# The Diophantine Equation

Authors: Ahmet Tekcan; Merve Tayat; Meltem E. Özbek
Journal: ISRN Combinatorics
 (2014)

---

## Abstract

Lett
≥
2 be an odd integer such that 2
t
2
-
1 is a prime. In this work, we determine all integer solutions of the Diophantine equation D
t
:
8
x
2
-
y
2
+
8
x
(
1
+
t
)
+
(
2
t
+
1
)
2
=
0 and then we deduce the general terms of all t-balancing numbers.


---

## Body

## 1. Introduction

Balancing numbers were first considered by Behera and Panda in [1] when they considered the integer solutions of the Diophantine equation
(1)
1
+
2
+
⋯
+
(
n
-
1
)
=
(
n
+
1
)
+
(
n
+
2
)
+
⋯
+
(
n
+
r
)
for some positive integers n and r. In this case n is called abalancing numberwithbalancer (orcobalancing number) r. For example, 6,35,204,1189 and 6930 are balancing numbers with cobalancing numbers which are 2,14,84,492 and 2870, respectively.Thenth balancing number is denoted by B
n and the nth cobalancing number is denoted by b
n. They satisfy the recurrence relations B
n
+
1
=
6
B
n
-
B
n
-
1 and b
n
+
1
=
6
b
n
-
b
n
-
1
+
2 for n
≥
2, with initial values B
1
=
1
,
B
2
=
6 and b
1
=
0
,
b
2
=
2.From (1), one has
(2)
r
=
-
(
2
n
+
1
)
+
8
n
2
+
1
2
,
n
=
2
r
+
1
+
8
r
2
+
8
r
+
1
2
.
So B
n is a balancing number if and only if 8
B
n
2
+
1 is a perfect square and b
n is a cobalancing number if and only if 8
b
n
2
+
8
b
n
+
1 is a perfect square. Set C
n
=
8
B
n
2
+
1 and c
n
=
8
b
n
2
+
8
b
n
+
1. Then C
n is called the nthLucas-balancing number and c
n is called the nthLucas-cobalancing number. Binet formulas for balancing, cobalancing, Lucas-balancing, and Lucas-cobalancing numbers are B
n
=
(
α
2
n
-
β
2
n
)
/
4
2, b
n
=
(
(
α
2
n
-
1
-
β
2
n
-
1
)
/
4
2
)
-
(
1
/
2
)
,
C
n
=
(
α
2
n
+
β
2
n
)
/
2, and c
n
=
(
α
2
n
-
1
+
β
2
n
-
1
)
/
2, respectively, where α
=
1
+
2 and β
=
1
-
2 (for further details see also [2–7]).Recently, there are many studies on balancing numbers. In [8], the authors generalized the theory of balancing numbers to numbers defined as follows. Let y
,
k
,
l
∈
ℤ
+ such that y
≥
4. Then a positive integer x such that x
≤
y
-
2 is called a (
k
,
l
)-power numerical center for y if
(3)
1
k
+
⋯
+
(
x
-
1
)
k
=
(
x
+
1
)
l
+
⋯
+
(
y
-
1
)
l
.
Later in [9], the authors extended the concept of balancing numbers to the (
a
,
b
)-balancing numbers defined as follows. Let a
>
0 and let b
≥
0 be coprime integers. If for some positive integers, n and r, one has
(4)
(
a
+
b
)
+
⋯
+
(
a
(
n
-
1
)
+
b
)
=
(
a
(
n
+
1
)
+
b
)
+
⋯
+
(
a
(
n
+
r
)
+
b
)
,
then the number a
n
+
b is called an (
a
,
b
)-balancing number and is denoted by B
n
(
a
,
b
).Like in(
a
,
b
)-balancing numbers, in [10], Dash et al. defined the t-balancing number and derived some results on it. Let t
≥
0 be an integer. Then a positive integer n is called a t-balancing numberif
(5)
1
+
2
+
⋯
+
n
=
(
n
+
1
+
t
)
+
(
n
+
2
+
t
)
+
⋯
+
(
n
+
r
+
t
)
for some positive integer r which is called t-balancer (or t-cobalancing number). 0- and 1-balancing numbers can be given in terms of balancing numbers; indeed, B
n
0
=
b
n
+
1
,
b
n
0
=
B
n
,
C
n
0
=
c
n
+
1
,
c
n
0
=
C
n and B
n
1
=
B
n
+
1
-
1
,
b
n
1
=
b
n
+
1
,
C
n
1
=
C
n
+
1
,
c
n
1
=
c
n
+
1. So it is assumed that t
≥
2.Thenth t-balancing number is denoted by B
n
t and the nth t-cobalancing number is denoted by b
n
t. From (5) we see that
(6)
r
=
-
(
2
n
+
2
t
+
1
)
+
8
n
2
+
8
n
(
1
+
t
)
+
(
2
t
+
1
)
2
2
,
n
=
(
2
r
-
1
)
+
8
r
2
+
8
t
r
+
1
2
.
Hence, B
n
t is a t-balancing number if and only if 8
(
B
n
t
)
2
+
8
B
n
t
(
1
+
t
)
+
(
2
t
+
1
)
2 is a perfect square and b
n
t is a t-cobalancing number if and only if 8
(
b
n
t
)
2
+
8
t
b
n
t
+
1 is a perfect square. Set
(7)
C
n
t
=
8
(
B
n
t
)
2
+
8
B
n
t
(
1
+
t
)
+
(
2
t
+
1
)
2
,
c
n
t
=
8
(
b
n
t
)
2
+
8
t
b
n
t
+
1
.
Then C
n
t is called the nthLucas
t-balancing number and c
n
t is called the nthLucas
t-cobalancing number.


