Fibonacci Diophantine Triples

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1 Fibonacci Diophantine Triples Florian Luca Instituto de Matemáticas Universidad Nacional Autonoma de México C.P , Morelia, Michoacán, México László Szalay Institute of Mathematics and Statistics University of West Hungary 9400, Sopron, Erzsébet utca 9, Hungary May 10, 009 Abstract In this paper, we show that there are no three distinct positive integers a, b, c such that ab + 1, ac + 1, bc + 1 are all three Fibonacci numbers. 1 Introduction A Diophantine m-tuple is a set of {a 1,..., a m } of positive rational numbers or integers such that a i a j +1 is a square for all 1 i < j m. Diophantus found the rational quadruple {1/16, 33/16, 17/4, 105/16}, while Fermat found the integer quadruple {1, 3, 8, 10}. Infinitely many Diophantine quadruples of integers are known and it is conjectured that there is no Diophantine quintuples. This was almost proved by Dujella [5], who showed that there can be at most finitely many Diophantine quintuples and all of them are, at least in 1

2 theory, effectively computable. In the rational case, it is not known that the size m of the Diophantine m-tuples must be bounded and a few examples with m = 6 are known by the work of Gibbs [8]. We also note that some generalization of this problem for squares replaced by higher powers of fixed, or variable exponents were treated by many authors see [1], [9], [], [13] and [10]. In the paper [7], the following variant of this problem was treated. Let r and s be nonzero integers such that = r + 4s 0. Let u n n 0 be a binary recurrence sequence of integers satisfying the recurrence u n+ = ru n+1 + su n for all n 0. It is well-known that if we write α and β for the two roots of the characteristic equation x rx s = 0, then there exist constants γ, δ IK = Q[α] such that u n = γα n + δβ n holds for all n 0. 1 Assume further that the sequence u n n 0 is nondegenerate, which means that γδ 0 and α/β is not root of unity. We shall also make the convention that α β. A Diophantine triples with values in the set U = {u n : n 0}, is a set of three distinct positive integers {a, b, c}, such that ab + 1, ac + 1, bc + 1 are all in U. Note that if u n = n + 1 for all n 0, then there are infinitely many such triples namely, take a, b, c to be any distinct powers of two. The main result in [7] shows that the above example is representative for the sequences u n n 0 with real roots for which there exist infinitely many Diophantine triples with values in U. The precise result proved there is the following. Theorem 1. Assume that u n n 0 is a nondegenerate binary recurrence sequence with > 0 such that there exist infinitely many sextuples of nonnegative integers a, b, c; x, y, z with 1 a < b < c such that ab + 1 = u x, ac + 1 = u y, bc + 1 = u z. Then β {±1}, δ {±1}, α, γ Z. Furthermore, for all but finitely many of the sextuples a, b, c; x, y, z as above one has δβ z = δβ y = 1 and one of the following holds: i δβ x = 1. In this case, one of δ or δα is a perfect square;

3 ii δβ x = 1. In this case, x {0, 1}. No finiteness result was proved for the case when < 0. The case δβ z = 1 is not hard to handle. When δβ z 1, results from Diophantine approximations relying on the Subspace Theorem, as well as on the finiteness of the number of solutions of nondegenerate unit equations with variables in a finitely generated multiplicative group and bounds for the greatest common divisors of values of rational functions at units points in the number fields setting, allow one to reduce the problem to elementary considerations concerning polynomials. The Fibonacci sequence F n n 0 is the binary recurrent sequence given by r, s = 1, 1, F 0 = 0 and F 1 = 1. It has α = 1+ 5/ and β = 1 5/. According to Theorem 1, there should be only finitely many triples of distinct positive integers {a, b, c} such that ab+1, ac+1, bc+1 are all three Fibonacci numbers. Our main result here is that in fact there are no such triples. Theorem. There do not exist positive integers a < b < c such that ab + 1 = F x, ac + 1 = F y, bc + 1 = F z, 3 where x < y < z are positive integers. Let us remark that since the values n = 1,, 3 and 5 are the only positive integers n such that F n = k + 1 holds with some suitable integer k see [6], it follows from Theorem that all the solutions of equation 4 under the more relaxed condition 0 < a b c are { 1, 1, Ft 1; 3, t, t, t 3; a, b, c; x, y, z =,, F t 1/; 5, t, t, t 4, t 0 mod 3; Note also that there are at least two rational solutions 0 < a < b < c, namely a, b, c; x, y, z = /3, 3, 18; 4, 7, 10, 9/, /3, 1; 9, 10, 11. It would be interesting to decide whether equation 3 has only finitely many rational solutions a, b, c; x, y, z with 0 < a < b < c, and in the affirmative case whether the above two are the only ones. 3

