1. Introduction

Consider a natural mechanical system with Hamiltonian

$$ H=\frac{p_{1}^{2}+p_{2}^{2}}{2}+V(x,y), $$
(1.1)

where \( V \) is a real-analytic periodic function on the plane \( 𝕉^{2} \) with some periodic lattice \( \Lambda\subset 𝕉^{2} \). This system is completely integrable if there exists an additional first integral functionally independent with Hamiltonian (1.1) almost everywhere. Recall that a function \( F(x,y,p_{1},p_{2}) \) is a first integral of a Hamiltonian system if the function is preserved along the trajectories of the system:

$$ \frac{dF}{dt}=\{F,H\}=\frac{\partial F}{\partial x}\frac{\partial H}{\partial p_{1}}-\frac{\partial F}{\partial p_{1}}\frac{\partial H}{\partial x}+\frac{\partial F}{\partial y}\frac{\partial H}{\partial p_{2}}-\frac{\partial F}{\partial p_{2}}\frac{\partial H}{\partial y}\equiv 0. $$

The following integrable cases are known for a natural system with Hamiltonian (1.1) (see, for instance, [1]).

1. If \( V(x,y)=V(\alpha x+\beta y) \), where \( \alpha,\beta\in 𝕉 \); then there exists a linear integral of the form \( F_{1}=\alpha p_{2}-\beta p_{1} \).

2. If \( V(x,y)=V_{1}(\alpha_{1}x+\beta_{1}y)+V_{2}(\alpha_{2}x+\beta_{2}y) \), where \( \alpha_{i},\beta_{i}\in 𝕉 \) are some constants consistent with the lattice \( \Lambda \) and \( \alpha_{1}\alpha_{2}+\beta_{1}\beta_{2}=0 \); then there exists a quadratic integral of the form

$$ F_{2}=(d_{1}+d_{2})p_{1}^{2}+4p_{1}p_{2}-(d_{1}+d_{2})p_{2}^{2}+2(d_{1}-d_{2})(V_{1}-V_{2}), $$

where \( d_{i}=\alpha_{i}/\beta_{i} \).

The question of whether this system can have an integral of degree higher than 2 polynomial in momenta is unsolved so far in the general case. Basically, only some results are known concerning low-degree integrals. In particular, the absence of irreducible integrals of degrees 3 and 4 was proved in [1,2,3,4]. In was actually proved in [5] (for details, see Remark 3 in Section 2 below) that an integral of degree 5 of this system reduces always to a linear integral excluding possibly one case when up to a rotation in the plane \( x,y \) the potential \( V(x,y) \) looks as

$$ V(x,y)=\widetilde{V}_{1}(x+z_{1}y)+\widetilde{V}_{2}(x-z_{1}y)+\widetilde{V}_{3}(x+z_{2}y)+\widetilde{V}_{4}(x-z_{2}y)+\widetilde{V}_{5}(y), $$
(1.2)

where

$$ z_{1,2}=\sqrt{1\pm 2/\sqrt{5}}. $$

Namely, it remains unclear whether the Hamiltonian system with Hamiltonian (1.1) and an analytic periodic potential (1.2) can have an irreducible integral of degree 5. In Section 3 of the present article, we will prove that if all functions \( \widetilde{V}_{i} \), \( i=1,\dots,5 \), in (1.2) are nonconstant then this is impossible. Therefore, we have

Theorem 1

Suppose that the natural mechanical system on the two-dimensional torus with Hamiltonian (1.1) and analytic periodic potential \( V(x,y) \) has a polynomial integral of degree \( 5 \) independent of the energy integral. As shown in \( [5] \), in this case we may assume without loss of generality that the potential has the form (1.2). Then at least one of the functions \( \widetilde{V_{i}} \), \( i=1,\dots,5 \), is constant.


If the potential \( V(x,y) \) is a trigonometric polynomial and the natural system with Hamiltonian (1.1) is integrable; then, as known, an additional high degree integral in momenta reduces to an integral of degree 1 or 2 (see [6]).

