Non-Spectrality of Certain Self-Affine Measures on the Generalized Spatial Sierpinski Gasket

Yongli Hu, Zhicheng Zhang, Qi Wang

School of Mathematics and Data Science, Shaanxi University of Science & Technology, Xi’an, China.

DOI: 10.4236/jamp.2024.1211241                

Abstract

Let ${\mu }_{M\mathrm{,}D}$ be a self-affine measure associated with an expanding integer matrix $M=\left[p_1\mathrm{,0,0<mo>;</mo>}{p}_4\mathrm{,}{p}_2\mathrm{,0<mo>;</mo>}{p}_5\mathrm{,0,}{p}_3\right]$ and the digit set $D=\left\{\mathrm{0,}{e}_1\mathrm{,}{e}_2\mathrm{,}{e}_3\right\}$ in the space $R^3$ , where $p_1\mathrm{,}{p}_2\mathrm{,}{p}_3\in Z\backslash \left\{\mathrm{0,}\pm 1\right\}$ , $p_4\mathrm{,}{p}_5\in Z$ and $e_1\mathrm{,}{e}_2\mathrm{,}{e}_3$ are the standard basis of unit column vectors in $R^3$ . In this paper, we mainly consider the case $p_1\mathrm{,}{p}_2\mathrm{,}{p}_3\in 2Z+\mathrm{1,}{p}_2\ne p_3\mathrm{,}{p}_4=l\left(p_1-p_2\right)\mathrm{,}{p}_5=l\left(p_3-p_1\right)\mathrm{,}$ where $l\in 2Z$ . We prove that ${\mu }_{M\mathrm{,}D}$ is a non-spectral measure, and there are at most 4-element ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials, and the number 4 is the best. The results here generalize the known results.

Keywords

Sierpinski Gasket, Non-Spectrality, Orthogonal Exponentials, Digit Set

1. Introduction

Let $M\in M_n\left(Z\right)$ be an expanding integer matrix (that is, all the eigenvalues $\left|{\lambda }_i\left(M\right)\right|>1$ ) and $D\subset Z^n$ be a finite subset of cardinality $\left|D\right|$ . The unique probability measure $\mu \mathrm{<mo>:</mo>}={\mu }_{M\mathrm{,}D}$ satisfying the self-affine identity

$\mu =\frac{1}{\left|D\right|}{\displaystyle \sum _{d\in D}}\mu \circ {\varphi }_d^{-1}$ (1.1)

with equal weight, where ${\varphi }_d\left(\xi \right)=M^{-1}\left(\xi +d\right)\left(\xi \in R^n\right)$ forms an affine iterated function system (IFS) ${\left\{{\varphi }_d\right\}}_{d\in D}$ . ${\mu }_{M\mathrm{,}D}$ is also called invariant measure or self-similar measure. Such μ is supported on an invariant set $T\left(M\mathrm{,}D\right)$ (see [1]), which is a unique nonempty compact set satisfying

$MT={\displaystyle \underset{d\in D}{\cup }}\left(T+d\right)\mathrm{.}$

If there exists a set $\Lambda \subset R^n$ such that $E\left(\Lambda \right)\mathrm{<mo>:</mo>}=\left\{{\text{e}}^{2\pi i\langle \lambda \mathrm{,}x\rangle }\mathrm{<mo>:</mo>}\lambda \in \Lambda \right\}$ forms an orthogonal basis (Fourier basis) for the Hilbert space $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , we call ${\mu }_{M\mathrm{,}D}$ a spectral measure. The set Λ is then called a spectrum for ${\mu }_{M\mathrm{,}D}$ and the pair $\left(\mu \mathrm{,}\Lambda \right)$ is called a spectrum pair. The study of spectral measures dates back to the work of Fuglede [2], whose famous spectral-tiling conjecture relation is difficult to establish in most cases. The research on the spectrality or non-spectrality of ${\mu }_{M\mathrm{,}D}$ become a hot topic, which has its origin in number theory, harmonic analysis, fractal geometry and dynamical systems [3]-[6].

Jorgensen and Pedersen found the first non-atomic, singular continuous spectral measure, which showed that the Fourier transform theory can be applied to certain classes of fractals [3] [7]. In all these studies, the Fourier transform ${\widehat{\mu }}_{M\mathrm{,}D}$ of ${\mu }_{M\mathrm{,}D}$ plays an important role. From (1.1), ${\widehat{\mu }}_{M\mathrm{,}D}\left(\xi \right)$ is given by

${\widehat{\mu }}_{M\mathrm{,}D}\left(\xi \right)={\displaystyle \int {\text{e}}^{2\pi i\langle x\mathrm{,}\xi \rangle }\text{d}{\mu }_{M\mathrm{,}D}\left(x\right)}={\displaystyle \prod _{j=1}^{\infty }}{m}_D\left(M^{\ast -j}\xi \right)\mathrm{,}\left(\xi \in R^n\right)$

where $M^{\ast }$ denotes the transposed conjugate of M and

$m_D\left(\xi \right)=\frac{1}{\left|D\right|}{\displaystyle \sum _{d\in D}}{\text{e}}^{2\pi i\langle d\mathrm{,}\xi \rangle }\mathrm{,}\left(\xi \in R^n\right)$

denotes the symbol function of D. It is known that there are several methods to deal with the spectrality of self-affine measure (see [8]-[11] and references cited therein). Compared with spectral self-affine measures, there are a lot of non-spectral self-affine measures.

Usually, the orthogonal exponentials in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ is called ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials. Generally, the non-spectral problem can be divided into the following two classes.

Class (I): There are at most a finite number of orthogonal exponentials in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ (but the supremum of these finite numbers may be infinite, see [12] [13] and references cited therein), that is, ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials contain at most finite elements. The main questions here are to estimate the number of orthogonal exponentials in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ and to find them (see [14] [15]).

Class (II): There are natural infinite families of orthogonal exponentials in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , but none of them forms an orthogonal basis in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ . The main question here is whether some of these families can be combined to form larger collections of orthogonal exponentials (see [16]-[18]).

For the generalized spatial Sierpinski gasket corresponding to

$M=\left[\begin{array}{ccc}{p}_1& 0& 0\\ p_4& p_2& 0\\ p_5& p_6& p_3\end{array}\right]\text{and}D=\left\{\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}\mathrm{,}$ (1.2)

where $p_1\mathrm{,}{p}_2\mathrm{,}{p}_3\in Z\backslash \left\{\mathrm{0,}\pm 1\right\}$ and $p_4\mathrm{,}{p}_5\mathrm{,}{p}_6\in Z$ . The previous results on the non-spectrality of self-affine measures can be summarized as the following.

Theorem 1.1. For self-affine measure ${\mu }_{M\mathrm{,}D}$ given by (1.2) and $p_4=p_5=p_6=0$ , the following spectrality and non-spectrality hold (see [12] [19]-[22]):

1) If $p_1\mathrm{,}{p}_2\mathrm{,}{p}_3\in \left(2Z+1\right)\backslash \left\{\pm 1\right\}$ , then ${\mu }_{M\mathrm{,}D}$ is a non-spectral measure, and there exist at most 4 mutually orthogonal exponential functions in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , and the number 4 is the best;

2) If $p_1\in 2Z\backslash \left\{0\right\}$ and $p_2=p_3\in \left(2Z+1\right)\backslash \left\{\pm 1\right\}$ , then there exist at most 4 mutually orthogonal exponential functions in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , and the number 4 is the best;

3) If $p_1\in 2Z\backslash \left\{0\right\}$ and $p_2=-p_3\in \left(2Z+1\right)\backslash \left\{\pm 1\right\}$ , then there exist at most 8 mutually orthogonal exponential functions in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , where the number 8 is the best possible;

4) If $p_1\in 2Z\backslash \left\{0\right\}$ and $p_2\ne \pm p_3\in \left(2Z+1\right)\backslash \left\{\pm 1\right\}$ , then for any $l\in N$ , there exist $2l+6$ mutually orthogonal exponential functions in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ .

Theorem 1.2. [23] Let ${\mu }_{M\mathrm{,}D}$ corresponding to (1.2), if $p_1\mathrm{,}{p}_2\mathrm{,}{p}_3\in \left(2Z+1\right)\backslash \left\{\pm 1\right\}$ and

$p_4=p_1-p_2,\gamma p_6=\eta \left(p_3-p_2\right),\gamma \left(p_5-p_1+p_3\right)=\eta \left(p_2-p_1\right)$ (1.3)

for some $\beta \mathrm{,}\gamma \in 2Z+1$ and $\eta \in 2Z$ , then ${\mu }_{M^{\mathrm{<mo>*</mo>}}\mathrm{,}D}$ is a non-spectral measure, and there exist at most 4 mutually orthogonal exponential functions in $L^2\left({\mu }_{M^{\mathrm{<mo>*</mo>}}\mathrm{,}D}\right)$ , where the number 4 is the best.