4 Proof of Theorem.1 Preliminary results In the sequel, we suppose that 1 a < b < c and 4 x < y < z. We write L n n 0 for the companion sequence of the Fibonacci numbers given by L 0 =, L 1 = 1 and L n+ = L n+1 + L n for all n 0. It is well-known see, for example, Ron Knott s excellent web-site on Fibonacci numbers [11], or Koshy s monograph [1], that the formulae F n = αn β n α β and L n = α n β n hold for all n 0, where α = 1 + 5/ and β = 1 5/. We shall need the following statements. Lemma 3. The following divisibilities hold: i gcdf u, F v = F gcdu,v ; { u Lgcdu,v, if ii gcdl u, L v = 1 or, otherwise; { u Lgcdu,v, if iii gcdf u, L v = 1 or, otherwise. gcdu,v gcdu,v v gcdu,v v gcdu,v 1 mod ; 1 mod ; Proof. This is well-known see, for instance, the proof of Theorem VII. in [3]. Lemma 4. The following formulae hold: F u 1 L u+1, if u 1 mod 4; F u+1 L u 1, if u 3 mod 4; F u 1 = F u L u+, if u mod 4;, if u 0 mod 4. F u+ L u Proof. This too is well-known see, for example, Lemma in [14]. Lemma 5. Let u 0 be a positive integer. Put u0 β ε i = log α i 1, δ i = log α α 1 + β 1i 1 α 5 u0 4

5 for i = 1,, respectively. Here, log α is the logarithm in base α. Then for all integers u u 0, the two inequalities α u+ε L u α u+ε 1 4 and hold. α u+δ F u α u+δ 1 5 Proof. Let c 0 = 1, or 5, according to whether u n = L n or u n = F n, respectively. Obviously, } L u αu + β u 0 α u 1 + β u 0 α u α u 1 + u0 β α, F u c 0 c 0 c 0 which prove the upper bounds from the formulae 4 and 5, respectively. Similarly, } L u αu β u 0 α u 1 β u 0 α u α u 1 u0 β α F u c 0 c 0 c 0 lead to the lower bounds from the formulae 4 and 5, respectively. Lemma 6. Suppose that a > 0 and b 0 are real numbers, and that u 0 is a positive integer. Then for all integers u u 0, the inequality aα u + b α u+κ holds, where κ = log α a + Proof. This is obvious. b α u 0. Lemma 7. Assume that a, b, z are integers. Furthermore, suppose that all the expressions appearing inside the gcd s below are also integers. Then the following hold: i If a b, then z+b 4 ; gcd z+a, z+b 4 a b. Otherwise, 5 gcd z+a, z+b 4 =