## 2. Main Results

To determine the general terms of allt-balancing numbers, we have to determine all integer (in fact only positive) solutions of some specific Diophantine equations. Let us explain this as follows. We see as above that 8
x
2
+
8
x
(
1
+
t
)
+
(
2
t
+
1
)
2 is a perfect square for a t-balancing number x. So we set 8
x
2
+
8
x
(
1
+
t
)
+
(
2
t
+
1
)
2
=
y
2 for some integer y
≠
0. Hence, we get the Diophantine equation:
(8)
D
t
:
8
x
2
-
y
2
+
8
x
(
1
+
t
)
+
(
2
t
+
1
)
2
=
0
.
If we make the change of variables x
=
v
+
h and y
=
u
+
k for some h and k, then we get
(9)
D
t
~
:
8
(
v
+
h
)
2
-
(
u
+
k
)
2
+
8
(
v
+
h
)
(
1
+
t
)
+
(
2
t
+
1
)
2
.
Here we obtain h
=
-
(
t
+
1
)
/
2 and k
=
0. Thus (9) becomes
(10)
D
t
~
:
u
2
-
8
v
2
=
2
t
2
-
1
which is a Pell equation (see [11–14]). Before considering its integer solutions, we need some notations. Now let Δ be a nonsquare discriminant. Then the Δ-order O
Δ is defined for nonsquare discriminants Δ to be the ring O
Δ
=
{
x
+
y
ρ
Δ
:
x
,
y
∈
ℤ
}, where ρ
Δ
=
Δ
/
4 if Δ
≡
0
(mod 4), or (
1
+
Δ
)
/
2 if Δ
≡
1
(mod 4). So O
Δ is a subring of ℚ
(
Δ
)
=
{
x
+
y
Δ
:
x
,
y
∈
ℚ
}. The unit group O
Δ
* is defined to be the group of units of the ring O
Δ.For the quadratic formF
(
x
,
y
)
=
a
x
2
+
b
x
y
+
c
y
2, we can write F
=
(
x
a
+
y
(
(
b
+
Δ
)
/
2
)
)
(
x
a
+
y
(
(
b
-
Δ
)
/
2
)
)
/
a. So the module M
F of F is the O
Δ-module M
F
=
{
x
a
+
y
(
(
b
+
Δ
)
/
2
)
:
x
,
y
∈
ℤ
}
⊂
ℚ
(
Δ
). Therefore, we get (
u
+
v
ρ
Δ
)
(
x
a
+
y
(
(
b
+
Δ
)
/
2
)
)
=
x
′
a
+
y
′
(
(
b
+
Δ
)
/
2
), where
(11)
(
x
′
,
y
′
)
=
{
(
x
,
y
)
(
u
-
b
2
v
a
v
-
c
v
u
+
b
2
v
)
if
Δ
≡
0
(
mod
⁡
4
)
(
x
,
y
)
(
u
+
1
-
b
2
v
a
v
-
c
v
u
+
1
+
b
2
v
)
if
Δ
≡
1
(
mod
⁡
4
)
.
So there is a bijection Ψ
:
{
(
x
,
y
)
:
F
Δ
(
x
,
y
)
=
m
}
→
{
γ
∈
M
F
:
N
(
γ
)
=
a
m
} for solving the Pell equation F
(
x
,
y
)
=
m; that is, a
x
2
+
b
x
y
+
c
y
2
=
m. The action of O
Δ
,
1
*
=
{
α
∈
O
Δ
*
:
N
(
α
)
=
1
} on the set Ω
=
{
(
x
,
y
)
:
F
Δ
(
x
,
y
)
=
m
} of integral solutions of the equation F
(
x
,
y
)
=
m is most interesting when Δ is a positive nonsquare since O
