On sets of integers whose shifted products are powers

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1 On sets of integers whose shifted products are powers C.L. Stewart 1 Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, Canada, NL 3G1 Abstract Let N be a positive integer and let A be a subset of {1,...,N} with the property that aa + 1 is a pure power whenever a and a are distinct elements of A. We prove that A, the cardinality of A, is not large. In particular, we show that A log N /3 log log N 1/3. Key words: pure powers, extremal graph theory, linear forms in logarithms 1991 MSC: primary 11B75, 11D99; secondary 05D10, 05C38 1. Introduction Diophantus initiated the study of sets of positive rational numbers with the property that the product of any two of them is one less than the square of a rational number. For instance, he found the set { 1, 33, 17, } Fermat was apparently the first to find a set of four positive integers with the property that the product of any two of the integers plus one is a square. His example was {1, 3, 8, 10}. Dujella [6] has recently shown that there are no sets of six integers and that there are only finitely many sets of five positive integers with the above property. In [3] Bugeaud and Dujella considered the analogous property with the squares replaced by the set of k-th powers. For each integer k larger than one let Ck denote the largest cardinality of a set of integers for which the product of any two of the integers plus one is a k-th power of an integer. They proved that C3 7, C4 5, Ck 4 for 5 k 176 and that Ck 3 for k > address: cstewart@uwaterloo.ca C.L. Stewart. 1 This research was supported in part by the Canada Research Chairs Program and by Grant A358 from the Natural Sciences and Engineering Research Council of Canada. Preprint submitted to Elsevier Science 7 August 007

2 Let V denote the set of pure powers, that is, the set of positive integers of the form x k with x and k positive integers and k > 1. In [9], Gyarmati, Sárközy and Stewart asked how large a set of positive integers A can be if aa +1 is in V whenever a and a are distinct elements of A. They conjectured that there is an absolute bound for A and recently Luca [1] has shown that this follows as a consequence of the abc conjecture. While Gyarmati, Sárközy and Stewart could not establish an absolute bound for A they were able to show that A cannot be very dense. In particular, they proved that if N is a positive integer and A is a subset of {1,..., N} with the property that aa +1 is in V whenever a and a are distinct integers from A then, for N sufficiently large, log N A < 340 log log N. 1 The proof of 1 depends on a gap principle, the result of Dujella and two results from extremal graph theory. By means of an improved gap principle, which depends on the work in [3], Bugeaud and Gyarmati [4] proved that for N sufficiently large, 1 may be replaced with A < log N. log log N Recently Luca [1] introduced estimates for linear forms in the logarithms of rational numbers into the mix to efficiently treat the large powers which might occur and as a consequence he has shown that there is a positive number c 0 such that for N sufficiently large 3/ log N A < c 0. log log N The linear forms which Luca employs consist of 4 terms. In [8] Gyarmati and Stewart introduced a modification of Luca s argument which allows one to deal with linear forms in only terms. They were able to prove that there is a positive number c 1 such that for N sufficiently large A < c 1 log N. 3 Somewhat earlier, Dietmann, Elsholtz, Gyarmati and Simonovits [5] had proved that for N sufficiently large A < 8000 log N loglog N, 4 under the additional assumption that any two terms of the form aa + 1 are coprime.

3 Our objective in this note is to establish the following improvement of 3 and 4. Theorem. Let N be an integer with N 3 and let A be a subset {1,...,N} with the property that aa + 1 is in V whenever a and a are distinct integers from A. There exists an effectively computable positive real number c such that A < clog N /3 log log N 1/3. 5 We shall follow Luca [1] and Gyarmati and Stewart [8] by making use of estimates for linear forms in the logarithms of rational numbers in order to treat the large powers. For powers of intermediate size we appeal to an estimate for simultaneous linear forms in the logarithms of algebraic numbers due to Loxton [11]. As in [9] we shall also employ a gap principle and results from extremal graph theory.. Preliminary Lemmas Lemma 1 There is no set of six positive integers {a 1,...,a 6 } with the property that a i a j + 1 is a square for 1 i < j 6. Proof. This is Theorem of [6]. Lemma Let n and r be integers with 3 r n. Let G be a graph on n vertices with at least r r 1 n edges. Then G contains a complete subgraph on r edges. Proof. This follows from Turán s graph theorem, see [15] or Lemma 3 of [4]. Lemma 3 Let G be a graph with n > 1 vertices and e edges and suppose that e > 1 n 3/ + n n 1/. Then G contains a cycle of length 4. Proof. This is a special case of Theorem.3, Chapter VI of [] and is due to Kövári, Sós and Turán [10]. 3

