Abstract

We consider the Diophantine equation for some natural numbers x, k, and r, and we call as kth order 2-gap balancing number. It was also proved that there are infinitely many first order 2-gap balancing numbers. In this paper, we show that the only second order 2-gap balancing number is 1.

1. Introduction

In [1], Finkelstein defined th power numerical center for as solutions of the Diophantine equation: For , it coincides with the notion of balancing numbers introduced by Behera and Panda [2]. Finkelstein conjectured that if then there is no integer greater than with th power numerical center. Ingram in [3] proved Finkelstein's conjecture for . Further, in [4] Panda studied (1) slightly differently and called the solution of (1) as th order balancing number.

The concept of gap balancing numbers was introduced by Panda and Rout [5] in connection with the Diophantine equation: They call a 2-gap balancing number or balancing number for some . Motivated by higher order balancing numbers [4] and 2-gap balancing numbers [5], we introduce higher order -gap balancing number as follows.

Let be the fixed odd positive integer. We call the positive integer an th order -gap balancing number if Equation (3) is equivalent to (1) when . Similarly for fixed even positive integer , we call the positive integer an th order -gap balancing number if In this paper, we prove the following theorem.

Theorem 1. The only positive integer possessing second order 2-gap balancing number is 1.

2. Background

Before we prove the main result of this paper, it is better to look into the special cases corresponding to . For and (4) is equivalent to (2). Further, (2) reduces to the Diophantine equation which again reduces to the Pell's equation , ensuring infinitude of the first order 2-gap balancing numbers. Now consider the case for and arbitrary. Like the previous case (4) reduces to the Pell like equations for even and for odd which also shows infinitude of solutions of first order -gap balancing numbers.

To solve (4) for and , we need the following results.

Theorem 2 (see [6]). Let be nonzero integers. Then the equation has only finitely many solutions in integers .

Theorem 3 (see [7]). Let be a cubic field over the field of rational numbers, and let be an integer in the ring . Suppose , where is an odd rational prime, and . Further, suppose that , where , and are rational integers, and . Then if , is never zero for any .

Theorem 4 (see [8]). The Diophantine equation ( or ; if ; ; positive integers) has at most one solution in nonzero integers . There is the unique exception for the equation which has exactly two integral solutions and .

Theorem 5 (see [9]). If and are integers, then where the last inequality holds unless and .

Since we are dealing with Diophantine equations of degree three, we need to discuss cubic field , where (see [10]). The necessary information for our problem is as follows.(1)The integers of are of the form , where , , and are rational integers.(2)The ring of integers of is a unique factorization domain.(3)By Dirichlet's theorem on units, there is only one fundamental unit of the field, which we designate by of , with , is given by  All the units of the field are given by , where is any rational integer. Any such power of is of the form , where , , and are rational integers. Norm of is given by . All units of norm 1 in is given by .

First, we have to find the number of equivalence classes of associated primes of norm , , and .

Since , 3 is a perfect cube in , apart from unit factors. So and 3 is the cube of a prime of norm 3 times an unit factor. Hence, there is only one equivalence class of associated primes of norm 3 in , as any integer of norm 3 in must divide 3, apart from unit factors and there is only one such integer.

Furthermore, is a rational prime of the form . So 5 can split into two primes in . That is where the norm of first factor is 5 and norm of second factor is 25. Hence there is only one equivalence class of associated primes of norm 5 in , as any integer in with norm 5 must divide 5, and apart from unit factors, there is only one such integer.

Lastly, since is a rational prime of the form , it splits into two primes in : Thus the norm of first factor of 71 is 71 and the norm of second factor is 5041. Hence there is only one equivalence class of associated primes of norm 71 in .

3. Proof of Theorem 1

Let be a second order 2-gap balancing number. Equation (4) with and , simplifies to Simplification of the above equation gives Setting and , we get We shall now prove several lemmas which together imply that the only solution of (12) subject to the conditions is .

Lemma 6. All the integral solutions of (12) satisfying the condition (13) correspond to the integral solutions of the equations:

Proof. Let be any integral solution of (12) subject to the conditions (13). Let . Letting and and substituting in (12), we get where Let Then or equivalently, Substituting (18) in (17) we get and using (17) and (19), we obtain
Therfore , hence as . So the possible values of are 1, 3, 5, 15, 25, 71, 75, 213, 355, 1065, 1775, and . Thus, solving (12) subject to (13) can be reduced to solving the set of equations: However, the equation has solutions by Theorem 4 which violates the condition in (16). Hence Lemma 6 follows.

