POLYNOMIAL FIRST INTEGRALS FOR WEIGHT-HOMOGENEOUS PLANAR POLYNOMIAL DIFFERENTIAL SYSTEMS OF WEIGHT DEGREE 4

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1 ROCKY MOUNTAIN JOURNAL OF MATHEMATICS Volume 46, Number 5, 2016 POLYNOMIAL FIRST INTEGRALS FOR WEIGHT-HOMOGENEOUS PLANAR POLYNOMIAL DIFFERENTIAL SYSTEMS OF WEIGHT DEGREE 4 JAUME LLIBRE AND CLAUDIA VALLS ABSTRACT. We classify all of the weight-homogeneous planar polynomial differential systems of weight degree 4 having a polynomial first integral. 1. Introduction and statement of the main result. In this paper, we deal with polynomial differential systems of the form: 1.1) dx dt = ẋ = Px), x = x, y) C2, with Px) = P 1 x), P 2 x)) and P i C[x, y] for i = 1, 2. As usual, Q +, R and C will denote the sets of non-negative rational, real and complex numbers, respectively, and C[x, y] denotes the polynomial ring over C in the variables x, y. Here, t is real or complex. System 1.1) is weight homogeneous or quasi-homogeneous if there exist s = s 1, s 2 ) N 2 and d N such that, for arbitrary α R + = {a R, a > 0}, 1.2) P i α s 1 x, α s 2 y) = α s i 1+d P i x, y), for i = 1, 2. We call s = s 1, s 2 ) the weight exponent of system 1.1) and d the weight degree with respect to the weight exponent s. In the particular case where s = 1, 1), system 1.1) is called a homogeneous polynomial differential system of degree d AMS Mathematics subject classification. Primary 34A05, 34A34, 34C14. Keywords and phrases. Polynomial first integrals, weight-homogeneous polynomial differential systems. The first author is partially supported by MINECO/FEDER, grant Nos. MTM and MTM P, ICREA Academia, grant Nos. FP7- PEOPLE-2012-IRSES and , AGAUR, grant No. 2014SGR-568, and grant No. UNAB13-4E The second author is supported by Portuguese national funds through FCT, Fundação para a Ciência e a Tecnologia, grant No. PEst- OE/EEI/LA0009/2013 CAMGSD). Received by the editors on October 9, 2012, and in revised form on January 9, DOI: /RMJ Copyright c 2016 Rocky Mountain Mathematics Consortium 1619

2 1620 JAUME LLIBRE AND CLAUDIA VALLS Recently, such systems have been investigated by several authors. Labrunie [12] and Ollagnier [15] characterized all polynomial first integrals of the three-dimensional a, b, c) Lotka-Volterra systems. Maciejewski and Strelcyn [14] proved that the so-called Halphen system has no algebraic first integrals. However, some of the best results for general weight homogeneous polynomial differential systems have been provided by Furta [10] and Goriely [11]. For quadratic homogeneous polynomial differential systems, we refer the reader to [13, 16]. Additionally, we refer the reader to [1 5]. A non-constant function Hx, y) is a first integral of system 1.1) if it is constant on all solution curves xt), yt)) of system 1.1), i.e., Hxt), yt))is constant for all values of t for which the solution xt), yt)) is defined. If H is C 1, then H is a first integral of system 1.1) if and only if H 1.3) P 1 x + P H 2 y = 0. The function Hx, y) is weight homogeneous of weight degree m with respect to the weight exponent s if it satisfies Hα s 1 x, α s 2 y) = α m Hx, y), for all α R +. Given H C[x, y], we can split it into the form H = H m + H m H m+l, where H m+i is a weight homogeneous polynomial of weight degree m+i with respect to the weight exponent s, i.e., H m+i α s 1 x, α s 2 y) = α m+i H m+i x, y). The following well-known proposition see [13] for proof) reduces the study of the existence of analytic first integrals of a weighthomogeneous polynomial differential system 1.1) to the study of the existence of a weight-homogeneous polynomial first integral. Proposition 1.1. Let H be an analytic function, and let H = i H i be its decomposition into weight-homogeneous polynomials of weight degree i with respect to the weight exponent s. Then, H is an analytic first

3 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1621 integral of the weight-homogeneous polynomial differential system 1.1) if and only if each weight-homogeneous part H i is a first integral of system 1.1) for all i. The main goal of this paper is to classify all analytic first integrals of the weight-homogenous planar polynomial differential systems of weight degree 4. In view of Proposition 1.1, we only need to classify all polynomial first integrals of the weight-homogenous planar polynomial differential systems of weight degree 4. The classification of all polynomial first integrals and hence, of all analytic first integrals) of the weight-homogenous planar polynomial differential systems of weight degree 1 is straightforward and trivial. The classification of all polynomial first integrals and hence, of all analytic first integrals) of the weighthomogenous planar polynomial differential systems of weight degree 2 was given in [8, 13] and for systems of weight degree 3 in [6, 8]. Kowalevskaya exponents are used in the classification of all polynomial first integrals for weight-homogenous planar polynomial differential systems of weight degrees 2 and 3. However, it was shown in [6, Theorem 4] that these exponents are useless for classifying the polynomial first integrals for weight-homogenous planar polynomial differential systems of weight degrees larger than 3. Proposition 1.2. The systems with weight degree 4 in C 2 and their corresponding values of s can be written as follows: s = 1, 1) : ẋ = a 40 x 4 + a 31 x 3 y + a 22 x 2 y 2 + a 13 xy 3 + a 04 y 4, ẏ = b 40 x 4 + b 31 x 3 y + b 22 x 2 y 2 + b 13 xy 3 + b 04 y 4 ; s = 1, 2) : ẋ = a 40 x 4 + a 21 x 2 y + a 02 y 2, ẏ = b 50 x 5 + b 31 x 3 y + b 12 xy 2 ; s = 1, 3) : ẋ = a 40 x 4 + a 11 xy, ẏ = b 60 x 6 + b 31 x 3 y + b 02 y 2 ; s = 1, 4) : ẋ = a 40 x 4 + a 01 y, ẏ = b 70 x 7 + b 31 x 3 y; s = 2, 3) : ẋ = a 11 xy, ẏ = b 30 x 3 + b 02 y 2 ; s = 2, 5) : ẋ = a 01 y, ẏ = b 40 x 4 ; s = 3, 3) : ẋ = a 20 x 2 + a 11 xy + a 02 y 2, ẏ = b 20 x 2 + b 11 xy + b 02 y 2 ; s = 6, 9) : ẋ = a 01 y, ẏ = b 20 x 2.

