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Calculation of hyperelliptic systems of sequences of rank 4 A. A. Illarionov National Research University "Higher School of Economics", Moscow
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     § 1. IntroductionA sequence $h\colon \mathbb{Z} \to \mathbb{Q}$ satisfying
$$
\begin{equation*}
h_{m+n}h_{m-n} = h_{m+1}h_{m-1}h^2_n - h_{n+1}h_{n-1}h^2_m
\end{equation*}
\notag
$$
is said to be elliptic. It is known (see [1]) that, except for some degenerate cases, such a sequence is of the form $h(n) = \sigma (nv) \sigma^{-n^2}(v)$, where $\sigma$ is the Weierstrass sigma function, and $v\in \mathbb{C}$. Bykovskii (see [2]) set out the following construction. Definition 1.1. Let $A,B\colon \mathbb{Z}\to\mathbb{C}$ be sequences, not identically zero, such that (for all $n,m\in\mathbb{Z}$) the expansions hold for some sequences $a_j,b_j,\widetilde a_j,\widetilde b_j\colon \mathbb{Z}\to\mathbb{C}$ and the smallest possible $N_0, N_1\in\mathbb{Z}_+ = \mathbb{N}\cup\{0\}$. In this case, the pair $(A,B)$ is called a hyperelliptic system of sequences of rank $R(A,B)= R_0(A,B)+R_1(A,B)$, where Hyperelliptic sequences are closely related to Somos sequences (see [3]), which have many remarkable properties and have been extensively studied (see [2]–[14] and the references there). The problem of finding sequences satisfying (1.1) is equivalent to the solution of the functional equation in the class of 1-periodic functions $f,g,\phi_j,\psi_j\colon \mathbb{C}\to\mathbb{C}$ (see [3]). This equation is closely related to elliptic functions. For example, the addition formula
$$
\begin{equation}
\sigma(x+y)\sigma(x-y) = \sigma^2(x)\sigma^2(y)(\wp(y)- \wp(x)) \qquad (\wp = -(\ln \sigma)'')
\end{equation}
\tag{1.3}
$$
implies that the pair $(f,g) = (\sigma,\sigma)$ satisfies (1.1) with $N = 2$. Also, it is known that all non-elementary solutions of equation (1.2) for $N=2$ (see [15]) and for $N=3$ (see [16]) can be expressed in terms of the Weierstrass sigma function. In addition, equation (1.2) is a special case, for $s=1$) of the relations of the form
$$
\begin{equation}
f_1(x_1+z)\cdots f_s(x_s+z)g(x_1+\dots +x_s-z) = \sum_{j=1}^{m}\varphi_j(x_1,\dots,x_s)\psi_j(z),
\end{equation}
\tag{1.4}
$$
which play an important role in the analysis of $(s+1)$-linear functional differential equations and addition theorems (see [17]–[19]). In turn, the analysis of (1.4) can be reduced to that of expansion (1.2) (see [16], [20]–[22]). The functional equation (1.2) has been extensively studied (see [15], [16], [20], [21], [23]–[29] and the references there). However, its general solution is known only for $N=1,2$ (see [15], [20], [26]) and $N=3$ (see [16], [28], [29]). The question of the solvability of equation (1.2) for $N\geqslant 4$ remains open. Earlier, in [3] the author of the present paper described all hyperelliptic systems of sequences of rank $\leqslant 3$, and also of rank $4$ under the additional condition $A=B$. In the present paper, we describe all hyperelliptic systems of sequences of rank 4. As a corollary, we obtain all 1-periodic solutions of the functional equation (1.2) for $N=4$. Our analysis is based on the methods and results of [3]. We also employ the ideas and results from [7], [14], [21], [28]. § 2. PreliminariesAll the results of this section stated without a reference can be found in [3]. 2.1. Relation between hyperelliptic systems of sequences and functionsThroughout, we will use the notation Definition 2.1. Let entire non identically zero functions $f,g\colon \mathbb{C}\to\mathbb{C}$ satisfy (1.2) together with some $\phi_j,\psi_j\colon \mathbb{C}\to\mathbb{C}$ with smallest possible $N\in \mathbb{Z}_+$. Such a pair $(f,g)$ will be called a solution of the functional equation (1.2), written $R(f,g)=N$. Theorem 2.1. Let $f,g$ be $1$-periodic entire functions $\mathbb{C}\to\mathbb{C}$ and let $A,B\colon \mathbb{Z}\to \mathbb{C}$ be the sequences of Fourier coefficients in the expansions Then $R(f,g)=R(A,B)$. Lemma 2.1 (see [2]). For each hyperelliptic system of sequences $(A,B)$, there exists a positive real constant $\lambda_0 \in (0,+\infty)$ such that Consider any $\lambda\in \mathbb{C}$ such that $\operatorname{Im} \lambda > \lambda_0/2\pi$ and define $\widetilde A_n = A_n e(n^2\lambda)$, $\widetilde B_n = B_n e(n^2\lambda)$. The series
$$
\begin{equation}
f(z)=\sum_{n\in\mathbb{Z}}\widetilde A_n e(nz),\qquad g(z)=\sum_{n\in\mathbb{Z}}\widetilde B_n e(nz)
\end{equation}
\tag{2.2}
$$
are absolutely uniformly convergent on each compact set. Hence $f, g$ are $1$-periodic entire functions, and $R(f, g) = R (\widetilde A, \widetilde B)= R(A,B)$. 2.2. Equivalence relationsDefinition 2.2. Solutions $(f,g)$, $(\widetilde f,\widetilde g)$ of equation (1.2) are called equivalent (written $(f,g)\sim (\widetilde f, \widetilde g)$) if either $(\widetilde f,\widetilde g) = (f_1, g_1)$ or $(\widetilde f,\widetilde g) = (g_1,f_1)$, where with some $\alpha,\beta_j,\gamma_j, z_j \in \mathbb{C}$, $j=1,2$, $z_0\in \mathbb{C}\setminus\{0\}$. It is easy tho check that $R(f,g)=R(\widetilde f, \widetilde g)$ for $(f,g)\sim (\widetilde f, \widetilde g)$ (see [20]). Consider the following transformations of the system of sequences $(A,B)$: I) $(A,B)\to (B,A)$; II) $(A,B)\to (A,\widehat B)$, where $\widehat B_n = B_{-n}$; III) $(A,B)\to (A,\widehat B)$, where $\widehat B_n = B_n e(\beta n+\gamma)$, $\beta,\gamma\in \mathbb{C}$; IV) $(A,B)\to (A,\widehat B)$, where $\widehat B_n = B_{n+n_0}$, $n_0\in \mathbb{Z}$; V) $(A,B)\to (\widehat A,\widehat B)$, where
$$
\begin{equation*}
\widehat A_n = \begin{cases} A_{n/d} &\text{for } n\in d\mathbb{Z}, \\ 0 &\text{otherwise}, \end{cases}\quad \widehat B_n = \begin{cases} B_{n/d} &\text{for } n\in d\mathbb{Z}, \\ 0 &\text{otherwise}, \end{cases}\quad d\in \mathbb{N};
\end{equation*}
\notag
$$
VI) $(A,B)\to (\widehat A,\widehat B)$, where $\widehat A_n = A_{dn}$, $\widehat B_n = B_{dn}$, $d\in \mathbb{N}$, provided $A_n = B_n = 0$ for all $n\notin d\mathbb{Z}$; VII) $(A,B)\to (\widehat A,\widehat B)$, where $\widehat A_n= A_n e(\alpha n^2)$, $\widehat B_n= B_n e(\alpha n^2)$, $\alpha \in \mathbb{C}$. Definition 2.3. If the system of sequences $(\widetilde A,\widetilde B)$ is obtained from a system $(A,B)$ by a composition of transformations of type I)–VII) (of type I)–VI, respectively)), then $(\widetilde A,\widetilde B)$, $(A,B)$ are called equivalent (strongly equivalent) systems, written $(A,B)\sim (\widetilde A, \widetilde B)$ ($(A,B)\stackrel{s}{\sim} (\widetilde A, \widetilde B)$, respectively). The following properties hold. 1. The binary relations $\sim$ and $\stackrel{s}{\sim}$ are equivalence relations on the set of hyperelliptic systems of sequences. 2. Let $f$, $g$, $\widetilde f$, $\widetilde g $ be $1$-periodic entire functions. Then $(f,g)\sim (\widetilde f, \widetilde g)$ if and only if $\Phi(f,g)\stackrel{s}{\sim} \Phi(\widetilde f, \widetilde g)$. Here, $\Phi(f,g) = (A,B)$, where $A$ and $B$ are the sequences of Fourier coefficients (from expansions (2.1) of functions $f$ and $g$, respectively. 3. If a pair $(\widetilde A,\widetilde B)$ is obtained from a pair $(A,B)$ by transformations of type I), II), III), VII), then $R_j(A,B)= R_j(\widetilde A,\widetilde B)$, $j=0,1$. 4. If $(A,B)\sim (\widetilde A,\widetilde B)$, then $R(A,B)= R(\widetilde A,\widetilde B)$. Remark 2.1. If the system of sequences $(\widetilde A,\widetilde B)$ is obtained from $(A,B)$ by means of a transformation of type V) or VI), then $R_j(A,B)\neq R_j(\widetilde A,\widetilde B)$, $j=0,1$ in general. If a transformation of type IV) is used, then $R_j(A,B) = R_j(\widetilde A,\widetilde B)$, $j=0,1$ if $n_0$ is odd and $R_0(A,B) = R_1(\widetilde A,\widetilde B)$, $R_1(A,B) = R_0(\widetilde A,\widetilde B)$ if it is even. 2.3. Determinant equationsGiven sequences $A, B: \mathbb{Z}\to \mathbb{C}$ and integers $n_0,\dots, n_k$, $m_0,\dots, m_k$, consider the determinants
$$
\begin{equation*}
\begin{aligned} \, D_{A,B}\begin{pmatrix} n_0 & \dots & n_k \\ m_0 & \dots & m_k \end{pmatrix} &=\begin{vmatrix} A_{n_0+m_0} B_{n_0-m_0} & \dots & A_{n_0+m_k}B_{n_0-m_k} \\ \vdots & & \vdots \\ A_{n_k+m_0} B_{n_k-m_0} & \dots & A_{n_k+m_k} B_{n_k-m_k} \end{vmatrix}, \\ \widetilde D_{A,B}\begin{pmatrix} n_0 & \dots & n_k \\ m_0 & \dots & m_k \end{pmatrix} &= \begin{vmatrix} A_{n_0+m_0+1} B_{n_0-m_0} & \dots & A_{n_0+m_k+1}B_{n_0-m_k} \\ \vdots & & \vdots \\ A_{n_k+m_0+1} B_{n_k-m_0} & \dots & A_{n_k+m_k+1} B_{n_k-m_k} \end{vmatrix}. \end{aligned}
\end{equation*}
\notag
$$
In what follows, we will write $D(\,{\dots}\,)$ and $\widetilde D(\,{\dots}\,)$ in place of $D_{A,B}(\,{\dots}\,)$ and $\widetilde D_{A,B}(\,{\dots}\,)$, respectively, if no confusion will ensue. Lemma 2.2. Let $A, B \colon \mathbb{Z}\to \mathbb{C}$, $A,B \not\equiv 0$, and $k\in \mathbb{Z}_+$. Then 1) the inequality $R_0(A,B)\leqslant k$ is equivalent to the relation
$$
\begin{equation*}
D\begin{pmatrix} n_0 & \dots & n_k \\ m_0 & \dots & m_k \end{pmatrix} =0
\end{equation*}
\notag
$$
for all $n_0,\dots, n_k$, $m_0,\dots, m_k$; 2) the inequality $R_1(A,B)\leqslant k$ is equivalent to the relation for all $n_0,\dots, n_k$, $m_0,\dots, m_k$.2.4. A special caseLemma 2.3. Let $c_1,c_2\in \mathbb{C}$, $c_1c_2\neq 0$, $\beta\in \mathbb{C}$, Then $R_j(A,B)\geqslant 2$, $j=0,1$ if and only if either $\beta \notin \frac{1}{4}\mathbb{Z}$ or $c_1\neq \pm c_2$ and $e(\beta) c_1c_2 \neq \pm 1$. Proof. 1. Let $\beta \in \frac{1}{4}\mathbb{Z}$, and let $c_1 = \pm c_2$ or $c_1c_2 e(\beta)= \pm 1$. Note that $(A,B)\sim (\widetilde A,\widetilde B)$, $\widetilde A_{2k} = c_1$, $\widetilde B_{2k} = c_2$, $\widetilde A_{2k+1} = \widetilde B_{2k+1} = c_0$, where $c_0 = e(\beta)$ for $\beta \in \frac{1}{2} \mathbb{Z}$, and $c_0 = e(\beta-1/8)$ for $\beta \notin \frac{1}{2} \mathbb{Z}$. Actually, if $\beta \in \frac{1}{2} \mathbb{Z}$, then $(A,B)=(\widetilde A,\widetilde B)$. If $\beta \notin \frac{1}{2} \mathbb{Z}$, then is suffices to put $\widetilde A_n = A_n S_n$, $\widetilde B_n = B_n S_n$, where $S_n = e(n(n-2)/8)$.
