# On Second Order Gap Balancing Numbers

Authors: S. S. Rout
Journal: Journal of Numbers
 (2014)

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## Abstract

We consider the Diophantine equation1k+2k+⋯+x-1k=x+2k+x+3k+⋯+x+rk for some natural numbers x, k, and r, and we call 2x+1 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.


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## Body

## 1. Introduction

In [1], Finkelstein defined mth power numerical center x for y as solutions of the Diophantine equation:
(1)1m+2m+⋯+xm=xm+(x+1)m+(x+2)m+⋯+ym.
For m=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:
(2)1+2+⋯+(x-1)=(x+2)+(x+3)+⋯+y.
They call 2x+1 a 2-gap balancing number or g2 balancing number for some y. Motivated by higher order balancing numbers [4] and 2-gap balancing numbers [5], we introduce higher order k-gap balancing number as follows.Letk be the fixed odd positive integer. We call the positive integer x an mth order k-gap balancing number if
(3)1m+2m+⋯+(x-k+12)m=(x+k+12)m+(x+k+32)m+⋯+ym.
Equation (3) is equivalent to (1) when k=1. Similarly for fixed even positive integer k, we call the positive integer 2x+1 an mth order k-gap balancing number if
(4)1m+2m+⋯+(x-k2)m=(x+k2+1)m+(x+k+32)m+⋯+ym.
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 tom=1. For m=1 and k=2 (4) is equivalent to (2). Further, (2) reduces to the Diophantine equation 2x2+7=y2 which again reduces to the Pell's equation 8z2+1=w2, ensuring infinitude of the first order 2-gap balancing numbers. Now consider the case for m=1 and k arbitrary. Like the previous case (4) reduces to the Pell like equations 2x2+2k2-1=y2 for even k and 8x2+2k2-1=y2 for odd k which also shows infinitude of solutions of first order k-gap balancing numbers.To solve (4) for m=2 and k=2, we need the following results.Theorem 2 (see [6]).
Leta,b,c be nonzero integers. Then the equation ax3+by3=c has only finitely many solutions in integers (x,y).Theorem 3 (see [7]).
LetK(ρ) be a cubic field over the field of rational numbers, and let α=Aρ2+Bρ+C be an integer in the ring (1,ρ,ρ2). Suppose A≡B≡0(mod⁡pk), where p is an odd rational prime, and (α,p)=1. Further, suppose that tA+sB≢0(mod⁡p2k), where t,s, and k are rational integers, and k>0. Then if αn=Anρ2+Bnρ+Cn, tAn+sBn is never zero for any n≠0.Theorem 4 (see [8]).
The Diophantine equationAx3+By3=C (C=1 or 3; 3∤AB if C=3; A>B; A,B positive integers) has at most one solution in nonzero integers (x,y). There is the unique exception for the equation 2x3+y3=3 which has exactly two integral solutions (x,y)=(1,1) and (x,y)=(4,-5).Theorem 5 (see [9]).
Ifx and y are integers, then
(5)|x3-2y3|⩾max⁡{|x|,|y|}0.53,|x3-3y3|⩾max⁡{|x|,|y|}0.24,|x3-6y3|⩾max⁡{|x|,|y|}0.65,
where the last inequality holds unless |x|=467 and |y|=257.Since we are dealing with Diophantine equations of degree three, we need to discuss cubic fieldK(θ), where θ3=2 (see [10]). The necessary information for our problem is as follows.(1)
The integers ofK(θ) are of the form α=A+Bθ+Cθ2, where A, B, and C are rational integers.(2)
The ring of integers ofK(θ) is a unique factorization domain.(3)
By Dirichlet's theorem on units, there is only one fundamental unit of the field, which we designate byϵ0 of K(θ), with 0<ϵ0<1, is given by
(6)ϵ0=-1+θ.All the units of the field are given by±ϵ0m, where m is any rational integer. Any such power of ϵ0 is of the form a+bθ+cθ2, where a, b, and c are rational integers. Norm of α=A+Bθ+Cθ2 is given by N(α)=A3+2B3+4C3-6ABC. All units of norm 1 in K(θ) is given by ϵ0m.First, we have to find the number of equivalence classes of associated primes of norm3, 5, and 71.Sincex3-2≡(x-2)3(mod⁡3), 3 is a perfect cube in K(θ), apart from unit factors. So
(7)3=(θ+1)3(θ-1),
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 K(θ), as any integer of norm 3 in K(θ) must divide 3, apart from unit factors and there is only one such integer.Furthermore,5 is a rational prime of the form 3r+2. So 5 can split into two primes in K(θ). That is
(8)5=(-3+2θ2)(9+8θ+6θ2),
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 K(θ), as any integer in K(θ) with norm 5 must divide 5, and apart from unit factors, there is only one such integer.Lastly, since71 is a rational prime of the form 3r+2, it splits into two primes in K(θ):
(9)71=(5-3θ)(25+15θ+9θ2).
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 K(θ).


