# Nonspectrality of Certain Self-Affine Measures onℝ

Authors: Gui-Bao Gao
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We will determine the nonspectrality of self-affine measureμ
B
,
D corresponding to B
=
diag
[
p
1
,
p
2
,
p
3
]
(
p
1
∈
(
2
ℤ
+
1
)
∖
{
±
1
}, p
2
∈
2
ℤ
∖
{
0
}
), and D
=
{
0
,
e
1
,
e
2
,
e
3
} in the space ℝ
3 is supported on T
(
B
,
D
), where e
1
,
e
2, and e
3 are the standard basis of unit column vectors in ℝ
3, and there exist at most 4 mutually orthogonal exponential functions in L
2
(
μ
B
,
D
), where the number 4 is the best. This generalizes the known results on the spectrality of self-affine measures.


---

## Body

## 1. Introduction

LetB
∈
M
n
(
ℤ
) be an expanding integer matrix; that is, all the eigenvalues of the integer matrix B have modulus greater than 1. Associated with a finite subset D
⊂
ℤ
n, there exists a unique nonempty compact set T
∶
=
T
(
B
,
D
)
⊂
ℝ
n such that T
=
∪
d
∈
D
ϕ
d
(
T
). More precisely, T
(
B
,
D
) is an attractor (or invariant set) of the affine iterated function system (IFS) {
ϕ
d
(
x
)
}
d
∈
D. Denote by |
D
| the cardinality of D. Relating to the IFS {
ϕ
d
(
x
)
}
d
∈
D, there also exists a unique probability measure μ
∶
=
μ
B
,
D satisfying
(1)
μ
=
1
|
D
|
∑
d
∈
D
‍
μ
∘
ϕ
d
-
1
.For a given pair(
B
,
D
), the spectrality or nonspectrality of μ
B
,
D is directly connected with the Fourier transform μ
^
B
,
D
(
ξ
). From (1), we get
(2)
μ
^
B
,
D
(
ξ
)
∶
=
∫
‍
e
2
π
i
〈
x
,
ξ
〉
d
μ
B
,
D
(
x
)
=
∏
j
=
1
∞
‍
m
D
(
B
*
-
j
ξ
)
,
(
ξ
∈
ℝ
n
)
,
where
(3)
m
D
(
x
)
=
1
|
D
|
∑
d
∈
D
‍
e
2
π
i
〈
d
,
x
〉
,
(
x
∈
ℝ
n
)
.The self-affine measureμ
B
,
D has received much attention in recent years. The previous research on such measure and its Fourier transform revealed some surprising connections with a number of areas in mathematics, such as harmonic analysis, number theory, dynamical systems, and others; see [1, 2] and references cited therein.Theμ
B
,
D and T
(
B
,
D
) are all determined by the pair (
B
,
D
). So, for n
=
1, in the way of examples, there are Cantor set and Cantor measure on the line. And for n
=
2 there is a rich variety of geometries, see Li [3–6], of which the best known example is the Sierpinski gasket. But for n
=
3, it is more complex.The problem considered below started with a discovery in an earlier paper of Jorgensen and Pedersen [7] where it was proved that certain IFS fractals have Fourier bases, and furthermore that the question of counting orthogonal Fourier frequencies (or orthogonal exponentials in L
2
(
μ
B
,
D
)) for a fixed fractal involves an intrinsic arithmetic of the finite set of functions making up the IFS {
ϕ
d
(
x
)
}
d
∈
D under consideration. For example, if (
B
,
D
)
=
(
3
,
{
0,2
}
) is the middle-third Cantor example on the line, there cannot be more than two orthogonal Fourier frequencies [7, Theorem 6.1], while a similar Cantor example, using instead a subdivision scale 4 (i.e., (
B
,
D
)
=
(
4
,
{
0,2
}
)), turns out to have an ONB in L
2
(
μ
B
,
D
) consisting of Fourier frequencies [7, Theorem 3.4].With the effort of Jorgensen and Pedersen [7, Example 7.1], Strichartz [8], Li [3], and Yuan [9], the related conclusions discussed that the diagonal elements of B are all even or odd, If one of the diagonal elements is even, what about the result? The general case on the spectrality or nonspectrality of the self-affine measure μ
B
,
D is not known.The present paper is motivated by these earlier results; we will determine the nonspectrality of self-affine measureμ
B
,
D; the main result of the present paper is the following.Theorem 1.
Letμ
B
,
D correspond to B
=
diag
[
p
1
,
p
1
,
p
2
]
(
p
1
∈
(
2
ℤ
+
1
)
∖
{
±
1
}
,
p
2
∈
2
ℤ
∖
{
0
}
) and D
=
{
0
,
e
1
,
e
2
,
e
3
}; then the self-affine measure μ
B
,
D is a nonspectral measure, and there exist at most 4 mutually orthogonal exponential functions in L
2
(
μ
B
,
D
), where the number 4 is the best.By the proof of Theorem1, we get a more general case.Theorem 2.
Letμ
B
,
D correspond to
(4)
B
=
[
p
1
0
p
4
0
p
1
p
3
0
0
p
2
]
(
p
1
∈
(
2
ℤ
+
1
)
∖
{
±
1
}
;
p
2
,
p
3
,
p
4
∈
ℤ
)
,
D
=
{
(
0
0
0
)
,
(
1
0
0
)
,
(
0
1
0
)
,
(
0
0
1
)
}
;
then the self-affine measure μ
B
,
D is a nonspectral measure, and there exist at most 4 mutually orthogonal exponential functions in L
2
(
μ
B
,
D
), where the number 4 is the best.