4 For the proof of 3 Gyarmati and Stewart [8] modified Lemma 3 to treat graphs whose edges are coloured and which do not contain cycles of length four consisting of two monochromatic paths of length two. We shall need to deal with a slightly more complicated situation where we replace a cycle of length four with several monochromatic paths of length two which have the same starting vertex and the same finishing vertex. The result which we shall establish below is a generalization of Lemma.4 of [8] and is proved by a minor alteration of the proof of Theorem.3, Chapter VI, of [] which gives a bound for the number of edges of a graph on n vertices which does not contain a Kl,, a bipartite graph which contains all edges between a vertex set of size l and a vertex set of size. Lemma 4 Let G be a graph with n vertices and e edges coloured by k different colours. Suppose that G does not contain distinct vertices a, c, b 1,, b l, with the property that for j = 1,..., l the edges ab j and cb j are in G and of the same colour. Then e l 1kn 3 1/ + kn. Proof. We first count the number of monochromatic paths of G of length. Let a 1,..., a n be the vertices of G and let d i,j denote the number of edges emanating from a i of colour j. The number of monochromatic paths of G of length is exactly n k di,j. i=1 j=1 since for every pair However this number is less than or equal to l 1 n a, c there exist at most l 1 vertices b 1,...,b l 1 such that ab j and cb j have the same colour for j = 1,...,l 1. Thus Since n i=1 kj=1 d i,j = e, n k di,j n l 1. i=1 j=1 n k i=1 j=1 d i,j 1 e l 1nn. By the Cauchy-Schwarz inequality, hence ni=1 kj=1 d i,j kn e l 1 nn 1, e kne l 1n n 1k/. 4

5 Thus whence e kn + k n + 4l 1n n 1k 1/ /4 e l 1n 3 k 1/ + kn. Lemma 5 Let k be an integer with k and let a 1, a, a 3 and a 4 be positive integers with a 1 < a 3 and a < a 4. If a 1 a + 1, a 1 a 4 + 1, a a 3 + 1, a 3 a are k-th powers then a 3 a 4 > a 1 a k 1. Proof. This follows from the proof of Theorem 1 of [7]. For any non-zero rational number α, where α = a/b with a and b coprime integers, we put Hα = max{ a, b }. Further, for any real number x let x denote the smallest integer greater than or equal to x. Recall that non-zero rational numbers α 1,...,α t are said to be multiplicatively dependent if there exist integers l 1,..., l t, not all zero, for which α l 1 1 α lt t = 1. They are said to be multiplicatively independent otherwise. Our next lemma will be used in the proof of our main theorem to produce a set of multiplicatively independent rational numbers which satisfy the hypotheses of Lemma 9. Lemma 6 Let ε be a positive real number with 0 < ε < 1 and let t be an integer with t. Let α 1,...,α k be non-zero rational numbers with the property that any t of them are multiplicatively dependent. Suppose that t 1 t 1 k t 1 +. ε Then there exist distinct integers i 0,...,i t for which H αi0 α i1 Hα i Hα it ε/t 1. Proof. This follows from Theorem 1 of [14] on replacing ε with ε/t 1. Lemma 7 Let B 1 and B be non-zero integers and let α 1 and α be positive rational numbers with Hα 1 A 1 and Hα A. Put Λ = B 1 log α 1 + 5

6 B log α. There exists an effectively computable positive number c such that if Λ 0 then Λ > exp c loga 1 loga 1 + log B loga 1 + B 1. loga Proof. This follows from Theorem. of Philippon and Waldschmidt [13]. Lemma 8 Let M be an integer with M 16 and let a 1, a, a 3 and a 4 be distinct integers with M 1/ a i M for i = 1,, 3, 4 and put a 5 = a 1. Suppose that x 1, x, x 3, x 4 and b 1, b, b 3, b 4 are positive integers and that a i a i = x b i i, for i = 1,, 3, 4. Put b = min{b 1, b, b 3, b 4 } and t = max 1 i 4 {b i b}. There exists an effectively computable positive number C such that t + 1 > C b log M. 6 Proof. Following the proof of Lemma 3.1 of [1] we observe that a 1 a a 3 a 4 = x b 1 1 1x b = x b 1x b hence or, equivalently, x b 1 1 1x b x b 1x b = 0 x b 1 1 x b 3 3 x b x b 4 4 = x b x b 3 3 x b x b Since x b x b 3 3 x b x b 4 4 = a 1 a 3 a a 4 and since the a i s are distinct we find that x b 1 1 x b 3 3 x b x b Thus, if we put x x 4 Λ = b log + log x 1 x 3 x b b x b 4 b 4 x b 1 b 1 x b 3 b 3, 8 we see that Λ 0. We may assume, without loss of generality, that x b 1 1 x b i i for i =, 3, 4. Thus, by 7, x b x b 4 4 x b 1 1 x b x b