Δ
,
1
* is infinite. Therefore, the orbit of each solution will be infinite and so the set Ω is either empty or infinite. Since O
Δ
,
1
* can be explicitly determined, the set Ω is satisfactorily described by the representation of such a list, called a set of representatives of the orbits. Let ɛ
Δ be the smallest unit of O
Δ, that is, greater than 1, and let τ
Δ
=
ɛ
Δ if N
(
ɛ
Δ
)
=
1, or ɛ
Δ
2 if N
(
ɛ
Δ
)
=
-
1. Then every O
Δ
,
1
* orbit of integral solutions of F
(
x
,
y
)
=
m contains a solution (
x
,
y
)
∈
ℤ
2 such that 0
≤
y
≤
U, where U
=
|
a
m
τ
Δ
/
Δ
|
1
/
2
(
1
-
(
1
/
τ
Δ
)
) if a
m
>
0 or U
=
|
a
m
τ
Δ
/
Δ
|
1
/
2
(
1
+
(
1
/
τ
Δ
)
) if a
m
<
0. So for finding a set of representatives of the O
Δ
,
1
* orbits of integral solutions of F
(
x
,
y
)
=
m, we must find, for each integer y such that 0
≤
y
≤
U, all integers x that satisfy F
(
x
,
y
)
=
m. If F
(
x
,
y
)
=
m, then Δ
y
2
+
4
a
m
=
(
2
a
x
+
b
y
)
2 and so x
=
±
(
Δ
y
2
+
4
a
m
-
b
y
)
/
2
a.For the Pell equationD
t
~, there are one, two, three, or more sets of representatives (or sets of solutions) depending on t. For example, for t
=
3, the set of representatives is {
(
±
5,1
)
}; for t
=
9, the set of representatives is {
(
±
13,1
)
,
(
±
17,4
)
}; for t
=
89, the set of representatives is {
(
±
127,6
)
,
(
±
129,10
)
,
(
±
143,24
)
,
(
±
177,44
)
}; and for t
=
93, the set of representatives is {
(
±
133,7
)
,
(
±
155,29
)
,
(
±
186,46
)
}. So we cannot determine for which values of t there are one, two, three, or more sets of representatives. Consequently, we cannot determine all integer solutions of D
t
~ in one way. To determine all integer solutions of D
t
~, we have to put some restrictions on t. From now on, t is assumed to be odd such that 2
t
2
-
1 is a prime.Thus for the Pell equationD
t
~, we get τ
32
=
3
+
8 and 32
y
2
+
8
t
2
-
4 is a square only for y
=
(
t
-
1
)
/
2 in the range 0
≤
y
≤
U. Hence, x
=
±
(
2
t
-
1
). Therefore, there is exactly one O
32,1
* set of representatives of the orbits and that {
(
±
(
2
t
-
1
)
,
(
t
-
1
)
/
2
)