Theorem 1.3. [24] Let ${\mu }_{M\mathrm{,}D}$ corresponding to (1.2) and $p_6=0$ , if

$p_1\mathrm{,}{p}_2=p_3\in \left(2Z+1\right)\backslash \left\{\mathrm{0,}\pm 1\right\}\mathrm{,}$ (1.4)

then ${\mu }_{M\mathrm{,}D}$ is a non-spectral measure, and there exist at most 4 mutually orthogonal exponential functions in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , where the number 4 is the best.

Theorem 1.4. [25] Let ${\mu }_{M\mathrm{,}D}$ corresponding to (1.2) and $p_4=0$ , $p_5=-p_6=\xi \in R$ , if

$p_1\mathrm{<mo>=</mo>}{p}_2\mathrm{,}{p}_3\in \left\{\frac{p}{q}\mathrm{<mo>:</mo>}p\mathrm{,}q\in 2Z+1\right\}\mathrm{,}$ (1.5)

then ${\mu }_{M^{\mathrm{<mo>*</mo>}}\mathrm{,}D}$ is a non-spectral measure, and there exist at most 4 mutually orthogonal exponential functions in $L^2\left({\mu }_{M^{\mathrm{<mo>*</mo>}}\mathrm{,}D}\right)$ , where the number 4 is the best.

In the case $p_6=0$ , one can rewrite (1.3) as follows

$p_4=p_1-p_2,p_5=p_1-p_3.$ (1.6)

From Theorem 1.2, we know that ${\mu }_{M\mathrm{,}D}$ is a non-spectral measure in the case (1.6), and there exist at most 4 mutually orthogonal exponential functions in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , where the number 4 is the best. In Theorem 1.3, Yuan is given the exact number of ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials in the case (1.5). This naturally leads to an open problem: How about the spectrality or non-spectrality of ${\mu }_{M\mathrm{,}D}$ in the case

$p_1\mathrm{,}{p}_2\mathrm{,}{p}_3\in \left(2Z+1\right)\backslash \left\{\mathrm{0,}\pm 1\right\}\mathrm{,}{p}_2\ne p_3?$ (1.7)

Motivated by this, in the present paper, we resolve the above question in some cases. The main result of the paper is the following.

Theorem 1.5. Let the self-affine measure ${\mu }_{M\mathrm{,}D}$ corresponding to (1.2) (1.7) and $p_6=0$ , if

$p_4=l\left(p_1-p_2\right),p_5=l\left(p_3-p_1\right),$ (1.8)

where $l\in 2Z$ , then ${\mu }_{M\mathrm{,}D}$ is a non-spectral measure, there are at most 4-element ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials, where the number 4 is the best.

2. The Zero Set $Z\left({\widehat{\mu }}_{M,D}\right)$ of ${\mu }_{M,D}$

For the given digit set D in (1.2), it is known [19] that the zero set

$Z\left(m_D\right)\mathrm{<mo>:</mo>}=\left\{x\in R^3\mathrm{<mo>:</mo>}{m}_D\left(x\right)=0\right\}=Z_1\cup Z_2\cup Z_3\mathrm{,}$

where

$Z_1=\left\{\left(\begin{array}{c}\frac{1}{2}+k_1\\ a+k_2\\ \frac{1}{2}+a+k_3\end{array}\right)\mathrm{<mo>:</mo>}a\in R\mathrm{,}{k}_1\mathrm{,}{k}_2\mathrm{,}{k}_3\in Z\right\}\subset R^3\mathrm{<mo>;</mo>}$

$Z_2=\left\{\left(\begin{array}{c}\frac{1}{2}+a+k_1\\ \frac{1}{2}+k_2\\ a+k_3\end{array}\right)\mathrm{<mo>:</mo>}a\in R\mathrm{,}{k}_1\mathrm{,}{k}_2\mathrm{,}{k}_3\in Z\right\}\subset R^3\mathrm{<mo>;</mo>}$

$Z_3=\left\{\left(\begin{array}{c}a+k_1\\ \frac{1}{2}+a+k_2\\ \frac{1}{2}+k_3\end{array}\right)\mathrm{<mo>:</mo>}a\in R\mathrm{,}{k}_1\mathrm{,}{k}_2\mathrm{,}{k}_3\in Z\right\}\subset R^3\mathrm{.}$

Then the zero set of Fourier transform ${\widehat{\mu }}_{M\mathrm{,}D}$ can be presented as

$Z\left({\widehat{\mu }}_{M,D}\right)={\displaystyle \underset{j=1}{\overset{\infty }{\cup }}}{M}^{\ast j}Z\left(m_D\right):=A_1\cup A_2\cup A_3,$(2.1)

where

$A_1={\displaystyle \underset{j=1}{\overset{\infty }{\cup }}}\left\{\left(\begin{array}{c}m\\ \left(a+k_2\right)p_2^j\\ \left(\frac{1}{2}+a+k_3\right)p_3^j\end{array}\right)\mathrm{<mo>:</mo>}a\in R\mathrm{,}{k}_1\mathrm{,}{k}_2\mathrm{,}{k}_3\in Z\right\}\subset R^3\mathrm{<mo>;</mo>}$ (2.2)

$A_2={\displaystyle \underset{j=1}{\overset{\infty }{\cup }}}\left\{\left(\begin{array}{c}n\\ \left(\frac{1}{2}+k_2\right)p_2^j\\ \left(a+k_3\right)p_3^j\end{array}\right)\mathrm{<mo>:</mo>}a\in R\mathrm{,}{k}_1\mathrm{,}{k}_2\mathrm{,}{k}_3\in Z\right\}\subset R^3\mathrm{<mo>;</mo>}$ (2.3)

$A_3={\displaystyle \underset{j=1}{\overset{\infty }{\cup }}}\left\{\left(\begin{array}{c}p\\ \left(\frac{1}{2}+a+k_2\right)p_2^j\\ \left(\frac{1}{2}+k_3\right)p_3^j\end{array}\right):a\in R\mathrm{,}{k}_1\mathrm{,}{k}_2\mathrm{,}{k}_3\in Z\right\}\subset R^3$ (2.4)

and

$m=\left(\frac{1}{2}+k_1\right)p_1^j+\left(a+k_2\right){\displaystyle \sum _{i=0}^{j-1}}{p}_1^i{p}_2^{j-1-i}{p}_4+\left(\frac{1}{2}+a+k_3\right){\displaystyle \sum _{i=0}^{j-1}}{p}_1^i{p}_3^{j-1-i}{p}_5;$

$n=\left(\frac{1}{2}+a+k_1\right)p_1^j+\left(\frac{1}{2}+k_2\right){\displaystyle \sum _{i=0}^{j-1}}{p}_1^i{p}_2^{j-1-i}{p}_4+\left(a+k_3\right){\displaystyle \sum _{i=0}^{j-1}}{p}_1^i{p}_3^{j-1-i}{p}_5;$

$p=\left(a+k_1\right)p_1^j+\left(\frac{1}{2}+a+k_2\right){\displaystyle \sum _{i=0}^{j-1}}{p}_1^i{p}_2^{j-1-i}{p}_4+\left(\frac{1}{2}+k_3\right){\displaystyle \sum _{i=0}^{j-1}}{p}_1^i{p}_3^{j-1-i}{p}_5.$

Combined with (1.7) and (1.8), one can verify the following propositions hold.

Proposition 2.1. The sets $A_1\mathrm{,}{A}_2\mathrm{,}{A}_3$ given by (2.2), (2.3) and (2.4) hold the following statements:

(i) $\zeta \in A_j$ if and only if $-\zeta \in A_j$ , $j=1,2,3$ ;

(ii) $Z\left({\widehat{\mu }}_{M\mathrm{,}D}\right)\cap Z^3=Z\left({\widehat{\mu }}_{M\mathrm{,}D}\right)\cap \left\{{\left(x\mathrm{,}y\mathrm{,}z\right)}^t\mathrm{<mo>:</mo>}x\mathrm{,}y\mathrm{,}z\in \frac{1}{2}+Z\right\}=\varnothing$ ;

(ii) If $\zeta ={\left({\zeta }_1\mathrm{,}{\zeta }_2\mathrm{,}{\zeta }_3\right)}^t\in A_1$ , then ${\zeta }_1+l{\zeta }_2-l{\zeta }_3\in \frac{1}{2}+Z$ ;

(iv) If $\zeta ={\left({\zeta }_1\mathrm{,}{\zeta }_2\mathrm{,}{\zeta }_3\right)}^t\in A_2$ , then ${\zeta }_2\in \frac{1}{2}+Z$ ;

(v) If $\zeta ={\left({\zeta }_1\mathrm{,}{\zeta }_2\mathrm{,}{\zeta }_3\right)}^t\in A_3$ , then ${\zeta }_3\in \frac{1}{2}+Z$ .

