On Second Order Gap Balancing Numbers

Form = 1, it coincides with the notion of balancing numbers introduced by Behera and Panda [2]. Finkelstein conjectured that if m > 1 then there is no integer greater than 1 with mth power numerical center. Ingram in [3] proved Finkelstein’s conjecture for m = 5. Further, in [4] Panda studied (1) slightly differently and called the solution x of (1) as mth order balancing number. The concept of gap balancing numbers was introduced by Panda and Rout [5] in connection with the Diophantine equation:

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  = 1.Similarly for fixed even positive integer , we call the positive integer 2+1 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.

Background
Before we prove the main result of this paper, it is better to look into the special cases corresponding to  = 1.For  = 1 and  = 2 (4) is equivalent to (2).Further, (2) reduces to the Diophantine equation 2 2 + 7 =  2 which again reduces to the Pell's equation 8 2 + 1 =  2 , ensuring infinitude of the first order 2-gap balancing numbers.Now consider the case for  = 1 and  arbitrary.Like the previous case (4) reduces to the Pell like equations 2 2 + 2 2 − 1 =  2 for even  and 8 2 + 2 2 − 1 =  2 for odd  which also shows infinitude of solutions of first order -gap balancing numbers.
To solve (4) for  = 2 and  = 2, we need the following results.
Since we are dealing with Diophantine equations of degree three, we need to discuss cubic field (), where  3 = 2 (see [10]).The necessary information for our problem is as follows.
(1) The integers of () are of the form  = ++ 2 , 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  0 of (), with 0 <  0 < 1, is given by All the units of the field are given by ±  0 , where  is any rational integer.Any such power of  0 is of the form ++ 2 , where , , and  are rational integers.Norm of  =  +  +  2 is given by () =  3 + 2 3 +4 3 −6.All units of norm 1 in () is given by   0 .
First, we have to find the number of equivalence classes of associated primes of norm 3, 5, and 71.Since  3 − 2 ≡ ( − 2) 3 (mod 3), 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, 5 is a rational prime of the form 3 + 2. 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 71 is a rational prime of the form 3 + 2, 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 ().

Proof of Theorem 1
Let  be a second order 2-gap balancing number.Equation (4) with  = 2 and  = 2, simplifies to Simplification of the above equation gives Setting  = −(2 + 1) and  = (2 + 1), we get We shall now prove several lemmas which together imply that the only solution of (12) subject to the conditions  is negative and odd,  is positive and odd ( 13) is (, ) = (−3, 5).

Lemma 8. The only integral solution of the equation
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 5−3.Let   0 =   +  +    2 be a unit of ().Our requirement is that the coefficient of  2 in be zero.This gives We claim that ( 26) is impossible for  ̸ = 0. Now let Since we have  +3 0 Since from ( 27) we have   = 5  − 3  by (26).Also from ( 29) Now  3 0 = 1 + 3 − 3 2 which shows that  3 ≡  3 ≡ 0 (mod 3) , Hence by Theorem 3,   is never zero for any  ̸ = 0 which completes the proof of the lemma.

Lemma 9. The only integral solution of the equation
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  2 in we claim that (35) is never zero.As in previous lemma Hence by Theorem 3,   is never zero for any  ̸ = 0.
Proof.We consider the integer 1 − 3 + 2 2 whose norm is 15 and any other integer of norm 15 in () must be of the form (1 − 3 + 2 2 )(  0 ), as all primes of norm 5 in () are associated and also all primes of norm 3. Let   0 =   +    +    2 and Hence We seek all integers of () with norm 15 of the form  + .
Equation ( 14) is equivalent to  3 − 2 3 =  by setting  = − and  = V.Theorem 5 give a lower bound for the absolute value of  3 − 2 3 .This lower bound also immediately gives the proof of Theorem 1.