If $c_1=\pm c_2$, then
$$
\begin{equation*}
\widetilde A_{n+m+1} \widetilde B_{n-m}\equiv \begin{cases} c_0c_1 &\text{for } c_1=c_2, \\ c_0c_2 (-1)^n (-1)^m &\text{for } c_1=-c_2. \end{cases}
\end{equation*}
\notag
$$
Hence $R_1(A,B)=R_1(\widetilde A,\widetilde B)=1$. If $c_1c_2 e(\beta)= \pm 1$, then $c_1c_2 = \pm c_0^2$, and
$$
\begin{equation*}
\widetilde A_{n+m}\widetilde B_{n-m}\equiv \begin{cases} c_0^2 &\text{for } c_1c_2=c_0^2, \\ c_1 c_2 (-1)^n (-1)^m &\text{for } c_1c_2=-c_0^2. \end{cases}
\end{equation*}
\notag
$$
Therefore, $R_0(A,B)=R_1(\widetilde A,\widetilde B)=1$.
2. Let $R_0(A,B)<2$ or $R_1(A,B)<2$. Then, for each odd $n$,
$$
\begin{equation*}
D\begin{pmatrix} n & 0 \\ n & 0 \end{pmatrix} = c_1^2c_2^2 - e(2n\beta) =0 \quad \text{or}\quad \widetilde D\begin{pmatrix} n & 0 \\ n & 0 \end{pmatrix} = c_2^2e(2\beta(n+1))-c_1^2 = 0.
\end{equation*}
\notag
$$
The first case is possible only for $\beta \in \frac{1}{4}\mathbb{Z}$, $c_1c_2 e(\beta)= \pm 1$, and the second one, for $\beta \in \frac{1}{4}\mathbb{Z}$, $c_1=\pm c_2$. This proves Lemma 2.3. § 3. The main resultsLet $L$ be a lattice in $\mathbb{C}$ (a discrete additive subgroup of the field $\mathbb{C}$) and $\sigma_L\colon \mathbb{C}\,{\to}\, \mathbb{C}$ be the Weierstrass sigma function associated with (possibly degenerate) lattice $L$. By $\mathbb{Z}_d^*$ we denote the reduced residue system modulo $d$. Theorem 3.1. Let at least one of sequences $A$ or $B$ have infinitely many non-zero terms. Then $R(A,B)=4$ if and only if $(A,B)\sim (\widetilde A,\widetilde B)$, where the system $(\widetilde A,\widetilde B)$ is one of the following five types: 1) $\widetilde A_n = \sigma_L(nv+u_1)$, $\widetilde B_n = \sigma_L(nv+u_2)$, where $v\in \mathbb{C}\setminus (\frac{1}{2} L)$, $u_1,u_2\in \mathbb{C}$; 2) $\widetilde A\equiv 1$, $\widetilde B_n = n +c$, where $c\in \mathbb{C}$; 3) $\widetilde A\equiv 1$, $\widetilde B_n = 1+c e(\beta n)$, where $c\in \mathbb{C}\setminus\{0\}$, $\beta\in \mathbb{C}\setminus \mathbb{Z}$; 4) $\displaystyle \widetilde A_n = \begin{cases} c_1 &\text{if } n\equiv 0 \ (\operatorname{mod} d), \\ e(\beta n) &\text{if } n\equiv t \ (\operatorname{mod} d), \\ 0 &\text{otherwise}, \end{cases}\quad \widetilde B_n = \begin{cases} c_2 &\text{if } n\equiv 0 \ (\operatorname{mod} d), \\ e(\beta n) &\text{if } n\equiv t \ (\operatorname{mod} d), \\ 0 &\text{otherwise}, \end{cases}$ where $d\in \mathbb{N}$, $d\geqslant 2$, $t\in \mathbb{Z}_d^*$, $t\not\equiv d \pmod 2$, $c_1,c_2 \in \mathbb{C}\setminus\{0\}$, $\beta \in \mathbb{C}$, and if $d=2$, $\beta\in \frac{1}{4}\mathbb{Z}$, then $c_1\neq \pm c_2$, $c_1c_2 e(\beta)\neq \pm 1$; 5) $\displaystyle \widetilde A_n = \begin{cases} c &\text{if } n\equiv 0 \ (\operatorname{mod} d), \\ e(\beta n) &\text{if } n\equiv t \ (\operatorname{mod} d), \\ 0 &\text{otherwise}, \end{cases}\quad \widetilde B_n = \begin{cases} 1 &\text{if } n\equiv 0 \ (\operatorname{mod} d), \\ 0 & \text{otherwise}, \end{cases}$ where $c\in \mathbb{C}\setminus\{0\}$, $\beta\in \mathbb{C}$, $d\in \mathbb{Z}$, $d\geqslant 2$, $t\in \mathbb{Z}_d^*$, $t\not\equiv d \pmod 2$. Let
$$
\begin{equation*}
\operatorname{supp} A = \{n\in \mathbb{Z}\colon A_n\neq 0\}, \qquad \operatorname{supp}B= \{m\in \mathbb{Z}\colon B_m\neq 0\}.
\end{equation*}
\notag
$$
In what follows, $|X|$ is the cardinality of $X$. Lemma 3.1. Let $R(A,B)=4$, and let $\operatorname{supp}B$ (or $\operatorname{supp} A$) be bounded from above or below. Then $|{\operatorname{supp} A}|\leqslant 4$ and $|{\operatorname{supp}B}|\leqslant 4$. In view of Lemma 3.1 the case where $A$ or $B$ has a finite number of non-zero terms is devoid of interest. Theorem 3.1 and Lemma 3.1 will be proved in § 7. A function $Q\colon \mathbb{C} \to \mathbb{C}$ is called a quasi-polynomial of rank $r$ if
$$
\begin{equation*}
Q(z) = \sum_{j=1}^s P_j(z) e(\lambda_j z),\qquad r = \sum_{j=1}^s (1+ \deg P_j),
\end{equation*}
\notag
$$
where $\lambda_j$ are different complex numbers and $P_j$ are polynomials of degree $\deg P_j$. If a function $Q$ is $1$-periodic, then $\deg P_j = 0$, $\lambda_j\in \mathbb{Z}$. It is easy to see that each pair of quasi-polynomials is a solution of the functional equation (1.2) for some $N$. Definition 3.1. A solution $(f,g)$ of the functional equation (1.2) is said to be elementary if $(f,g)= (h Q_1,h Q_2)$, where $Q_1$, $Q_2$ are quasi-polynomials, and $h(z)=e(\alpha z^2)$, $\alpha\in \mathbb{C}$. Let
$$
\begin{equation*}
\tau\in \mathbb{C}, \quad \Omega = \begin{pmatrix}\omega_{11} & \omega_{12} \\ \omega_{12} & \omega_{22}\end{pmatrix},\qquad \omega_{11}, \omega_{12}, \omega_{22}\in \mathbb{C},
\end{equation*}
\notag
$$
where $\operatorname{Im} \tau>0$ and $\operatorname{Im} \Omega$ is positive definite. Recall that the theta functions $\theta =\theta(\,{\cdot}\,;\tau)\colon \mathbb{C}\to \mathbb{C}$ and $\Theta(\,{\cdot}\,; \Omega)\colon \mathbb{C}^2 \to \mathbb{C}$ are defined by the Fourier series
$$
\begin{equation*}
\begin{aligned} \, \theta(z;\tau) &= \sum_{k\in \mathbb{Z}} \exp(2\pi i k z + \pi ik^2 \tau) \equiv \sum_{k\in \mathbb{Z}} e(k z) e\biggl(\frac{k^2 \tau}2\biggr), \\ \Theta(z_1,z_2; \Omega) &= \sum_{n_1,n_2\in \mathbb{Z}} e\biggl( n_1z_1+ n_2z_2 + \frac{\omega_{11} n_1^2+ 2\omega_{12} n_1n_2+\omega_2 n_2^2}{2} \biggr). \end{aligned}
\end{equation*}
\notag
$$
The following result will be obtained in § 7 via Theorems 2.1 and 3.1. Corollary 3.1. Let $N=4$. Then, up to an equivalence, the set of $1$-periodic non-elementary solutions of equation (1.2) coincides with the set of pairs $(f,g)$ of the following three types § 4. The case $R_0(A,B)=1$ or $R_1(A,B)=1$Lemma 4.1 (see [3]). Let $A,B\colon \mathbb{Z}\to \mathbb{C}$. Then $R(A,B)=1$ if and only if $|{\operatorname{supp} A}|=|{\operatorname{supp}B}|=1$. Lemma 4.2 (see [3]). Let $A,B\colon \mathbb{Z}\to \mathbb{C}$. Then $R(A,B)=2$ if and only if $(A,B)$ satisfies, up to an equivalence, one of the following three conditions: 1) $|T_{A}|=2$, $|T_{B}|=1$; 2) $T_{A}=T_{B}$, and $|T_{A}|=2$; 3) $A\equiv B\equiv 1$. Lemma 4.3 (see[3]). Let $A,B\colon \mathbb{Z}\to \mathbb{C}$. Then $R(A,B)=3$ if and only if $(A,B)$ satisfies, up to an equivalence, one of the following three conditions: 1) $|T_{A}|=3$, $|T_{B}|=1$; 2) $T_{A}=T_{B} = \{0, \pm s\}$, where $s\in \mathbb{N}$; 3) $A_n=B_n = \begin{cases} c_0 & \text{ for even } n, \\ c_1 & \text{ for odd } n, \end{cases}$ where $c_0,c_1\in \mathbb{C}\setminus \{0\}$, $c_0\neq c_1$. Lemma 4.4. Let $R_0(A,B)=1$. Then the system $(A,B)$ satisfies, up to an equivalence, one of the following conditions: a) $\operatorname{supp} A = \{0\}$, $\operatorname{supp}B \cap 2\mathbb{Z} = \{0\}$; b) $\operatorname{supp} A =\operatorname{supp}B = \{0, s\}$, and $s\equiv 1 \pmod{2}$; c) $\operatorname{supp} A = \operatorname{supp}B= \{0,\pm s\}$, and $s\equiv 1\pmod 2$; d) $A_n=B_{n+1} = \begin{cases} c_0 &\text{if }n\text{ is even}, \\ c_1 &\text{if }n\text{ is odd}, \end{cases}$ where $c_0,c_1\in \mathbb{C}\setminus \{0\}$. Proof. Since $R_0(A,B)=1$, there exist sequences $a,b\colon \mathbb{Z}\to \mathbb{C}$ such that Hence $a(n+m)b(n-m)=A_{2n}B_{2m}$, $a(n+m+1)b(n-m)= A_{2n+1}B_{2m+1}$, and, therefore, $R_j(a,b)\leqslant 1$, $j=0,1$. Consequently, $R(a,b)\leqslant 2$. In view of Lemmas 4.1 and 4.2 there are four cases to consider.
1. Each of the sequences $a$ and $b$ contains precisely one non-zero term. Assume that $a(s)b(t)\neq 0$. Setting $\widehat a(n) = a(n+s)$, $\widehat b(m) = b(m+t)$, $\widehat A_n = A_{n+s+t}$, $\widehat B_m = B_{m+s-t}$, we have
$$
\begin{equation}
\widehat A_{n+m}\widehat B_{n-m} = \widehat a(n)\widehat b(m),
\end{equation}
\tag{4.2}
$$
where $\widehat a(n)\widehat b(m)\neq 0$ only if $n=m= 0$. Hence $\widehat A(0)\widehat B(0)\neq 0$. Putting $m=n$ ($m=-n$) in (4.2) $m=n$ ($m=-n$), we conclude that $\widehat A_{2m}=\widehat B_{2m}=0$ for any $m\neq 0$. If there exist $n,m$ such that $\widehat A_{2n+1}\widehat B_{2m+1}\neq 0$, then $\widehat A_{2n+1}\widehat B_{2m+1} = \widehat a(n+m+1)\widehat b(n-m) \neq 0$ according to (4.2). Therefore, $n+m+1 = n-m =0$, which is impossible. So, up to an equivalence, the system $(A,B)$ satisfies condition a).
2. One of the sequences $a$ or $b$ has precisely two non-zero terms, and the other one has precisely one non-zero term. Arguing as in the previous case, we can assume that $a(s)a(0)b(0)\neq 0$. From (4.1) it follows that $A_0 A_sB_0B_s\neq 0$. If $s\equiv 0 \pmod 2$, then $a(s/2)b(s/2) = A_s B_0\neq 0$. Hence $s\equiv 1 \pmod 2$. Using (4.1), this gives
$$
\begin{equation*}
\begin{gathered} \, A_{2n} B_0 = a(n)b(n), \qquad A_0 B_{2m} = a(m)b(-m), \\ A_{2n+s} B_s = a(n+s)b(n), \qquad A_s B_{2n+s} = a(n+s)b(-n), \end{gathered}
\end{equation*}
\notag
$$
which shows that $\operatorname{supp} A = \operatorname{supp}B = \{0,s\}$. Thus, we arrive at case b).
