EFFECTIVE RESOLUTION OF DIOPHANTINE EQUATIONS OF THE FORM u n +u m = wp z1

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1 EFFECTIVE RESOLUTION OF DIOPHANTINE EQUATIONS OF THE FORM u n +u m = wp z1 1 pzs s arxiv: v1 [math.nt] 16 Apr 016 ISTVÁN PINK AND VOLKER ZIEGLER Abstract. Let u n be a fixed non-degenerate binary recurrence sequence with positive discriminant, w a fixed non-zero integer and p 1,p,...,p s fixed, distinct prime numbers. In this paper we consider the Diophantine equation u n + u m = wp z 1 1 pzs s and prove under mild technical restrictions effective finiteness results. In particular we give explicit upper bounds for n, m and z 1,...,z s. Furthermore, we provide a rather efficient algorithm to solve Diophantine equations of the described type and we demonstrate our method by an example. 1. Introduction Recently several authors considered the problem to find all indices n and m and exponents z such that u n +u m = z, where u n is a fixed recurrence. In particular Bravo and Luca considered the cases, where u n is the Fibonacci sequence [4] and the Lucas sequence [5] respectively. Also the case where u n is the generalized k-fibonacci sequence has been considered independently by Bravo, Gómez and Luca[3] and Marques[9]. In[] Bertók, Hajdu, Pink and Rábai solved completely equations of the form u n = a +3 b +5 c, where u n is one of the Fibonacci, Lucas, Pell and associated Pell sequences, respectively. In the present paper we aim to generalize the results due to Bravo and Luca [5, 4] and consider the more general Diophantine equation 1) u n +u m = wp z1 1 pzs s in non-negative integer unknowns n,m,z 1,...,z s, where {u n } n 0 is a binary nondegenerate recurrence sequence, p 1,...,p s are distinct primes and w is a non-zero integer with p i w for 1 i s. For reasons of symmetry we assume that n m. Although the ideas of Bravo, Gómez, Luca and Marques [5, 4, 3, 9] can be easily extended to this case, we provide two new aspects treating the title Diophantine equation. Firstly, the main argument due to Bravo et.al. is to consider two linear forms in complex) logarithms and obtain by a clever trick upper bounds for n,m and z. In our approach we replace one of the two linear forms in complex logarithms by linear forms in p i -adic logarithms for every prime p i with 1 i s. The advantage in doing this is that instead of considering a linear form in s+ logarithms we only have to consider a linear form in two p-adic logarithms which reduces the upper bound for n,m,z 1,...,z s drastically for large s. 010 Mathematics Subject Classification. 11D61,11B39,11Y50. Key words and phrases. Lucas sequences, S-units, automatic resoltion. The research was granted by the Austrian science found FWF) under the project P 4801-N6. 1

2 I. PINK AND V. ZIEGLER The second novelty in treating this kind of problems is a consequence to our p-adic approach by using a p-adic reduction method due to Pethő and de Weger [1] instead of real and complex) approximation lattices or the method of Baker and Davenport [1]. This approach leads to a better performance of the reduction step and a rather efficient algorithm. Before we state our main result, let us fix some notations. We call the sequence {u n } n 0 = {u n A,B,u 0,u 1 )} n 0 a binary linear recurrence sequence defined over the integers if the relation ) u n = Au n 1 +Bu n n ) holds, where A,B Z with AB 0 and u 0,u 1 are fixed rational integers with u 0 + u 1 > 0. The polynomial fx) = x Ax B attached to recurrence ) is the so-calledcompanionpolynomialofthe sequence{u n } n 0 and we denote by = A +4B thediscriminantoff. Letαandβ betherootsofthecompanionpolynomial f and assume that 0, then it is well known that there exist constants a = u 1 u 0 β and b = u 1 u 0 α such that the following formula holds 3) u n = aαn bβ n. α β The sequence {u n } n 0 is called non-degenerate, if abαβ 0 and α/β is not a root of unity. Throughout the paper we will assume that u n is non-degenerate and that > 0. The last assumption implies that the sequence {u n } n 0 possesses a dominant root, which means that without loss of generality α > β. Under these assumptions and notations the main result of our paper is now as follows: Theorem 1. Let {u n } n 0 be a non-degenerate binary recurrence with > 0. Let us assume that p i gcda,b) for all 1 i s and furthermore let us assume that none of the following two conditions hold β = ±1 and m = logβm b/a) logα. β = 1 and there exists an positive odd integer x and integers t 1,...,t s such that wα+1) 4) aα x +1) = p t1 1...ps ts. Then there exists an effectively computable constant C depending only on {u n } n 0, w, s, p 1,...,p s such that all solutions n,m,z 1,...,z s ) to equation 1) satisfy max{n,m,z 1,...,z s } < C. Let us stress out that during the course of proof of Theorem 1 we give a very precise description of how to compute this bound C see for instance Proposition 3 and Section 10). Moreover, we present an easy implementable algorithm to solve Diophantine equation 1) completely see Section 8) under the assumptions of Theorem 1. Let us discuss the technical assumptions made in Theorem 1. First, the assumption that p i gcda,b) for all 1 i s is to avoid technical difficulties. Using instead of lower bounds for linear forms of p-adic logarithms lower bounds for linear forms of complex logarithms in Section 3 one can avoid these difficulties, e.g. by adopting the method due to Bravo and Luca [4]. However we do not intend to further discuss this case in detail.