## 2. Proof of Theorem1

Firstly, we know from (2) that the zero set Z
(
μ
^
B
,
D
(
ξ
)
) of the Fourier transform μ
^
B
,
D
(
ξ
) is
(5)
Z
(
μ
^
B
,
D
(
ξ
)
)
=
⋃
j
=
1
∞
‍
B
*
j
Z
(
m
D
(
ξ
)
)
∶
=
B
1
∪
B
2
∪
B
3
.
For the given digit set D in (4), we have
(6)
Z
(
m
D
(
x
)
)
∶
=
{
x
∈
ℝ
3
:
m
D
(
x
)
=
0
}
=
A
1
∪
A
2
∪
A
3
,
where
(7)
m
D
(
x
)
=
1
4
{
1
+
e
2
π
i
x
1
+
e
2
π
i
x
2
+
e
2
π
i
x
3
}
,
x
=
(
x
1
,
x
2
,
x
3
)
t
∈
ℝ
3
,
and A
1, A
2, and A
3 are given by
(8)
A
1
=
{
(
1
2
+
k
1
a
+
k
2
1
2
+
a
+
k
3
)
:
a
∈
ℝ
,
k
1
,
k
2
,
k
3
∈
ℤ
}
⊂
ℝ
3
,
A
2
=
{
(
1
2
+
a
+
k
1
1
2
+
k
2
a
+
k
3
)
:
a
∈
ℝ
,
k
1
,
k
2
,
k
3
∈
ℤ
}
⊂
ℝ
3
,
A
3
=
{
(
a
+
k
1
1
2
+
a
+
k
2
1
2
+
k
3
)
:
a
∈
ℝ
,
k
1
,
k
2
,
k
3
∈
ℤ
}
⊂
ℝ
3
.So(9)
B
1
=
⋃
j
=
1
∞
‍
B
*
j
A
1
=
⋃
j
=
1
∞
‍
{
(
(
1
2
+
k
1
)
p
1
j
(
a
+
k
2
)
p
1
j
(
1
2
+
a
+
k
3
)
p
2
j
)
:
a
∈
ℝ
,
k
1
,
k
2
∈
ℤ
}
⊂
ℝ
3
,
B
2
=
⋃
j
=
1
∞
‍
B
*
j
A
2
=
⋃
j
=
1
∞
‍
{
(
(
1
2
+
a
+
k
1
)
p
1
j
(
1
2
+
k
2
)
p
1
j
(
a
+
k
3
)
p
2
j
)
:
a
∈
ℝ
,
k
1
,
k
2
∈
ℤ
}
⊂
ℝ
3
,
B
3
=
⋃
j
=
1
∞
‍
B
*
j
A
3
=
⋃
j
=
1
∞
‍
{
(
(
a
+
k
1
)
p
1
j
(
1
2
+
a
+
k
2
)
p
1
j
(
1
2
+
k
3
)
p
2
j
)
:
a
∈
ℝ
,
k
1
,
k
2
∈
ℤ
}
⊂
ℝ
3
.Sincep
1
∈
(
2
ℤ
+
1
)
∖
{
±
1
}, we can verify directly that the following two lemmas hold.Lemma 3.
The setsB
j (j
=
1,2
,
3) given by (9) satisfy the following properties:(a)
ξ
∈
B
j
⇔
-
ξ
∈
B
j (j
=
1,2
,
3);
(b)
Z
(
μ
^
B
,
D
(
ξ
)
)
⋂
‍
(
ℤ
×
ℤ
×
ℝ
)
=
∅;
(c)
ifξ
=
(
ξ
1
,
ξ
2
,
ξ
3
)
t
∈
B
1
±
B
1, then ξ
1
∈
ℤ;
(d)
ifξ
=
(
ξ
1
,
ξ
2
,
ξ
3
)
t
∈
B
2
±
B
2, then ξ
2
∈
ℤ;
(e)
ifξ
=
(
ξ
1
,
ξ
2
,
ξ
3
)
t
∈
B
3
±
B
3 and ξ
∈
Z
(
μ
^
B
,
D
(
ξ
)
), then ξ
∈
B
1
∪
B
2 and ξ
1
,
ξ
2
∈
(
1
/
2
)
+
ℤ.
By proving (e), the others can be checked directly.
In (e), ifξ
=
(
ξ
1
,
ξ
2
,
ξ
3
)
t
∈
B
3
±
B
3, then
(10)
ξ
=
(
ξ
1
,
ξ
2
,
ξ
3
)
t
=
⋃
j
=
1
∞
‍
{
(
(
a
1
+
k
11
)
p
1
j
1
(
1
2
+
a
1
+
k
12
)
p
1
j
1
(
1
2
+
k
13
)
p
2
j
)
±
(
(
a
2
+
k
21
)
p
1
j
2
(
1
2
+
a
2
+
k
22
)
p
1
j
2
(
1
2
+
k
23
)
p
2
j
)
}
=
⋃
j
=
1
∞
‍
{
(
(
a
1
+
k
11
)
p
1
j
1
±
(
a
2
+
k
21
)
p
1
j
2
(
1
2
+
a
1
+
k
12
)
p
1
j
1
±
(
1
2
+
a
2
+
k
22
)
p
1
j
2
(
1
2
+
k
13
)
p
2
j
±
(
1
2
+
k
23
)
p
2
j
)
}
,
where a
i
∈
ℝ and k
i
1
,
k
i
2
∈
ℤ, i
=
1,2. From (10) one has ξ
2
-
ξ
1
=
(
1
/
2
)
(
p
1
j
1
±
p
1
j
2
)
±
(
k
12
-
k
11
)
p
1
j
1
±
(
k
22
-
k
21
)
p
1
j
2
∈
ℤ (since k
i
1
,
k
i
2
∈
ℤ, i
=
1,2 and p
1
∈
(
2
ℤ
+
1
)
∖
{
±
1
}); then ξ
∈
B
1
∪
B
2. In fact, if ξ
∈
Z
(
μ
^
B
,
D
(
ξ
)
) and ξ
∈
B
3, then ξ
2
-
ξ
1
∈
(
1
/
2
+
ℤ
) is a contradiction. Since ξ
∈
B
1
∪
B
2 and ξ
2
-
ξ
1
∈
ℤ, then ξ
1
,
ξ
2
∈
(
1
/
2
)
+
ℤ.Lemma 4.
Letξ
=
(
ξ
1
,
ξ
2
,
ξ
3
)
t
∈
Z
(
μ
^
B
,
D
(
ξ
)
)
=
B
1
∪
B
2
∪
B
3. Then the following statements hold:(i)
ifξ
∈
B
j, then ξ
j
∈
(
1
/
2
)
+
ℤ, where j
=
1,2;
(ii)
ifξ
1
∈
ℤ, then ξ
∈
B
2
∪
B
3 and ξ
2
∈
(
1
/
2
)
+
ℤ;
(iii)
ifξ
2
∈
ℤ, then ξ
∈
B
1
∪
B
3 and ξ
1
∈
(
1
/
2
)
+
ℤ.Suppose thatλ
j
(
j
=
1,2
,
3,4
,
5
)
∈
ℝ
3 are such that the five exponential functions
(11)
e
2
π
i
〈
λ
1
,
x
〉
,
e
2
π
i
〈
λ
2
,
x
〉
,
e
2
π
i
〈
λ
3
,
x
〉
,
e
2
π
i
〈
λ
4
,
x
〉
,
e
2
π
i
〈
λ
5
,
x
〉
are mutually orthogonal in L
2
(
μ
B
,
D
), so the differences λ
j
-
λ
k
(
1
≤
j
≠
k
≤
5
) are in the zero set Z
(
μ
^
B
,
D
(
ξ
)
). Then we get
(12)
λ
j
-
λ
k
∈
Z
(
μ
^
B
,
D
(
ξ
)
)
=
B
1
∪
B
2
∪
B
3
(
1
≤
j
≠
k
≤
5
)
.Denoteλ
j
-
λ
k by
(13)
λ
j
-
λ
k
=
(
x
j
,
k
,
y
j
,
k
,
z
j
,
k
)
t
∈
ℝ
3
for
1
≤
j
≠
k
≤
5
.
We will apply the above two lemmas to get the contradiction below.The following ten differences(14)
λ
2
-
λ
1
,
λ
3
-
λ
1
,
λ
4
-
λ
1
,
λ
5
-
λ
1
λ
2
-
λ
1
,
λ
3
-
λ
2
,
λ
4
-
λ
2
,
λ
5
-
λ
2
λ
2
-
λ
1
,
λ
3
-
λ
1
,
λ
4
-
λ
3
,
λ
5
-
λ
3
λ
2
-
λ
1
,
λ
3
-
λ
1
,
λ
4
-
λ
1
,
λ
5
-
λ
4
belong to B
1
∪
B
2
∪
B
3. Then
(15)
λ
2
-
λ
1
,
λ
3
-
λ
1
,
λ
4
-
λ
1
,
λ
5
-
λ
1
∈
B
1
∪
B
2
∪
B
3
.Claim 1.
The setB
1 or B
2 or B
3 cannot contain any three differences of the form: λ
j
1
-
λ
j, λ
j
2
-
λ
j, λ
j
3
-
λ
j, where 1
≤
j
1
≠
j
2
≠
j
3
≠
j
≤
5.Claim1 can be checked easily. For example, if λ
j
1
-
λ
j, λ
j
2
-
λ
j, λ
j
3
-
λ
j
∈
B
1, so, by applying Lemma 3(c) and (12), we get
(16)
λ
j
1
-
λ
j
3
=
(
λ
j
1
-
λ
j
)
-
(
λ
j
3
-
λ
j
)
∈
(
B
1
-
B
1
)
∩
Z
(
μ
^
M
,
D
(
ξ
)
)
,
λ
j
2
-
λ
j
3
=
(
λ
j
2
-
λ
j
)
-
(
λ
j
3
-
λ
j
)
∈
(
B
1
-
B
1
)
∩
Z
(
μ
^
B
,
D
(
ξ
)
)
,
λ
j
1
-
λ
j
2
=
(
λ
j
1
-
λ
j
)
-
(
λ
j
2
-
λ
j
)
∈
(
B
1
-
B
1
)
∩
Z
(
μ
^
B
,
D
(
ξ
)
)
,
x
j
1
,
j
3
,
x
j
2
,
j
3
,
x
j
1
,
j
2
∈
ℤ
,
and by Lemma 4(ii),
(17)
λ
j
1
-
λ
j
3
,
λ
j
2
-
λ
j
3
,
λ
j
1
-
λ
j
2
∈
B
2
∪
B
3
,
which shows that at least one of the two sets B
2 and B
3 contains two differences. If B
2 contains two differences, say λ
j
2
-
λ
j
3 and λ
j
1
-
λ
j
2, then
(18)
λ
j
1
-
λ
j
3
=
(
λ
j
1
-
λ
j
2
)
+
(
λ
j
2
-
λ
j
3
)
∈
B
2
+
B
2
.
This shows (by Lemma 3(d)) that y
j
1
,
j
3
∈
ℤ, a contradiction with Lemma 3(b). If B
3 contains two differences, say λ
j
2
-
λ
j
3 and λ
j
1
-
λ
j
2, then
(19)
λ
j
1
-
λ
j
3
=
(
λ
j
1
-
λ
j
2
)
+
(
λ
j
2
-
λ
j
3
)
∈
B
3
+
B
3
;
by Lemma 3(e), x
j
1
,
j
3
∈
(
1
/
2
)
+
ℤ, a contradiction with (16).From (15) and Claim 1, we only need to deal with the following four typical cases.Case 1.
λ
2
-
λ
1
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1
∈
B
3, and λ
5
-
λ
1
∈
B
1.Case 2.
λ
2
-
λ
1
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1
∈
B
3, and λ
5
-
λ
1
∈
B
3.Case 3.
λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1 and λ
4
-
λ
1
,
λ
5
-
λ
1
∈
B
2.Case 4.
λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1 and λ
4
-
λ
1, λ
5
-
λ
1
∈
B
3.Note that Case1 denotes the 2
-
1
-
1 or 1
-
2
-
1 distribution in (15), Case 2 denotes the 1
-
1
-
2 distribution in (15), and Case 3 denotes the 2
-
2
-
0 distribution in (15), while Case 4 denotes the 2
-
0
-
2 (or 0
-
2
-
2) distribution in (15). If the four cases can be proved, then the other cases can be proved similarly.


