# The Exponential Diophantine Equation4

Authors: Juanli Su; Xiaoxue Li
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Letm be a positive integer. In this paper, using some properties of exponential diophantine equations and some results on the existence of primitive divisors of Lucas numbers, we prove that if m
>
90 and 3
|
m, then the equation 4
m
2
+
1
x
+
5
m
2
-
1
y
=
(
3
m
)
z has only the positive
integer solution (
x
,
y
,
z
)
=
(
1
,
1
,
2
).


---

## Body

## 1. Introduction

Letℤ, ℕ be the sets of all integers and positive integers, respectively. Given a triple (
a
,
b
,
c
) of coprime positive integers with min
⁡
{
a
,
b
,
c
}
>
1, there are many papers that investigated the equation
(1)
a
x
+
b
y
=
c
z
,
x
,
y
,
z
∈
ℕ
(see [1–8]). Recently, Terai [8] proved that if a, b, c satisfy
(2)
a
=
4
m
2
+
1
,
b
=
5
m
2
-
1
,
c
=
3
m
,
m
∈
ℕ
,
then (1) has only the solution (
x
,
y
,
z
)
=
(
1,1
,
2
), provided that m
≢
3
(
mod
⁡
6
) or m
≤
20. The proof of this result is based on elementary methods and Baker’s method. In this paper, using some properties of exponential diophantine equations and some results on the existence of primitive divisors of Lucas numbers, we prove a general result as follows.Theorem 1.
Leta
,
b
,
c be positive integers satisfying (2). If m
>
90 and 3
∣
m, then (1) has only the solution (
x
,
y
,
z
)
=
(
1,1
,
2
).Some Note. Combining Terai [8] and our Theorem, we know that only the values 20
<
m
≤
90, 3
∣
m are left to investigate (2). In this case, the equation (
4
m
2
+
1
)
x
+
(
5
m
2
-
1
)
y
=
(
3
m
)
z has only finitely many solutions in (
x
,
y
,
z
). Moreover, they are not only effectively but also practically solvable. With some computer assistance, we would be able to solve completely the equation.


## 2. Preliminaries

In this section, we assume thata
,
b
,
c are positive integers satisfying (2). Then, (1) can be written as
(3)
(
4
m
2
+
1
)
x
+
(
5
m
2
-
1
)
y
=
(
3
m
)
z
,
x
,
y
,
z
∈
ℕ
.
Further, let (
x
,
y
,
z
) be a solution of (3) with (
x
,
y
,
z
)
≠
(
1,1
,
2
).Lemma 2.
Ifm
>
90, then 2
∤
m, 2
∤
y and
(4)
z
>
1
5
m
2
.Proof.
Sincemin
⁡
{
x
,
y
}
≥
1, by (3), we have z
≥
2 and 1
+
(
-
1
)
y
≡
(
4
m
2
+
1
)
x
+
(
5
m
2
-
1
)
y
≡
(
3
m
)
z
≡
0
(
mod
⁡
m
2
). Since m
>
1, we get
(5)
2
∤
y
.
Since(
x
,
y
,
z
)
≠
(
1,1
,
2
), we have max
⁡
{
x
,
y
}
≥
2 and (
3
m
)
z
>
min
⁡
{
(
4
m
2
+
1
)
2
,
(
5
m
2
-
1
)
2
}
>
16
m
4. It implies that
(6)
z
≥
4
.
Hence, by (3), (5), and (6), we get
(7)
4
x
+
5
y
≡
0
(
mod
⁡
m
2
)
.
If2
∣
m, then from (7) we get 2
∣
y, which contradicts (5). So we have
(8)
2
∤
m
.
Since4
x
+
5
y is a positive integer, by (7), we have
(9)
4
x
+
5
y
≥
m
2
.
On the other hand, by (3), we have (
3
m
)
z
>
(
4
m
2
+
1
)
x and (
3
m
)
z
>
(
5
m
2
-
1
)
y. It implies that
(10)
z
2
>
x
log
⁡
(
4
m
2
+
1
)
log
⁡
(
9
m
2
)
,
z
2
>
y
log
⁡
(
5
m
2
-
1
)
log
⁡
(
9
m
2
)
;
whence we obtain
(11)
z
>
4
x
+
5
y
2
log
⁡
(
9
m
2
)
/
log
⁡
(
4
m
2
+
1
)
+
5
log
⁡
(
9
m
2
)
/
2
log
⁡
(
5
m
2
-
1
)
.
Sincem
>
90, we have (
log
⁡
(
9
m
2
)
)
/
log
⁡
(
5
m
2
-
1
)
<
(
log
⁡
(
9
m
2
)
)
/
log
⁡
(
4
m
2
+
1
)
<
1.08. Thus, by (9) and (11), we get (4). The lemma is proved.
LetD
1, D
2, k be positive integers such that min
⁡
{
D
1
,
D
2
,
k
}
>
1 and gcd
(
D
1
,
D
2
)
=
gcd
(
k
,
2
D
1
D
2
)
=
1.Lemma 3 ([9, Lemmas 1 and 6]).
If(
X
,
Y
,
Z
) is a fixed solution of the equation
(12)
D
1
X
2
+
D
2
Y
2
=
k
Z
,
X
,
Y
,
Z
∈
ℤ
,
gcd