3 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 3 In the case that the first exceptional case of Theorem 1 holds, i.e. that β = ±1 and m = logβm b/a) logα, equation 1) can be rewritten as aα n α±1 = wpz1 1...pzs s. It is easy to see that this Diophantine equation may have infinitely many solutions which are easy to determine. In particular, it is possible to determine all solutions to such an equation by slightly adapting our method. Note that α has to be a rational integer if β = ±1. For instance, let us choose a = b = β = w = s = 1 and α = p 1 =. Then equation 1) turns into n m 1 1 = z1 which has infinitely many solutions of the form m = 1 and n = z 1. Let us turn to the second exceptional case of Theorem 1. If β = 1 and n m is odd and let us write x = n m > 0, then Diophantine equation 1) turns into aα n +b 1) n + aαm +b 1) m = aαm α x +1) α+1 α+1 α+1 If we insert now the assumed relation 4) we get α m = p z1 t1 1...ps zs ts = wp z1 1...pzs s. and it is easy to see to see that this Diophantine equation may have infinitely many solutions. Moreover, also in this case it is easy to determine all solutions to such an equation. In view of Theorem 1 only the question how to find all solutions to equation 4) remains to solve Diophantine equation 1) completely in any case. However, Diophantine equation 4) is nothing else than u x = w p t1 1...pts s with w = w/a. This type of Diophantine equation has been studied by Pethő and de Weger [1] and they gave a practical algorithm how to solve such equations. Moreover, we will also discuss this kind of Diophantine equation in Section 4. Let us provide an example with infinitely many solutions in this case. For instance let a = b = w = 1, β = 1 and α = p 1...p s. Then it is easy to see that equation 1) has infinitely many solutions of the form n = m+1 and z 1 = = z s = m. Finally let us outline the plan of the paper. The next three sections will provide a proof of Theorem 1. First, we introduce some further notations and prove some auxiliary results in Section. In Section 3 we use lower bounds of linear forms in two p-adic logarithms due to Bugeaud and Laurent [6] in order to prove some kind ofgapprinciple. In the Sections 4 and 5we exploit ideas due to Pethőand de Weger [1] and Bravo and Luca [5, 4] respectively. This leads us to our upper bound C in Theorem 1. Due to the use of Baker s method the constant C in Theorem 1 is usually very large. By using the LLL-algorithm and ideas due to de Weger [7] and Pethő and de Weger [1] it is possible to reduce these bounds considerably in concrete examples. In Section 7 we will discuss the method of de Weger [7] and Pethő and de Weger [1] and show how to apply them to our problem. All together this provides an algorithm to solve Diophantine equations of type 1) completely see Section 8). In order to demonstrate the efficiency of our algorithm we solve the Diophantine equation u n +u m = z1 3 z 199 z46

4 4 I. PINK AND V. ZIEGLER completely, where u n is either the Fibonacci sequence or the Lucas sequence see Section 9). Since it is hard to track all the constants which appear in the proof of Theorem 1, we provide in the final section a list of all constants and their explicit determination.. Notations and auxiliary results In this section we keep the notations of the introductory section. However, before we start with the proof of Theorem 1 we need to introduce some more notations. For a positive real number x > 0 we define the function log by log : R >0 R log x := max{0,logx}. By ϕ = 1+ 5 we denote the golden ratio. Finally we write K = Qα,β) = Q ) for the number field corresponding to our binary sequence {u n } n 0 and define d K = [K : Q]. In the rest of the paper there will appear constants c 1,c, and also occasionally constants of slightly different form, e.g. c 8,i, which are all explicitly computable. Sometimes we do not state them explicitly in our results or proofs for aesthetic reasons. However, Section 10 provides a list of all constants and their explicit determination. We advise readers who wish to keep track of all constants and their explicit determination to keep a bookmark at Section 10. The first lemma is an elementary result due to Pethő and de Weger [1]. It will be used over and over again in the proof of Theorem 1. For a proof of Lemma 1 we refer to [14, Appendix B]. Lemma1. Let u,v 0,h 1 and x R be the largest solution of x = u+vlogx) h. Then x < max{ h u 1/h +v 1/h logh h v)) h, h u 1/h +e ) h }. Let us assume from now on that n,m,z 1,...,z s ) is a solution to Diophantine equation 1) satisfying the assumptions made in Theorem 1. The next lemma gives lower and upper bounds for the quantity u n + u m, as well as, upper and lower bounds for the exponents z i in terms of n. These bounds will be utilized in the proof of Theorem 1. Lemma. There exist constants c 1,...,c 5 such that the following holds: i) u n +u m < c 1 α n. ii) If n > c, then we have z i < log α logp i n for i = 1,...,s. In particular, for all 1 i s we have z i < log α log n. iii) Provided that n > c 3, we have c 4 α n < u n +u m. iv) Provided that n > c 3, we have s i=1 n < z ilogp i +c 5. log α Proof. Let us start with i). We note that since α > β and α β =, we get by the triangle inequality u n = aα n bβ n α β < a + b α n.

5 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 5 Thus, the above inequality leads to u n +u m u n + u m < a + b α n + a + b α m, which by n m gives u n +u m < c 1 α n, with c 1 = a + b. Next, we prove ii). By combining equation 1) and the just proved part i) of the lemma we may write for every 1 i s w p zi i w p z1 1...pzs s = u n +u m < c 1 α n. Thus, by taking logarithms the above inequality implies z i logp i nlog α 1+ c ), n c 1 w with c = log log α. Assuming that n > c we get 5) z i < nlog α logp i and the first inequality of ii) follows. The second inequality of ii) is a simple consequence of 5) by noting that p i for every 1 i s. For proving iii) we note that since α and β are roots of the quadratic monic polynomial x Ax B with A,B Z and = A +4B > 0 we have A +4 B α A± A 4B A + Since we assume that n m we find 1+ 1 hence we obtain by formula 3) 6) α n m u n +u m = 1 aα n ϕ, α n m 1 a α n 1 1 ϕ 1+ 5 ) bβ n +bβ m ) ) b β n + β m ). Let us distinguish between the case that β 1 and β > 1. In the case that β 1 inequality 6) yields u n +u m 1 a α n 1 1 ) b ϕ = = 1 ϕ) a α n b ) a α 1 n 1. ϕ Let us assume that n is large enough such that 1 4 b ϕ log a ϕ 1) log α with c 4 = a ϕ 1) ϕ.. Thus, we obtain u n +u m > 1 a α n = ϕ. b a α n 1 1 ϕ) > 1, i.e. n > c = 1 1 ) = c 4 α n, ϕ

6 6 I. PINK AND V. ZIEGLER Now let us assume that β > 1, then by n m and inequality 6) we get u n +u m 1 a α n 1 1 ) b β n = ϕ = 1 a α n 1 1 ϕ) 1 If we assume that n is large enough, say n > c = log b β n 1 > 1. Thus also in this case we have a α n 1 1 ϕ) b β n ) a α 1 n 1. ϕ 4 b ϕ a ϕ 1) log α β u n +u m > 1 a α n 1 1 ) = c 4 α n, ϕ, then we get that which proves iii) in the case that β > 1. All together we have proved iii) assuming that n > c 3 = max{c,c }. Finally we turn to the proof of iv). Let us note that by assuming that n is large enough, i.e. n > c 3, we may combine the inequality from iii) with equation 1). Thus we get and by taking logarithms this yields c 4 α n < u n +u m = w p z1 1...pzs s, nlog α < log w c 4 + s z i logp i i=1 which proves iv), with c 5 = log w c 4 log α. Before stating the next result let us introduce some further notations. Let L be a number field and η L. We denote by hη) as usual the absolute logarithmic Weil height of η, i.e. hη) = 1 d log a 0 + log d η ) ) i), where a 0 is the leading coefficient of the minimal polynomial of η over Z and the η i) -s are the conjugates of η in C. We will use the following well known properties of the absolute logarithmic height without special reference: i=1 hη ±γ) hη)+hγ)+log, hηγ ±1 ) hη)+hγ), hη l ) = l hη), for l Z. In general it is a very hard problem to find lower bounds for the height of elements in a number field of given degree. However, in the case of quadratic fields the problem can be solved easily: Lemma 3. For an algebraic number α of degree two we have hα) 0.4 or α is a root of unity.