4 1622 JAUME LLIBRE AND CLAUDIA VALLS The proof of Proposition 1.2 is provided in Section 2. In what follows, we state our main result, i.e., we classify when the systems of Proposition 1.2 exhibit a polynomial first integral. The systems with weight exponent 1, 1) having a polynomial first integral are given in Section 3. The systems with weight exponent 3, 3) having a polynomial first integral are studied inside the systems with weight exponent 1, 2). The polynomial first integrals for the other systems of Proposition 1.2 are provided in this introduction. We introduce the change X, Y ) = x 2, y) in the planar weight homogeneous polynomial differential systems 1.1) of weight degree 4 with weight exponent 1, 2). With these new variables X, Y ) the system with weight exponent 1, 2) becomes, after introducing the new independent variable dτ = x dt, 1.4) X = 2a 40 X 2 + 2a 21 XY + 2a 02 Y 2, Y = b 50 X 2 + b 31 XY + b 12 Y 2, where the prime denotes the derivative with respect to τ. System 1.4) is a homogeneous quadratic planar polynomial system with s = 1, 1). It is well known see [9]) that, for each quadratic homogeneous system, there exist some linear transformation and a rescaling of time which transform system 1.4) into systems in 1.5). 1.5) ẋ = 2xy xp 1x + p 2 y), ẋ = 2xy xp 1x + p 2 y), ẋ = x xp 1x + p 2 y), ẋ = 2 3 xp 1x + p 2 y), ẋ = 2 3 xp 1x + p 2 y), ẏ = x 2 + y yp 1x + p 2 y), ẏ = x 2 + y yp 1x + p 2 y), ẏ = 2xy yp 1x + p 2 y), ẏ = x yp 1x + p 2 y), ẏ = 2 3 yp 1x + p 2 y). We prove the following theorem which characterizes all polynomial first integrals for the systems in 1.5). Theorem 1.3. The homogeneous polynomial systems in 1.5) have a polynomial first integral H if and only if one of the following conditions hold. a) The first system in 1.5) with p 1 = 0, p 2 = 31 q)/1 + 2q) with q = n/m Q + and, in this case, H = x m 3y 2 x 2 ) n.

5 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1623 b) The second system in 1.5) with p 1 = 0, p 2 = 31 q)/1 + 2q) with q = n/m Q + and, in this case, H = x m 3y 2 + x 2 ) n. c) The third system, in 1.5) with p 1 = 0 and p 2 = 31 2q)/21+q)) with q = n/m Q + and, in this case, H = x m y n. d) The fourth system in 1.5) with p 1 = p 2 = 0 and, in this case, H = x. We note that systems with weight exponent 3, 3) coincide with systems 1.4), and hence, it can be written into systems in 1.5). Therefore, Theorem 1.3 applies to those systems. The proof of Theorem 1.3 is given in Section 4. Theorem 1.4. The weight homogeneous polynomial differential systems with weight exponent 1, 3) and weight degree 4 have a polynomial first integral H if and only if the following conditions hold: a) a 11 = a 40 = 0 with H = x. b) b 60 = b 31 = b 02 = 0 with H = y. c) 3a 11 b 02 )3a 11 2b 02 ) 0, a 40 = a 11 b 31 /3a 11 2b 02 ), 3a 11 /6a 11 2b 02 ) = m/n Q + and m/n < 1 with H = x 3n m) 3a 11 2b 02 )b 60 x a 11 b 02 )b 31 x 3 y 9a a 11 b b 2 02)y 2 ) m. d) b 31 /a 40 = m/n and m/n Q + with H = x 3m b 60 x 3 + b 31 3a 40 )y) 3n. e) b 02 = 0, a 11 0, a 40 = b 31 /3 and b 06 = 3a 40 b 31 ) 2 /12a 11 ) with H = b 31 x 3 3a 11 y. f) 3a 11 b 02 )3a 11 2b 02 ) 0, a 40 = a 11 b 31 /3a 11 2b 02 ), b 06 = 3a 40 b 31 ) 2 /43a 11 b 02 )), b 02 0 and 3a 11 /b 02 = n/m Q +, with H = x 3n b 31 x 3 + 2b 02 3a 11 )y) m. The proof of Theorem 1.4 is given in Section 5. Theorem 1.5. The weight homogeneous polynomial differential systems with weight exponent 1, 4) and weight degree 4 have a polynomial first integral H if and only if the following conditions hold:

6 1624 JAUME LLIBRE AND CLAUDIA VALLS a) a 40 = a 01 = 0 and H = x. b) b 70 = b 31 = 0 and H = y. c) b 31 = 4a 40, and 4a a 01 b 70 0 with H = b 70 x 8 + 8a 40 x 4 y + 4a 01 y 2. d) b 31 = 4a 40, a 40 b 70 0 and 4a a 01 b 70 = 0 with H = b 70 x 4 4a 40 y. The proof of Theorem 1.5 is given in Section 6. Theorem 1.6. The weight homogeneous polynomial differential systems with weight exponent 2, 3) and weight degree 4 have a polynomial first integral H if and only if a 11 = 0, in which case H = x, or b 30 = b 02 = 0, in which case, H = y, or a 11 3a 11 2b 02 ) 0 and 2b 02 /a 11 = n/m Q +, in which case, H = x n 2b 30 x 3 3a 11 y 2 + 2b 02 y 2 ) m. The proof of Theorem 1.6 is given in Section 7. Theorem 1.7. The weight homogeneous polynomial differential systems with weight exponent 2, 5) and weight degree 4 have the polynomial first integral H = 2b 40 x 5 5a 01 y 2. The proof of Theorem 1.7 is given in Section 8. Theorem 1.8. The weight homogeneous polynomial differential systems with weight exponent 6, 9) and weight degree 4 have the polynomial first integral H = 2b 20 x 3 3a 01 y 2. The proof of Theorem 1.8 is given in Section Proof of Proposition 1.2. From the definition of weight homogeneous polynomial differential systems 1.1) with weight degree 4, the exponents u i and v i of any monomial x u i y v i of P i for i = 1, 2, are such that they satisfy respectively the relations 2.1) s 1 u 1 + s 2 v 1 = s and s 1 u 2 + s 2 v 2 = s 2 + 3,