Lemma 7. The equation is impossible in rational integers .

Proof. We consider the integer whose norm is 5 and any other integer of norm 5 in must be of the form , as all primes of norm 5 in are associated. Let and Hence We seek all integers of with norm 5 of the form . Hence must be zero, and thus the congruence must be solvable for every modulus . We shall show that Therefore we only need to check for . Using (23), we have the values of to modulo 7 as 2, −2, −1, 4, 3, −1, and −2 and none of these is zero. This completes the proof of Lemma 7.

Lemma 8. The only integral solution of the equation is .

Proof. We seek all the integers of which are of the form . Since all primes of norm 71 in are associated, any such prime must be an associate of . Let be a unit of . Our requirement is that the coefficient of in be zero. This gives We claim that (26) is impossible for . Now let Since we have , from which it follows that Since from (27) we have by (26). Also from (29) Now which shows that Hence by Theorem 3, is never zero for any which completes the proof of the lemma.

Lemma 9. The only integral solution of the equation is .

Proof. In this case we seek all integers of the form with norm 25. Here we employ the same method of proof as in Lemma 8. Observe that We need to show that the coefficient of in is zero. That is we claim that (35) is never zero. As in previous lemma Hence by Theorem 3, is never zero for any .

Lemma 10. The equation is impossible in rational integers .

Proof. We consider the integer whose norm is 15 and any other integer of norm 15 in must be of the form , as all primes of norm 5 in are associated and also all primes of norm 3. Let and Hence We seek all integers of with norm 15 of the form . Hence must be zero, and thus the congruence must be solvable for every modulus . We shall now show that Also we have by manual verification , , and . Therefore satisfies the conditions , and . Therefore we only need to check for . Using (31) and (38) we have the values of to , modulo 31 is −5, 3, 10, −10, −2, 14, 16, 1, −6, and 19 and none of these is zero. This completes the proof of Lemma 10.

Lemma 11. The equation is impossible in rational integers .

Proof. We are interested to find all elements of the form such that . The integer has norm . Any other integer of norm must be of this form . Therefore Hence The values of to modulo 31 are −2, −4, 21, 9, −1, −3, −10, −19, 25, 12 and none of these is zero. This completes the proof of Lemma 11.

Lemma 12. The equations are impossible in rational integers .

Proof. Using the norm of 3, 5, and 71 and multiplicative property of the norm, the integers having norms , , , , and are , , , , and , respectively. Also we know that all primes of norm 3, 5, and 71 in are associated. Therefore the integers whose norms are , , , , and , respectively, can be represented by Hence, But we seek all integers of the form . That means for any modulus and for . We want to show that for some . Also satisfy the congruences , , and . Using these congruences the values of modulo 31 are as follows. The values of to modulo 31 are 5, −2, −12, 16, −14, 13, 19, 2, 15, and −13, the values of to modulo 31 are 10, -1, −4, 21, −2, 20, 2, −6, 1, and −14, the values of to modulo 31 are 6, 17, −14, −15, 11, −2, 20, −12, 5, and 10, the values of to modulo 31 are 3, −17, −10, 9, 4, 7, 7, −7, 7, and 7, and the values of to modulo 31 are 12, 8, −3, 5, −2, 3, 2, −15, 17, and 2 and none of these is zero. This completes the proof of Lemma 12

Till now we get the solutions of (14) satisfying the conditions (16). We need to find the solutions of (12) for which the exact value of must be calculated. It follows from Lemma 8 to Lemma 12 that the only integral solution of (14) is . In both the cases, both and are relatively prime and odd; is negative and is positive. Therefore from (15), we have In either case, .

Thus, the only integral solution of (12) satisfying conditions (13) is . Hence the only integral solution of (10) is . This completes the proof of Theorem 1.

Equation (14) is equivalent to by setting and . Theorem 5 give a lower bound for the absolute value of . This lower bound also immediately gives the proof of Theorem 1.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author is grateful to Professor G. K. Panda for his useful suggestions during the preparation of the paper.