Note also that on surfaces of genus \( g>1 \), the natural system with Hamiltonian (1.1) cannot be completely integrable in the class of analytic integrals [7]. On the other hand, quite a few integrable examples are known on the two-dimensional sphere in which the additional integral can be a polynomial in momenta of degree greater than 2.

2. Preliminary Statements

Suppose that the Hamiltonian system

$$ \dot{x}=\frac{\partial H}{\partial p_{1}},\quad\dot{y}=\frac{\partial H}{\partial p_{2}},\quad\dot{p_{1}}=-\frac{\partial H}{\partial x},\quad\dot{p_{2}}=-\frac{\partial H}{\partial y} $$
(2.1)

on the two-dimensional torus with Hamiltonian (1.1) has a first integral polynomial in momenta of an arbitrary degree \( N \) of the form

$$ F=F_{N}+F_{N-1}+\dots+F_{1}+F_{0}, $$

where \( F_{j} \) is a polynomial homogeneous in momenta of degree \( j \) with analytic periodic coefficients

$$ F_{j}=\sum\limits_{k=0}^{j}f_{jk}(x,y)p_{1}^{k-j}p_{2}^{j}. $$

The Poisson bracket of two homogeneous polynomials is a homogeneous polynomial as well. Thus, if \( F \) is a first integral then so are the sums

$$ G_{0}=F_{N}+F_{N-2}+F_{N-4}+\cdots,\quad G_{1}=F_{N-1}+F_{N-3}+F_{N-5}+\cdots. $$

Therefore, we will assume henceforth that a first integral of degree \( N \) looks as

$$ \begin{gathered}\displaystyle F=G_{0}=F_{N}+F_{N-2}+F_{N-4}+\cdots\\ \displaystyle=\sum\limits_{k=0}^{N}a_{k}(x,y)p_{1}^{N-k}p_{2}^{k}+\sum\limits_{k=0}^{N-2}b_{k}(x,y)p_{1}^{N-2-k}p_{2}^{k}+\cdots.\end{gathered} $$

It is easy to show (see [3, 8]) that the coefficients at the higher degrees \( a_{k}(x,y) \), \( k=0,\dots,N \), do not depend on the coordinates, i.e., they are constant.

Remark 1

As shown by Poincaré (see [9]), if

$$ F_{N}=\sum\limits_{k=0}^{N}a_{k}p_{1}^{N-k}p_{2}^{k},\quad H_{2}=\frac{p_{1}^{2}+p_{2}^{2}}{2} $$

are functionally dependent then the integral \( F \) is reducible; i.e., in this case, there necessarily exists an integral of a lesser degree. It was proved in [10] that these functions are dependent if and only if the Birkhoff sums are equal to zero; i.e.,

$$ a_{0}-a_{2}+a_{4}-\dots=0,\quad a_{1}-a_{3}+a_{5}-\dots=0. $$
(2.2)

Remark 2

As shown in [5], if the natural system with Hamiltonian (1.1) admits a polynomial integral of degree 5 then the potential \( V(x,y) \) can be reduced either to the form (1.2) or to the form

$$ V(x,y)=V_{1}(x+zy)+V_{2}(x-zy)+V_{3}(y),\quad z=\sqrt{1/3}. $$

However, both Birkhoff sums are zero in the last case, which is easy to see by a straightforward check. Consequently, the first integral is reducible in this case.


Following [5] (see also [1]), expand the potential \( V \) in the Fourier series:

$$ V(x,y)=\sum\limits_{\lambda^{*}\in\Lambda^{*}}{v_{\lambda^{*}}e^{i(\lambda_{1}^{*}x+\lambda_{2}^{*}y)}}, $$

where \( \lambda^{*}=(\lambda_{1}^{*},\lambda_{2}^{*}) \) is an element of the lattice \( \Lambda^{*} \) dual to \( \Lambda \):

$$ \Lambda^{*}=\{\lambda^{*}\mid(\lambda^{*},\lambda)\in 2\pi 𝕑,\ \lambda\in\Lambda\}. $$

The set \( S=\{\lambda^{*}\mid v_{\lambda^{*}}\neq 0\} \) is the spectrum of the potential \( V \). As known (see [11]), if there exists an integral linear in momenta then the spectrum of the potential lies on one straight line passing through the origin. In the case of a quadratic integral, the spectrum lies on two straight lines intersecting at the origin orthogonally (with respect to the inner metric defined by the kinetic energy) (see [8]). In the general case, if a natural system admits a polynomial integral of an arbitrary degree \( N \) then the spectrum of the potential lies on at most \( N \) straight lines intersecting at the origin (see [1, 3]).

