Some Diophantine Problems Related to k -Fibonacci Numbers

: Let k ≥ 1 be an integer and denote ( F k , n ) n as the k -Fibonacci sequence whose terms satisfy the recurrence relation F k , n = kF k , n − 1 + F k , n − 2 , with initial conditions F k ,0 = 0 and F k ,1 = 1. In the same way, the k -Lucas sequence ( L k , n ) n is deﬁned by satisfying the same recursive relation with initial values L k ,0 = 2 and L k ,1 = k . The sequences ( F k , n ) n ≥ 0 and ( L k , n ) n ≥ 0 were introduced by Falcon and Plaza, who derived many of their properties. In particular, they proved that F 2 k , n + F 2 k , n + 1 = F k ,2 n + 1 and F 2 k , n + 1 − F 2 k , n − 1 = kF k ,2 n , for all k ≥ 1 and n ≥ 0. In this paper, we shall prove that if k > 1 and F sk , n + F sk , n + 1 ∈ ( F k , m ) m ≥ 1 for inﬁnitely many positive integers n , then s = 2. Similarly, that if F sk , n + 1 − F sk , n − 1 ∈ ( kF k , m ) m ≥ 1 holds for inﬁnitely many positive integers n , then s = 1 or s = 2. This generalizes a Marques and Togbé result related to the case k = 1. Furthermore, we shall solve the Diophantine equations F k , n = L k , m , F k , n = F n , k and L k , n = L n , k .

In particular, this naive identity asserts that the sum of the square of two consecutive Fibonacci numbers is always a Fibonacci number. In a recent paper, Marques and Togbé [5] searched for similar identities in higher powers. However, they proved that if F s n + F s n+1 is a Fibonacci number for all sufficiently large n, then s = 1 or s = 2.
The Fibonacci sequence was generalized in many different ways, some of them you can find in [6][7][8][9][10][11][12][13][14][15][16][17][18]. By keeping its order (which is 2), we have a general Lucas sequence (C n ) n = (C n (a, b)) n which is defined by the recurrence C n = aC n−1 + bC n−2 , for n ≥ 2, and with C i = i, for i ∈ {0, 1} (moreover, the integer parameters a and b must be such that if x 2 − ax − b = (x − r)(x − s), then r/s is not a root of unity). In particular, (C(1, 1)) n is the Fibonacci sequence. Recently, Falcon and Plaza [11,19,20] studied the case (C(k, 1)) n which was called the k-Fibonacci sequence and denoted by F (k) = (F k,n ) n . So, this sequence satisfies the recurrence relation with initial conditions F k,0 = 0 and F k,1 = 1. In the same way, the companion k-Lucas sequence L (k) = (L k,n ) n is defined by satisfying the same recursive relation with initial values L k,0 = 2 and L k,1 = k. In particular, they proved that the following identities hold for all n ≥ 0 and In this paper, we shall work on some Diophantine problems related to these sequences. Our first result concerns the search for higher power identities related to (2) and (3) in the spirit of the Marques and Togbé paper ( [5]). More precisely, we prove that Theorem 1. Let k ≥ 2 be any integer. If s ≥ 1 satisfies that F s k,n + F s k,n+1 ∈ (F k,m ) m≥1 for infinitely many positive integers n, then s = 2.
Theorem 2. Let k ≥ 1 be any integer. If s ≥ 1 satisfies that F s k,n+1 − F s k,n−1 ∈ (kF k,m ) m≥1 for infinitely many positive integers n, then s = 1 or s = 2.
However, some related questions are still open problems, as for instance, it is unknown if the sets P ∩ F and P ∩ R are infinite. The usual method is to solve some special Diophantine equations (see e.g., [26][27][28][29][30][31]).
In the next results, we shall find the intersection F (k) ∩ L (k) as well as we shall solve the symmetric equations F k,n = F n,k and L k,n = L n,k . More precisely,  Now, we give a brief overview of the methods which will be used here. For the proof of Theorems 1 and 2, we shall apply the same approach as in [32] (i.e., asymptotic results together with Galois' theory). For Theorems 3-5, we shall use a plenty of inequalities together with some (combinatorial) identities for F k,n and L k,n .