Proposition 2.2. Let $\zeta ={\left({\zeta }_1\mathrm{,}{\zeta }_2\mathrm{,}{\zeta }_3\right)}^t\in Z\left({\widehat{\mu }}_{M\mathrm{,}D}\right)$ . Then the following statements hold:

(a) If $\zeta \in A_1\pm A_1$ , then $\zeta \in A_2\cup A_3$ and ${\zeta }_1+l{\zeta }_2-l{\zeta }_3\in Z$ ;

(b) If $\zeta \in A_2\pm A_2$ , then $\zeta \in A_1\cup A_3$ and ${\zeta }_2\in Z\mathrm{,}{\zeta }_3\notin Z$ ;

(c) If $\zeta \in A_3\pm A_3$ , then $\zeta \in A_1\cup A_2$ and ${\zeta }_3\in Z\mathrm{,}{\zeta }_2\notin Z$ ;

(d) If $\zeta \in A_1$ and ${\zeta }_2\in Z$ , then ${\zeta }_3\notin Z$ ; if $\zeta \in A_1$ and ${\zeta }_3\in Z$ , then ${\zeta }_2\notin Z$ ;

(e) If $\zeta \in A_2$ and ${\zeta }_3\in Z$ , then ${\zeta }_1\notin Z$ ;

(f) If $\zeta \in A_3$ and ${\zeta }_2\in Z$ , then ${\zeta }_1\notin Z$ .

3. Proof of Theorem 1.5

Assume that if ${\lambda }_j\left(j=\mathrm{1,2,}\cdots \mathrm{,5}\right)\in R^3$ are such that the five functions

${\text{e}}^{2\pi i\langle {\lambda }_1\mathrm{,}x\rangle }\mathrm{,}{\text{e}}^{2\pi i\langle {\lambda }_2\mathrm{,}x\rangle }\mathrm{,}{\text{e}}^{2\pi i\langle {\lambda }_3\mathrm{,}x\rangle }\mathrm{,}{\text{e}}^{2\pi i\langle {\lambda }_4\mathrm{,}x\rangle }\mathrm{,}{\text{e}}^{2\pi i\langle {\lambda }_5\mathrm{,}x\rangle }$

are mutually orthogonal in $L^2\left({\mu }_{M\mathrm{,}D}\right)$ , then

${\lambda }_i-{\lambda }_j\in Z\left({\widehat{\mu }}_{M\mathrm{,}D}\right)=A_1\cup A_2\cup A_3\mathrm{,}\left(1\le i\ne j\le 5\right)\mathrm{.}$

Equivalently, the following 10 differences

$\begin{array}{r}\hfill {\lambda }_2-{\lambda }_1\mathrm{,}{\lambda }_3-{\lambda }_1\mathrm{,}{\lambda }_4-{\lambda }_1\mathrm{,}{\lambda }_5-{\lambda }_1\mathrm{<mo>;</mo>}\\ \hfill {\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{<mo>;</mo>}\\ \hfill {\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_3\mathrm{<mo>;</mo>}\\ \hfill {\lambda }_5-{\lambda }_4\end{array}$

belong to $A_1\cup A_2\cup A_3$ . In particular, we have

${\lambda }_2-{\lambda }_1\mathrm{,}{\lambda }_3-{\lambda }_1\mathrm{,}{\lambda }_4-{\lambda }_1\mathrm{,}{\lambda }_5-{\lambda }_1\in A_1\cup A_2\cup A_3\mathrm{.}$ (3.1)

For convenience, define

${\lambda }_j-{\lambda }_k={\left(x_{j\mathrm{,}k}\mathrm{,}{y}_{j\mathrm{,}k}\mathrm{,}{z}_{j\mathrm{,}k}\right)}^t\in R^3\text{for}j\mathrm{,}k\ge 1\text{and}j\ne k\mathrm{.}$

From Proposition 2.1 and 2.2, one can obtain the following claims.

Claim 3.1. The set $A_1$ (or $A_2$ or $A_3$ ) can not contain three differences of the form ${\lambda }_{j_1}-{\lambda }_j\mathrm{,}{\lambda }_{j_2}-{\lambda }_j\mathrm{,}{\lambda }_{j_3}-{\lambda }_j$ , where $j_1\mathrm{,}{j}_2\mathrm{,}{j}_3\in \left\{\mathrm{1,2,}\cdots \mathrm{,5}\right\}\backslash \left\{j\right\}$ are three different numbers.

In fact, if

${\lambda }_{j_1}-{\lambda }_j\mathrm{,}{\lambda }_{j_2}-{\lambda }_j\mathrm{,}{\lambda }_{j_3}-{\lambda }_j\in A_1\mathrm{,}$ (3.2)

then ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}\in A_2\cup A_3$ (by Proposition 2.2 (a)). If ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}\in A_2$ , then (by Proposition 2.1 (iv)), $y_{j_2\mathrm{,}{j}_1}\mathrm{,}{y}_{j_3\mathrm{,}{j}_1}\in \frac{1}{2}+Z$ and $y_{j_3\mathrm{,}{j}_2}\in Z$ , a contradiction. The same reason illustrates that ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}$ can not in the set $A_3$ simultaneously. Without loss of generality, we may assume that ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\in A_2\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}\in A_3$ . By Proposition 2.1 (iv), Proposition 2.2 (a) and (3.2), we have $y_{j_2\mathrm{,}{j}_1}\mathrm{,}{y}_{j_3\mathrm{,}{j}_1}\mathrm{,}{z}_{j_3\mathrm{,}{j}_2}\in \frac{1}{2}+Z$ and $x_{j_3\mathrm{,}{j}_2}+l{y}_{j_3\mathrm{,}{j}_2}-l{z}_{j_3\mathrm{,}{j}_2}\in Z$ , then $x_{j_3\mathrm{,}{j}_2}\mathrm{,}{y}_{j_3\mathrm{,}{j}_2}\in Z$ , a contradiction (by Proposition 2.2 (f)). Hence, the set $A_1$ can not contain three differences of the form ${\lambda }_{j_1}-{\lambda }_j\mathrm{,}{\lambda }_{j_2}-{\lambda }_j\mathrm{,}{\lambda }_{j_3}-{\lambda }_j$ . If

${\lambda }_{j_1}-{\lambda }_j\mathrm{,}{\lambda }_{j_2}-{\lambda }_j\mathrm{,}{\lambda }_{j_3}-{\lambda }_j\in A_2\mathrm{,}$ (3.3)

then ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}\in A_1\cup A_3$ (by Proposition 2.2 (b)). If ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}\in A_1$ , then (by Proposition 2.1 (iii))

$\begin{array}{l}{x}_{j_2\mathrm{,}{j}_1}+l{y}_{j_2\mathrm{,}{j}_1}-l{z}_{j_2\mathrm{,}{j}_1}\mathrm{,}{x}_{j_3\mathrm{,}{j}_1}+l{y}_{j_3\mathrm{,}{j}_1}-l{z}_{j_3\mathrm{,}{j}_1}\in \frac{1}{2}+Z\\ \text{and}{x}_{j_3\mathrm{,}{j}_2}+l{y}_{j_3\mathrm{,}{j}_2}-l{z}_{j_3\mathrm{,}{j}_2}\in Z\mathrm{,}\end{array}$

a contradiction of Proposition 2.1 (iii). If ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}\in A_3$ , then $z_{j_2\mathrm{,}{j}_1}\mathrm{,}{z}_{j_3\mathrm{,}{j}_1}\in \frac{1}{2}+Z$ and $z_{j_3\mathrm{,}{j}_2}\in Z$ (by Proposition 2.1 (v)), a contradiction. Without loss of generality, let

${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\in A_1\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_2}\in A_3\mathrm{.}$ (3.4)

Then from Proposition 2.1 (v), Proposition 2.2 (b) and (3.3) (3.4), we have $y_{j_3\mathrm{,}{j}_2}\in Z$ , $z_{j_3\mathrm{,}{j}_2}\in \frac{1}{2}+Z$ and $x_{j_2\mathrm{,}{j}_1}+l{y}_{j_2\mathrm{,}{j}_1}-l{z}_{j_2\mathrm{,}{j}_1}$ , $x_{j_3\mathrm{,}{j}_1}+l{y}_{j_3\mathrm{,}{j}_1}-l{z}_{j_3\mathrm{,}{j}_1}\in \frac{1}{2}+Z$ , which yields $x_{j_3\mathrm{,}{j}_2}\in Z$ , a contradiction of Proposition 2.2 (f). If ${\lambda }_{j_2}-{\lambda }_{j_1}\mathrm{,}{\lambda }_{j_3}-{\lambda }_{j_1}\in A_3$ , ${\lambda }_{j_3}-{\lambda }_{j_2}\in A_1$ , from Proposition 2.1 (v), Proposition 2.2 (b) and (3.3), we have $z_{j_2\mathrm{,}{j}_1}\mathrm{,}{z}_{j_3\mathrm{,}{j}_1}\in \frac{1}{2}+Z$ , then $y_{j_3\mathrm{,}{j}_2}\mathrm{,}{z}_{j_3\mathrm{,}{j}_2}\in Z$ , a contradiction of Proposition 2.2 (d). Hence, the set $A_2$ can not contain three differences of the form ${\lambda }_{j_1}-{\lambda }_j\mathrm{,}{\lambda }_{j_2}-{\lambda }_j\mathrm{,}{\lambda }_{j_3}-{\lambda }_j$ . This proof is also suitable for the set $A_3$ .