3. The sequence $a$ has precisely two non-zero terms with numbers $n_1,n_2$ and the sequence $b$ has precisely two non-zero terms with numbers $m_1,m_2$, where $n_2-n_1 = m_2-m_1>0$. Without loss of generality we can assume that $n_1=m_1 =0$ and $n_2=m_2 = s$. As in the previous case, using (4.1), we find that $\operatorname{supp} A = \{0,s,2s\}$, $\operatorname{supp}B= \{-s,0,s\}$, where $s\equiv 1\pmod 2$. 4. Let $(a,b)\sim (1,1)$. Then $\operatorname{supp} A$, $T_b$ are arithmetic progressions with the same difference. Without loss of generality we can assume that $\operatorname{supp} A = T_b = d\mathbb{Z}$, $d\in \mathbb{N}$. Using (4.1) and arguing as in the previous case, we obtain $d\equiv 1 \pmod 2$ and $\operatorname{supp} A = \operatorname{supp}B = d\mathbb{Z}$. Hence $(A,B)\sim (\widetilde A,\widetilde B)$, where $\widetilde A_n= A_{nd}$, $\widetilde B_m= B_{md}$. Since $R_0(\widetilde A,\widetilde B)=1$, Lemma 2.2 shows that, for any integer $n$,
$$
\begin{equation*}
D_{\widetilde A, \widetilde B}\begin{pmatrix} n & 0 \\ 1 & 0 \end{pmatrix} = \begin{vmatrix} \widetilde A_{n+1}\widetilde B_{n-1} & \widetilde A_n\widetilde B_n \\ \widetilde A_1\widetilde B_{-1} & \widetilde A_0\widetilde B_0 \end{vmatrix}=0,
\end{equation*}
\notag
$$
that is, $\widetilde A_{n+1}\widetilde B_{n-1}=c \widetilde A_n\widetilde B_n$, where $c= \widetilde A_1\widetilde B_{-1} \widetilde A_0^{-1}\widetilde B_0^{-1}$. Therefore, $\widetilde A_{n+1} = \widetilde B_n c^{n} \widetilde A_1 \widetilde B_0^{-1}$. Changing an equivalent system, it can be assumed that $\widetilde A_{n+1}= \widetilde B_n$. In this case, $R_1 (\widetilde A,\widetilde A)=1$. So, $R(\widetilde A,\widetilde A)\leqslant 3$ by Lemma 5.5 in [3]. Using Lemmas 4.1–4.3 and taking into account that $\widetilde A$ has no zero terms, we conclude that either $(\widetilde A,\widetilde A)\sim (1,1)$ (and hence $(A,B)\sim (1, 1)$), or $\widetilde A_n=c_0 \mu^{n^2}\lambda^n$ for ${n\equiv 0 \pmod 2}$, and $\widetilde A_n=c_1 \mu^{n^2}\lambda^n $ for $n\equiv 1\pmod 2$. In both cases, assertion d) holds up to an equivalence. This proves Lemma 4.4. Lemma 4.5. Let $R(A,B)=4$, where $R_0(A,B)=1$ or $R_1(A,B)=1$. Then, up to an equivalence, $\operatorname{supp} A =\{0\}$ and $\operatorname{supp}B= \{0,n_1,n_2,n_3\}$, where $n_j\equiv 1\pmod 2$. Proof. Let $R_0(A,B)=1$. By Lemma 4.4, one of assertions a)–d) is satisfied. If b), c) or d) holds, then $R(A,B)\leqslant 3$ by Lemmas 4.2, 4.3.
Let a) be satisfied. Consider functions (2.2), where $\widetilde A_n = A_n e(n^2\alpha)$, $\widetilde B_n = B_n e(n^2\alpha)$, and the constant $\alpha$ is chosen so that the series converges. Hence $R(f,g)= 4$ by Theorem 2.1. Since $f \equiv A_0$, it follows that $g$ is a quasi-polynomial of rank $4$ (see [30]). So, $|\operatorname{supp}B|=4$. Therefore, $\operatorname{supp}B= \{0,n_1,n_2,n_3\}$, where $n_j\equiv 1\pmod 2$. Let $R_1(A,B)=1$. Then $(A,B)\sim (\widetilde A, B)$, where $\widetilde A_n = A_{n+1}$, and $R_0(\widetilde A,B)=1$. This case was considered above. This proves Lemma 4.5. § 5. The structure of the sets $\operatorname{supp} A$ and $\operatorname{supp}B$ with $R_0(A,B)=R_1(A,B)=2$The aim of this section is to prove the following lemma. Lemma 5.1. Let $R_0(A,B)=R_1(A,B)=2$, where 1) there are no $d\in \mathbb{Z}\cap [2,+\infty)$ and $t_1,t_2\in \mathbb{Z}$ such that $\operatorname{supp} A\subset (d\mathbb{Z}+t_1)$, $\operatorname{supp}B\subset (d\mathbb{Z}+t_2)$; 2) the sets $\operatorname{supp} A$ and $\operatorname{supp}B$ are not bounded from below and above. Then, up to an equivalence, the system $(A,B)$ satisfies one of the following conditions: a) $\operatorname{supp} A =\mathbb{Z}$; b) there exist $n,m\in \mathbb{Z}$ such that $A_{n-1}A_n A_{n+1}B_{m-1}B_m B_{m+1}\neq 0$; c) $\operatorname{supp} A = d\mathbb{Z} \cup(d\mathbb{Z} + t)$, $\operatorname{supp}B = d \mathbb{Z}$, where $d\not\equiv t \pmod 2$, $\operatorname{\textrm{gcd}}(d,t)=1$; d) $\operatorname{supp} A = \operatorname{supp}B= d\mathbb{Z} \cup(d\mathbb{Z} + t)$, where $d\not\equiv t \pmod 2$, $\operatorname{\textrm{gcd}}(d,t)=1$. All the remaining results in this section are auxiliary, and will not be used in the following sections. Remark 5.1. In this section, we will use three types of transformation of a system $(A,B)$ to an equivalent system 1) $(A,B)\to (B,A)$; 2) $(A,B)\to (A,\widetilde B)$, where $\widetilde B_n = B_{n+n_0}$; 3) $(A,B)\to (A,\widetilde B)$, where $\widetilde B_n = B_{-n}$. Elements of the set $\operatorname{supp} A$ (the set $\operatorname{supp}B$) will be denoted by $n_i$ (by $m_j$), where
$$
\begin{equation*}
n_{-1}< n_0 <n_1<\cdots, \qquad m_{-1}< m_0 <m_1<\cdots.
\end{equation*}
\notag
$$
The choice of the points $n_0$, $m_0$ is immaterial, but the important point here is that
$$
\begin{equation*}
A_{n_i}\neq 0, \quad A_n = 0\text{ for }n_i<n< n_{i+1},\quad B_{m_j}\neq 0, \quad B_m = 0 \text{ for } m_j <m< m_{j+1}
\end{equation*}
\notag
$$
for all $i,j\in \mathbb{Z}$. Lemma 5.2. Let $R_0(A,B)=R_1(A,B)=2$, and let $n_i\equiv n_{i+1}$, $m_j\equiv m_{j+1}\pmod 2$. Then $n_{i+1}-n_i=m_{j+1}-m_j$. Proof. Without loss of generality we can assume that $n_i=m_j=0$ (otherwise we change to an equivalent system $(\widetilde A,\widetilde B)$, where $\widetilde A_n = A_{n+n_i}$, $\widetilde B_m = B_{m+m_j}$). We have $n_{i+1}=2s$, $m_{j+1}=2t$, where $s,t\in \mathbb{N}$. By definition, $B_t = A_s=0$. We claim that $s=t$. Assume that $s\neq t$ (for example, $s>t$). Hence $0<s+t < 2s$, and, therefore, $A_{s+t}=0$. Hence
$$
\begin{equation*}
D\begin{pmatrix} 0 & t & s \\ 0 & -t & s \end{pmatrix} = \begin{vmatrix} A_0B_0 & 0 & 0 \\ \ast & A_0 B_{2t} & 0 \\ \ast & \ast & A_{2s} B_0 \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
This contradicts Lemma 2.2. Here and in what follows, “$\ast$” will denote the elements whose values have no effect on the determinant of the corresponding matrix. Therefore, $s=t$, completing the proof. Lemma 5.3. Under the hypotheses of Lemma 5.1, let the sequence $\operatorname{supp} A$ contain three neighbouring terms of the same parity. Then assertion c) of Lemma 5.1 is satisfied up to an equivalence. Proof. Without loss of generality we can assume that $-2t$, $0$, $2q$ are three neighbouring terms of the sequence $\operatorname{supp} A$; $t,q\in \mathbb{N}$. Note that $A_{-t} = A_{q}=0$. We can also assume without loss of generality that $B_0\neq 0$.
1. We claim that $q=t$. Indeed, otherwise $A_{q-t}=0$ and
$$
\begin{equation*}
D\begin{pmatrix} -t & 0 & q \\ -t & 0 & q \end{pmatrix} = \begin{vmatrix} A_{-2t}B_0 & 0 & 0 \\ \ast & A_0 B_0 & 0 \\ \ast & \ast & A_{2q} B_0 \end{vmatrix} \neq 0,
\end{equation*}
\notag
$$
but this contradicts Lemma 2.2. Hence $q=t$, and so
2. We claim that $\operatorname{supp}B$ does no contain three neighbouring terms such that
$$
\begin{equation*}
m_j\equiv m_{j+1}\not\equiv m_{j+2} \pmod 2 \quad\text{or}\quad m_j\not\equiv m_{j+1}\equiv m_{j+2} \pmod 2.
\end{equation*}
\notag
$$
Assume on the contrary that, for example, $m_j\equiv m_{j+1}\not\equiv m_{j+2} \pmod 2$ (the second case is reduced to this by the substitution $(A,B)\to (A,\widetilde B)$, where $\widetilde B_n = B_{-n}$). We have $m_{j+1}-m_j = 2t$ by Lemma 5.2. Without loss of generality we can assume that $m_{j+1}=0$. Hence
$$
\begin{equation*}
m_j=-2t, \quad m_{j+1}=0, \quad m_{j+2} = 2s+1 \qquad (s\in \mathbb{Z}_+).
\end{equation*}
\notag
$$
Since $-2t< 2s+1 - 2t < 2s+1$ and $2s+1 - 2t \neq 0$, we have $B_{2s+1 - 2t}=0$,
$$
\begin{equation*}
\widetilde D\begin{pmatrix} s-t & s & t+s \\ -s-t-1 & -s-1 & t-s-1 \end{pmatrix} = \begin{vmatrix} A_{-2t}B_{2s+1} & 0 & 0 \\ \ast & A_0 B_{2s+1} & 0 \\ \ast & \ast & A_{2t} B_{2s+1} \end{vmatrix} \neq 0,
\end{equation*}
\notag
$$
which contradicts Lemma 2.2.
3. Let us verify that $\operatorname{supp}B$ does not contain three neighbouring terms of the same parity. Assume on the contrary that there exist $m_j\equiv m_{j+1}\equiv m_{j+2}\pmod 2$. Hence $m_{j+1}-m_j = m_{j+2}-m_{j+1}=2t$ by Lemma 5.2. Without loss of generality we can assume that $m_{j+1}=0$. Hence $m_{j}=-2t$, $m_{j+2}=2t$. We claim that $\operatorname{supp} A, \operatorname{supp}B \subset 2\mathbb{Z}$. Indeed, let $2n+1$ be the smallest (in absolute value) odd number from $\operatorname{supp} A\cup \operatorname{supp}B$. By symmetry we can assume that $2n+1 \in \operatorname{supp} A\,\cap \, \mathbb{N}$. Since $2n+1> 2t$, we have $A_{2n+1-2t}=0$. Next $B_{\pm t} =0$, and so
$$
\begin{equation*}
\widetilde D\begin{pmatrix} n-t & n & n+t \\ n+t & n & n-t \end{pmatrix} = \begin{vmatrix} A_{2n+1}B_{-2t} & 0 & 0 \\ \ast & A_{2n+1} B_0 & 0 \\ \ast & \ast & A_{2n+1} B_{2t} \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
Therefore, $\operatorname{supp} A, \operatorname{supp}B \subset 2\mathbb{Z}$, which contradicts condition 1) of Lemma 5.1.
4. From steps 2, 3 of the proof, it follows that the terms of the sequence $\operatorname{supp}B$ have alternating parity, that is, $m_j\not\equiv m_{j+1}\pmod 2$ for all $j$. We claim that $m_{j+2}-m_j = 2t$ for all $j$. Without loss of generality we can assume that
$$
\begin{equation*}
m_j =0, \quad m_{j+1}= 2k+1, \quad m_{j+2}= 2s, \qquad 0\leqslant k < s.