7 Since a 3 and a 4 are at least M 1/ in size it follows that and, therefore, by 8 and 9, x b 3 3 > M, e Λ 1 M. Notice that if y is a real number and e y 1 1/8 then y < 1/. Further e y 1 y / for y < 1/ and so, since M 16, Λ < 4 M, hence log Λ < 1 log M. 10 We now apply Lemma 7 with α 1 = x x 4 /x 1 x 3, α = x b b x b 4 b 4 / x b 1 b 1 x b 3 b 3, B 1 = b and B = 1. Note that log x i log M/b for i = 1,, 3, 4. Therefore log Hα 1 log x 1 + log x + log x 3 + log x 4 8 log M b and log Hα 6t + 1 log M. b Put log A 1 = 8 log M/b and log A = 6t + 1 log M/b. Let C 1, C,... denote effectively computable positive numbers. By Lemma 7 t + 1log M b log Λ > C log 1 + b t + 1 log M and so, by 10, b t + 1 log M 1 + log 1 + b 1 < C. t + 1 log M Therefore and Lemma 8 now follows. b t + 1 log M < C 3, In 1986 Loxton gave an estimate for simultaneous linear forms in the logarithms of algebraic numbers. We shall state his result for the special case when the algebraic numbers are rationals. Let n and t be integers with n 7

8 and t 1 and let α 1,...,α n be non-zero multiplicatively independent rational numbers. Let b i,j for i = 1,...,t and j = 1,...,n be rational numbers and suppose that the matrix b i,j formed by the b ij s has rank t. Put Λ i = b i,1 log α b i,n log α n, for i = 1,...,t, where the logarithms are principal. We shall suppose that Hα j A j with A j 4 for j = 1,..., n and that Hb i,j B with B 4 for i = 1,..., t and j = 1,...,n. Put Ω = log A 1 log A n. Building on an estimate of Baker [1] for the case t = 1, Loxton [11] proved the following result. Lemma 9 max Λ i > exp CΩ log Ω 1/t logbω, 1 i t where C = 16n 00n. 3. Proof of Main Theorem Let A be a subset of {1,...,N} with the property that aa +1 is in V whenever a and a are distinct integers from A. We may suppose that A > log N /3 log log N 1/3, 11 since otherwise our result holds. Let c 1, c,... denote positive numbers which are effectively computable. We shall suppose that There is an integer m with /3 1 log N > log log N 1 m loglog N/ log, 13 log such that A has more than A 3/loglog N/ log / log elements from { m, m + 1,..., m+1 1}. Let us denote the set of these elements by A m and put n = A m and M = m+1. Then, by 11, for N > c 1, Further, by 11 and 14, n > M > n > A log log N. 14 log N/3 15 log log N /3 8

9 and since 1 holds, M > Form the complete graph G whose vertices are the elements of A m. Associate to each edge between two vertices a and a the smallest prime p for which aa + 1 is a perfect p-th power. For any real number x let [x] denote the greatest integer less than or equal to x. We next define a sequence t 1, t,... inductively by putting t 1 = 1 + [ log M log log M 1/3] 17 and [ ] Ct t i+1 = t i 1 + i, 18 log M for i = 1,, 3,..., where C is the positive number given in Lemma 8. By 15, 17 and 18 we see that for N greater than c, as we shall suppose, the sequence t 1, t,... is strictly increasing. We are now in a position to give a colouring of G. If the edge between a and a is associated with a prime p which is smaller than t 1 we colour the edge with the prime p. On the other hand if p t 1 we colour the edge with the integer t i for which t i p < t i+1. For any real number x let πx denote the number of prime numbers less than or equal to x. Notice that the total number of colours of G is bounded from above by πt 1 + h 19 where h counts the number of colours t i for which t i is smaller than log M/ log. In order to estimate h we first estimate the number of t i s in each interval of the form k, k+1 ]. By 17 we need only consider such intervals for which k + 1 > 1 3 log loglog M log log M. 0 Further, if a and a are distinct elements of A m then aa + 1 is at most M and so we may also suppose that k < log/ log log M/ log. By 18, if t i is in k, k+1 ] then t i+1 t i > C k / log M and so, by 15 and 0, we see that for N greater than c 3, t i+1 t i > Ck log M. 9