(
X
,
Y
)
=
1
,
Z
>
0
,
then there exists a unique positive integer l such that
(13)
l
≡
-
D
1
X
Y
(
mod
⁡
k
)
,
l
<
k
.
The positive integer l is called the characteristic number of the solution (
X
,
Y
,
Z
) and is denoted by the symbol 〈
X
,
Y
,
Z
〉.Lemma 4 ([9, Theorems 1 and 3]).
For a fixed characteristic numberl, let S
(
l
) be the set of all solutions (
X
,
Y
,
Z
) of (13) with 〈
X
,
Y
,
Z
〉
≡
±
l
(
mod
⁡
k
). Then one has the following.(i)
S
(
l
) has a unique solution (
X
1
,
Y
1
,
Z
1
) satisfying X
1
>
0, Y
1
>
0 and Z
1
≤
Z, where Z through all solutions (
X
,
Y
,
Z
) of S
(
l
). Such (
X
1
,
Y
1
,
Z
1
) is called the least solution of S
(
l
).
(ii)
Every element(
X
,
Y
,
Z
) of S
(
l
) can be expressed as
(14)
Z
=
Z
1
t
,
t
∈
ℕ
,
2
∤
t
,
X
D
1
+
Y
-
D
2
=
λ
1
(
X
1
D
1
+
λ
2
Y
1
-
D
2
)
t
,
h
h
h
h
h
h
h
h
h
h
h
λ
1
,
λ
2
∈
{
±
1
}
.Lemma 5.
Ifmin
⁡
{
D
1
,
D
2
}
≥
4 and 3
∤
D
1
D
2, then the equality
(15)
D
1
r
D
1
+
D
2
s
-
D
2
=
λ
1
(
D
1
+
λ
2
-
D
2
)
t
,
h
h
h
h
h
h
h
i
λ
1
,
λ
2
∈
{
±
1
}
,
r
,
s
,
t
∈
ℤ
,
r
≥
0
,
s
≥
0
,
t
>
1
,
2
∤
t
cannot hold.Proof.
If (15) holds, then
(16)
D
1
r
=
λ
1
∑
i
=
0
(
t
-
1
)
/
2
‍
(
-
i
)
i
(
t
2
i
)
D
1
(
t
-
1
)
/
2
-
i
D
2
i
,
D
2
s
=
λ
1
λ
2
∑
i
=
0
(
t
-
1
)
/
2
‍
(
-
i
)
i
(
t
2
i
+
1
)
D
1
(
t
-
1
)
/
2
-
i
D
2
i
,
(17)
D
1
r
D
1
-
D
2
s
-
D
2
=
λ
1
(
D
1
-
λ
2
-
D
2
)
t
.
By (15) and (17), we have
(18)
D
1
2
r
+
1
+
D
2
2
s
+
1
=
(
D
1
+
D
2
)
t
.
We may assume thatD
1
2
r
+
1
<
D
2
2
s
+
1; then, by (18), we have
(19)
2
D
2
2
s
+
1
>
(
D
1
+
D
2
)
t
>
D
2
t
.
Since D
2
≥
4 and t
≥
3, we see from (19) that s
≥
1. Further, since gcd
(
D
1
,
D
2
)
=
1, by (16), we get
(20)
D
2
∣
t
.
It implies that 2
∤
D
2.
Letp be an odd prime divisor of D
2. Since 3
∤
D
2, we have p
≥
5. Further let
(21)
p
α
∥
D
2
,
p
β
∥
t
,
p
γ
j
∥
2
j
+
1
,
j
=
1
,
…
,
t
-
1
2
.
Since p
≥
5, by (21), we have
(22)
γ
j
≤
log
⁡
(
2
j
+
1
)
log
⁡
p
<
j
,
j
=
1
,
…
,
t
-
1
2
.
Further, by (21) and (22), we get
(23)
(
-
1
)
j
(
t
2
j
+
1
)
D
1
(
t
-
1
)
/
2
-
j
D
2
j
≡
(
-
1
)
j
t
(
t
-
1
2
j
)
D
1
(
t
-
1
)
/
2
-
j
D
2
j
2
j
+
1
≡
0
(
mod
⁡
p
β
+
1
)
,
j
=
1
,
…
,
t
-
1