7 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 7 Proof. Assume that the conjugate of α is β and that α β. If α is not an algebraic integer, then the leading coefficient a 0 of its minimal polynomial satisfies a 0 and we obtain hα) 1 log a 0 log > On the other hand if α is an algebraic integer, then α ϕ and hα) 1 log α +log β ) logϕ > The Application of p-adic methods Our next goal is to prove an upper bound for n in terms of n m and logn. Therefore we utilize linear forms in p-adic logarithms. In particular, we use the results due to Bugeaud and Laurent [6]. This result has the drawback that we have to assume that the algebraic numbers involved in the p-adic logarithms are linearly independent in contrast to the results due to Yu [16, 17] which can be applied without this assumption. Thus dealing with the linearly dependent case causes some technical difficulties. Nevertheless we use the result due to Bugeaud and Laurent [6] since their result yields rather small upper bounds and dealing with the technical difficulties will pay off when we come to the LLL-reduction step. For aprime number p denote by Q p the field ofp-adicnumberswith the standard p-adic valuation ord p. Further, let α 1,α be algebraic numbers over Q and we regard them as elements of the field L = Q p α 1,α ). We equip the field L with the ultrametric absolute value x p = p νpx), where ν p denotes the unique extension to L of the standard p-adic valuation ord p over Q p normalized by ν p p) = 1 we set ν p 0) = + ). Note, that for every non-zero δ L we have ν p δ) = ord pn L/Qp δ)) d L, where d L = [L : Q p ] is the degree of the field extension L/Q p and N L/Qp δ) is the norm of δ with respect to Q p. Denote by e the ramification index of the local field extension L/Q p and by f the residual degree of this extension. Put and let D = [Qα 1,α ) : Q] f { h α i ) max hα i ), logp }, i = 1,). D Before stating the above mentioned result of Bugeaud and Laurent we introduce p-adic logarithms see e.g. [15, Chapter 5.1] for a short but thorough treatment of this topic). Let K be a field complete with respect to p and containing Q p. There exists a function log p x) defined on all of K such that log p xy) = log p x)+log p y). Moreover, for every ξ K with ξ 1 p < p 1/p 1) we can compute the p-adic logarithm by 1 ξ) i log p ξ) =. i i=1

8 8 I. PINK AND V. ZIEGLER Moreover we have 7) log p ξ) p = ξ 1 p or equivalently 8) ν p log p ξ)) = ν p ξ 1) provided that ξ 1 p < p 1/p 1). Note that in the case that ξ 1 p = p 1/p 1) we can replace in equation 7) the = by a sign, or equivalently in equation 8) the = by a sign). With these notations at hand, let us state a result due to Bugeaud and Laurent [6, Corollary 1]): Theorem. Let b 1,b be positive integers and suppose that α 1 and α are multiplicatively independent algebraic numbers such that ν p α 1 ) = ν p α ) = 0. Put 9) b := Then we have b 1 Dh α ) + b Dh α 1 ). 10) ν p α b1 1 αb 1) 4ppf 1)D 4 p 1)logp) 4B h α 1 )h α ), with 11) B := max { logb +loglogp+0.4,10, 10logp }. D We only use this result in the case where L = Q p ), i.e. 1 d L,D,e,f and obtain as a simple Corollary Corollary 1. Let B = max{b 1,b,p 10,e 10 } and with the notations and assumptions of Theorem in force we have where ν p α b1 1 αb 1) < C 1p)h α 1 )h α )logb ), C 1 p) = 947pf logp) 4. Proof. First let us note that since f,d one easily deduces that 4pp f 1)D 4 p 1)logp) 4 < 768pf log 4 p. Moreover, b b 1 = Dh α ) + b Dh α 1 ) b 1 +b logp, and since we assume that B max{p 10,e 10 } we get { B = max logb +loglogp+0.4,10, 10logp } loge 0.4 B ) 1.11logB. D Putting everything together yields the Corollary. Note that it is rather easy to compute e and f. Indeed, let 0 be the squarefree ) part of, then we) have e = if and only if p 0 and f = if and only if = 1, where denotes the Legendre symbol. 0 p 0 p

9 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 9 Proposition 1. Let us assume that n > m > 3, P = max 1 i s {p i } and that n > c 6 := max { c 3,17.5log α max{log aα,log bβ }+0.4),P 10,e 10}. Then there exist constants c 7 and c 8,i with i = 1,...,s such that with 1 i s. n < c 7 n m)logn) and z i < c 8,i n m)logn), Proof. First, let us note that the first inequality is a direct consequence of the second inequality and Lemma iv) and c 7 can be easily computed once we have found the constants c 8,i see Section 10). Therefore we are left to prove the second inequality for each index 1 i s. Let us fix the index i. In order to avoid an overloadof indices we drop the index i for the rest of the proof, i.e. we write p = p i, c 8 = c 8,i and so on. Further, we will work in the field L = Q p α,β) = Q p ). Since we assume p gcda,b) we have that ν p α) = 0 or ν p β) = 0. Without loss of generality we may assume that ν p α) = 0, since we make no use of the fact that α > β in the whole proof of the proposition. Let us consider equation 1) and rewrite it as follows 1) bβ n m +1) aα n m +1) β α ) m 1 = wα β)α m a 1 α n m +1) 1 p z1 1...pzs s LetusdenotebyΛthelefthandsideof 1)andletuscomputeν p Λ)byconsidering the right hand side of equation 1). Since by assumption ν p α) = ν p w) = 0 we obtain ν p Λ) = z ν p aα n m +1))+ν p α β). Let us note that α β is an algebraic integer, hence ν p α β) 0. Moreover, we have ν p aα n m +1)) log a +log+n m)log α logp n m) log aα := c 9 n m). logp and obtain 13) ν p Λ) z c 9 n m). Next, we will apply Corollary 1 in order to bound ν p Λ) from above. Therefore let usestimate the heightof bβn m +1) aα n m +1). Note that α and β resp. a andb) areeither rational integers or conjugate ) algebraic integers in a real quadratic field. Therefore we conclude that h max{log α,log β } provided that α and β are either α β rational integers or conjugate algebraic integers in a real quadratic field. Indeed this is clear in the rational case. In the real quadratic case under the assumption that α β we get h α β ) = 1 ) α log a 0 +log β +log β α 1 log α +log β +log α log β ) = log α.

