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

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.

• 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

$\mu =\frac{1}{\left|D\right|}{\displaystyle \underset{d\in D}{\sum }}\mu \circ {\varphi }_d^{-1}$(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 ${\hat{\mu }}_{M\mathrm{,}D}$ of ${\mu }_{M\mathrm{,}D}$ plays an important role. From (1.1), ${\hat{\mu }}_{M\mathrm{,}D}\left(\xi \right)$ is given by

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

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

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

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 ( μ ^ M , D ) of μ 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 ${\hat{\mu }}_{M\mathrm{,}D}$ can be presented as

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

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

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

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

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

(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 $\zeta ={\left({\zeta }_1\mathrm{,}{\zeta }_2\mathrm{,}{\zeta }_3\right)}^t\in Z\left({\hat{\mu }}_{M\mathrm{,}D}\right)$. 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

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

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).

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