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We consider the polynomial vector fields of arbitrary degree in R3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb R ^3$$\end{document} having the 2-dimensional algebraic torus T2(l,m,n)={(x,y,z)∈R3:(x2l+y2m-r2)2+z2n-1=0},\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathbb T ^2(l,m,n)=\{(x,y,z)\in \mathbb R ^3 : (x^{2l}+y^{2m}-r^2)^2+z^{2n}-1=0\}, \end{aligned}$$\end{document}where l,m\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$l,m$$\end{document}, and n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n$$\end{document} positive integers, and r∈(1,∞)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r\in (1,\infty )$$\end{document}, invariant by their flow. We study the possible configurations of invariant meridians and parallels that these vector fields can exhibit on T2(l,m,n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb T ^2(l,m,n)$$\end{document}. Furthermore, we analyze when these invariant meridians or parallels are limit cycles.


Introduction and Statement of the Main Results
In 1878 Darboux published two works, [5] and [6], about polynomial vector fields or equivalently autonomous polynomial differential equations on R n or C n . There, he showed that if a polynomial vector field has a sufficient number of invariant algebraic hypersurfaces then it has a first integral. If we have a polynomial vector field in R n or C n with a first integral, then we can reduces its study in one dimension; of course, in the planar case we can describe completely its phase portrait, which is the main goal of the qualitative theory of differential equations.
Darboux's work was improved by Jouanolou [12] in 1979, and recently in [18] the authors improve the Darboux's classical result and the new one of Jouanolou taking into account the multiplicity of the invariant algebraic hypersurfaces. Since Darboux's result, it has existed a big interest in the study of invariant algebraic hypersurfaces of polynomial vector fields, in particular in the planar case (see for instance [3,4] and references there in). Recently, several papers have been published looking for particular invariant algebraic sets, for instance invariant straight lines (see [1,15,[20][21][22]), invariant hyperplanes in R n (see [13,17,23]), invariant circles and limit cycles for polynomial vector fields on the sphere (see [2,10,16]).
The study of limit cycles of polynomial vector fields on the plane has been intensively treated since 1900 due to the second part of the 16th Hilbert problem, which ask about the configuration, i.e. the number and relative position, of the limit cycles that a planar polynomial vector field of degree d can exhibit. In particular, a part of this problem is: Provide an upper bound, depending only on d, for the maximum number of limit cycles that any polynomial vector field of degree d can have, for more details see [8,9,11].
In this work we consider the 2-dimensional algebraic torus where l, m, and n are positive integers, and r is a real number greater than one. We are going to study the polynomial vector fields of arbitrary degree in R 3 having the 2-dimensional algebraic torus T 2 (l, m, n) invariant by their flow. We will study the possible configurations of invariant parallels and meridians that these vector fields can exhibit on T 2 (l, m, n). Additionally we will consider when these invariant parallels or meridians can be (algebraic) limit cycles. Our goal is to generalize to all l, m, and n the results given in [14] where the authors consider the particular case l = m = n = 1. On T 2 (l, m, n) we define parallels and meridians as the curves obtained by the intersection of this algebraic surface with the planes orthogonal to the z-axis and the planes containing the z-axis, respectively. More precisely, a parallel of T 2 (l, m, n) is a connected component of the intersection {z − z 0 = 0} ∩ T 2 (l, m, n), with z 0 ∈ [−1, 1]. If z 0 = ±1 such intersection has two parallels, otherwise it has only one parallel. A meridian of T 2 (l, m, n) is a connected component of the intersection {ax + by = 0} ∩ T 2 (l, m, n), with a, b ∈ R and a 2 + b 2 = 0. So meridians always are in pairs.
If a polynomial vector field X in R 3 has the algebraic torus T 2 (l, m, n) invariant by its flow, then a parallel of T 2 (l, m, n) is called invariant of X if it is formed by orbits of X . In similar way we define invariant meridian. We note that a parallel or a meridian which is invariant of X is an invariant algebraic curve of X .
Let X d (l, m, n) be the set of polynomial vector fields of degree ≤ d having T 2 (l, m, n) invariant by their flow. The set X d (l, m, n) is a R-vector space of finite dimension. If X d (l, m, n) is different from the zero polynomial vector field, then we can identify it with the affine space R q for some q ≥ 1, by associating to each element X of X d (l, m, n) a point of R q whose coordinates are all the different free coefficients of X . We can give to X d (l, m, n) the topology induced by this identification. Thus we say that a subset of X d (l, m, n) is generic if its image in R q is an open and dense subset.
In our study we can assume that l ≥ m, otherwise we take the change of coordinates (x, y, z) → (y, x, z) to obtain the desired condition.
Our main result about the maximum number of invariant parallels and invariant meridians is the following.