Thus, Theorem 1 implies that for \( N=5 \) the spectrum of potential (1.2) lies on at most four straight lines.

In Section 4, we study the integrals of degree 7, basically along the ideas of [5], and prove the following:

Theorem 2

Suppose that the natural mechanical system (2.1) on the two-dimensional torus with Hamiltonian looks like (1.1) admits an irreducible polynomial integral of degree 7. Then the spectrum of the potential lies exactly on three straight lines intersecting at the origin.

3. Integrals of Degree 5

Let us prove Theorem 1. The Hamiltonian looks like \( H=\frac{p_{1}^{2}+p_{2}^{2}}{2}+V(x,y) \). By the results of [5] and Remark 2 in Section 2, we may assume without loss of generality that the potential \( V(x,y) \) has the form (1.2). Thus, the spectrum of \( V(x,y) \) lies on five straight lines intersecting at the origin and having the equal angles \( \frac{\pi}{5} \) between each other. This agrees completely with the well-known general result (see [1]), in accordance with which if there is a polynomial integral of degree \( n \) independent of the energy integral then the \( n \) straight lines of the spectrum of the potential have angles \( \frac{\pi}{n},\ \frac{2\pi}{n},\dots,\frac{(n-1)\pi}{n} \) between each other.

Suppose that all functions \( \widetilde{V_{i}} \), \( i=1,\dots,5 \), are nonconstant. Then, due to the periodicity of \( V(x,y) \), there exist nonzero constants \( \alpha_{1},\dots,\alpha_{5} \) such that the potential can be represented as

$$ \begin{gathered}\displaystyle V(x,y)=V_{1}\bigg{(}\alpha_{1}\bigg{(}\frac{x}{z_{1}}+y\bigg{)}\bigg{)}+V_{2}\bigg{(}\alpha_{2}\bigg{(}\frac{x}{z_{1}}-y\bigg{)}\bigg{)}+V_{3}\bigg{(}\alpha_{3}\bigg{(}\frac{x}{z_{2}}+y\bigg{)}\bigg{)}\\ \displaystyle+V_{4}\bigg{(}\alpha_{4}\bigg{(}\frac{x}{z_{2}}-y\bigg{)}\bigg{)}+V_{5}(\alpha_{5}y),\quad z_{1,2}=\sqrt{1\pm 2/\sqrt{5}},\end{gathered} $$

and all \( V_{j} \)’s are \( 2\pi \)-periodic:

$$ V_{j}(z)=V_{j}(z+2\pi),\quad j=1,\dots,5. $$

We show that this is impossible.

The periodicity of the potential implies that on the plane with coordinates \( x,y \), there is a pair of linearly independent vectors \( e_{1}=(u_{1},v_{1}) \), \( e_{2}=(u_{2},v_{2}) \) such that \( V(x,y)=V(x+u_{1},y+v_{1})=V(x+u_{2},y+v_{2}) \). Write down the potential as

$$ V(x,y)=\sum\limits_{j=1}^{5}V_{j}(a_{j}x+b_{j}y). $$

Then the condition of the \( 2\pi \)-periodicity of each of the summands \( V_{j} \) takes the form

$$ a_{j}u_{1}+b_{j}v_{1}=2\pi k_{j1},\quad a_{j}u_{2}+b_{j}v_{2}=2\pi k_{j2},\quad k_{j1},k_{j2}\in 𝕑,\ j=1,\dots,5. $$
(3.1)

