# On the Odd Prime Solutions of the Diophantine Equationx

Authors: Yuanyuan Deng; Wenpeng Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Using the elementary method and some properties of the least solution of Pell’s equation, we prove that the equationx
y
+
y
x
=
z
z has no positive integer solutions (x
,
y
,
z) with x and y being odd primes.


---

## Body

## 1. Introduction

LetZ, N be the sets of all integers and positive integers, respectively. In recent years, there are many authors who investigated the various properties of exponential diophantine equation with circulating form (see [1–4]). Recently, Zhang et al. [5] are interested in the equation
(1)
x
y
+
y
x
=
z
z
,
x
,
y
,
z
∈
N
,
min
⁡
{
x
,
y
,
z
}
>
1
.
Using the p-adic lower bound of the log-linear model method, they proved all solutions (
x
,
y
,
z
) of (1) satisfying z
<
2.8
×
1
0
9. Meanwhile, they proposed a conjecture as follows.Conjecture 1.
Equation (1) has no positive integer solution (
x
,
y
,
z
).Using the method in [5], it seems to be a very difficult problem to improve the upper bound estimate for z. In this paper, we use the elementary method and some properties of the least solution of Pell’s equation to solve the conjecture partly. That is, we will prove the following.Theorem 2.
Equation (1) has no positive integer solution (
x
,
y
,
z
) with x and y being odd primes.


## 2. Several Lemmas

LetD be a nonsquare positive integer, and let h
(
4
D
) denote the class number of binary quadratic primitive forms with discriminant 4
D. Then we have the following.Lemma 3.
For the equation(2)
u
2
-
D
v
2
=
1
,
u
,
v
∈
Z
,
there is a solution (
u
,
v
) with u
v
≠
0, and there is a unique positive integer solution (
u
1
,
v
1
) satisfying 0
<
u
1
+
v
1
D
≤
u
+
v
D, where (
u
,
v
) pass through all positive integer solutions of (2). We call (
u
1
,
v
1
) as the least solution of (2). Every solution (
u
,
v
) of (2) can be expressed as
(3)
u
+
v
D
=
±
(
u
1
+
v
1
D
)
s
,
s
∈
Z
.Proof.
See Section 10.9 of [6].Lemma 4.
LetD be an odd prime satisfying D
≡
1
(
mod
⁡
4
) and D
<
1
0
11, and then every least solution (
u
1
,
v
1
) of (2) satisfies v
1
≢
0
(
mod
⁡
D
).Proof.
See [7].Lemma 5.
Considerh
(
4
D
)
<
D.Proof.
According to Table II in Chapter 16 of [6], we know that Lemma 5 holds for D
≤
25, and when D
>
25, by Theorems 12.10.1 and 12.13.3 of [6], we have
(4)
h
(
4
D
)
<
2
D
(
1
+
log
⁡
2
+
(
1
/
2
)
log
⁡
D
)
log
⁡
(
u
1
+
v
1
D
)
,
where (
u
1
,
v
1
) is the least solution of (2). If h
(
4
D
)
≥
D, and v
1
≥
1, u
1
=
D
v
1
2
+
1
≥
D
+
1, then by (4), we have
(5)
5
<
D
<
4
+
4
log
⁡
2
+
2
log
⁡
D
2
log
⁡
2
+
log
⁡
D
=
2
+
4
2
log
⁡
2
+
log
⁡
D
<
3
,
a contradiction. This proves Lemma 5.Lemma 6.
Letp be an odd prime with p
∤
D. If the equation
(6)
X
2
-
D
Y
2
=
p
Z
,
X
,
Y
,
Z
∈
Z
,
gcd
(
X
,
Y
)
=
1
,
Z
>
0
,
has the solution (
X
,
Y
,
Z
), then every solution (
X
,
Y
,
Z
) can be expressed as
(7)
Z
=
Z
1
t
,
t
∈
N
,
X
+
Y
D
=
(
X
1
+
λ
Y
1
D
)
t
(
u
+
v
D
)
,
λ
∈
{
±
1
}
,
where X
1
,
Y
1
,
Z
1 are positive integers satisfying
(8)
X
1
2
-
D
Y
1
2
=
p
Z
1
,
gcd
(
X
1
,
Y
1
)
=
1
,
Z
1
∣
h
(
4
D
)
,
and (
u
,
v
) is a solution of (2).Proof.
See Lemma 4 of [8].