Theorem 1
Suppose that X ∈ X d (l, m, n) has finitely many invariant parallels and invariant meridians.
(a) If 2n ≥ 2l + 1 then the maximum number of invariant parallels of X is 2(d − 2l − 1), and the maximum number of invariant meridians of X is 2(d − 2l + 1). Moreover, there are l, m, n, d, and X ∈ X d (l, m, n) such that X has either 2(d − 2l − 1) invariant parallels, or 2(d − 2l + 1) invariant meridians. (b) If 2n < 2l + 1 then the maximum number of invariant parallels of X is 2(d − 2n), and the maximum number of invariant meridians of X is 2(d − 2n + 2). Moreover, there are l, m, n, d, and X ∈ X d (l, m, n) such that X has either 2(d − 2l − 1) invariant parallels, The study of the distribution of invariant parallels and invariant meridians of X ∈ X d (l, m, n) for arbitrary l, m, n, and d is a difficult problem. Hence in order to obtain satisfactory results we need to consider some restrictions on l, m, n, and d. A first result in this direction is the following where l, m, and n have arbitrary values, but we consider a restriction on the degree of the vector fields.
3) The sum of invariant parallels and invariant meridians of X is at least four and at most 2(d − 2(m + n) + 3).
Under the condition d < 2l − 1 Theorem 2 gives a complete description about all the possible configurations of invariant parallels, invariant meridians, and shows that they cannot be limit cycles.
The following two results generalize to the case l = m = n the main results of [14].

Theorem 3
Assume that X ∈ X d (n, n, n) has finitely many invariant meridians and invariant parallels. (a) There exists X ∈ X d (n, n, n) having exactly ν invariant parallels and 2k invariant meridians. (b) There exists X ∈ X d (n, n, n) having exactly either ν invariant parallels or 2k invariant meridians which are algebraic limit cycles.
Theorems 3 and 4 give all the possible configurations of invariant parallels and invariant meridians and show when they are limit cycles.
At the moment we cannot be able to provide complete results as above for arbitrary l, m, n, and d; however, under generic conditions on the vector fields on T 2 (l, m, n) and some restrictions on l, m, n, and d we can give all the possible configurations of invariant parallels and invariant meridians and show when they are limit cycles, the result is the following. The paper is organized as follows. In Sect. 2 we will give the necessary definitions about invariant algebraic surfaces of polynomial vector fields in R 3 . Also we will recall the concept and properties of the extactic polynomial, which we will be the main tool for our study. In Sect. 3 we are going to obtain general properties and conditions about the polynomial vector fields on T 2 (l, m, n), which allow us to give the proofs of the main results. Theorems 1 and 2 will be proved in Sects. 4 and 5 respectively. In Sect. 6 we will give the proof of Theorems 3 and 4. Finally, in Sect. 7 we will give the proof of Theorem 5.