10 10 I. PINK AND V. ZIEGLER Note that the leading coefficient a 0 of the minimal polynomial of α/β satisfises a 0 αβ = B. Therefore we obtain bβ n m ) +1) h aα n m max{log a +log α n m +1,log b +log β n m +1 } +1) 14) max{log a +n m)log α +log, log b +n m)log β +log } hence bβ h n m ) +1) aα n m +1) n m)max{log aα,log bβ }, n m)max{log aα,log bβ,logp} := c 10 n m). Furthermore note that h α/β) max{log α,logp}. We have to distinguish now between two cases, namely whether bβn m +1) aα n m +1) and α/β are multiplicatively independent or not. Let us deal with the dependent case first. In this case there exist co-prime integers r and s such that aα n m ) r/s +1) α bβ n m +1) =. β Let us find upper bounds for r and s. First, we note that δ := α/β) 1/s has to be still of degree two and is not a root of unity. Indeed α/β) r/s K therefore α/β) r K s. Since by assumption r and s are coprime we obtain that α/β) K s hence δ := α/β) 1/s K. Hence 0.4 < hα/β) s 1 s log α and s < 4.log α by Lemma 3. On the other hand we deduce from the upper bound 14) that bβ n m ) +1) 4.n m)max{log aα,log bβ }log α s h aα n m +1) ) α = r h r 0.4 β i.e. In particular, we have r < 17.5n m)log α max{log aα,log bβ }. ms r m s + r 17.5nlog α max{log aα,log bβ }+0.4) Therefore rewriting 1) we obtain ν p δ ms r 1 ) =ν p wα β)α m a 1 α n m +1) 1 ) p z1 1...pzs s >z c 9 n m) Now, on supposing that z 3 +c 9n m) we may use property 8) of the p-adic logarithm to obtain ν p log p δ)+ν p ms+r) > z c 9 n m).

11 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 11 Ifwe assume that n > 17.5log α max{log aα,log bβ }+0.4)we get by a very crude estimate z <ν p log p δ)+ logn logp +c 9n m) < ν p log p δ)+ ) logp +c 9 n m)logn Note that :=c 11 n m)logn ν p log p δ)) = ν p 1 s log p )) )) α α ν p log β p. β Therefore we may assume that bβn m +1) aα n m +1) and α/β are multiplicatively independent. However, before applying Corollary 1, we have to ensure that ν p bβ n m +1) aα n m +1) ) ) β = ν p = 0. α Let us assume for the moment that one of the p-adicvaluations is not zero. But this would imply that the p-adic valuation of the left hand side of 1) is zero, hence z = 0. Therefore we may apply Corollary 1 and we obtain 15) ν p Λ) C 1 p)max{log α,logp}c 10 n m)max{logn,10logp,10}. Comparing upper and lower bounds of ν p Λ) i.e. inequalities 13) and 15) yields C 1 p)max{log α,logp}c 10 n m)logn) > z c 9 n m) or by solving for z followed by a crude estimate we obtain our upper bound for z: z < C 1 p)max{log α,logp}c 10 +c 9 )n m)logn) := c 8 n m)logn). 4. The case n = m Before we continue with the main line of the proof of Theorem 1 let us consider the special case n = m, i.e. the Diophantine equation u n = wp z1 1...pzs s. Of course this equation has only a solution if w is even or one of the primes, say p 1 =. Thus we are reduced to consider Diophantine equations of the type 16) u n = w p z1 1...pzs s Let us note that this type of equation has been considered by Pethő and de Weger [1]. They gave a practical method to solve this equation completely for a givenbinary, non-degeneratesequenceu n withdiscriminant > 0, non-zerointeger w and primes p 1,...,p s by using p-adic techniques. In particular, they use linear forms in p-adic logarithms and a p-adic version of the Baker-Davenport method [1]. We also want to refer to the paper of Mignotte and Tzanakis [11], where also the case of non-degenerate recurrences of arbitrary order is discussed. Proposition Pethő and de Weger [1]). Let n,z 1,...,z s be a solution to 16), then there exist explicitly computable constants c 1,i with 1 i s and c 13 such that z i < c 1,i for all 1 i s and n < c 13.

12 1 I. PINK AND V. ZIEGLER The constant c 13 has been stated explicitly by Pethő and de Weger [1, Theorem 4.1]. However they used a version due to Schinzel [13] for lower bounds in linear forms in p-adic logarithms that has been succeeded by newer developments e.g. the results due to Yu [16, 17] and in particular by Bugeaud and Laurent [6]. For the sake of completeness and with respect to the newer developments we give a proof of Proposition. Proof. Since the proof due to Pethő and de Weger [1] and the similarity to the proof of Proposition 1 we only sketch the proof. As in the proof of Proposition 1 we fix an index i and drop it for the rest of the proof. We may assume without loss of generality that ν p α) = ν p b/a) = ν p β/α) = 0. We rewrite equation 16) and obtain 17) b a ) n β 1 = w α β)a 1 α n p z1 1 α pzs s. Let us assume for the moment that b a and β α are multiplicatively independent n and that n > max{p 10,e 10 }. Let us put Λ = b a β α) 1. Estimating the p-adic valuation on the right hand side of 17) yields ν p Λ) z log a logp and an application of Corollary 1 yields ) ) b β ν p Λ) C 1 p)h h max{logn,10logp,10}, a α with and ) b h max{log a,log b,logp} a ) β h max{log α,log β,logp}. α Thus we get z < ) ) b β C 1 p)h h + log ) a max{logn,10logp,10} a α logp =c 14 max{logn,10logp,10}. If we assume that n > c 3 we may apply Lemma iv) and obtain either an absolute bound for n or an inequality of the form n < c 15 logn) and Lemma 1 yields an absolute upper bound for n. In any case we obtain the Proposition. We are left with the case that a b and α β are co-prime integers s and r such that a b are multiplicatively dependent, i.e. there ) ) s r 1/s. = α β and let δ = α β) As in the proof of Proposition 1 we deduce s < 4.log α and by a similar computation we obtain r < 17.5log α max{log a,log b,1}. Hence we obtain from 17) the inequality ν p δ ns r 1) z log a logp.