\end{equation*}
\notag
$$
Let us show that $s=t$. Suppose the contrary. If $t<s$, then $B_{2t}=0$,
$$
\begin{equation*}
D\begin{pmatrix} -t & 0 & t \\ -t & 0 & t \end{pmatrix} = \begin{vmatrix} A_{-2t}B_0 & \ast & \ast \\ 0 & A_0 B_0 & \ast \\ 0 & 0 & A_{2t} B_0 \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
Next, if $t>s$, then $A_{t-s}=A_{2t-s}=0$,
$$
\begin{equation*}
D\begin{pmatrix} 0 & t & t+s \\ 0 & t & t-s \end{pmatrix} = \begin{vmatrix} A_0B_0 & 0 & 0 \\ \ast & A_{2t} B_0 &0 \\ \ast & \ast & A_{2t} B_{2s} \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
Both cases are impossible. Hence $t=s$, $m_{j+2}-m_j = 2t$ for all $j$.
5. By step 4 it can be assumed that $\operatorname{supp}B = 2t\mathbb{Z} \,{\cup}\, (2t\mathbb{Z}+2k+1)$, where $0\leqslant k< t$, and $-2t,0,2t$ are three neighbouring terms of $\operatorname{supp} A$. 6. We claim that $\operatorname{supp} A\subset 2\mathbb{Z}$. Assume the contrary. Let $2n+1$ be the smallest (in absolute value) odd number from $\operatorname{supp} A$. Without loss of generality we can assume that $n\geqslant 0$. Hence $2n+1 > 2t$ and $A_i=0$ for all odd $i$ such that $|i|<2n+1$. Since $0<n-k< n+k+1< 2n+1$, $n-k\not\equiv n+k+1 \pmod 2$, we have $A_{n-k}A_{n+k+1}=0$. A similar analysis shows that $A_{t+n-k}A_{n+k+1+t}=0$. Therefore,
$$
\begin{equation*}
\begin{aligned} \, &D\begin{pmatrix} 0 & t & n+k+1 \\ 0 & t & n-k \end{pmatrix} \\ &\qquad= \begin{vmatrix} A_0B_0 & 0 & A_{n-k}B_{k-n} \\ 0 & A_{2t}B_0 & A_{t+n-k}B_{t-n+k} \\ A_{n+k+1}B_{n+k+1} & A_{n+k+1+t}B_{n+k+1-t} & A_{2n+1} B_{2k+1} \end{vmatrix} \neq 0. \end{aligned}
\end{equation*}
\notag
$$
7. Since $\operatorname{supp} A\subset 2\mathbb{Z}$, we have $\operatorname{supp} A = 2t \mathbb{Z}$ by step 1 of the proof. It remains to take into account that $\operatorname{\text{gcd}}(2t,2k+1)=1$ by condition 1) of Lemma 5.1. This proves Lemma 5.3. Lemma 5.4. Under the hypotheses of Lemma 5.1, let the sequence $\operatorname{supp}B$ do not contain three neighbouring terms of the same parity, and there exist integer $i$ and $j$ such that Then a) if condition (5.1) implies that
$$
\begin{equation}
m_{j+1}-m_{j-1} = m_{j+2}-m_j = n_{i+1}-n_i = n_{i+2}-n_{i+1},
\end{equation}
\tag{5.2}
$$
then assertion c) of Lemma 5.1 holds up to an equivalence; b) otherwise, assertion a) or b) of Lemma 5.1 holds. Proof. a) Let condition a) be satisfied. Without loss of generality we can assume that $n_i=m_j=0$. We have where $d= 2k+1$, $k,l\in \mathbb{N}$, $k\geqslant l$.
1. We claim that $\operatorname{supp} A = d\mathbb{Z}$. In view of condition a), to this end it suffices to show that the parity of the elements of the sequence $\operatorname{supp} A$ alternates. Assume the contrary. Let $n_q$ be the smallest (in absolute value) number from $\operatorname{supp} A$ such that $n_q\equiv n_{q+1}\pmod 2$. Changing to an equivalent system it can be assumed that $n_q=n_{i+2}$. Hence $n_{i+3}-n_{i+2}= 2l$ by Lemma 5.2, that is, $n_{i+3}= 2(l+ d)$. Since
$$
\begin{equation*}
d< l+d\leqslant k+d < 2d, \qquad d< d+2l < 2d, \qquad 0<l< 2l,
\end{equation*}
\notag
$$
we have $A_{l+d}=A_{d+2l}= B_l =0$. Therefore,
$$
\begin{equation*}
D\begin{pmatrix} 0 & l & l+d\\ 0 & -l & l+d \end{pmatrix} = \begin{vmatrix} A_0B_0 & 0 & 0 \\ \ast & A_0B_{2l} & 0 \\ \ast & \ast & A_{2l+2d} B_0 \end{vmatrix}\neq 0,
\end{equation*}
\notag
$$
which contradicts Lemma 2.2. Therefore, the parity of the elements of $\operatorname{supp} A$ alternates. So, $\operatorname{supp} A = d\mathbb{Z}$.
2. We claim that the sequence $\operatorname{supp}B\pmod 2$ has period $0,0,1,1$. Assume the contrary. Since $\operatorname{supp}B$ does not have three neighbouring terms of the same parity and $m_j\equiv m_{j+1} \pmod 2$, there exists a number $l$ such that
$$
\begin{equation*}
m_l\equiv m_{l+1}\not\equiv m_{l+2}\not\equiv m_{l+3}\quad \text{or} \quad m_l\not\equiv m_{l+1}\not\equiv m_{l+2}\equiv m_{l+3} \pmod 2.
\end{equation*}
\notag
$$
Let $m_l\equiv m_{l+1}\not\equiv m_{l+2}\not\equiv m_{l+3} \pmod 2$ (the second case is reduced to this one by the substitution $\widehat B_n = B_{-n}$). By condition (5.2) we have $m_{l+2}-m_l = d$. Without loss of generality we can assume that $m_l=0$. Hence
$$
\begin{equation*}
\begin{gathered} \, n_i =0, \qquad n_{i+1}= d, \qquad n_{i+2} = 2d, \\ m_{l} = 0,\qquad m_{l+1} = 2l, \qquad m_{l+2} =d, \qquad m_{l+3} = 2s, \end{gathered}
\end{equation*}
\notag
$$
where $d\in \mathbb{N}$ is odd, $l,s\in \mathbb{N}$, $2l<d < 2s$.
If $B_{s+l}\neq 0$, then $s+l=d$. Hence $l-s = 2l -d \notin d \mathbb{Z}$, and $A_{l-s}=0$. Therefore, $B_{s+l}A_{l-s}=0$. Since $B_l= B_{d+s}=0$, we have
$$
\begin{equation*}
D\begin{pmatrix} l & d & s \\ -l & 0 & -s \end{pmatrix} = \begin{vmatrix} A_0B_{2l} & 0 & 0 \\ \ast & A_dB_d & 0 \\ \ast & \ast & A_0 B_{2s} \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
3. By the previous step and condition a), we have $m_j-m_{j+2}= d$ for all $j$. Hence $\operatorname{supp}B = d\mathbb{Z} \cup (d\mathbb{Z} + 2l)$. So, the pair $(B,A)$ satisfies assertion c) of Lemma 5.1. b) Let condition a) be not met. Then there are numbers $i$, $j$, satisfying (5.1) and for which at least one of the relations in (5.2) is not true. Since $\operatorname{supp}B$ does not contain three neighbouring terms of the same parity, we have
$$
\begin{equation*}
m_{j-1}\not\equiv m_j \equiv m_{j+1}\not \equiv m_{j+2}\pmod 2.
\end{equation*}
\notag
$$
We can assume that $n_{i+1}-n_i \neq m_{j+1}-m_{j-1}$ (the remaining cases can be reduced to this). In addition, we can assume that $n_{i+1}=m_{j}=0$. Hence
$$
\begin{equation*}
\begin{gathered} \, n_i=-2k-1, \qquad n_{i+1}= 0, \qquad n_{i+2} = 2t+1, \\ m_{j-1}=-2l-1, \qquad m_j = 0, \qquad m_{j+1}=2s, \end{gathered}
\end{equation*}
\notag
$$
where $k,t,l\in \mathbb{Z}_+$, $s\in\mathbb{N}$, and $k\neq s+l$.
1. We claim that $k=l$. Assume on the contrary that $k\neq l$. Then $A_{l-k}B_{k-l}=0$. Since $B_s=0$, we have
$$
\begin{equation*}
D\begin{pmatrix} -k-l-1 & 0 & s \\ l-k & 0 & -s \end{pmatrix} = \begin{vmatrix} A_{-2k-1}B_{-2l-1} & \ast & A_{-k-l-s-1}B_{s-k-l-1} \\ 0 & A_0B_0 & 0 \\ A_{s+l-k}B_{s-l+k} & \ast & A_0 B_{2s} \end{vmatrix} = 0.
\end{equation*}
\notag
$$
Hence $A_{s+l-k}B_{s-k-l-1}B_{s-l+k}\neq 0$. However, if $k>s+l$, then $A_{s+l-k}=0$, and if $k<s+l$, then $-2l-1<s-k-l-1< s-l+k<2s$, so $B_{s-k-l-1}B_{s-l+k} = 0$, a contradiction. Hence $k=l$.
2. Replacing $A_n$ to $\widehat A_n = A_{-n}$ and using the results of the previous step, we get $t=l$. So, $t=l=k$. 3. We claim that $m_{j+1}-m_j = 2 (m_j - m_{j-1})$, that is, $s=2k+1$. Assume on the contrary that $s\neq 2k+1$. Since $-2k-1<2s-2k-1< 2s$ and $2s-2k-1 \neq 0$, we have $B_{2s-2k-1}=0$. Next, since $-2k-1< s-2k-1 < 2s$, $s\neq 2k+1$, we have $B_{s-2k-1}=0$. Further, $B_s=0$, and so
$$
\begin{equation*}
\widetilde D\begin{pmatrix} -k-1 & s-k-1 & k+s \\ -k-1 & -k-s-1 & k-s \end{pmatrix} = \begin{vmatrix} A_{-2k-1}B_0 & 0 & 0 \\ \ast & A_{-2k-1} B_{2s} & 0 \\ \ast & \ast & A_{2k+1} B_{2s} \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
Hence $s=2k+1$, that is, $m_{j+1}-m_j = 2 (m_j - m_{j-1})$.
4. From the previous steps of the proof it follows that
$$
\begin{equation*}
n_i=-d, \quad n_{i+1}= 0, \quad n_{i+2} = d, \qquad m_{j-1}=-d, \quad m_j = 0, \quad m_{j+1}=2d,
\end{equation*}
\notag
$$
where $d=2k+1$, $k\geqslant 0$. We claim that $\operatorname{supp} A,\operatorname{supp}B\subset d \mathbb{Z}$. Assume the contrary. Let $n$ be the smallest (in absolute number) number from $\operatorname{supp} A\cup \operatorname{supp}B$ not divisible by $d$. We will consider only the case $n\geqslant 0$. In this case, $n>d$, and $A_q=B_q=0$ for all $|q|<n$, $q\not\equiv 0 \pmod d$.
Let $n$ be even, that is, $n=2m$. Then $m\not\equiv 0 \pmod d$. Hence $A_m = A_{m+d} = 0$. In addition, $B_d=0$. If $2m\in \operatorname{supp} A$, then
$$
\begin{equation*}
D\begin{pmatrix} -d & d & m \\ 0 & d & m \end{pmatrix} = \begin{vmatrix} A_{-d}B_{-d} & \ast & \ast \\ 0 & A_{2d} B_0 & \ast \\ 0 & 0 & A_{2m} B_0 \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
If $2m\in \operatorname{supp}B$, then
$$
\begin{equation*}
D\begin{pmatrix} -d & d & m \\ 0 & d & -m \end{pmatrix} \neq 0.
\end{equation*}
\notag
$$
Let $n$ be odd, that is, $n=2m+1$. Since $2m+1 \not\equiv 0 \pmod d$, the numbers $m-k$, $m+k+1$ are not divisible by $d$. Therefore, $A_{m-k} = A_{m+k+1}=0$. Moreover, $B_{2k+1}=0$. Hence
$$
\begin{equation*}
\begin{alignedat}{2} \widetilde D\begin{pmatrix} -k-1 & k & m \\ -k-1 & k & m \end{pmatrix} &\neq 0 &\quad &\text{if}\quad 2m+1 \in \operatorname{supp} A, \\ \widetilde D\begin{pmatrix} -k-1 & k & m \\ -k-1 & k & -m-1 \end{pmatrix} &\neq 0 &\quad &\text{if}\quad 2m+1 \in \operatorname{supp}B. \end{alignedat}
\end{equation*}
\notag
$$
Both cases are impossible. So, $\operatorname{supp} A,\operatorname{supp}B\subset d \mathbb{Z}$.