Vector Fields of Degree d
Let R[x, y, z] be the ring of the polynomials in the variables x, y, and z with real coefficients. Recall that a polynomial vector field X of degree d in R 3 is an expression of the form where P, Q, R ∈ R[x, y, z], and d = max{deg(P), deg(Q), deg(R)}.
If F ∈ R[x, y, z], then the algebraic surface {F = 0} ⊂ R 3 is called an invariant algebraic surface of the vector field (1) if there exists K ∈ R[x, y, z] such that where F x , F y , and F z denote the partial derivatives of F respect to x, y, and z, respectively. The polynomial K is called the cofactor of the invariant algebraic surface {F = 0}. Moreover, if X is of degree d, then the cofactor K is of degree at most d − 1. Also from (2) we get that if an orbit of (1) has a point in the algebraic surface {F = 0}, then the whole orbit is contained in {F = 0}, i.e. X has to {F = 0} invariant by its flow. This justifies the name of invariant surface. Let F ∈ R[x, y, z] be a fixed polynomial. A polynomial vector field (1) that satisfies (2) is called a polynomial vector field on the algebraic surface {F = 0}. If X(F) denotes the set of polynomial vector fields (1) on {F = 0}, then from (2) we get that X(F) is an R-vector space. Now we will introduce a polynomial which will allow to detect when an algebraic surface is invariant by the flow of a polynomial vector field.
Let X be the polynomial vector field (1) and let W ⊂ R[x, y, z] be a R-vector space of finite dimension p > 1. The extactic polynomial of X associated to W is the polynomial E W (X ) given by the determinant of the matrix ⎛ The definition of the extactic polynomial E W (X ) is independent of the chosen basis of W because of the properties of the determinant and of the derivation.

Example 1 Consider the polynomial vector field in
where X (P) = P P x + Q P y + R P z and X (Q) = P Q x + Q Q y + R Q z . A simple computation shows that We remark that the previous matrix already appears in the work of Lagutinskii (see Dobrovol'skii et al. [7]). For a geometric explanation of the meaning of such a matrix see Pereira [19]. Christopher et al. [4] used the extactic polynomial to study the algebraic multiplicity of invariant algebraic curves of planar polynomial vector fields.
An application of the extactic polynomial is given by the following result.

Proposition 6 Let X be the polynomial vector field (1) and let W be a R-vector subspace of
If F ∈ W , then it can be chosen as the first element of a basis of W . If {F = 0} is an invariant algebraic surface of X , then (2) holds. By using this fact it is easy to see that This and the properties of determinant imply that F is a factor of E W (X ). This is the argument for the proof of Proposition 6.
Our idea is to apply Proposition 6 for studying the invariant meridians and invariant parallels of the polynomial vector field (1) having the algebraic torus T 2 (l, m, n) invariant by its flow. Such application is as follows. The maximum number of factors of the form z − z 0 of the extactic polynomial E {1,z} (X ) gives an upper bound for the number of invariant planes {z − z 0 = 0} of X , and this allow to obtain an upper bound for the number of its invariant parallels. Analogously, the maximum number of factors of the form ax + by of the extactic polynomial E {x,y} (X ) allow to obtain an upper bound for the number of invariant meridians of X .
From the definition of the extactic polynomial we get that and Remark 1 In general Proposition 6 gives a necessary condition but not a sufficient one. Indeed, in Example 1 we see that the plane y = 2 is a factor of the extactic polynomial, but it is no a invariant plane of the vector field. However, in the case of invariant planes {ax + by = 0} and {z − z 0 = 0} the converse statement is true. This is, we have the following lemma.
Lemma 7 Let X be the polynomial vector field (1).
. Since x and y have no factors in common, there is a polynomial S such that Q − aT = yS and bT + P = x S. Therefore, PM x + QM y + RM z = a P +bQ = a(x S−bT )+b(yS+aT ) = S(ax + by) = SM. Thus, M satisfies (2). This completes the proof of statement (b).
If X is polynomial vector field having T 2 (l, m, n) invariant by its flow, then Proposition 6 and statement (a) of Lemma 7 transform the study of invariant parallels of X to the study of factors of the form z − z 0 of E {1,z} (X ). Analogously, Proposition 6 and statement (b) of Lemma 7 transform the study of invariant meridians of X to the study of factors of the form ax + by of E {x,y} (X ).