13 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 13 Nowonsupposing z 3 + log a log p we may use the properties of the p-adic logarithms e.g. see formula 8)) and we obtain z < ν p log p δ)+ log4.nlog α +17.5log α max{log a,log b,1})+log a. logp Assuming that n > 17.5log α max{log a,log b,1} we obtain z < ν p log p δ)+ logp + log a logp assumption n > c 3 we obtain s i=1 νpi log pi δ)logp i ++log a ) ) n < log α ) logn, and by Lemma iv) and the logn+c 5 = c 16 logn+c 5 and once again applying Lemma 1 yields an upper bound c 13 for n for an explicit determination see Section 10). 5. The case n > m In order to solve the case n > m, we have to use lower bounds for linear forms in complex logarithms. The currently best result suitable for our purpose is the following theorem due to Matveev [10]. Theorem 3 Matveev [10]). Denote by η 1,...,η n algebraic numbers, not 0 or 1, by logη 1,...,logη n determinations of their logarithms, by D the degree over Q of the number field L = Qη 1,...,η n ), and by b 1,...,b n rational integers. Define B = max{ b 1,..., b n }, and A i = max{dhη i ), logη i,0.16} 1 i n), where hη) denotes the absolute logarithmic Weil height of η. Assume that the number Λ = b 1 logη 1 + +b n logη n does not vanish. Then log Λ Cn,κ)D A 1 A n loged)logeb )}, where κ = 1 if K R and κ = otherwise and { ) κ } 1 1 Cn,κ) = min κ en 30 n+3 n 3.5, 6n+0. Since the field K = Q ) of our interest is real and of degree at most and all the α i will be positive we obtain 18) log Λ C n)hη 1 ) hη n )logb, where C n) = n+3 n 4.5 provided that B 3. Note that since Lemma 3 we can choose A i = hη i ). The main purpose of this section is to find absolute upper bounds for n and the exponents z i with 1 i s. For technical reasons we will assume that n m is not to small. In particular, we may assume that )) log 1+ b a n m > { }) = c 17. log min α β, α Indeed, due to Proposition 1 the case that n m c 17 immediately implies n < c 7 n m)logn) < c 7 c 17 logn). Thus by Lemma 1 we get that n 4c 7 c 17 log4c 7 c 17 ).

14 14 I. PINK AND V. ZIEGLER Hence we assume for the rest of this section that n m > c 17. Let us rewrite equation 1) as 19) p z1 1 pzs s wα β)a 1 α n 1 = bβn +aα m bβ m aα n Let us estimate the right hand side of 19). We obtain bβ n +aα m bβ m aα n bmax{β n,β m } aα n + α m n b { } β n a max β m α n, α n + α m n { α b ) } m n a max, α m n + α m n β 1+ b ) { } n m β max a α, 1. α Note that our assumption n m c 17 was chosen such that the right hand side of equation 19) is smaller than 1/. Since log x+1 x provided 0 x 1/ we obtain by taking logarithms on both sides of equation 19) the inequality ) 0) Λ := z 1 logp 1 + +z s logp s +log γ nlog α < { } n m, min α β, α 1+ b a where γ = w a. In order to apply Matveev s Theorem we have to ensure that Λ does not vanish. However this is established by the following Lemma: Lemma 4. If Λ = 0 then there exists a constant c 18 such that n < c 18. In order to complete the proof of Theorem 1 we postpone the proof of Lemma 4 to the next section. Now by Lemma 4 we may assume that Λ does not vanish and we may apply Matveev s Theorem. Before we apply Matveev s Theorem let us note that B = max{z 1,...,z s,1,n} n log α < n, log provided that n > max{c, log α log }. Thus by comparing the upper bound from inequality 0) with the lower bound of Λ from Matveev s bound 18) we obtain the inequality C s+)logp 1...logp s log α hγ)logn > { }) α n m)log max β, α log 1+ b ), a i.e. there exists a constant c 19 see Section 10 for an explicit form) such that n m < c 19 logn. In combination with Proposition 1 we obtain n < c 7 c 19 logn) 3, provided that n > c 6. Once again using Lemma 1 yields an upper bound c 0 for n and Lemma yields an upper bound z i < c 1,i for each 1 i s. With the notations from above in force we have proved so far the following proposition, which clearly implies Theorem 1:

15 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 15 Proposition 3. Under the assumptions of Theorem 1 we have for i = 1,...,s. max{n,m} < c 0 and z i < c 1,i 6. Proof of Lemma 4 First, let us note that Λ = 0 implies that 1) bβ n aα m +bβ m = 0. In the case that the > 0 is not a perfect square α and β respectively a and b are algebraic conjugate and we obtain by conjugating the left side of equation 1) aα n +bβ m aα m = 0 which yields together with the original equation 1) the relation bβ n = aα n. Therefore n < log b/a log α/β = c. Therefore we may assume that is a perfect square and that α,β,a,b are all rational integers. Assume for the moment that β = 1. Then 1) turns into aα m = 0 or aα m = b. The first case would imply that u n is degenerate and the second case has been excluded. Thus we may assume that β. Under these assumptions we obtain from 1) by dividing through bβ n the inequality a α m b β n 1 β m n and by taking logarithms we get ) Λ := log a +mlog α nlog β < b β n m Let us assume for the moment that Λ = 0. Since a,b,α,β are integers Λ = 0 implies that aα m = ±bβ n and in combination with equation 1) this leads either to bβ m = 0 or β n m = 1. But, both cases are contradictions in view of β and n > m. Hence we may suppose that Λ 0 and therefore we may apply Matveev s Theorem 3 to the left side of inequality ) and obtain log Λ > C 3)max{0.16,log max{ a, b }}log α log β logn. On the other hand we have log Λ < log n m)log β. Hence we get that n m < c 3 logn see Section 10 for an explicit determination of c 3 ). Combining this with the results of Proposition 1 we get n < c 7 c 3 logn) 3 provided that n > c 6. Thus we find an absolute upper bound c 18 for n by applying Lemma 1. Let us note that in Section 8 we explain in more detail how to handle this case in practice. 7. Reduction of our bounds This section is devoted to the problem of reducing the rather large bounds obtained by Theorem 1 and Proposition 3 respectively. In this paper we will make use of the LLL-algorithm due to Lenstra, Lenstra and Lovász [8] to reduce our upper bounds for n m and z 1,...,z s. In the p-adic case we use instead of approximation lattices an idea due to Pethő and de Weger [1, Algorithm A].