## 3. Proof of Theorem2

Let(
x
,
y
,
z
) be one of the solutions of (1). Without loss of generality we may assume that x
≥
y, since x and y are symmetrical in (1). By [5], we know that x
,
y
,
z are coprime, and x
>
z
>
y
>
1. When x and y are odd primes, z must be even. Note that y
≥
3; by (1) we have z
z
>
y
x; then z
log
⁡
z
>
x
log
⁡
y
≥
x
log
⁡
3
>
x, so by the result z
<
2.8
×
1
0
9 in [5], we get
(9)
3
≤
y
<
z
<
x
<
z
log
⁡
z
<
6.2
×
1
0
10
.By (9), we know that z
>
2; then from (1) we get 0
≡
z
z
≡
x
y
+
y
x
≡
x
+
y
(
mod
⁡
4
). Therefore,
(10)
x
≡
ɛ
(
mod
⁡
4
)
,
y
≡
-
ɛ
(
mod
⁡
4
)
,
ɛ
∈
{
±
1
}
.Ifɛ
=
1, by (10), x is an odd prime with x
≡
1
(
mod
⁡
4
). We see from (1) that the equation
(11)
X
2
-
x
Y
2
=
y
Z
,
X
,
Y
,
Z
∈
Z
,
gcd
(
X
,
Y
)
=
1
,
Z
>
0
,
has the solution
(12)
(
X
,
Y
,
Z
)
=
(
z
z
/
2
,
x
(
y
-
1
)
/
2
,
x
)
.
Since y is an odd prime with y
∤
x, applying Lemma 6 to (11) and (12), we have
(13)
x
=
Z
1
t
,
t
∈
N
,
(14)
z
z
/
2
+
x
(
y
-
1
)
/
2
x
=
(
X
1
+
λ
Y
1
x
)
t
(
u
+
v
x
)
,
λ
∈
{
±
1
}
,
where X
1
,
Y
1
,
Z
1 are positive integers satisfying
(15)
X
1
2
-
x
Y
1
2
=
y
Z
1
,
gcd
(
X
1
,
Y
1
)
=
1
,
(16)
Z
1
∣
h
(
4
x
)
,
(
u
,
v
) is a solution of the equation
(17)
u
2
-
x
v
2
=
1
,
u
,
v
∈
Z
,
and h
(
4
x
) denotes the class number of binary quadratic primitive forms with discriminant 4
x.Sincex is an odd prime, we know from (13) that t
=
1 or x. If t
=
1, Z
1
=
x by (13), so from (16) we have x
∣
h
(
4
x
) and x
≤
h
(
4
x
). But, by Lemma 5, this is impossible; thus t
=
x.Sincet
=
x, by (13) we know that Z
1
=
1, so that (14) and (15) read
(18)
z
z
/
2
+
x
(
y
-
1
)
/
2
x
=
(
X
1
+
λ
Y
1
x
)
x
×
(
u
+
v
x
)
,
λ
∈
{
-
1,1
}
,
(19)
X
1
2
-
x
Y
1
2
=
y
,
gcd
(
X
1
,
Y
1
)
=
1
.Further,z
z
/
2
+
x
(
y
-
1
)
/
2
x
>
0, and from (19) we know X
1
+
λ
Y
1
x
>
0, since u
+
v
x
>
0 by Lemma 3. Thus, according to Lemma 3 there is s
∈
Z such that
(20)
u
+
v
x
=
(
u
1
+
v
1
x
)
s
,
s
∈
Z
,
where (
u
1
,
v
1
) is the least solution of (17). For the integer s, there exist integers q and r satisfying
(21)
s
=
x
q
+
r
,
0
≤
r
<
x
.
Let
(22)
X
2
+
Y
2
x
=
(
X
1
+
λ
Y
1
x
)
(
u
1
+
v
1
x
)
q
.
From (17), (19), and (22), we know that X
2 and Y
2 are integers satisfying
(23)
X
2
2
-
x
Y
2
2
=
y
,
gcd
(
X
2
,
Y
2
)
=
1
.
And from (18), (20), (21), and (22), we have
(24)
z
z
/
2
+
x
(
y
-
1
)
/
2
x
=
(
X
2
+
Y
2
x
)
x
(
u
1
+
v
1
x
)
r
.Ifr
=
0 in (21), then, from (24), we have
(25)
x
(
y
-
1
)
/
2
=
Y
2
∑
i
=
0
(
x
-
1
)
/
2
‍
(
x
2
i
+
1
)
X
2
x
-
2
i
-
1
(
x
Y
2
2
)
i
.
However, since x
>
y from (9), and by (23), we have X
2
2
>
x. According to (25), we get x
(
y
-
1
)
/
2
>
x
X
2
x
-
1
>
x
(
x
+
1
)
/
2
>
x
(
y
+
1
)
/
2, which is impossible. Thus, from (21), we have 0
<
r
<
x and
(26)
x
∤
r
.Let(27)
X
′
+
Y
′
x
=
(
X
2
+
Y
2
x
)
x
,
u
′
+
v
′
x
=
(
u
1
+
v
1
x
)
r
.
Then X
′
,
Y
′
,
u
′
,
v
′ are integers with gcd