6 ii If 3a b, then gcd z+a z+a 8., 3z+b 8 3a b. Otherwise, gcd z+a, 3z+b 8 = Proof. This is an easy applications of the Euclidean algorithm. Lemma 8. Assume that z 8 is an integer. Then the following hold: i If z is odd then L < F z ; ii If z is even then L z Proof. For i, note that < F z. L = L + 1 L + = F z + F z +, and the right hand side above is easily seen to be smaller than F z when z 8. For ii, we similarly have L z L z + = F z 3 + F + < F z, where the last inequality is equivalent to F z 3 + < F z, or < F z 4, which is fulfilled for z 8. Lemma 9. All positive integer solutions of the system 3 satisfy z y. Proof. The last two equations of system 3 imply that c divides both F y 1 and F z 1. Consequently, c gcdf y 1, F z 1. 6 Obviously, F z = bc+1 < c ; hence, F z < c. From 6, we obtain F z < F y. Clearly, α z 1 < F z < F y < αy Since y 5 entails α y + 1 < 4 5 α y, we get α z 1 < α y, which easily leads to the conclusion that y z. 6

7 . The Proof of Theorem By Lemma 9, we have Fz < gcdf z 1, F y 1. 8 Applying Lemma 4, we obtain gcd F z i gcdf z 1, F y 1 = gcd, F y j gcd F z i, L y+j F z i L z+i gcd L z+i, F y j, F y j L y+j gcd 9 L z+i, L y+j where i, j {±1, ±}. The values i and j depend on the residue classes of z and y modulo 4, respectively. In what follows, we let d 1, d, d 3 and d 4 denote suitable positive integers which will be defined shortly. Lemma 3 yields m 1 = F gcd z i, y j = F z i. 10 d 1 The second factor m on the right hand of 9 can be 1,, or The third factor m 3 is again 1,, or, m = L gcd z i, y+j = L z i d. 11 m 3 = L gcd z+i, y j = L z+i d 3. 1 Finally, if the fourth factor m 4 is neither 1 nor, then We now distinguish two cases. Case 1. z 150. m 4 = L gcd z+i, y+j = L z+i d In this case, we ran an exhaustive computer search to detect all positive integer solutions of system 3. Observe that we have F x 1F y 1 a =, 4 x < y < z 150. F z 1 Going through all the eligible values for x, y and z, and checking if the above number a is an integer, we found no solution to system 3. 7

8 Case. z > 150. In this case, Lemma 5 gives < δ 1 for F z. Hence, α z < F z. If d k 5 holds for all k = 1,, 3, 4, then the subscripts z±i d k of the Fibonacci and Lucas numbers appearing in are at most z±i 10 now gives that ε < 0.5 and δ < 1 hold for L z±i 10 because z±i > 14. Now formulae 8 13 lead to 10 α z < F z < α z i z i 10 and F z i 10 each. Lemma 5, respectively, z+i z+i , 14 which implies that z < z , contradicting the fact that z > 150. From now on, we analyze those cases when at least one of the numbers d k for k = 1,, 3, 4, which we will denote by d, is less than five. First assume that d = 4. Then either z+η 1i = y+η j, or z+η 1i = y+η j, where η 1, η {±1}. If the first equality holds, then Lemma 9 leads to z = 4y+4η j η 1 i y. Thus, z y η 1 i 4η j 10, contradicting the fact that z > 150. The second equality leads to y = 3z+3η 1i 4η j. In this case, 4 y + η j = 3z + 3η 1i + tj, 15 8 where t = 4η η {±8, 0} for η {±1}. Applying Lemma 7, we get z + η gcd 1 i, y + η j z + η = gcd 1 i, 3z + 3η 1i + tj 8 3η 1 η 1 i tj 14, 16 for all η 1, η η 1, η {±1}. For the last inequality above, we used Lemma 7 together with the fact that 3η 1 η 1 tj 0. Indeed, if 3η 1 η 1 tj = 0, then 3 tj, and since t {±8, 0}, j {±1, ±}, we get that t = 0, therefore η = η. Since also 3η 1 η 1 = tj = 0, we get η 1 = η 1, therefore η 1, η = η 1, η, which is not allowed. 8

