Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 14, pp. 1–14. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu EXISTENCE OF RATIONAL SOLUTIONS FOR q-DIFFERENCE PAINLEVÉ EQUATIONS HONG YAN XU, JIN TU Abstract. This article studies properties of meromorphic solutions for sev- eral types of q-difference Painlevé equations. We obtain conditions for the existence, and the form of rational solutions for two classes of q-difference Painlevé equations. Also for a solution f we obtain results about the fixed points, the exponents of convergence of poles of f,∆qf, (∆qf)/f . Our results extend previous theorems given in the references. 1. Introduction and statement of main results Painlevé equations have been an important research subject in the field of the mathematics and physics, and they occur in many physical situations: plasma physics, statistical mechanics, nonlinear waves, etc. They appear as differential Painlevé equation, discrete Painlevé equation, difference Painlevé, and so on; see [3, 7, 8]. Around 2006, with the development of Nevanlinna theory, Halburd-Korhonen [11] and Chiang-Feng [5] established independently important results about the complex difference and difference operators. By utilizing these results, Halburd- Korhonen [10, 11, 12] discussed the equation f(z + 1) + f(z − 1) = R(z, f), (1.1) where R(z, f) is rational in f and meromorphic in z. They pointed out that this equation can be transformed into difference Painlevé I equations f(z + 1) + f(z − 1) = az + b f(z) + c, (1.2) f(z + 1) + f(z) + f(z − 1) = az + b f(z) + c, (1.3) and into difference Painlevé II equations f(z + 1) + f(z − 1) = (az + b)f(z) + c 1− f(z)2 . (1.4) In 2010, Ronkainen [21] further investigated the meromorphic solutions of the equation f(z + 1)f(z − 1) = R(z, f) (1.5) 2010 Mathematics Subject Classification. 39A13, 30D35. Key words and phrases. Rational solution; q-difference Painlevé equation; fixed point. c©2020 Texas State University. Submitted January 13, 2019. Published February 5, 2020. 1 2 H. Y. XU, J. TU EJDE-2020/14 where R(z, f) is a rational and irreducible in f and meromorphic in z. He proved that either f satisfies the difference Riccati equation f(z + 1) = A(z)f(z) +B(z) f(z) + C(z) , or equation (1.5) can be transformed to one of the following equations f(z + 1)f(z − 1) = η(z)f(z)2 − λ(z)f(z) + µ(z) (f(z)− 1)(f(z)− υ(z)) , f(z + 1)f(z − 1) = η(z)f(z)2 − λ(z)f(z) f(z)− 1 , f(z + 1)f(z − 1) = η(z)(f(z)− λ(z)) (f(z)− 1) , f(z + 1)f(z − 1) = h(z)f(z)m, where η(z), λ(z), υ(z) satisfy certain conditions. Generally speaking, the above four equation can be called as the difference Painlevé III equations. In the past two decades, many mathematicians paid consideration attention to the value distribution of solutions for complex difference equations, and obtained lots of important results on the properties of solutions for difference Painlevé I-III equations (see [2, 3, 10, 11, 12, 18, 19, 33]). In 2010, Chen-Shon [4] considered the difference Painlevé I equation (1.2) and obtained the following theorem. Theorem 1.1 (see [4, Theorem 4]). Let a, b, c be constants, where a, b are not both equal to zero. Then (i) if a 6= 0, then (1.2) has no rational