2
.
Therefore, by (16), (21), and (23), we obtain
(24)
α
s
≥
β
.
Putting p through all odd prime divisors of D
2, we get from (24) that D
2
s
∣
t and
(25)
t
≥
D
2
s
.
By (19) and (25), we have D
2
2
s
+
2
>
2
D
2
2
s
+
1
>
D
2
t
≥
D
2
D
2
s and
(26)
2
s
+
2
>
D
2
s
≥
4
s
.
But (26) is impossible for any positive integer s. Thus, (15) is false. The lemma is proved.Letα
,
β be algebraic integers. If α
+
β and α
β are nonzero coprime integers and α
/
β is not a root of unity, then (
α
,
β
) is called a Lucas pair. Further, let A
=
α
+
β and C
=
α
β. Then we have
(27)
α
=
1
2
(
A
+
λ
B
)
,
β
=
1
2
(
A
-
λ
B
)
,
λ
∈
{
±
1
}
,
where B
=
A
2
-
4
C. We call (
A
,
B
) the parameters of the Lucas pair (
α
,
β
). Two Lucas pairs (
α
1
,
β
1
) and (
α
2
,
β
2
) are equivalent if α
1
/
α
2
=
β
1
/
β
2
=
±
1. Given a Lucas pair (
α
,
β
), one defines the corresponding sequence of Lucas numbers by
(28)
L
n
(
α
,
β
)
=
α
n
-
β
n
α
-
β
,
n
=
0,1
,
2
,
…
.
For equivalent Lucas pairs (
α
1
,
β
1
) and (
α
2
,
β
2
), we have L
n
(
α
1
,
β
1
)
=
±
L
n
(
α
2
,
β
2
) for any n. A prime p is called a primitive divisor of L
n
(
α
,
β
)
(
n
>
1
) if p
∣
L
n
(
α
,
β
) and p
∤
B
L
1
(
α
,
β
)
,
…
,
L
n
-
1
(
α
,
β
). A Lucas pair (
α
,
β
) such that L
n
(
α
,
β
) has no primitive divisors will be called an n-defective Lucas pair. Further, a positive integer n is called totally nondefective if no Lucas pair is n-defective.Lemma 6 (see [10]).
Letn satisfy 4
<
n
≤
30 and n
≠
6. Then, up to equivalence, all parameters of n-defective Lucas pair are given as follows:(i)
n
=
5, (
A
,
B
)
=
(
1,5
), (
1
,
-
7
), (
2
,
-
40
), (
1
,
-
11
), (
1
,
-
15
), (
12
,
-
76
), (
12
,
-
1364
),
(ii)
n
=
7, (
A
,
B
)
=
(
1
,
-
7
)
,
(
1
,
-
19
),
(iii)
n
=
8, (
A
,
B
)
=
(
2
,
-
24
)
,
(
1
,
-
7
),
(iv)
n
=
10, (
A
,
B
)
=
(
2
,
-
8
)
,
(
5
,
-
3
)
,
(
5
,
-
47
),
(v)
n
=
12, (
A
,
B
)
=
(
1,5
)
,
(
1
,
-
7
)
,
(
1
,
-
11
)
,
(
2
,
-
56
)
,
(
1
,
-
15
)
,
(
1
,
-
19
),
(vi)
n
∈
{
13,18,30
}, (
A
,
B
)
=
(
1
,
-
7
).Lemma 7 (see [11]).
Ifn
>
30, then n is totally nondefective.
LetD
,
k be positive integers such that min
⁡
{
D
,
k
}
>
1 and gcd