5. By condition 1) of Lemma 5.1, $d=1$. Hence $A_{-1}A_0A_1\neq 0$, $B_{-1}B_0B_2\neq 0$, $B_1=0$. If $B_{-2}\neq 0$, then assertion b) of Lemma 5.1 holds. Let $B_{-2}=0$. It remains to prove that $\operatorname{supp} A =\mathbb{Z}$. Suppose the contrary. Let $k$ be the smallest (in absolute value) number such that $A_k =0$. Without loss of generality we can assume that $k>0$. Hence $A_{k-1}A_{k-2}A_{k-3}\neq 0$. Let $k=2$ (otherwise we need to consider $\widetilde A_n = A_{n+k-3}$). We have
$$
\begin{equation*}
D\begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \end{pmatrix} = \begin{vmatrix} A_1B_{-1} & \ast & \ast \\ 0 & A_{-1} B_{-1} & \ast \\ 0 & 0 & A_0 B_2 \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
Therefore, $\operatorname{supp} A =\mathbb{Z}$, and we have case a) of Lemma 5.1. Proof of Lemma 5.1. According to Lemmas 5.3, 5.4 it remains to consider the case where the sequences $\operatorname{supp} A\pmod 2$, $\operatorname{supp}B \pmod 2$ have the same period $1,1,0,0$ or $1,0$. Indeed, let the conditions of Lemmas 5.3, 5.4 be not met and let one of the sequences, say $\operatorname{supp}B \pmod 2$, be not periodic with period $0,1$. Then $\operatorname{supp}B$ contains two neighbouring terms of the same parity. If $\operatorname{supp} A$ has three neighbouring terms of the same parity, then we are under the hypotheses of Lemma 5.3. If the sequence $\operatorname{supp} A$ has three neighbouring terms of alternating parity, then we arrive at the conditions of Lemma 5.4. So, $\operatorname{supp} A\pmod 2$ has period $0,0,1,1$. By changing $A$ and $B$, and repeating the above arguments, we conclude that the sequence $\operatorname{supp}B \pmod 2$ has period $0,0,1,1$.
So, let $\operatorname{supp} A\pmod 2$, $\operatorname{supp}B \pmod 2$ have the same period $1,1,0,0$ or $1,0$. Let us prove that assertion d) of Lemma 5.1 holds up to an equivalence. To this end, it suffices to verify that, for all $i$, $j$,
$$
\begin{equation}
\begin{gathered} \, n_{i+2}-n_i = m_{j+2}-m_j, \\ n_{i+1}-n_i = m_{j+1}-m_{j}\quad \text{or}\quad n_{i+1}-n_i = m_{j+2}-m_{j+1}. \end{gathered}
\end{equation}
\tag{5.3}
$$
Consider arbitrary $i$, $j$.
1. Let the sequences $\operatorname{supp} A\pmod 2$, $\operatorname{supp}B \pmod 2$ have period $1,1,0,0$. Then $n_i\equiv n_{i+1}\not\equiv n_{i+2}\pmod 2$ or $n_i\not\equiv n_{i+1}\equiv n_{i+2}\pmod 2$. Let $n_i\equiv n_{i+1}\not\equiv n_{i+2}\pmod 2$ (otherwise we need to consider $\widetilde A_n = A_{-n}$). By the same reason we can assume that $m_j\equiv m_{j+1}\not\equiv m_{j+2}\pmod 2$. Hence $n_{i+1}-n_i = m_{j+1}-m_j$ by Lemma 5.2. We can consider only the case
$$
\begin{equation*}
n_i=0, \quad n_{i+1} = 2k, \quad n_{i+2} = 2s+1, \qquad m_j=0, \quad m_{j+1} = 2k, \quad m_{j+2} = 2t+1.
\end{equation*}
\notag
$$
It is remains to prove that $s=t$. Assume the contrary. Let, for example, $s>t$. Note that $A_k=B_k=0$. Besides, we have and hence $A_{k+s-t}A_{s+t-k+1}=0$, $A_{s-t}A_{s+t+1}=0$. Therefore,
$$
\begin{equation*}
\begin{aligned} \, D\begin{pmatrix} 0 & k & s+t+1 \\ 0 & -k & s-t \end{pmatrix} &= \begin{vmatrix} A_0B_0 & 0 & A_{s-t}B_{t-s} \\ 0 & A_0 B_{2k} & A_{k+s-t}B_{k-s+t} \\ A_{s+t+1}B_{s+t+1} & A_{s+t+1-k}B_{s+t+k+1} & A_{2s+1} B_{2t+1} \end{vmatrix} \\ &= A_0B_0A_0B_{2k}A_{2s+1}B_{2t+1} \neq 0. \end{aligned}
\end{equation*}
\notag
$$
So, $s=t$. Hence conditions (5.3) are satisfied.
2. Let the terms of the sequences $\operatorname{supp} A$, $\operatorname{supp}B$ have alternating parities. Without loss of generality we can assume that
$$
\begin{equation*}
n_i=0, \quad n_{i+1} = 2k+1, \quad n_{i+2} = 2t, \qquad m_j=0, \quad m_{j+1} = 2l+1, \quad m_{j+2} = 2s,
\end{equation*}
\notag
$$
where $t\leqslant s$.
2.1. Let us prove that $n_{i+1}-n_i = m_{j+1}-m_j$ or $n_{i+2}-n_{i+1} = m_{j+1}-m_j$. Assume the contrary. Then $k\neq l$, $t\neq k+l+1$. Hence $A_tB_t = A_{k+l+1}B_{k+l+1}=0$. Furthermore, $B_{t-k+l}=0$, since $0< t-k+l < 2s$, $t-k+l \neq 2l+1$. So,
$$
\begin{equation*}
D\begin{pmatrix} 0 & k+l+1 & t \\ 0 & k-l & t \end{pmatrix} = \begin{vmatrix} A_0B_0 & \ast & \ast \\ 0 & A_{2k+1} B_{2l+1} & \ast \\ 0 & 0 & A_{2t} B_0 \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
2.2. We next assume that $n_{i+1}-n_i = m_{j+1}-m_j$, that is, $l=k$ (otherwise we need to consider $\widetilde A = A_{-n}$). We claim that $t=s$. Assume the opposite. Then $t<s$. Since $2k+1<t+s < 2s$, we have $B_{t+s}=0$. If $s\neq 2k+1$, then $B_s=A_tB_{2s-t}=0$,
$$
\begin{equation*}
D\begin{pmatrix} 0 & s & t+s \\ 0 & s & t-s \\ \end{pmatrix} = \begin{vmatrix} A_0B_0 & \ast & \ast \\ 0 & A_{2s} B_0 & 0 \\ 0 & \ast & A_{2t} B_{2s} \end{vmatrix} \neq 0,
\end{equation*}
\notag
$$
and if $s=2k+1$, then $t\neq 2k+1$, $A_t=A_{2t-s}=0$,
$$
\begin{equation*}
D\begin{pmatrix} 0 & t & t+s \\ 0 & t & t-s \\ \end{pmatrix} = \begin{vmatrix} A_0B_0 & \ast & \ast \\ 0 & A_{2t} B_0 & 0 \\ 0 & \ast & A_{2t} B_{2s} \end{vmatrix} \neq 0.
\end{equation*}
\notag
$$
Therefore, $t=s$. Conditions (5.3) are satisfied. This verifies assertion d) of Lemma 5.1, and, therefore, Lemma 5.1. § 6. The case $R_0(A,B)=R_1(A,B)=2$If the conditions of Lemma 5.1 are met, then, up to an equivalence, one of the following assertions holds: i) $\operatorname{supp} A =\operatorname{supp}B =\mathbb{Z}$; ii) $\operatorname{supp} A\neq \mathbb{Z}$, $\operatorname{supp}B \neq \mathbb{Z}$ and there exist $n,m\in \mathbb{Z}$ such that iii) $\operatorname{supp} A =\mathbb{Z}$, $\operatorname{supp}B\neq \mathbb{Z}$, $\operatorname{supp}B\neq 2\mathbb{Z}$, $\operatorname{supp}B\neq 2\mathbb{Z} + 1$; iv) $\operatorname{supp} A = d\mathbb{Z} \cup(d\mathbb{Z} + t)$, $\operatorname{supp}B = d \mathbb{Z}$, where $d\geqslant 2$, $d\not\equiv t \pmod 2$, and $\operatorname{\text{gcd}}(d,t)=1$; v) $\operatorname{supp} A = \operatorname{supp}B= d\mathbb{Z} \cup(d\mathbb{Z} + t)$, where $d\geqslant 3$, $d\not\equiv t \pmod 2$, $\operatorname{\text{gcd}}(d,t)=1$. The case where $\operatorname{supp} A =\mathbb{Z}$, and $\operatorname{supp}B =2\mathbb{Z}$ or $\operatorname{supp}B =2\mathbb{Z} +1$ is contained in assertion iv) with $d=2$. Let us consider separately the cases of fulfillment of each of the above assertions. 6.1. Fulfillment of assertion i)Theorem 6.1. Assume that the sequences $A$, $B$ contain no zero terms, and 1) there are $\alpha,\gamma\in \mathbb{C}\setminus\{0\}$, $\beta,\delta \in \mathbb{C}$ such that, for all $n\in \mathbb{Z}$,
$$
\begin{equation*}
A_{n+2}B_{n-2} = \alpha A_{n+1}B_{n-1}+\beta A_n B_n,\quad A_{n-2}B_{n+2} = \gamma A_{n-1}B_{n+1}+\delta A_n B_n;
\end{equation*}
\notag
$$
2) the following relations hold
$$
\begin{equation*}
D\begin{pmatrix} 1 & 0 & -1 \\ 1 & 0 & -1 \end{pmatrix} = \widetilde D\begin{pmatrix} 1 & 0 & -1 \\ 1 & 0 & -1 \end{pmatrix} = \widetilde D\begin{pmatrix} 1 & 0 & -1 \\ 0 & -1 & -2 \end{pmatrix}=0;
\end{equation*}
\notag
$$
3) the inequality $\Delta_1\Delta_2\Delta_3\neq 0$ holds, where
$$
\begin{equation*}
\Delta_1= \begin{vmatrix} A_2B_0 & A_1B_1 \\ A_1 B_{-1} & A_0B_0 \end{vmatrix}, \quad \Delta_2= \begin{vmatrix} A_0B_2 & A_1B_1 \\ A_{-1} B_1 & A_0B_0 \end{vmatrix}, \quad \Delta_3= \begin{vmatrix} A_1B_0 & A_0B_1 \\ A_0 B_{-1} & A_{-1}B_0 \end{vmatrix}.
\end{equation*}
\notag
$$
Then $(A,B)\sim (\widetilde A,\widetilde B)$, where $\widetilde A_n = \sigma_L(nv+u_1)$, $\widetilde B_n = \sigma_L(nv+u_2)$ ($u_1,u_2,v\in \mathbb{C}$, and $L$ is any lattice) or $\widetilde A\equiv 1$, $\widetilde B$ is a quasi-polynomial of rank $2$. Theorem 6.1 was obtained in [14]. Note that the proof (and the result in [14] contains two inaccuracies. First, the condition $\alpha \gamma \neq 0$ is missing. If $\alpha \gamma = 0$, then appear new solutions (see [7] or step 3 of the proof of Lemma 6.1). Second, the case when one of the sequences $A$ or $B$ is constant was not considered (in the language of [14] this occurs when one of the points $z_1$, $z_2$ is a point) at infinity. For example, if $A\equiv c$, then $B$ is a quasi-polynomial of rank $2$ (see [31]). Lemma 6.1. Let $\operatorname{supp} A =\operatorname{supp}B = \mathbb{Z}$ and $R_0(A,B)=R_1(A,B)=2$. Then $(A,B)\sim (\widetilde A,\widetilde B)$, where the system $(\widetilde A, \widetilde B)$ is one of the types 1)–4) in Theorem 3.1. Proof. If $(A,B)\sim (A,A)$, then the required assertion follows from Theorem 7.1 in [3]. So, in what follows, we will assume that $(A,B)\not\sim (A,A)$.
Condition 2) of Theorem 6.1 is secured by Lemma 2.2. Since
$$
\begin{equation*}
D\begin{pmatrix} n & 1 & 0 \\ 2 & 1 & 0 \end{pmatrix} =0, \qquad D\begin{pmatrix} n & 1 & 0 \\ -2 & -1 & 0 \end{pmatrix} =0
\end{equation*}
\notag
$$
we have, for some $h_1,h_2,g_1,g_2\in \mathbb{C}$,
$$
\begin{equation*}
\begin{aligned} \, \Delta_1 A_{n+2}B_{n-2}- h_1 A_{n+1}B_{n-1}+ g_1 A_n B_n &=0, \\ \Delta_2 A_{n-2}B_{n+2}- h_2 A_{n-1}B_{n+1}+ g_2 A_n B_n &=0. \end{aligned}
\end{equation*}
\notag
$$
If $h_1h_2\Delta_1\Delta_2\Delta_3\neq 0$, then the required result follows from Theorem 6.1. Note that if $v\in L$, then $A_n\equiv\sigma_L(u_1)$, $B_n\equiv\sigma_L(u_2)$ and $R(A,B)\leqslant 2$. If $v\in (\frac{1}{2}L) \setminus L$, then by the quasi-periodicity of the Weierstrass sigma function, $(A,B)\sim (\widetilde A, \widetilde B)$, where $\widetilde A_{2n}=\widetilde B_{2n}=1$, $\widetilde A_{2n+1}=c_1$, $\widetilde B_{2n+1}=c_2$. In addition, $c_1\neq \pm c_2$, $c_1c_2 \neq \pm 1$ (see Lemma 2.3 for $\beta=0$). If $v\in \frac{1}{2} L$, then we get the case 4 of Theorem 3.1, where $d=2$, $t=1$, $\beta=0$.