solution; (ii) if a = 0, and b 6= 0, then (1.2) has a nonzero constant solution w(z) = A, where A satisfies 2A2 − cA− b = 0. The other rational solution is w(z) = P (z) Q(z) +A, where P (z) and Q(z) are relatively prime polynomials and satisfy degP < degQ. In 2014, Zhang-Yang [30] studied the difference Painlevé III equations with the constant coefficients, and obtained the following result. Theorem 1.2 ([30]). If f is a transcendental finite-order meromorphic solution of f(z+ 1)f(z− 1)(f(z)− 1) = ηw(z) or f(z+ 1)f(z− 1)(f(z)− 1) = f(z)2 − λw(z), where η(6= 0), λ( 6= 0, 1) are constants, then (i) λ(f) = σ(f); (ii) f has at most one non-zero Borel exceptional value for σ(f) > 0. The Logarithmic Derivative Lemma on q-difference operators was established by Barnett, Halburd, Korhonen and Morgan [1] in 2007. Then the interest in studying the properties on the existence and value distribution of solutions has increased considerably for some q-difference equation which are formed by replacing the q- difference f(qz), q ∈ C \ {0, 1} with f(z + c) of meromorphic function in some expression concerning complex difference equations; see [6, 9, 14, 15, 16, 20, 22, 23, 24, 25, 26, 29, 31, 32]. In 2015, Qi-Yang [20] considered the equations f(qz) + f (z q ) = az + b f(z) + c, (1.6) EJDE-2020/14 q-DIFFERENCE PAINLEVÉ EQUATIONS 3 which can be seen as q-difference analogues of (1.2), and obtained the following result. Theorem 1.3 ([20, Theorem 1.1]). Let f(z) be a transcendental meromorphic solution with zero order of equation (1.6), and let a, b, c be constants such that a, b cannot vanish simultaneously. Then (i) f(z) has infinitely many poles. (ii) If a 6= 0 and any d ∈ C, then f(z)− d has infinitely many zeros. (iii) If a = 0 and f(z) takes a finite value A finitely often, then A is a solution of 2z2 − cz − b = 0. In 2018, Liu-Zhang [17] studied the difference equation Y (ωz) + Y (z) + Y ( z ω ) = V (z) Y (z) + c, (1.7) which is a q-difference analogues of (1.3), and obtained the following result. Theorem 1.4 ([17, Thereom 1.2]). Let c ∈ C\{0}, |ω| 6= 1, and V (z) = X(z) B(z) be an irreducible rational function, where X(z) and B(z) are polynomials with degX(z) = x and degB(z) = b. (i) Suppose that x ≥ b and x−b is zero or an even number. If (1.7) has an irre- ducible rational solution Y (z) = I(z) J(z) , where I(z) and J(z) are polynomials with deg I(z) = i and deg J(z) = j, then i− j = x−b 2 . (ii) Suppose that x < b. If (1.7) has an irreducible rational solution Y (z) = I(z) J(z) , then Y (z) satisfies one of the following two cases: (1) Y (z) = I(z) J(z) = c 3 + T (z) D(z) , where T (z) and D(z) are polynomials with deg T (z) = t and degD(z) = d, and b− x = d− t. (2) i− j = x− b. Motivated by the idea [17] and [20, 30], we investigate some properties of mero- morphic solutions of the following two equations f(qz)f( z q )f(z)(f(z)− 1) = µ, (1.8) f(qz)f( z q )(f(z)− 1)2 = (f(z)− λ)2, (1.9) which can be seen as q-difference Painlevé III equations. Before stating our main theorems, let us introduce some basic notation in the theory of Nevanlinna value distribution (see Hayman [13], Yang [27] and Yi and Yang [28]). We denote σ(f), λ(f) and λ( 1 f ) by the