Since \( V_{5}=V_{5}(\alpha_{5}y) \), we may assume without loss of generality that one of the vectors \( e_{1} \) and \( e_{2} \) is vertical. Suppose, for instance, that \( e_{2}=(0,v_{2}) \). Then, by (3.1), for \( j=5 \) we obtain \( v_{2}=2\pi k_{5}/\alpha_{5} \), \( k_{5}\in 𝕑 \). Owing to this, relations (3.1) for \( e_{2} \) give \( \alpha_{j}k_{5}=\alpha_{5}k_{j} \), with \( k_{j}\in 𝕑 \) for \( j=1,\dots,4 \), which in particular implies that \( \alpha_{i}/\alpha_{j}\in 𝕈 \) for all \( i,j=1,\dots,5 \).

Furthermore, from (3.1) for \( e_{1} \) we have

$$ u_{1}/z_{1}+v_{1}=2\pi m_{1}/\alpha_{1},\quad u_{1}/z_{1}-v_{1}=2\pi m_{2}/\alpha_{2},\quad m_{1},m_{2}\in 𝕑. $$

Summing up these two equalities, we obtain \( u/z_{1}=\pi(m_{1}\alpha_{2}+m_{2}\alpha_{1})/(\alpha_{1}\alpha_{2}) \). Likewise, \( u/z_{2}=\pi(m_{3}\alpha_{4}+m_{4}\alpha_{3})/(\alpha_{3}\alpha_{4}) \), where \( m_{3},m_{4}\in 𝕑 \). Consequently,

$$ \frac{z_{2}}{z_{1}}=\frac{\alpha_{3}}{\alpha_{1}}\frac{\alpha_{4}}{\alpha_{2}}\frac{\big{(}m_{1}\frac{\alpha_{2}}{\alpha_{1}}+m_{2}\big{)}}{\bigl{(}m_{3}\frac{\alpha_{4}}{\alpha_{1}}+m_{4}\frac{\alpha_{3}}{\alpha_{1}}\big{)}}. $$

The previous arguments imply that the right-hand side of the above equality is a rational. We come to a contradiction since the reals \( z_{1}=\sqrt{1+2/\sqrt{5}} \) and \( z_{2}=\sqrt{1-2/\sqrt{5}} \) are not commensurable. Therefore, at least one of the functions \( V_{j} \), \( j=1,\dots,5 \), must be constant. Theorem 1 is proved.

4. Integrals of Degree 7

Let us prove Theorem 2. The proof relies essentially on the ideas of [5]. Hence, we will only give a sketch of the proof. For \( N=7 \), the first integral takes the form

$$ F=\sum\limits_{k=0}^{7}a_{k}p_{1}^{7-k}p_{2}^{k}+\sum\limits_{k=0}^{5}b_{k}(x,y)p_{1}^{5-k}p_{2}^{k}+\sum\limits_{k=0}^{3}c_{k}(x,y)p_{1}^{3-k}p_{2}^{k}+\sum\limits_{k=0}^{1}d_{k}(x,y)p_{1}^{1-k}p_{2}^{k}. $$

It is easy to show (see [1, 3]) that the potential has the form

$$ V(x,y)=\sum\limits_{m=1}^{7}V_{m}(\alpha_{m}x+\beta_{m}y); $$

i.e., the spectrum of the potential lies on at most seven straight lines intersecting at the origin. Here the quantities \( z_{m}=\frac{\alpha_{m}}{\beta_{m}} \) are roots of the equation

$$ \begin{gathered}\displaystyle a_{6}z^{7}+(7a_{7}-2a_{5})z^{6}+(3a_{4}-6a_{6})z^{5}+(5a_{5}-4a_{3})z^{4}\\ \displaystyle+(5a_{2}-4a_{4})z^{3}+(3a_{3}-6a_{1})z^{2}+(7a_{0}-2a_{2})z+a_{1}=0.\end{gathered} $$
(4.1)