Based on Claim 3.1, we only need to consider the following four cases:

Case 1. 2-2-0 (2-0-2) distribution. In this case, we may assume (without loss of generality) that

${\lambda }_2-{\lambda }_1\mathrm{,}{\lambda }_3-{\lambda }_1\in A_1\mathrm{,}{\lambda }_4-{\lambda }_1\mathrm{,}{\lambda }_5-{\lambda }_1\in A_2\mathrm{.}$ (3.5)

Then (by Proposition 2.1 (iv) and Proposition 2.2 (a) (b))

$x_{\mathrm{3,2}}+l{y}_{\mathrm{3,2}}-l{z}_{\mathrm{3,2}}\in Z\mathrm{,}{y}_{\mathrm{4,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in \frac{1}{2}+Z$ (3.6)

and

${\lambda }_3-{\lambda }_2\in A_2\cup A_3\mathrm{,}{\lambda }_5-{\lambda }_4\in A_1\cup A_3\mathrm{.}$ (3.7)

We can divide (3.7) into four cases.

Case 1.1. ${\lambda }_3-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_4\in A_1$ . In this case, from Proposition 2.1 (iii) (iv), we have

$y_{\mathrm{3,2}}\in \frac{1}{2}+Z\mathrm{,}{x}_{\mathrm{5,4}}+l{y}_{\mathrm{5,4}}-l{z}_{\mathrm{5,4}}\in \frac{1}{2}+Z\mathrm{.}$ (3.8)

If ${\lambda }_5-{\lambda }_2\in A_1$ , then (by Proposition 2.1 (iii) and (3.6) (3.8))

$x_{\mathrm{5,2}}+l{y}_{\mathrm{5,2}}-l{z}_{\mathrm{5,2}}\in \frac{1}{2}+Z\text{and}{x}_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\mathrm{,}{x}_{\mathrm{4,3}}+l{y}_{\mathrm{4,3}}-l{z}_{\mathrm{4,3}}\in Z\mathrm{,}$ (3.9)

which yields ${\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\in A_2\cup A_3$ . In the case ${\lambda }_4-{\lambda }_2\in A_2$ , from (3.8) and Proposition 2.2 (b), we have $y_{\mathrm{4,3}}\in Z$ . Hence, ${\lambda }_4-{\lambda }_3\in A_3$ and $z_{\mathrm{4,3}}\in \frac{1}{2}+Z$ . Combined with (3.9), we get $x_{\mathrm{4,3}}\in Z$ , a contradiction. In the case ${\lambda }_4-{\lambda }_2\in A_3$ , from Proposition 2.1 (v), we have

$z_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.10)

If ${\lambda }_4-{\lambda }_3\in A_2$ , then $y_{\mathrm{4,2}}\mathrm{,}{x}_{\mathrm{4,2}}\in Z$ (by Proposition 2.2 (b) and (3.8) (3.9)), a contradiction. If ${\lambda }_4-{\lambda }_3\in A_3$ , then $z_{\mathrm{3,2}}\mathrm{,}{x}_{\mathrm{3,2}}\in Z$ (by Proposition 2.2 (c) and (3.8) (3.10)), a contradiction. Hence, ${\lambda }_5-{\lambda }_2\notin A_1$ . The same reason illustrates that ${\lambda }_5-{\lambda }_3\mathrm{,}{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\notin A_1$ . Hence

${\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2\cup A_3\mathrm{.}$

Based on Claim 3.1, we only need to consider the following three cases.

Case 1.1.1 ${\lambda }_4-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_2\in A_3$ . In this case, from Proposition 2.2 (c) and (3.6), we have $y_{\mathrm{5,4}}\mathrm{,}{z}_{\mathrm{5,4}}\in Z$ and ${\lambda }_5-{\lambda }_4\in A_1$ , a contradiction (by Proposition 2.2 (d)).

Case 1.1.2 ${\lambda }_4-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_2\in A_3$ . In this case, from Proposition 2.2 (b) and (3.6), we have $y_{\mathrm{2,1}}\in Z$ , $y_{\mathrm{5,2}}\in \frac{1}{2}+Z$ . Combined with ${\lambda }_3-{\lambda }_2\in A_2$ , we get $y_{\mathrm{5,3}}\in Z$ and ${\lambda }_5-{\lambda }_3\in A_3$ . Now, ${\lambda }_3-{\lambda }_2\mathrm{<mo>=</mo>}\left({\lambda }_5-{\lambda }_2\right)-\left({\lambda }_5-{\lambda }_3\right)\in A_3-A_3$ , we have $x_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction.

Case 1.1.3 ${\lambda }_4-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2$ . This case can be proved similarly to Case 1.1.2.

Case 1.2. ${\lambda }_3-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_4\in A_1$ . This case can be proved similarly to Case 1.1.

Case 1.3. ${\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_4\in A_3$ . In this case, by Proposition 2.1 (v) and (3.6), we have

$z_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{5,4}}\in \frac{1}{2}+Z\mathrm{,}{y}_{\mathrm{5,4}}\in Z\mathrm{.}$ (3.11)

We first show that the following Claim holds.

Claim 3.2. The set $\left\{{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\right\}$ has at least one element in $A_1$ .

In fact, if

${\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2\cup A_3\mathrm{,}$ (3.12)

then, from Claim 3.1, we need to divide (3.12) into the following three cases.

Case A: ${\lambda }_4-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_2\in A_2$ . In this case, from (3.6) Proposition 2.1 (iii) and Proposition 2.2 (b), we have

$y_{\mathrm{4,2}}\mathrm{,}{y}_{\mathrm{5,2}}\in \frac{1}{2}+Z\text{and}{y}_{\mathrm{2,1}}\in Z\mathrm{.}$ (3.13)

In the case ${\lambda }_4-{\lambda }_3\in A_2$ , from (3.6) (3.11) (3.13) and Proposition 2.2 (b), we have $x_{\mathrm{3,2}}\mathrm{,}{y}_{\mathrm{3,2}}\in Z$ , a contradiction. In the case ${\lambda }_4-{\lambda }_3\in A_3$ , from (3.11) and (3.12), we have ${\lambda }_5-{\lambda }_3\in A_2$ and $y_{\mathrm{5,3}}\in \frac{1}{2}+Z$ . Combined with (3.6) (3.11) (3.13), we get $x_{\mathrm{3,2}}\mathrm{,}{y}_{\mathrm{3,2}}\in Z$ , a contradiction.

Case B: ${\lambda }_4-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_2\in A_3$ . In this case, from Proposition 2.2 (b) and (3.6), we have $y_{\mathrm{2,1}}\in Z$ and $y_{\mathrm{5,2}}\in \frac{1}{2}+Z$ . For ${\lambda }_5-{\lambda }_3=\left({\lambda }_5-{\lambda }_2\right)-\left({\lambda }_3-{\lambda }_2\right)\in A_3-A_3$ and (3.12) (3.13), we get $y_{\mathrm{5,3}}\in \frac{1}{2}+Z$ and $y_{\mathrm{3,2}}\in Z$ . Combined with (3.6) and (3.9), we have $x_{\mathrm{3,2}}\in Z$ , a contradiction.

Case C: ${\lambda }_4-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_2\in A_2$ . This case can be proved similarly to Case B.

Therefore, the set $\left\{{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\right\}$ has at least one element in $A_1$ .

From Claim 3.2, without loss of generality, we may assume that ${\lambda }_4-{\lambda }_2\in A_1$ , then

$x_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.14)

By Proposition 2.2 (a) (f) and (3.11), we get that ${\lambda }_5-{\lambda }_2\notin A_1$ and ${\lambda }_5-{\lambda }_3\notin A_1$ . Then ${\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2\cup A_3$ . Based on Claim 3.1, we need to consider the following three cases.