(
k
,
2
D
)
=
1.Lemma 8 (see [9, Theorems 1 and 3]).
Every solution(
X
,
Y
,
Z
) of the equation
(29)
X
2
+
D
Y
2
=
k
Z
,
X
,
Y
,
Z
∈
ℤ
,
gcd

(
X
,
Y
)
=
1
,
Z
>
0
can be expressed as
(30)
Z
=
Z
1
t
,
t
∈
ℕ
,
(31)
X
+
Y
-
D
=
λ
1
(
X
1
+
λ
2
Y
1
-
D
)
t
,
λ
1
,
λ
2
∈
{
±
1
}
,
where X
1
,
Y
1
,
Z
1 are positive integers satisfying
(32)
X
1
2
+
D
Y
1
2
=
k
Z
1
,
gcd

(
X
1
,
Y
1
)
=
1
,
h
(
-
4
D
)
≡
0
(
mod
⁡
Z
1
)
,
where h
(
-
4
D
) is the class number of positive binary quadratic primitive forms of discriminant -
4
D.
For any positive integera, let P
(
a
) denote the set of distinct prime divisors of a.Lemma 9.
If(
X
,
Y
,
Z
) is a solution of (29) with P
(
|
Y
|
)
⊆
P
(
D
), then h
(
-
4
D
)
≡
0
(

mod
Z
), except the possibility of the following cases:(i)
t
∈
{
2,3
,
4,6
}, where t is defined as in (30),
(ii)
(
D
,
k
,
X
,
Y
,
Z
)
=
(
10,11
,
±
401
,
±
5,5
), (
19,55
,
±
22434
,
±
1,5
), (
341,377
,
±
2759646
,
±
1,5
).Proof.
Let(
X
,
Y
,
Z
) be a solution of (29) with P
(
|
Y
|
)
⊆
P
(
D
). By Lemma 8, X
,
Y, and Z satisfy (30) and (31), where X
1
,
Y
1, and Z
1 satisfy (32). Let
(33)
α
=
X
1
+
Y
1
-
D
,
β
=
X
1
-
Y
1
-
D
.
By (32) and (33), we have that α
+
β
=
2
X
1, α
-
β
=
2
Y
1
-
D, α
β
=
k
Z
1, and α
/
β satisfies k
Z
1
(
α
/
β
)
2
-
2
(
X
1
2
-
D
Y
1
2
)
(
α
/
β
)
+
k
Z
1
=
0. It implies that (
α
,
β
) is a Lucas pair with parameters (
2
X
1
,
-
4
D
Y
1
2
). Let L
n
(
α
,
β
)
(
n
=
0,1
,
2
,
…
) denote the corresponding Lucas numbers. By (28), (31), and (33), we get
(34)
Y
=
Y
1
|
α
t
-
β
t
α
-
β
|
=
Y
1
|
L
t
(
α
,
β
)
|
.
Since P
(
|
Y
|
)
⊆
P
(
D
), by the definition of primitive divisors, we see from (34) that either t
=
1 or t
>
1 and the Lucas number L
t
(
α
,
β
) has no primitive divisor.
Ift
=
1, then from (30) and (32) we get h
(
-
4
D
)
≡
0
(

mod
Z
). If t
>
1, by Lemmas 6 and 7, using an easy computation, the solution (
X
,
Y
,
Z
) satisfies the case (i) or (ii). Thus, the lemma is proved.Lemma 10 ([12, Theorems 12.10.1 and 12.14.3]).
For any positive integerD, one has
(35)
h
(
-
4
D
)
<
4
π
D
log
⁡
(
2
e
D
)
.