Polynomial Vector Fields on T 2 (l, m, n)
From now on we will consider the polynomials f = x 2l + y 2m − r 2 and Hence T 2 (l, m, n) = {F = 0}. Thus, a polynomial vector field (1) of degree d that satisfies (2) will be called a polynomial vector field of degree d on T 2 (l, m, n). Hence X d (l, m, n) denotes the set of all polynomial vector field of degree d on T 2 (l, m, n).
We recall that l ≥ m, hence deg( f ) = 2l. Let X be a polynomial vector field in X d (l, m, n), then by using (5) Eq. (2) can be written as the equation 4 f lx 2l−1 P + my 2m−1 Q +2nz 2n−1 R = K f 2 + z 2n − 1 or equivalently as In order to obtain general properties on any polynomial vector field on T 2 (l, m, n), we note that every polynomial Y ∈ R[x, y, z] of degree d can be written as thus, if we write the polynomials P, Q, R, and K in the previous notation, then the left hand side of (6) becomes and the right side of (6) is We must note: first, in (7) the term corresponding to Therefore, as (7) is equal to (8), then we get the following systems of equations and for i = 0, 1, . . . , 2n − 2, and since f and f 2 − 1 do not have common factors, for every and From (16) we get that and that the homogeneous parts of degree less than 2m − 1 vanish identically.

Proof of Theorem 1
We will study invariant parallels and invariant meridians through the extactic polynomials E {1,z} (X ) and E {x,y} (X ), respectively. About invariant parallels we will prove the following result.
If ax + by is a factor of E {x,y} (X ) then it is a factor of G i for all i = 0, 1, . . . , d. Hence we will study the maximum number of factors of the form ax + by that G i can have. A result in this direction is the following.
On the other hand the polynomial vector field where T is a polynomial of degree d − 2l + 1, is of degree d and satisfies (6) taking K ≡ 0, thus X ∈ X d (l, m, n). In addition, has d − 2l + 1 different factors of the form ax + by. Hence from Lemma 7(b) follows that X has exactly 2(d − 2l + 1) invariant meridians. This property is for all n, in particular for 2n ≥ 2l + 1. This complete the proof of statement (a).
has d − 2n + 2 different factors of the form ax + by. Hence from Lemma 7(b) follows that X has exactly 2(d − 2n + 2) invariant meridians. This completes the proof of statement (b).
Let X ∈ X d (l, m, n). From statement (b) of Lemma 7 the number of invariant parallels of X is bounded by two times the maximum number of factors of the form z − z 0 (z 0 ∈ [−1, 1]) of the extactic polynomial E {1,z} (X ) = R. We note that z − z 0 is a factor of R if and only if z − z 0 is a factor of R(x 0 , y 0 , z) for a point (x 0 , y 0 ) ∈ R 2 . Then we will prove Lemma 8 by finding the maximum number of factors of the form z − z 0 of R(x 0 , y 0 , z), where (x 0 , y 0 ) is a particular point. (6) holds for all (x, y, z) ∈ R 3 , in particular it holds for the points (0, 0, z). In such a case (6) becomes

Proof of Lemma 9 We have that
By reordering terms the last equation is equivalent to the following Since x 2l−1 and y 2m−1 do not have common factors, there exist a polynomial T i such that Now, as deg(  (18) we get that S i = lx 2l + my 2m V i / r 4 − 1 . Therefore (19) reduces to