Case 1.3.1 ${\lambda }_5-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2$ . In this case, from (3.6) (3.11) and Proposition 2.2 (b), we have $x_{\mathrm{3,2}}\mathrm{,}{y}_{\mathrm{3,2}}\in Z$ , a contradiction.

Case 1.3.2 ${\lambda }_5-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_3\in A_3$ . In this case, for ${\lambda }_5-{\lambda }_2=\left({\lambda }_5-{\lambda }_3\right)-\left({\lambda }_2-{\lambda }_3\right)\in A_3-A_3$ and Proposition 2.2 (b), we get $y_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{5,2}}\in Z$ . Combined with (3.6) (3.11) (3.14), we have $x_{\mathrm{4,2}}\mathrm{,}{y}_{\mathrm{4,2}}\mathrm{,}{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z$ , a contradiction of Proposition 2.1 (ii).

Case 1.3.3 ${\lambda }_5-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2$ . This case can be proved similarly to Case 1.3.2.

Case 1.4 ${\lambda }_3-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_4\in A_3$ . In this case, from (3.6) and Proposition 2.1 (iv) (v), we have

$y_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{5,4}}\in \frac{1}{2}+Z\mathrm{.}$ (3.15)

With the same method as Claim 3.2, we can prove that the set $\left\{{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\right\}$ has at least one element in $A_3$ . Without loss of generality, we may assume that ${\lambda }_4-{\lambda }_2\in A_3$ , then (by Proposition 2.1 (v))

$z_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.16)

By Proposition 2.2 (c) and (3.15), we get $z_{\mathrm{5,2}}\in Z$ and ${\lambda }_5-{\lambda }_2\in A_1\cup A_2$ .

If ${\lambda }_5-{\lambda }_2\in A_1$ , then $x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z$ (by Proposition 2.2 (a) and (3.6)). In the case ${\lambda }_4-{\lambda }_3\in A_1$ , from Proposition 2.2 (a) and (3.6), we have $x_{\mathrm{4,1}}+l{y}_{\mathrm{4,1}}-l{z}_{\mathrm{4,1}}\in Z$ , $x_{\mathrm{5,4}}+l{y}_{\mathrm{5,4}}-l{z}_{\mathrm{5,4}}\in Z$ . Combined with (3.6) (3.15), we get $x_{\mathrm{5,4}}\in Z$ , a contradiction. In the case ${\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.1 (iv) and (3.6) (3.15), we have $y_{\mathrm{4,3}}\in \frac{1}{2}+Z$ and $y_{\mathrm{4,2}}\in Z$ . Combined with (3.15) (3.16), we have $z_{\mathrm{5,2}}\in Z$ , a contradiction. In the case ${\lambda }_4-{\lambda }_3\in A_3$ , for ${\lambda }_3-{\lambda }_2=\left({\lambda }_4-{\lambda }_2\right)-\left({\lambda }_4-{\lambda }_3\right)\in A_3-A_3$ and (3.6), we get $x_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction.

If ${\lambda }_5-{\lambda }_2\in A_2$ , then $y_{\mathrm{2,1}}\in Z$ and $y_{\mathrm{4,2}}\in \frac{1}{2}+Z$ (by Proposition 2.2 (b) and (3.6)). In the case ${\lambda }_4-{\lambda }_3\in A_1$ , from Proposition 2.2 (a) and (3.5), we get

$x_{\mathrm{4,1}}+l{y}_{\mathrm{4,1}}-l{z}_{\mathrm{4,1}}\in Z\mathrm{.}$ (3.17)

By (3.5) (3.16) and (3.17), we have $x_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z$ and $x_{\mathrm{4,2}}\in \frac{1}{2}+Z$ , a contradiction of Proposition 2.1 (ii). In the case ${\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.1 (iv) and (3.15), we have $y_{\mathrm{2,1}}\in \frac{1}{2}+Z$ and $y_{\mathrm{5,2}}\in Z$ , a contradiction. In the case ${\lambda }_4-{\lambda }_3\in A_3$ , from Proposition 2.1 (v) and (3.6) (3.15) (3.16), we get $x_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction.

Similarly, we can prove 2-0-2 distribution does not hold.

Case 2 2-1-1 distribution. In this case, we may assume that

${\lambda }_2-{\lambda }_1\mathrm{,}{\lambda }_3-{\lambda }_1\in A_1\mathrm{,}{\lambda }_4-{\lambda }_1\in A_2\mathrm{,}{\lambda }_5-{\lambda }_1\in A_3\mathrm{,}$ (3.18)

then (by Proposition 2.1 (iv)(v) and Proposition 2.2 (a))

$y_{\mathrm{4,1}}\mathrm{,}{z}_{\mathrm{5,1}}\in \frac{1}{2}+Z\text{and}{x}_{\mathrm{3,2}}+l{y}_{\mathrm{3,2}}-l{z}_{\mathrm{3,2}}\in Z$ (3.19)

and

${\lambda }_3-{\lambda }_2\in A_2\cup A_3\mathrm{.}$ (3.20)

We can divide (3.20) into two cases.

Case 2.1. ${\lambda }_3-{\lambda }_2\in A_2$ . In this case, we have (by Proposition 2.1 (iv))

$y_{\mathrm{3,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.21)

With the same method as Claim 3.2, we can prove that the set $\left\{{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\right\}$ has at least one element in $A_1$ . Without loss of generality, we may assume that ${\lambda }_4-{\lambda }_2\in A_1$ , then (by (3.6) and Proposition 2.2 (a))

$x_{\mathrm{4,1}}+l{y}_{\mathrm{4,1}}-l{z}_{\mathrm{4,1}}\in Z\mathrm{.}$ (3.22)

If ${\lambda }_5-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.6)),

$x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z\mathrm{.}$ (3.23)

By Proposition 2.1 (iii) and (3.22) (3.23), we get ${\lambda }_5-{\lambda }_4\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_4\in A_2$ , from Proposition 2.2 (b) and (3.19) (3.23), we have $x_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_4\in A_3$ , from Proposition 2.2 (c) and (3.19) (3.22), we have $x_{\mathrm{4,1}}\mathrm{,}{z}_{\mathrm{4,1}}\in Z$ , a contradiction. Hence, ${\lambda }_5-{\lambda }_2\notin A_1$ . The same reason illustrates that ${\lambda }_5-{\lambda }_3\notin A_1$ . Then

${\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2\cup A_3\mathrm{.}$

Based on Claim 3.1, we need to consider the following three cases.

Case 2.1.1 ${\lambda }_5-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_3\in A_3$ . From Proposition 2.2 (c), we have ${\lambda }_3-{\lambda }_2=\left({\lambda }_5-{\lambda }_2\right)-\left({\lambda }_5-{\lambda }_3\right)\in A_3-A_3$ . Combined with (3.19) (3.21), we have $x_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction.

Case 2.1.2 ${\lambda }_5-{\lambda }_2\in A_2\mathrm{,}{\lambda }_5-{\lambda }_3\in A_3$ . From Proposition 2.1 (v) Proposition 2.2 (c) and (3.21), we have

$z_{\mathrm{3,1}}\mathrm{,}{y}_{\mathrm{5,3}}\in Z\mathrm{,}{z}_{\mathrm{5,3}}\in \frac{1}{2}+Z\mathrm{.}$ (3.24)

By Claim 3.1 and (3.22) (3.33), we know that ${\lambda }_5-{\lambda }_4\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_4\in A_2$ , from Proposition 2.2 (b) and (3.19) (3.23), we have $x_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_4\in A_3$ , from Proposition 2.2 (c) and (3.19) (3.22), we have $x_{\mathrm{4,1}}\mathrm{,}{y}_{\mathrm{4,1}}\in Z$ , a contradiction.

Case 2.1.3 ${\lambda }_5-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_3\in A_2$ . This case can be proved similarly to Case 2.1.2.

Case 2.2. ${\lambda }_3-{\lambda }_2\in A_3$ . This case can be proved similarly to Case 2.1.

Case 3 0-2-2 distribution. In this case, we may assume that

${\lambda }_2-{\lambda }_1\mathrm{,}{\lambda }_3-{\lambda }_1\in A_2\mathrm{,}{\lambda }_4-{\lambda }_1\mathrm{,}{\lambda }_5-{\lambda }_1\in A_3\mathrm{,}$ (3.25)

then (by Proposition 2.1 (iv) (v) and Proposition 2.2 (b) (c))

$y_{\mathrm{2,1}}\mathrm{,}{y}_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{4,1}}\mathrm{,}{z}_{\mathrm{5,1}}\in \frac{1}{2}+Z\text{and}{y}_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{5,4}}\in Z$ (3.26)

and

${\lambda }_3-{\lambda }_2\in A_1\cup A_3\text{and}{\lambda }_5-{\lambda }_4\in A_1\cup A_2\mathrm{.}$ (3.27)

We can divide (3.27) into four cases.