7 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1625 respectively. Moreover, we can assume that P 1 and P 2 are coprime, and, without loss of generality, we can also assume that s 1 s 2. We consider different values of s 1. Case s 1 = 1. If s 2 = 1, then in view of 2.1), we must have u 1 + v 1 = 4 and u 2 + v 2 = 4, that is, u i, v i ) = 0, 4), u i, v i ) = 1, 3), u i, v i ) = 2, 2), u i, v i ) = 3, 1) and u i, v i ) = 4, 0), for i = 1, 2. If s 2 = 2, then, in view of 2.1), we must have u 1 + 2v 1 = 4 and u 2 + 2v 2 = 5, that is, u 1, v 1 ) = 0, 2), u 1, v 1 ) = 2, 1) and u 1, v 1 ) = 4, 0), while u 2, v 2 ) = 1, 2), u 2, v 2 ) = 3, 1), and finally, u 2, v 2 ) = 5, 0). If s 2 = 3, then, in view of 2.1), we must have u 1 + 3v 1 = 4 and u 2 + 3v 2 = 6, that is, u 1, v 1 ) = 1, 1), u 1, v 1 ) = 4, 0), while u 2, v 2 ) = 0, 2), u 2, v 2 ) = 3, 1), and finally, u 2, v 2 ) = 6, 0). If s 2 = 4, then, in view of 2.1), we must have u 1 + 4v 1 = 4 and u 2 + 4v 2 = 7, that is, u 1, v 1 ) = 0, 1), u 1, v 1 ) = 4, 0), while u 2, v 2 ) = 3, 1), and finally, u 2, v 2 ) = 7, 0). If s 2 = 4 + l with l 1, then equation 2.1) becomes 2.2) u l)v 1 = 4 and u l)v 2 = 7 + l. From the first equation of 2.2), we obtain v 1 = 0 and u 1 = 4. By the second equation of 2.2), it follows that v 2 {0, 1}. If v 2 = 0, then u 2 = 7 + l; if v 2 = 1, then u 2 = 3. In both cases, P 1 and P 2 are not coprime. Thus, this case is not considered. Case s 1 = 2. Now, we have s 2 2. If s 2 = 2, then in view of 2.1), we must have 2u 1 + 2v 1 = 5 and 2u 2 + 2v 2 = 5, which is not possible because 5 is not an even number. If s 2 = 3, then, in view of 2.1), we must have 2u 1 + 3v 1 = 5 and 2u 2 + 3v 2 = 6, that is, u 1, v 1 ) = 1, 1), while u 2, v 2 ) = 0, 2) and u 2, v 2 ) = 3, 0). If s 2 = 4, then, in view of 2.1), we must have 2u 1 + 4v 1 = 5 and 2u 2 + 4v 2 = 7, which is not possible because 5 is not even. If s 2 = 5, then, in view of 2.1), we must have 2u 1 + 5v 1 = 5 and 2u 2 + 5v 2 = 8, that is, u 1, v 1 ) = 0, 1) and u 2, v 2 ) = 4, 0). If s 2 = 5 + l with l 1, then equation 2.1) becomes 2.3) 2u l)v 1 = 5 and 2u l)v 2 = 8 + l.

8 1626 JAUME LLIBRE AND CLAUDIA VALLS The first equation of 2.3) is not possible because 5 is not an even number, 5 + l 6 and u 1 and v 1 are non-negative integers. Case s 1 = 3. Now, we have s 2 3. If s 2 = 3, then in view of 2.1), we must have 3u 1 + 3v 1 = 6 and 3u 2 + 3v 2 = 6, that is, u i, v i ) = 0, 2), u i, v i ) = 1, 1) and u i, v i ) = 2, 0), for i = 1, 2. If s 2 = 3 + l with l 1, then in view of 2.1), we must have 2.4) 3u l)v 1 = 6 and 3u l)v 2 = 6 + l. From the first equation of 2.4) we have that v 1 = 6 3u l and, using l 1, then v 1 {0, 1} l, When v 1 = 0, then 3u 1 = 6; thus, u 1 = 2. Then, from the second equation of 2.4) we obtain that v 2 {0, 1}. If v 2 = 0, then u 2 1, and, if v 2 = 1, then u 2 = 1. In both cases, we have that P 1 and P 2 are not coprime. When v 1 = 1, then 3u 1 = 3 l, which is not possible since u 1 is an integer and l 1. Case s 1 = 3 + l with l 1. Now, we have s l with l 1, and equation 2.1) becomes 2.5) and 3 + l)u 1 + s 2 v 1 = 6 + l = 3 + l) l)u 2 + s 2 v 2 = 3 + s 2. From the first equation of 2.5), and taking into account that l 1, we obtain that u 1 {0, 1}. When u 1 = 0, we must have s 2 v 1 = 6 + l, and since s l, we obtain v 1 = 6 + l 3 + l) + 3 = s l 3 + l. Since v 1 0 and l 1, we must have v 1 = 1. Then s 2 = 6 + l. Now, the second equation of 2.5) becomes 2.6) 3 + l)u l)v 2 = 6 + l) + 3.

9 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1627 Then v 2 = 0. From equation 2.6), we have u 2 = 1 + 6/l + 3). Thus, l = 3 and u 2 = 2, and we obtain the systems with weight exponent 6, 9). When u 1 = 1, we must have s 2 v 1 = 3, and, since s 2 3+l, we obtain 3 + l)v 2 3, which is not possible because l 1. This concludes the proof of the proposition. 3. Weight exponent s = 1, 1). A weight homogeneous polynomial system, ẋ = P 1 x, y); ẏ = P 2 x, y), with weight exponent 1, 1) and weight degree d is integrable, and its inverse integrating factor is V x, y) = xp 2 x, y) yp 1 x, y). See [7] for more details. As P 1 x, y), P 2 x, y) and V x, y) are homogeneous polynomials, if the degree of P 1 x, y) and P 2 x, y) is d, then, of course, the degree of V x, y) is d + 1. Thus, for d = 4, we can write the homogeneous polynomials as follows: 3.1) and P 1 x, y) = p 1 a 1 )x 4 + p 2 4a 2 )x 3 y + p 3 6a 3 )x 2 y 2 + p 4 4a 4 )xy 3 a 5 y 4, P 2 x, y) = a 0 x 4 + 4a 1 + p 1 )x 3 y + 6a 2 + p 2 )x 2 y 2 + 4a 3 + p 3 )xy 3 + a 4 + p 4 )y 4, V x, y) = a 0 x 5 + 5a 1 x 4 y + 10a 2 x 3 y a 3 x 2 y 3 + 5a 4 xy 4 + a 5 y 5. Thus, the first integral is Hx, y) = P1 x, y) dy + gx), V x, y) satisfying H/ x = P 2 /V. The canonical forms appear in the factorization of V. Assume that V x, y) factorizes as: i) five simple real roots: a 0 x r 1 y)x r 2 y)x r 3 y)x r 4 y)x r 5 y), ii) one double and three simple real roots: a 0 x r 1 y) 2 x r 2 y)x r 3 y)x r 4 y),

