Miskolc Mathematical Notes Vol. 14 (2013), No. 3, pp. 951–957
HU e-ISSN 1787-2413
BALANCING WITH BALANCING POWERS NURETTIN IRMAK Received 21 March, 2013 l l Abstract. In this paper, the Diophantine equation B1k C B2k C C Bnk 1 D BnC1 C BnC2 C l C BnCr for the positive integer unknowns n 2; k; l and r is studied in certain cases, where Bn denotes the nt h term of the balancing sequence.
2010 Mathematics Subject Classification: 11B39 Keywords: balancing numbers, powers
1. I NTRODUCTION The first definition of balancing numbers is essentially due to Finkelstein [5], though he named them numerical centers. A positive integer n 2 is called balancing number if 1 C 2 C C .n
1/ D .n C 1/ C .n C 2/ C C .n C r/
(1.1)
holds for some positive integer r which is called balancer corresponding to the balancing number n: The mt h term of the sequence of balancing numbers is denoted by Bm : It is known that the balancing numbers satisfy the recurrence Bn D 6Bn
1
Bn
(1.2)
2
where initial contidions are defined by B0 D 0 and B1 D 1: The first few terms of the balancing sequence are 0; 1; 6; 35; 204; 1189; 6930; 40391; : : : : The Binet formula of the balancing numbers is given by p n p n n n 3 C 2 2 3 2 2 ˛ ˇ D ; Bn D p ˛ ˇ 4 2 where ˛ and ˇ are the roots of the chaeacteristic polynomial x 2 recursive sequence (1.2). Behera et al. [4] showed that the diophantine equation F1k C F2k C C Fnk
1
l l l D FnC1 C FnC2 C C FnCr
6x C 1 of the
(1.3)
c 2013 Miskolc University Press
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NURETTIN IRMAK
has no solution in the positive integers n; r; k; l with n 2 in the case k l and .k; l/ D .2; 1/ ; .3; 1/ ; .3; 2/ :
(1.4)
nt h
Here Fn denotes the Fibonacci number. They also conjectured in [4] that only the quadruple .n; r; k; l/ D .4; 3; 8; 2/ satisfies (1.3). Their conjecture was proved by Alvarado et al. [2]. In this paper, we focus on the diophantine equation B1k C B2k C C Bnk
l l l D BnC1 C BnC2 C C BnCr :
1
(1.5)
In the specific cases (1.4) we found no solutions. Based on Theorems 6-9 and an extended computer search, we conjectured the following. Conjecture 1. There is no quadruple .n; r; k; l/ of positive integers .n 2/ which satisfies (1.5). 2. L EMMAS The proofs of the theorems use several statements collected in this section. Lemma 1. For any positive integer m (a): BmC1 Bm 1 D .Bm C 1/ .Bm 1/ ; 2 2 (b): B2m 1 D Bm Bm 1; 2; (c): B1 C B3 C C B2m 1 D Bm 2 2 2 (d): B1 C B2 C C Bm 1 < BmC1 C BmC2 C C B2m 3 for m 5; 2 2 2 3 (e): B2mC5 C B2mC6 C C B3mC4 < B13 C B23 C C B2mC3 for m 2; (f): 4Bm < BmC1 Bm < 5Bm hold. Proof. The first three identities appear in [3] and [6]. The fourth property (d) can be justified by induction on m: Clearly, B12 C B22 C B32 C B42 < B6 C B7 : Now suppose that, 2 B12 C B22 C C Bm
1
< BmC1 C BmC2 C C B2m
3:
2 1 C Bm
< BmC1 C BmC2 C C B2m
2 3 C Bm
Evaluating 2 B12 C B22 C C Bm 2 0 and b 0 are real numbers and u0 is a positive real number. Then a˛ u C b ˛ uC holds for any u u0 where D log˛ a C ˛bu0 : Proof. For the proof, see Lemma 2.4 in [4].
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3. THE RESULTS Here we present four theorems and their proofs which confirm Conjecture 1. l l C C BnC2 Theorem 1. The diophantine equation B1k C B2k C C Bnk 1 D BnC1 l C BnCr has no solution for any positive integers r and n 2 if k l:
Proof. Applying Lemma 2(a) for k l; we get B1k
C B2k
C C Bnk 1
.B1 C B2 C C Bn
Bn
Bn 4 which completes the proof of Theorem 1.