9 Continuing with the case d = 4, since F 14 < L 14 = 843 and z±i > 18, we 8 get that ε < 0.5 and δ < 0.5, where these values correspond to L z±i and 8, respectively. It now follows that F z±i 8 α z < α z±i L α z Thus, z < log α 843 < 116, which completes the analysis for this case. Consider now the case d = 3. The only possibility is z+η 1i = y+η j for 6 some η 1, η {±1}. Together with Lemma 9, we get z = 3y+3η j η 1 i y. Consequently, z y η 1i 3η j 8, which is impossible. Assume next that d =. Then z+η 1i = y+η j for some η 4 1, η {±1}. We get that y = z+η 1i η j. Thus, y+η j = z+η 1i+tj with t = η 4 η {±4, 0}. By Lemma 7, we have z + η gcd 1 i, y + η j z + η = gcd 1 i, z + η 1i + tj 4 η 1 η 1 i tj 1. The argument works assuming that the last number above is not zero for η 1, η η 1, η {±1}. Assume that it is. Then η 1 η 1 i = tj. Clearly, tj is always a multiple of 4. If it is zero, then t = 0, so η = η. Then also η 1 η 1i = tj = 0, therefore η 1 = η 1. Hence, η 1, η = η 1, η, which is not allowed. Assume now that t 0. Then η 1 η 1i 0, so η 1 = η 1. Also, t 0, therefore η = η. We get that η 1 i = 4η j, therefore η 1 i = η j. Thus, i = ±1 and j = ±. In particular, z is odd and y is even. Now z + η 1 i/ is divisible by a larger power of than y + η j/. A quick inspection of formulae defining m 1, m, m 3 and m 4 together with Lemma 3 ii and iii, shows that the only interesting cases are when k = 1 or since m 3 and m 4. Thus, η 1, η = 1, 1 or 1, 1. Hence, η 1, η = 1, 1 or 1, 1, and here we have that m 3 and m 4 anyway. This takes care of the case when η 1 η 1 i tj = 0. Continuing with d =, since z±i 37, Lemma 5 yields ε 4, δ < 0.1. We then get the estimate α z < α z±i L α z , leading to z < log α 3 < 150.5, which is a contradiction. Finally, we assume that d = 1. The equality z±i = y±j leads to z = y i ± j. Obviously, here i ± j must be positive, otherwise we would 9

10 get z y. Note that in the application of Lemma 4, both z and y are classified according to their congruence classes modulo 4. The following table summarizes the critical cases of d = 1. Only 6 layouts in Table 1 below need further investigations the sign abbreviates a contradiction. z, y 4 i, j possible equalities consequence 1 1, 1 1, 1 z = y + : x y mod 4 1, 1, = y+ z = y + 3 : d must be even z+1 1, = y+ z = y + 1 : z y 1 mod 4 3 1, 3 1, 1 z = y + : d 1 must be even 4 1, 0 1, = y+ z = y + 3 : x y + 1 mod 4 z+1 1, = y+ z = y + 1 is possible d 3 = 1 5 3, 1 1, 1 z = y + is possible d 4 = 1 z+1 6 3, 1, = y+ z = y + 1 : d must be even 1, = y+ z = y + 3 : x y + 1 mod 4 7 3, 3 1, 1 z = y + : x y mod 4 z+1 8 3, 0 1, = y+ z = y + 1 : x y 1 mod 4 1, = y+ z = y + 3 is possible d 3 = 1 z 9, 1, 1 = y 1 z = y + 1 : d 1 must be even z, 1 z = y + 3 : x y + 1 mod 4 z 10,, = y+ z = y + 4 : d must be even z 11, 3, 1 z = y + 3 : d 1 must be even z, 1 = y 1 z = y + 1 : x y 1 mod 4 z 1, 0, = y+ z = y + 4 : x y + mod 4 z 13 0, 1, 1 = y 1 z = y + 1 : x y 1 mod 4 z, 1 z = y + 3 is possible d 4 = 1 z 14 0,, = y+ z = y + 4 : x y + mod 4 z 15 0, 3, 1 z = y + 3 : x y + 1 mod 4 z, 1 = y 1 z = y + 1 is possible d 4 = 1 z 16 0, 0, = y+ z = y + 4 is possible d 3 = 1 Table 1. The case d = 1. In what follows, we consider separately the 6 exceptional cases. The common treatment of all these cases is to go back to the system 3. In the 10