Case 3.1. ${\lambda }_3-{\lambda }_2\in A_1\mathrm{,}{\lambda }_5-{\lambda }_4\in A_1$ . In this case, we have

$x_{\mathrm{3,2}}+l{y}_{\mathrm{3,2}}-l{z}_{\mathrm{3,2}}\mathrm{,}{x}_{\mathrm{5,4}}+l{y}_{\mathrm{5,4}}-l{z}_{\mathrm{5,4}}\in \frac{1}{2}+Z\mathrm{.}$ (3.28)

With the same method as Claim 3.2, we can prove that the set $\left\{{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\right\}$ has at least one element in $A_1$ . Without loss of generality, we may assume that ${\lambda }_4-{\lambda }_2\in A_1$ , then

$x_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.29)

From Claim 3.1 and (3.28) (3.29), we have ${\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\in A_2\cup A_3$ .

In the case ${\lambda }_5-{\lambda }_2\in A_2\mathrm{,}{\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.26) (3.28), we have $y_{\mathrm{4,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ and $y_{\mathrm{5,4}}\in Z$ , a contradiction.

In the case ${\lambda }_5-{\lambda }_2\in A_3\mathrm{,}{\lambda }_4-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.26), we have $z_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ and $y_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction.

In the case ${\lambda }_5-{\lambda }_2\in A_3\mathrm{,}{\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) (c) and (3.26) (3.29), we have $z_{\mathrm{2,1}}\mathrm{,}{y}_{\mathrm{4,1}}\in Z$ and $x_{\mathrm{4,2}}\mathrm{,}{y}_{\mathrm{4,2}}\mathrm{,}{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z$ , a contradiction.

In the case ${\lambda }_5-{\lambda }_2\in A_2\mathrm{,}{\lambda }_4-{\lambda }_3\in A_3$ , from Proposition 2.2 (b) (c) and (3.26) (3.28) (3.29), we have $z_{\mathrm{3,1}}\mathrm{,}{y}_{\mathrm{5,1}}\mathrm{,}{x}_{\mathrm{5,2}}+l{y}_{\mathrm{5,2}}-l{z}_{\mathrm{5,2}}\in Z$ and $x_{\mathrm{5,3}}+l{y}_{\mathrm{5,3}}-l{z}_{\mathrm{5,3}}\in \frac{1}{2}+Z$ . Combined with (3.26), we get $x_{\mathrm{5,3}}\mathrm{,}{y}_{\mathrm{5,3}}\mathrm{,}{z}_{\mathrm{5,3}}\in \frac{1}{2}+Z$ , a contradiction.

Case 3.2. ${\lambda }_3-{\lambda }_2\in A_1\mathrm{,}{\lambda }_5-{\lambda }_4\in A_2$ . In this case, from Proposition 2.1 (iii) (iv), we have

$y_{\mathrm{5,4}}\in \frac{1}{2}+Z\mathrm{,}{x}_{\mathrm{3,2}}+l{y}_{\mathrm{3,2}}-l{z}_{\mathrm{3,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.30)

With the same method as Claim 3.2, we can prove that the set $\left\{{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\right\}$ has at least one element in $A_1$ . Without loss of generality, we may assume that ${\lambda }_4-{\lambda }_2\in A_1$ , then (by Proposition 2.1 (iii))

$x_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.31)

From Claim 3.1 and Proposition 2.2 (a), we have ${\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_2\in A_2$ , ${\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.26) (3.30), we have $y_{\mathrm{4,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ and $y_{\mathrm{5,4}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_3$ , ${\lambda }_4-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.26), we have $z_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ and $y_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_3$ , ${\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) (c) and (3.26) (3.31), we have $z_{\mathrm{2,1}}\mathrm{,}{y}_{\mathrm{4,1}}\in Z$ and $x_{\mathrm{4,2}}\mathrm{,}{y}_{\mathrm{4,2}}\mathrm{,}{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_2$ , ${\lambda }_4-{\lambda }_3\in A_3$ , from ${\lambda }_4-{\lambda }_2={\lambda }_5-{\lambda }_2-\left({\lambda }_5-{\lambda }_4\right)\in A_2-A_2$ and (3.26), we have $y_{\mathrm{4,2}}\in Z$ , $y_{\mathrm{4,1}}\in \frac{1}{2}+Z$ and $y_{\mathrm{4,3}}\in Z$ . Combined with ${\lambda }_4-{\lambda }_3={\lambda }_4-{\lambda }_2-\left({\lambda }_3-{\lambda }_2\right)\in A_1-A_1$ and Proposition 2.2 (a), we get $x_{\mathrm{4,3}}\in Z$ , a contradiction.

Case 3.3. ${\lambda }_3-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_4\in A_1$ . This case can be proved similarly to Case 3.2.

Case 3.4. ${\lambda }_3-{\lambda }_2\in A_3\mathrm{,}{\lambda }_5-{\lambda }_4\in A_2$ . In this case, from Proposition 2.1 (iv) (v), we have

$z_{\mathrm{3,2}}\mathrm{,}{y}_{\mathrm{5,4}}\in \frac{1}{2}+Z\mathrm{.}$ (3.32)

With the same method as Claim 3.2, we can prove that the set $\left\{{\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_5-{\lambda }_3\right\}$ has at least one element in $A_1$ . Without loss of generality, we may assume that ${\lambda }_4-{\lambda }_2\in A_1$ , then

$x_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.33)

From Claim 3.1 and Proposition 2.2 (a), we have ${\lambda }_5-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_2\in A_2$ , ${\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.26), we have $y_{\mathrm{4,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ and $y_{\mathrm{5,4}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_2$ , ${\lambda }_4-{\lambda }_3\in A_3$ , from Proposition 2.2 (b) (c) and (3.32), we have $y_{\mathrm{4,2}}\mathrm{,}{z}_{\mathrm{4,2}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_3$ , ${\lambda }_4-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) (c) and (3.32), we have $y_{\mathrm{5,3}}\mathrm{,}{z}_{\mathrm{5,3}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_3$ , ${\lambda }_4-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.25), we have $z_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{2,1}}\in Z$ , a contradiction of (3.32).

Case 4. 1-2-1 distribution. In this case, we may assume that

${\lambda }_2-{\lambda }_1\in A_1\mathrm{,}{\lambda }_3-{\lambda }_1\mathrm{,}{\lambda }_4-{\lambda }_1\in A_2\mathrm{,}{\lambda }_5-{\lambda }_1\in A_3\mathrm{,}$ (3.34)

then (by Proposition 2.1 (iii) (iv) (v))

$y_{\mathrm{3,1}}\mathrm{,}{y}_{\mathrm{4,1}}\mathrm{,}{z}_{\mathrm{5,1}}\mathrm{,}{x}_{\mathrm{2,1}}+l{y}_{\mathrm{2,1}}-l{z}_{\mathrm{2,1}}\in \frac{1}{2}+Z$ (3.35)

and

${\lambda }_4-{\lambda }_3\in A_1\cup A_3\mathrm{.}$ (3.36)

We can divide (3.36) into two cases.

Case 4.1. ${\lambda }_4-{\lambda }_3\in A_1$ . In this case, we have

$x_{\mathrm{4,3}}+l{y}_{\mathrm{4,3}}-l{z}_{\mathrm{4,3}}\in \frac{1}{2}+Z\mathrm{.}$ (3.37)

We can prove the following Claim hold.

Claim 3.3. The set $\left\{{\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_2\right\}$ has at least one element in $A_1$ .

In fact, if

${\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_2\in A_2\cup A_3\mathrm{.}$

Then, based on Claim 3.1, we need to consider the following three cases.

Case (a): ${\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_2\in A_3$ . In this case, from (3.35) and Proposition 2.2 (c), we have $y_{\mathrm{4,3}}\mathrm{,}{z}_{\mathrm{4,3}}\in Z$ , a contradiction.