## 3. Proof of Theorem

We now assume that(
x
,
y
,
z
) is a solution of (3) with (
x
,
y
,
z
)
≠
(
1,1
,
2
). By Lemma 2, we have
(36)
2
∤
m
,
2
∤
y
.We first consider the case that2
∤
x. Then, by (3) and (36), the equation
(37)
(
4
m
2
+
1
)
X
2
+
(
5
m
2
-
1
)
Y
2
=
(
3
m
)
Z
,
X
,
Y
,
Z
∈
ℤ
,
gcd
(
X
,
Y
)
=
1
,
Z
>
0
has a solution
(38)
(
X
,
Y
,
Z
)
=
(
(
4
m
2
+
1
)
(
x
-
1
)
/
2
,
(
5
m
2
-
1
)
(
y
-
1
)
/
2
,
z
)
.
Let l
=
〈
(
4
m
2
+
1
)
(
x
-
1
)
/
2
,
(
5
m
2
-
1
)
(
y
-
1
)
/
2
,
z
〉. Since 3
∣
m, we have 3
m
∣
m
2. Hence, by Lemma 3, we get
(39)
l
≡
-
(
4
m
2
+
1
)
(
x
-
1
)
/
2
(
5
m
2
-
1
)
(
y
-
1
)
/
2
≡
(
-
1
)
(
y
+
1
)
/
2
(
mod
⁡
3
m
)
.In addition, (37) has another solution
(40)
(
X
,
Y
,
Z
)
=
(
1,1
,
2
)
.
Let l
′
=
〈
1,1
,
2
〉. We have
(41)
l
′
≡
-
(
4
m
2
+
1
)
≡
-
1
(
mod
⁡
3
m
)
.
By (39) and (41), we get l
′
≡
±
l
(
mod
⁡
3
m
). It implies that the solutions (38) and (40) belong to a same class S
(
l
) of solutions of (37). Further, since (
4
m
2
+
1
)
X
2
+
(
5
m
2
-
1
)
Y
2
≥
(
4
m
2
+
1
)
+
(
5
m
2
-
1
)
=
(
3
m
)
2, (40) is the least solution of S
(
l
). Therefore, applying Lemma 4 to (38), we get 2
∣
z, 2
∤
z
/
2, and
(42)
(
4
m
2
+
1
)
(
x
-
1
)
/
2
4
m
2
-
1
+
(
5
m
2
-
1
)
(
y
-
1
)
/
2
-
(
5
m
2
-
1
)
=
λ
1
(
4
m
2
+
1
+
λ
2
-
(
5
m
2
-
1
)
)
z
/
2
,
hhi
λ
1
,
λ
2
∈
{
±
1
}
.
However, since (
x
,
y
,
z
)
≠
(
1,1
,
2
) and 3
∣
m, we have z
/
2
>
1 and 3
∤
(
4
m
2
+
1
)
(
5
m
2
-
1
). By Lemma 5, (42) is false.We finally consider the case that2
∣
x. Then the equation
(43)
X
2
+
(
5
m
2
-
1
)
Y
2
=
(
3
m
)
Z
,
X
,
Y
,
Z
∈
ℤ
,
gcd
(
X
,
Y
)
=
1
,
Z
>
0
has a solution
(44)
(
X
,
Y
,
Z
)
=
(
(
4
m
2
+
1
)
x
/
2
,
(
5
m
2
-
1
)
(
y
-
1
)
/
2
,
z
)
.
Since 2
∤
3
m, applying Lemma 8 to (44), we have
(45)
z
=
Z
1
t
,
t
∈
ℕ
,
(46)
(
4
m
2
+
1
)
x
/
2
+
(
5
m
2
-
1
)
(
y
-
1
)
/
2
-
(
5
m
2
-
1
)
=
λ
1
(
X
1
+
λ
2
Y
1
-
(
5
m
2
-
1
)
)
t
,
where λ
1
,
λ
2
∈
{
±
1
}, X
1
,
Y
1
,
Z
1 are positive integers satisfying
(47)
X
1
2
+
(
5
m
2
-
1
)
Y
1
2
=
(
3