10 1628 JAUME LLIBRE AND CLAUDIA VALLS iii) two double roots and one simple real root: a 0 x r 1 y) 2 x r 2 y) 2 x r 3 y), iv) one triple and two simple real roots: a 0 x r 1 y) 3 x r 2 y)x r 3 y), v) one triple and one double real roots: a 0 x r 1 y) 3 x r 2 y) 2, vi) one quadruple and one simple real roots: a 0 x r 1 y) 4 x r 2 y), vii) one quintuple real root: a 0 x ry) 5, viii) three real and one couple of conjugate complex roots: a 0 x r 1 y)x r 2 y)x r 3 y)x 2 + bxy + cy 2 ) with b 2 4c < 0, ix) one double, one simple real and one couple of conjugate complex roots: a 0 x r 1 y) 2 x r 2 y)x 2 + bxy + cy 2 ) with b 2 4c < 0, x) one triple real and one couple of conjugate complex roots: a 0 x r 1 y) 3 x 2 + bxy + cy 2 ) with b 2 4c < 0, xi) one simple real and two couples of conjugate complex roots: a 0 x ry)x 2 + b 1 xy + c 1 y 2 )x 2 + b 2 xy + c 2 y 2 ) with b 2 1 4c 1 < 0, b 2 2 4c 2 < 0, xii) one simple real and one double couple of conjugate complex roots: a 0 x ry)x 2 + bxy + cy 2 ) 2 with b 2 4c < 0. Now, we shall compute the first integral for each case and obtain the conditions in order to show that it is a polynomial. We define the function: fr) = 5p 4 + p 3 r + p 2 r 2 + p 1 r 3 ). Case i). A first integral H is x r 1 y) γ 1 x r 2 y) γ 2 x r 3 y) γ 3 x r 4 y) γ 4 x r 5 y) γ 5, where γ i = fr i ) + a 0 5 r i r j ) j=1 j i 5 r i r j ) j=1 j i. We note that an integer power of H is a polynomial if and only if γ i Q for i = 1, 2, 3, 4, 5, if they all have the same sign.

11 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1629 Case ii). A first integral H is x r 1 y) γ 1 x r 2 y) γ 2 x r 3 y) γ 3 x r 4 y) γ 4 ) fr 1 )x exp, r 1 r 1 r 2 )r 1 r 3 )r 1 r 4 )x r 1 y) with γ 1 = A 1 /B 1 = A 1 /[r 1 r 2 ) 2 r 1 r 3 ) 2 r 1 r 4 ) 2 ], A 1 = 5p 1 r 2 + r 3 + r 4 )r 2 1 2r 3 r 4 + r 2 r 3 + r 4 ))r 1 + 3r 2 r 3 r 4 )r p 2 r 3 1 r 3 r 4 + r 2 r 3 + r 4 ))r 1 + 2r 2 r 3 r 4 )r 1 + 5p 3 2r 3 1 r 2 + r 3 + r 4 )r r 2 r 3 r 4 ) + 5p 4 3r 2 1 2r 2 + r 3 + r 4 )r 1 + r 3 r 4 + r 2 r 3 + r 4 )) 2a 0 B 1, for i = 2,..., 4 and γ i = A i B i = fr i ) + a 0 B i ). 4 r 1 r i ) 2 r i r j ) We note that an integer power of H is a polynomial if and only if fr 1 ) = 0 and γ i Q for i = 1, 2, 3, 4, if they all have the same sign. Case iii). A first integral H is x r 1 y) γ 1 x r 2 y) γ 2 x r 3 y) γ 3 exp j=2 j i 2 i=1 ) fr i )x r i r 1 r 2 ) 2, r i r 3 )r i y x) with γ i = A i /B i = A i /[r 1 r 2 ) 3 r i r 3 ) 2 ] for i = 1, 2, γ 3 = A 3 /B 3 = fr 3 ) + a 0 B 3 )/[r 1 r 3 ) 2 r 2 r 3 ) 2 ], A i = 2a 0 B i + 1) i+1 5p 4 3r i r j 2r 3 ) 5p 3 r 1 + r 2 )r 3 2r 2 i ) + 5p 2 r i r i r 1 + r 2 ) 2r j r 3 ) + 5p 1 r 2 i 3r j r 3 + r i 2r j + r 3 ))), for i, j = 1, 2 and i j. We note that an integer power of H is a polynomial if and only if fr i ) = 0 for i = 1, 2 and γ i Q for i = 1, 2, 3, if they all have the same sign. Case iv). A first integral H is ) 5βx x r 1 y) γ1 x r 2 y) γ2 x r 3 y) γ3 exp 2r1 2r 1 r 2 ) 2 r 1 r 3 ) 2 r 1 y x) 2,

12 1630 JAUME LLIBRE AND CLAUDIA VALLS with β = p 1 r 2 1 3r 2 r 1 3r 3 r 1 +5r 2 r 3 )r 3 1 p 2 r 2 1 +r 2 r 1 +r 3 r 1 3r 2 r 3 )r 2 1 p 3 3r 2 1 r 2 r 1 r 3 r 1 r 2 r 3 )r 1 +p 4 5r r 2 r 1 +3r 3 r 1 r 2 r 3 ))x + 2r 1 p 1 r 1 r 2 2r 3 r 2 + r 1 r 3 )r p 3 2r 1 r 2 r 3 )r p 2 r 2 1 r 2 r 3 )r p 4 3r 2 1 2r 2 r 1 2r 3 r 1 + r 2 r 3 ))y, γ 1 = A 1 /B 1 = A 1 /[r 1 r 2 ) 3 r 1 r 3 ) 3 ], A 1 = 3a 0 B 1 5r 2 p 2 + p 1 r 2 )r r 1 3r 2 )p 2 + p 1 r 2 )r 3 r p 2 r p 1 r 1 r 2 1 3r 2 r 1 + 3r 2 2))r 2 3) + 5p 3 r 3 1 3r 2 r 3 r 1 + r 2 r 3 r 2 + r 3 )) + 5p 4 3r 2 1 3r 2 + r 3 )r 1 + r r r 2 r 3 ), γ i = A i /B i = 1) i fr i ) + a 0 B i )/[r 1 r i ) 3 r 2 r 3 )] for i = 2, 3. We note that an integer power of H is a polynomial if and only if β = 0 and γ i Q for i = 1, 2, 3, if they all have the same sign. Case v). A first integral H is where β = 2r 1 r 2 )xfr 2 )r 2 1 r 2 r 2 y x) and x r 1 y) γ 1 x r 2 y) γ 2 expβ), + r 1 r 2 ) 2 x 2 fr 1 ) x r 1 y) r 1 r 2 )2p 3 + 2p 1 r 1 r 2 + p 2 r 1 + r 2 ))r1 2 + p 4 3r 1 r 2 ))x, r 1 y x γ 1 = 2r 2 1 3a0 r 1 r 2 ) p 4 + 5p 3 r 1 + 2r 2 ) + 5r 2 3p 1 r 1 r 2 + p 2 2r 1 + r 2 )) ), γ 2 = 2r 2 1 2a0 r 1 r 2 ) 4 15p 4 5p 3 r 1 + 2r 2 ) 5r 2 3p 1 r 1 r 2 + p 2 2r 1 + r 2 )) ). We note that an integer power of H is a polynomial if and only if β = 0 and γ i Q for i = 1, 2, if they all have the same sign. Case vi). A first integral H is x r 1 y) γ 1 x r 2 y) γ 2 expβ),

