A Quadratic Diophantine Equation Involving Generalized Fibonacci Numbers

: The sequence of the k -generalized Fibonacci numbers ( F ( k ) n ) n is deﬁned by the recurrence F ( k ) n = ∑ kj = 1 F ( k ) n − j beginning with the k terms 0, . . . , 0, 1. In this paper, we shall solve the Diophantine equation F ( k ) n = ( F ( l ) m ) 2 + 1, in positive integers m , n , k and l .

Like any very studied object in Mathematics, the Fibonacci sequence admits many generalizations (in several distinct ways). Among these generalizations, we are interested in the k-generalized Fibonacci sequence (F (k) n ) n≥−(k−2) which is defined by the recurrence relation with initial values F (k) j = 0 (for j ∈ [−(k − 2), 0]) and F (k) 1 = 1. For instance, if k = 2, we have the usual Fibonacci numbers (F (2) n ) n , for k = 3, (F (3) n ) n the sequence is called the Tribonacci sequence and so on (Kessler and Schiff [7] remarked the appearance of these numbers in probability theory and in certain sorting algorithms).
In the past few years, k-Fibonacci numbers are in the mainstream of many works. For example, in 2013, two related conjectures were proved. The first one (proposed by Marques [8]) was proved by Bravo and Luca [9] and is related to repdigits among k-Fibonacci numbers. The second conjecture (proposed by Noe and Post [10]) concerns the intersection between these sequences, and was solved (independently) by Marques [11] and Bravo, Luca [12]. In addition, Chaves and Marques [13] solved the equation (F (k) m and then Bednařík et al. [14] generalized this study to the equation (F (k) m . In 2019, Trojovský [15] proved that the Diophantine equation F (k) m = m t , with t > 1 and m > k + 1, has only the solutions F We remark that the problem of determining all the perfect powers among Fibonacci numbers was settled in a seminal work due to Bugeaud, Mignotte, and Siksek [16]. However, the problem of solving completely the equation F (k) n = y t , for k > 2 and t > 1, is still far from being solved. Indeed, the particular case (k, t) = (3, 2) (i.e., to find all Tribonacci numbers which are perfect squares) is a known open problem which appeared as Problem 1 in a paper due to Pethő [17].
In this paper, we are interested in this kind of problem. Indeed, our goal is to study when a term of a k-generalized Fibonacci sequence is near to a perfect square, whose basis is also a generalized Fibonacci number (possibly of another order). More precisely, we have the Diophantine equation Thus, in this paper, we shall solve this equation for c = 1 by proving that Theorem 1. The solutions of Equation (1), for c = 1, in m, n, k and l, with min{m, n} ≥ 1 and min{k, l} ≥ 2, are (n, m, k, l) ∈ {(3, 1, k, l), (3, 2, k, l), (5, 3, 2, l)}.

Remark 1.
We point out that the method presented here can be used to obtain all solutions of Equation (1), for any previously fixed value of c (the choice of c = 1 has nothing of special). See a more detailed discussion (on this fact) in Section 8. In addition, we remark that it is well-known that (for any given c) this equation has only finitely many solutions (by a result of Nemes and Pethő [18]).

Remark 2.
The Mandelbrot set is the set of complex numbers c for which the sequence (z n ) n defined by a nonlinear recurrence z n+1 = z 2 n + c, with z 0 = 0, does not diverge. Thus, the problem of solving the Diophantine Equation (1) can be rephrased as: For which values of t > 0, a pair of consecutive z t 's belongs to (F (k) For proving our main result, we shall apply Baker's theory, a Dujella-Pethö reduction method, some key arguments due to Bravo-Luca, and a combinatorial lemma to deal with an extremal case.
As mentioned before, we also shall use lower bounds for linear forms in logarithms. Among the several results on this topic, we decided to use one due to Bugeaud, Mignotte and Siksek [16] (Theorem 9.4). Lemma 1. Let γ 1 , . . . , γ t be nonzero real algebraic numbers and let b 1 , . . . , b t be nonzero integers. Let D = [Q[γ 1 , . . . , γ t ] : Q] and let A j be a real number satisfying Take B ≥ max{|b 1 |, . . . , |b t |}.
In the previous lemma, the logarithmic height of an n-degree algebraic number γ is defined as where a is the leading coefficient of the minimal polynomial of γ and (γ (j) ) 1≤j≤n are the algebraic conjugates of γ. Some basic properties of the logarithmic height are: , for all r ∈ Q * (nonzero rational numbers) and α ∈ Q (algebraic numbers); After establishing an upper bound for one of our variables (which is in general too large to perform the necessary computations), the next step makes it substantially smaller. For this purpose, our next ingredient is a theorem due to Dujella and Pethö [25]. Recall that, for a real number x, the Nint function at x is x := min{|x − n| : n ∈ Z}. Lemma 2. Let M ∈ Z >0 and let γ, µ ∈ R, such that γ is irrational. Let p/q be a convergent of the continued fraction expansion γ with q > 6M, and let A, B be real numbers with A > 0 and B > 1. If := µq −M γq is positive, then the Diophantine inequality does not have solution in integers m, n, and k with Our last ingredient is a combinatorial argument which will be essential to deal with the extremal case n = 2m − 2.
Lemma 3. Let k, m be any integers. For all k ≥ 2 and m > k + 1, we have Proof. It is well-known that F Now, we are ready to start the proof of our main result. We shall split it in some sections in order to make the text more readable.