## 3. Proof of Theorem1

Letx be a second order 2-gap balancing number. Equation (4) with m=2 and k=2, simplifies to
(10)(x-1)x(2x-1)6=y(y+1)(2y+1)6-(x+1)(x+2)(2x+3)6.
Simplification of the above equation gives
(11)2(2x+1)3+11(2x+1)=(2y+1)3-(2y+1).
Setting A=-(2x+1) and B=(2y+1), we get
(12)-2[A3+11A]=B3-B.
We shall now prove several lemmas which together imply that the only solution of (12) subject to the conditions
(13)Aisnegativeandodd,Bispositiveandodd
is (A,B)=(-3,5).Lemma 6.
All the integral solutions of (12) satisfying the condition (13) correspond to the integral solutions of the equations:
(14)2u3+v3=5,15,25,71,75,213,355,1065,1775,and5325.Proof.
Let(A,B) be any integral solution of (12) subject to the conditions (13). Let gcd(A,B)=d. Letting A=du and B=dv and substituting in (12), we get
(15)d2(v3+2u3)=v-22u,
where
(16)gcd(u,v)=1,uisnegativeandodd,vispositiveandodd}.
Let
(17)2u3+v3=c.
Then v-22u=cd2 or equivalently,
(18)v=22u+cd2.
Substituting (18) in (17) we get
(19)d6c3+66uc2d4+1452u2d2c+10650u3=c
and using (17) and (19), we obtain
(20)d6c3+66uc2d4+1452u2d2c+5324c=5325v3.
Therforec∣5325v3, hence c∣5325 as gcd(c,v)=1. So the possible values of c are 1, 3, 5, 15, 25, 71, 75, 213, 355, 1065, 1775, and 5325. Thus, solving (12) subject to (13) can be reduced to solving the set of equations:
(21)2u3+v3=1,3,5,15,25,71,75,213,355,1065,1775,and5325.
However, the equation 2u3+v3=1,3 has solutions (u,v)=(1,-1),(1,1),(4,-5) by Theorem 4 which violates the condition in (16). Hence Lemma 6 follows.Lemma 7.
The equation2u3+v3=5 is impossible in rational integers (u,v).Proof.
We consider the integer-3+2θ2 whose norm is 5 and any other integer of norm 5 in K(θ) must be of the form (-3+2θ2)(ϵ0m), as all primes of norm 5 in K(θ) are associated. Let ϵ0m=am+bmθ+cmθ2 and
(22)Xm+Ymθ+Zmθ2=(-3+2θ2)(am+bmθ+cmθ2).
Hence
(23)Zm=-3cm+2am.
We seek all integers of K(θ) with norm 5 of the form a+bθ. Hence Zm must be zero, and thus the congruence Zm≡0(mod⁡t) must be solvable for every modulus t. We shall show that
(24)Zm≢0(mod⁡7).
Therefore we only need to check Zi≢0(mod⁡7) for i=0,1,…,6. Using (23), we have the values of Z0 to Z6 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 equation2u3+v3=71 is (u,v)=(-3,5).Proof.
We seek all the integers ofK(θ) which are of the form a+bθ. Since all primes of norm 71 in K(θ) are associated, any such prime must be an associate of 5-3θ. Let ϵ0m=am+bmθ+cmθ2 be a unit of K(θ). Our requirement is that the coefficient of θ2 in
(25)(5-3θ)(am+bmθ+cmθ2)
be zero. This gives
(26)5cm-3bm=0
We claim that (26) is impossible for m≠0. Now let
(27)xm+ymθ+zmθ2=(5-3θ)(ϵ0m).
Since
(28)ϵ03+3ϵ02+3ϵ0-1=0,
we have ϵ0m+3+3ϵ0m+2+3ϵ0m+1-ϵ0m=0, from which it follows that
(29)zm+3+3zm+2+3zm+1-zm=0.
Since from (27)
(30)xm+ymθ+zmθ2=(5-3θ)(am+bmθ+cmθ2),
we have zm=5cm-3bm by (26). Also from (29)
(31)zm+3≡zm(mod⁡3).
Now ϵ03=1+3θ-3θ2 which shows that
(32)b3≡c3≡0(mod⁡3),z3≡5c3-3b3≡3(mod⁡9).
Hence by Theorem 3, zm is never zero for any m≠0 which completes the proof of the lemma.Lemma 9.
The only integral solution of the equation2u3+v3=25 is (u,v)=(-1,3).Proof.
In this case we seek all integers of the forma+bθ with norm 25. Here we employ the same method of proof as in Lemma 8. Observe that
(33)25=(3-θ)(9+3θ+θ2).
We need to show that the coefficient of θ2 in
(34)(3-θ)(am+bmθ+cmθ2)
is zero. That is
(35)3cm-bm=0
we claim that (35) is never zero. As in previous lemma
(36)b3≡c3≡0(mod⁡3),z3≡3c3-b3≡3(mod⁡9)
Hence by Theorem 3, zm is never zero for any m≠0.Lemma 10.
The equation2u3+v3=15 is impossible in rational integers (u,v).Proof.