(
X
′
,
Y
′
)
=
gcd
(
u
′
,
v
′
)
=
1, and
(28)
Y
′
=
Y
2
∑
i
=
0
(
x
-
1
)
/
2
‍
(
x
2
i
+
1
)
X
2
x
-
2
i
-
1
(
x
Y
2
2
)
i
,
(29)
v
′
=
v
1
∑
j
=
0
[
(
r
-
1
)
/
2
]
‍
(
r
2
j
+
1
)
u
1
r
-
2
j
-
1
(
x
v
1
2
)
j
,
where [
(
r
-
1
)
/
2
] is the integral part of (
r
-
1
)
/
2. From (29), we have
(30)
Y
′
≡
0
(
mod
⁡
x
)
,
v
′
≡
r
u
1
r
-
1
v
1
(
mod
⁡
x
)
.
Applying (27) to (24), we get
(31)
x
(
y
-
1
)
/
2
=
X
′
v
′
+
Y
′
u
′
.
From (17) and (26), gcd
(
r
u
1
r
-
1
,
x
)
=
1, by (30), v
1
≡
0
(
mod
⁡
x
). However, we get from (9) that x is an odd prime satisfying x
≡
1
(
mod
⁡
4
) and x
<
6.2
×
1
0
10; then from Lemma 4, we know it is impossible. Thus, the theorem holds for ɛ
=
1.Similarly, ifɛ
=
-
1, by (10) y is an odd prime with y
≡
1
(
mod
⁡
4
). We see from (1) that the equation
(32)
X
2
-
y
Y
2
=
x
Z
,
X
,
Y
,
Z
∈
Z
,
gcd
(
X
,
Y
)
=
1
,
Z
>
0
,
has the solution
(33)
(
X
,
Y
,
Z
)
=
(
z
z
/
2
,
y
(
x
-
1
)
/
2
,
y
)
.
Applying Lemmas 5 and 6 to (11) and (12), we have
(34)
z
z
/
2
+
y
(
x
-
1
)
/
2
y
=
(
X
1
+
λ
Y
1
y
)
y
(
u
+
v
y
)
,
λ
∈
{
±
1
}
,
where X
1
,
Y
1
,
Z
1 are positive integers satisfying
(35)
X
1
2
-
y
Y
1
2
=
x
,
gcd
(
X
1
,
Y
1
)
=
1
,
and (
u
,
v
) is a solution of the equation
(36)
u
2
-
y
v
2
=
1
,
u
,
v
∈
Z
.
Applying Lemma 3 to (34) and (35), we have
(37)
u
+
v
y
=
(
u
1
+
v
1
y
)
s
,
s
∈
Z
,
where (
u
1
,
v
1
) is the least solution of (36). In addition, the integer s can be expressed as
(38)
s
=
y
q
+
r
,
q
,
r
∈
Z
,
0
≤
r
<
y
.
Let
(39)
X
2
+
Y
2
y
=
(
X
1
+
λ
Y
1
y
)
(
u
1
+
v
1
y
)
q
.
From (35) and (36), we know that X
2 and Y
2 are integers satisfying
(40)
X
2
2
-
y
Y
2
2
=
x
,
gcd
(
X
2
,
Y
2
)
=
1
.
And from (34), (37), (38), and (39), we get
(41)
z
z
/
2
+
y
(
x
-
1
)
/
2
y
=
(
X
2
+
Y
2
y
)
y
(
u
1
+
v
1
y
)
r
.Ifr
=
0 in (38), then, from (41), we have
(42)
y
(
x
-
1
)
/
2
=
Y
2
∑
i
=
0
(
y
-
1
)
/
2
‍
(
y
2
i
+
1
)
X
2
y
-
2
i
-
1
(
y
Y
2
2
)
i
.
Since y is an odd prime, x
>
y
≥
5, by (42), we know
(43)
0
≡
y
(
x
-
1
)
/
2
≡
y
X
2
y
-
1
Y
2
(
mod
⁡
y
2
)
.
From (40), we know gcd
(
X
2
,
y
)
=
1; then from (43) we get y
∣
Y
2. Let y
α
∥
Y
2, and y
≥
5, so
(44)
y
α
+
1
∥
Y
2
(
y
1
)
X
2
y
-
1
,
(45)
Y
2
(
y
2
i
+
1
)
X
2
y
-
2
i
-
1
(
y
Y
2
2
)
i
≡
y
Y
2
(
y
-
1
2
i
)
X
2
y
-
2
i
-
1
(
y
Y
2
2
)
i
2
i
+
1
≡
0
(
mod
⁡
y
α
+
2
)
,
i
≥
1
.
By (45),
(46)
y
α
+
1
∥
Y
2
∑
i
=
0
(
y
-
1
)
/
2
‍
(
y
2
i
+
1
)
X
2
y
-
2
i
-
1
(
y
Y
2
2
)
i
.
Combining (42) and (46) we may immediately get
(47)
α
+
1
=
x
-
1
2
.
However, from (42) and (47), we get |
y
Y
2
|
≥
y
α
+
1
=
y
(
x
-
1
)
/
2
>
|
y
X
2
y
-
1
Y
2
|
>
|
y
Y
2
|, but it is impossible. Therefore, we have 0
<
r
<
y and
(48)
y
∤
r
.Now, using the similarly proof withɛ
=
1, from (41) and (48) can obtain contradiction.This completes the proof of our theorem.



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*Source: 186416-2014-07-13.xml*