It remains to consider the cases $h_1h_2\Delta_1\Delta_2\Delta_3= 0$. 1. Let $\Delta_1=0$. From the relation
$$
\begin{equation*}
D\begin{pmatrix} n & 1 & 0 \\ m & 1 & 0 \end{pmatrix} = \begin{vmatrix} A_{n+m}B_{n-m} & A_{n+1}B_{n-1} & A_nB_n \\ A_{1+m}B_{1-m} & A_2 B_0 & A_1B_1 \\ A_m B_{-m} & A_1B_{-1} & A_0B_0 \end{vmatrix} =0
\end{equation*}
\notag
$$
it follows that $\lambda(m) A_{n+1}B_{n-1} = \mu(m) A_n B_n$, where
$$
\begin{equation*}
\lambda(m) = \begin{vmatrix} A_{1+m}B_{1-m} & A_1B_1 \\ A_m B_{-m} & A_0B_0 \end{vmatrix},\qquad \mu(m) = \begin{vmatrix} A_{1+m}B_{1-m} & A_2 B_0\\ A_m B_{-m} & A_1B_{-1} \end{vmatrix}.
\end{equation*}
\notag
$$
Assume that there exists a‘number $m$ such that $\lambda(m)\neq 0$. Then $A_{n+1}B_{n-1} = c A_n B_n$, where $c = \mu(m)/\lambda(m)$. Hence $A_{n+1} = B_n c^n A_1 B_0^{-1}$, and, therefore, $(A,B)\sim (A,A)$. Let $\lambda(m)= 0$ for all $m$. Then, for $\widetilde B_m = B_{-m}$ we obtain an equation of the type $A_{m+1}\widetilde B_{m-1} = c A_m \widetilde B_m$. So, we have arrived at the case considered above. 2. The case $\Delta_2=0$ (or $\Delta_3=0$) is reduced to the case 1 by the substitution $(A,B)\to (B, A)$ (or by the substitution $(A,B)\to (\widetilde A, B)$, where $\widetilde A_n = A_{n-1}$). 3. Let $h_1h_2 =0$, $\Delta_1\Delta_2\neq 0$. We claim that in this case assertion 4) of Theorem 3.1 is satisfied, where $d=2$. Let, for example, $h_1=0$. Then $A_{n+2}B_{n-2}= c A_n B_n$, $c= - g_1/\Delta_1$. Without loss of generality we can assume that $c=1$ (otherwise we consider the equivalent system $(\widetilde A,B)$, where $A_n = \widetilde A_n c^{n/2}$). The equation $A_{n+2}B_{n-2}= A_n B_n$ splits into the system of two linear first-order difference equations (for $A_{2k}$ and $A_{2k+1}$), which have the unique solution
$$
\begin{equation}
A_{2k} = \lambda_0 B_{2k-2} \quad \biggl(\lambda_0 = \frac{A_0}{B_{-2}}\biggr), \qquad A_{2k+1} = \lambda_1 B_{2k-1} \quad \biggl(\lambda_1 = \frac{A_1}{B_{-1}}\biggr).
\end{equation}
\tag{6.1}
$$
If $\lambda_0=\pm\lambda_1$, then $A_n=\lambda_0 B_{n-2}$ for $\lambda_0=\lambda_1$ and $A_n=\lambda_0 (-1)^n B_{n-2}$ for $\lambda_0 = {-}\lambda_1$. Hence $(A,B)\sim (A,A)$. Therefore, $\lambda_0\neq \pm\lambda_1$. Since $R_1(A,B)=2$, the second equation in (1.1) is satisfied with $N_1=2$. Choosing $m=0,1,2$, we obtain three relations, whose right-hand sides are linearly dependent. So, there exist $(\alpha_0,\alpha_1,\alpha_2)\in \mathbb{C}^3\setminus\{0\}$ such that
$$
\begin{equation*}
\alpha_2 A_{n+3}B_{n-2}= \alpha_1 A_{n+2}B_{n-1}+ \alpha_0 A_{n+1} B_n.
\end{equation*}
\notag
$$
Putting $n=2k$ and $n=2k+1$ in the last equality and using (6.1), we obtain
$$
\begin{equation*}
\begin{gathered} \, \alpha_2 \lambda_1 B_{2k+1}B_{2k-2} = (\alpha_1\lambda_0+ \alpha_0\lambda_1)B_{2k}B_{2k-1}, \\ \alpha_2 \lambda_0 B_{2k+2}B_{2k-1} = (\alpha_1\lambda_1+ \alpha_0\lambda_0)B_{2k+1}B_{2k}. \end{gathered}
\end{equation*}
\notag
$$
If $\alpha_2 = 0$, then we get from these equalities that $\lambda_0=\pm\lambda_1$. Hence $\alpha_2\neq 0$, and, therefore, $B_{2k+1}B_{2(k-1)} =\gamma_0 B_{2k} B_{2k-1}$, $B_{2(k+1)}B_{2k-1} =\gamma_1 B_{2k} B_{2k+1}$, where $\gamma_0,\gamma_1\in \mathbb{C}$. Consequently,
$$
\begin{equation*}
\begin{aligned} \, &B_{2k+1} = B_1 B_0^{-1} \gamma_0^k B_{2k}, \quad B_{2k} = B_0 B_{-1}^{-1} \gamma_1^k B_{2k-1} \\ &\Longrightarrow \quad B_{2k+1} = B_1 \biggl(\frac{B_1}{B_{-1}}\biggr)^k (\gamma_0\gamma_1)^{k(k+1)/2}, \quad B_{2k} = B_0 \biggl(\frac{B_1\gamma_1}{B_{-1}}\biggr)^{k} (\gamma_0\gamma_1)^{k(k-1)/2}. \end{aligned}
\end{equation*}
\notag
$$
In view of (6.1) one can easily check that $(A,B)\sim (\widetilde A,\widetilde B)$, where $(\widetilde A,\widetilde B)$ has a type specified in Lemma 2.3. Using this lemma, we arrive at case 4) of Theorem 3.1, in which $d=2$. This proves Lemma 6.1. 6.2. Fulfillment of assertion ii)Lemma 6.2. Let $R_0(A,B)\leqslant 2$, $A_0=B_0 =0$, $A_1A_2A_3 B_1B_2B_3 \neq 0$. Then a) the sequences $A$, $B$ are uniquely determined by their terms with numbers from $-1$ to $4$; b) if $A_{-1}B_{-1}=0$, then $A_k=B_k = 0$ for $k\notin \{1,2,3,4\}$; c) if $R_1(A,B)\leqslant 2$ and $A_{\pm 1}=B_{\pm 1}=\pm 1$, then $A\equiv B$ or $A_n = B_n (-1)^{n+1}$. Proof. Suppose we know the terms of the sequences $A$, $B$ with numbers from $-2k+1$ to $2k+2$, where $k\in \mathbb{N}$. According to Lemma 2.2,
$$
\begin{equation}
D\begin{pmatrix} n+1 & 2 & 1 \\ n-1 & 1 & 0 \end{pmatrix} = \begin{vmatrix} A_{2n}B_2 & A_{n+2}B_n & A_{n+1}B_{n+1} \\ A_{n+1}B_{3-n} & A_3 B_1 & A_2B_2 \\ A_n B_{2-n} & 0 & A_1B_1 \end{vmatrix} =0,
\end{equation}
\tag{6.2}
$$
$$
\begin{equation}
D\begin{pmatrix} n+1 & 2 & 1 \\ -n+1 & -1 & 0 \end{pmatrix} = \begin{vmatrix} A_2B_{2n} & A_nB_{n+2} & A_{n+1}B_{n+1} \\ A_{3-n}B_{n+1} & A_1 B_3 & A_2B_2 \\ A_{2-n} B_n & 0 & A_1B_1 \end{vmatrix} =0,
\end{equation}
\tag{6.3}
$$
$$
\begin{equation}
D\begin{pmatrix} n & 2 & 1 \\ n-1 & 1 & 0 \end{pmatrix} = \begin{vmatrix} A_{2n-1}B_1 & A_{n+1}B_{n-1} & A_nB_n \\ A_{n+1}B_{3-n} & A_3 B_1 & A_2B_2 \\ A_n B_{2-n} & 0 & A_1B_1 \end{vmatrix} =0,
\end{equation}
\tag{6.4}
$$
$$
\begin{equation}
D\begin{pmatrix} n & 2 & 1 \\ -n+1 & -1 & 0 \end{pmatrix} = \begin{vmatrix} A_1B_{2n-1} & A_{n-1}B_{n+1} & A_nB_n \\ A_{3-n}B_{n+1} & A_1 B_3 & A_2B_2 \\ A_{2-n} B_n & 0 & A_1B_1 \end{vmatrix} =0 .
\end{equation}
\tag{6.5}
$$
Taking $n=-k$ in (6.2), (6.3), we evaluate $A_{-2k}$ and $B_{-2k}$, respectively. Choosing $n=k+2$ in (6.4), (6.5), and then $n=-k$, we evaluate $A_{2k+3}$ and $B_{2k+3}$, as well as $A_{-2k-1}$ and $B_{-2k-1}$. Setting $n=k+2$ in (6.2), (6.3), we evaluate $A_{2k+4}$ and $B_{2k+4}$. Now assertion a) easily follows by induction.
Let us verify b). Let, for instance, $A_{-1}=0$. Choosing in (6.5) $n=0$, we get $B_{-1}=0$. Choosing $n=-1$ in (6.2)–(6.5), we find that $A_{-2}=0$, $B_{-2}=0$, $A_{-3}=0$, $B_{-3}=0$. Continuing the process, we conclude that $A_n=B_n=0$ for $n\notin\{1,2,3,4\}$. Let us prove c). Choosing $n=2$ and $n=3$ in the equality
$$
\begin{equation*}
D\begin{pmatrix} n & 1 & 0 \\ 1 & -1 & 0 \end{pmatrix} = \begin{vmatrix} A_{n+1}B_{n-1} & A_{n-1}B_{n+1} & \ast \\ 0 & 0 & 1 \\ -1 & -1 & \ast \end{vmatrix} = A_{n+1}B_{n-1}- A_{n-1}B_{n+1} =0,
\end{equation*}
\notag
$$
we obtain $A_3=B_3$ and $A_4B_2 = B_4A_2$, respectively. Since
$$
\begin{equation*}
\widetilde D\begin{pmatrix} 2 & 1 & 0 \\ 1 & -1 & 0 \end{pmatrix} = \begin{vmatrix} \ast & A_2B_3 & A_3B_2 \\ \ast & B_2 & A_2 \\ -A_2 & 0 & 0 \end{vmatrix} = -A_2 (A_2^2B_3 -B_2^2A_3)=0,
\end{equation*}
\notag
$$
we have $A_2^2 = B_2^2$. There are two cases to consider.
1. Let $A_2=B_2$. Then $A_n = B_n$ for $-1\leqslant n\leqslant 4$. Since $R_0(A, B)=R_0(B,A)\leqslant 2$ and $(A_n, B_n) = (B_n,A_n)$ for $-1\leqslant n\leqslant 4$, we have $(A,B)=(B,A)$ by assertion a). So, $A\equiv B$. 2. Let $A_2=-B_2$. Then $A_n=B_n$ for $n=\pm 1, 3$ and $A_n=-B_n$ for $n=2,4$. Putting $\widetilde B_n = B_n(-1)^{n+1}$, we have $R_0(A,\widetilde B)\leqslant 2$ and $A_n = \widetilde B_n$ for $-1\leqslant n \leqslant 4$. As in the previous case, we have $A_n=\widetilde B_n$, proving Lemma 6.2. Lemma 6.3. Let $R_0(A,B)=R_1(A,B)=2$ and let assertion ii) hold. Then $(A,B)\sim (\widetilde A, \widetilde B)$, where $(\widetilde A, \widetilde B)$ satisfies one of the conditions: 1) $\widetilde A_n=\widetilde B_n =0$ for $n\notin\{1,2,3,4\}$, $\widetilde A_1\widetilde A_2\widetilde A_3\widetilde B_1 \widetilde B_2\widetilde B_3\neq 0$, and $\widetilde A_4\neq 0$ or $\widetilde B_4\neq 0$; 2) $\widetilde A_n=\widetilde B_n = \sigma_L(nv)$, where $v\notin \frac{1}{2}L$; 3) assertion 4) of Theorem 3.1 holds, where $d=2$ and $c_1=c_2$. Proof. By ii), there exist $s$, $t$ such that Without loss of generality we can assume that $s=t=0$.