### 2.1. Case1

In this case, we have(20)
λ
5
-
λ
2
=
(
λ
5
-
λ
1
)
-
(
λ
2
-
λ
1
)
∈
B
1
-
B
1
.
Applying Lemma 3(c), (12), and Lemma 4(ii), we get
(21)
λ
5
-
λ
2
=
(
x
5,2
,
y
5,2
,
z
5,2
)
t
∈
B
2
∪
B
3
,
x
5,2
∈
ℤ
,
y
5,2
∈
1
2
+
ℤ
.
From λ
5
-
λ
2
∈
B
2
∪
B
3, Case 1 can be divided into two subcases.Case
1.1. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
2
∈
B
3.Step 1.
We will prove Case 1.1; sinceλ
3
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 1.1 can be divided into three cases.Case
1.1.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.1.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2, λ
3
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.1.3. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
3
-
λ
2
∈
B
3.We will give a method to deal with each case by considering the remainder differences in (14), and each case is concluded with a contradiction.Case
1.1.1. In this case, we have
(22)
λ
3
-
λ
1
=
(
λ
3
-
λ
2
)
+
(
λ
2
-
λ
1
)
∈
B
1
+
B
1
;
by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(23)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
ℤ
,
y
3,1
∈
1
2
+
ℤ
.By Lemma3(a) and Claim 1 we know that λ
4
-
λ
2
∉
B
1. So λ
4
-
λ
2
∈
B
2 or λ
4
-
λ
2
∈
B
3, so Case 1.1.1 can be divided into two subcases.Case
1.1.1.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2, λ
4
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.1.1.2. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
4
-
λ
2
∈
B
3.By considering the remainder differences in (14), we apply Lemmas 3 and 4 to deal with each case.(I)
In Case 1.1.1.1, we have(24)
λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
)
∈
B
2
-
B
2
; by applying Lemma 3(d), (12) and Lemma 4(iii), we have
(25)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
3
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
ℤ
.
(i)
Ifλ
5
-
λ
4
∈
B
1, so λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1; by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(26)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
ℤ
,
y
4,1
∈
1
2
+
ℤ
.It follows from (23), (26), and λ
4
-
λ
3
=
(
λ
4
-
λ
1
)
-
(
λ
3
-
λ
1
) that
(27)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
,
x
4,3
∈
ℤ
,
y
4,3
∈
ℤ
,
which shows a contradiction with Lemma 3(b).(ii)
Ifλ
5
-
λ
4
∈
B
3, so λ
5
-
λ
1
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
1
)
∈
B
3
+
B
3; by applying Lemma 3(e) and (12), we get
(28)
λ
5
-
λ
1
=
(
x
5,1
,
y
5,1
,
z
5,1
)
t
∈
B
1
,
x
5,1
∈
1
2
+
ℤ
,
y
5,1
∈
1
2
+
ℤ
.Now, consider the remainder difference λ
4
-
λ
3 in (14): by Lemma 3(a) and Claim 1, we have λ
4
-
λ
3
∉
B
3, so λ
4
-
λ
3
∈
B
1
∪
B
2. If λ
4
-
λ
3
∈
B
1, then
(29)
λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
1
+
B
1
;
by applying Lemma 3(c), (12), and Lemma 4(ii) we get
(30)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
2
,
x
4,2
∈
ℤ
,
y
4,2
∈
1
2
+
ℤ
.
It follows from (21), (30), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
) that x
5,4
∈
ℤ
,
y
5,4
∈
ℤ, a contradiction with Lemma 3(b). If λ
4
-
λ
3
∈
B
2, then
(31)
λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2
;
by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(32)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
.
It follows from (25), (32), and λ
5
-
λ
1
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
1
) that x
5,1
∈
ℤ and y
5,1
∈
ℤ, a contradiction with Lemma 3(b).From parts (i), (ii), and (25), Case 1.1.1.1 is proved.(II)
In Case 1.1.1.2, we have(33)
λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3
; by applying Lemma 3(e) and (12), we get
(34)
λ
2
-
λ
1
=
(
x
2,1
,
y
2,1
,
z
2,1
)
t
∈
B
1
,
x
2,1
∈
1
2
+
ℤ
,
y
2,1
∈
1
2
+
ℤ
.We consider the remainder three differencesλ
4
-
λ
3, λ
5
-
λ
3, and λ
5
-
λ
4 in (14). By Claim 1, we have λ
4
-
λ
3
∉
B
3, so λ
4
-
λ
3
∈
B
1
∪
B
2.(i)
Ifλ
4
-
λ
3
∈
B
1, we have
(35)
λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
1
+
B
1
; by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(36)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
3
,
x
4,2
∈
ℤ
,
y
4,2
∈
1
2
+
ℤ
.
It follows from (21), (36), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
) that
(37)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
,
x
5,4
∈
ℤ
,
y
5,4
∈
ℤ
,
a contradiction with Lemma 3(b).(ii)
Ifλ
4
-
λ
3
∈
B
2, we have
(38)
λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2
; by applying Lemma 3(d), (12), and Lemma 4(iii) we have
(39)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
.
We consider the remainder difference λ
5
-
λ
4 in (14). By Claim 1, we have λ
5
-
λ
4
∉
B
3, so λ
5
-
λ
4
∈
B
1
∪
B
2. If λ
5
-
λ
4
∈
B
1, we have
(40)
λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1
,
by applying Lemma 3(c), (12), and Lemma 4(ii), we get x
4,1
∈
ℤ, a contradiction with (39). If λ
5
-
λ
4
∈
B
2, we have
(41)
λ
4
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
4
)
∈
B
2
-
B
2
;
by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(42)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
3
,
x
4,2
∈
1
2
+
ℤ
,
y
4,2
∈
ℤ
.
It follows from (39), (42), and λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
) that x
2,1
∈
ℤ, a contradiction with (34).The above (i) and (ii) in (II) illustrate that Case 1.1.1.2 is proved. Hence Case 1.1.1 is proved.Case
1.1.2. In this case, we have
(43)
λ
2
-
λ
1
=
(
λ
3
-
λ
1
)
-
(
λ
3
-
λ
2
)
∈
B
2
-
B
2
,
λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
2
-
B
2
;
by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(44)
λ
2
-
λ
1
=
(
x
2,1
,
y
2,1
,
z
2,1
)
t
∈
B
1
,
x
2,1
∈
1
2
+
ℤ
,
y
2,1
∈
ℤ
,
(45)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
∈
B
1
∪
B
3
,
x
5,3
∈
1
2
+
ℤ
,
y
5,3
∈
ℤ
.
By applying Lemma 3(a) and Claim 1, we get
(46)
λ
4
-
λ
2
∉
B
2
,
λ
4
-
λ
3
∉
B
2
.Ifλ
4
-
λ
2
∈
B
3, then λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3; by Lemma 3(e), and (12), we get x
2,1
∈
(
1
/
2
)
+
ℤ, y
2,1
∈
(
1
/
2
)
+
ℤ, a contradiction with (44); hence
(47)
λ
4
-
λ
2
∉
B
3
,
λ
4
-
λ
2
∈
B
1
.
We consider λ
4
-
λ
3; if λ
4
-
λ
3
∈
B
1, we get λ
3
-
λ
2
=
(
λ
4
-
λ
2
)
-
(
λ
4
-
λ
3
)
∈
B
1
-
B
1; by Lemma 3(c), (12), and Lemma 4(ii), we have
(48)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
2
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
,
which, combined with (21) and λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
), shows that x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction with Lemma 3(b); hence
(49)
λ
4
-
λ
3
∉
B
1
,
λ
4
-
λ
3
∈
B
3
.Sinceλ
3
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
3
)
∈
B
3
-
B
3, by Lemma 3(e) and (12), we get
(50)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
1
2
+
ℤ
,
y
3,1
∈
1
2
+
ℤ
.Combined with (21), (44), and λ
5
-
λ
1
=
(
λ
5
-
λ
2
)
+
(
λ
2
-
λ
1
), we get
(51)
λ
5
-
λ
1
=
(
x
5,1
,
y
5,1
,
z
5,1
)
t
,
x
5,1
∈
1
2
+
ℤ
,
y
5,1
∈
1
2
+
ℤ
,
which, combined with (50), (51), and λ
5
-
λ
3
=
(
λ
5
-
λ
1
)
-
(
λ
3
-
λ
1
), shows that x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction of (45). This proves Case 1.1.2.Case
1.1.3. From λ
4
-
λ
2
∈
B
1
∪
B
2
∪
B
3, this Case can be divided into the three cases.Case
1.1.3.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
4
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
1.1.3.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2, λ
4
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
1.1.3.3. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, λ
3
-
λ
2, and λ
4
-
λ
2
∈
B
3.The above three cases denote the3
-
2
-
2 or 2
-
3
-
2 or 2
-
2
-
3 distribution. In each case, we have (21). The discussion of the first two cases is similar. In the following, we use the above method to deal with Cases 1.1.3.1 and 1.1.3.3.In Case 1.1.3.1, since(52)
λ
4
-
λ
1
=
(
λ
4
-
λ
2
)
+
(
λ
2
-
λ
1
)
∈
B
1
+
B
1
,
by applying Lemma 3(c), (12), and Lemma 4(ii), we have
(53)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
ℤ
,
y
4,1
∈
1
2
+
ℤ
.We consider the remainder three differencesλ
4
-
λ
3, λ
5
-
λ
3, and λ
5
-
λ
4. From (21) and (53) we have
(54)
λ
4
-
λ
3
∉
B
2
or
λ
4
-
λ
3
∈
B
1
∪
B
3
,
(55)
λ
5
-
λ
3
∉
B
3
or
λ
5
-
λ
3
∈
B
1
∪
B
2
.Ifλ
4
-
λ
3
∈
B
2, by Lemma 3, so λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2; we get y
4,1
∈
ℤ, a contradiction with (53). If λ
5
-
λ
3
∈
B
3, so λ
5
-
λ
2
=
(
λ
5
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
3
+
B
3 by Lemma 3(e); we get x
5,2
∈
(
1
/
2
)
+
ℤ, a contradiction with (21). Hence (54) and (55) hold.According to (54), we deal with the following two cases.(i)
Ifλ
4
-
λ
3
∈
B
1, then, from λ
3
-
λ
2
=
(
λ
4
-
λ
2
)
-
(
λ
4
-
λ
3
)
∈
B
1
-
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(56)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
3
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
.From (21), (56), and λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
), we get
(57)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
,
x
5,3
∈
ℤ
,
y
5,3
∈
ℤ
,
a contradiction with Lemma 3(b).(ii)
Ifλ
4
-
λ
3
∈
B
3, since λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
3
+
B
3 and λ
3
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
3
)
∈
B
3
-
B
3, by applying Lemma 3(e) and (12), we get
(58)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
1
2
+
ℤ
,
y
3,1
∈
1
2
+
ℤ
.We consider (55); if λ
5
-
λ
3
∈
B
1, by Lemma 3 and λ
3
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
3
)
∈
B
1
-
B
1, we get x
3,1
∈
ℤ, a contradiction with (58). If λ
5
-
λ
3
∈
B
2, since λ
3
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
3
)
∈
B
2
-
B
2, by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(59)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
3
,
x
3,2
∈
1
2
+
ℤ
,
y
3,2
∈
ℤ
.
From (58), (59) and λ
2
-
λ
1
=
(
λ
3
-
λ
1
)
-
(
λ
3
-
λ
2
), we have x
2,1
∈
ℤ, a contradiction with Lemma 4(i) (for λ
2
-
λ
1
∈
B
1 gives x
2,1
∈
(
1
/
2
)
+
ℤ).The above (i), (ii), and (54) indicae that Case 1.1.3.1 is proved.In Case 1.1.3.3, we get(60)
λ
4
-
λ
3
=
(
λ
4
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
3
-
B
3
,
λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3
;
by applying Lemma 3(e) and (12), we have
(61)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
∈
B
1
∪
B
2
,
x
4,3
∈
1
2
+
ℤ
,
y
4,3
∈
1
2
+
ℤ
,
(62)
λ
2
-
λ
1
=
(
x
2,1
,
y
2,1
,
z
2,1
)
t
∈
B
1
,
x
2,1
∈
1
2
+
ℤ
,
y
2,1
∈
1
2
+
ℤ
.
By Lemma 3(a) and λ
1
-
λ
4, λ
2
-
λ
4
∈
B
3, we know from Claim 1 that λ
5
-
λ
4
∉
B
3. So λ
5
-
λ
4
∈
B
1 or λ
5
-
λ
4
∈
B
2.From (61), we consider the following two cases.(i)
Ifλ
4
-
λ
3
∈
B
1, we consider λ
5
-
λ
4, if λ
5
-
λ
4
∈
B
1. From λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
)
∈
B
1
+
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(63)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
∈
B
2
∪
B
3
,
x
5,3
∈
ℤ
,
y
5,3
∈
1
2
+
ℤ
.From (21), (63), and λ
3
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
3
), we get x
2,1
∈
ℤ
,
y
2,1
∈
ℤ, a contradiction with Lemma 3(b). If λ
5
-
λ
4
∈
B
2, since λ
4
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
4
)
∈
B
2
-
B
2, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(64)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
1
∪
B
3
,
x
4,2
∈
1
2
+
ℤ
,
y
4,2
∈
ℤ
.
From (21), (64), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
), we get that x
5,4
∈
(
1
/
2
)
+
ℤ
,
y
5,4
∈
(
1
/
2
)
+
ℤ, combined with (61) and λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
), yields x
5,4
∈
ℤ
,
y
5,4
∈
ℤ, a contradiction with Lemma 3(b).(ii)
Ifλ
4
-
λ
3
∈
B
2, so λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2; by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(65)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
. If λ
5
-
λ
4
∈
B
1, by Lemma 3(c) and λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1, we get x
4,1
∈
ℤ, a contradiction with (65). If λ
5
-
λ
4
∈
B
2, so λ
4
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
4
)
∈
B
2
-
B
2, by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(66)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
3
,
x
4,2
∈
1
2
+
ℤ
,
y
4,2
∈
ℤ
.From (65), (66), and λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
), we get x
2,1
∈
ℤ, a contradiction with (62).The above (i), (ii), and (61) illustrate that Case 1.1.3.3 is proved and Case 1.1 is proved.Step 2.
In Case 1.2, fromλ
3
-
λ
2
∈
B
1
∪
B
2
∪
B
3, this case can be divided into three cases.Case
1.2.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
2
∈
B
3.Case
1.2.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
3
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
2
∈
B
3.Case
1.2.3. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, and λ
3
-
λ
2
∈
B
3.Case 1.2.2 is similar to Case 1.1.3; we will deal with the other two cases by considering the remainder differences in (14), and each case is concluded with a contradiction.Case
1.2.1. In this case, we get
(67)
λ
3
-
λ
1
=
(
λ
3
-
λ
2
)
+
(
λ
2
-
λ
1
)
∈
B
1
+
B
1
;
by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(68)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
ℤ
,
y
3,1
∈
1
2
+
ℤ
.By Lemma3(a) and λ
1
-
λ
2
,
λ
3
-
λ
2
∈
B
1, we know from Claim 1 that λ
4
-
λ
2
∉
B
1. So λ
4
-
λ
2
∈
B
2 or λ
4
-
λ
2
∈
B
3, so Case 1.2.1 can be divided into two cases.Case
1.2.1.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
4
-
λ
2
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2
∈
B
3.Case
1.2.1.2. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, λ
4
-
λ
2
∈
B
3.The above two cases denote the3
-
2
-
2 distribution and 3
-
1
-
3 distribution. Case 1.2.1.1 is similar to Case 1.1.3.1. So we only need to deal with Case 1.2.1.2. By considering the remainder differences in (14), we apply Lemmas 3 and 4 to deal with the case.In Case 1.2.1.2, we get(69)
λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3
;
by applying Lemma 3(e) and (12), we get
(70)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
2
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
1
2
+
ℤ
.By Claim1, λ
4
-
λ
3
∉
B
3; then λ
4
-
λ
3
∈
B
1 or λ
4
-
λ
3
∈
B
2.From (70), we consider the following two cases.(i)
Ifλ
5
-
λ
4
∈
B
1, since λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii) we get
(71)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
ℤ
,
y
4,1
∈
1
2
+
ℤ
.It follows from (68), (71), and λ
4
-
λ
3
=
(
λ
4
-
λ
1
)
-
(
λ
3
-
λ
1
) that
(72)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
,
x
4,3
∈
ℤ
,
y
4,3
∈
ℤ
,
which shows a contradiction with Lemma 3(b).(ii)
Ifλ
5
-
λ
4
∈
B
2, we consider λ
4
-
λ
3, if λ
4
-
λ
3
∈
B
1. Since λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
1
+
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(73)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
2
∪
B
3
,
x
4,2
∈
ℤ
,
y
4,2
∈
1
2
+
ℤ
.From (21), (73), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
), we get x
5,4
∈
ℤ, y
5,4
∈
ℤ, a contradiction with Lemma 3(b). If λ
4
-
λ
3
∈
B
2, then λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2; by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(74)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
.It follows from (68), (74), and λ
4
-
λ
3
=
(
λ
4
-
λ
1
)
-
(
λ
3
-
λ
1
) that we get
(75)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
∈
B
2
,
x
4,3
∈
1
2
+
ℤ
,
y
4,3
∈
1
2
+
ℤ
.From (70), (75), and λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
), we get x
5,4
∈
ℤ, y
5,4
∈
ℤ, a contradiction with Lemma 3(b).Parts (i), (ii), and (70) indicate that Case 1.2.1.2 is proved and Case 1.2.1 is proved.Case
1.2.3. In this case, since
(76)
λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
3
-
B
3
,
by applying Lemma 3(e) and (12), we get
(77)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
∈
B
1
∪
B
2
,
x
5,3
∈
1
2
+
ℤ
,
y
5,3
∈
1
2
+
ℤ
.From (77) Case 1.2.3 can be divided into two cases.Case
1.2.3.1. λ
2
-
λ
1
,
λ
5
-
λ
1
,
λ
5
-
λ
3
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, and λ
3
-
λ
2
∈
B
3.Case
1.2.3.2. λ
2
-
λ
1
,
λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1
,
λ
5
-
λ
3
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, and λ
3
-
λ
2
∈
B
3.The above two cases denote the3
-
1
-
3 distribution and 2
-
2
-
3 distribution. Case 1.2.3.1 is similar to Case 1.2.1.2. Case 1.2.3.2 is similar to Case 1.1.3.3. So Case 1.2.3 is proved, and so Case 1.2 is proved.Thus, the proof of Case1 is completed.