} is a set of representatives; that is, there are two classes of solutions. Here we see that(1)
(
2
t
-
1
,
(
t
-
1
)
/
2
)
M
n generates the solutions (
u
2
n
+
1
,
v
2
n
+
1
) for n
≥
0,
(2)
(
1
-
2
t
,
(
1
-
t
)
/
2
)
M
n generates the solutions (
-
u
2
n
+
1
,
-
v
2
n
+
1
) for n
≥
0,
(3)
(
2
t
-
1
,
(
1
-
t
)
/
2
)
M
-
n generates the solutions (
u
2
n
+
1
,
-
v
2
n
+
1
) for n
≥
0,
(4)
(
1
-
2
t
,
(
t
-
1
)
/
2
)
M
-
n generates the solutions (
-
u
2
n
+
1
,
v
2
n
+
1
) for n
≥
0,
(5)
(
2
t
-
1
,
(
1
-
t
)
/
2
)
M
n generates the solutions (
u
2
n
,
v
2
n
) for n
≥
1,
(6)
(
1
-
2
t
,
(
t
-
1
)
/
2
)
M
n generates the solutions (
-
u
2
n
,
-
v
2
n
) for n
≥
1,
(7)
(
2
t
-
1
,
(
t
-
1
)
/
2
)
M
-
n generates the solutions (
u
2
n
,
-
v
2
n
) for n
≥
1,
(8)
(
1
-
2
t
,
(
1
-
t
)
/
2
)
M
-
n generates the solutions (
-
u
2
n
,
v
2
n
) for n
≥
1,whereM
=
(
3
1
8
3
) by (11). Thus, we can give the following theorem.Theorem 1.
The set of all positive integer solutions ofD
t
~ is Ω
(
D
t
~
)
=
{
(
u
2
n
,
v
2
n
)
,
(
u
2
n
+
1,v
2
n
+
1
)
}, where
(12)
(
u
2
n
+
1
,
v
2
n
+
1
)
=
(
2
t
-
1
,
t
-
1
2
)
M
n
f
o
r
n
≥
0
,
(
u
2
n
,
v
2
n
)
=
(
2
t
-
1
,
1
-
t
2
)
M
n
f
o
r
n
≥
1
.To determine all positive integer solutions ofD
t
~, we have to formulate the nth power of M. So we can give the following theorem which can be proved by induction on n.Theorem 2.
Thenth power of M is
(13)
M
n
=
(
C
n
B
n
8
B
n
C
n
)
,
where B
n is the nth balancing number and C
n is the nth Lucas-balancing number.Consequently from the above two theorems, we can give the following main theorem.Theorem 3.
The set of all positive integer solutions ofD
t
~ is Ω
(
D
t
~
)
=
{
(
u
2
n
,
v
2
n
)
,
(
u
2
n
+
1,v
2
n
+
1
)
}, where
(14)
(
u
2
n
+
1
,
v
2
n
+
1
)
=
(
(
2
t
-
1
)
C
n
+
4
(
t
-
1
)
B
n
,
(
2
t
-
1
)
B
n
+
t
-
1
2
C
n
)
f
o
r
n
≥
0
(
u
2
n
,
v
2
n
)
=
(
(
2
t
-
1
)
C
n
+
4
(
1
-
t
)
B
n
,
(
2
t
-
1
)
B
n
+
1
-
t
2
C
n
)
f
o
r
n
≥
1
.After formulating all integer solutions ofD
t
~, we can determine all integer solutions of D
t. Notice that x
=
v
-
(
(
t
+
1
)
/
2
) and y
=
u. So applying Theorem 3, we get