Auxiliary Results
In this section, we shall provide some useful tools which will be very useful in order to prove our theorems.
First, we observe that, similarly to the Fibonacci and Lucas sequences, their k versions also satisfy the Binet's formulas where By using Binet's formula of k-Fibonacci numbers, it is a simple matter to prove that lim n→∞ F k,n+ F k,n = σ 1 .
In addition, we shall need some lower and upper bounds for the terms of the k-Fibonacci and k-Lucas sequences.

Lemma 2.
For all k ≥ 1 and n ≥ 1, it holds that The proof of this lemma can be found in ( [32], Lemma 2.2).

Lemma 3. Let l, m be any positive integers. Then
Proof. For m = 1 we can rewrite inequality (5) as 4 ≤ (l + 1) 2 , thus (5) holds for every positive integer l. Now, we show that the following inequality, slightly stronger than (5), holds for every m ≥ 2 and every positive integer l. Indeed, this inequality follows from the fact that m ≥ 2/l and so Thus, (l + 2) m+2 ≤ l(l + 4) m+1 as desired.
Now, we are ready to deal with the proof of theorems.

Proof of Theorem 1
Suppose that F s k,n + F s k,n+1 = F k,t(n) for infinitely many positive integers n (say n belonging to an infinity set S), where t(n) is an arithmetic function.
By using the estimates in (2), we have Thus, ns − s + 1 < t(n) < ns + 3 for all n ∈ S. Therefore, t(n) = ns + t, for all n ∈ S (for some infinite set S ⊆ S), where t is a constant (depending only on s).

Proof of Theorem 2
The proof of Theorem 2 is similar to the proof of Theorem 1, so we will not formulate it completely, but we will only state the steps in which we use different identities.
Suppose that F s k,n+1 − F s k,n−1 = kF k,t(n) for infinitely many positive integers n (say n belonging to an infinite set S), where t(n) is an arithmetic function.
By using the estimates in Lemma 2, we have Thus, sn − 2s < t(n) < sn + 3 for all n ∈ S. Therefore, t(n) = ns + t, for all n ∈ S ⊆ S, where S is an infinite set and t is a constant, which depends only on s. Then, we consider the equation Hence, we obtain the Diophantine equation Again, we shall use a little of Galois theory. Note that σ 1 , σ 2 ∈ K := Q( √ k 2 + 4), then by conjugating the relation (12) by the automorphism ψ : σ 1 → σ 2 of Gal(K/Q), we obtain By multiplying (12) by (13) and by using that σ 1 σ 2 = −1, we obtain which can be rewritten, by using Binet's formula for k-Lucas numbers and using a clear condition that t has to be even, by the following way For s = 1 and s = 2 we have L k,2 = k 2 k 2 + 4 0 + 2 or L k,4 = k 2 k 2 + 4 1 + 2, thus the last equality holds for any k. For s ≥ 3, we deduce from Lemmas 1 and 3 that there is no solution. The proof is then complete.

Proof of Theorem 3
By using the estimates in Lemma 2, we get Then m < n < m + 3 which yields n ∈ {m + 1, m + 2}. Now, we have two cases to consider: In this case, we have the equation F k,m+1 = L k,m and by using Binet's formulas, we obtain and after a straightforward calculation, we get σ m−1 . If m = 1, then we have the family of solutions F k,2 = L k,1 = k, for all k ≥ 1. If m > 1, then k < σ m−1 1 = σ m−1 2 < 1 arriving at an absurdity.