### 2.2. Case2

In this case, since(78)
λ
5
-
λ
4
=
(
λ
5
-
λ
1
)
-
(
λ
4
-
λ
1
)
∈
B
3
-
B
3
;
by applying Lemma 3(e) and (12), we get
(79)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
2
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
1
2
+
ℤ
.
From λ
5
-
λ
4
∈
B
1
∪
B
2, the discussion here can be divided into two cases: λ
5
-
λ
4
∈
B
1 and λ
5
-
λ
4
∈
B
2. That is, we have the following two subcases.Case
2.1. λ
2
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.Case
2.2. λ
2
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.The discussion of Case 2.1 is analogous to Case 2.2; it denotes the2
-
1
-
2 or 1
-
2
-
2 distribution. So we only need to deal with Case 2.1.Fromλ
3
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 2.1 can be divided into three cases.Case
2.1.1. λ
2
-
λ
1, λ
5
-
λ
4, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.Case
2.1.2. λ
2
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
3
-
λ
1, λ
3
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.Case
2.1.3. λ
2
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.The above three cases denote the3
-
1
-
2 distribution, 2
-
2
-
2 distribution, and 2
-
1
-
3 distribution. Case 2.1.1 is similar to Case 1.2.1, Case 2.1.2 is similar to Case 1.1.3, and Case 2.1.3 is similar to Case 1.2.3.So the proof of Case2 is completed.