m
)
Z
1
,
gcd
(
X
1
,
Y
1
)
=
1
,
h
(
-
4
(
5
m
2
-
1
)
)
≡
0
(
mod
⁡
Z
1
)
.
If 2
∣
t, then from (46) we get
(48)
(
4
m
2
+
1
)
x
/
2
+
(
5
m
2
-
1
)
(
y
-
1
)
/
2
-
(
5
m
2
-
1
)
=
λ
1
(
X
2
+
Y
2
-
(
5
m
2
-
1
)
)
2
,
where X
2
,
Y
2 are integers satisfying
(49)
X
2
2
+
(
5
m
2
-
1
)
Y
2
2
=
(
3
m
)
Z
/
2
,
gcd
(
X
2
,
Y
2
)
=
1
.
By (49), we have
(50)
(
4
m
2
+
1
)
x
/
2
=
λ
1
(
X
2
2
-
(
5
m
2
-
1
)
Y
2
2
)
,
(
5
m
2
-
1
)
(
y
-
1
)
/
2
=
2
λ
1
X
2
Y
2
.
Further, since gcd
(
4
m
2
+
1,5
m
2
-
1
)
=
1, we see from (50) that |
X
2
|
=
±
1, |
Y
2
|
=
(
5
m
2
-
1
)
(
y
-
1
)
/
2
/
2 and
(51)
(
4
m
2
+
1
)
x
/
2
=
1
4
(
5
m
2
-
1
)
y
-
1
.
Furthermore, by (51), we get 1
≡
(
4
m
2
+
1
)
x
/
2
≡
(
1
/
4
)
(
5
m
2
-
1
)
y
-
1
≡
-
1
/
4
-
1
(
mod
⁡
m
2
); whence we obtain m
2
∣
9. But, since m
>
90, it is impossible. So we have 2
∤
t.Ift
=
3, then from (46) we get
(52)
(
4
m
2
+
1
)
x
/
2
=
λ
1
X
1
(
X
1
2
-
3
(
5
m
2
-
1
)
Y
1
2
)
,
(53)
(
5
m
2
-
1
)
(
y
-
1
)
/
2
=
λ
1
λ
2
Y
1
(
3
X
1
2
-
(
5
m
2
-
1
)
Y
1
2
)
.
Since gcd
(
3
X
1
,
5
m
2
-
1
)
=
1, by (47), we see from (53) that Y
1
=
(
5
m
2
-
1
)
(
y
-
1
)
/
2 and
(54)
3
X
1
2
-
(
5
m
2
-
1
)
y
=
±
1
.
Further, since 3
∣
m and 2
∤
y, by (54), we have
(55)
3
X
1
2
-
(
5
m
2
-
1
)
y
=
1
.
But, since 2
∤
m, we get from (55) that 4
∣
5
m
2
-
1, 2
∤
X
1, and 3
≡
3
X
1
2
-
(
5
m
2
-
1
)
y
≡
1
(
mod
⁡
4
), a contradiction. So we have t
≠
3.Notice that the solution (44) satisfies P
(
|
Y
|
)
⊆
P
(
5
m
2
-
1
). Therefore, since 2
∤
t and t
≠
3, by Lemma 9, we have h
(
-
4
(
5
m
2
-
1
)
)
≡
0
(
mod
⁡
z
) and
(56)
z
≤
h
(
-
4
(
5
m
2
-
1
)
)
.
Further, applying Lemma 10 to (56), we get
(57)
z
<
4
π
5
m
2
-
1
log
⁡
(
2
e
5
m
2
-
1
)
.
On the other hand, by Lemma 2, z satisfies (4). The combination of (4) and (57) yields
(58)
m
2
<
20
π
5
m
2
-
1
log
⁡
(
2
e
5
m
2
-
1
)
.
But (58) is false for m
>
90. Thus, the solution (
x
,
y
,
z
) with (
x
,
y
,
z
)
≠
(
1,1
,
2
) does not exist. The theorem is proved.



---

*Source: 670175-2014-04-09.xml*