16 16 I. PINK AND V. ZIEGLER 7.1. Real approximation lattices. Let us start with gathering some basic facts on LLL-reduced bases and approximation lattices. Therefore let L R k be a k- dimensional lattice with LLL-reduced basis b 1,...,b k and let B be the matrix with columns b 1,...,b k. Moreover, we denote by b 1,...,b k the orthogonal basis of Rk which we obtain by applying the Gram-Schmidt process to the basis b 1,...,b k. In particular, we have that i 1 b i = b i µ i,j b j, µ i,j = b i,b j b j,b j. j=1 Further, let us define ll,y) = { min x L { x y }, y L min 0 x L { x }, y L, where denotes the euclidian norm on R k. It is well known, that by applying the LLL algorithm it is possible to give in a polynomial time a lower bound for ll,y) c 1 see e.g. [14, Section 5.4]). Lemma 5. Let y R k, z = B 1 y and if y L let i 0 be the largest index such that z i0 0. Put σ = {z i0 }, where { } denotes the distance to the nearest integer, and in case that y L we put σ = 1. Moreover, let Then we have c = max 1 j k { b 1 b j } ll,y) c 1 σ b 1 = c 1. In our applications suppose we are given η 0,η 1,...,η k real numbers linearly independent over Q and two positive constants c 3, c 4 such that 3) η 0 +x 1 η 1 + +x k η k c 3 exp c 4 H), where the integers x i with 1 i k are bounded by x i X i with X i given upper bounds for all 1 i k. Set X 0 = max 1 i s {X i }. The basic idea in such a situation, due to de Weger [7], is to approximate the linear form 3) by an approximation lattice. Namely, we consider the lattice L generated by the columns of the matrix Cη 1 Cη... Cη k 1 Cη k where C is a large constant usually of the size about X k 0. Let us assume that we have an LLL-reduced basis b 1,...,b k of L and that we have a lower bound ll,y) c 1 with y = 0,0,..., Cη 0 ). Then we have with these notations the following Lemma concerning inequality 3) c.f. [14, Lemma VI.1]):.

17 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 17 Lemma 6. Assume that S = k 1 i=1 X i and T = 1+ k i=1 Xi. If c 1 T +S, then we have either x 1 = x = = x k 1 = 0 and x k = Cη0 ) 4) H 1 c 4 logc c 3 ) log Cη k ) or c 1 S T )). We will apply Lemma 6 to inequality 0), i.e. k = s + 1 and x i = z i for i = 1,...,s and x s+1 = n. For the coefficients η i with 0 i s + 1 we take η 0 = logγ, η i = logp i for 1 i s and η s+1 = log α. Furthermore we have c 3 = 1+ b a ) and c 4 = log min { α β, α }). For sure η 1,...,η s are linearly independent over Q and it is easy to find a multiplicative dependence of γ and/or α of the p i with i = 1,...,s. However, let us assume for the moment that log γ,logp 1,...,logp s and log α are linearly independent overqa similar argumentholds if this is not the case). Let us alsoassume that C was chosen large enough such that c 1 > s s c i=1 1,i + c ) 1,i +c 0 = T +S. i=1 Then we get by Lemma 6 either a new bound for H = n m or z 1 =...z s = 0 and our Diophantine equation 1) reduces to u n +u m = w wich can be easily solved by brute force if w is not too large. Let us note that a new bound for n m immediately yields new upper bounds for z i with 1 i s and n by Proposition 1. We can apply the same trick once again with these new bounds and obtain again a further reduction of the bounds for z i with 1 i s and n. We can repeat this as long as our reduction method yields smaller upper bounds for n m. 7.. The p-adic reduction method. In the p-adic case it is also possible to use so-called p-adic approximation lattices see e.g. [14, Section 5.6]). However we are only interested in a very special case namely in the situation appearing in inequality 1) under the assumption that n m = t is a fixed small) integer. The following is based on an idea due to Pethő and de Weger [1, Algorithm A]. We reproduce theire idea and fit it into our frame work. Let us fix an index i with 1 i s. In order to avoid an overloaded notation we drop the index i for the rest of this section. Furthermore let us assume that N and Z are given upper bounds for n and z respectively. Let us consider the p-adic valuation of the left and right side of 1). Then we obtain as in Section 3) ) m β ν p τt) 1) = z ν p aα t +1))+ν p α β) = z z 0, α where τt) = bβt +1) aα t +1) and z 0 is easily computable for a fixed t. Let us assume that z 3/+ z 0 = c 5, then we may take p-adic logarithms and obtain 5) ν p logp τt)) mlog p α/β) ) = z z 0. We distinguish now between two cases whether log p τt)) = 0 or not.

18 18 I. PINK AND V. ZIEGLER Let us discuss the rather unlikely case that log p τt)) = 0 first. In this case equation 5) turns into i.e. ν p mlog p α/β)) = ν p m)+ν p log p α/β)) = z z 0, z < logm logp +ν plog p α/β))+z 0 < logn logp +ν plog p α/β))+z 0 = c 6 and max{ c 5, c 6 } is a new upper bound for z. Let us turn to the case that log p τt)) 0. Since α and β are conjugate in Q ) Q p ) also log p α) and log p β) are conjugate, hence log p α/β) = log p α) log p β) Q p. Similarly we get log p τt)) Q p, since τt) is the quotient of conjugates. In particular, we get ζ = log p τt))/log p α/β) = u 0 +u 1 p+u p + Q p. Let r be the smallest possible exponent such that p r > N and let 0 m 0 < p r be the unique integer such that m 0 ζ mod p r. Moreover, let R be the smallest index r such that u R 0, if such an index exists. Then we get z z 0 =ν p logp τt)) mlog p α/β) ) ν p logp τt)) m 0 log p α/β) ) =ν p log p τt))+ν p 1 ζ u R p R...) log pα/β) log p τt)) ) logp α/β) =ν p log p τt))+r+ν p. log p τt)) ) Therefore we get a new upper bound for z namely ) logp α/β) z ν p log p τt))+r+ν p. log p τt)) Let us discuss the case, in which R does not exist. In this case we would obtain that ζ = m 0 is an integer, hence and 5) turns into log p τt)) = m 0 log p α/β)), ν p logp τt)) mlog p α/β) ) = ν p m m0 )log p α/β) ) = z z 0. Therefore we are in a similar situation as in the case that log p τt)) = 0 and we also get in this case a new upper bound for z. In any case we get for each index i with 1 i s a new upper bound z i and therefore by Lemma iv) a new upper bound for n. We can repeat the procedure with these new, small upper bounds as long as we get smaller upper bounds for n. 8. The Algorithm At this point the inclined reader may have already a farely well idea how to solve Diophantine equations of type 1) in practice. However, let us summerize the key steps and give some comments on the practical implementation. The algorithm to solve Diophantine equation 1) can be splitted up into six key steps:

19 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 19 Step I - First upper bounds for n and z i for all 1 i s: Compute one after the other the constants c 1, until you have found the upper bounds n < c 0 and z i < c 1,i with 1 i s. Using the explicit determination of the constants given in Section 10 this is straight forward. Step II - The case that n = m: If w and all p i with 1 i s are odd, this case has no solution. Otherwise in case that w is even we replace w by w/ or if p 1 = we replace z 1 by z 1 1 and compute upper bounds for n < c 13 and z i < c 1,i for 1 i s. With these upper bounds we use our p-adic reduction procedure from Section 7. with t = 1 and τt) = b/a and obtain a small bound for n, which is in most cases small enough to performe a brute force search. Step III - The case that bβ n aα m +bβ m = 0: If is not a perfect square or β = ±1 there are no solutions in this case and we may omitt this step. Otherwise we may assume that α,β,a and b are rational integers and that β. In the case that β α we obtain the equation a b α β ) m = β n m +1 We may assume that α/β = P/Q with Q > 1 and P,Q are coprime integers. Then m logq/loga since otherwise the left hand side cannot be an integer. This presumably small bound for m yields also a small bound for n and a brute force search will resolve this case. Therefore we may assume that α/β = P is an integer and let us assume that p is the largest prime factor dividing P and let us assume for the moment that m > 1. Then we have that m ν p b) =ν p β n m 1) =ν p log p β))+ν p n m) <ν p log p β))+logc 3 logc 18 ). Usually this will yield a reasonable small bound for m and n to perform a brute force search. If the new upper bound max{n,m} < C is still large one may use once again the above inequality to obtain m ν p b) =ν p log p β))+ν p n m) <ν p log p β))+ log C) logp. This may be repeated until no further improvement on the bound for max{n,m} is possible and one has to perform a brute force search for possible solutions to Diophantine equation 1). Step IV - First reduction of n: Perform the reduction step described in Section 7.1 to obtain upper bounds n < N, z i < Z i for 1 i s and n m T 0. Step V - Second reduction of n: Perform the reduction step described in Section7.foreach1 t T 0 andobtainforeachtupper boundsn < Ñt andz i < Z i,t for 1 i s. Thus we get new upper bounds for n < Ñ = max 1 t T 0 {Ñt} and

20 0 I. PINK AND V. ZIEGLER z i < Z i = max 1 t T0 { Z i,t } for 1 i s. Step VI - Brute force search: The upper bounds Ñ and Z i for 1 i s obtained in Step V are usually rather small. However replacing c 3 by Ñ and c,i by Z i for 1 i s we can go back to Step IV and after performing the reduction steps IV and V again maybe we obtain sharper bounds for n and z i with 1 i s. This can be repeated until now further improvement is possible. And we have to check the remaing cases by a brute force search. For instance we compute for all 1 n m Ñ the values of u n + u m and write them into a list L. For each element from the list L we perform a trial division including the primes p 1,...,p s. If Ñ and P = max{p 1,...,p s } are not unusually large, say Ñ,P < 10000this brute force search can be done within a reasonable time see Section 9). 9. An example In this section we illustrate our algorithm by two examples. We completely solve Diophantine equation 1) in the case that w = 1, p 1,...,p 46 are all primes smaller than 00 and u n is the Fibonacci sequence or the Lucas sequence, respectively. We have the following theorem: Theorem 4. i) Let {F n } n 0 be the Fibonacci sequence defined by F 0 = 0,F 1 = 1 and F n = F n 1 +F n for n. Consider the equation 6) F n +F m = z1 3 z z46 in non-negative integer unknowns n,m,z 1,...,z 46 with n m. Then there are 35 solutions n,m,z 1,...,z 46 ) and there exists no solution with max{n,m} > 59. ii) Let {L n } n 0 be the Lucas sequence defined by L 0 =,L 1 = 1 and L n = L n 1 +L n for n. Consider the equation 7) L n +L m = z1 3 z z46 in non-negative integer unknowns n,m,z 1,...,z 46 with n m. Then there are 84 solutions n,m,z 1,...,z 46 ) and there exists no solution with max{n,m} > 63. We refrain from giving a list of all 35 and 84 solutions to Diophatine equations 6) and 7), respectively. Let us note that a brute force computer search for all solutions to 6) and 7) with m n 59 and m n 63 respectively is a matter of a few seconds on a usual PC. However looking through the solutions to Diophatine equations 6) and 7) respectively we noticed the following interesting facts. z 1 8 and z 6 in both cases. In the case of equation 6) we have that z 3,z 5 3 and for all other indices 1,,3,5 we have that z i. In the case of equation 7) we have that z 3 3 and for all other indices i 4 we have that z i. For all solutions to equation 6) the exponents of 79,83,131,139,163 and 167 are always zero. For all solutions to equation 7) the exponent of 73,149,157,173,181,191, 193 and 197 are always zero.

21 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 1 Proof of Theorem 4. We will give the computational details of our algorithm only in the resolution of Diophantine equation 6). The resolution of Diophantine equation 7) is similar and of the same computational time. In the proof follow the stepsgivenintheprecedingsection. Aneasycomputationshowsthatifu n isthefibonacci sequence, then we have a = b = 1,α = 1+ 5,β = 1 5, = α β = 5. Step I - First upper bounds for n and z i for all 1 i s: By using the explicit determination of the constants given in Section 10 we obtain by a straight forward computation that 8) max{n,z i } < , 1 i 46). Step II - The case that n = m: It is clear, that in this case we may suppose that z 1 1, since otherwise we do not have a solution. Now, we replace z 1 with z 1 1 and compute the initial upper bounds n < c 13 and z i < c 1,i 1 i 46) and get max{n,z i } < N := For every p = p i 1 i 46) we perform the p-adic reduction procedure described in Section 7. with t = 1 and τt) = b/a = 1. Since α β = 5 and a = 1 we obtain that z 0 = ν p aα t +1)) ν p α β) 1 and log p τt)) = log p 1) = 0 holds for every p = p i 1 i 46). Further, ν p log p α/β)) and hence we get that z = z i < log ) logp Note that the values Z 0,i form a vector ++1 = Z 0,i 1 i 46). Z 0 = 79,51,36,30,5,3,1,1,19,18,18,17,17,17,16,16,16,15,15,15,15,15, 15,14,14,14,14,14,14,14,14,13,13,13,13,13,13,13,13,13,13,13,13,13,13,13). Since in our case c 5 < 6, we infer by iv) of Lemma that n < 46 i=1 Z 0,ilogp i ) ) log 1+ 5 By repeating the above reduction step once more, now with N = 6100, we obtain n Step III - The case that bβ n aα m + bβ m = 0: One can easily see that this case cannot occur if {u n } n 0 = {F n } n 0. Step IV - First reduction of n: We may suppose that n > m. We apply the reduction method described in Section 7.1 with the initial upper bound provided by Step I. Namely max{n,z i } < X 00 := and choose C := Further, put c 3 = 6, c 4 = log 1+ 5, k = 47, x i = z i with 1 i 46, x 47 = n and η 0 = log 5), η i = logp i with 1 i 46. After performing the LLL algorithm we found that c ) 1 := 1/5000 and σ := 0.49 is an appropriate choice. Thus, we may apply Lemma 6 and we obtain a reduced bound n m T 1 = Inserting this bound for n m into Proposition 1 we are able to derive a new, better bound for max{n,z i }. An easy calculation shows that max{n,z i } <

22 I. PINK AND V. ZIEGLER X 01 := is an appropriate choice. By repeating once more the above procedure, with X 01 := ,C := , c ) 1 := 1/81550 and σ := 0.49 we obtain n m T = Step V - Second reduction of n: We perform the reduction step described in Section 7. for each 1 t T = Since max{n,z i } < X 01 := we apply the reduction step described in Section 7. for each prime p = p i with 1 i 46 and for each 1 t T = In particular, we obtain that z = z i Z 1,i with 1 i 46, where the Z 1,i form a vector Z 1 = 9,57,41,37,4,7,5,19,,0,17,1,18,0,18,0,16,16,17,14,17,14, 15,15,15,13,16,16,15,17,14,13,14,14,13,1,14,14,14,14,1,13,13,14,15,14). Since in our case c 5 < 6 we infer by iv) of Lemma that n < 46 i=1 Z 1,ilogp i ) log 1+ 5 ) By repeating the above reduction step once more, now with X 0 = 6600, we obtain n 300. Step VI - Brute force search: Now, we have proved that m n 300, a value which is small enough to provide a brute force search. Namely, we used trial division with primes up to 00 for each quantity F n +F m. We found 34 solutions. Finally, we note that the total computational time of our algorithm for the sequence {F n } n 0 was about four and a half hours on a computer with an Intel Core 5 M330 processor. The most time consuming step was the first LLL-algorithm in Step IV which took almost four hours. Let us denote by P = max 1 i s {p i }. Constants appearing in Lemma : 10. Appendix - Explicit bounds c 1 = a + b, c = log c1 w log α, c 3 = max log 4 b ϕ a ϕ 1) log α w c 4 c 4 = a ϕ 1) ϕ, c 5 = log log α, log 4 b ϕ a ϕ 1) log α β The constant C 1 p) from Bugeaud and Laurent s lower bound of linear forms in two p-adic logarithms: C 1 p) = 947pfp log 4 p.,

23 DIOPHANTINE EQUATIONS OF THE FORM u n + u m = wp z 1 1 pzs s 3 Constants appearing in Proposition 1 and its proof: c 6 =max{c 3,17.5log α max{log aα,log bβ }+0.4),P 10,e 10 }, c 9 = log aα logp, c 10 =max{log aα,log bβ,logp}, c 11 =ν p log p α/β))+ logp +c 9, c 8,i =max{c 1 p i )max{log α,logp i }c 10 +c 9,c 11 }, s i=1 c 7 = c 8,ilogp i +c 5, log α where P = max 1 i s {p i }. Constants appearing in Proposition and its proof: where ) b β c 14 = C 1 p)h h a α + log ) a logp s i=1 c 15 = c 14logp i +c 5, log α s i=1 νpi log pi α/β))logp i ++log a ) ) c 16 = log α c 13 = max{4c 15 log4c 15 ),c 5 +logc 16 ),c 3,c,P 10,e 10 }, c 1,i = log α logp i c 13. h a ) ) β max{log a,log b,logp}, and h max{log α,log β,logp}. b α The constant C n) from Matveev s lower bound of linear forms in complex logarithms: C n) = n+3 n 4.5. Constants appearing in the proof of Lemma 4 c = log b/a log α/β, c 3 = max{0.16,log max{ a, b }}log α, c 18 =max{8c 7 c 3 log7c 7 c 3 ) 3,c 6 }.

24 4 I. PINK AND V. ZIEGLER Constants appearing in the rest of Section 5: )) log 1+ b a c 17 = { }), log min α β, α C s+)logp 1...logp s hw a 1 )log α +log c 19 = log max c 0 =max{8c 7 c 19 log7c 7 c 19 ) 3,c 18,c 6,c }, c 1,i = log α logp i c 0. References 1+ b a { α β, α }), [1] A. Baker and H. Davenport. The equations 3x = y and 8x 7 = z. Quart. J. Math. Oxford Ser. ), 0:19 137, [] C. Bertók, L. Hajdu, I. Pink, and Z. Rábai. Linear combinations of prime powers in binary recurrence sequences. to appear in International Journal of Number Theory. [3] J. J. Bravo, C. A. Gómez, and F. Luca. Powers of two as sums of two k-fibonacci numbers. available at arxiv.org/abs/ [4] J. J. Bravo and F. Luca. On the Diophantine equation F n+f m = a. to appear in Questiones Mathematicae. [5] J. J. Bravo and F. Luca. Powers of two as sums of two Lucas numbers. J. Integer Seq., 178):Article , 1, 014. [6] Y. Bugeaud and M. Laurent. Minoration effective de la distance p-adique entre puissances de nombres algébriques. J. Number Theory, 61):311 34, [7] B. M. M. de Weger. Solving exponential Diophantine equations using lattice basis reduction algorithms. J. Number Theory, 63):35 367, [8] A. K. Lenstra, H. W. Lenstra, Jr., and L. Lovász. Factoring polynomials with rational coefficients. Math. Ann., 614): , 198. [9] D. Marques. Powers of two as sum of two generalized Fibonacci numbers. available at [10] E. M. Matveev. An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers. II. Izv. Ross. Akad. Nauk Ser. Mat., 646):15 180, 000. [11] M. Mignotte and N. Tzanakis. Arithmetical study of recurrence sequences. Acta Arith., 574): , [1] A. Pethö and B. M. M. de Weger. Products of prime powers in binary recurrence sequences. I. The hyperbolic case, with an application to the generalized Ramanujan-Nagell equation. Math. Comp., 47176):713 77, [13] A. Schinzel. On two theorems of Gelfond and some of their applications. Acta Arith, 13:177 36, 1967/1968. [14] N. P. Smart. The algorithmic resolution of Diophantine equations, volume 41 of London Mathematical Society Student Texts. Cambridge University Press, Cambridge, [15] L. C. Washington. Introduction to cyclotomic fields, volume 83 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, [16] K. Yu. p-adic logarithmic forms and group varieties. II. Acta Arith., 894): , [17] K. Yu. p-adic logarithmic forms and a problem of Erdős. Acta Math., 11):315 38, 013. I. Pink Institute of Mathematics, University of Debrecen H-4010 Debrecen, P.O. Box 1, Hungary and University of Salzburg Hellbrunnerstrasse 34/I A-500 Salzburg, Austria address: pinki@science.unideb.hu; istvan.pink@sbg.ac.at )