If $A_{-1}B_{-1}=0$, then $A_n=B_n =0$ for $n\notin\{1,2,3,4\}$ by Lemma 5.2. Furthermore, $A_4\neq 0$ or $B_4\neq 0$, for otherwise $R(A,B)=3$ according to Lemma 4.3. Let $A_{-1}B_{-1}\,{\neq}\, 0$. We can assume that $A_{\pm 1} = B_{\pm 1} = \pm 1$ (otherwise we can consider an equivalent system $(\widetilde A,\widetilde B)$, where $\widetilde A_n = A_ne(\alpha_1 n +\beta_1)$, $B_m= B_me(\alpha_2 m + \beta_2)$, the coefficients $\alpha_1$, $\alpha_2$, $\beta_1$, $\beta_2$ are chosen so that $\widetilde A_{\pm 1}=\widetilde B_{\pm 1}= \pm 1$). From Lemma 6.2, c), we have $A_n= B_n$, up to an equivalence. Therefore, $R_0(A,A)=R_1(A,A)=2$. It remains to invoke Theorem 7.1 in [3], completing the proof. 6.3. Fulfillment of assertion iii)Lemma 6.4. Let $R_0(A,B)\leqslant 2$ and $\operatorname{supp} A =\mathbb{Z}$, $B_0=0$. Then the set $Z_B = \{m\in \mathbb{Z}\colon B_m=0\}$ is a subgroup of the group $(\mathbb{Z},+)$. Proof. By the conditions, there exist $a_1,a_2,b_1,b_2\colon \mathbb{Z}\to \mathbb{C}$ such that where $\vec{a}=(a_1,a_2)$, $\vec{b}=(b_1,b_2)$, $\vec{a}(n)\cdot \vec{b}(m) = a_1(n)b_1(m)+a_2(n)b_2(m)$.
Since $A$ has no zero terms, we have $\vec{a}(n)\neq 0$ and $\vec{b}(n)\neq 0$ for all integer $n$. Let $t\in Z_B$ be arbitrary. Then $\vec{a}(0)\vec{b}(0) =\vec{a}(0)\vec{b}(-t) =0$. Since $\vec{a}(0)\neq 0$, the vectors $\vec{b}(0)$, $\vec{b}(-t)$ are linearly dependent (over $\mathbb{C}$). Hence $c_0 \vec{b}(0) = c_1 \vec{b}(-t)$, where $c_0c_1\neq 0$. Now the equality $c_1A_{n-t}B_{n+t} = c_0 A_nB_n$ holds by (6.6). Putting $n= s-t$ in this equality, where $s\in Z_B$ is arbitrary, we obtain $c_0 A_{s-t}B_{s-t}=0$, that is, $s-t\in Z_B$. Hence $(Z_B,+)$ is a group. This proves Lemma 6.4. Lemma 6.5. Let $R_j(A,B)\leqslant 2$, $j=0,1$, and $\operatorname{supp} A =\mathbb{Z}$, $B_0=0$. Then the sequences $A$, $B$ are uniquely defined by their terms $A_n$, $B_m$ with $-2\leqslant n\leqslant 2$, $-1\leqslant m\leqslant 2$. Proof. The element $B_{-2}$ can be found from the formula
$$
\begin{equation}
D\begin{pmatrix} m & 1 & 0 \\ 1 & -1 & 0 \end{pmatrix} = \begin{vmatrix} A_{m+1}B_{m-1} & A_{m-1}B_{m+1} & A_m B_m \\ 0 & A_0 B_2 & A_1 B_1 \\ A_1 B_{-1} & A_{-1} B_1 & 0 \end{vmatrix}= 0,
\end{equation}
\tag{6.7}
$$
in which we put $m=-1$. Note that $B_{\pm 1}\neq 0$ by Lemma 6.4. By symmetry, it suffices to show how $A_n,B_n$ are evaluated for $n>0$. Suppose that we have evaluated all $A_n,B_n$ for $1\leqslant n\leqslant m$ ($m\geqslant 2$). Then the element $B_{m+1}$ is obtained from the equality
$$
\begin{equation*}
\widetilde D\begin{pmatrix} m-1 & 0 & -1 \\ -2 & -1 & 0 \end{pmatrix} = \begin{vmatrix} A_{m-2}B_{m+1} & A_{m-1}B_m & A_m B_{m-1} \\ A_{-1}B_2 & A_0 B_1 & 0 \\ A_{-2} B_1 & 0 & A_0 B_{-1} \end{vmatrix}= 0.
\end{equation*}
\notag
$$
If $B_{m-1}\,{\neq}\, 0$, then the element $A_{m+1}$ can be found from (6.7). If $B_{m-1} = 0$, then $B_{m-2}\,{\neq}\, 0$ (by Lemma 6.4) and the element $A_{m+1}$ can be evaluated from the relation
$$
\begin{equation*}
\widetilde D\begin{pmatrix} m-1 & 1 & 0 \\ 1 & -1 & 0 \end{pmatrix} = \begin{vmatrix} A_{m+1}B_{m-2} & A_{m-1}B_m & A_m B_{m-1} \\ 0 & A_1 B_2 & A_2 B_1 \\ A_2 B_{-1} & A_0B_1 & 0 \end{vmatrix}= 0.
\end{equation*}
\notag
$$
This proves Lemma 6.5. Let $\wp_L = -(\ln \sigma_L)''$ be the Weierstrass $\wp$-function. Lemma 6.6. Let $c_{-1},c_0,c_1, k\in \mathbb{C}\setminus\{0\}$, and $c_0\neq c_{-1}$ or $c_0\neq c_1$. Then there exist a lattice $L$ (possible degenerate) and points $u\in \mathbb{C}\setminus L$, $v\in \mathbb{C}\setminus \frac{1}{2}L$ such that Proof. For brevity, we put $\wp_L = \wp$. Let us show that the quantities $(\xi_1,\nu_1)= (\wp(u),\wp'(u))$, $(\xi_2,\nu_2)= (\wp(v),\wp'(v))$ and the parameters of the elliptic curve $E$, associated with the lattice $L$ can be uniquely determined from (6.8).
It is clear that $\nu_2 = k$. Since $c_{-1}-c_1 = \wp(u+v)-\wp(u-v)$, the formula
$$
\begin{equation*}
\wp(u+v)-\wp(u-v) = \frac{\wp'(u)\wp'(v)}{(\wp(u)-\wp(v))^2}= \frac{ \nu_1 k}{c_0^2}
\end{equation*}
\notag
$$
(see [32]) implies that $\nu_1 = (c_{-1}-c_1)c_0^2 k^{-1}$. Using the addition formula for the $\wp$-function, we have
$$
\begin{equation*}
\wp(u+v) = c_2 -\wp(u)-\wp(v), \qquad c_2 = \frac{1}{4} \biggl(\frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)} \biggr)^2 = \frac{(\nu_1-k)^2}{4c_0^2}.
\end{equation*}
\notag
$$
Substituting this relation into the third equation in (6.8), we find that $2\wp(v) +\wp (u) = c_1+c_2$, that is, $2\xi_2 +\xi_1 = c_1+c_2$. Considering the second equation in (6.8), we have
$$
\begin{equation*}
\xi_1 = \frac{c_1-2 c_0+c_2}{3}, \qquad \xi_2 = \frac{c_1+ c_0+c_2}{3}.
\end{equation*}
\notag
$$
Let the elliptic curve $E=E(g_2,g_3)$ be given by the equation $y^2= 4x^3-g_2 x-g_3$. Then the parameters $g_2$, $g_3$ form a unique solution of the system
$$
\begin{equation*}
\nu_1^2= 4 \xi_1^3-g_2 \xi_1 - g_3, \qquad \nu_2^2= 4 \xi_2^3-g_2 \xi_2 - g_3.
\end{equation*}
\notag
$$
The converse result also holds: if
$$
\begin{equation*}
(\wp(u),\wp'(u))=(\xi_1,\nu_1), \qquad (\wp(v),\wp'(v))=(\xi_2,\nu_2), \qquad E=E(g_1,g_2),
\end{equation*}
\notag
$$
where $\xi_j$, $\nu_j$, $g_j$ are determined by the above formulas, then conditions (6.8) are met. It remains to choose points $u$, $v$. There exist (possible infinite) points $u$, $v$ such that $(\wp(u),\wp'(u))=(\xi_1,\nu_1)$, $(\wp(v),\wp'(v))=(\xi_2,\nu_2)$. If the curve $E$ is non-degenerate, then $u,v\in \mathbb{C}$. Let $E$ be a degenerate curve. Since $\wp'(v)=k\neq 0$, we have $v\in \mathbb{C}$. If $u$ is infinite, then from (6.8) we have $c_{-1}=c_0=c_1$. Lemma 6.7. Let $R_0(A,B)=R_1(A,B)=2$ and let assertion iii) hold. Then $(A,B)\sim (\widetilde A,\widetilde B)$, where $(\widetilde A,\widetilde B)$ is of the form of 1), 2) or 3) in Theorem 3.1. Proof. Consider the sequence $c_n = A_{n-1}A_{n+1} A_n^{-2}$.
1. Let $\{c_n\}$ be non-constant. Then there is a number $t$ such that $c_t\neq c_{t+1}$. Without loss of generality we can assume that $t=0$ and $B_0 =0$, $B_{\pm 1} = \pm 1$. By Lemma 6.6, there exist a lattice $L$ and points $u\in\mathbb{C}\setminus L$, $v\in \mathbb{C}\setminus \frac{1}{2} L$, satisfying (6.8), where $k= -B_2$. Note that $B_2\neq 0$ by Lemma 6.4 and the condition $\operatorname{supp}B \neq 2\mathbb{Z} +1$. Consider the sequences
$$
\begin{equation*}
\Phi_n = \frac{A_0}{\sigma(u)} \biggl(\frac{\sigma(v)\sigma(u) A_1}{\sigma(u+v) A_0}\biggr)^n \frac{\sigma(nv+u)}{\sigma(v)^{n^2}} , \qquad \Psi_m = \frac{\sigma(mv)}{\sigma(v)^{m^2}}.
\end{equation*}
\notag
$$
Here and below, $\sigma = \sigma_L$, $\wp = \wp_L$. We have $\Phi_0= A_0$, $\Phi_1= A_1$, $\Psi_{\pm 1}= \pm 1$, $\Psi_0 = 0$. From (1.3) we have
$$
\begin{equation*}
\sigma(2v+u) = \sigma((v+u)+v) = \frac{\sigma^2(v)\sigma^2(u+v)}{\sigma(u)} \bigl(\wp(v)-\wp(v+u)\bigr),
\end{equation*}
\notag
$$
using which in view of (1.3) and (6.8) we have
$$
\begin{equation*}
\begin{aligned} \, \Phi_{-1} &= \frac{A_0^2}{A_1}\frac{\sigma(u+v)\sigma(u-v)} {\sigma(u)^2\sigma(v)^2} =\frac{A_0^2}{A_1} \bigl(\wp(v)-\wp(u)\bigr)= \frac{A_0^2}{A_1} c_{-1} = A_{-1}, \\ \Phi_2 &= \frac{A_1^2}{A_0}\frac{\sigma(u)\sigma(u+2v)} {\sigma(v)^2 \sigma(u+v)^2} = \frac{A_1^2}{A_0} \bigl(\wp(v)-\wp(u+v)\bigr)= \frac{A_1^2}{A_0} c_0 = A_2, \\ \Phi_{-2} &= \frac{A_0^3}{A_1^2}\frac{\sigma(u+v)^2\sigma(u-2v)} {\sigma(u)^3 \sigma(v)^6} =\frac{A_0^3}{A_1^2} \frac{\sigma(u+v)^2 \sigma(v)^2\sigma(u-v)^2} {\sigma(u)^4 \sigma(v)^6}\bigl(\wp(v)-\wp(u-v)\bigr) \\ &=\frac{A_0^3}{A_1^2} \bigl(\wp(v)-\wp(u)\bigr)^2\bigl(\wp(v)- \wp(u-v)\bigr)= \frac{A_0^3}{A_1^2} c_0^2 c_{-1} = A_{-2}. \end{aligned}
\end{equation*}
\notag
$$
Furthermore, $\Psi_2 = \sigma_L(2v)/\sigma_L(v)^4 = -\wp'(v) = B_2$. Thus, $A_n = \Phi_n$, $B_m = \Psi_m$ for $-2\leqslant n\leqslant 2$, $-1\leqslant m \leqslant 2$. From (1.3) it follows that $R_j(\Phi,\Psi)\leqslant 2$, $j=0,1$. Hence $(\Phi,\Psi) = (A,B)$ by Lemma 6.5. We have arrived at case 1 of Theorem 3.1.
