA, Gy€ ory K, Pinter A Compositio Math (to appear) [4] Bennett MA, Skinner C (2004) Ternary Diophantine equations via Galois representations and modular forms. Canadian J Math 56: 23–54 [5] Borevich ZI, Shafarevich IR (1972) Number Theory, 2nd edn. Moscow: Nauka (in Russian) [6] Darmon H, Merel L (1997) Winding quotient and some variants of Fermat’s Last Theorem. J Reine Angew Math 490: 81–100 [7] Erd€os P (1951) On a diophantine equation. J London Math Soc 26: 176–178 [8] Erd€os P, Selfridge JL (1975) The product of consecutive integers is never a power. Illionis J Math 19: 292–301 [9] Gandhi JM (1972) On generalized Fermat’s last theorem II. J Reine Angew Math 256: 163–167


33

[10] Gebel J, Peth€o A, Zimmer HG (1994) Computing integral points on elliptic curves. Acta Arith 68: 171–192 € ber die diophantische Gleichung xp þ yp ¼ czp . Publ Math Debrecen 13: [11] Gy€ory K (1966) U 301–305 [12] Gy€ory K (1997) On the diophantine equation ðnkÞ ¼ xl . Acta Arith 80: 289–295 [13] Gy€ory K (1998) On the diophantine equation nðn þ 1Þ ðn þ k 1Þ ¼ bxl : Acta Arith 83: 87–92 [14] Gy€ory K (1999) Power values of products of consecutive integers and binomial coefficients. In: Kanemitsu S, Gy€ ory K (eds) Number Theory and Its Applications, pp 145–156. Amsterdam: Kluwer [15] Hanrot G, Saradha N, Shorey TN (2001) Almost perfect powers in consecutive integers. Acta Arith 99: 13–25 [16] Laurent M, Mignotte M, Nesterenko Yu (1995) Formes lineaires en deux logarithmes et determinants d’interpolation. J Number Theory 55: 285–321 [17] Mignotte M (1996) A note on the equation axn byn ¼ c. Acta Arith 75: 287–295 [18] Mignotte M (1998) A corollary to a theorem of Laurent-Mignotte-Nesterenko. Acta Arith 86: 101–111 [19] Peth€o A (1987) On the resolution of Thue inequalities. J Symbolic Comput 4: 103–109 [20] Ribet KA (1997) On the equation ap þ 2 b p þ cp ¼ 0. Acta Arith 79: 7–16 [21] Saradha N (1997) On perfect powers in products with terms from arithmetic progression. Acta Arith 82: 147–172 [22] Saradha N, Shorey TN (2001) Almost perfect powers in arithmetic progression. Acta Arith 99: 363–388 [23] Saradha N, Shorey TN (2003) Almost squares and factorization in consecutive integers. Compositio Math 138: 113–124 [24] Selmer E (1951) The diophantine equation ax3 þ by3 þ cz3 ¼ 0. Acta Math 85: 205–362 [25] Shorey TN (2002) Powers in arithmetic progression. In: Wu }stholz G (ed) A Panorama in Number Theory, pp 325–336. Cambridge: University Press [26] Shorey TN (2002) Powers in arithmetic progression (II). In: New Aspects of Analytic Number Theory, pp 202–214. Kyoto: Kyoto Univ [27] Shorey TN, Tijdeman R (1997) Some methods of Erd€ os applied to finite arithmetic progressions. In: Graham RL (ed) The Mathematics of Paul Erd€ os I, pp 251–267. Berlin Heidelberg New York: Springer [28] Stroeker RJ, Tzanakis N (1994) Solving elliptic diophantine equations by estimating linear forms in elliptic logarithms. Acta Arith 67: 177–196 Authors’ address: Number Theory Research Group of the Hungarian Academy of Sciences, University of Debrecen, Institute of Mathematics, P.O. Box 12, H-4010 Debrecen, Hungary, e-mails: gyory@ math.klte.hu, [email protected]