Proof of Theorem 2
We will prove Theorem 2 by assuming the following result where T is a polynomial of degree at most d − 2(m + n) + 1.
We will prove this lemma after the proof of the theorem.
Proof of Theorem 2 Let X be a polynomial vector field in X d (l, m, n). Proof of statement (a). If d < 2(m + n) − 1 then from Lemma 10, T ≡ 0, hence X ≡ 0. Proof of statement (b). From Lemma 10 we get that X = 0 if and only if T = 0. Hence the extactic polynomials do not vanish identically. Therefore X has finitely many invariant parallels and invariant meridians. In addition, E {1,z} (X ) has at most (d − 2(m + n) + 1) factors of the form z − z 0 with z 0 = ±1 and has the two factors z − 1 and z + 1, which implies that X has at least two and at most 2(d − 2(m + n) + 2) invariant parallels. On the other hand E {x,y} (X ) has at most d − 2(m + n) + 1 factors of the form ax + by with b = 0 and has the factor x, which implies that X has at least two and at most 2(d − 2(m + n) + 2) invariant meridians. Therefore the sum of invariant parallels and meridians is at least four and at most 2(deg(T )) + 4 = 2(d − 2(m + n) + 3).
Finally, statement (d) follows from the property that T is an arbitrary polynomial of degree d − 2(m + n) + 1 and that P 1 and P 2 , and M 1 and M 2 , are always invariant parallels and invariant meridians, respectively.
Proof of Lemma 10 Let X ∈ X d (l, m, n) be a polynomial vector field (1) of degree d on T 2 (l, m, n). We are assuming that d < 2l − 1 then from (17) Hence from (15) we get that K d−1−i ≡ 0 for i = 0, 1, . . . , 2n − 2, and from (16) we get that Form the previous paragraph we have that Then (6) reduces to 2n R − z 2n K = K f 2 − 1 − 4 f lx 2l−1 P + my 2m−1 Q , which can be written as If 4l P − x K does not vanish identically then the right hand side of (21) is a polynomial of degree at least 4l −1 > d, while the left hand side has degree at most d. Hence 4l P −x K ≡ 0, thus there existsK ∈ R[x, y] such that and Eq. (21) reduces to If 4lnK r 2 − y 2m + 4my 2m−1 Q does not vanish identically, then the left hand side of the previous equation is a polynomial of degree at least 2l > d, while the right hand side has degree at most d. Hence 4lnK r 2 − y 2m = −4my 2m−1 Q. Since y 2m−1 and r 2 − y 2m does not have common factors, there exists T ∈ R[x, y] such that By using (20), (22), and (23) we prove that It is clear that deg(T ) ≤ d − 2(n + m) + 1.

Proof of Theorems 3 and 4
In this section we are assuming that l = m = n. This is f = x 2n + y 2n − r 2 and F = f 2 + z 2n − 1.
Proof of Theorem 3 Let X be a polynomial vector field in X d (n, n, n).
Statement (b) follows from statement (b) of Theorem 1.
Proof of statement (c). For i = 0, 1, . . . , d the polynomial G d−i has degree d − i + 1, then for i = 2n, 2n + 1, . . . , d the degree of G d−i is at most d − 2n + 1. Moreover, from Lemma 9 we know that for i = 0, 1, . . . , 2n − 2 the polynomial G d−i has at most d − 2n + 1 − i ≤ d − 2n + 1 factors of the form ax + by. Therefore, to complete the proof we only need to prove that G d−2n+1 = x Q d−2n+1 − y P d−2n+1 has at most d − 2n + 1 factors of the form ax + by. Next we will prove this fact.
Following the previous idea we can prove that if we consider X with H 0 1 the X has ν invariant parallels which are (algebraic) limit cycles.

Proof of Theorem 5
The idea to prove Theorem 5 is as follows. From the Case 1 of proof of Lemma 8 we know that if R(0, 0, z) does not vanish identically then X has at most 2(d − 2n) invariant parallels. Then we will prove the following lemma.

Lemma 11
If l = m and d ≥ 2n ≥ 2l, or if l > m and d ≥ 2n ≥ 2(l + m), then the subset R d ⊂ X d (l, m, n) of polynomial vector fields such that R(0, 0, z) ≡ 0 is generic.
On the other hand, if X ∈ X d (l, m, n) and G d ≡ 0, then from Lemma 9 we know that G d has at most d − 2l + 1 factors of the form ax + by, which implies that X has at most 2(d − 2l + 1) invariant meridians. Then we will prove the following lemma.
As a result we will obtain that R d ∩ G d will be the desired generic subset for the proof of the first two statements of Theorem 5. For proving the next two statements we will construct a polynomial vector field X having the desired properties. That construction is as follows: Let d, l, m, and n be positive integers such that either d ≥ 2n ≥ 2l if l = m, or d ≥ 2n ≥ 2(l + m) if l > m. For every k ∈ {0, 1, . . . , d − 2l + 1} we define the nonnegative number Consider the polynomial vector field where

