Solutions of the Diophantine Equations and

Abstract

Let , , and be the balancing, Jacobsthal, and Lucas balancing numbers, respectively. In this paper, the diophantine equations and are completely solved. The solutions rely basically on Matveev’s theorem on linear forms in logarithms of algebraic numbers and a procedure of reducing the upper bound due to Dujella and Pethö.

1. Introduction

Balancing numbers are generated by the equation for with . So, the initial terms are as follows:

Lucas balancing numbers are strongly related to balancing numbers and defined by , and for . Its first terms are as follows:

The initial terms of the Jacobsthal sequence are and , and it follows the equation for . So, we have the following equation:

Balancing numbers and associated sequences have been considered in many papers concerning diophantine equations. In [1], Ray solved some diophantine equations that involve balancing and Lucas balancing numbers. In [2], Dey and Rout found the perfect powers in the sequences of balancing and Lucas balancing numbers and identified the Lucas balancing numbers which are products of a power of 3 and a perfect power. In addition, they proved that many diophantine equations that contains balancing and Lucas balancing numbers have no solutions. In [3], Rayaguru and Panda found all the repdigits that exist in the product of consecutive balancing or Lucas balancing numbers, and in [4], they explored the repdigits that are expressible as products of balancing and Lucas balancing numbers with their indices in arithmetic progressions. In [5], Erduvan and Keskin studied and determined Fibonacci numbers which are products of two balancing numbers. In [6], Rayaguru et al. found the factoriangular numbers in the sequences of balancing and Lucas balancing numbers. In [7], Ddamulira obtained all the repdigits that can be written as sums of three balancing numbers. In [8], Patra and Panda solved the equation for . In [9], Nansoko et al. solved completely the diophantine equation in positive integers .

In this paper, our purpose is to solve the following equations:and

All the solutions of the two equations are given as follows:

Theorem 1. Let and let be a non-negative solution of equation (4). Then, the solutions are as follows:

Theorem 2. Let and let be a non-negative solution of equation (5). Then, the solutions are as follows:

The key idea is to use a variant of Baker’ theory due to Matveev to find an upper bound for all the implied variables in terms of a single variable. The obtained upper bound is usually too large to be investigated by computer calculations. Therefore, we apply a reduction method of Dujella and Pethö to cut down the upper bound. Lastly, we use Sage to determine all the solutions.

2. Preliminary Results

This section gathers the relevant background material that will be used throughout the paper.

2.1. Balancing and Lucas Balancing Sequences

The sequences of balancing and Lucas balancing numbers are characterized by the following equation:

Let and be the solutions. Their Binet formulas are as follows:

One can show thatand

For more details concerning balancing and Lucas balancing numbers, see [10, 11].

2.2. Jacobsthal Sequence

The Jacobsthal numbers obey the following Binet formula:

For , one can show that

We refer to [12, 13] for more details. The next inequality of linear forms is fundamental. Bugeaud, Mignotte, and Siksek deduced it, see [14], from Matveev’s theorem [15].

2.3. A Theorem of Matveev

Consider an algebraic number . Suppose the minimal polynomial (over ) of has degree and let ’s be the conjugates of . Then, the minimal polynomial can be written as follows:where is positive integer. The logarithmic Weil height (over algebraic real field) of is given by the following equation:

The function of logarithmic height satisfies the following properties (see [16] for proofs):

Theorem 3 (Matveev). Suppose that are positive real algebraic numbers in a real algebraic number field of degree and that are nonzero integers such that the quantity

Let , for and . Then,

2.4. Reduction Lemma

Let the distance between a real number and the closest integer. The subsequent result is due to Dujella and Pethö, see [17].

Lemma 4. Let be given real numbers. Let be a positive integer. Assume is a convergent of with . If and satisfywith , then

2.5. Legendre Theorem

Legendre proved the following essential result in his book [18]. We will use this theorem in some cases of our investigation of balancing numbers. The interested reader can see [19] for more details.