### 2.3.  Case3

In this case, sinceλ
3
-
λ
2
=
(
λ
3
-
λ
1
)
-
(
λ
2
-
λ
1
)
∈
B
1
-
B
1, and λ
5
-
λ
4
=
(
λ
5
-
λ
1
)
-
(
λ
4
-
λ
1
)
∈
B
2
-
B
2, by Lemmas 3 and 4, we get
(80)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
2
∪
B
3
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
,
(81)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
3
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
ℤ
.Fromλ
3
-
λ
2
∈
B
2
∪
B
3, Case 3 can be divided into the following two cases.Case
3.1. λ
2
-
λ
1
,
λ
3
-
λ
1
∈
B
1 and λ
4
-
λ
1
,
λ
5
-
λ
1
,
λ
3
-
λ
2
∈
B
2.Case
3.2. λ
2
-
λ
1
,
λ
3
-
λ
1
∈
B
1
,
λ
4
-
λ
1
,
λ
5
-
λ
1
∈
B
2 and λ
3
-
λ
2
∈
B
3.Case 3.2 is similar to Case 1.1, so we only need to deal with Case 3.1.Fromλ
5
-
λ
4
∈
B
1
∪
B
3, the Case 3.1 can be divided into the following two cases.Case
3.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1 and λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
2.Case
3.1.2. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
2, and λ
5
-
λ
4
∈
B
3.Case 3.1.2 is similar to Case 1.1.2, so we only need to deal with Case 3.1.1.Consider the remainder differenceλ
4
-
λ
2 in (14). From λ
4
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 3.1.1 can be divided into three cases.Case
3.1.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4, λ
4
-
λ
2
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
2.Case
3.1.1.2. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2, and λ
4
-
λ
2
∈
B
2.Case
3.1.1.3. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
2, and λ
4
-
λ
2
∈
B
3.Case 3.1.1.2 is similar to Case 3.1.1.1, and Case 3.1.1.3 is similar to Case 1.1.1.1. So we only need to deal with Case 3.1.1.1. In this case, sinceλ
5
-
λ
2
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
2
)
∈
B
1
+
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(82)
λ
5
-
λ
2
=
(
x
5,2
,
y
5,2
,
z
5,2
)
t
∈
B
2
∪
B
3
,
x
5,2
∈
ℤ
,
y
5,2
∈
1
2
+
ℤ
.
Combined with (80), (82) yields
(83)
λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
)
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
,
x
5,3
∈
ℤ
,
y
5,3
∈
ℤ
,
a contradiction with Lemma 3(b). Case 3.1.1.1 is proved.So the proof of Case3 is completed.


### 2.4.  Case4

In this case, sinceλ
3
-
λ
2
=
(
λ
3
-
λ
1
)
-
(
λ
2
-
λ
1
)
∈
B
1
-
B
1, and λ
5
-
λ
4
=
(
λ
5
-
λ
1
)
-
(
λ
4
-
λ
1
)
∈
B
3
-
B
3, by Lemmas 3 and 4, we get
(84)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
2
∪
B
3
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
,
(85)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
2
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
1
2
+
ℤ
.Fromλ
3
-
λ
2
∈
B
2
∪
B
3, Case 4 can be divided into the following two cases.Case
4.1. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1 and λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
3.Case
4.2. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1, λ
3
-
λ
2
∈
B
2, and λ
4
-
λ
1, λ
5
-
λ
1
∈
B
3.Case 4.2 is similar to Case 1.2, so we only need to deal with Case 4.1.Fromλ
5
-
λ
4
∈
B
1
∪
B
2, Case 4.1 can be divided into the following two cases.Case
4.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1 and λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
4.1.2. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1, λ
5
-
λ
4
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case 4.1.2 is similar to Case 1.2.3, so we only need to deal with Case 4.1.1.Consider the remainder differenceλ
4
-
λ
2 in (14). According to λ
4
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 4.1.1 can be divided into three cases.Case
4.1.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4, λ
4
-
λ
2
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
4.1.1.2. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
2
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
4.1.1.3. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2, and λ
4
-
λ
2
∈
B
3.Case 4.1.1.2 is similar to Case 1.2.1.2. We will deal with the other two cases by considering the remainder differences in (14), and each case is concluded with a contradiction.Case
4.1.1.1. In this case, since λ
5
-
λ
2
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
2
)
∈
B
1
+
B
1, by Lemma 3(c), (12), and Lemma 4(ii), we get
(86)
λ
5
-
λ
2
=
(
x
5,2
,
y
5,2
,
z
5,2
)
t
∈
B
2
∪
B
3
,
x
5,2
∈
ℤ
,
y
5,2
∈
1
2
+
ℤ
,
combined with (84), (86), and λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
) yields x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction with Lemma 3(b).Case
4.1.1.3. In this case, since λ
4
-
λ
3
=
(
λ
4
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
3
-
B
3, by Lemma 3(e) and (12), we get
(87)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
∈
B
1
∪
B
2
,
x
4,3
∈
1
2
+
ℤ
,
y
4,3
∈
1
2
+
ℤ
,
combined with (85), (87), and λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
), yields x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction with Lemma 3(b). So Case 4.1.1.3 is proved, and Case 4.1.1 is proved.So the proof of Case4 is completed.Summing up the above discussion, we know that there exist at most4 mutually orthogonal exponential functions in L
2
(
μ
B
,
D
). We can find many such orthogonal systems which contain 4 elements, for example, the exponential function system E
(
S
) with S given by
(88)
S
=
{
(
0
0
0
)
,
(
p
1
2
p
1
2
0
)
,
(
0
p
1
2
p
1
2
)
,
(
p
1
2
0
p
1
2
)
}
.
This shows that the number 4 is the best. The proof of Theorem 1 is complete.The proof of Theorem2 is similar, it is so omitted.Corollary 5.
For the self-affine measureμ
B
,
D corresponding to
(89)
B
=
[
p
0
m
0
p
0
0
0
p
]
(
p
∈
ℤ
∖
{
0
,
±
1
}
)
,
m
∈
ℤ
,
D
=
{
(
0
0
0
)
,
(
1
0
0
)
,
(
0
1
0
)
,
(
0
0
1
)
}
,
if p
∈
(
2
ℤ
+
1
)
∖
{
±
1
}, then μ
B
,
D is a nonspectral measure, and there exist at most 4 mutually orthogonal exponential functions in L
2
(
μ
B
,
D
), where the number 4 is the best.The corollary improved Yuan [9, Theorem 1].