13 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1631 where β = 2r 1 r 2 ) 3 fr 1 )x 3 r 3 1 r 1y x) 3 and + 30r 1 r 2 )p 3 +r 2 p 2 + p 1 r 2 ))r p 4 3r 2 1 3r 2 r 1 + r 2 2))x r 3 1 r 1y x) + 15r 1 r 2 ) 2 p 4 3r 1 2r 2 )+r 1 p 2 +p 1 r 2 )r p 3 2r 1 r 2 )))x 2 r 3 1 x r 1y) 2, γ 1 = 6 4a 0 r 1 r 2 ) 4 + 5p 4 + 5r 2 p 3 + r 2 p 2 + p 1 r 2 ))), γ 2 = 6a 0 r 1 r 2 ) 4 + 5p 4 + 5r 2 p 3 + r 2 p 2 + p 1 r 2 ))). We note that an integer power of H is a polynomial if and only if β = 0 and γ i Q for i = 1, 2, if they all have the same sign. Case vii). A first integral H is x ry) γ 1 exp where βx x ry) 4 β = rxrx p 2 x + 3p 1 rx + 4p 2 ry) + p 3 x 2 4ryx + 6r 2 y 2 )) 3p 4 x 2ry)x 2 2ryx + 2r 2 y 2 ), and γ 1 = 12a 0 r 4. We note that x ry is a polynomial first integral if and only if β = 0. Case viii). A first integral H is x r 1 y) γ 1 x r 2 y) γ 2 x r 3 y) γ 3 x 2 + bxy + cy 2 ) γ 4 βx bx + 2cy exp 3 i=1 c + b r i + ri 2) 4c b 2 )x arctan )), 2 4c b2 )x 2 where β = 52p 1 c 3 bp 2 p 1 r 1 + r 2 + r 3 )) + 2p 3 + p 2 r 1 + r 2 + r 3 ) + p 1 r 2 r 3 + r 1 r 2 + r 3 ))))c 2 + p 3 + p 1 r 1 r 2 + p 1 r 1 + r 2 )r 3 )b 2 + 3p 4 3p 1 r 1 r 2 r 3 + p 3 r 1 + r 2 + r 3 ) p 2 r 2 r 3 + r 1 r 2 + r 3 )))b + 2p 2 r 1 r 2 r 3 + p 4 r 1 + r 2 + r 3 ) + p 3 r 2 r 3 + r 1 r 2 + r 3 ))))c ),

14 1632 JAUME LLIBRE AND CLAUDIA VALLS bp 4 b + r 1 )b + r 2 )+p 1 r 1 b 3 p 2 r 1 b 2 p 4 b + p 3 r 1 b 2p 4 r 1 )r 2 bp 4 b + r 1 ))r 3 ), and γ i = A i /B i = fr i ) + a 0 B i )/[c + br i + r 2 i ) 3 j=1, j i r i r j )] for i = 1, 2, 3, γ 4 = A 4 /B 4 = A 4 /[2 3 i=1 c + b r i + r 2 i )] A 4 = a 0 B 4 + 5p 2 + p 1 r 1 + r 2 + r 3 ))c 2 p 4 + p 1 r 1 r 2 r 3 + p 3 r 1 + r 2 + r 3 ) + p 2 r 2 r 3 + r 1 r 2 + r 3 )) + bp 3 p 1 r 2 r 3 + r 1 r 2 + r 3 ))))c + p 4 b + r 1 )b + r 2 ) + p 4 b + r 1 ) + p 4 + p 1 b 2 p 2 b + p 3 )r 1 )r 2 )r 3 ). We note that an integer power of H is a polynomial if and only if β = 0 and γ i Q for i = 1, 2, 3, 4, if they all have the same sign. Case ix). A first integral H is x r 1 y) γ1 x r 2 y) γ2 x 2 +bxy+cy 2 ) γ3 fr 1 )x exp r 1 r 1 r 2 )c+r 1 b+r1 2)r 1y x) βx bx + 2cy + 2 i=1 c + b r i + ri 2)3 i 4c b 2 )x arctan )), 2 4c b2 )x 2 where β = 52p 1 c 3 bp 2 p 1 2r 1 + r 2 )) + 2p 3 + p 2 2r 1 + r 2 ) + p 1 r 1 r 1 + 2r 2 )))c 2 + p 3 + p 1 r 1 r 1 + 2r 2 ))b 2 + 3p 4 + p 3 2r 1 + r 2 ) r 1 3p 1 r 1 r 2 + p 2 r 1 + 2r 2 )))b + 2p 4 2r 1 + r 2 ) + r 1 p 2 r 1 r 2 + p 3 r 1 + 2r 2 ))))c bp 4 b + r 1 ) 2 + p 4 b 2 2p 4 r 1 b+bp 1 b 2 p 2 b + p 3 ) 2p 4 )r 2 1)r 2 ), A 1 γ 1 = A 1 = B 1 r 1 r 2 ) 2 c + b r 1 + r1 2, )2 A 1 = 2a 0 c 2 r 1 r 2 ) 2 + 2a 0 b 2 r 2 1r 1 r 2 ) 2 + c 5p 4 + r 1 4a 0 b + r 1 )r 1 r 2 ) 2 + 5p 1 r 1 2r 1 3r 2 ) + 5p 2 r 1 2r 2 )) 5p 3 r 2 ) + b4a 0 r 1 r 1 r 2 ) 2 5p 3 + 5p 1 r 1 r 1 2r 2 ) 5p 2 r 2 )r p 4 r 2 2r 1 ))+r 1 5p 4 2r 2 3r 1 ) + r 1 2a 0 r 1 r 2 ) 2 5p 2 5p 1 r 2 )r p 3 r 2 2r 1 ))),