2. Let $c_n=c$ for any $n$. The transformation of sequences of the type $A_n \to A_n e(\alpha n^2 +\beta n)$ does not change the values $c_{n+1}/c_n$. Therefore, we assume that $A_{-1}=A_0=A_1$ (otherwise we consider an equivalent system with this property). We have $A_{n-1}A_{n+1}A_n^{-2} \equiv 1$ and $A\equiv A_0$. Since $R_0(A,B)=2$, $B$ is a quasi-polynomial of rank $2$ (see, for example, [21]). Hence assertion 2) or 3) of Theorem 3.1 is satisfied. 6.4. Fulfillment of assertions iv) or v)Lemma 6.8. Let $R_0(A,B)=R_1(A,B)=2$ and let assertion iv) or v) hold. Then $(A,B)\sim (\widetilde A, \widetilde B)$, where $(\widetilde A, \widetilde B)$ is of type 4) or 5) in Theorem 3.1. Proof. We set $A'_n = A_{nd}$, $A''_n = A_{nd+t}$, $B'_n = B_{nd}$, $B''_n = B_{nd+t}$. Let us show that
$$
\begin{equation}
R(A',B') = R(A'',B') = 2, \qquad R(A',B'') \leqslant 2.
\end{equation}
\tag{6.9}
$$
We can consider only the case where $d$ is even (the case of odd $d$ is dealt with similarly). Hence $B_n=0$ for $n\equiv d/2 \pmod d$, and, therefore,
$$
\begin{equation*}
\begin{gathered} \, D\begin{pmatrix} n_0d & n_1d & d/2 \\ m_0d & m_1d & d/2 \end{pmatrix} = A_d B_0\cdot D_{A',B'} \begin{pmatrix} n_0 & n_1 \\ m_0 & m_0 \end{pmatrix} =0, \\ D\begin{pmatrix} n_0d + d/2 & n_1d+ d/2 & 0 \\ m_0d+d/2 & m_1d+ d/2 & 0 \end{pmatrix} = A_0 B_0\cdot \widetilde D_{A',B'} \begin{pmatrix} n_0 & n_1 \\ m_0 & m_0 \end{pmatrix} =0. \end{gathered}
\end{equation*}
\notag
$$
As a result, $R_0(A',B')=R_1(A',B')=1$. So, $R(A',B')=2$. Applying the above arguments to the equivalent systems $(\widetilde A,B)$, $(A,\widetilde B)$, where $\widetilde A_n = A_{n+t}$, $\widetilde B_n = B_{n+t}$, we obtain the remaining relations in (6.9).
Using (6.9) and Lemma 4.2 we have $A'_n= e(P_1(n))$, $A''_n= e(P_2(n))$, $B'_n= e(P_3(n))$; $B''_n= 0$ for $\operatorname{supp}B = d\mathbb{Z}$ and $B''_n = e(P_4(n))$, otherwise. Here $P_j(n)=\alpha n^2+\beta_jn+\gamma_j$. Therefore, the system $(A,B)$ is equivalent to a system of type 4) (where $d\geqslant 3$) or type 5) in Theorem 3.1. § 7. Proofs of the main resultsLemma 7.1. Let $R_0(A,B)=R_1(A,B)=2$, and let $\operatorname{supp}B$ be bounded from below or above. Then a) $\operatorname{supp} A$ has at most two elements of equal parity; b) the estimates $|\operatorname{supp} A|\leqslant 4$, $|{\operatorname{supp}B}|\leqslant 4$ hold. Proof. Suppose that the set $\operatorname{supp}B$ is bounded from below, and the set $\operatorname{supp} A$ has three terms of equal parity. Changing, if necessary, to an equivalent system, we can assume that $0 =\min \operatorname{supp}B$, $0,2n,2m \in \operatorname{supp} A$, $0<n<m$. Hence $B_{-n}=B_{-m}= B_{n-m}=0$, and a contradiction. Therefore, assertion a) holds. Accordingly, $|{\operatorname{supp} A}|\leqslant 4$. Applying assertion a) to the pair $(B,A)$ we conclude that $|{\operatorname{supp}B}|\leqslant 4$, whence the claim. Proof of Lemma 3.1. Changing, if necessary, to the equivalent system we can assume that condition 1) of Lemma 5.1 holds. If $R_0(A,B) = 0$ or $R_1(A,B)=0$, then by Lemma 5.3 in [3], the sets $\operatorname{supp} A$, $\operatorname{supp}B$ are contained in an arithmetic progression with common difference $2$. Therefore, $R_j(A,B)\geqslant 1$, $j=0,1$. It remains to apply Lemmas 4.5 and 7.1. This proves Lemma 3.1. Proof of Theorem 3.1. Suppose that $R(A,B)=4$ and one of the sequences $A$ or $B$ has infinitely many non-zero elements. Then the sequences $\operatorname{supp} A$ and $\operatorname{supp}B$ are not bounded from below and above by Lemma 3.1. Without loss of generality we can assume that condition 1) of Lemma 5.1 holds. Then $R_j(A,B)\geqslant 1$, $j=0,1$, by Lemma 5.3 in [3]. Moreover, $R_j(A,B)\neq 1$ by Lemma 4.5, that is, $R_j(A,B)=2$. In view of Lemma 5.1, one of assertions i)–v) of the previous section holds up to an equivalence. An application of Lemmas 6.1, 6.3, 6.8, 6.7 gives that $(A,B)\sim (\widetilde A, \widetilde B)$, where the system $(\widetilde A, \widetilde B)$ has one of the types 1)–5) from Theorem 3.1. It remains to prove that $R(\widetilde A, \widetilde B)=4$ for each systems of type 1)–5).
If $(A,B)$ is of type 1), then $R_j(A,B)\leqslant 2$. To check this, it is enough to use the definition of a hyperelliptic system and employ the addition formula (1.3). Since $v\notin \frac{1}{2} L$, we have $R_j(A,B)\geqslant 2$. This is verified in the same way as in the corresponding result in [3] (see the proof of Theorem 7.1 there). If 2) or 3) is satisfied, then, obviously, $R_j(A,B) = 2$. Let $(A,B)$ be of type 4). If $d\geqslant 3$, then $R(A,B)\geqslant 4$ by Lemmas 4.1–4.3. If $d=2$, then $R(A,B)\geqslant 4$ by Lemma 2.3. We claim that $R(A,B)\leqslant 4$. By Lemma 2.1, there exists $\alpha\in \mathbb{C}$ such that the series
$$
\begin{equation}
f(z)= \sum_{n\in \mathbb{Z}} A_n e(nz) e\biggl(\frac{n^2\alpha}2\biggr), \qquad g(z)= \sum_{m\in \mathbb{Z}} B_m e(mz) e\biggl(\frac{m^2\alpha}2\biggr)
\end{equation}
\tag{7.1}
$$
converge absolutely and uniformly on each compact set. In view of Theorem 2.1 it remains to verify that $R(f,g)\leqslant 4$. It is clear that
$$
\begin{equation}
\begin{aligned} \, &\sum_{n\equiv t \,(\operatorname{mod} d)} e(nz)e(\beta n) e\biggl(\frac{\alpha n^2}2\biggr) \nonumber \\ &\qquad= e\biggl(tz+ \beta t + \frac{\alpha t^2}2\biggr) \sum_{k\in \mathbb{Z}} e\bigl(kd(z+\beta + \alpha t)\bigr) e\biggl(\frac{\alpha d^2 k^2}2\biggr) \nonumber \\ &\qquad= e\biggl(\beta t + \frac{\alpha t^2}2\biggr) e(tz) \theta \bigl(dz+ d(\beta +\alpha t); \tau\bigr), \qquad \tau = \alpha d^2. \end{aligned}
\end{equation}
\tag{7.2}
$$
Therefore, $f(z) = c_1\theta (dz;\tau) + c_0 e(tz) \theta (dz+z_0)$, $g(z) = c_2\theta (dz;\tau) + c_0 e(tz) \theta (dz+z_0)$, where $c_0 = e(\beta t + \alpha t^2)$, $z_0 = d(\beta + \alpha t)$, $\tau = \alpha d^2$.
Using the addition formula $\theta(x+y)\theta(x-y) = a_1(x)b_1(y)+a_2(x)b_2(y)$ (the explicit form of the functions $a_1,a_2,b_1,b_2$ is immaterial), one easily finds that $R(f,g)\leqslant 4$. The case where $(A,B)$ is of type 5) is considered similarly to the previous one. Theorem 3.1 is proved. Proof of Corollary 3.1. According to Theorem 2.1 and Lemma 2.1, the set of 1-periodic solutions of equations (1.2) coincides, up to an equivalence, with the set of systems $(f,g)$ of type (7.1), where $R(A,B)=4$, and $\alpha$ is an arbitrary complex constant for which the series (7.1) is convergent (see Lemma 3.2 in [3]). In addition, a 1-periodic function is a quasi-polynomial if and only if the sequence of Fourier coefficients of this function contains a finite number of non-zero terms. Therefore, by Theorem 3.1, the required set of solutions coincides, up to an equivalence, with the set of systems $(f,g)$ of type (7.1), where the system $(A,B)$ is one of types 1)–5) in Theorem 3.1.
1. Let the system $(A,B)$ be of type 1). 1.1. Let the lattice $L$ be non-degenerate. Then there are $a,b,\tau,c_2,c_1,c_0\in \mathbb{C}$ such that $\sigma_L(z) = \theta(az+b;\tau)e(c_1z^2 +c_2 z+c_3)$ (see [32], Ch. 21). Therefore, up to an equivalence $A_n = \theta(nw+y_1;\tau)$, $B_n = \theta(nw+y_2;\tau)$, and $w\notin \frac{1}{2}\mathbb{Z} + \frac{\tau}{2}\mathbb{Z}$. Hence
$$
\begin{equation*}
\begin{aligned} \, f(z)&= \sum_{n\in \mathbb{Z}}\theta(nw+y_1;\tau) e\biggl(\frac{n^2\alpha}2 + nz\biggr) =\sum_{k,n\in \mathbb{Z}} e\biggl(nk w+ y_1 k +\frac{k^2\tau}2 +\frac{n^2\alpha}2 + nz\biggr) \\ &= \Theta (z,y_1;\Omega), \qquad\text{where} \quad \Omega =\begin{pmatrix} \alpha & w \\ w & \tau \end{pmatrix}. \end{aligned}
\end{equation*}
\notag
$$
Similarly $g(z)= \Theta (z,y_2;\Omega)$. We get case (3.1). 1.2. If $L=\{0\}$, then $\sigma_L(z)= z$, $ A_n= nv+u_1$, $B_n=nv+u_2$,
$$
\begin{equation*}
\begin{aligned} \, f(z) &= \sum_{n\in \mathbb{Z}} (nv+u_1) e\biggl(\frac{n^2\alpha}2 + nz\biggr) \\ &= \frac{v}{2\pi i}\,\frac{\partial}{\partial z}\sum_{n\in \mathbb{Z}} e\biggl(\frac{n^2\alpha}2 + nz\biggr)+ u_1 \sum_{n\in \mathbb{Z}} e\biggl(\frac{n^2\alpha}2 + nz\biggr) = c \theta'(z;\alpha)+ u_1 \theta(z;\alpha). \end{aligned}
\end{equation*}
\notag
$$
A similar analysis gives $g(z)= c \theta'(z;\alpha)+ u_2 \theta(z;\alpha)$. We thus have case (3.2).
1.3. If $L= \{m \omega\colon m\in \mathbb{Z}\}$, $\omega\in \mathbb{C}\setminus\{0\}$, then $\sigma_L(z) = c_1 \sin(\pi \omega^{-1} z) \exp (c_2 z^2)$, where $c_1=\omega \pi^{-1}$, $c_2 = \pi^2 \omega^{-2} 6^{-1}$. Therefore, up to an equivalence,
$$
\begin{equation*}
\begin{gathered} \, A_n = 2i \sin \biggl(\frac{\pi(nv+u_1)}{\omega}\biggr) = c_1 e(\beta n) - c_1^{-1} e(-\beta n), \quad B_n = c_2 e(\beta n) - c_2^{-1} e(-\beta n), \\ c_j = e\biggl(\frac{u_j}{2\omega}\biggr), \qquad \beta = \frac{v}{2\omega}\notin \frac{1}{4}\mathbb{Z}, \\ f(z) = c_1 \theta(z+\beta;\alpha)- c_1^{-1} \theta(z-\beta;\alpha),\qquad g(z) = c_2 \theta(z+\beta;\alpha)- c_2^{-1} \theta(z-\beta;\alpha). \end{gathered}
\end{equation*}
\notag
$$
In this case, $(f,g)$ is equivalent to a system of type (3.3).
2. Let the system $(\widetilde A,\widetilde B)$ be of type 2). Arguing as in case 1.2, we arrive at case (3.2), where $c_1=0$, $c_2=c_4=1$. 3. Let the system $(\widetilde A,\widetilde B)$ be of type 3), 4) or 5). Using (7.2), we arrive at case (3.3), in which $d=1$ if 3) holds, and $d\geqslant 2$ in other cases. This proves Corollary 3.1. 
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