Diophantine Triples and k-Generalized Fibonacci Sequences

We show that if k≥2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$k\ge 2$$\end{document} is an integer and (Fn(k))n≥0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\big (F_n^{(k)}\big )_{n\ge 0}$$\end{document} is the sequence of k-generalized Fibonacci numbers, then there are only finitely many triples of positive integers 1<a<b<c\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$1<a<b<c$$\end{document} such that ab+1,ac+1,bc+1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ab+1,~ac+1,~bc+1$$\end{document} are all members of {Fn(k):n≥1}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\big \{F_n^{(k)}: n\ge 1\big \}$$\end{document}. This generalizes a previous result where the statement for k=3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$k=3$$\end{document} was proved. The result is ineffective since it is based on Schmidt’s subspace theorem.


Introduction
There are many papers in the literature concerning Diophantine m-tuples which are sets of m distinct positive integers {a 1 , . . . , a m } such that a i a j + 1 is a square for all 1 ≤ i < j ≤ m (see [4], for example). A variation of this classical problem is obtained if one changes the set of squares by some different subsets of positive integers like kpowers for some fixed k ≥ 3, or perfect powers, or primes, or members of some linearly recurrent sequence, etc. (see [7,[12][13][14]18]). In this paper, we study this problem with the set of values of k-generalized Fibonacci numbers for some integer k ≥ 2. Recall that these numbers denoted by F (k) n satisfy the recurrence n+k−1 + · · · + F (k) n and start with 0, 0, . . . , 0 (k − 1 times) followed by 1. Notationwise, we assume that F hold for some integers x, y, z.
Our result generalizes the results obtained in [8], [11] and [13], where this problem was treated for the cases k = 2 and k = 3. In [13] it was shown that there does not exist a triple of positive integers a, b, c such that ab + 1, ac + 1, bc + 1 are Fibonacci numbers. In [11] it was shown that there is no Tribonacci Diophantine quadruple, that is a set of four positive integers {a 1 , a 2 , a 3 , a 4 } such that a i a j + 1 is a member of the Tribonacci sequence (3-generalized Fibonacci sequence) for 1 ≤ i < j ≤ 4, and in [8] it was proved that there are only finitely many Tribonacci Diophantine triples. In the current paper, we prove the same result for all such triples having values in the sequence of k-generalized Fibonacci numbers.
For the Proof of Theorem 1, we proceed as follows. In Sect. 2, we recall some properties of the k-generalized Fibonacci sequence F (k) n which we will need and we prove two lemmata. The first lemma shows that any k − 1 roots of the characteristic polynomial are multiplicatively independent. In the second lemma, the greatest common divisor of F (k) x − 1 and F (k) y − 1 for 2 < y < x is estimated. In Sect. 3, we assume that the Theorem 1 is false and give, using the Subspace theorem, a finite expansion of infinitely many solutions. In Sect. 4, we use a parametrization lemma which is proved by using results about finiteness of the number of non-degenerate solutions to S-unit equations. Applying it to the finite expansion, this leads us to a condition on the leading coefficient, which turns out to be wrong. This contradiction is obtained by showing that a certain Diophantine equation has no solutions; this last Diophantine equation has been treated in particular cases in [2] and [16].