Without loss of generality, we may assume (see [1, 5]) that \( a_{1}=a_{3}=0 \). Moreover, we have the relations (see [5])

$$ 35a_{0}+5a_{2}+3a_{4}+5a_{6}=0,\quad a_{5}+7a_{7}=0. $$

As above, if the spectrum lies on two straight lines then either there are no additional integrals at all or there exists a quadratic integral. Therefore, we will assume henceforth that the number of straight lines in the spectrum is at least three. Following [5], rotate the plane \( x,y \) to make another straight line in the spectrum vertical. This allows us (see [5] for details) to obtain the system of equations in \( z \); i.e.,

$$ \begin{gathered}\displaystyle\bigg{(}21a_{0}+\frac{6a_{4}}{5}+2a_{6}\bigg{)}z^{6}-(35a_{0}+7a_{4}+5a_{6})z^{4}-35a_{7}z^{3}\\ \displaystyle+(3a_{4}-6a_{6})z^{2}+21a_{7}z+a_{6}=0,\end{gathered} $$
(4.2)
$$ \begin{gathered}\displaystyle(35a_{0}+7a_{4}+5a_{6})z^{6}+70a_{7}z^{5}-(175a_{0}+30a_{4})z^{4}-175a_{7}z^{3}\\ \displaystyle+(70a_{0}+18a_{4}-5a_{6})z^{2}+35a_{7}z-a_{4}=0.\end{gathered} $$
(4.3)

Thus, the question of the number of straight lines in the spectrum of the potential reduces to the question of how many common real roots two equations (4.2) and (4.3) of degree 6 may have; the coefficients of these equations are parametrized by the four real parameters \( a_{0} \), \( a_{4} \), \( a_{6} \), and \( a_{7} \).

Lemma 1

Equations (4.2) and (4.3) have at most four common real roots.

Proof

Multiply (4.2) by 5 and extract (4.3) from the result. The obtained equality has the form

$$ (z^{2}+1)S_{4}=0, $$
(4.4)

where

$$ \begin{gathered}\displaystyle S_{4}=(70a_{0}-a_{4}+5a_{6})z^{4}-70a_{7}z^{3}\\ \displaystyle-(70a_{0}+4a_{4}+30a_{6})z^{2}+70a_{7}z+(a_{4}+5a_{6}).\end{gathered} $$

This immediately proves Lemma 1.


Put

$$ \begin{gathered}\displaystyle S_{6}=\bigg{(}21a_{0}+\frac{6a_{4}}{5}+2a_{6}\bigg{)}z^{6}\\ \displaystyle-(35a_{0}+7a_{4}+5a_{6})z^{4}-35a_{7}z^{3}+(3a_{4}-6a_{6})z^{2}+21a_{7}z+a_{6}.\end{gathered} $$

Then (4.2) and (4.4) (and hence (4.2) and (4.3)) are equivalent to the system

$$ S_{4}=0,\quad S_{6}=0. $$
(4.5)

Lemma 2

If (4.5) have exactly four common real roots then the first integral \( F \) is reducible.

Proof

Without loss of generality, we may assume that the coefficient at the higher degree \( z^{4} \) in \( S_{4} \) is equal to \( 1 \) (otherwise, there would be at most three common roots); i.e., \( a_{4}=70a_{0}+5a_{6}-1 \). Divide \( S_{6} \) by \( S_{4} \) with a remainder: \( S_{6}=S_{4}\cdot S_{2}+S_{3} \), where \( S_{2} \) and \( S_{3} \) are polynomials in \( z \) of degrees \( 2 \) and \( 3 \) respectively. If system (4.5) has four common roots then \( S_{3} \) is identical zero; i.e., all its coefficients must be zero. We obtain a rather cumbersome system of four algebraic equations on the three coefficients \( a_{0} \), \( a_{6} \), and \( a_{7} \), which we omit here. Using the program “Singular” (see [12]), we found all solutions to this overdetermined system: They have the form

$$ a_{7}=\pm\frac{2i}{35},\quad a_{0}=-\frac{1}{140},\quad a_{6}=\frac{1}{4}; $$
(1)
$$ a_{7}=0,\quad a_{0}=\frac{1}{60},\quad a_{6}=-\frac{1}{20}; $$
(2)
$$ a_{7}=0,\quad a_{0}=\frac{1}{100},\quad a_{6}=\frac{1}{20}. $$
(3)