We consider the integer1-3θ+2θ2 whose norm is 15 and any other integer of norm 15 in K(θ) must be of the form (1-3θ+2θ2)(ϵ0m), as all primes of norm 5 in K(θ) are associated and also all primes of norm 3. Let ϵ0m=am+bmθ+cmθ2 and
(37)Xm+Ymθ+Zmθ2=(1-3θ+2θ2)(am+bmθ+cmθ2).
Hence
(38)Zm=cm-3bm+2am.
We seek all integers of K(θ) with norm 15 of the form a+bθ. Hence Zm must be zero, and thus the congruence Zm≡0(mod⁡t) must be solvable for every modulus t. We shall now show that
(39)Zm≢0(mod⁡31).
Also we have by manual verification ϵ010≡-6(mod⁡31), ϵ020≡5(mod⁡31), and ϵ030≡1(mod⁡31). Therefore Zm satisfies the conditions Zm+10≡-6Zm(mod⁡31), Zm+20≡5Zm(mod⁡31) and Zm+30≡Zm(mod⁡31). Therefore we only need to check Zi≢0(mod⁡31) for i=0,1,…,9. Using (31) and (38) we have the values of Z0 to Z9, 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 equation2u3+v3=75 is impossible in rational integers (u,v).Proof.
We are interested to find all elements of the forma+bθ such that N(a+bθ)=75. The integer 11-6θ-2θ2 has norm 75. Any other integer of norm 75 must be of this form (11-6θ-2θ2)ϵ0m. Therefore
(40)Xm+Ymθ+Zmθ2=(11-6θ-2θ2)(am+bmθ+cmθ2).
Hence
(41)Zm=11cm-6bm-2am.
The values of Z0 to Z9 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 equations2u3+v3=213,355,1065,1775,5325 are impossible in rational integers (u,v).Proof.
Using the norm of 3, 5, and 71 and multiplicative property of the norm, the integers having norms213, 355, 1065, 1775, and 5325 are -11+3θ+5θ2, -27+9θ+10θ2, 45+11θ-37θ2, 15-14θ+3θ2, and -43+20θ+12θ2, respectively. Also we know that all primes of norm 3, 5, and 71 in K(θ) are associated. Therefore the integers whose norms are 213, 355, 1065, 1775, and 5325, respectively, can be represented by
(42)Xm1+Ym1θ+Zm1θ2=(-11+3θ+5θ2)(am1+bm1θ+cm1θ2),Xm2+Ym2θ+Zm2θ2=(-27+9θ+10θ2)(am2+bm2θ+cm2θ2),Xm3+Ym3θ+Zm3θ2=(45+11θ-37θ2)(am3+bm3θ+cm3θ2),Xm4+Ym4θ+Zm4θ2=(15-14θ+3θ2)(am4+bm4θ+cm4θ2),Xm5+Ym5θ+Zm5θ2=(-43+20θ+12θ2)(am5+bm5θ+cm5θ2).
Hence,
(43)Zm1=-11cm1+3bm1+5am1,Zm2=-27cm2+9bm2+10am2,Zm3=45cm3+11bm3-37am3,Zm4=15cm4-14bm4+3am4,Zm5=-43cm5+20bm5+12am5.
But we seek all integers of the form a+bθ. That means Zmi≡0(mod⁡t) for any modulus t and for i=1,2,…,5. We want to show that Zmi≢0(mod⁡t) for some t. Also Zmi satisfy the congruences Zm+10i≡-6Zmi(mod⁡31), Zm+20i≡5Zmi(mod⁡31), and Zm+30i≡Zmi(mod⁡31). Using these congruences the values of Zmi modulo 31 are as follows. The values of Z01 to Z91 modulo 31 are 5, −2, −12, 16, −14, 13, 19, 2, 15, and −13, the values of Z02 to Z92 modulo 31 are 10, -1, −4, 21, −2, 20, 2, −6, 1, and −14, the values of Z03 to Z93 modulo 31 are 6, 17, −14, −15, 11, −2, 20, −12, 5, and 10, the values of Z04 to Z94 modulo 31 are 3, −17, −10, 9, 4, 7, 7, −7, 7, and 7, and the values of Z05 to Z95 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 12Till now we get the solutions(u,v) of (14) satisfying the conditions (16). We need to find the solutions (A,B) of (12) for which the exact value of d must be calculated. It follows from Lemma 8 to Lemma 12 that the only integral solution of (14) is (u,v)=(-1,3),(-3,5). In both the cases, both u and v are relatively prime and odd; u is negative and v is positive. Therefore from (15), we have
(44)25d2=25,71d2=71.
In either case, d=1.Thus, the only integral solution of (12) satisfying conditions (13) is (A,B)=(-1,3),(-3,5). Hence the only integral solution of (10) is (x,y)=(1,2). This completes the proof of Theorem 1.Equation (14) is equivalent to x3-2y3=c by setting y=-u and x=v. Theorem 5 give a lower bound for the absolute value of x3-2y3. This lower bound also immediately gives the proof of Theorem 1.



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*Source: 216738-2014-10-08.xml*