Lemma 13
The polynomial vector field X given by (24) satisfies that X ∈ R d ∩ G d .
Next we are going to give the proof of Theorem 5 by assuming the three previous lemmas whose proofs will be given after the proof of the theorem.
Also we have that Then from Lemma 7(b) it follows that X has k invariant planes of the form ax + by because lx 2l + my 2m is irreducible, hence X has 2k invariant meridians. This complete the proof of statement (c).
To finish the proof of theorem we will provide the proof of statement (d).
If we consider H 0 1 = 1 in X , then X does not have any invariant meridian because E {x,y} (X ) has no factors of the form ax + by. Hence, for every ν ∈ {0, 1, 2, . . . , 2(d − 2n)} we can choose H 2 in such a way that X has ν different invariant parallels which are periodic orbits since X does not have any singular point on T 2 (l, m, n). In addition, these invariant parallels are stable or unstable (algebraic) limit cycles because the sign of R is constant between two consecutive parallels and it changes when we cross one parallel.
If we consider H 2 = 1 in X , then X does not have any invariant parallel because z − z 0 is no a factor of R for any constant z 0 . Therefore, X has 2k different invariant meridians which are periodic orbits since X does not have any singular point on T 2 (l, m, n). On the other hand the differential system associated to X in cylindrical coordinates satisfies thaṫ θ = x Q − y P r 2 = lr 2l cos 2l θ + mr 2m sin 2m θ) r 2 (z − z 0 ) α k H k 1 (r cos θ, r sin θ).
Since |z 0 | > 1, between two consecutive invariant meridians the sign(θ) = sign(H k 1 (r cos θ, r sin θ) ∈ {−1, 1}, and these signs change when we cross one meridian. So the periodic orbits on the invariant meridians are stable or unstable (algebraic) limit cycles alternately.
Proof of Lemma 11 From Lemma 13, the polynomial vector field (24) is an element of R d , so R d = ∅. As the number of free parameters in H 2 is d − 2n + 1 then R d can be identify with a real linear space of dimension at least d − 2n + 1 ≥ 1.
On the other hand, recall that X d (l, m, n) can be identified with R q , for some q ≥ 1. The previous argument implies that q ≥ d − 2n + 1 ≥ 1. On the other hand, R(0, 0, z) is a polynomial in R[z] whose coefficients are linear combinations of the coefficients of R, then the subset X d (l, m, n) \ R d of polynomial vector fields such that R(0, 0, z) ≡ 0 has codimension 1 in X d (l, m, n). Therefore R d is generic.
Proof of Lemma 12 First we prove that G d = ∅. We are assuming that d ≥ 2l − 1. Then we consider an arbitrary homogeneous polynomial H 1 ∈ R[x, y] of degree d − 2l + 1, which has d − 2l + 2 independent real coefficients. It is easy to see that the polynomial vector field is of degree d and satisfies (6)  This implies that X ∈ G d . So G d can be identify with a real linear space of dimension at least d − 2l + 2 ≥ 1.
On the other hand, recall that X d (l, m, n) can be identified with R q , for some q ≥ 1. The previous argument implies that q ≥ d − 2l + 2 ≥ 1. On the other hand, G d = x Q d − y P d is a polynomial in R[x, y] whose coefficients are linear combinations of the coefficients of P d and Q d , then the subset X d (l, m, n) \ G d of polynomial vector fields such that G d = x Q d − y P d ≡ 0 has codimension 1 in X d (l, m, n). Therefore G d is generic. Lemma 13 In is easy to see that X satisfies (6) by taking K = 4lnz 2n−1 H 2 . Now we will see that X is of degree d in both cases: l = m and l > m. Case 1. l = m. Hence deg(P) = deg(Q) = d. In this case R = 2l z 2n − 1 − r 2 f H 2 , then deg(R) = max{2n, 2l} + d − 2n and since 2n ≥ 2l, deg(R) = d. Therefore X ∈ X d (l, l, n). Case 2. l > m. Hence deg(P) = deg(Q) = d and deg(R) = max{2n, 2l + 2m} + d − 2n, as in this case 2n ≥ 2l + 2m then deg(R) = d. Therefore X ∈ X d (l, m, n).