order, the exponent of conver- gence of zeros and the exponent of convergence of poles of meromorphic function f(z), respectively, and τ(f) by the exponent of convergence of fixed points of f(z), which is defined as τ(f) = lim sup r→+∞ logN(r, 1 f(z)−z ) log r . In addition, let S(r, f) be any quantity satisfying S(r, f) = o(T (r, f)) for all r on a set F of logarithmic density 1, the logarithmic density of a set F is defined as lim sup r→∞ 1 log r ∫ [1,r]∩F 1 t dt. 4 H. Y. XU, J. TU EJDE-2020/14 Our main results in this paper are the following. Theorem 1.5. Let q(6= 0) ∈ C, |q| 6= 1, and µ(6= 0) ∈ C, and suppose that f(z) is a nonconstant rational solution of equation (1.8). Then f(z) can be represented in the form f(z) = a(zn + b)2 (zn + q−nb)(zn + qnb) , and a = q2n + qn + 1 (qn + 1)2 , µ = a3(a− 1) = −q n(q2n + qn + 1)3 (qn + 1)8 , where b is an any nonzero constant and n ∈ N+; Example 1.6. Let f(z) = 7(z + 1)2 9(2z + 1)( z 2 + 1) , then f(z) satisfies the equation f(2z)f( z 2 )f(z)(f(z)− 1) = −2 73 94 . This example shows that our conclusion about the form of rational solutions for equation (1.8) is sharp. Theorem 1.7. Let q ∈ C − {0, 1} and µ(6= 0) ∈ C, and suppose that f(z) is a transcendental meromorphic solution with zero order of equation (1.8). Then (i) f(ηz) has infinitely many fixed-points and τ(f(ηz)) = σ(f) for any η ∈ C− {0, 1}; (ii) f(z) has infinitely many zeros and poles, and ∆qf, (∆qf)/f have infinitely many poles, and λ(f) = λ ( 1 f ) = λ ( 1 ∆qf ) = λ ( 1 (∆qf)/f ) . Theorem 1.8. Let q, λ ∈ C − {0, 1} and |q| 6= 1. If (1.9) has a nonconstant rational solution f(z) = R(z) = P (z) Q(z) = apz p + ap−1z p−1 + · · ·+ a1z + a0 btzt + bt−1zt−1 + · · ·+ b1z + b0 , then p = t and λ = a2, where a = R(∞) = ap/bp. Example 1.9. Let a = 1/9, λ = 1/81 and f(z) = 1 9 (z + 1 z − 1 )2 , then f(z) satisfies the difference equation f(2z)f( z 2 )[f(z)− 1]2 = [ f(z)− 1 81 ]2 . This example shows that our conclusion about the form of rational solutions for equation (1.9) is sharp to a certain extent. Theorem 1.10. Let q, λ ∈ C − {0, 1}. Suppose that f(z) is a nonconstant mero- morphic solution with zero order of equation (1.9). Then (i) f(ηz) has infinitely many fixed-points and τ(f(ηz)) = σ(f) for any η ∈ C− {0, 1}; EJDE-2020/14 q-DIFFERENCE PAINLEVÉ EQUATIONS 5 (ii) f(z) has infinitely many zeros and poles, and ∆qf, ∆qf f have infinitely many poles, and λ(f) = λ ( 1 f ) = λ ( 1 ∆qf ) = λ ( 1 (∆qf)/f ) . 2. Proof of Theorem 1.5 Proof. Let f(z) = P (z)/Q(z) be a nonconstant rational solution of (1.8), where P (z), Q(z) are relatively prime polynomials with degrees p and t respectively. In view of (1.8), it follows that P (qz) Q(qz) P ( z q ) Q( z q ) P (z) Q(z) P (z)−Q(z) Q(z) = µ. (2.1) Without loss of generality, we assume that the coefficients of the highest degree terms of P (z) and Q(z) are a(6= 0) and 1 respectively, and set s = p− t. If s > 0, then P (z)/Q(z) = azs(1 + o(1)) as |z| = r → ∞. Thus, by virtue of (2.1), it follows that a3z3s(1 + o(1))(azs(1 + o(1))− 1) = µ, r →∞, this is impossible for a 6= 0. If s < 0, then as r →∞, it follows that P (z) Q(z) = o(1) and P (qz) Q(qz) = o(1), P ( z q ) Q( z q ) = o(1). Substituting these into (2.1), we get o(1) = µ as r →∞, this is a contradiction for µ 6= 0. Thus, it yields that s = 0 and p = t. From the assumptions of this theorem, we know that the zeros of Q(z) are not the zeros of P (z) and P (z)−Q(z). Hence, in view of (2.2), it follows that all the zeros of Q2(z) are the zeros of P (qz)P ( z q ). Since degz[Q(z)2] = degz[P (qz)P ( z q )] = 2p, then it yields from (2.1) that P (qz)P ( z q ) = a2Q(z)2, (2.2) P (z)(P (z)−Q(z)) = a(a− 1)Q(qz)Q( z q ). (2.3) Next, we confirm that the orders of all the zeros of P (z) are even. Let z0 be a zero of P (z) with the order k. If z0 6= 0 and k is an odd integer. Then P (z) has the term (z − z0)k, and P (qz)P ( z q ) has the term (z − qz0)k ( z − z0 q )k . (2.4) It means that qz0 and z0 q are both zeros of P (qz)P ( z q ) with the order at least k. In addition, since P (z) and Q(z) are relatively prime polynomials, in view of (2.3), it follows that Q(qz)Q( z q ) has the term (z − z0)k. Suppose that Q(qz) and Q( z q ) have the terms (z − z0)m and (z − z0)l respectively, where m, l ∈ N and m+ l = k. Obviously, in view of (2.4), we have m 6= 0 and l 6= 0. Thus, Q(z) has the term (z− qz0)m(z− z0 q )l, that is, Q(z)2 has the term (z− qz0)2m(z− z0 q )2l. So, in view of (2.3), it follows that P (qz)P ( z q ) has the term (z − qz0)2m(z − z0 q )2l. 6 H. Y. XU, J. TU EJDE-2020/14 In view of m+ l = k and k is an odd integer, without loss of generality, assume that m < l. Thus, 2m < k and 2l > k. Thus, qz0 is a zero of P (qz)P ( z q ) with the order 2m < k, this is a contradiction with (2.4). Thus, any nonzero zeros of P (z) have even orders. If 0 is a zero of P (z), by combining with (2.2), then 0 is also a zero of Q(z), this is a contradiction with P (z), Q(z) being relatively prime polynomials. Therefore, all the zeros of P (z) are nonzero with even orders. Let P (z) = ar(z)2, where r(z) = zn +An−1z n−1 +An−2z n−2 + · · ·+A1z +A0, and A0, A1, . . . , An−1 are constants. Since 0 is not the zero of P (z), then A0 6= 0. In view of (2.2) and (2.3), it yields Q(z) = r(qz)r( z q ) and ar(z)2 − r(qz)r( z q ) = (a− 1)r(q2z)r(q−2z). Denote ϕ(z) = ar(z)2 − r(qz)r(z q )− (a− 1)r(q2z)r(q−2z). Thus, ϕ(z) ≡ 0. Substituting r(z) into ϕ(z), then we give the coefficients of term z2n−1, z2n−2, z2n−3, z2n−4, . . . , zn+1 as follows B2n−1 = −An−1[a(q + q−1 + 2)− (q + q−1 + 1)](q + q−1 − 2), (2.5) B2n−2 = −An−2[a(q2 + q−2 + 2)− (q2 + q−2 + 1)](q2 + q−2 − 2), (2.6) B2n−3 = −An−2[a(q2 + q−2 + 2)− (q2 + q−2 + 1)](q2 + q−2 − 2) +An−1An−2[a(q + q−1 + 2)− (q + q−1 + 1)](q + q−1 − 2), (2.7) B2n−4 = −An−4 [ a(q4 + q−4 + 2)− (q4 + q−4 + 1)](q2 + q−2 − 2) +An−1An−3[a(q2 + q−2 + 2)− (q2 + q−2 + 1) ] (q2 + q−2 − 2), (2.8) . . . B2n−i =−An−i[a(qi + q−i + 2)− (qi + q−i + 1)](qi + q−i − 2) +An−1An−i+1[a(qi−2 + q−(i−2) + 2)− (qi−2 + q−(i−2) + 1)] × (qi−2 + q−(i−2) − 2) + · · ·+An−[ i 2 ]An−[ i 2 ]+1 [ a(q2 + q−2 + 2) − (q2 + q−2 + 1) ] (q2 + q−2 − 2), (2.9) . . . Bn+1 = −A1[a(qn−1 + q−(n−1) + 2)− (qn−1 + q−(n−1) + 1)] × (qn−1 + q−(n−1) − 2) +An−1A2[a(qn−3 + q−(n−3) + 2) − (qn−3 + q−(n−3) + 1)](qn−3 + q−(n−3) − 2) + . . . . (2.10) Note that if i is an even integer in (2.9), then An−[ i 2 ]An−[ i 2 ]+1 should be replaced by An− i 2−1An− i 2 +1. In view of (2.5)-(2.10), we conclude that there are at most one of A1, A2, . . . , An−1 can be equal to 0. Otherwise, if there exist two integers i, j ∈ N+ such that i 6= j, Aj 6= 0, Ai 6= 0 and At = 0 for t = 1, 2, . . . , n−1, t 6= i, t 6= j. From (2.5)-(2.10), we have a(qi + q−i + 2)− (qi + q−i + 1) ≡ 0, a(qj + q−j + 2)− (qj + q−j + 1) ≡ 0. EJDE-2020/14 q-DIFFERENCE PAINLEVÉ EQUATIONS 7 This is impossible as |q| 6= 1. Thus, without loss of generality, we assume that An−1 6= 0 and Ai = 0 for j = 1, 2, . . . , n − 2; j 6= n − 1. Then, in view of (2.5), it follows that a = q2 + q + 1 (q + 1)2 . (2.11) Thus, r(z), P (z) can be represented in the form r(z) = zn +An−1z n−1 +A0, P (z) = a[zn +An−1z n−1 +A0]2, (2.12) where n 6= 1. It leads to Q(z) = (qnzn +An−1q n−1zn−1 +A0)(q−nzn +An−1q −n+1zn−1 +A0). (2.13) Substituting (2.12) and (2.13) into ϕ(z), and analyzing the coefficients of the term zn, we have Bn = −A0[a(qn + q−n + 2)− (qn + q−n + 1)](qn + q−n − 2). (2.14) In view of (2.11), |q| 6= 1 and n 6= 1, it follows that a(qn + q−n + 2)− (qn + q−n + 1) 6= 0. Thus, by combining with A0 6= 0, this is a contradiction with ϕ(z) ≡ 0. Therefore, A1 = A2 = · · · = An−1 ≡ 0; that is, r(z) = zn + b, P (z) = a(zn + b)2, Q(z) = (qnzn + b)(q−nzn + b), (2.15) where b is an any nonzero constant. Substituting (2.15) into ϕ(z), we have a = q2n + qn + 1 (qn + 1)2 . (2.16) Thus, substituting (2.15) and (2.16) into (2.1), we obtain µ = a3(a− 1) = −q n(q2n + qn + 1)3 (qn + 1)8 . This completes the proof of Theorem 1.5. � 3. Proof of Theorem 1.7 The following lemmas are necessary. Lemma 3.1 ([1, Theorem 2.5]). Let f be a nonconstant zero-order meromorphic solution of Pq(z, f) = 0, where Pq(z, f) is a q-difference polynomial in f(z). If Pq(z, a) 6≡ 0 for slowly moving target a(z), then m ( r, 1 f − a ) = S(r, f). Lemma 3.2 ([29, Theorem 1.1 and 1.3]). Let f(z) be a nonconstant zero-order meromorphic function and q ∈ C \ {0}. Then T (r, f(qz)) = (1 + o(1))T (r, f(z)), N(r, f(qz)) = (1 + o(1))N(r, f(z)), on a set of lower logarithmic density 1. 8 H. Y. XU, J. TU EJDE-2020/14 Lemma 3.3 ([15, Theorem 2.5]). Let f be a transcendental meromorphic solution of order zero of a q-difference equation of the form Uq(z, f)Pq(z, f) = Qq(z, f), where Uq(z, f), Pq(z, f) and Qq(z, f) are q-difference polynomials such that the total degree deg Uq(z, f) = n in f(z) and its q-shifts, whereas degQq(z, f) ≤ n. More- over, we assume that Uq(z, f) contains just one term of maximal total degree in f(z) and its q-shifts. Then m(r, Pq(z, f)) = S(r, f). Remark 3.4. For q ∈ C\{0, 1}, a polynomial in f(z) and finitely many of its q-shifts f(qz), . . ., f(qnz) with meromorphic coefficients in the sense that their Nevanlinna characteristic functions are o(T (r, f)) on a set F of logarithmic density 1, can be called as a q-difference polynomial of f . Lemma 3.5 (Valiron-Mohon’ko [28]). Let f(z) be a meromorphic function. Then for all irreducible rational functions in f , R(z, f(z)) = ∑m i=0 ai(z)f(z)i∑n j=0 bj(z)f(z)j , with meromorphic coefficients ai(z), bj(z), the characteristic function of R(z, f(z)) satisfies T (r,R(z, f(z))) = dT (r, f) +O(Ψ(r)), where d = max{m,n} and Ψ(r) = maxi,j{T (r, ai), T (r, bj)}. Lemma 3.6 ([1, Theorem 1.1]). Let f(z) be a nonconstant zero order meromorphic function and q ∈ C \ {0}. Then m ( r, f(qz) f(z) ) = S(r, f). Proof of Theorem 1.7. (i) Let f(z) be a transcendental meromorphic function of zero order. For any η ∈ C− {0, 1}, substituting ηz into (1.8), we have f(qηz)f( ηz q )f(ηz)(f(ηz)− 1) = µ. (3.1) Denoting g(z) = f(ηz), equation (3.1) can be represented as g(qz)g( z q )g(z)(g(z)− 1) = µ. Let P1(z, g) := g(qz)g (z q ) g(z)(g(z)− 1)− µ = 0. It follows that P1(z, z) = z3(z − 1)− µ 6≡ 0. In view of P1(z, z) 6≡ 0, by Lemma 3.1 we have m ( r, 1 g(z)− z ) = S(r, g). Since f is of zero order, from Lemma 3.2, it follows that N ( r, 1 f(ηz)− z ) = N ( r, 1 g(z)− z ) = T (r, g) + S(r, g) = T (r, f(ηz)) + S(r, f(ηz)) = T (r, f) + S(r, f). EJDE-2020/14 q-DIFFERENCE PAINLEVÉ EQUATIONS 9 Therefore, f(ηz) has infinitely many fixed points, and τ(f(ηz)) = σ(f) for any η ∈ C− {0, 1}. (ii) Since f is a transcendental meromorphic solution of zero order. In view of µ 6= 0, by Lemmas 3.2-3.6, 4T (r, f) = T ( r, µ f3(f − 1) ) +O(1) = T ( r, f(qz)f( z q ) f(z)2 ) +O(1) ≤ T ( r, f(qz) f(z) ) + T ( r, f( z q ) f(z) ) +O(1) ≤ 2T ( r, f(qz) f(z) ) + S(r, f) = 2T ( r, ∆qf f ) + S(r, f); that is, 2T (r, f) ≤ T ( r, ∆qf f ) + S(r, f). (3.2) Thus, from Lemma 3.6 and (3.2), we conclude that N ( r, ∆qf f ) = T ( r, ∆qf f ) −m ( r, ∆qf f ) ≥ 2T (r, f) + S(r, f). This means that ∆qf f has infinitely many poles, and λ ( 1 ∆qf f ) = σ(f). Also, we can rewrite equation (1.8) as f(qz)f( z q ) = (∆qf + f)(∆q−1f + f) = µ f(f − 1) ; that is, ∆qf∆q−1f + (∆qf + ∆q−1f)f = µ− f4 + f3 f(f − 1) . (3.3) Thus, in view of Lemmas 3.2 and 3.5, it follows that 4T (r, f) = T ( r, µ− f4 + f3 f(f − 1) ) +O(1) = T (r,∆qf∆q−1f) + (∆qf + ∆q−1f)f) +O(1) ≤ T (r, f) + 2T (r,∆qf) + 2T (r,∆q−1f) +O(1) ≤ T (r, f) + 4T (r,∆qf) + S(r, f); that is, 3 4 T (r, f) ≤ T (r,∆qf) + S(r, f). (3.4) On the other hand, (1.8) can be represented as f(qz)f( z q )f(z)2 = µ+ f(qz)f( z q )f(z). By Lemma 3.3, we obtain m(r, f) = S(r, f). Thus, we can conclude from Lemma 3.6 that N(r,∆qf) = T (r,∆qf)−m(r,∆qf) ≥ T (r,∆qf)− [ m(r, f) +m ( r, ∆qf f )] 10 H. Y. XU, J. TU EJDE-2020/14 ≥ 1 2 T (r, f) + S(r, f), which implies that ∆qf has infinitely many poles and λ( 1 ∆qf ) = σ(f). Since m ( r, 1 f ) = m ( r, f(qz)f( z q )(f − 1) µ ) = m ( r, f(qz)f( z q ) f2 f2(f − 1) µ ) , by combining with m(r, f) = S(r, f) and Lemma 3.6, it follows that m ( r, 1 f ) = S(r, f), which yields N ( r, 1 f ) = T (r, f) + S(r, f), N(r, f) = T (r, f) + S(r, f), which implies that f has infinitely many poles and zeros, and λ(f) = λ( 1 f ) = σ(f). This completes the proof � 4. Proof of Theorem 1.8 Suppose that f(z) = P (z)/Q(z) is a nonconstant rational solution of equation (2.2), where P (z), Q(z) are relatively prime polynomials with degz P (z) = p and degz Q(z) = t. Substituting this into (1.9), we have P (qz) Q(qz) P ( z q ) Q( z q ) (P (z) Q(z) − 1 )2 = (P (z) Q(z) − λ )2 . (4.1) By using the same argument as in the proof of Theorem 1.5 (i), we obtain p = t. Suppose that λ 6= a2. Without loss of generality we assume that the coefficients of the highest degree terms of P (z), Q(z) are a and 1, respectively. In view of (1.9), letting |z| = r → +∞ yields a2(a− 1)2 = (a− λ)2. (4.2) Since λ 6= 0, 1, then a 6= 0, 1, λ. Now, we rewrite (1.9) in the form P (qz) Q(qz) P ( z q ) Q( z q ) = (P (z)− λQ(z) P (z)−Q(z) )2 . Since degz[P (z)− λQ(z)] = degz[P (z)−Q(z)] = p, it follows that (a− λ)2P (qz)P (z q ) = a2(P (z)− λQ(z))2, (4.3) (a− 1)2Q(qz)Q (z q ) = (P (z)−Q(z))2. (4.4) In view of (4.3) and (4.4), it is easy to see that 0 is not the zero of P (z), Q(z). Otherwise, if 0 is a zero of P (z), from (4.3), we can get that 0 is also a zero of Q(z), this is a contradiction with the hypothesis of P (z), Q(z) being relatively prime polynomials; if 0 is a zero of Q(z), from (4.4), we can also get a contradiction. Now, suppose that z0(6= 0) is a zero of P (z) with order k, and k is an odd integer. Then z0/q is a zero of P (qz) with order k. However, in view of (4.3), it yields that the orders of the zeros of P (qz)P (z/q) are all even integers, thus, z0/q must be a zero of P (z/q) with order l, and l is an odd integer. Hence, z0/q 2 is a zero of P (z) with the odd order l. Thus, continue this process, we obtain that z0/q m are the zeros of P (z) for any integer m. This is impossible as degz P (z) = p and |q| 6= 1. EJDE-2020/14 q-DIFFERENCE PAINLEVÉ EQUATIONS 11 Therefore, all the zeros of P (z) have even orders. Similarly, all the zeros of Q(z) have even orders. Thus, set P (z) = aα(z)2 and Q(z) = β(z)2, where α(z) = zn +An−1z n−1 + · · ·+A1z +A0, (4.5) β(z) = zn +Bn−1z n−1 + · · ·+B1z +B0, (4.6) and A0, A1, . . . , An−1, B0, B1, . . . , Bn−1 are constants. Obviously, A0, B0 can not be equal to 0 simultaneously. Then, in view of (4.3) and (4.4), we have (a− λ)α(qz)α (z q ) = aα(z)2 − λβ(z)2, (4.7) (a− 1)β(qz)β (z q ) = aα(z)2 − β(z)2. (4.8) Substituting (4.5) and (4.6) into the above equations, and analyzing the coeffi- cients of terms z2n−1, z2n−2, . . ., we can deduce that (a− λ)(q + q−1)An−1 = 2aAn−1 − 2λBn−1, (4.9) (a− 1)(q + q−1)Bn−1 = 2aAn−1 − 2Bn−1, (4.10) (a− λ)[(q2 + q−2)An−2 +A2 n−1] = a(2An−2 +A2 n−1)− λ(2Bn−2 +B2 n−1), (4.11) (a− 1)[(q2 + q−2)Bn−2 +B2 n−1] = a(2An−2 +A2 n−1)− (2Bn−2 +B2 n−1), (4.12) . . . , (a− λ)[(qi + q−i)An−i + (qi−2 + q−(i−2))An−1An−i+1 + . . . + (q + q−1)An− i−1 2 An− i+1 2 ] = a(2An−i + 2An−1An−i+1 + · · ·+ 2An− i−1 2 An− i+1 2 ) − λ(2Bn−i + 2Bn−1Bn−i+1 + · · ·+ 2Bn− i−1 2 Bn− i+1 2 ), (4.13) (a− 1)[(qi + q−i)Bn−i + (qi−2 + q−(i−2))Bn−1Bn−i+1 + . . . + (q + q−1)Bn− i−1 2 Bn− i+1 2 ] = a(2An−i + 2An−1An−i+1 + · · ·+ 2An− i−1 2 An− i+1 2 ) − (2Bn−i + 2Bn−1Bn−i+1 + · · ·+ 2Bn− i−1 2 Bn− i+1 2 ), (4.14) . . . , (a− λ)[(qj + q−j)An−j + (qj−2 + q−(j−2))An−1An−j+1 + · · ·+A2 n− j 2 ] = a(2An−j + 2An−1An−j+1 + · · ·+A2 n− j 2 ) − λ(2Bn−1Bn−j+1 + · · ·+ 2B2 n− j 2 ), (4.15) (a− 1)[(qj + q−j)Bn−j + (qj−2 + q−(j−2))Bn−1Bn−j+1 + · · ·+B2 n− j 2 ] = a(2An−j + 2An−1An−j+1 + · · ·+A2 n− j 2 ) − (2Bn−1Bn−j+1 + · · ·+ 2B2 n− j 2 ), (4.16) . . . , where i is an odd integer, and j is an even integer. Assume that there exist a positive integer i ∈ N+, i < n satisfying An−i 6= 0 and An−j = 0 for any non-negative integer j < i. Without loss of generality, we let 12 H. Y. XU, J. TU EJDE-2020/14 i = 1, that is, An−1 6= 0, thus, Bn−1 6= 0. Otherwise, if Bn−1 = 0, then