15 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1633 γ 2 = A 2 B 2 = fr 2 ) + a 0 B 2 r 1 r 2 ) 2 c + b r 2 + r2 2), γ 3 = A 3 A 3 = B 3 2c + b r 1 + r1 2)2 c + b r 2 + r2 2), A 3 = a 0 B 3 + 5p 2 + p 1 2r 1 + r 2 ))c 2 p 4 + p 3 2r 1 + r 2 ) + r 1 p 1 r 1 r 2 + p 2 r 1 + 2r 2 )) + bp 3 p 1 r 1 r 1 + 2r 2 )))c + p 4 b + r 1 ) 2 + p 1 b 2 p 2 b + p 3 )r p 4 r 1 + bp 4 )r 2 ). We note that an integer power of H is a polynomial if and only if fr 1 ) = 0, β = 0 and γ i Q for i = 1, 2, 3, if they all have the same sign. Case x). A first integral H is x r 1 y) γ 1 x 2 + bxy + cy 2 ) γ fr1 2 )c + b r 1 + r 2 exp 1) 2 x 2 r1 2x r 1y) 2 β 1c + b r 1 + r1)x 2 β 2 x bx + 2cy r1 2x r + 1y) 4c b2 )x arctan )), 2 4c b2 )x 2 where and β 1 = 10p 2 bp 1 )r p 3 cp 1 )r cp 2 + bp 3 + 3p 4 )r bp 4 r 1 + cp 4 ), β 2 = 10p 1 r 3 1 p 4 )b 3 r 1 p 2 r p 4 )b 2 + r 2 1p 3 r 1 3p 4 )b 2p 4 r c 3 p 1 c 2 2p 3 + bp 2 3p 1 r 1 ) + 6r 1 p 2 + p 1 r 1 )) + c3p 1 r p 3 )b 2 + 3p 4 + r 1 p 3 r 1 p 2 + p 1 r 1 )))b + 2r 1 3p 4 + r 1 3p 3 + p 2 r 1 )))), γ 1 = 23a 0 c + r 1 b + r 1 )) 3 + 5p 1 b 2 p 2 b + p 3 )r p 4 r bp 4 r 1 + b 2 p 4 + c 2 p 2 + 3p 1 r 1 ) cp 4 + bp 3 3p 1 r 2 1) + r 1 3p 3 + r 1 3p 2 + p 1 r 1 ))))), γ 2 = 5p 1 b 2 p 2 b + p 3 )r p 4 r bp 4 r 1 + b 2 p 4 + c 2 p 2 + 3p 1 r 1 ) cp 4 + bp 3 3p 1 r 2 1) + r 1 3p 3 + r 1 3p 2 + p 1 r 1 )))) 2a 0 c + r 1 b + r 1 )) 3.

16 1634 JAUME LLIBRE AND CLAUDIA VALLS We note that an integer power H is a polynomial if and only if fr 1 ) = 0, β 1 = β 2 = 0 and γ i Q for i = 1, 2, if they all have the same sign. Case xi). A first integral H is x 2 + b 1 xy + c 1 y 2 ) γ 1 x 2 + b 2 xy + c 2 y 2 ) γ 2 x r 1 y) γ 3 2 β i x exp 4ci b arctan bi x + 2c i y )), 2 i )x2 4ci b 2 i )x2 i=1 with β i = α i /δ i for i = 1, 2, where α i = 5 b i c 2 i p 2 + c i c j p 2 + b j c i c i p 1 + p 3 ) 3c i p 4 + c j p 4 ) + b i c 2 i p 1 + c j p 3 + c i 3c j p 1 + b j p 2 + p 3 ) + b j p 4 )r 1 + b 3 i p 4 +c j p 1 r 1 )+b 2 i c i c j p 1 +c i p 3 +b j p 4 b j c i p 1 +c j p 2 +p 4 )r 1 ) + 2c 3 i p 1 c j p 4 r 1 c 2 i c j p 1 + p 3 + p 2 r 1 b j p 2 + p 1 r 1 )) + c i p 4 r 1 + c j p 3 + p 2 r 1 ) b j p 4 + p 3 r 1 )))), δ i = b 2 2c 1 + c 1 c 2 ) 2 + b 2 1c 2 b 1 b 2 c 1 + c 2 ))c i + b i r 1 + r 2 1), for i, j = 1, 2 and i j, γ i = 2a 0 b 2 jc i + c i c j ) 2 + b 2 i c j b i b j c i + c j ))c i + r 1 b i + r 1 )) + 5b 2 i p 4 + c j p 1 r 1 ) + b i c i c j p 1 c i p 3 c j p 2 r 1 + p 4 r 1 ) + c i c j ) p 4 p 3 r 1 + c i p 2 + p 1 r 1 )) b j c 2 i p 1 + p 4 b i + r 1 ) c i p 3 + b i p 1 + p 2 )r 1 ))), for i, j = 1, 2 and i j. Finally, γ 3 = A 3 = fr 1) + a 0 B 3 ). B 3 2 c i + b i r 1 + r1) 2 i=1 We note that an integer power of H is a polynomial if and only if β 1 = β 2 = 0 and γ i Q for i = 1, 2, 3, if they all have the same sign. Case xii). A first integral H is x ry) γ 1 x 2 +bxy+cy 2 ) γ 2 exp β 1 x 3 arctan [4c b 2 )x 2 ] 3/2 ) bx + 2cy 4c b2 )x 2

17 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS β 2 c + b r + r 2 ) )x b 2 4c)cx 2 + bxy + cy 2, ) where β 1 = 10p 4 + rp 3 rp 2 + p 1 r)))b 3 + 4p 3 r 2 b 2 + 2r 2 p 3 r 3p 4 )b 4p 4 r 3 + 4c 3 p 1 + c 2 4p 3 + rp 2 + 3p 1 r)) 2bp 2 3p 1 r)) 2c2p 2 rb 2 + 3p 4 rp 3 + r3p 1 r p 2 )))b + 2r3p 4 + rp 3 + p 2 r)))), β 2 = p 4 yb 3 + cp 1 rx + p 3 y) p 4 x + ry))b 2 + p 1 x + ry) p 2 y)c 2 + p 2 rx + 3p 4 y + p 3 x + ry))c p 4 rx)b + 2cp 1 yc 2 p 1 rx + p 3 y + p 2 x + ry))c + p 3 rx + p 4 x + ry)), γ 1 = 2a 0 c + br + r 2 ) 2 + fr)), γ 2 = 4a 0 c + br + r 2 ) 2 fr). We note that an integer power of H is a polynomial if and only if β 1 = β 2 = 0 and γ i Q for i = 1, 2, if they all have the same sign. 4. Weight exponent s = 1, 2). In this section, we prove Theorem 1.3. Since systems in equation 1.5) are homogeneous, we know that they are integrable because they have the inverse integrating factor V = xẏ yẋ. The strategy will be to obtain such first integrals and to determine which of them are polynomials. Denoting systems in equation 1.5) by ẋ = P x, y) and ẏ = Qx, y), the first integral is Hx, y) = satisfying H/ x = Qx, y)/v x, y). P x, y) dy + gx), V x, y) The first system in equation 1.5) has the first integral H = x 3 2p 2 3y 2 x 2 ) 3+p 2 exp 2 x 3p 1 arctanh )). 3y Note that an integer power of H is a polynomial if and only if p 1 = 0 and p 2 = 31 q)/1 + 2q) with q Q +.