11 all exceptional cases we have z = y + s, where s {1,, 3, 4}. Hence, ab + 1 = F x, ac + 1 = F z s, bc + 1 = F z, 17 and, as previously, c gcdf z s 1, F z 1. Table 1, Row 4. z 1, y 0 mod 4, z = y + 1 and Clearly, gcd while F z+1, F F 1 = F z+1 gcdl z 3 = 1, gcd, F = L z 3 L z 3, F z 1 = F, L z+1 = 1, gcd { Lgcd z 3, = L 1 = 1 1 or L z+1. F z+1 }, L z+1. = 1,, Therefore c 4, and we arrived at a contradiction because F z = bc contradicts z > 150. Table 1, Row 5. z 3, y 1 mod 4, z = y + and Since F z 1 = F z 3 gcd L F z 3, F z 1 = F z+1, F z+1 = 1, L. we get c gcdf z 1, F z 1 = L. Consequently, by the proof of Lemma 9, By Lemma 8, we now have L = c 1 c > c 1 Fz. c 1 < L <. Fz Hence, c 1 = 1, therefore c = L. In view of equation 17, we get a = F z 3, b = F z+1, and so F x = F z 3 F z = F = F

12 By the work of Cohn [4], we get that 18 is not possible for z > 150. Table 1, Row 8. z 3, y 0 mod 4, z = y + 3 and F z 3 1 = F L z 5 It follows easily, by Lemma 3, that gcd F, F z+1 = 1, gcd L z 5, L and gcd L z 5, F z+1 =, F z 1 = F z+1 = 1, gcd { Lgcd z+1, z 5 L 3 = 4 1 or L. F }, L 4. = 1,, Thus, c gcdf z 3 1, F z 1 8. However, the inequalities a < b < c 8 contradict the fact that z > 150. Since Table 1, Row 13. z 0, y 1 mod 4, z = y + 3 and F z 3 1 = F z 4 gcd F z 4, F z+ L z, F z 1 = F z+ = F gcd z 4, z+ F 3 =, L z. 19 we have c gcdf z 1, F z 1 = L, or c gcdf z 1, F z 1 = L. In the first case, we get L z = c c > c Fz, and applying Lemma 8 we arrive at which is a contradiction. In the second case, put c < L < 1, Fz L z = c 3 c > c 3 Fz. Again by Lemma 8, we obtain c 3 < L Fz <. 1

13 Thus, c 3 = 1, therefore c = L. System 17 and relations 19 lead to a = F z 4, b = F z+, and F x = 1 4 F z 4 F z On the one hand, since z > 150, by Lemma 5, we get α x 1.67 > F x > 1 4 α z α z > α , therefore x > z On the other hand, by combining Lemma 5 and Lemma 6 with κ < 0.01, we get α x 1.68 < F x < 1 4 α z α z < α , leading to x < z But the interval z 5.48, z 5.53 does not contain any integer, which takes care of this case. Since Table 1, Row 15. z 0, y 3 mod 4, z = y + 1 and F 1 = F z L z we get c gcdf 1, F z 1 = L z 9, it follows that, F z 1 = F z+ gcdf z, F z+ = 1, L z L z.. Consequently, by the proof of Lemma = c 4 c > c 4 Fz. Now Lemma 8 leads to the contradiction c 4 < L z < 1. Fz Table 1, Row 16. z 0, y 0 mod 4, z = y + 4 and Obviously, gcdf z while, F z+ F z 4 1 = F z L z 6 = 1, gcdl z 6 gcdl z 6, F z+ =, F z 1 = F z+, L z L z. = 1, gcdf z { Lgcd z 6, z+ L 4 = 7 1 or Thus, c 14, which leads to a contradiction with z > 150. The proof of the Theorem is now complete. 13 }, L z = 1,, 7.

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