10 Thus the number of t i s in the interval k, k+1 ] is at most 1+ log M/C k. We may assume that N exceeds c 1, c and c 3, since the result is immediate otherwise. Therefore h 1 + log M 1 3 log loglog M log log M 1<k< 1 log C log M k log log and so h c 4 log M/ log log M 1/3. 1 Thus, by 19, 1 and the prime number theorem, the number of colours of G is at most c 5 log M/ log log M /3. By Lemma, if the number of edges of G coloured with exceeds /5n then there is a complete subgraph of G on 6 vertices coloured with and this is impossible by Lemma 1. Therefore the number of edges of G coloured with something other than is at least n /5n = n /10 n/. Let G 1 be the subgraph of G consisting of the vertices of G together with the edges of G which are coloured with a prime p for which log M 3/10 < p < t 1 and let G be the subgraph of G consisting of the vertices of G together with the edges of G which are coloured with an integer t i with 1 i h or with a prime p satisfying < p log M 3/10. We shall suppose first that G has at least n /0 n/ edges. The number of colours of G is at most πlog M 3/10 + h. Thus by 1 the number of colours of G is at most c 6 log M/ log log M 1/3. Accordingly, there is a colour of G which appears on n /0 n/ c 6 log M loglog M 1/3 different edges. Since M N we see from 14 that if A > c 7 log N log log N 1/3, 3 then there is a colour which appears on more than n 3/ + n n 1/ / edges and so, by Lemma 3, G contains a monochromatic cycle of length 4. By 16, 18 and Lemma 8 there is no cycle of length 4 with a colour t i in G, hence in G, and thus G must contain a cycle of length 4 coloured with a prime 10

11 p, satisfying. In particular there exist integers a 1, a, a 3 and a 4 with a 1 < a 3 and a < a 4 for which a 1 a + 1, a 1 a 4 + 1, a a and a 3 a are p-th powers. Thus, by Lemma 5, a 3 a 4 > a 1 a. 4 But a 1, a, a 3 and a 4 are in { m,..., m+1 1} and so a 3 a 4 < m+ a 1 a, which contradicts 4. Thus either the supposition 3 is false, in which case our result follows, or G has fewer than n /0 n/ edges. We may assume the latter possibility and therefore G 1 contains at least n /0 edges. The number of colours of G 1 is at most the number of colours of G hence is at most c 5 log M/ log log M /3. Since N > M the number of colours of G is at most c 5 log N/ log log N /3. Thus we may apply Lemma 4 to the graph G 1. Since G 1 has at least n /0 edges we see by 14, that if A > c 8 log N log log N 1/3, 5 then there are integers a 1, a,...,a 870 with the property that a 1 a j and a a j are edges of G 1 and are of the same colour for j = 3,...,870. Put γ j = a 1a j + 1 for j = 3,...,870. a a j + 1 We claim that we can find a subset of {γ 3,..., γ 870 } consisting of 4 multiplicatively independent numbers. If this is not so then we may apply Lemma 6 with ε = 1/ and t = 4. Since = 868 there exist distinct integers i 0,...,i 4 for which But so Notice that and so H γi0 γ i1 H γi0 γ i1 Hγ i Hγ i3 Hγ i4 1/6. Hγ is < M, for s =, 3, 4, H γi0 γ i1 γ i0 γ i1 = a 1a i0 + 1a a i1 + 1 a a i0 + 1a 1 a i1 + 1, < M. 6 a 1 a a i0 a i1 gcda 1 a i0 + 1a a i1 + 1, a a i0 + 1a 1 a i

