3. Proof of Theorem <xref reftype="statement" rid="thm1.1">2</xref>
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
}
.
If
ɛ
=
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
.
Since
x
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
.
Since
t
=
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
.
If
r
=
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, if
ɛ
=

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
.
If
r
=
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 with
ɛ
=
1
, from (41) and (48) can obtain contradiction.
This completes the proof of our theorem.