Case (b): ${\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_2\in A_2$ . In this case, from (3.35) and Proposition 2.2 (b), we have

$y_{\mathrm{2,1}}\in Z\mathrm{.}$ (3.38)

If ${\lambda }_5-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_4\in A_2$ , then $y_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,2}}\in Z$ and ${\lambda }_5-{\lambda }_2\in A_1\cup A_3$ (by Proposition 2.1 (iv) and (3.38)). In the case ${\lambda }_5-{\lambda }_2\in A_1$ , from Proposition 2.2 (a) and (3.35), we have $x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z$ and $x_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_3$ , from Proposition 2.2 (c) and (3.38), we get $y_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{2,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_4\in A_3$ , then (by Proposition 2.2 (c) and (3.38)) $y_{\mathrm{4,3}}\mathrm{,}{z}_{\mathrm{4,3}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_3\in A_2$ , ${\lambda }_5-{\lambda }_4\in A_3$ , then, $y_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,2}}\in Z$ and ${\lambda }_5-{\lambda }_2\in A_1\cup A_3$ (by Proposition 2.2 (c) and (3.38)). In the case ${\lambda }_5-{\lambda }_2\in A_1$ , from Proposition 2.2 (a) and (3.35), we have $x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z$ and $x_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_3$ , from Proposition 2.2 (c) and (3.38), we get $y_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{2,1}}\in Z$ , a contradiction. Hence the set $\left\{{\lambda }_5-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_4\right\}$ has at least one element in $A_1$ . In this case, without loss of generality, we may assume that

${\lambda }_5-{\lambda }_3\in A_1\mathrm{.}$ (3.39)

It follows from Claim 3.1 that ${\lambda }_5-{\lambda }_2\in A_1\cup A_3$ . If ${\lambda }_5-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.35) (3.38))

$x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z\mathrm{.}$ (3.40)

From Claim 3.1, we know that ${\lambda }_5-{\lambda }_4\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_4\in A_2$ , from Proposition 2.2 (b) and (3.35) (3.40), we have $x_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_4\in A_3$ , from (3.37) (3.39) (3.40) and Proposition 2.2 (c), we get $x_{\mathrm{4,1}}+l{y}_{\mathrm{4,1}}-l{z}_{\mathrm{4,1}}\in Z$ and $x_{\mathrm{4,1}}\mathrm{,}{z}_{\mathrm{4,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_3$ , then $y_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{2,1}}\in Z$ ( by Proposition 2.2 (c) and (3.35) (3.38)), a contradiction.

Case (c): ${\lambda }_3-{\lambda }_2\in A_2\mathrm{,}{\lambda }_4-{\lambda }_2\in A_3$ . In this case, from (3.35) and Proposition 2.2 (b), we have

$y_{\mathrm{2,1}}\in Z\text{and}{y}_{\mathrm{4,2}}\mathrm{,}{z}_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.41)

With the same method as Case (b), we can prove that the set $\left\{{\lambda }_5-{\lambda }_3\mathrm{,}{\lambda }_5-{\lambda }_4\right\}$ has at least one element in $A_1$ . In this case, without loss of generality, we may assume that

${\lambda }_5-{\lambda }_3\in A_1\mathrm{.}$ (3.42)

Combined with (3.37), we have ${\lambda }_5-{\lambda }_4\in A_2\cup A_3$ . If ${\lambda }_5-{\lambda }_4\in A_2$ , from Proposition 2.2 (b) and (3.35) (3.41), we have

$y_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,2}}\in Z\text{and}{\lambda }_5-{\lambda }_2\in A_1\cup A_3\mathrm{.}$ (3.43)

In the case ${\lambda }_5-{\lambda }_2\in A_1$ , from Proposition 2.2 (a) and (3.35) (3.43), we have $x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z$ and $x_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_3$ , from Proposition 2.2 (c) and (3.41), we get $y_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{2,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_4\in A_3$ , then (by Proposition 2.2 (c) and (3.35) (3.37) (3.42))

$z_{\mathrm{4,1}}\mathrm{,}{z}_{\mathrm{5,2}}\in Z\mathrm{,}{z}_{\mathrm{5,4}}\mathrm{,}{z}_{\mathrm{2,1}}\in \frac{1}{2}+Z\text{and}{x}_{\mathrm{5,4}}+l{y}_{\mathrm{5,4}}-l{z}_{\mathrm{5,4}}\in Z$ (3.44)

and ${\lambda }_5-{\lambda }_2\in A_1\cup A_2$ . In the case ${\lambda }_5-{\lambda }_2\in A_1$ , from Proposition 2.2 (a) and (3.35) (3.44), we have $x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}$ , $x_{\mathrm{4,1}}+l{y}_{\mathrm{4,1}}-l{z}_{\mathrm{4,1}}\in Z$ and $x_{\mathrm{4,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_2\in A_2$ , from Proposition 2.1 (iv) and (3.35) (3.41) (3.44), we have $y_{\mathrm{5,1}}\in \frac{1}{2}+Z$ and $x_{\mathrm{5,4}}\mathrm{,}{y}_{\mathrm{5,4}}\in Z$ , a contradiction.

Hence the set $\left\{{\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_2\right\}$ has at least one element in $A_1$ . In this case, without loss of generality, we may assume that ${\lambda }_3-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.37))

$x_{\mathrm{3,1}}+l{y}_{\mathrm{3,1}}-l{z}_{\mathrm{3,1}}\in Z\text{and}{x}_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\in Z$ (3.45)

and ${\lambda }_4-{\lambda }_2\in A_2\cup A_3$ .

If ${\lambda }_4-{\lambda }_2\in A_2$ , then

$y_{\mathrm{2,1}}\in Z\mathrm{.}$ (3.46)

If ${\lambda }_5-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.35) (3.45))

$x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\mathrm{,}{x}_{\mathrm{5,3}}+l{y}_{\mathrm{5,3}}-l{z}_{\mathrm{5,3}}\in Z$ (3.47)

and ${\lambda }_5-{\lambda }_3\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.35) (3.47), we get $x_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.35) (3.45), we have $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_2$ , then (by (3.35) (3.46))

$y_{\mathrm{5,1}}\in \frac{1}{2}+Z\mathrm{,}{y}_{\mathrm{5,3}}\in Z$ (3.48)

and ${\lambda }_5-{\lambda }_3\in A_1\cup A_3$ . In the case ${\lambda }_5-{\lambda }_3\in A_1$ , from Proposition 2.1 (iii) and (3.35) (3.37) (3.45) (3.48), we have

$x_{\mathrm{5,4}}+l{y}_{\mathrm{5,4}}-l{z}_{\mathrm{5,4}}\mathrm{,}{y}_{\mathrm{5,4}}\in Z\mathrm{.}$ (3.49)

Hence, ${\lambda }_5-{\lambda }_4\in A_3$ and $z_{\mathrm{5,4}}\in \frac{1}{2}+Z$ . Combined with (3.49), we get $x_{\mathrm{5,4}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_3\in A_3$ , from (3.35) (3.45), we have $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_3$ , then (by Proposition 2.2 (c) and (3.46)) $z_{\mathrm{2,1}}\in Z$ , a contradiction.

If ${\lambda }_4-{\lambda }_2\in A_3$ , then (by Proposition 2.1 (v) and (3.35) (3.45))

$z_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{,}{x}_{\mathrm{4,2}}+l{y}_{\mathrm{4,2}}-l{z}_{\mathrm{4,2}}\in Z\text{and}{x}_{\mathrm{4,1}}+l{y}_{\mathrm{4,1}}-l{z}_{\mathrm{4,1}}\in \frac{1}{2}+Z\mathrm{.}$ (3.50)

If ${\lambda }_5-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.35) (3.45))

$x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z$ (3.51)

and ${\lambda }_5-{\lambda }_3\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.35) (3.51), we have $x_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.35) (3.45), we get $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_2$ , then (by Proposition 2.1 (iv))

$y_{\mathrm{5,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.52)

If ${\lambda }_5-{\lambda }_3\in A_1$ , then (by Proposition 2.2 (a) and (3.37)) $x_{\mathrm{5,4}}+l{y}_{\mathrm{5,4}}-l{z}_{\mathrm{5,4}}\in Z$ and ${\lambda }_5-{\lambda }_4\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_4\in A_2$ , from Proposition 2.2 (b) and (3.50) (3.52), we get $x_{\mathrm{4,2}}\mathrm{,}{y}_{\mathrm{4,2}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_4\in A_3$ , from Proposition 2.2 (c) (3.50) (3.52) and ${\lambda }_5-{\lambda }_2=\left({\lambda }_5-{\lambda }_3\right)-\left({\lambda }_2-{\lambda }_3\right)\in A_1-A_1$ , we have $z_{\mathrm{4,1}}\in Z\mathrm{,}{z}_{\mathrm{2,1}}\in \frac{1}{2}+Z$ and $x_{\mathrm{5,2}}\mathrm{,}{z}_{\mathrm{5,2}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_3\in A_2$ , then (by Proposition 2.2 (b) and (3.35) (3.52)) $y_{\mathrm{3,2}}\in Z$ and $y_{\mathrm{2,1}}\in \frac{1}{2}+Z$ . Combined with (3.35) and (3.50), we have $x_{\mathrm{4,2}}\mathrm{,}{y}_{\mathrm{4,2}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_3\in A_3$ , then (by Proposition 2.2 (c) and (3.35) (3.45)) $x_{3.1}\mathrm{,}{z}_{3.1}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_3$ , then (by Proposition 2.2 (c) and (3.35)) $z_{\mathrm{2,1}}\in Z$ . Combined with (3.35) and (3.50), we have $x_{\mathrm{4,1}}\mathrm{,}{y}_{\mathrm{4,1}}\mathrm{,}{z}_{\mathrm{4,1}}\in \frac{1}{2}+Z$ , a contradiction.