12 But a 1 a i0 + 1a a i1 + 1 a a i0 + 1a 1 a i1 + 1 = a 1 a a i0 a i1 and therefore gcda 1 a i0 + 1a a i1 + 1, a a i0 + 1a 1 a i1 + 1 maxa 1, a maxa i0, a i1. Accordingly H γi0 γ i1 mina 1, a mina i0, a i1 and, since a 1, a, a i0 and a i1 are all at least M 1/, H γi0 γ i1 M. 7 Since 6 and 7 are incompatible there exists a subset of {γ 3,...,γ 870 } consisting of 4 multiplicatively independent numbers. Without loss of generality we may suppose that γ 3,...,γ 6 are multiplicatively independent. We have with p 3, p 4, p 5, p 6 primes satisfying γ i = x p i i, for i = 3, 4, 5, 6, log M 3/10 < p i log M /3 log log M 1/3, for i = 3, 4, 5, 6. Further x 3, x 4, x 5 and x 6 are multiplicatively independent since γ 3, γ 4, γ 5 and γ 6 are multiplicatively independent. We have a 1 a j + 1 = y p j j, 8 a a j + 1 = z p j j 9 and x j = y j /z j for j = 3, 4, 5, 6. Then, as in the proof of Lemma 8, so Again, since y p z p j j distinct, we find that a 1 a a 3 a j = y p 3 3 1z p j j 1 = y p j j 1z p y p 3 3 z p j j y p j j z p 3 3 = y p z p j j y p j j z p y p j j z p 3 3 = a 1 a a 3 a j and since the a i s are y3 p3 zj pj z 3 y j 1

13 For j = 4, 5, 6, put and note, by 31, that Λ j 0. Put Λ j = p 3 log x 3 p j log x j, g j = x p 3 3 x p j j and G j = maxg j, gj 1, for j = 4, 5, 6. It follows from 30, as in the proof of Lemma 8, that G j 1 M, for j = 4, 5, 6. But G j = e Λ j for j = 4, 5, 6. As before, we remark that if y is a real number and e y 1 1/8 then y < 1/. Therefore, by 16, Λ j < 1/ for j = 4, 5, 6. Since M > 16 and since e y 1 y / for y < 1/ we have Λ j < 4 M, hence for j = 4, 5, 6. log Λ j < 1 log M, 3 We now apply Lemma 9 with α j = x j+ for j = 1,, 3, 4 and with b j = p j+ for j = 1,, 3, 4. Note that and so, by 8 and 9, log Hα j = log Hx j+ logmax y j+, z j+ log Hα j log M p j+, 33 for j = 1,, 3, 4. Further B = max{p 3, p 4, p 5, p 6 } and so B log M log. 34 Furthermore, the matrix p 3 p p 3 0 p 5 0 p p 6 13

14 has rank 3. Thus, by 3, 33, 34 and Lemma 9 with t = 3, hence log M 4 1/3 log M < c 9 log log M log log M p 3 p 4 p 5 p 6 p 3 p 4 p 5 p 6 < c 10 log Mlog log M 4, 35 But p 3, p 4, p 5 and p 6 exceed log M 3/10 and this is incompatible with 35 for M larger than c 11, hence by 15, for N larger than c 1. Thus 5 does not hold and the result follows. References [1] A. Baker, The theory of linear forms in logarithms, A. Baker and D.W. Masser eds., Transcendence Theory, Advances and Applications, Academic Press, 1977, pp [] B. Bollobás, Extremal Graph Theory, London Mathematical Society Monographs No. 11, Academic Press, London, New York, San Francisco, [3] Y. Bugeaud and A. Dujella, On a problem of Diophantus for higher powers, Math. Proc. Camb. Phil. Soc [4] Y. Bugeaud and K. Gyarmati, On generalizations of a problem of Diophantus, Illinois J. Math [5] R. Dietmann, C. Elsholtz, K. Gyarmati and M. Simonovits, Shifted products that are coprime pure powers, J. Comb. Theory, Ser. A [6] A. Dujella, There are only finitely many Diophantine quintuples, J. reine angew. Math [7] K. Gyarmati, On a problem of Diophantus, Acta Arith [8] K. Gyarmati and C.L. Stewart, On powers in shifted products, Glasnik Mat., to appear. [9] K. Gyarmati, A. Sárközy and C.L. Stewart, On shifted products which are powers, Mathematika [10] T. Kövári, V. Sós and P. Turán, On a problem of K. Zarankiewicz, Colloq. Math [11] J.H. Loxton, Some problems involving powers of integers, Acta Arith [1] F. Luca, On shifted products which are powers, Glasnik Mat

15 [13] P. Philippon and M. Waldschmidt, Lower bounds for linear forms in logarithms, New Advances in Transcendence Theory, ed. A. Baker, Cambridge University Press, 1988, pp [14] C.L. Stewart, On heights of multiplicatively dependent algebraic numbers, Acta Arith., to appear. [15] P. Turán, On an extremal problem in graph theory in Hungarian, Mat. Fiz. Lapak