(15)
x
2
n
=
(
2
t
-
1
)
B
n
+
1
-
t
2
C
n
-
t
+
1
2
,
y
2
n
=
(
2
t
-
1
)
C
n
+
4
(
1
-
t
)
B
n
for n
≥
1 and
(16)
x
2
n
+
1
=
(
2
t
-
1
)
B
n
+
t
-
1
2
C
n
-
t
+
1
2
y
2
n
+
1
=
(
2
t
-
1
)
C
n
+
4
(
t
-
1
)
B
n
for n
≥
0. Here we note that (
x
2
n
,
y
2
n
) and (
x
2
n
+
1
,
y
2
n
+
1
) are positive integer solutions. By symmetry, (
x
2
n
,
-
y
2
n
) and (
x
2
n
+
1
,
-
y
2
n
+
1
) are also integer solutions. Apart from them, we get from (
1
-
2
t
,
(
1
-
t
)
/
2
)
M
n generates the solutions (
-
u
2
n
+
1
,
-
v
2
n
+
1
) for n
≥
0 that
(17)
X
2
n
=
(
1
-
2
t
)
B
n
+
t
-
1
2
C
n
-
t
+
1
2
for n
≥
1 and from (
1
-
2
t
,
(
t
-
1
)
/
2
)
M
n generates the solutions (
-
u
2
n
,
-
v
2
n
) for n
≥
1 that
(18)
X
2
n
+
1
=
(
1
-
2
t
)
B
n
+
1
-
t
2
C
n
-
t
+
1
2
for n
≥
0. So (
X
2
n
,
±
Y
2
n
) and (
X
2
n
+
1
,
±
Y
2
n
+
1
) are also integer solutions, where Y
2
n
=
y
2
n and Y
2
n
+
1
=
y
2
n
+
1. Thus, we proved the following theorem.Theorem 4.
The set of all integer solutions ofD
t is Ω
(
D
t
)
=
{
(
x
2
n
,
±
y
2
n
)
,
(
x
2
n
+
1
,
±
y
2
n
+
1
),(
X
2
n
,
±
Y
2
n
)
,
(
X
2
n
+
1
,
±
Y
2
n
+
1
)
}, where
(19)
x
2
n
+
1
=
(
2
t
-
1
)
B
n
+
t
-
1
2
C
n
-
t
+
1
2
,
y
2
n
+
1
=
(
2
t
-
1
)
C
n
+
4
(
t
-
1
)
B
n
,
X
2
n
+
1
=
(
1
-
2
t
)
B
n
+
1
-
t
2
C
n
-
t
+
1
2
,
Y
2
n
+
1
=
y
2
n
+
1
for n
≥
0 and
(20)
x
2
n
=
(
2
t
-
1
)
B
n
+
1
-
t
2
C
n
-
t
+
1
2
,
y
2
n
=
(
2
t
-
1
)
C
n
+
4
(
1
-
t
)
B
n
,
X
2
n
=
(
1
-
2
t
)
B
n
+
t
-
1
2
C
n
-
t
+
1
2
,
Y
2
n
=
y
2
n
for n
≥
1.Example 5.
Lett
=
3. Then we get
(21)
x
2
n
+
1
∈
{
-
1,6
,
45,272,1595
,
…
}
,
y
2
n
+
1
∈
{
5,23,133,775,4517
,
…
}
,
X
2
n
+
1
∈
{
-
3
,
-
10
,
-
49
,
-
276
,
-
1599
,
…
}
,
x
2
n
∈
{
0,11,74,441,2580
,
…
}
,
y
2
n
∈
{
7,37,215,1253,7303
,
…
}
,
X
2
n
∈
{
-
4
,
-
15
,
-
78
,
-
445
,
-
2584
,
…
}
.
So the set of all integer solutions of D
3
:
8
x
2
-
y
2
+
32
x
+
49
=
0 is
(22)
Ω
(
D
3
)
=
{
…
,
(
-
2584
,
±
7303
)
,
(
-
1599
,
±
4517
)
,
(
-
445
,
±
1253
)
,
(
-
276
,
±
775
)
,
(
-
78
,
±
215
)
,
(
-
49
,
±
133
)
,
(
-
15
,
±
37
)
,
(
-
10
,
±
23
)
,
(
-
4
,
±
7
)
,
(
-
3
,
±
5
)
,
(
-
1
,
±
5
)
,
(