18 1636 JAUME LLIBRE AND CLAUDIA VALLS The second system in equation 1.5) has the first integral H = x 3 2p 2 3y 2 + x 2 ) 3+p 2 exp 2 x 3p 1 arctan )). 3y Note that an integer power of H is a polynomial if and only if p 1 = 0 and p 2 = 31 q)/1 + 2q) with q Q +. The third system in equation 1.5) has the first integral H = x 23+p1) y 3+2p1 exp 2p 2y x. Note that an integer power of H is a polynomial if and only if p 2 = 0 and p 1 = 31 2q)/21 + q)) with q Q +. The fourth system in equation 1.5) has the first integral H = x exp y 2p ) 1x + p 2 y 3x 2. Note that an integer power of H is a polynomial if and only if p 1 = p 2 = 0. The fifth system in equation 1.5) has the first integral x/y which is never a polynomial. 5. Weight exponent s = 1, 3). Performing a change of variables X, Y ) = x 3, y), the planar weight homogeneous systems of weight degree 4 and weight exponent 1, 3) become 5.1) Ẋ = 3a 40 X 2 + 3a 11 XY, Ẏ = b 60 X 2 + b 31 XY + b 02 Y 2. Again, we shall use the inverse integrating factor V = XẎ Y Ẋ for computing the first integrals of system 5.1). It is clear that, if a 11 = a 40 = 0, then a polynomial first integral is X, and, if b 60 = b 31 = b 02 = 0, then a polynomial first integral is Y. Now, we consider the other cases. Case i). 3a 11 b 02 0 and R = 3a 40 b 31 ) a 11 + b 02 )b In this case, system 5.1) has the first integral ) 6 3a40 X b 31 X +6a 11 Y 2b 02 Y a 11 3a 40 + b 31 ) 2a 40 b 02 ) arctan R RX

19 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1637 ) Y 3a40 X b 31 X +3a 11 Y b 02 Y ) +23a 11 b 02 ) log X +3a 11 log X 2 b 60. Here, log A always means log A and, as usual, log is the logarithm in base e. Since this first integral must be a polynomial, we must have 5.2) a 11 3a 40 + b 31 ) 2a 40 b 02 = 0. Now, we consider different subcases. If 3a 11 2b 02 0, then from equation 5.2), we obtain a 40 = a 11b 31 3a 11 2b 02. Therefore, using the exponential of the previous first integral, we obtain that the first integral is where H = X 1 3a11/6a11 2b02) px, Y ) 3a11/6a11 2b02), px, Y ) = 3a 11 2b 02 )b 60 X a 11 b 02 )b 31 XY 9a a 11 b b 2 02)Y 2. Note that, since a 11 b 02 0 and 3a 11 2b 02 0, we have 9a a 11 b b Therefore, an integer power of H is a polynomial first integral if and only if 3a 11 /6a 11 2b 02 ) = m/n Q +, and m/n < 1. In this case, the first integral H is X n m 3a 11 2b 02 )b 60 X a 11 b 02 )b 31 XY 9a a 11 b b 2 02)Y 2 ) m. If 3a 11 2b 02 = 0, that is, b 02 = 3a 11 /2. In this case, from equation 5.2), we obtain a 11 b 31 = 0. Hence, either a 11 = 0 or b 31 = 0. However, if a 11 = 0, then b 02 = 0, which contradicts the fact that 3a 11 2b Therefore, this case is not possible and we must have b 31 = 0. Then, the first integral is which is never a polynomial. H = 2b 60X 2 + 6a 40 XY + 3a 11 Y 2, X

20 1638 JAUME LLIBRE AND CLAUDIA VALLS Case ii). b 02 = 3a 11 and b 31 3a In this case, system 5.1) has the first integral b 31 3a 40 ) 2 log X + 3a 113a 40 b 31 )Y X + 3 3a b 31 a 40 a 11 b 60 ) log b ) 60X 3a 40 Y + b 31 Y. X In order for the first integral to be a polynomial, we must have a 11 3a 40 b 31 ) = 0, that is, a 11 = 0 and hence, b 02 = 0). Then, using the exponential of the previous first integral, we obtain the following first integral X 1/3 b60 X 3a 40 Y + b 31 Y X ) a40/3a 40 b 31). Then, we must have b 31 /a 40 = m/n with m/n Q +. In this case, the previous first integral becomes H = X m b 60 X 3a 40 Y + b 31 Y ) 3n. Case iii). b 02 = 3a 11 and b 31 = 3a 40. System 5.1) has the first integral 2b 60 X 2 log X + 6a 40 Y X + 3a 11 Y 2 6X 2. Since the case a 40 = a 11 = 0 has been studied, we have that in this case the first integral is never a polynomial. Case iv). 3a 11 b 02 0 and R = 0. Then, b 06 = 3a 40 b 31 ) 2 43a 11 b 02 ), and system 5.1) has the first integral 3 3a 11 a 40 +2b 02 a 40 a 11 b 31 )X 3a 40 X b 31 X +6a 11 Y 2b 02 Y +3a 11 b 02 ) log36a 11 X 12b 02 X) + 3a 11 log 3a ) 40X b 31 X + 6a 11 Y 2b 02 Y. X In order to show that it is a polynomial, we must have 5.3) 2a 40 b 02 a 11 3a 40 + b 31 ) = 0. We will now consider two different subcases.