We can omit the first case since we are interested in the real coefficients of the integral \( F \). In the second case, there are four common roots but only two of them are real: \( z_{1,2}=\pm 1/\sqrt{3} \) and \( z_{3,4}=\pm i \). Finally, in the third case, all four common roots are real: \( z_{1,2,3,4}=\pm\sqrt{1\pm 2/\sqrt{5}} \). It remains to notice that in the second and third cases, the corresponding Birkhoff sums (2.2) vanish. Thus, either the number of common real roots of system (4.5) is at most three or \( F \) is reducible. Lemma 2 is proved.


The following lemma was proved by V.Yu. Gubarev:

Lemma 3

The polynomials \( S_{4} \) and \( S_{6} \) can have exactly three common real roots for no values of \( a_{0} \), \( a_{4} \), \( a_{6} \), and \( a_{7} \).


Expose the idea of the proof of Lemma 3. For simplifying calculations, by the polynomial \( S_{6} \) we will henceforth mean \( 5S_{6} \). Thus, \( S_{4} \) and \( S_{6} \) belong to the ring of polynomials of the parameters \( a_{0} \), \( a_{4} \), \( a_{6} \), and \( a_{7} \) and the variable \( z \) with integer coefficients; i.e., \( S_{4},S_{6}\in 𝕑[a_{0},a_{4},a_{6},a_{7}][z] \).

Let us first list all possible cases depending on whether the higher coefficients are zero in \( S_{4} \) and \( S_{6} \) or not.

Case 1: The higher coefficient of \( S_{6} \) is nonzero.

1(a) The higher coefficient of \( S_{4} \) is nonzero.

1(b) The higher coefficient of \( S_{4} \) is zero.

Case 2: The higher coefficient of \( S_{6} \) is zero. Then we have two polynomials \( S_{4} \) and \( S_{6} \) of degree at most \( 4 \). The following cases are possible:

2(a) Both coefficients at \( z^{4} \) for \( S_{4} \) and \( S_{6} \) are nonzero.

2(b) The coefficient at \( z^{4} \) is zero for \( S_{4} \) and is nonzero for \( S_{6} \).

2(c) The coefficient at \( z^{4} \) is zero for \( S_{6} \) and is nonzero for \( S_{4} \).

2(d) Both coefficients at \( z^{4} \) for \( S_{6} \) and \( S_{4} \) are zero.

Examine in detail case 1(a) which seems to be the most meaningful. By analogy with the proof of Lemma 2, divide \( S_{6} \) by \( S_{4} \) with a remainder \( S_{3} \) of degree at most \( 3 \). Note that \( S_{3} \) is no longer a polynomial of \( a_{i} \) and \( z \) and has the form

$$ (1/Q)\sum\limits_{i=0}^{3}R_{i}z^{i}, $$

where \( Q,R_{i}\in 𝕑[a_{0},a_{4},a_{6},a_{7}] \). The arising expressions are rather cumbersome; therefore, we do not write them explicitly but just confine presentation by the scheme of the proof.

If \( R_{3}=0 \) then the greatest common divisor \( (S_{4},S_{6}) \) has degree at most \( 2 \) in \( z \). Consequently, \( S_{4} \) and \( S_{6} \) cannot have three common roots. Let \( R_{3}\neq 0 \). Introduce the polynomial \( T_{3}=QS_{3} \). Divide \( S_{6} \) and \( S_{4} \) by \( T_{3} \) with remainders \( S_{2} \) and \( T_{2} \) respectively. For \( S_{4} \) and \( S_{6} \) to have exactly three common roots, it is necessary that \( S_{2}=T_{2}=0 \). As above, \( S_{2} \) and \( T_{2} \) are not polynomials but rational fractions. The numerators of the fractions at the degrees \( z^{2},z^{1} \), and \( z^{0} \) in \( S_{2} \) and \( T_{2} \) give the system