0
,
±
7
)
,
(
6
,
±
23
)
,
(
11
,
±
37
)
,
(
45
,
±
133
)
,
(
74
,
±
215
)
,
(
272
,
±
775
)
,
(
441
,
±
1253
)
,
(
1595
,
±
4517
)
,
(
2580
,
±
7303
)
,
…
}
.After determining all integer solutions ofD
t, we can determine all t-balancing numbers.Theorem 6.
The general terms of allt-balancing numbers are
(23)
B
2
n
-
1
t
=
(
B
n
+
b
n
+
1
)
t
-
(
b
n
+
1
+
1
)
,
b
2
n
-
1
t
=
(
B
n
+
b
n
)
t
-
B
n
,
C
2
n
-
1
t
=
(
4
b
n
+
1
+
2
)
t
-
c
n
+
1
,
c
2
n
-
1
t
=
4
t
B
n
-
C
n
for n
≥
1 and
(24)
B
2
n
-
2
t
=
(
B
n
-
1
+
b
n
)
t
+
b
n
,
b
2
n
-
2
t
=
(
B
n
-
1
+
b
n
-
1
)
t
+
B
n
-
1
,
C
2
n
-
2
t
=
(
4
b
n
+
2
)
t
+
c
n
,
c
2
n
-
2
t
=
4
t
B
n
-
1
+
C
n
-
1
for n
≥
2.Proof.
Sincex
=
B
n
t, we get from Theorem 4 that
(25)
B
2
n
-
1
t
=
(
2
t
-
1
)
B
n
+
t
-
1
2
C
n
-
t
+
1
2
=
(
B
n
+
b
n
+
1
)
t
-
(
b
n
+
1
+
1
)
for n
≥
1 since 2
B
n
+
(
(
C
n
-
1
)
/
2
)
=
B
n
+
b
n
+
1 and B
n
+
(
(
C
n
+
1
)
/
2
)
=
b
n
+
1
+
1.
Note thatb
n
=
(
(
α
2
n
-
1
-
β
2
n
-
1
)
/
4
2
)
-
(
1
/
2
) and c
n
=
(
α
2
n
-
1
+
β
2
n
-
1
)
/
2. So we get from (7) that
(26)
C
2
n
-
1
t
=
8
(
B
2
n
-
1
t
)
2
+
8
B
2
n
-
1
t
(
1
+
t
)
+
(
2
t
+
1
)
2
=
8
[
(
B
n
+
b
n
+
1
)
t
-
(
b
n
+
1
+
1
)
]
2
+
8
[
(
B
n
+
b
n
+
1
)
t
-
(
b
n
+
1
+
1
)
]
(
1
+
t
)
+
(
2
t
+
1
)
2
=
(
α
2
n
+
1
(
2
t
-
1
)
-
β
2
n
+
1
(
2
t
+
1
)
2
)
2
=
[
4
(
α
2
n
+
1
-
β
2
n
+
1
4
2
-
1
2
)
+
2
]
t
-
(
α
2
n
+
1
+
β
2
n
+
1
2
)
=
(
4
b
n
+
1
+
2
)
t
-
c
n
+
1
.
Thus from (6), we obtain
(27)
b
2
n
-
1
t
=
(
B
n
+
b
n
)
t
-
B
n
and hence
(28)
c
2
n
-
1
t
=
8
(
b
2
n
-
1
t
)
2
+
8
t
b
2
n
-
1
t
+
1
=
8
[
(
B
n
+
b
n
)
t
-
B
n
]
2
+
8
t
[
(
B
n
+
b
n
)
t
-
B
n
]
+
1
=
(
α
2
n
(
2
t
-
1
)
-
β
2
n
(
2
t
+
1
)
2
)
2
=
α
2
n
(
2
t
-
1
)
-
β
2
n
(
2
t
+
1
)
2
=
4
t
(
α
2
n
-
β
2
n
4
2
)
-
(
α
2
n
+
β
2
n
2
)
=
4
t
B
n
-
C
n
since B
n
=
(
α
2
n
-
β
2
n
)
/
4
2 and C
n
=
(
α
2
n
+
β
2
n
)
/
2.
Similarly we deduce that(29)
B
2
n
-
2
t
=
(
B
n
-
1
+
b
n
)
t
+
b
n
,
C
2
n
-
2
t
=
(
4
b
n
+
2
)
t
+
c
n
,
b
2
n
-
2
t
=
(
B
n
-
1
+
b
n
-
1
)
t
+
B
n
-
1
,
c
2
n
-
2
t
=
4
t
B
n
-
1
+
C
n
-
1
for n
≥
2.



---

*Source: 897834-2014-03-20.xml*