## 2.1. Case1

In this case, we have(20)
λ
5
-
λ
2
=
(
λ
5
-
λ
1
)
-
(
λ
2
-
λ
1
)
∈
B
1
-
B
1
.
Applying Lemma 3(c), (12), and Lemma 4(ii), we get
(21)
λ
5
-
λ
2
=
(
x
5,2
,
y
5,2
,
z
5,2
)
t
∈
B
2
∪
B
3
,
x
5,2
∈
ℤ
,
y
5,2
∈
1
2
+
ℤ
.
From λ
5
-
λ
2
∈
B
2
∪
B
3, Case 1 can be divided into two subcases.Case
1.1. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
2
∈
B
3.Step 1.
We will prove Case 1.1; sinceλ
3
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 1.1 can be divided into three cases.Case
1.1.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.1.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2, λ
3
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.1.3. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
3
-
λ
2
∈
B
3.We will give a method to deal with each case by considering the remainder differences in (14), and each case is concluded with a contradiction.Case
1.1.1. In this case, we have
(22)
λ
3
-
λ
1
=
(
λ
3
-
λ
2
)
+
(
λ
2
-
λ
1
)
∈
B
1
+
B
1
;
by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(23)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
ℤ
,
y
3,1
∈
1
2
+
ℤ
.By Lemma3(a) and Claim 1 we know that λ
4
-
λ
2
∉
B
1. So λ
4
-
λ
2
∈
B
2 or λ
4
-
λ
2
∈
B
3, so Case 1.1.1 can be divided into two subcases.Case
1.1.1.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2, λ
4
-
λ
2
∈
B
2, and λ
4
-
λ
1
∈
B
3.Case
1.1.1.2. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
4
-
λ
2
∈
B
3.By considering the remainder differences in (14), we apply Lemmas 3 and 4 to deal with each case.(I)
In Case 1.1.1.1, we have(24)
λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
)
∈
B
2
-
B
2
; by applying Lemma 3(d), (12) and Lemma 4(iii), we have
(25)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
3
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
ℤ
.
(i)
Ifλ
5
-
λ
4
∈
B
1, so λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1; by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(26)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
ℤ
,
y
4,1
∈
1
2
+
ℤ
.It follows from (23), (26), and λ
4
-
λ
3
=
(
λ
4
-
λ
1
)
-
(
λ
3
-
λ
1
) that
(27)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
,
x
4,3
∈
ℤ
,
y
4,3
∈
ℤ
,
which shows a contradiction with Lemma 3(b).(ii)
Ifλ
5
-
λ
4
∈
B
3, so λ
5
-
λ
1
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
1
)
∈
B
3
+
B
3; by applying Lemma 3(e) and (12), we get
(28)
λ
5
-
λ
1
=
(
x
5,1
,
y
5,1
,
z
5,1
)
t
∈
B
1
,
x
5,1
∈
1
2
+
ℤ
,
y
5,1
∈
1
2
+
ℤ
.Now, consider the remainder difference λ
4
-
λ
3 in (14): by Lemma 3(a) and Claim 1, we have λ
4
-
λ
3
∉
B
3, so λ
4
-
λ
3
∈
B
1
∪
B
2. If λ
4
-
λ
3
∈
B
1, then
(29)
λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
1
+
B
1
;
by applying Lemma 3(c), (12), and Lemma 4(ii) we get
(30)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
2
,
x
4,2
∈
ℤ
,
y
4,2
∈
1
2
+
ℤ
.
It follows from (21), (30), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
) that x
5,4
∈
ℤ
,
y
5,4
∈
ℤ, a contradiction with Lemma 3(b). If λ
4
-
λ
3
∈
B
2, then
(31)
λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2
;
by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(32)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
.
It follows from (25), (32), and λ
5
-
λ
1
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
1
) that x
5,1
∈
ℤ and y
5,1
∈
ℤ, a contradiction with Lemma 3(b).From parts (i), (ii), and (25), Case 1.1.1.1 is proved.(II)
In Case 1.1.1.2, we have(33)
λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3
; by applying Lemma 3(e) and (12), we get
(34)
λ
2
-
λ
1
=
(
x
2,1
,
y
2,1
,
z
2,1
)
t
∈
B
1
,
x
2,1
∈
1
2
+
ℤ
,
y
2,1
∈
1
2
+
ℤ
.We consider the remainder three differencesλ
4
-
λ
3, λ
5
-
λ
3, and λ
5
-
λ
4 in (14). By Claim 1, we have λ
4
-
λ
3
∉
B
3, so λ
4
-
λ
3
∈
B
1
∪
B
2.(i)
Ifλ
4
-
λ
3
∈
B
1, we have
(35)
λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
1
+
B
1
; by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(36)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
3
,
x
4,2
∈
ℤ
,
y
4,2
∈
1
2
+
ℤ
.
It follows from (21), (36), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
) that
(37)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
,
x
5,4
∈
ℤ
,
y
5,4
∈
ℤ
,
a contradiction with Lemma 3(b).(ii)
Ifλ
4
-
λ
3
∈
B
2, we have
(38)
λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2
; by applying Lemma 3(d), (12), and Lemma 4(iii) we have
(39)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
.
We consider the remainder difference λ
5
-
λ
4 in (14). By Claim 1, we have λ
5
-
λ
4
∉
B
3, so λ
5
-
λ
4
∈
B
1
∪
B
2. If λ
5
-
λ
4
∈
B
1, we have
(40)
λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1
,
by applying Lemma 3(c), (12), and Lemma 4(ii), we get x
4,1
∈
ℤ, a contradiction with (39). If λ
5
-
λ
4
∈
B
2, we have
(41)
λ
4
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
4
)
∈
B
2
-
B
2
;
by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(42)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
3
,
x
4,2
∈
1
2
+
ℤ
,
y
4,2
∈
ℤ
.
It follows from (39), (42), and λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
) that x
2,1
∈
ℤ, a contradiction with (34).The above (i) and (ii) in (II) illustrate that Case 1.1.1.2 is proved. Hence Case 1.1.1 is proved.Case
1.1.2. In this case, we have
(43)
λ
2
-
λ
1
=
(
λ
3
-
λ
1
)
-
(
λ
3
-
λ
2
)
∈
B
2
-
B
2
,
λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
2
-
B
2
;
by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(44)
λ
2
-
λ
1
=
(
x
2,1
,
y
2,1
,
z
2,1
)
t
∈
B
1
,
x
2,1
∈
1
2
+
ℤ
,
y
2,1
∈
ℤ
,
(45)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
∈
B
1
∪
B
3
,
x
5,3
∈
1
2
+
ℤ
,
y
5,3
∈
ℤ
.
By applying Lemma 3(a) and Claim 1, we get
(46)
λ
4
-
λ
2
∉
B
2
,
λ
4
-
λ
3
∉
B
2
.Ifλ
4
-
λ
2
∈
B
3, then λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3; by Lemma 3(e), and (12), we get x
2,1
∈
(
1
/
2
)
+
ℤ, y
2,1
∈
(
1
/
2
)
+
ℤ, a contradiction with (44); hence
(47)
λ
4
-
λ
2
∉
B
3
,
λ
4
-
λ
2
∈
B
1
.
We consider λ
4
-
λ
3; if λ
4
-
λ
3
∈
B
1, we get λ
3
-
λ
2
=
(
λ
4
-
λ
2
)
-
(
λ
4
-
λ
3
)
∈
B
1
-
B
1; by Lemma 3(c), (12), and Lemma 4(ii), we have
(48)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
2
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
,
which, combined with (21) and λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
), shows that x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction with Lemma 3(b); hence
(49)
λ
4
-
λ
3
∉
B
1
,
λ
4
-
λ
3
∈
B
3
.Sinceλ
3
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
3
)
∈
B
3
-
B
3, by Lemma 3(e) and (12), we get
(50)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
1
2
+
ℤ
,
y
3,1
∈
1
2
+
ℤ
.Combined with (21), (44), and λ
5
-
λ
1
=
(
λ
5
-
λ
2
)
+
(
λ
2
-
λ
1
), we get
(51)
λ
5
-
λ
1
=
(
x
5,1
,
y
5,1
,
z
5,1
)
t
,
x
5,1
∈
1
2
+
ℤ
,
y
5,1
∈
1
2
+
ℤ
,
which, combined with (50), (51), and λ
5
-
λ
3
=
(
λ
5
-
λ
1
)
-
(
λ
3
-
λ
1
), shows that x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction of (45). This proves Case 1.1.2.Case
1.1.3. From λ
4
-
λ
2
∈
B
1
∪
B
2
∪
B
3, this Case can be divided into the three cases.Case
1.1.3.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
4
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
1.1.3.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2, λ
4
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
1.1.3.3. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
2
∈
B
2, λ
4
-
λ
1, λ
3
-
λ
2, and λ
4
-
λ
2
∈
B
3.The above three cases denote the3
-
2
-
2 or 2
-
3
-
2 or 2
-
2
-
3 distribution. In each case, we have (21). The discussion of the first two cases is similar. In the following, we use the above method to deal with Cases 1.1.3.1 and 1.1.3.3.In Case 1.1.3.1, since(52)
λ
4
-
λ
1
=
(
λ
4
-
λ
2
)
+
(
λ
2
-
λ
1
)
∈
B
1
+
B
1
,
by applying Lemma 3(c), (12), and Lemma 4(ii), we have
(53)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
ℤ
,
y
4,1
∈
1
2
+
ℤ
.We consider the remainder three differencesλ
4
-
λ
3, λ
5
-
λ
3, and λ
5
-
λ
4. From (21) and (53) we have
(54)
λ
4
-
λ
3
∉
B
2
or
λ
4
-
λ
3
∈
B
1
∪
B
3
,
(55)
λ
5
-
λ
3
∉
B
3
or
λ
5
-
λ
3
∈
B
1
∪
B
2
.Ifλ
4
-
λ
3
∈
B
2, by Lemma 3, so λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2; we get y
4,1
∈
ℤ, a contradiction with (53). If λ
5
-
λ
3
∈
B
3, so λ
5
-
λ
2
=
(
λ
5
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
3
+
B
3 by Lemma 3(e); we get x
5,2
∈
(
1
/
2
)
+
ℤ, a contradiction with (21). Hence (54) and (55) hold.According to (54), we deal with the following two cases.(i)
Ifλ
4
-
λ
3
∈
B
1, then, from λ
3
-
λ
2
=
(
λ
4
-
λ
2
)
-
(
λ
4
-
λ
3
)
∈
B
1
-
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(56)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
3
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
.From (21), (56), and λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
), we get
(57)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
,
x
5,3
∈
ℤ
,
y
5,3
∈
ℤ
,
a contradiction with Lemma 3(b).(ii)
Ifλ
4
-
λ
3
∈
B
3, since λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
3
+
B
3 and λ
3
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
3
)
∈
B
3
-
B
3, by applying Lemma 3(e) and (12), we get
(58)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
1
2
+
ℤ
,
y
3,1
∈
1
2
+
ℤ
.We consider (55); if λ
5
-
λ
3
∈
B
1, by Lemma 3 and λ
3
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
3
)
∈
B
1
-
B
1, we get x
3,1
∈
ℤ, a contradiction with (58). If λ
5
-
λ
3
∈
B
2, since λ
3
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
3
)
∈
B
2
-
B
2, by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(59)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
3
,
x
3,2
∈
1
2
+
ℤ
,
y
3,2
∈
ℤ
.
From (58), (59) and λ
2
-
λ
1
=
(
λ
3
-
λ
1
)
-
(
λ
3
-
λ
2
), we have x
2,1
∈
ℤ, a contradiction with Lemma 4(i) (for λ
2
-
λ
1
∈
B
1 gives x
2,1
∈
(
1
/
2
)
+
ℤ).The above (i), (ii), and (54) indicae that Case 1.1.3.1 is proved.In Case 1.1.3.3, we get(60)
λ
4
-
λ
3
=
(
λ
4
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
3
-
B
3
,
λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3
;
by applying Lemma 3(e) and (12), we have
(61)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
∈
B
1
∪
B
2
,
x
4,3
∈
1
2
+
ℤ
,
y
4,3
∈
1
2
+
ℤ
,
(62)
λ
2
-
λ
1
=
(
x
2,1
,
y
2,1
,
z
2,1
)
t
∈
B
1
,
x
2,1
∈
1
2
+
ℤ
,
y
2,1
∈
1
2
+
ℤ
.
By Lemma 3(a) and λ
1
-
λ
4, λ
2
-
λ
4
∈
B
3, we know from Claim 1 that λ
5
-
λ
4
∉
B
3. So λ
5
-
λ
4
∈
B
1 or λ
5
-
λ
4
∈
B
2.From (61), we consider the following two cases.(i)
Ifλ
4
-
λ
3
∈
B
1, we consider λ
5
-
λ
4, if λ
5
-
λ
4
∈
B
1. From λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
)
∈
B
1
+
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(63)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
∈
B
2
∪
B
3
,
x
5,3
∈
ℤ
,
y
5,3
∈
1
2
+
ℤ
.From (21), (63), and λ
3
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
3
), we get x
2,1
∈
ℤ
,
y
2,1
∈
ℤ, a contradiction with Lemma 3(b). If λ
5
-
λ
4
∈
B
2, since λ
4
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
4
)
∈
B
2
-
B
2, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(64)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
1
∪
B
3
,
x
4,2
∈
1
2
+
ℤ
,
y
4,2
∈
ℤ
.
From (21), (64), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
), we get that x
5,4
∈
(
1
/
2
)
+
ℤ
,
y
5,4
∈
(
1
/
2
)
+
ℤ, combined with (61) and λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
), yields x
5,4
∈
ℤ
,
y
5,4
∈
ℤ, a contradiction with Lemma 3(b).(ii)
Ifλ
4
-
λ
3
∈
B
2, so λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2; by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(65)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
. If λ
5
-
λ
4
∈
B
1, by Lemma 3(c) and λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1, we get x
4,1
∈
ℤ, a contradiction with (65). If λ
5
-
λ
4
∈
B
2, so λ
4
-
λ
2
=
(
λ
5
-
λ
2
)
-
(
λ
5
-
λ
4
)
∈
B
2
-
B
2, by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(66)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
3
,
x
4,2
∈
1
2
+
ℤ
,
y
4,2
∈
ℤ
.From (65), (66), and λ
2
-
λ
1
=
(
λ
4
-
λ
1
)
-
(
λ
4
-
λ
2
), we get x
2,1
∈
ℤ, a contradiction with (62).The above (i), (ii), and (61) illustrate that Case 1.1.3.3 is proved and Case 1.1 is proved.Step 2.
In Case 1.2, fromλ
3
-
λ
2
∈
B
1
∪
B
2
∪
B
3, this case can be divided into three cases.Case
1.2.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
2
∈
B
3.Case
1.2.2. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
3
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
2
∈
B
3.Case
1.2.3. λ
2
-
λ
1, λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, and λ
3
-
λ
2
∈
B
3.Case 1.2.2 is similar to Case 1.1.3; we will deal with the other two cases by considering the remainder differences in (14), and each case is concluded with a contradiction.Case
1.2.1. In this case, we get
(67)
λ
3
-
λ
1
=
(
λ
3
-
λ
2
)
+
(
λ
2
-
λ
1
)
∈
B
1
+
B
1
;
by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(68)
λ
3
-
λ
1
=
(
x
3,1
,
y
3,1
,
z
3,1
)
t
∈
B
2
,
x
3,1
∈
ℤ
,
y
3,1
∈
1
2
+
ℤ
.By Lemma3(a) and λ
1
-
λ
2
,
λ
3
-
λ
2
∈
B
1, we know from Claim 1 that λ
4
-
λ
2
∉
B
1. So λ
4
-
λ
2
∈
B
2 or λ
4
-
λ
2
∈
B
3, so Case 1.2.1 can be divided into two cases.Case
1.2.1.1. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1, λ
4
-
λ
2
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2
∈
B
3.Case
1.2.1.2. λ
2
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, λ
4
-
λ
2
∈
B
3.The above two cases denote the3
-
2
-
2 distribution and 3
-
1
-
3 distribution. Case 1.2.1.1 is similar to Case 1.1.3.1. So we only need to deal with Case 1.2.1.2. By considering the remainder differences in (14), we apply Lemmas 3 and 4 to deal with the case.In Case 1.2.1.2, we get(69)
λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
)
∈
B
3
-
B
3
;
by applying Lemma 3(e) and (12), we get
(70)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
2
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
1
2
+
ℤ
.By Claim1, λ
4
-
λ
3
∉
B
3; then λ
4
-
λ
3
∈
B
1 or λ
4
-
λ
3
∈
B
2.From (70), we consider the following two cases.(i)
Ifλ
5
-
λ
4
∈
B
1, since λ
4
-
λ
1
=
(
λ
5
-
λ
1
)
-
(
λ
5
-
λ
4
)
∈
B
1
-
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii) we get
(71)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
ℤ
,
y
4,1
∈
1
2
+
ℤ
.It follows from (68), (71), and λ
4
-
λ
3
=
(
λ
4
-
λ
1
)
-
(
λ
3
-
λ
1
) that
(72)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
,
x
4,3
∈
ℤ
,
y
4,3
∈
ℤ
,
which shows a contradiction with Lemma 3(b).(ii)
Ifλ
5
-
λ
4
∈
B
2, we consider λ
4
-
λ
3, if λ
4
-
λ
3
∈
B
1. Since λ
4
-
λ
2
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
2
)
∈
B
1
+
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(73)
λ
4
-
λ
2
=
(
x
4,2
,
y
4,2
,
z
4,2
)
t
∈
B
2
∪
B
3
,
x
4,2
∈
ℤ
,
y
4,2
∈
1
2
+
ℤ
.From (21), (73), and λ
5
-
λ
4
=
(
λ
5
-
λ
2
)
-
(
λ
4
-
λ
2
), we get x
5,4
∈
ℤ, y
5,4
∈
ℤ, a contradiction with Lemma 3(b). If λ
4
-
λ
3
∈
B
2, then λ
4
-
λ
1
=
(
λ
4
-
λ
3
)
+
(
λ
3
-
λ
1
)
∈
B
2
+
B
2; by applying Lemma 3(d), (12), and Lemma 4(iii), we get
(74)
λ
4
-
λ
1
=
(
x
4,1
,
y
4,1
,
z
4,1
)
t
∈
B
3
,
x
4,1
∈
1
2
+
ℤ
,
y
4,1
∈
ℤ
.It follows from (68), (74), and λ
4
-
λ
3
=
(
λ
4
-
λ
1
)
-
(
λ
3
-
λ
1
) that we get
(75)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
∈
B
2
,
x
4,3
∈
1
2
+
ℤ
,
y
4,3
∈
1
2
+
ℤ
.From (70), (75), and λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
), we get x
5,4
∈
ℤ, y
5,4
∈
ℤ, a contradiction with Lemma 3(b).Parts (i), (ii), and (70) indicate that Case 1.2.1.2 is proved and Case 1.2.1 is proved.Case
1.2.3. In this case, since
(76)
λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
3
-
B
3
,
by applying Lemma 3(e) and (12), we get
(77)
λ
5
-
λ
3
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
∈
B
1
∪
B
2
,
x
5,3
∈
1
2
+
ℤ
,
y
5,3
∈
1
2
+
ℤ
.From (77) Case 1.2.3 can be divided into two cases.Case
1.2.3.1. λ
2
-
λ
1
,
λ
5
-
λ
1
,
λ
5
-
λ
3
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, and λ
3
-
λ
2
∈
B
3.Case
1.2.3.2. λ
2
-
λ
1
,
λ
5
-
λ
1
∈
B
1, λ
3
-
λ
1
,
λ
5
-
λ
3
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
2, and λ
3
-
λ
2
∈
B
3.The above two cases denote the3
-
1
-
3 distribution and 2
-
2
-
3 distribution. Case 1.2.3.1 is similar to Case 1.2.1.2. Case 1.2.3.2 is similar to Case 1.1.3.3. So Case 1.2.3 is proved, and so Case 1.2 is proved.Thus, the proof of Case1 is completed.