21 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1639 If 3a 11 2b 02, then condition 5.3) becomes a 40 = a 11b 31 3a 11 2b 02. Using the exponential of the previous first integral, we obtain the following first integral: X 1/36a11 12b02) b31 X 3a 11 Y + 2b 02 Y X From this first integral, we obtain the first integral: Xb 31 X 3a 11 Y + 2b 02 Y ) 3a11/b02. ) a11 /4b 02 3a 11 ) 2. So, if b 02 0, then 3a 11 /b 02 = m/n with m/n Q +, and the polynomial first integral is X n b 31 X 3a 11 Y + 2b 02 Y ) m. If b 02 = 0, then H = b 31 X 3a 11 Y is a polynomial first integral. This concludes the proof of Theorem Weight exponent s = 1, 4). We introduce the change X, Y ) = x 4, y) in the planar weight homogeneous polynomial differential systems 1.1) of weight degree 4 with weight exponent 1, 4). In these new variables X, Y ), the systems with weight exponent 1, 4) become, after introducing the new independent variable dτ = x 3 dt, as follows: 6.1) X = 4a 40 X + a 01 Y ), Y = b 70 X + b 31 Y, where the prime denotes the derivative with respect to τ. If a 40 = a 01 = 0, then a polynomial first integral is X, and, if b 70 = b 31 = 0, then a polynomial first integral is Y. Now, we consider the other cases. Case i). R = 4a 40 b 31 ) 2 16a 01 b Now, a first integral of system 6.1) is: ) 24a 40 + b 31 ) 4a40 b 31 )X + 8a 01 Y arctan R RX + log b 70 X 2 + 4a 40 b 31 )XY + 4a 01 Y 2 ).

22 1640 JAUME LLIBRE AND CLAUDIA VALLS Since it must be a polynomial, we must have b 31 = 4a 40 such that the polynomial first integral is H = b 70 X 2 + 8a 40 XY + 4a 01 Y 2. Case ii). R = 0. We consider different subcases. First, we study when b Then, from R = 0, we obtain 6.2) a 01 = 4a 40 b 31 ) 2 16b 70. If b 31 4a 40 0, then the first integral is: 24a 40 + b 31 )b 70 X 2b 70 X + 4a 40 b 31 )Y + 4a 40 b 31 ) log2b 70 X + 4a 40 + b 31 )Y ), which is a polynomial if and only if b 31 = 4a 40. The polynomial first integral is b 70 X 4a 40 Y. If b 31 = 4a 40, then a 40 0 otherwise, b 31 = 0 and, from equation 6.2), we also have a 01 = 0, which has already been considered), and the first integral of equation 6.1) is which is never a polynomial. H = Y X b 70 log X 4a 40, If b 70 = 0, then from R = 0, we obtain b 31 = 4a 40. We only consider the case a 40 0, where the first integral of equation 6.1) is: a 40X Y This completes the proof of Theo- which is never a polynomial. rem a 01 log Y, 7. Weight exponent s = 2, 3). We prove Theorem 1.6. The planar weight homogeneous polynomial differential systems 1.1) with weight degree 4 and weight-exponent 2, 3) are: 7.1) ẋ = a 11 xy, ẏ = b 30 x 3 + b 02 y 2. If a 11 3a 11 2b 02 ) 0, then the first integral of system 7.1) is: H = x 2b 02/a 11 2b 30 x 3 3a 11 y 2 + 2b 02 y 2).

23 WEIGHT-HOMOGENEOUS POLYNOMIAL SYSTEMS 1641 Then, an integer power of H is a polynomial first integral if and only if 2b 02 /a 11 Q +. If a 11 = 0, then H = x is a polynomial first integral of system 7.1). If b 30 = b 02 = 0, then H = y is a polynomial first integral of system 7.1). If a 11 0 and 3a 11 = 2b 02, then the first integral of system 7.1) is: This completes the proof of Theo- which is never a polynomial. rem 1.6. H = y4 x 3 2b 30 a 11 log x, 8. Weight exponent s = 2, 5). Here, we prove Theorem 1.7. The planar weight homogeneous polynomial differential systems 1.1) of weight degree 4 with weight exponent 2, 5) are: ẋ = a 01 y, ẏ = b 40 x 4. It is straightforward to prove that H = 2b 40 x 5 5a 01 y 2 is a polynomial first integral. 9. Weight exponent s = 6, 9). Now, we prove Theorem 1.8. The planar weight homogeneous polynomial differential systems 1.1) of weight degree 4 with weight exponent 6, 9) are: ẋ = a 01 y, ẏ = b 20 x 2. It is straightforward to prove that H = 2b 20 x 3 3a 01 y 2 is a polynomial first integral. This completes the proof of Theorem 1.8. REFERENCES 1. A. Algaba, E. Freire, E. Gamero and C. García, Monodromy, center-focus and integrability problems for quasi-homogeneous polynomial systems, Nonlin. Anal ), A. Algaba, N. Fuentes and C. García, Centers of quasi-homogeneous polynomial planar systems, Nonlin. Anal. Real World Appl ), A. Algaba, C. García and M. Reyes, Integrability of two dimensional quasihomogeneous polynomial differential systems, Rocky Mountain J. Math ), , Rational integrability of two-dimensional quasi-homogeneous polynomial differential systems, Nonlin. Anal ),

24 1642 JAUME LLIBRE AND CLAUDIA VALLS 5. A. Algaba, C. García and M.A. Teixeira, Reversibility and quasi-homogeneous normal forms of vector fields, Nonlin. Anal ), C. Cairó and J. Llibre, Polynomial first integrals for weight-homogeneous planar polynomial differential systems of weight degree 3, J. Math. Anal. Appl ), J. Chavarriga, H. Giacomini, J. Giné and J. Llibre, Darboux integrability and the inverse integrating factor, J. Diff. Eq ), A. Cima and J. Llibre, Algebraic and topological classification of the homogeneous cubic vector fields in the plane, J. Math. Anal. Appl ), T. Date and M. Lai, Canonical forms of real homogeneous quadratic transformations, J. Math. Anal. Appl ), S.D. Furta, On non-integrability of general systems of differential equations, Z. Angew. Math. Phys ), A. Goriely, Integrability, partial integrability, and nonintegrability for systems of ordinary differential equations, J. Math. Phys ), S. Labrunie, On the polynomial first integrals of the a, b, c) Lotka-Volterra system, J. Math. Phys ), J. Llibre and X. Zhang, Polynomial first integrals for quasi-homogeneous polynomial differential systems, Nonlinearity ), A.J. Maciejewski and J.M. Strelcyn, On algebraic non-integrability of the Halphen system, Phys. Lett ), J. Moulin-Ollagnier, Polynomial first integrals of the Lotka-Volterra system, Bull. Sci. Math ), A. Tsygvintsev, On the existence of polynomial first integrals of quadratic homogeneous systems of ordinary differential equations, J. Phys. Math. Gen ), Departament de Matemàtiques, Universitat Autònoma de Barcelona, Bellaterra, Barcelona, Catalonia, Spain address: Departamento de Matemática, Instituto Superior Técnico, Lisboa, Portugal address: