A DIOPHANTINE EQUATION RELATED TO THE SUM OF POWERS OF TWO CONSECUTIVE GENERALIZED FIBONACCI NUMBERS

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1 A DIOPHANTINE EQUATION RELATED TO THE SUM OF POWERS OF TWO CONSECUTIVE GENERALIZED FIBONACCI NUMBERS ANA PAULA CHAVES AND DIEGO MARQUES Abstract. Let F n) n 0 be the Fibonacci sequence given by F m+ = F m+1 + F m, for m 0, where F 0 = 0 and F 1 = 1. There are several interesting identities involving this sequence such as Fm +Fm+1 = F m+1, for all m 0. In 011, Luca and Oyono [8] proved that if Fm s +Fm+1 s is a Fibonacci number, with m, then s = 1 or. One of the mosnown generalization of the Fibonacci sequence, is the k-generalized Fibonacci sequence F n k) ) n which is defined by the initial values 0, 0,..., 0, 1 k terms) and such that each term afterwards is the sum of the k preceding terms. In this paper, we generalize Luca and Oyono s method to prove that the Diophantine equation F m k) ) s + F k) m+1 )s = F n k) has no solution in positive integers n, m, k and s, if 3 k min{m, log s}. 1. Introduction Let F n ) n 0 be the Fibonacci sequence given by F n+ = F n+1 + F n, for n 0, where F 0 = 0 and F 1 = 1. A few terms of this sequence are 0, 1, 1,, 3, 5, 8, 13, 1, 34, 55, 89, 144, 33,.... The Fibonacci numbers are well-known for possessing wonderful and amazing properties consult [7] together with their very extensive annotated bibliography for additional references and history). Among the several pretty algebraic identities involving Fibonacci numbers, we are interested in the following one F n + F n+1 = F n+1, for all n ) In particular, this naive identity which can be proved easily by induction) tell us that the sum of the square of two consecutive Fibonacci numbers is still a Fibonacci number. In order to check if the sum of higher powers of two consecutive Fibonacci numbers could also belong to this sequence, Marques and Togbé [9] showed that, for a fixed s, if Fm s + Fm+1 s is a Fibonacci number for infinitely many m, then s = 1 or. In 011, Luca and Oyono [8] solved this problem completely, showing that the Diophantine equation F s m + F s m+1 = F n 1.) has no solutions m, n, s) with m and s 3. Le and denote F k) := F n k) ) n k ), the k-generalized Fibonacci sequence whose terms satisfy the recurrence relation F k) n+k = F k) n+k 1 + F k) n+k + + F k) n, 1.3) with initial conditions 0, 0,..., 0, 1 k terms) and such that the first nonzero term is F k) 1 = 1. The above sequence is one among the several generalizations of Fibonacci numbers. Such a sequence is also called k-step Fibonacci sequence, the Fibonacci k-sequence, or k-bonacci 1

2 ANA PAULA CHAVES AND DIEGO MARQUES sequence. Clearly for k =, we obtain the well-known Fibonacci numbers F n ) = F n, for k = 3, the Tribonacci numbers F n 3) = T n and for k = 4, the Tetranacci numbers F n 3) = Q n. The aim of this paper is to study a generalization of the equation 1.) in the k-generalized Fibonacci context. More precisely, we have the following result Theorem 1.1. The Diophantine equation F k) m ) s + F k) m+1 )s = F k) n 1.4) has no solution in positive integers m, n, k and s, with 3 k {m, log s}. We recall that for k ) m 1 there are trivial solutions to the Eq. 1.4) for all k. Our method follows roughly the following steps: First, we use Matveev s result [11] on linear forms in logarithms to obtain an upper bound for s in terms of m. When m is small, say m 1394, we use Dujella and Pethö s result [1] to decrease the range of possible values and then let the computer check the non existence of solutions in this case. To the case where m 1395, we use again linear forms in logarithms to obtain an upper bound for s, now in terms of k, which, combined with the hypothesis k < log s, gives us an absolute upper bound for s. In the final step, we use continued fractions to lower the bounds and then let the computer cover the range of possible values, showing that there are no solutions also in this case, which completes the proof.. Auxiliary results We know that the characteristic polynomial of F n k) ) n is ψ k x) := x k x k 1 x 1 and it is irreducible over Q[x] with just one zero outside the unit circle. That single zero is located between 1 k ) and as can be seen in [10]). Also, it was proved in [1, Lemma 1] that α n F k) n α n 1, for all n 1,.1) where α is the dominant root of ψ k x). Recall that for k =, one has the useful Binet s formula F n = αn β n 5, where α = 1 + 5)/ = β 1. There are many closed formulas representing these k- generalized Fibonacci numbers, as can be seen in [3, 4, 5, 6]. However, we are interested in the simplified Binet-like formula due to G. Dresden [, Theorem 1] : F k) n = k α i 1 + k + 1)α i ) αn 1 i,.) for α = α 1,..., α k being the roots of ψ k x). Also, the contribution of the roots inside the unit circle in formula.) is almost trivial. More precisely, it was proved in [] that E n k) < 1,.3) where E n k) := F k) n gα, k)α n 1 and gx, y) := x 1)/ + y + 1)x )). We shall use a few times a result due to Matveev [11], which states the following

3 THE DIOPHANTINE EQUATION F n k) ) s + F k) n+1 )s = F m k) 3 Theorem.1 Matveev). Let α 1,..., α t real algebraic numbers, b 1,..., b t nonzero integers and Λ := α b 1 1 αbt t 1. Let D = [Qα 1,..., α t ) : Q] and A 1,..., A t positive real numbers satisfying Let B max{ b 1,..., b t }. Also define A j max{dhα j ), log α j, 0.16}, for j = 1,..., t. C t,d := t+3 t 4.5 D 1 + log D). If Λ 0, then Λ > exp C t,d 1 + log B)A 1 A t ). Where for an algebraic number η we write hη) for its logarithmic or Weil s) height whose formula is ) hη) = 1 d log a 0 + logmax{ η i), 1}), d with d being the degree of η over Q, and fx) := a 0 d X η i) ) Z[X] the minimal primitive polynomial over the integers having positive leading coefficient and η = η 1). Another result which will play an important role in our proof is due to Dujella and Pethö [1] Lemma.. Let M be a positive integer, let p/q be a convergent of the continued fraction of the irrational γ such that q > 6M, and let µ be some real number. Let ɛ := µq M γq. If ɛ > 0, then there is no solution to the inequality in positive integers n and s with 0 < nγ s + µ < AB n logaq/ɛ) log B n M. The next theorem about continued fractions is due to Legendre, and will help us to finish the demonstration. Theorem.3. Let ξ be a real number. If a rational number a/b is such that then it is a convergent of ξ. ξ a < 1 b b, Now, we are ready to deal with the proof of Theorem 1.1.

4 4 ANA PAULA CHAVES AND DIEGO MARQUES and 3. Proof of Theorem 1.1 First, observe that by using the estimates in.1), we obtain α n 1 > F k) n α n < F k) n = F k) m ) s + F k) m+1 )s > α m )s + α m 1)s = α m )s 1 + α s ) > α m 1)s, = F k) m ) s + F k) m+1 )s < α m 1)s + α ms = α m 1)s 1 + α s ) < α ms+1, where we used that 1 + α s < α s+1 for all k 3, which is an immediate consequence of α s α 1) > 7/4) 3 7/4 1) = 109/56 > 1. Now, the estimate α n < F n k) < α n 1 together with the previous estimates yield m 1)s + 1 < n < ms + 3. In conclusion, we have proved that if m, n, k, s) is a solution of Eq. 1.4), then n {m 1)s +, m 1)s + 1,..., ms + } An inequality for s in terms of m. Using formula.), we rewrite 1.4) as F k) m+1 )s gα n 1 = F k) m ) s E n k). 3.1) Since E n k) < 1/, we have that F k) m+1 )s gα n 1 [F m k) ) s 1/, F m k) ) s + 1/], which is positive. Now applying the absolute value and the triangle inequality in 3.1), we obtain Dividing by F k) m+1 )s to get gα n 1 F k) m+1 )s < 1 k) + F m ) s < F m k) ) s. ) 1 m+1 )s < F m k) s F k). 3.) m+1 gα n 1 F k) In order to give an upper bound to the right side of 3.), we use that Thus, for k 3 F k) m+1 = F m k) + F k) m F k) k) m k+1 = F m F k) m k. F k) m+1 F m k) = F k) m k F m k) 1 > ) k Now, using estimate 3.3) in 3.), we get our firsey inequality: gα n 1 F k) 1 m+1 )s < 1.65 s. 3.4) In a first application of Matveev s result, take t := 3, γ 1 := F k) m+1, γ := α, γ 3 := g, e b 1 := s, b := n 1, b 3 := 1. Now, consider Λ 1 := gα n 1 F k) m+1 ) s 1,

5 THE DIOPHANTINE EQUATION F n k) ) s + F k) n+1 )s = F m k) 5 which is positive an immediate consequence of what we observed right after 3.1)). The algebraic number field containing γ 1, γ and γ 3 is K := Qα), whose degree is D := [K : Q] = k. For the value of B, note that max{ b 1, b, b 3 } = max{s, n 1, 1} = max{s, n 1}, but since s + m 1)s + n, then s < s + 1 n 1, and we can take B := n 1. Now we need to estimate the logarithmic heights hγ 1 ), hγ ) and hγ 3 ). Since hγ 1 ) = log F k) m+1 < m log, and max{dhγ 1), log γ 1, 0.16} < km log, we can take A 1 := km log. Similarly, since hγ ) = hα) = log α/k < log /k, where we used the fact that α is the only root of ψ k x) outside the unit circle, and max{dhγ ), log γ, 0.16} < log, we can take A := log. For hγ 3 ), we have ) hγ 3 ) = hg) = 1 k log a 0 + logmax{ g i, 1}). d First, note that applying the conjugation over K) on g = gα, k), we obtain g i = gα i, k) for all 1 i k, and since [Qα) : Qg)] = 1, we have d = k. Put Gx) = k ) α i 1 x Q[x]. + α i )k + 1) The leading coefficient a 0 of the minimal polynomial of g over the integers divides k + α i )k + 1)). But, k k + α i )k + 1)) = k + 1)k Since ) k + 1 α i = k + 1)k ψ k y) < max{y k, y k y + 1} < k, for all 0 < y <, we get the following upper bound for a 0 : a 0 k + 1) k ψ k ) < k k + 1) k. k + 1 ψ k ), k + 1 For all k 3, we have k + 1) < k 3 and then log a 0 < klogk + 1) + log ) < 3k log k. Now, we need to estimate g i. If i k, we have α i < 1, which gives us Hence, + k + 1)α i ) k + 1) α i > k 1. α i 1 + k + 1)α i ) α i + 1 < 1 g i < 1. In the case of the dominant root, α, is easy to see that g < 1, as follows g = α 1 + k + 1)α ) < 1 k + 1 1, k 1

6 6 ANA PAULA CHAVES AND DIEGO MARQUES where we used in the above inequality tha+1 k 1 for k 3. Therefore, max{ g i, 1} = 1, and we finally obtain ) hγ 3 ) = 1 k log a 0 + log max{ g i, 1}) < 1 3k log k) < 3 log k. k k So we can take A 3 := 3k log k. Now, applying Theorem.1) to get a lower bound for Λ 1, Λ 1 > exp k 1 + log k)1 + logn 1)) 3k log k)log )km log )) Λ 1 > exp k 4 m log k log k)1 + logn 1))). Using the previous estimates for n in terms of s and m, is easy to see that n 8, where the inequality 1 + logn + 1) < logn 1) holds, and n 1 ms + 1. Then, Λ 1 > exp mk 4 log k) logms + 1)). Comparing the above inequality with 3.4), we get 1.65 s > exp mk 4 log k) logms + 1)). Now, taking logarithms in the previous inequality, we have which leads to log s log 1.65 > mk 4 log k) logms + 1), s < mk 4 log k) logms + 1). Since m, s 3, then log m, log s 1 and so logms + 1) < logms) < 4 log m log s. Thus, we obtain s log s < m 5 log m) 3, 3.5) where we also used the hypothesis k m. Now, we are going to use the following argument showed by Luca and Oyono [8], for x > e, x < A x < A log A, 3.6) log x whenever A 3. Thus, taking A := m 5 log m) 3, inequality 3.5) yields s < m 5 log m) 3 ) log m 5 log m) 3 ) < m 5 log m) log m + 3 log log m) < m 5 log m) log m), In the last chain of inequalities, we have used that log log m < log m and log m < 34.9 log m holds for all m 3. Hence, we have the following result. Lemma 3.1. If m, n, k, s) is a nontrivial solution in positive integers of Eq. 3 k min{m, log s}, then 1.4) with s < m 5 log m) )

7 THE DIOPHANTINE EQUATION F n k) ) s + F k) n+1 )s = F m k) The case of small m. Following our plan, we next consider the cases when m [3, 1394], and after finding an upper bound for n, the next step is to reduce it and then let the computer handle with the possible solutions. To do that, first observe that in this case s < ) 5 log 44) 4 s < Thus, we obtain the following upper bounds for k and n: n ms + n < n < , k log s k log ) k 77. Also note that n < ms +, gives us s > n )/1394. Now, in order to use the reduction method due to Dujella and Pethö [1], take ) 1 Γ 1 := n 1) log α log s log F k) m+1 g. 3.8) Then Λ 1 = e Γ 1 1 > 0, since Λ 1 > 0, and from 3.4) we have 0 < Γ 1 < e Γ 1 1 = Λ 1 < 1.65 s. Dividing both sides of the previous inequality by log F k) m+1, and using that s > n )/1394, we obtain ) ) log α logα/g) 0 < n log F k) s m+1 log F k) < ) n m+1 With, γ m,k := log α log F k) m+1 the previous inequality yields, µ m,k := logα/g) log F k) m+1, A :=.01 and B := 1.65) , 0 < nγ m,k s µ m,k < AB n. 3.9) Let us show that γ m,k is an irrational number. Indeed, if γ m,k Q, we have α q = F m+1) k) p for some p, q Q, with q > 0. Conjugating this relation over K), taking the product and then the absolute value, we get k q α i = F k) kp m+1) 1. On the other hand, we already know that this module is equal to one, since α, α,, α k are the roots of ψ k x), which contradicts the relation above. Thus, γ m,k Q. Take M := Let q t,m,k be the denominator of the t-th convergent to γ m,k. To do the following calculation, we have used the Mathematica 9 software on a OSX , 1.8 GHz Intel Core i5 with 4GB of RAM. Calculating the smallest value of q 700,m,k, for 4 m 1394 and 3 k min{m, 77}, we have that q 700,m,k > > 6M, and for the same range, ɛ 700,m,k > , which means that ɛ 700,m,k is always positive this is not true for ɛ 600,m,k ). Hence, by Lemma., there are no integer solutions for 3.9) when max 3 k 77 3 m 1394 logaq 700,m,k /ɛ 700,m,k ) log B n

8 8 ANA PAULA CHAVES AND DIEGO MARQUES log / ) n log1.65) ) n Therefore, we have n and consequently s 75756, since s n )/m 1). Also, using tha log s, we ge 13. A computer search with Mathematica revealed no solutions to the equation 1.1) in the range 3 m 1394, 3 k 13, 1 s and m 1)s + n ms +. This finishes the case m [3, 1394] Finding absolute upper bounds. From now on, we assume that m Set Lemma 3.1 gives us where we used that X m := E mk) s gα m 1. X m < E mk) m 5 log m) 4 gα m 1 < m 5 log m) 4 < ) m , 3.10) α m 1)/ < α m 1, holds for m In particular, X m < α 697 < 7/4) 697 < Similarly, X m+1 = E m+1k) s gα m+1 < 1/) m 5 log m) 4 1/)α m+1 < 1 α m 1)/. We now write F m k) ) s = g s α m 1)s 1 + E ) s mk) gα m 1 and F k) m+1 )s = g s α ms 1 + E ) s m+1k) gα m. 3.11) If E m k) > 0, then 1 < 1 + E ) s mk) gα m 1 = 1 + E ) s mk) gα m 1 < e s Emk) /gαm 1 ) < 1 + Xm, because e x < 1 + x, for 0 < x < 1.5, while if E m k) < 0, then 1 > 1 + E ) s mk) gα m 1 = 1 E ) s mk) gα m 1 = exp s log 1 E )) mk gα m 1 > 1 X m, now because log1 x) > x, for 0 < x < The same inequalities are true if we replace m by m + 1. Combining these two facts with 3.11), we can see how F m k) ) s is well approximated by g s α m 1)s, as follows ) s F m k) ) s = g s α m 1)s 1 + Emk) < g s α m 1)s 1 + X gα m 1 m ) F k) m ) s g s α m 1)s < X m g s α m 1)s, 3.1) F k) m ) s = g s α m 1)s 1 + Emk) gα m 1 ) s > g s α m 1)s 1 X m ) F k) m ) s g s α m 1)s > X m g s α m 1)s, 3.13)

9 thus THE DIOPHANTINE EQUATION F n k) ) s + F k) n+1 )s = F m k) 9 F k) m ) s g s α m 1)s < X m g s α m 1)s and F k) m+1 )s g s α ms < X m+1 g s α ms. 3.14) We now go back to 1.4) and rewrite it as which gives us gα n 1 + E n k) = F k) n = F m k) ) s + F k) m+1 )s = g s α m 1)s + g s α ms gα n 1 g s α m 1)s 1 + α s ) F k) m ) s g s α m 1)s ) Dividing both sides by g s α ms, g 1 s α n ms+1) 1 + α s ) < + + F m k) ) s g s α m 1)s) F k) m+1 )s g s α ms), + F k) m+1 )s g s α ms ) + E n k) < X m g s α m 1)s + X m+1 g s α ms g s α ms X m + X m ) where 3.15) holds since α s > 7/4) 3 > 5.35, and then /α s < /5.35 < Now, we need a lower bound to g s α ms in terms of α m 1 : ) g s m 1 ms 1 s ) 7 s ) 7 m )s m 1 α > > 49 3 ) ) 607 > > 10 3, therefore g s α ms ) 1 < 0.001/α m 1)/. Using it in 3.15) jointly with 3.10), we obtain Hence, we conclude that g 1 s α n ms+1) 1 + α s ) < α m α m 1 g 1 s α n ms+1) 1 < 1 α s +.39 α m 1 + α m <.39 α m 1, 3.16) < 3.39 α l, 3.17) where we put l := min{s, m 1 }. Having in mind to use Matveev s result once more, we now set Λ := g 1 s α n ms+1) 1, 3.18) but before this, we must show that Λ 0. Indeed, if Λ = 0, then conjugating over Qα) and taking the product of all conjugates, we have k s 1 k g i ) = g i ) s 1 k n ms+1) = α i = 1, but on the other hand, we saw that g i < 1 for all 1 i k which contradicts the above identity. Thus, Λ 0.

10 10 ANA PAULA CHAVES AND DIEGO MARQUES So, we take t =, λ 1 := g, λ := α and c 1 := 1 s, c := n ms+1). Again, K := Qα) and D := [K : Q] = k. Now, to choose the value of B max{ c 1, c } = max{s 1, n ms+1) }, observe that n m 1)s + n ms + 1) s 1) n ms + n ms + 1) s 1. Hence n ms + 1) < s 1, and we can take B := s 1. As in the previous application of Theorem.1, we can take A 1 := 3k log k and A := log. We thus get that Λ > exp k 1 + log k)1 + logs 1)) 3k log k)log )) > exp k 3 log k1 + logs 1))). Combining the last inequality with 3.17) we obtain If l = s, then 3.19) becomes and using tha < log s, 3.39 α l > exp k 3 log k1 + logs 1))) log 3.39 l < log α log α 108 k 3 log k1 + logs 1)) l < k 3 log k1 + logs 1)). 3.19) s < k 3 log k1 + logs 1)), s < log s) 3 log log s1 + logs 1)), which is valid only for s If l = m 1)/, we use Lemma 3.1 and 3.19) to get m 1 < k 3 log k1 + log m 5 log m) 4 )) < k 3 log k log m + 4 log log m), thus, m log m < k 3 log k, 3.0) where we used that, for m 1395, the inequality log log m < 0.31 log m holds, giving log m + 4 log log m < log m < 13 log m, and that m 1 > m/ Now, using again 3.6), we gain an upper bound for m in terms of k: m < k 3 log k) log k 3 log k) < k 3 log k). 3.1) Again, by Lemma 3.1, now combined with 3.1), we have an upper bound for s in terms of k, which will give us, as in the previous case, an absolute upper bound for s: s < k 3 log k) ) 5 log k 3 log k) )) 4 < k 5 log k) log k + log log k) 4 s < k 5 log k) 14,

11 and since k log s, we have THE DIOPHANTINE EQUATION F n k) ) s + F k) n+1 )s = F m k) 11 s < log s) 5 log log s) 14, which is true only for s < , so k < log ) < 3. In both cases, we have s < and k < 3, which are still very high to let the computer do the calculation. In order to reduce these bounds, we use a criterion due to Legendre Theorem.3) about the convergents in a continued fraction. To use it, we go back to 3.17) to get, using that s 0 and m 1394, the following upper bound: Set, Λ < 1 α s +.39 α m 1)/ < α0 α 697 < ) Γ := s 1) logg 1 ) ms + 1 n) log α. Note that Λ = e Γ 1, then by the previous inequality, we have Now, since we have > Λ = e Γ 1 e Γ 1 e Γ < ) Γ e Γ e Γ 1 < ) Λ < ) 1 α s +.39 ) α m 1)/ 1 s 1) logg 1 ) ms + 1 n) log α < ) α s +.39 ) α m 1)/ then dividing it by s 1) log α, logg 1 ) log α ms + 1 n s 1 < ) s 1) log α 1 α s +.39 ) α m 1)/. 3.4) Assume next that s 9. Then α s > 7/4) 9 > s and, since m 1395, we have α m 1)/ > 7/4) 697 > s, which in 3.4) gives us logg 1 ) log α ms + 1 n s 1 < s 1). 3.5) By Theorem.3, inequality 3.5) implies that the rational number ms + 1 n)/s 1) is a convergent to β k = logg 1 ))/log α). Let [a k) 0, ak) 1, ak),...] be the continued fraction of β k and p k) t /q k) t its t-th convergent. Assume that ms + 1 n)/s 1) = p k) /q k) for some. Then s 1 = d k q k), for some positive integer d k, so s 1 q k). On the other hand, using again the Mathematica software, we get that min k {3,3} qk) 50 > > > s 1, therefore 1 50, for all 3 k 3. Also using Mathematica, we observe that a tk +1 max{a k) t } < , for k {3, 3} and t {1, 51}. From the properties of,

12 1 ANA PAULA CHAVES AND DIEGO MARQUES continued fractions, we have β k ms + 1 n s 1 = β k pk) q k) > 1 a tk +1 + )q k) ) which contradicts 3.5). So, s 91, and k < {3, 4, 5}. d k s 1) s 1), 3.4. The final step. Let s go back to 3.16). Divide it across by 1 + α s ) to obtain, α n ms+1) g 1 s 1 + α s ) 1 1 < ) α m 1)/ Now, set t := ms + 1 n. Using inequality 3.), we get the following relations and, g 1 s α t 1 < t > s 1) log g 1 log α > 0.68s 0.69, g 1 s α t 1 > t < s 1) log g 1 log α < 1.7s 1.6. log ) log α log ) log α Therefore, t [ 0.68s , 1.7s 1.6 ]. A computational search for the range 0 s 91, 3 k 5 and t in the previous interval, returned { α t g 1 s } min 1 + α s ) 1 > > m 33, α m 1)/ which contradicts the fact that m Hence, the theorem is proved. 4. acknowledgements The second author is supported by grants from DPP-UnB, FAP-DF and CNPq-Brazil. References [1] J. J. Bravo, F. Luca, On a conjecture about repdigits in k-generalized Fibonacci sequences, Publ. Math. Debrecen 8 Fasc ). [] G. P. Dresden, Z. Du, A simplified Binet formula for k-generalized Fibonacci numbers, Journal of Integer Sequences, 17 4) 014). [3] D. E. Ferguson, An expression for generalized Fibonacci numbers, Fibonacci Quart ), [4] I. Flores, Direct calculation of k-generalized Fibonacci numbers, Fibonacci Quart ), [5] H. Gabai, Generalized Fibonacci k-sequences, Fibonacci Quart ), [6] D. Kalman, Generalized Fibonacci numbers by matrix methods, Fibonacci Quart ), [7] D. Kalman, R. Mena, The Fibonacci numbers exposed, Math. Mag ), no. 3, [8] F. Luca, R. Oyono, An exponential Diophantine equation related to powers of two consecutive Fibonacci numbers. Proc. Japan Acad. Ser. A, ) p

13 THE DIOPHANTINE EQUATION F n k) ) s + F k) n+1 )s = F m k) 13 [9] D. Marques, A. Togbé, On the sum of powers of two consecutive Fibonacci numbers. Proc. Japan Acad. Ser. A, ) p [10] A. Wolfram, Solving generalized Fibonacci recurrences, Fibonacci Quart ), [11] E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in the logarithms of algebraic numbers, II, Izv. Ross. Akad. Nauk Ser. Mat. 64 6) 000) ; translation in: Izv. Math.64 6) 000) [1] A. Dujella, A. Pethő, Generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. ) 49195) 1998) MSC010: 11B39 Instituto de Matemática e Estatística, Universidade Federal de Goiás, Goiânia, , Brazil address: apchaves@ufg.br Departamento de Matemática, Universidade de Brasília, Brasília, , Brazil address: diego@mat.unb.br