by (4.10), it follows An−1 = 0, a contradiction. In view of (4.9)-(4.10), it yields (An−1 −Bn−1)[(a− λ)aAn−1 − λ(a− 1)Bn−1] = 0, (4.17) which leads to either An−1 = Bn−1 or (a − λ)aAn−1 − λ(a − 1)Bn−1 = 0. If An−1 = Bn−1, we can deduce from (4.10) that q1 + q−1 = 2, which implies a contradiction with |q| 6= 1. Thus, it yields that An−1 6= Bn−1 and (a− λ)aAn−1 = λ(a− 1)Bn−1. (4.18) Further, suppose that An−2 6= 0. Then Bn−2 6= 0. Indeed, if An−2 = 0, then from (4.11) and (4.12), we have Bn−2 = 0. Similarly, if Bn−2 = 0, then An−2 = 0. In view of (4.11) and (4.12), we have [2(An−2 −Bn−2) + (A2 n−1 −B2 n−1)][(a− λ)aAn−2 − λ(a− 1)Bn−2] = 0. (4.19) From (4.19), either 2(An−2−Bn−2) + (A2 n−1−B2 n−1) = 0 or (a−λ)aAn−2−λ(a− 1)Bn−2 = 0. If 2(An−2 − Bn−2) + (A2 n−1 − B2 n−1) = 0, we can deduce from (4.12) that q2 + q−2 = 2, which implies a contradiction with |q| 6= 1. Therefore (a− λ)aAn−2 = λ(a− 1)Bn−2. (4.20) It follows from (4.18) and (4.20) that (a− λ)2a2 = λ2(a− 1)2. Combining this with (4.2) yields λ = a2, a contradiction. Hence, An−2 = 0 and Bn−2 = 0. As in the above argument, it follows that An−3 = · · · = A1 = 0 and Bn−3 = · · · = B1 = 0. Thus, An−3 = · · · = A1 = 0 and Bn−3 = · · · = B1 = 0. Since 0 is not the zero of P (z), Q(z), it follows that A0 6= 0 and B0 6= 0. By analyzing the coefficients of the term zn, we deduce that (a− λ)aA0 = λ(a− 1)B0. (4.21) Thus, in view of (4.18),(4.21) and (4.2), it yields λ = a2, a contradiction. Hence, we conclude A1 = A2 = · · · = An−1 = 0 and B1 = B2 = · · · = Bn−1 = 0. In view of (4.9)-(4.16), it is easy to deduce that B1 = B2 = · · · = Bn−1 = 0. Thus, α(z) = zn +A0, β(z) = zn +B0. (4.22) Hence, substituting α, β into (4.7) and (4.8), by comparing the coefficients of the terms zn and constant, we have (a− λ)(qn + q−n)A0 = 2aA0 − 2λB0, (a− 1)(qn + q−n)B0 = 2aA0 − 2B0, (a− λ)A2 0 = aA2 0 − λB2 0 , (a− 1)B2 0 = aA2 0 −B2 0 . Then it follows that A0 = B0 or A0 = −B0. If A0 = B0, then qn + q−n = 2 is a contradiction. If A0 = −B0, then λ(a − 1)B0 − a(a − λ)A0 = 0. Thus, with a view of A0 6= 0, this yields λ = a2, a contradiction. This completes the proof of Theorem 1.8. For the proof of Theorem 1.10 we use the same argument as in the proof of Theorem 1.7 and the conclusion follows easily. EJDE-2020/14 q-DIFFERENCE PAINLEVÉ EQUATIONS 13 Acknowledgments. 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Chen; On properties of q-difference equations, Acta Math. Sci., 32B (2) (2012), 724-734. [32] X. M. Zheng, Z. X. Chen; Some properties of meromorphic solutions of q-difference equations, J. Math. Anal. Appl., 361 (2010), 472-480. [33] J. F. Zhong, H. F. Liu; The growth of meromorphic solutions of some type of nonlinear difference equations, J. Jiangxi Normal University (Natural Sciences), 43 (2019), 508-512. Hong Yan Xu School of Mathematics and Computer Science, Shangrao Normal University, Shangrao Jiangxi 334001, China Email address: xhyhhh@126.com Jin Tu Department of Mathematics, Jiangxi Normal University, Nanchan, Jiangxi 330022, China Email address: tujin2008@sina.com 1. Introduction and statement of main results 2. Proof of Theorem ?? 3. Proof of Theorem ?? 4. Proof of Theorem ?? Acknowledgments References