$$ S=\{f_{i}=0\mid i=1,\dots,6,\ f_{i}\in 𝕑[a_{0},a_{4},a_{6},a_{7}]\} $$

of six algebraic equations. Consider the ideal \( I \) in the ring \( K=𝕑[a_{0},a_{4} \), \( a_{6},a_{7}] \) generated by the set \( S \); this is the minimal subring in \( K \) including \( S \) and closed under multiplication by all elements of \( K \). Using Singular (see [12]), we found the radical \( R(I) \) of the ideal \( I \) (see [13]) consisting of seven generators, i.e.,

$$ R(I)=\{r\in K\mid\text{ there is }n\text{ such that }r^{n}\in I\}. $$

It is easy to check that \( R_{3}\in R(I) \), and hence the higher coefficient of \( S_{3} \) vanishes at each solution to \( S \). Hence, in this case, \( S_{4} \) and \( S_{6} \) cannot have three common roots. The remaining cases are studied in a similar way. Lemma 3 is proved.

By Lemmas 1–3, we arrive immediately at the claim of Theorem 2. Indeed, if there are no common roots then there is only one straight line in the spectrum (which is vertical); and then, as above, there is a first integral linear in momenta. If there is one common real root then the spectrum lies on two straight lines. Hence, either there are no additional first integrals at all or there is an integral of the second degree. Finally, two common real roots correspond to the presence of three straight lines in the spectrum of the potential. Theorem 2 is proved.

5. Concluding Remarks

Let us give several examples when, in the case of an integral of degree \( 7 \), the spectrum of the potential lies on three straight lines.

Example 1

Put

$$ a_{2}=-\frac{7a_{0}}{3},\quad a_{4}=-\frac{35a_{0}}{3},\quad a_{6}=\frac{7a_{0}}{3},\quad a_{7}=0. $$

If \( a_{0}\neq 0 \) then one of the Birkhoff sums is nonzero: \( a_{0}-a_{2}+a_{4}-a_{6}\neq 0 \). System (4.5) takes the form

$$ z^{2}(z^{2}-1)=0,\quad(5z^{4}+20z^{2}-1)(z^{2}-1)=0, $$

and its two common roots are \( z_{1,2}=\pm 1 \). Consequently, in this case, the potential looks as

$$ V(x,y)=V_{1}(x+y)+V_{2}(y)+V_{3}(x-y). $$

Example 2

Put \( a_{4}=a_{6}=0 \) and \( a_{0}=\sqrt{3}a_{7} \). If \( a_{7}\neq 0 \) then both Birkhoff sums are nonzero. System (4.5) has the form

$$ z(z^{2}-1)(\sqrt{3}z-1)=0,\quad z(3+z^{2}(-5+\sqrt{3}z(3z^{2}-5)))=0, $$

and its two common roots are \( z_{1}=0 \) and \( z_{2}=1/\sqrt{3} \). Consequently, in this case, the potential has the form

$$ V(x,y)=V_{1}(x)+V_{2}(y)+V_{3}(\sqrt{3}x+y). $$

In these examples, (4.1) is satisfied identically. But the analysis of the remaining relations that follow from the equality \( \{F,H\}=0 \) causes difficulties, and we were unable to complete it.

Let us also add that not much is known about the integrals of higher degrees of the natural system under consideration on the two-dimensional torus. Let us say that the case of integrals of odd degree is in general apparently somewhat simpler for study since there are additional relations for the higher coefficients of the integral (see [5]). However, the polynomial equations analogous to (4.2) and (4.3) have a higher degree and are noticeably more cumbersome already in the case of an integral of degree \( N=9 \), which significantly complicates the analysis similar to that of Section 4 for \( N=7 \).

We also note the recent work [14], in which a slightly different approach to the problem under consideration was developed and applied which bases on the study of the discrete symmetries of the straight lines of the spectrum of the potential. By this method, all possible admissible configurations of the straight lines of the spectrum are found in the case of an additional integral of degree 5 or 6. It is highly likely that we can similarly obtain analogous meaningful results on the integrals of a higher degree.