An Inequality for m in Terms of l
Our goal is to solve the Diophantine equation To avoid unnecessary repetitions, we shall consider l > k (the case l ≤ k can be handled in the same way). By the auxiliary results and Dresden-Du formula, we can rewrite (4) as where g := g(α, k) and h := g(β, l) and both |E n (k)| and |E m (l)| are smaller than 1/2, for all positive integers m and n. Thus, and, dividing by h 2 β 2m−2 , we get where we used hβ > 1.
Hence, from the useful fact that whenever x > e and A ≥ 3, we get the following upper bound for m in terms of l m < 4.9 × 10 14 × l 8 (log l) 3 .

The Case of Small l
Next, we treat the cases when l ∈ [3,238]. In this case, k < l ≤ 238, and inequality (9) implies that m < 8.27 × 10 35 and n < 4m < 3.31 × 10 36 . Now, write Suppose Λ 1 > 0 (the other case can be handled in the same way). Then, By dividing the above inequality by log β, we get where the numbers γ and µ are defined as γ := γ k,l = log α/ log β and µ := µ k,l = log(g/h 2 )/ log β. We claim that γ is irrational. Indeed, if γ = p/q, for some p, q ∈ Z >0 , we would obtain α q = β p , which is impossible by using the same argument as for Γ 1 = 0. Let us denote q (m,k,l) by the denominator of the m-th convergent of the continued fraction expansion of γ k,l .
In conclusion, there are no solutions of (4) for k < l ≤ 238 and m > 3 (and so for n > 3).

The Final Step
Now, we still have l ≥ 239. Then, it remains to verify the cases when k ≥ 1783. Thus, the following inequality holds: n < 4.42 × 10 118 k 32 (log k) 35 < 2 k/2 .
Using again the argument due to Bravo and Luca, we obtain where we used that 4k < 2 k/2 and 8k < 2 k are true for k ≥ 11. Combining (5), (12), and (19), we get If n ≤ k, then (4) becomes 2 n−2 = F (k) n = (F (l) m ) 2 + 1, which cannot happen for n ≥ 4, since a square plus 1 is never divisible by 4. The remaining cases, n ∈ {2, 3}, give us the solutions already known. It follows that n > k.

Further Comments: The Case of a General c
As mentioned in Remark 1, we only choose c = 1 in order to explicit all calculations. In the general case, the equation m ) 2 + c has infinitely many solutions (n, k, m, l, c) (this follows, clearly, because the linear dependence of equation in the variable c). For this reason, the more interesting case happens when c ≥ 1 is fixed. In this case, it seems reasonable to expect to deal with the case of an upper bound for all other variables (i.e., n, m, k and l) in terms of c. In fact, the proof is completely similar until we arrive at the inequality (6), which would be Now, we split the proof into two cases: • If β m/2 < c. In this case, we get directly the bounds l < m < 4 log c/ log 2 and k < n < 2 + 2 log(2c 4 )/ log 2, where the last inequality is obtained from • If β m/2 ≥ c. For this case, inequality (6) becomes which does not depend on c and thus, from this point on, we simply mimic the proof of Theorem 1.

Conclusions
In this paper, we study a Diophantine problem related to a higher order generalization of the Fibonacci sequence. In fact, the k-generalized Fibonacci numbers, denoted by (F (k) n ) n , are defined by the kth order recurrence F (k) n−j with initial values 0, . . . , 0, 1 (k terms), where F (k) 1 = 1. In particular, we solve completely the Diophantine equation F k n = (F (l) m ) 2 + 1 (which can be related to the problem of terms two (possibly distinct) generalized Fibonacci sequences as consecutive terms of an orbit in a quadratic dynamics related to the Mandelbrot set). The main tools in the proof are Baker's theory, reduction, and Bravo-Luca methods (combined with a combinatorial lemma and some Mathematica R routines).
Author Contributions: P.T. conceived the presented idea. Both authors have worked on the conceptualization, methodology, investigation, writing, reviewing, editing. The software was done by A.P.C. All authors have read and agreed to the published version of the manuscript. Acknowledgments: The authors express their debt of gratitude to the editor and reviewers for their helpful and detailed comments in improving the presentation and quality of the paper.

Conflicts of Interest:
The authors declare no conflict of interest.

Appendix A. Mathematica Commands
Below, we shall present the Mathematica commands used along the paper (the calculations in this paper took roughly four days on a 2.5 GHz Intel Core i5 4 GB Mac OSX.).