1
1
l
1/
k
D
Bn
Bn 4
1
1
k
< Bnl ;
Theorem 2. The diophantine equation B12 C B22 C C Bn2 C BnCr has no solution in positive integers r and n 2:
1
D BnC1 C BnC2 C
Proof. By the virtue of Lemma 2(a) and (b), the equation B12 C B22 C C Bn2
1
D BnC1 C BnC2 C C BnCr
(3.1)
is equvalent to B2n
1 C 8 .BnC1
Bn /
.2n
1/ D 8 .BnCrC1
BnCr / :
For brevity, we use the following labels: LS WD B2n
1 C 8 .BnC1
RS WD 8 .BnCrC1
Bn /
.2n
1/
BnCr / :
Using Lemma 1(f), B2n
1
< LS < B2n
1 C 40Bn
hold. The by Lemma 3, ˛ 2n
1:99
< LS < ˛ 2n
1:98
C 40˛ n
0:98
˛ 2n
1:98
C ˛ nC1:12 :
According to the Lemma 4, if a D b D 1 we obtain < 0:02: Thus ˛ 2n 1:99 < LS < ˛ nC1:12 ˛ n 3:1 C 1 < ˛ nC1:12 ˛ n 3:1 ˛ 0:02 : Consequently, ˛ 2n 1:99 < LS < ˛ 2n On the other hand, Lemma 1(f) implies
1:96
:
(3.2)
32BnCr < RS < 40BnCr ; and then ˛ nCr
0:99 1:95
˛
D ˛ nCrC0:96 < RS < ˛ nCrC1:12 D ˛ nCr
0:98 2:1
˛
(3.3)
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follow. Thus, from (3.2) and (3.3) we get 1:99; n C r C 0:96g < min f2n
max f2n
1:96; n C r C 1:12g
which yields the inequalities 1:99 < n C r C 1:12
2n and
n C r C 0:96 < 2n Hence, the positive integer n and r satisfy n that is r D n
1:96:
3:11 < r < n
3: If we plug r D n
2:92;
3 in (3.1), we obtain
B12 C B22 C C Bn2 1
D BnC1 C BnC2 C C B2n
(3.4)
3
which contradicts to Lemma 1(d). B13 C B23 C C Bn3 1
D BnC1 C BnC2 C Theorem 3. The diophantine equation C BnCr has no solution for any positive integers r and n 2: Proof. Obviously, n D 2 the case provides no solution. Therefore we may assume n 3; By Lemma 2(a) and (c), the equation in the theorem is equivalent to B3n B3.n
1/ C 1568BnC1
1715Bn C 147Bn
1 C 112 D 1568 .BnCrC1
BnCr / : (3.5)
Using the labels again, put LS WD B3n
B3.n
1/ C 1568BnC1
RS WD 1568 .BnCrC1
1715Bn C 147Bn
1 C 112;
BnCr / :
By Lemma 3, it is obvious that 196B3.n
1/
< LS < 197B3.n
1/ :
< LS < ˛ 3n
D ˛ 3n
Further we have ˛ 3n
3:99 2:98
˛
D ˛ 3n
1:01
0:98
3:98 3:00
˛
(3.6)
On the other hand, applying Lemma 1(f), ˛ nCrC3:96 D 1568 4 BnCr < RS < 1568 5BnCr D ˛ nCrC4:11
(3.7)
hold. From (3.6) and (3.7) we get max f3n
1:01; n C r C 3:96g < min f3n
0:98; n C r C 4:11g ;
and we deduce 2n 5:12 < r < 2n 4:94: Since n and r are integers in this inequality, only r D 2n n 3). But it contradicts to B3n
B3.n
1/ C 1568BnC1
1715Bn C 147Bn
1 C 112
5 is possible (recall that < 1568 .B3n
4
B3n
5/ :
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NURETTIN IRMAK
Therefore, there is no solution which satisfies (3.5).
2 2 C C BnC2 Theorem 4. The diophantine equation B13 C B23 C C Bn3 1 D BnC1 2 C BnCr has no solution for any positive integers r and n 2:
Proof. For 2 n 4; th statement is obvious. Assume now n 5: The application of Lemma 2(b)and (c) converts the equation B13 C B23 C C Bn3
1
2 2 2 D BnC1 C BnC2 C C BnCr
(3.8)
to 147 .Bn 32 196 which is equivalent to B3n
B3n
B3.n
B3.n
Bn
1/
147 .Bn
1/
Bn
1 / C 112
D
1 B2.nCr/C1 32
1 / C 196B2nC1 C 112
B2nC1
2r ;
D 196 B2.nCr/C1
2r :
Put LS WD B3n
B3.n
1/
RS WD 196 B2.nCr/C1
147 .Bn 2r :
Bn
1 / C 196B2nC1 C 112;
Now use Lemma 3 and Lemma 1(f) to obtain ˛ 2nC2rC3 < RS < ˛ 2nC2rC3:02 :
(3.9)
Similarly,
˛
196B3.n 1/ 3n 3 0:99 2:98 ˛
< LS < .197 C 196/ B3.n D˛
3n 1:01
< LS < ˛
1/ ; 3n 0:59
D ˛ 3n
3 0:98 3:39
˛
(3.10)
hold. From (3.9) and (3.10) max f2n C 2r C 3; 3n
1:01g < min f2n C 2r C 3:02; 3n
0:59g
follows, which yields n Clearly, only 2r D n
4:03 < 2r < n
3:59:
4 is possible. Inserting n D 2r C 4 to (3.8), we obtain
3 2 2 2 D B2rC5 C B2rC6 C C B3rC4 ; B13 C B23 C C B2rC3
and we arrived at a contradiction with Lemma 1(e). ACKNOWLEDGEMENT The authors would like to thank L´aszl´o Szalay for his helpful suggestions.
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R EFERENCES [1] M. Alp, N. Irmak, and L. Szalay, “Balancing diophantine triples,” Acta Univ. Sapientiae, Math., vol. 4, no. 1, pp. 11–19, 2012. [2] S. D. Alvarado, A. Dujella, and F. Luca, “On a conjecture regarding balancing with powers of Fibonacci numbers,” Integers, vol. 12, no. 6, pp. 1127–1158, a2, 2012. [3] A. Behera and G. K. Panda, “On the square roots of triangular numbers,” Fibonacci Q., vol. 37, no. 2, pp. 98–105, 1999. [4] A. Behera, K. Liptai, G. K. Panda, and L. Szalay, “Balancing with Fibonacci powers,” Fibonacci Q., vol. 49, no. 1, pp. 28–33, 2011. [5] R. Finkelstein, “The house problem,” Am. Math. Mon., vol. 72, pp. 1082–1088, 1965. [6] G. K. Panda, “Some fascinating properties of balancing numbers,” Congr. Numerantium, vol. 194, pp. 185–189, 2009.
Author’s address Nurettin Irmak Ni˘gde University, Art and Science Faculty, Mathematics Department, 51110 Ni˘gde, TURKEY E-mail address:
[email protected]