## 2.2. Case2

In this case, since(78)
λ
5
-
λ
4
=
(
λ
5
-
λ
1
)
-
(
λ
4
-
λ
1
)
∈
B
3
-
B
3
;
by applying Lemma 3(e) and (12), we get
(79)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
2
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
1
2
+
ℤ
.
From λ
5
-
λ
4
∈
B
1
∪
B
2, the discussion here can be divided into two cases: λ
5
-
λ
4
∈
B
1 and λ
5
-
λ
4
∈
B
2. That is, we have the following two subcases.Case
2.1. λ
2
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.Case
2.2. λ
2
-
λ
1
∈
B
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.The discussion of Case 2.1 is analogous to Case 2.2; it denotes the2
-
1
-
2 or 1
-
2
-
2 distribution. So we only need to deal with Case 2.1.Fromλ
3
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 2.1 can be divided into three cases.Case
2.1.1. λ
2
-
λ
1, λ
5
-
λ
4, λ
3
-
λ
2
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.Case
2.1.2. λ
2
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
3
-
λ
1, λ
3
-
λ
2
∈
B
2, λ
4
-
λ
1, and λ
5
-
λ
1
∈
B
3.Case
2.1.3. λ
2
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
3
-
λ
1
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.The above three cases denote the3
-
1
-
2 distribution, 2
-
2
-
2 distribution, and 2
-
1
-
3 distribution. Case 2.1.1 is similar to Case 1.2.1, Case 2.1.2 is similar to Case 1.1.3, and Case 2.1.3 is similar to Case 1.2.3.So the proof of Case2 is completed.