Preliminaries
There are already many results in the literature about (F (k) n ) n≥0 . We will only use what we need, which are the following properties. The sequence (F The polynomial Ψ k (X ) is separable and irreducible in Q[X ] and the Galois group thus acts transitively on the roots, which we denote by α 1 , . . . , α k . If k is even or prime, the Galois group is certainly S k (see [16] for these statements). The polynomial Ψ k (X ) has only one root, without loss of generality assume that this root is α 1 > 1, which is outside the unit disk (formally, this root depends also on k, but in what follows we shall omit the dependence on k on this and the other roots of Ψ k (X ) in order to avoid notational clutter). Thus, a representation which is sometimes useful. Furthermore, (see Lemma 3 in [3]). The Binet formula of (F (k) n ) n≥0 is given by (see Theorem 2 in [3]). We also need the fact that (see [1]). Furthermore, the following property is of importance; it follows from the fact that there is no non-trivial multiplicative relation between the conjugates of a Pisot number (cf. [15]). Since in our case it is rather easy to verify, we shall present a proof of what we need.
since for every finite place v we have |x| v = 1. This means that the image of λ lies in the hyperplane defined by X 1 + · · · + X d = 0 of R d . We will use the property that λ is a homomorphism (see e.g., [17]). So if we can prove that the k − 1 vectors λ(α 1 ), . . . , λ(α k−1 ) are linearly independent, this will prove the statement.
Hence the transpose of the matrix above is strictly diagonal-dominant since the diagonal entries, which are all equal to log |α 1 |, are positive, all other entries are negative, and each row-sum (in the transpose matrix) of all off-diagonal entries is in absolute value less than the corresponding diagonal entry. It follows that the matrix is regular, which is what we wanted to show. Finally, we prove the following result, which generalizes Proposition 1 in [8]. Observe that the upper bound depends now on k.
Proof We may assume that y ≥ 4, since for y = 3, we get F (k) y − 1 = 1, and there is nothing to prove. We put d = gcd F (k) From now on, we assume that y > κx + 1. Using (3) and (5), we write We where η is some algebraic integer in K. Note that the right-hand side above is not zero, for if it were, we would get α λ y − 1 ∈ Q, which is false for λ > 0. We compute norms from K to Q. Observe that In the above, we used the fact that (5)) as well as (2). Further, let σ i be any Galois automorphism that maps α 1 to α i . Then for i ≥ 2, we have We then have Hence, In order to balance (9) and (11), we choose κ such that κ = 1 − κ/k, giving κ = k/(k + 1), and the lemma is proved.

Parametrizing the Solutions
In order to simplify the notations, we shall from now onwards write F n instead of F (k) n ; we still mean the nth k-generalized Fibonacci number. The arguments in this section follow the arguments from [8]. We will show that if there are infinitely many solutions to (1), then all of them can be parametrized by finitely many expressions as given in (18) for c below.
We assume that there are infinitely many solutions to (1). Then, for each integer solution (a, b, c), we have we see that x + y > z − 1 and thus y ≥ z/2. In order to get a similar correspondence for x and z, we denote Then we use Lemma 2 to obtain and hence which we can write as x > C 1 z for some small constant C 1 < 1 (depending only on k), when z is sufficiently large. Next, we do a Taylor series expansion for c which was given by Using the power sum representations of F x , F y , F z , we get We then use the binomial expansion to obtain where O has the usual meaning, using estimates from [9] and where T is some index, which we will specify later. Since x < z and z < x/C 1 , the remainder term can also be written as O α −T x /C 1 1 , where x = max{x, y, z} = z. Doing the same for y and z likewise and multiplying those expression gives where the integer n depends only on T and where J is a finite set, d j are non-zero coefficients in L := Q(α 1 , . . . , α k ), and M j is a monomial of the form in which x = (x, y, z), and L i, j (x) are linear forms in x ∈ R 3 with integer coefficients which are all non-negative if i = 2, . . . , k and non-positive if i = 1. Note that each monomial M j is "small" that is there exists a constant κ > 0 (which we can even choose independently of k), such that This follows directly from Our aim of this section is to apply a version of the Subspace theorem given in [5] to show that there is a finite expansion of c involving terms as in (13); the version we are going to use can also be found in Sect. 3 of [10]. For the set-up-in particular the notion of heights-we refer to the mentioned papers.
We work with the field L = Q(α 1 , . . . , α k ) and let S be the finite set of infinite places (which are normalized so that the Product Formula holds, cf. [5]). Observe that α 1 , . . . , α k are S-units. According to whether −x + y + z is even or odd, we set = 0 or = 1 respectively, such that By going to a still infinite subset of the solutions, we may assume that is always either 0 or 1.
Using the fixed integer n (depending on T ) from above, we now define n +1 linearly independent linear forms in indeterminants (C, Y 0 , . . . , Y n ). For the standard infinite place ∞ on C, we set where ∈ {0, 1} is as explained above, and For all other places v in S, we define We will show that there is some δ > 0, such that the inequality is satisfied for all vectors The use of the correct ∈ {0, 1} guarantees that these vectors are indeed in L n . First, notice that the determinant in (16) Now notice that the last double product equals 1 due to the Product Formula and that An upper bound on the number of infinite places in L is k! and hence (6) and (5). And finally the first expression is just which, by (13), is smaller than some expression of the form Now we choose T (and the corresponding n) in such a way that For the height of our vector y, we estimate with suitable constants C 3 , C 4 , C 5 . For the second estimate, we used that and bounded it by the maximum of those expressions. Furthermore we have , which just changes our constant C 4 . Now finally, the estimate is satisfied when we pick δ small enough.
So all the conditions for the Subspace theorem are met. Since we assumed that there are infinitely many solutions (x, y, z) of (16), we now can conclude that all of them lie in finitely many proper linear subspaces. Therefore, there must be at least one proper linear subspace, which contains infinitely many solutions and we see that there exists a finite set J and (new) coefficients e j for j ∈ J in L such that we have with (new) non-zero coefficients e j and monomials M j as before. Likewise, we can find finite expressions of this form for a and b.

Proof of the Theorem
We use the following parametrization lemma: (1). Then there exists a line in R 3 given by

Lemma 3 Suppose, we have infinitely many solutions for
with rationals r 1 , r 2 , r 3 , s 1 , s 2 , s 3 , such that infinitely many of the solutions (x, y, z) are of the form (x(n), y(n), z(n)) for some integer n.
Proof Assume that (1) has infinitely many solutions. We already deduced in Sect.  y, z). In the same manner, we can write Since 1 + bc = F z = f 1 α z 1 + · · · + f k α z k , we get Substituting , we obtain an equation of the form j∈J e j α where again J is some finite set, e j are non-zero coefficients in L and L i, j are linear forms in x with integer coefficients. This is an S-unit equation, where S is the multiplicative group generated by {α 1 , . . . , α k , −1}. We may assume that infinitely many of the solutions x are nondegenerate solutions of (20) by replacing the equation by an equation given by a suitable vanishing subsum if necessary.
Therefore for i = j, the theorem on non-degenerate solutions to S-unit equations (see [6]) yields that the set of is contained in a finite set of numbers. By Lemma 1, α 1 , . . . , α k−1 are multiplicatively independent and thus the exponents (L 1,i − L 1, j )(x), . . . , (L k−1,i − L k−1, j )(x) take the same value for infinitely many x. Since we assumed that these linear forms are not all identically zero, this implies that there is some non-trivial linear form L defined over Q and some c ∈ Q with L(x) = c for infinitely many x. So there exist rationals r i , s i , t i for i = 1, 2, 3 such that we can parametrize x = r 1 p + s 1 q + t 1 , y = r 2 p + s 2 q + t 2 , z = r 3 p + s 3 q + t 3 with infinitely many pairs ( p, q) ∈ Z 2 .
We can assume that r i , s i , t i are all integers. If not, we define Δ as the least common multiple of the denominators of r i , s i (i = 1, 2, 3) and let p 0 , q 0 be such that for infinitely many pairs ( p, q) we have p ≡ p 0 mod Δ and q ≡ q 0 mod Δ. Then p = p 0 + Δλ, q = q 0 + Δμ and Since r i Δ, s i Δ and x, y, z are all integers, r i p 0 + s i q 0 + t i are integers as well. Replacing r i by r i Δ, s i by s i Δ and t i by r i p 0 + s i q 0 + t i , we can indeed assume that all coefficients r i , s i , t i in our parametrization are integers.
Using a similar argument as in the beginning of the proof, we get that our equation is of the form j∈J e j α where r := (λ, μ), J is a finite set of indices, e j are new non-zero coefficients in L and L i, j (r) are linear forms in r with integer coefficients. Again we may assume that we have (L 1,i (r), . . . , L k−1,i (r)) = (L 1, j (r), . . . , L k−1, j (r)) for any i = j.
Applying the theorem of non-degenerate solutions to S-unit equations once more, we obtain a finite set of numbers Λ, such that for some i = j, we have So every r lies on a finite collection of lines and since we had infinitely many r, there must be some line, which contains infinitely many solution, which proves our lemma.
Replacing n by m, r i by r i Δ and s i by rr i + s, we can even assume that r i , s i are integers. So we have This holds for infinitely many m, so we can choose a still infinite subset such that all of them are in the same residue class δ modulo 2 and we can write m = 2 + δ with fixed δ ∈ {0, 1}. Thus we have where η ∈ Z or η ∈ Z + 1/2. Using this representation, we can write (18) as for infinitely many , where as before and x = x( ) = (x(2 + δ), y(2 + δ), z(2 + δ)). From this, we will now derive a contradiction. First we observe that there are only finitely many solutions of (21) with c( ) = 0. This can be shown by using the fact that a simple non-degenerate linear recurrence has only finite zero-multiplicity (see [6] for an explicit bound). We will apply this statement here for the linear recurrence in ; it only remains to check that no quotient of two distinct roots of the form α is a root of unity or, in other words, that has no solutions in n ∈ Z/{0}, m 1 < 0 and m i > 0 for i = 2, . . . , k. Assume relation (22) holds. Replacing α 1 by (−1) k−1 (α 2 · · · α k ) −1 gives By squaring this equation and applying Lemma 1 we get 2(m 2 − m 1 ) = · · · = 2(m k − m 1 ) = 0 and thus m 1 = m 2 = · · · = m k , which is impossible because of the signs of m 1 and m 2 , . . . , m k . So we have confirmed that c( ) = 0 for still infinitely many solutions. We use (12) and write Then we insert the finite expansion (21) in for c into (23). Furthermore, we use the Binet formula (3) and write F x , F y , F z as power sums in x, y and z respectively. Using the parametrization (x, y, z) = (r 1 m + s 1 , r 2 m + s 2 , r 3 m + s 3 ) with m = 2 or m = 2 + 1 as above, we have expansions in on both sides of (3). Since there must be infinitely many solutions in , the largest terms on both sides have to grow with the same rate. In order to find the largest terms, we have to distinguish some cases: If we assume that e 0 = 0 for infinitely many of our solutions, then e 0 α (−x+y+z− )/2 1 is the largest term in the expansion of c and we have It follows that e 2 0 = f 1 α 1 . The case e 0 = 0 for infinitely many of our solutions is not possible, because then, the right-hand side of (23) would grow faster than the left-hand side so that (23) could be true for only finitely many of our . In the other cases, we have e 0 = f 1 α 1 , where ∈ {0, 1}. This now contradicts the following lemma, which turns out to be slightly more involved than in the special case on Tribonacci numbers (cf. [8]).
This leads to 2 k+1 k k − (k + 1) k+1 = w 2 for some integer w. But this equation has no integer solutions, which is proved in the theorem below. This concludes the proof.
In order to finish the proof, we have the following result, which might be of independent interest since particular cases were considered before in [2] and [16].

Theorem 2
The Diophantine equation (25) has no positive integer solutions (k, w) with k ≥ 2.
We then get Note that the two numbers in the right-hand side of (26) are coprime, for if p divides w 1 and (k + 1)/2, then p divides the left-hand side of (26). Thus p divides both k and (k + 1)/2, so also k − 2((k + 1)/2) = −1, a contradiction. Thus, the right-hand side is a sum of two coprime squares and therefore all odd prime factors of it must be 1 modulo 4 contradicting the fact that in the left-hand side we have k ≡ 3 (mod 4). This finishes the proof of this theorem.