Case 4.2. ${\lambda }_4-{\lambda }_3\in A_3$ . In this case, we have

$y_{\mathrm{4,3}}\in Z\mathrm{,}{z}_{\mathrm{4,3}}\in \frac{1}{2}+Z\mathrm{.}$ (3.53)

With the same method as Claim 3.3, we can prove that the set $\left\{{\lambda }_3-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_2\right\}$ has at least one element in $A_1$ . Without loss of generality, we may assume that ${\lambda }_3-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.35))

$x_{\mathrm{3,1}}+l{y}_{\mathrm{3,1}}-l{z}_{\mathrm{3,1}}\in Z\mathrm{.}$ (3.54)

If ${\lambda }_4-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.35) (3.53) (3.54)) $x_{\mathrm{4,1}}+l{y}_{\mathrm{4,1}}-l{z}_{\mathrm{4,1}}$ , $x_{\mathrm{4,3}}+l{y}_{\mathrm{4,3}}-l{z}_{\mathrm{4,3}}\in Z$ and $x_{\mathrm{4,3}}\in Z$ , a contradiction.

If ${\lambda }_4-{\lambda }_2\in A_2$ , then (by Proposition 2.2 (b) and (3.35))

$y_{\mathrm{2,1}}\in Z\mathrm{.}$ (3.55)

If ${\lambda }_5-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.35))

$x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z\mathrm{.}$ (3.56)

Combined with (3.54) and (3.55), we have $x_{\mathrm{5,3}}+l{y}_{\mathrm{5,3}}-l{z}_{\mathrm{5,3}}\in Z$ and ${\lambda }_5-{\lambda }_3\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.35) (3.56), we get $x_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.35) (3.54), we have $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_2$ , then (by Proposition 2.1 (iv) and (3.35) (3.55))

$y_{\mathrm{5,1}}\in \frac{1}{2}+Z\mathrm{,}{y}_{\mathrm{5,3}}\in Z$ (3.57)

and ${\lambda }_5-{\lambda }_3\in A_1\cup A_3$ . In the case ${\lambda }_5-{\lambda }_3\in A_1$ , from Proposition 2.1 (iii) and (3.54), we get

$x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in \frac{1}{2}+Z\mathrm{.}$ (3.58)

Combined with (3.35) and (3.57), we have $x_{\mathrm{5,1}}\in \frac{1}{2}+Z$ , a contradiction. In the

case ${\lambda }_5-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.54), we get $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_3$ , from Proposition 2.2 (c) and (3.55), we have $y_{\mathrm{2,1}}\mathrm{,}{z}_{\mathrm{2,1}}\in Z$ , a contradiction.

If ${\lambda }_4-{\lambda }_2\in A_3$ , then (by Proposition 2.1 (v))

$z_{\mathrm{4,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.59)

Combined with (3.53), we have

$z_{\mathrm{3,2}}\in Z\mathrm{.}$ (3.60)

If ${\lambda }_5-{\lambda }_2\in A_1$ , then (by Proposition 2.2 (a) and (3.35) (3.54))

$x_{\mathrm{5,1}}+l{y}_{\mathrm{5,1}}-l{z}_{\mathrm{5,1}}\in Z\text{and}{x}_{\mathrm{5,3}}+l{y}_{\mathrm{5,3}}-l{z}_{\mathrm{5,3}}\in Z\mathrm{.}$ (3.61)

Hence, ${\lambda }_5-{\lambda }_3\in A_2\cup A_3$ . In the case ${\lambda }_5-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.35) (3.61), we have $x_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{5,1}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_3\in A_3$ , from Proposition 2.2 (c) and (3.35) (3.54), we get $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_2$ , then (by Proposition 2.1 (iv))

$y_{\mathrm{5,2}}\in \frac{1}{2}+Z\mathrm{.}$ (3.62)

In the case ${\lambda }_5-{\lambda }_3\in A_1$ , for ${\lambda }_4-{\lambda }_2\mathrm{,}{\lambda }_4-{\lambda }_3\in A_3$ , Proposition 2.1 (i) and Claim 3.1, we have ${\lambda }_5-{\lambda }_4\in A_1\cup A_2$ . If ${\lambda }_5-{\lambda }_4\in A_1$ , then, for ${\lambda }_4-{\lambda }_3=\left({\lambda }_5-{\lambda }_3\right)-\left({\lambda }_5-{\lambda }_4\right)\in A_1-A_1$ and (3.53), we get $x_{\mathrm{4,3}}\mathrm{,}{y}_{\mathrm{4,3}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_4\in A_2$ , then (by Proposition 2.2 (b) and (3.35) (3.60) (3.62)), $y_{\mathrm{2,1}}\in \frac{1}{2}+Z$ and $y_{\mathrm{5,1}}\mathrm{,}{y}_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_3\in A_2$ , from Proposition 2.2 (b) and (3.35) (3.60) (3.62), we get $y_{\mathrm{3,2}}\mathrm{,}{z}_{\mathrm{3,2}}\in Z$ , a contradiction. In the case ${\lambda }_5-{\lambda }_3\in A_3$ , from (3.35) and (3.54), we have $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction. If ${\lambda }_5-{\lambda }_2\in A_3$ , then (by Proposition 2.2 (c) and (3.35) (3.59)) $z_{\mathrm{2,1}}\in Z\mathrm{,}{z}_{\mathrm{4,1}}\in \frac{1}{2}+Z$ . Combined with (3.53) and (3.54), we have $x_{\mathrm{3,1}}\mathrm{,}{z}_{\mathrm{3,1}}\in Z$ , a contradiction.

The above discussion shows that $L^2\left({\mu }_{M\mathrm{,}D}\right)$ can not contain five mutually orthogonal exponential functions. Take

$\Lambda =\left\{\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}\frac{\left(1-l\right)p_1^j+l{p}_3^j}{2}\\ 0\\ \frac{p_3^j}{2}\end{array}\right)\mathrm{,}\left(\begin{array}{c}\frac{\left(1-l\right)p_1^j-l{p}_2^j}{2}\\ \frac{p_2^j}{2}\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}\frac{l\left(p_2^j+p_3^j\right)}{2}\\ \frac{-p_2^j}{2}\\ \frac{p_3^j}{2}\end{array}\right)\right\}\mathrm{,}$

where $p_1\mathrm{,}{p}_2\mathrm{,}{p}_3\in \left(2Z+1\right)\backslash \left\{\mathrm{0,}\pm 1\right\}\mathrm{,}l\in 2Z$ . We can verify that $E\left(\Lambda \right)$ is a 4-element ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials, which yields the number 4 is the best.

Corollary 3.4. For the self-affine measure ${\mu }_{M\mathrm{,}D}$ corresponding to

$M=\left[\begin{array}{ccc}{P}_1& 0& 0\\ P_4& P_2& 0\\ P_5& 0& P_3\end{array}\right]\text{and}D=\left\{\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}d\\ 0\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}0\\ d\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}0\\ 0\\ d\end{array}\right)\right\}\mathrm{,}$

if $p_j\in \left(2Z+1\right)\backslash \left\{\mathrm{0,}\pm 1\right\}\left(j=\mathrm{1,2,3}\right)\mathrm{,}{p}_2\ne p_3\mathrm{,}d\ne 0$ and

$p_4=l\left(p_1-p_2\right),p_5=l\left(p_3-p_1\right),$

where $l\in 2Z$ , then there are at most 4-element ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials, and the number 4 is the best.

Example 3.5. Let

$M=\left[\begin{array}{ccc}5& 0& 0\\ -52& 31& 0\\ 36& 0& 23\end{array}\right]\mathrm{,}D=\left\{\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\mathrm{,}\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}\mathrm{,}$

then there are at most 4-element ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials, and the number 4 is the best.

Take $l\mathrm{<mo>=</mo>2}\in 2Z$ , one can verify that $p_1=5$ , $p_2=31$ , $p_3=23\in \left(2Z+1\right)\backslash \left\{0,\pm 1\right\}$ , $p_4=2\left(p_1-p_2\right)=-52$ , $p_5=2\left(p_3-p_1\right)=36$ , which shows that the condition (1.8) holds. Then there are at most 4-element ${\mu }_{M\mathrm{,}D}$ -orthogonal exponentials, and the number 4 is the best.

Funding

This work is supported by the National Natural Science Foundation of China (No.12001346).

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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