Proof of Theorem 4
Suppose, without loss of generality, that n > k (because the symmetry of the equation). If k = 1, then F 1,n = F n = 1 = F n,1 , only for n = 2, so we have the solution (n, k) = (2, 1). If k = 2, then F 2,n > 2 n−2 > n = F n,2 , for all n ≥ 5. For n = 1, 2, 3 and 4, we do not have any solution. For k ≥ 3, we have the following stronger result Proposition 1. If n > k ≥ 3, then F k,n > F n,k .
Proof. For proving this, we shall use the following combinatorial formula (see ([19], Proposition 7)): and then Since n > k, then the number of terms of the sum in F k,n is bigger than or equal to the one in F n,k (since (n − 1)/2 ≥ (k − 1)/2 ). Thus, it suffices to prove that for all 0 ≤ i ≤ (k − 1)/2 . To prove this, it is enough to show that Since n > k, then (n − i − 1) · · · (n − 2i) > (k − i − 1) · · · (k − 2i) and so, we only need to prove that k n−2i−1 > n k−2i−1 . In fact, we shall prove a stronger fact by showing that for any θ > 0, the function f (x) := (x − θ)/ log x is increasing, for x > e. For this, we observe that the its derivative is Since (x − θ)/x = 1 − θ/x < 1 < log x for x > e, then f (x) > 0 as desired. Now, observe that for n > k ≥ 3, then f (n) > f (k) (for θ = 2i + 1) and after a straightforward calculation we obtain k n−2i−1 > n k−2i−1 which completes the proof.

Proof of Theorem 5
In fact, after a straightforward computation (as done previously), we can deduce that there is no solution for L k,n = L n,k , when n > k and 1 ≤ k ≤ 5 (we can use Lemma 2, for example). So, it suffices to prove that Proposition 2. If n > k ≥ 6, then L k,n > L n,k .
Proof. We have that which follows from (4) together with the Binomial theorem. Thus (as in the proof of Proposition 1) in order to prove that L k,n > L n,k , it is enough to show that for n > k ≥ 5 and 1 ≤ i ≤ k/2 . This is equivalent (by applying the log function) to prove that the function is positive, for all x ≥ k + 1 (since n > k) and 1 ≤ i ≤ k/2 . Since g(k) = 0, we only need to show that g is an increasing function (i.e., that g (x) > 0, for all x ≥ k + 1). Indeed, the derivative of g(x) is Since i ≥ 1 and x − 1 ≥ k, then In conclusion, g is increasing for x ≥ k + 1. In particular, g(n) > g(k) = 0 and so (16) holds. This finishes the proof.

Conclusions
In this paper, we study some Diophantine problems related to two special generalizations of Fibonacci and Lucas numbers. Indeed, for a positive integer k, the k-Fibonacci and k-Lucas sequences (F k,n ) n and (L k,n ) n are defined by the same recurrence, namely C n = kC n−1 + C n−2 , with initial terms F k,i = i (for i ∈ {0, 1}) and L k,0 = 2 and L 1,k = k. The first kind of problem concerns the search for higher order identities similar to F 2 k,n + F 2 k,n+1 = F k,2n+1 and F 2 k,n+1 − F 2 k,n−1 = kF k,2n . In this case, we use some analytic and algebraic tools to conclude that if F s k,n + F s k,n+1 (resp., (F 2 k,n+1 − F 2 k,n−1 )/k) is a k-Fibonacci number for infinitely many positive integers n, then s = 2 (resp., s = 1 or s = 2). The second part of the work is devoted to problems related to the intersection between k-Fibonacci and k-Lucas sequences. More precisely, we make use of analytic, algebraic and combinatorial tools to solve completely the Diophantine equations F k,n = L k,m , F k,n = F n,k and L k,n = L n,k .
Author Contributions: P.T. conceived the presented idea, on the conceptualization, methodology, investigation, writing, reviewing and editing. The software was done by Š.H. Both authors have read and agreed to the published version of the manuscript.
Funding: The first author was supported by the grant no. 18-00496S of the Czech Science Foundation, Czech Republic. The second author was supported by the Project of Specific Research PrF UHK no. 2106/2020, University of Hradec Králové, Czech Republic.