## 2.3.  Case3

In this case, sinceλ
3
-
λ
2
=
(
λ
3
-
λ
1
)
-
(
λ
2
-
λ
1
)
∈
B
1
-
B
1, and λ
5
-
λ
4
=
(
λ
5
-
λ
1
)
-
(
λ
4
-
λ
1
)
∈
B
2
-
B
2, by Lemmas 3 and 4, we get
(80)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
2
∪
B
3
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
,
(81)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
3
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
ℤ
.Fromλ
3
-
λ
2
∈
B
2
∪
B
3, Case 3 can be divided into the following two cases.Case
3.1. λ
2
-
λ
1
,
λ
3
-
λ
1
∈
B
1 and λ
4
-
λ
1
,
λ
5
-
λ
1
,
λ
3
-
λ
2
∈
B
2.Case
3.2. λ
2
-
λ
1
,
λ
3
-
λ
1
∈
B
1
,
λ
4
-
λ
1
,
λ
5
-
λ
1
∈
B
2 and λ
3
-
λ
2
∈
B
3.Case 3.2 is similar to Case 1.1, so we only need to deal with Case 3.1.Fromλ
5
-
λ
4
∈
B
1
∪
B
3, the Case 3.1 can be divided into the following two cases.Case
3.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1 and λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
2.Case
3.1.2. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
2, and λ
5
-
λ
4
∈
B
3.Case 3.1.2 is similar to Case 1.1.2, so we only need to deal with Case 3.1.1.Consider the remainder differenceλ
4
-
λ
2 in (14). From λ
4
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 3.1.1 can be divided into three cases.Case
3.1.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4, λ
4
-
λ
2
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
2.Case
3.1.1.2. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2, and λ
4
-
λ
2
∈
B
2.Case
3.1.1.3. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
2, and λ
4
-
λ
2
∈
B
3.Case 3.1.1.2 is similar to Case 3.1.1.1, and Case 3.1.1.3 is similar to Case 1.1.1.1. So we only need to deal with Case 3.1.1.1. In this case, sinceλ
5
-
λ
2
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
2
)
∈
B
1
+
B
1, by applying Lemma 3(c), (12), and Lemma 4(ii), we get
(82)
λ
5
-
λ
2
=
(
x
5,2
,
y
5,2
,
z
5,2
)
t
∈
B
2
∪
B
3
,
x
5,2
∈
ℤ
,
y
5,2
∈
1
2
+
ℤ
.
Combined with (80), (82) yields
(83)
λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
)
=
(
x
5,3
,
y
5,3
,
z
5,3
)
t
,
x
5,3
∈
ℤ
,
y
5,3
∈
ℤ
,
a contradiction with Lemma 3(b). Case 3.1.1.1 is proved.So the proof of Case3 is completed.


## 2.4.  Case4

In this case, sinceλ
3
-
λ
2
=
(
λ
3
-
λ
1
)
-
(
λ
2
-
λ
1
)
∈
B
1
-
B
1, and λ
5
-
λ
4
=
(
λ
5
-
λ
1
)
-
(
λ
4
-
λ
1
)
∈
B
3
-
B
3, by Lemmas 3 and 4, we get
(84)
λ
3
-
λ
2
=
(
x
3,2
,
y
3,2
,
z
3,2
)
t
∈
B
2
∪
B
3
,
x
3,2
∈
ℤ
,
y
3,2
∈
1
2
+
ℤ
,
(85)
λ
5
-
λ
4
=
(
x
5,4
,
y
5,4
,
z
5,4
)
t
∈
B
1
∪
B
2
,
x
5,4
∈
1
2
+
ℤ
,
y
5,4
∈
1
2
+
ℤ
.Fromλ
3
-
λ
2
∈
B
2
∪
B
3, Case 4 can be divided into the following two cases.Case
4.1. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1 and λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2
∈
B
3.Case
4.2. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1, λ
3
-
λ
2
∈
B
2, and λ
4
-
λ
1, λ
5
-
λ
1
∈
B
3.Case 4.2 is similar to Case 1.2, so we only need to deal with Case 4.1.Fromλ
5
-
λ
4
∈
B
1
∪
B
2, Case 4.1 can be divided into the following two cases.Case
4.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1 and λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
4.1.2. λ
2
-
λ
1, λ
3
-
λ
1
∈
B
1, λ
5
-
λ
4
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case 4.1.2 is similar to Case 1.2.3, so we only need to deal with Case 4.1.1.Consider the remainder differenceλ
4
-
λ
2 in (14). According to λ
4
-
λ
2
∈
B
1
∪
B
2
∪
B
3, Case 4.1.1 can be divided into three cases.Case
4.1.1.1. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4, λ
4
-
λ
2
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
4.1.1.2. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
2
∈
B
2, λ
4
-
λ
1, λ
5
-
λ
1, and λ
3
-
λ
2
∈
B
3.Case
4.1.1.3. λ
2
-
λ
1, λ
3
-
λ
1, λ
5
-
λ
4
∈
B
1, λ
4
-
λ
1, λ
5
-
λ
1, λ
3
-
λ
2, and λ
4
-
λ
2
∈
B
3.Case 4.1.1.2 is similar to Case 1.2.1.2. We will deal with the other two cases by considering the remainder differences in (14), and each case is concluded with a contradiction.Case
4.1.1.1. In this case, since λ
5
-
λ
2
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
2
)
∈
B
1
+
B
1, by Lemma 3(c), (12), and Lemma 4(ii), we get
(86)
λ
5
-
λ
2
=
(
x
5,2
,
y
5,2
,
z
5,2
)
t
∈
B
2
∪
B
3
,
x
5,2
∈
ℤ
,
y
5,2
∈
1
2
+
ℤ
,
combined with (84), (86), and λ
5
-
λ
3
=
(
λ
5
-
λ
2
)
-
(
λ
3
-
λ
2
) yields x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction with Lemma 3(b).Case
4.1.1.3. In this case, since λ
4
-
λ
3
=
(
λ
4
-
λ
2
)
-
(
λ
3
-
λ
2
)
∈
B
3
-
B
3, by Lemma 3(e) and (12), we get
(87)
λ
4
-
λ
3
=
(
x
4,3
,
y
4,3
,
z
4,3
)
t
∈
B
1
∪
B
2
,
x
4,3
∈
1
2
+
ℤ
,
y
4,3
∈
1
2
+
ℤ
,
combined with (85), (87), and λ
5
-
λ
3
=
(
λ
5
-
λ
4
)
+
(
λ
4
-
λ
3
), yields x
5,3
∈
ℤ, y
5,3
∈
ℤ, a contradiction with Lemma 3(b). So Case 4.1.1.3 is proved, and Case 4.1.1 is proved.So the proof of Case4 is completed.Summing up the above discussion, we know that there exist at most4 mutually orthogonal exponential functions in L
2
(
μ
B
,
D
). We can find many such orthogonal systems which contain 4 elements, for example, the exponential function system E
(
S
) with S given by
(88)
S
=
{
(
0
0
0
)
,
(
p
1
2
p
1
2
0
)
,
(
0
p
1
2
p
1
2
)
,
(
p
1
2
0
p
1
2
)
}
.
This shows that the number 4 is the best. The proof of Theorem 1 is complete.The proof of Theorem2 is similar, it is so omitted.Corollary 5.
For the self-affine measureμ
B
,
D corresponding to
(89)
B
=
[
p
0
m
0
p
0
0
0
p
]
(
p
∈
ℤ
∖
{
0
,
±
1
}
)
,
m
∈
ℤ
,
D
=
{
(
0
0
0
)
,
(
1
0
0
)
,
(
0
1
0
)
,
(
0
0
1
)
}
,
if p
∈
(
2
ℤ
+
1
)
∖
{
±
1
}, then μ
B
,
D is a nonspectral measure, and there exist at most 4 mutually orthogonal exponential functions in L
2
(
μ
B
,
D
), where the number 4 is the best.The corollary improved Yuan [9, Theorem 1].



---

*Source: 294182-2014-01-08.xml*