Theorem 5. Let . Assume is a real number and . Ifthen is a convergent of the continued fraction of . Let S and t be integers (non-negative) such that and let . Then,

We assume, by symmetry of equations (4) and (5), that .

3. Solving the Equation

3.1. Bounding the Variables

Applying the inequalities (10) and (13), we get the following inequalities:

These imply that

The value of is approximately 0.39, so we can take . Binet formulas can be inserted into equation (4) to give the following equation:

Then,

This implies that

Thus,

Let

If , then . Let be the automorphism given by . Therefore, . Indeed, , a contradiction. So, .

Take . Then, . The logarithmic heights are as follows:

Settingand using the theorem of Matveev, we obtain the following inequality:where . So,

On the other hand, by inequality (28), one gets the following inequality:

A combination of inequalities (33) and (34) gives the following inequality:

Hence,

From equation (25),

Therefore,

Hence,

Let . Then,

Let

First, we show that . If , then . Again, let . This gives . Since , a contradiction, then we take . Immediately, we get the following equation:

Set

By Matveev’s Theorem, we get the following inequality:where .

Using equations (35), (36), and (40) with simple calculations give the following inequality:

By a simple Mathematica calculation, we find the following estimation:

3.2. Reducing the Upper Bound

It is known that for . Now, we aim to cut down the bound on . Let

Equation (28) gives, for ,which implies that

Then, . Therefore, we get

We observe that since . Then,

Using Lemma 4 with , , , , and , let , we find that . Compute

It follows, by Lemma 4, that . Set

Let . By equation (39), we have the following inequality:

This implies that

Then, . Therefore,

Then,

Applying Lemma 4 with , , , , and , we have . Consider in the following cases: Case I: and  By Lemma 4, we get , so and . Case II: . In this case the value of will be always negative and equation (4) becomes as follows:

Then, , and from equation (89), we get . As before, we can prove that

This gives, for , that

Since for , we get . Then, . So, is a convergent of . Using and some computations we find that

Therefore,

Thus,

Then, . Solving, using Sage, equation (102) for gives no solutions and for , , and gives the triples in Theorem 1.

4. Solving the Equation

4.1. Bounding the Variables

By equations (11) and (13), we have the following equation:

These imply that

We take . Using Binet formulas, equation (5) can be written as follows:

Then,

Therefore,

Then,

Thus,

Consider the following:

As before, we can prove that . Let . Then,

Setting

Taking , it follows thatwhere .

Consequently,

In addition, it follows from inequality (71) that

Comparing inequalities (76) and (77) entails that

Hence,

Equation (67) is equivalent to

So,

Then,

Let . Then,

Set

Again, . Let . Then,

Let

Then,where . Using equations (78), (79), and (83) and simple manipulations, it follows that

Solution by Mathematica gives the following equation:

4.2. Reducing the Upper Bound

For , let

From equation (71), we have the following equation:

So,

Then, . Therefore, we have the following equation:

Since , then

Applying Lemma 4 with , , , , and , the expansion of entails that . Computing

Thus, by Lemma 4, we get . Setand assume that . From equation (82), we deduce that

We conclude that

Then, . Therefore, we get

We have . So,

Applying Lemma 4 again with , , , , and , we get . Consider in two cases. Case I: If and  Hence, , , and . Case II: If , we get always negative. Solving equation (5) for . In this case equation (5) can be reduced to

By induction, we can prove that all the Lucas balancing numbers are odd. Therefore, we have no solutions in this case.

Solving equation (5) for , , and yields the solutions which appear in Theorem 2.

5. Conclusion

We determined all the balancing and Lucas balancing numbers that are sums of Jacobsthal numbers. We mainly used Matveev’s theorem. Also, we use a reduction lemma due to Dujella and Pethö to reduce the obtained upper bound. We revealed that there are four balancing numbers and three Lucas balancing numbers expressible as sums of two Jacobsthal numbers. In future, this work may be extended to investigate cobalancing and -balancing numbers that are expressible as sums of Jacobsthal numbers.

Data Availability

The data used in this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.