Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 26, pp. 1–21. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.26 ENTIRE SOLUTIONS TO FERMAT-TYPE DIFFERENCE AND PARTIAL DIFFERENTIAL-DIFFERENCE EQUATIONS IN Cn HONG YAN XU, GOUTAM HALDAR Abstract. In this article, we study the existence and the form of finite order transcendental entire solutions of systems of Fermat-type difference and par- tial differential-difference equations in several complex variables. Our results extend previous theorems given by Xu-Cao [49], Xu et al [52], and Zheng-Xu [55]. We give some examples to illustrate the content of this article. 1. Introduction and main results It is well known that Nevanlinna theory is an important tool to study value dis- tribution of entire and meromorphic solutions of complex differential equations (see [15, 21]). Initially, Fermat-type functional equations were investigated by Montel [35], Gross [6, 7], and Iyer [18], independently. In fact, Iyer [18] considered the Fermat type functional equation f(z)2 + g(z)2 = 1, (1.1) and proved that the entire solutions of (1.1) are of the form f(z) = cosα(z), g(z) = sinα(z), where α(z) is entire function. Many researchers pay considerable attention to study the existence of entire and meromorphic solutions of complex difference as well as complex differential- difference equations, and obtained a number of important results; see [29, 30, 32, 42]. Mainly utilizing difference analogues of Nevanlinna theory, which was developed by Halburd and Korhonen [8, 9], and Chiang and Feng [3], independently. In 2012, Liu et al. [30] proved that the Fermat-type difference equation f(z)2 + f(z+c)2 = 1 has the solutions of the form f(z) = sin(az+b), where c( ̸= 0), a, b ∈ C, a = (4k+1)π/2c, and k is an integer. In 2013, Liu and Yang [27] extended this result by considering the Fermat-type difference equation f(z)2 + P (z)2f(z + c)2 = Q(z) where P (z) and Q(z) are two non-zero polynomials. After that Liu [28], Liu and Dong [31] considered some variations of Fermat-type equations f(z)2 + [f(z + c)− f(z)]2 = 1, (1.2) a21f(z) 2 + [a2f(z + c) + a3f(z)] 2 = 1, (1.3) 2020 Mathematics Subject Classification. 39A45, 32H30, 30D35. Key words and phrases. Entire solutions; Fermat-type; partial differential equations; Nevanlinna theory. ©2024. This work is licensed under a CC BY 4.0 license. Submitted January 6, 2024. Published March 25, 2024. 1 2 H. Y. XU, G. HALDAR EJDE-2024/26 [a1f(z + c) + a2f(z)] 2 + [a3f(z + c) + a4f(z)] 2 = 1, (1.4) and obtained some remarkable results. Hereafter, we denote z + w = (z1 + w1, z2 + w2, . . . , zn + wn) for any z = (z1, z2, . . . , zn), w = (w1, w2, . . . , wn) and c = (c1, c2, . . . , cn), where z, w, c ∈ Cn except otherwise stated. Considering equations (1.2)–(1.4), in 2021, Zheng and Xu [55] extended the results due to Liu [28], Liu and Dong [31] from one complex variable to several complex variables, and obtained the following results. Theorem 1.1. [55] Let c = (c1, c2) ∈ C2 \ {(0, 0)}. Then there are no transcen- dental entire solutions f : C2 → P1(C) with finite order for equation (1.2). Theorem 1.2. [55] Let c = (c1, c2) ∈ C2 \ {(0, 0)} and a1, a2, a3 be nonzero con- stants in C. If (1.3) has a transcendental entire solution f : C2 → P1(C) with finite order, then a21 + a23 = a22 and f(z) = 1 a1 sin(L(z) + Φ(t) +A), where L(z) = α1z1 + α2z2, α1, α2, A ∈ C, Φ(t) is a polynomial in t := c2z1 − c1z2 in C, and L(z) satisfies L(c) = α1c1 + α2c2 = θ + kπ ± π 2 , tan θ = a3 a1 . Theorem 1.3. [55] Let c = (c1, c2) ∈ C2 \ {(0, 0)}, a1, a2, a3, a4 be nonzero con- stants in C, and let D := a1a4 − a2a3 ̸=0. If (1.4) has a transcendental entire solution f : C2 → P1(C) with finite order, then a21 + a23 = a22 + a24 and f(z) = 1 2D [ − (a3 + ia1)e L(z)+Φ(t)+A − (a3 − ia1)e −(L(z)+Φ(t)+A) ] , where L(z) = α1z1 + α2z2, α1, α2, A ∈ C, Φ(t) is a polynomial in t := c2z1 − c1z2 in C, and L(z) satisfies eL(c) = eα1c1+α2c2 = −a3 − ia1 a4 − ia2 = −a4 + ia2 a3 + ia1 . As far as our knowledge is concerned, although there are some remarkable results about the existence and forms of transcendental entire solutions of Fermat-type difference and partial differential-difference equations in several complex variables (see [17, 49, 50, 51, 55, 11, 12, 14, 45, 13, 47, 48, 46]), the number of results about the solutions of the system of Fermat-type equations in the literature (see [52]) are scanty. We would like to discuss some of these results which are relevant to the content of this article. Theorem 1.4. [52] Let c = (c1, c2) be constants in C2. Then any pair of tran- scendental entire solutions with finite order for the system of Fermat-type difference equations f1(z1, z2) 2 + (f2(z1 + c1, z2 + c2)) 2 = 1 f2(z1, z2) 2 + (f1(z1 + c1, z2 + c2)) 2 = 1, has the following form (f1(z), f2(z)) = (eL(z)+B1 + e−(L(z)+B1) 2 , A21e L(z)+B1 +A22e −(L(z)+B1) 2 ) , EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 3 where L(z) = α1z1 +α2z2, B1 is a constant in C, and c, A21, A22 satisfy one of the following cases (i) L(c) = 2kπi, A21 = −i and A22 = i, or L(c) = (2k + 1)πi, A21 = i and A22 = −i, here and below k is an integer; (ii) L(c) = (2k + 1/2)πi, A21 = −1 and A22 = −1, or L(c) = (2k − 1/2)πi, A21 = 1 and A22 = 1. Now, we consider the following systems Fermat-type functional equations on Cn. f1(z1, . . . , zn) 2 + (∆cf2(z1, . . . , zn)) 2 = 1 f2(z1, . . . , zn) 2 + (∆cf1(z1, . . . , zn)) 2 = 1, (1.5) where c = (c1, c2, . . . , cn) are constant in Cn. a21f1(z) 2 + (a2f2(z + c) + a3f2(z)) 2 = 1 a21f2(z) 2 + (a2f1(z + c) + a3f1(z)) 2 = 1, (1.6) (a1f1(z + c) + a2f1(z)) 2 + (a3f2(z + c) + a4f2(z)) 2 = 1 (a1f2(z + c) + a2f2(z)) 2 + (a3f1(z + c) + a4f1(z)) 2 = 1, (1.7) where fj : Cn → P1(C), j = 1, 2, c = (c1, c2, . . . , cn) are constants in Cn \ {0}, a1, a2, a3, a4 are nonzero constants in C, and ∆cf(z) = f(z1 + c1, . . . zn + cn) − f(z1, . . . zn) as defined in [20]. Inspired by Theorems 1.1–1.4, one may ask the following questions. What can be said about the existence and the forms of finite order transcendental entire solutions for the system of the Fermat-type functional equations (1.5)–(1.7)? Can we extend all the results stated above from C2 to Cn? Our main goal is to investigate the existence and form of finite order transcen- dental entire solutions of system (1.5)–(1.7) with the help of Nevanlinna theory andthe difference logarithmic lemma in several complex variables (see [2, 20]). We extend Theorems 1.1–1.4 from the complex Fermat-type difference equations to the Fermat-type system of difference equations. ow we list our main results. Theorem 1.5. There is no pair of transcendental entire solutions with finite order for the system of Fermat-type difference equation (1.5). Theorem 1.6. Let a1, a2, a3 be three non-zero complex constants in one variable and c = (c1, . . . cn) ∈ Cn \ {(0, 0, . . . , 0)}. If (f1, f2) is a pair of transcendental entire solution with finite order of simultaneous Fermat-type difference equation (1.6), then (f1, f2) takes one of the following forms (i) (f1, f2) = ( 1 a1 cos(L(z) + Φ(z) + A), 1 a1 cos(L(z) + Φ(z) + A + k) ) , where a22 = a21 + a23, e 2ik = 1, e2iL(c) = −a1 − ia3e −ik a1 + ia3eik , where L(z) = ∑n j=1 αjzj, αj , A ∈ C, j = 1, 2, . . . , n, and Φ(z) = n∑ i1,i2=1(i1 n and (ii) n > m ≥ 2, then (2.2) does not have any finite order transcendental entire solutions. In 2020, Xu et al. [52] considered the system of partial differential-difference equations and obtained the following result. Theorem 2.1. [52] Let c = (c1, c2) be a constant in C2, and mj , nj (j = 1, 2) be positive integers. If the system of Fermat-type partial differential-difference equa- tions (∂f1(z1, z2) ∂z1 )n1 + f2(z1 + c1, z2 + c2) m1 = 1(∂f2(z1, z2) ∂z1 )n2 + f1(z1 + c1, z2 + c2) m2 = 1, satisfies one of the conditions (i) m1m2 > n1n2 or (ii) mj > nj nj−1 , j = 1, 2, then the above system does not have any pair of transcendental entire solution with finite order. As far as we know, the Fermat-type mixed partial differential-difference equations in several complex variables has not been addressed in the literature before. To generalize Theorem 1.5, we consider the partial differential-difference equation (a∂If1(z1, z2) + b∂Jf1(z1, z2)) n1 + f2(z1 + c1, z2 + c2) m1 = 1, (a∂If2(z1, z2) + b∂Jf2(z1, z2)) n2 + f1(z1 + c1, z2 + c2) m2 = 1, (2.3) where ∂Ifj(z1, z2) = ∂|I|fj(z1, z2) ∂zα1 1 ∂zα2 2 and ∂Jfj(z1, z2) = ∂|J|fj(z1, z2) ∂zβ1 1 ∂zβ2 2 with I = (α1, α2) and J = (β1, β2) are multi-index with I ̸= J , where α1, α2, β1, and β2 are non-negative integers and a, b ∈ C, not both zero. We denote by | I | to denote the length of I, that is, | I |= α1 + α2. Similarly, for J also. As a matter of fact, we prove the next result for any order Fermat-type partial differential-difference equation (2.3). Theorem 2.2. Let c = (c1, . . . cn) be a constant in Cn and mj, nj be positive integers with j = 1, 2. If the Fermat-type simultaneous partial differential-difference equation (2.3) satisfies one of the following conditions: (i) m1m2 > n1n2; (ii) mj > nj nj−1 , for nj ≥ 2, j = 1, 2, EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 7 then (2.3) does not have any pair of finite order transcendental entire solutions of the form (f1, f2). The following example shows the existence of finite order transcendental entire solutions of system (2.3) when n1 = n2 = 2 and m1 = m2 = 1. Example 2.3. Let us consider the following particular type of system of equation of (2.3). (∂2f1 ∂z21 )2 + f2(z + c) = 1 (∂2f2 ∂z21 )2 + f1(z + c) = 1. (2.4) Let c = (c1, c2) ∈ C2 be such that c1 = 0 and ec2 = 1 3 . Let f1(z1, z2) = f2(z1, z2) = − 1 144 z41 + 1 2 z21e z2 + 1− 9e2z2 . Then one can easily verify that (f1, f2) is a solution of (2.4). Example 2.4. Let c = (c1, c2) ∈ C2 be such that c1 = 0 and ec2 = − 1 3 . Let f1(z1, z2) = − 1 144 z41− 1 2 z21e z2−9e2z2+1, f2(z1, z2) = − 1 144 z41+ 1 2 z21e z2−9e2z2+1. Then one can easily verify that (f1, f2) is a solution of (2.4). 3. Proof of main results First, we present here some lemmas which play key role to prove the main results. Lemma 3.1 ([16]). Let fj ̸≡ 0 (j = 1, 2, . . . ,m; m ≥ 3) be meromorphic functions on Cn such that f1, . . . , fm−1 are not constant, f1+f2+ · · ·+fm = 1 and such that m∑ j=1 { Nn−1 ( r, 1 fj ) + (m− 1)N(r, fj) } < λT (r, fj) +O(log+ T (r, fj)) holds for j = 1, . . . ,m − 1 and all r outside possibly a set with finite logarithmic measure, where λ < 1 is a positive number. Then fm = 1. Lemma 3.2 ([22, 38, 41]). For an entire function F on Cn, F (0) ̸≡ 0 and put ρ(nF ) = ρ < ∞. Then there exist a canonical function fF and a function gF ∈ Cn such that F (z) = fF (z)e gF (z). For the special case n = 1, fF is the canonical product of Weierstrass. Lemma 3.3 ([36]). If g and h are entire functions on the complex plane C and g(h) is an entire function of finite order, then there are only two possible cases: either (i) the internal function h is a polynomial and the external function g is of finite order; or (ii) the internal function h is not a polynomial but a function of finite order, and the external function g is of zero order. 8 H. Y. XU, G. HALDAR EJDE-2024/26 Lemma 3.4 ([1, 54]). Let f be a non-constant meromorphic function on Cn and let I = (α1, . . . , αn) be a multi-index with length |I| = ∑n j=1 αj. Assume that T (r0, f) ≥ e for some r0. Then m ( r, ∂If f ) = S(r, f) for all r ≥ r0 outside a set E ⊂ (0,+∞) of finite logarithmic measure, ∫ E dt t < ∞, where ∂If = ∂|I|f ∂z α1 1 ...∂z α2 2 . Lemma 3.5 ([2, 20]). Let f be a non-constant meromorphic function with finite order on Cn such that f(0) ̸= 0,∞, and let ϵ > 0. Then for c ∈ Cn, m ( r, f(z + c) f(z) ) +m ( r, f(z) f(z + c) ) = S(r, f) for all r ≥ r0 outside a set E ⊂ (0,+∞) of finite logarithmic measure, ∫ E dt t < ∞. Lemma 3.6 ([16]). Let fj (̸≡ 0), j = 1, 2, 3 be meromorphic functions on Cm such that f1 is not constant. If f1 + f2 + f3 = 1, and if m∑ j=1 { N2 ( r, 1 fj ) + 2N(r, fj) } < λT (r, fj) +O(log+ T (r, fj)), for all r outside possibly a set with finite logarithmic measure, where λ < 1 is a positive number, then either f2 = 1 or f3 = 1. Proof of Theorem 1.5. Suppose that (f1, f2) is a pair of transcendental entire func- tions with finite order satisfying system (1.5). We write (1.5) as follows: (f1(z) + i(f2(z + c)− f2(z)))(f1(z)− i(f2(z + c)− f2(z))) = 1 (f2(z) + i(f1(z + c)− f1(z)))(f2(z)− i(f1(z + c)− f1(z))) = 1. (3.1) Since f1, f2 are transcendental entire functions with finite order, there exist poly- nomials p1(z), p2(z) in Cn such that f1(z) + i(f2(z + c)− f2(z)) = ep1(z) f1(z)− i(f2(z + c)− f2(z)) = e−p1(z) f2(z) + i(f1(z + c)− f1(z)) = ep2(z) f2(z)− i(f1(z + c)− f1(z)) = e−p2(z). (3.2) In view of (3.2), we obtain f1(z) = 1 2 ( ep1(z) + e−p1(z) ) f2(z + c)− f2(z) = 1 2i ( ep1(z) − e−p1(z) ) f2(z) = 1 2 ( ep2(z) + e−p2(z) ) f1(z + c)− f1(z) = 1 2i ( ep2(z) − e−p2(z) ) . (3.3) After simple calculations, it follows from (3.3) that −iep2(z)+p1(z+c) − iep2(z)−p1(z+c) + iep2(z)+p1(z) + iep2(z)−p1(z) + e2p2(z) = 1 (3.4) EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 9 and −iep1(z)+p2(z+c) − iep1(z)−p2(z+c) + iep1(z)+p2(z) + iep1(z)−p2(z) + e2p1(z) = 1. (3.5) Now, we consider the following two possible cases. Case 1: Let p2(z) − p1(z) = k, where k is a constant in C. Then it follows from (3.4) and (3.5) that −iep1(z)+p1(z+c)+k − iep1(z)−p1(z+c)+k + ie2p1(z)+k + e2p1(z)+2k = 1− iek −iep1(z)+p1(z+c)+k − iep1(z)−p1(z+c)−k + ie2p1(z)+k + e2p1(z) = 1− ie−k. (3.6) First we consider that ek ̸= ±i. Then, we obtain from (3.6) that −iek 1− iek ep1(z)+p1(z+c) + −iek 1− iek ep1(z)−p1(z+c) + (i+ ek)ek 1− iek e2p1(z) = 1 −iek 1− ie−k ep1(z)+p1(z+c) + −ie−k 1− ie−k ep1(z)−p1(z+c) + 1 + iek 1− ie−k e2p1(z) = 1. (3.7) By Lemma 3.1, we obtain from (3.7) that −iek 1− iek ep1(z)−p1(z+c) = 1, −ie−k 1− ie−k ep1(z)−p1(z+c) = 1. (3.8) It follows from (3.7) and (3.8) that e−p1(z)+p1(z+c) = 1− iek e−p1(z)+p1(z+c) = 1− ie−k. (3.9) It follows from (3.8) and (3.9) that −ie−k = (1− ie−k)(1− iek), which yields that iek = 0, a contradiction. Next, suppose that ek = −i. Then from (3.6), we obtain that e−p1(z)+p1(z+c) + e−(p1(z)+p1(z+c)) = −2. This implies that T ( r, e−(p1(z)+p1(z+c)) ) = T ( r, e−p1(z)+p1(z+c) ) + S ( r, e−p1(z)+p1(z+c) ) . By the second fundamental theorem of Nevanlinna for several complex variables, we have T ( r, e−(p1(z)+p1(z+c)) ) ≤ N ( r, e−(p1(z)+p1(z+c)) ) +N ( r, 1 e−(p1(z)+p1(z+c)) ) +N ( r, 1 e−(p1(z)+p1(z+c)) + 2 ) + S ( r, e−(p1(z)+p1(z+c)) ) ≤ N ( r, 1 e−p1(z)+p1(z+c) ) + S ( r, e−p1(z)+p1(z+c) ) ≤ S ( r, e−(p1(z)+p1(z+c)) ) + S ( r, e−p1(z)+p1(z+c) ) , which implies that p1(z) is a constant, which is a contradiction. Similarly, we can get a contradiction for the cases ek = i. Case 2: Let p2(z)−p1(z) be non-constant. We consider the following two possible subcases: 10 H. Y. XU, G. HALDAR EJDE-2024/26 Subcase 2.1 Let p2(z) + p1(z) = k, where k ∈ C. Then, from (3.4) and (3.5), we obtain iekep1(z+c)−p1(z) + ieke−(p1(z)+p1(z+c)) − (i+ ek)eke−2p1(z) = iek − 1 iekep1(z)−p1(z+c) + ie−kep1(z)+p1(z+c) − (ie−k + 1)e2p1(z) = iek − 1. (3.10) Observe that ek ̸= −i. Otherwise, it follows from (3.10) that e2p1(z+c) = −1, which implies that p1(z) is constant, a contradiction. By Lemma 3.1, we obtain from (3.10) that iekep1(z+c)−p1(z) = iek − 1 iekep1(z)−p1(z+c) = iek − 1. (3.11) In view of (3.10) and (3.11), we obtain that ie−p1(z+c)+p1(z) = i+ ek ie−ke−p1(z)+p1(z+c) = ie−k + 1. (3.12) It follows from (3.11) and (3.12) that ek = −i/2 = −2i, which is not possible. Subcase 2.2 Suppose p2(z) + p1(z) is non-constant. Subcase 2.2.1 Let p2(z)− p1(z + c) = k1, a constant in C. Then (3.4) reduces to −iep2(z)+p1(z+c) + iep2(z)+p1(z) + iep2(z)−p1(z) + e2p2(z) = 1 + iek1 . (3.13) If 1 + iek1 = 0, then it follows from (3.13) that ep1(z+c)−p1(z) − e−2p1(z) − iep2(z)−p1(z) = 1. (3.14) By Lemma 3.1, it follows from (3.14) that ep1(z+c)−p1(z) = 1, which implies that p1(z + c) − p1(z) =constant. Then, we can assume that p1(z) = L(z) + Φ(z) + ξ, where L(z) = ∑n j=1 αjzj , Φ(z) is a polynomial defined in (i) in Theorem 1.6. Hence, p2(z) = L(z)+Φ(z)+L(c)+ξ+k1. But, then p2(z)−p1(z) = L(c)+k1 =constant, which is a contradiction. Hence, 1 + iek1 ̸= 0. In view of Lemma 3.1, it follows from (3.13) that −iep2(z)+p1(z+c) = 1 + iek1 . This implies that p2(z) + p1(z + c) = k2, a constant in C, say. But then p2(z) = (k1 + k2)/2 =constant, which is a contradiction. Subcase 2.2.2 Let p2(z) − p1(z + c) be non-constant. Then, by Lemma 3.1, it follows from (3.4) that −iep2(z)+p1(z+c) = 1, (3.15) which yields that p2(z) + p1(z + c) is constant, say k2. In view of (3.4) and (3.15) that −iep1(z)−p1(z+c) + ie2p1(z) + ep2(z)+p1(z) = −i. (3.16) By Lemma 3.1, it follows from (3.16) that −iep1(z)−p1(z+c) = −i, which yields that p1(z)− p1(z + c) =constant, say k3. But then, p2(z) + p1(z) = k1 + k3 is constant, which is a contradiction. This completes the proof of the theorem. □ Proof of Theorem 1.6. As we know the entire solutions of the functional equation f2+g2 = 1 are f = cosα(z) and g = sinα(z), where α(z) is an entire function. If f EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 11 and g are finite order entire functions, then p(z) must be a non-constant polynomial (see [6, 7, 35]). In view of the above fact, we easily obtain from (1.6) that a1f1(z) = 1 2 ( eip1(z) + e−ip1(z) ) a2f2(z + c) + a3f2(z) = 1 2i ( eip1(z) − e−ip1(z) ) a1f2(z) = 1 2 ( eip2(z) + e−ip2(z) ) a2f1(z + c) + a3f1(z) = 1 2i ( eip2(z) − e−ip2(z) ) , (3.17) where p1(z), p2(z) are two non-constant polynomials in Cn. After some simple calculations, from (3.17) we obtain − ia2 a1 ei(p2(z)+p1(z+c)) − ia2 a1 ei(p2(z)−p1(z+c)) − ia3 a1 ei(p2(z)+p1(z)) − ia3 a1 ei(p2(z)−p1(z)) + e2ip2(z) = 1 (3.18) and − ia2 a1 ei(p1(z)+p2(z+c)) − ia2 a1 ei(p1(z)−p2(z+c)) − ia3 a1 ei(p1(z)+p2(z)) − ia3 a1 ei(p1(z)−p2(z)) + e2ip1(z) = 1. (3.19) Now, we consider the following possible two cases. Case 1. Let p2(z)− p1(z) = k, a constant in C. It follows from (3.18) and (3.19) that − ia2e i(p1(z)+p1(z+c)) − ia2e i(p1(z)−p1(z+c)) − ( ia3 − a1e ik ) e2ip1(z) = a1e −ik + ia3 (3.20) and − ia2e i(p1(z)+p1(z+c)) − ia2e −2ikei(p1(z)−p1(z+c)) − ( ia3 − a1e −ik ) e2ip1(z) = ( a1 + ia3e −ik ) e−ik. (3.21) If eik ̸= ia1/a3, −ia3/a1, then by Lemma 3.1, it follows from (3.20) and (3.21) that −ia2e i(p1(z)−p1(z+c)) = a1e −ik + ia3 −ia2e i(p1(z)−p1(z+c)) = a1e ik + ia3. (3.22) In view of (3.20), (3.21) and (3.22), it follows that −ia2e i(−p1(z)+p1(z+c)) = ia3 − a1e ik −ia2e i(−p1(z)+p1(z+c)) = ia3 − a1e −ik. (3.23) In view of (3.23), we conclude that p1(z+ c)− p1(z) is constant, and hence we can assume that p1(z) = L(z) + Φ(z) + ξ, where L(z) = ∑n j=1 αjzj with αj , ξ ∈ C, j = 1, 2, . . . n, and Φ(z) is a polynomial defined in (i) in Theorem 1.6. 12 H. Y. XU, G. HALDAR EJDE-2024/26 Therefore, from (3.22) and (3.23), we obtain that −ia2e −iL(c) = a1e −ik + ia3, −ia2e −iL(c) = a1e ik + ia3, −ia2e iL(c) = ia3 − a1e ik, −ia2e iL(c) = ia3 − a1e −ik. (3.24) It follows from (3.24) that e2ik = 1, a22 = a21 + a23, e2iL(c) = −a1 − ia3e −ik a1 + ia3eik . It follows from (3.17) that the solution of the system (1.5) is (f1(z), f2(z)) = ( 1 a1 cos(L(z) + Φ(z) + ξ), 1 a1 cos(L(z) + Φ(z) + ξ + k) ) . Case 2. Let p2(z)− p1(z) be non-constant. We discuss the following two possible subcases: Subcase 2.1 Let p2(z) + p1(z) = k, a constant, k ∈ C. Therefore, from (3.18) and (3.19), we obtain − ia2e i(−p1(z)+p1(z+c)) − ia2e −i(p1(z)+p1(z+c)) − ( ia3 − a1e ik ) e−2ip1(z) = a1e −ik + ia3 (3.25) and − ia2e i(p1(z)−p1(z+c)) − ia2e −2ikei(p1(z)+p1(z+c)) − ( ia3e −ik − a1 ) e−ike2ip1(z) = a1e −ik + ia3. (3.26) If a1e −ik + ia3 = 0, then it follows from (3.25) that ei(p1(z)+p1(z+c)) + ei(p1(z)−p1(z+c)) = w, (3.27) where w = a1e ik−ia3 ia2 . Observe from (3.27) that N ( r, 1 ei(p1(z)+p1(z+c)) − w ) = N ( r, 1 ei(p1(z)−p1(z+c)) ) = S ( r, ei(p1(z)−p1(z+c)) ) . By the second fundamental theorem of Nevanlinna for several complex variables, we have T ( r, ei(p1(z)+p1(z+c)) ) ≤ N ( r, ei(p1(z)+p1(z+c)) ) +N ( r, 1 ei(p1(z)+p1(z+c)) ) +N ( r, 1 ei(p1(z)+p1(z+c)) − w ) + S ( r, ei(p1(z)+p1(z+c)) ) ≤ S ( r, ei(p1(z)+p1(z+c)) ) + S ( r, ei(p1(z)−p1(z+c)) ) . This implies that p1 is a constant, which is a contradiction. Therefore, a1e −ik + ia3 ̸= 0. By Lemma 3.1, we obtain from (3.25) and (3.26) that −ia2e i(−p1(z)+p1(z+c)) = a1e −ik + ia3, −ia2e i(p1(z)−p1(z+c)) = a1e −ik + ia3. (3.28) EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 13 In view of (3.25), (3.26) and (3.29), we have −ia2e i(p1(z)−p1(z+c)) = ia3 − a1e ik, −ia2e i(−p1(z)+p1(z+c)) = ia3 − a1e ik. (3.29) In view of (3.28), we conclude that p1(z + c) − p1(z) is constant, and hence we can assume that p1(z) = L(z) + Φ(z) + ξ, where L(z),Φ(z) are defined in Case 1. Therefore, p2(z) = −(L(z) + Φ(z) + ξ) + k It follows from (3.28) and (3.29) that −ia2e iL(c) = a1e −ik + ia3, −ia2e −iL(c) = a1e −ik + ia3, −ia2e −iL(c) = ia3 − a1e ik, −ia2e iL(c) = ia3 − a1e ik. (3.30) In view of (3.30), it follows that e2iL(c) = 1, e2ik = 1, a22 = (a1 ± a3) 2. In view of (3.17), the solution of the system (1.5) is (f1, f2) = ( 1 a1 cos[L(z) + Φ(z) + ξ], 1 a1 cos(−[L(z) + Φ(z) + ξ] + k) ) . Subcase 2.2 Let p2(z) + p1(z) be non-constant. Now, if p2(z)− p1(z + c) is non- constant, then by Lemma 3.1, we obtain from (3.18) that a2e i(p2(z)+p1(z+c)) = ia1, which yields that p2(z) + p1(z + c) is constant, say k ∈ C. It follows from (3.18) that ei(p1(z)−p2(z)) + e−i(p1(z)+p2(z)) = a1 − ia2e −ik ia3 . (3.31) Clearly, a1 − ia2e −ik ̸= 0. Otherwise, it follows from (3.31) that p1(z) is constant, which is a contradiction. In view of (3.31), we observe that N ( r, 1 e−i(p1(z)+p2(z)) − a1−ia2e−ik ia3 ) = S ( r, e−i(p1(z)+p2(z)) ) . By the second fundamental theorem of Nevanlinna for several complex variables, we have that [T ( r, e−i(p1(z)+p2(z)) ) ≤ N ( r, e−i(p1(z)+p2(z)) ) +N ( r, 1 e−i(p1(z)+p2(z)) ) +N ( r, 1 e−i(p1(z)+p2(z)) − a1−ia2e−ik ia3 ) + S ( r, e−i(p1(z)+p2(z)) ) ≤ S ( r, e−i(p1(z)+p2(z)) ) + S ( r, ei(p1(z)−p2(z)) ) . This implies that p1(z)+p2(z) is a constant in C, which contradicts our assumption. Thus, p2(z)− p1(z + c) is a constant in C. Let p2(z)− p1(z + c) = k, k ∈ C. Then (3.18) reduces to − ia2e i(p2(z)+p1(z+c)) − ia3e i(p2(z)+p1(z)) − ia3e i(p2(z)−p1(z)) + a1e 2ip2(z) = a1 + ia2e ik. (3.32) 14 H. Y. XU, G. HALDAR EJDE-2024/26 If a1 + ia2e ik ̸= 0, then using Lemma 3.1, we obtain from (3.32) that a1 + ia2e ik = −ia2e i(p2(z)+p1(z+c)), which implies that p2(z) + p1(z + c) = k1 ∈ C. But, then 2p2(z) = k + k1, a constant in C, which contradicts to the fact that p2(z) is non- constant. Thus, a1 + ia2e ik = 0. Then after simple computation, equation (3.32) reduces to equation (3.31). Then, by similar argument, we can get a contradiction. □ Proof of Theorem 1.11. Assume that (f1, f2) is a pair of transcendental entire solu- tion of (1.7) with each fj of finite order for j = 1, 2. Then, by an argument similar to the one in the proof of Theorem 1.6, we obtain a1f1(z + c) + a2f1(z) = 1 2 ( ep1(z) + e−p1(z) ) , a3f2(z + c) + a4f2(z) = 1 2i ( ep1(z) − e−p1(z) ) , a1f2(z + c) + a2f2(z) = 1 2 ( ep2(z) + e−p2(z) ) , a3f1(z + c) + a4f1(z) = 1 2i ( ep2(z) − e−p2(z) ) , where p1(z), p2(z) are two non-constant polynomials. Since D := a1a4 − a2a3 ̸= 0, solving the above system of equations, we obtain f1(z + c) = 1 2D ( a4 ( ep1(z) + e−p1(z) ) + ia2 ( ep2(z) − e−p2(z) )) , (3.33) f1(z) = 1 −2D ( a3 ( ep1(z) + e−p1(z) ) + ia1 ( ep2(z) − e−p2(z) )) , (3.34) f2(z + c) = 1 2D ( a4 ( ep2(z) + e−p2(z) ) + ia2 ( ep1(z) − e−p1(z) )) , (3.35) f2(z) = 1 −2D ( a3 ( ep2(z) + e−p2(z) ) + ia1 ( ep1(z) − e−p1(z) )) . (3.36) It follows from (3.33) and (3.34) that a3e p1(z+c)+p2(z) + a3e −p1(z+c)+p2(z) + ia1e p2(z+c)+p2(z) − ia1e −p2(z+c)+p2(z) + a4e p2(z)+p1(z) + a4e −p1(z)+p2(z) + ia2e 2p2(z) = ia2. (3.37) From (3.35) and (3.36), we obtain that a3e p2(z+c)+p1(z) + a3e −p2(z+c)+p1(z) + ia1e p1(z+c)+p1(z) − ia1e −p1(z+c)+p1(z) + a4e p1(z)+p2(z) + a4e p1(z)−p2(z) + ia2e 2p1(z) = ia2. (3.38) Now, we consider the following two possible cases. Case 1. Suppose p2(z) − p1(z) = k, where k is a constant in C. Then (3.37) and (3.38), respectively, yield ek ( a3 + ia1e k ) ep1(z)+p1(z+c) + ( a3e k − ia1 ) ep1(z)−p1(z+c) + ek ( a4 + ia2e k ) e2p1(z) = ( ia2 − a4e k ) (3.39) and ( a3e k + ia1 ) ep1(z)+p1(z+c) + ( a3e −k − ia1 ) ep1(z)−p1(z+c) + ( a4e k + ia2 ) e2p1(z) = ( ia2 − a4e −k ) . (3.40) EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 15 Now, we show that all of a3 + ia1e k, a3e k − ia1, a4 + ia2e k, ia2 − a4e k, a3e k + ia1, a3e −k − ia1, a4e k + ia2, and ia2 − a4e −k are non-zero. Suppose a3 + ia1e k = 0. Then, clearly, a4 + ia2e k ̸= 0. It follows from (3.39) that( a3e k − ia1 ) ep1(z)−p1(z+c) + ek ( a4 + ia2e k ) e2p1(z) = ia2 − a4e k. (3.41) Since a4 + ia2e k ̸= 0, it follows from (3.41) that a3e k − ia1 ̸= 0. Otherwise, p1(z) would be constant, which is not possible. Also, it is clear that ia2 − a4e k is non-zero. Otherwise, then we must have from (3.41) that − (a3ek − ia1 a4 + ia2ek ) e−(p1(z)+p1(z+c)) = 1, which implies that p1(z) + p1(z + c), and hence p1(z) is constant, which is a con- tradiction. In view of (3.41), T ( r, ep1(z)−p1(z+c) ) = T ( r, e2p1(z) ) + S ( r, e2p1(z) ) . Since p1(z) is a polynomial, it is easy to see that N ( r, 1 ep1(z)−p1(z+c) ) = N ( r, ep1(z)−p1(z+c) ) = N ( r, 1 e2p1(z) ) = S ( r, ep1(z) ) . Then, in view of (3.41) and using the second fundamental theorem of Nevanlinna in several complex variables, we obtain T ( r, ep1(z)−p1(z+c) ) ≤ N ( r, 1 ep1(z)−p1(z+c) ) +N ( r, ep1(z)−p1(z+c) ) +N ( r, 1 ep1(z)−p1(z+c) − α ) + S ( r, ep1(z)−p1(z+c) ) ≤ N ( r, 1 e2p1(z) ) + S ( r, ep1(z)−p1(z+c) ) ≤ S ( r, ep1(z)−p1(z+c) + S ( r, e2p1(z) )) where α = (ia2 − a4e k)/(a3e k − ia1). This implies T ( r, e2p1(z) ) = o ( T ( r, e2p1(z) )) , which is not possible as ep1(z) is transcendental entire. We conclude that a3+a1e k ̸= 0. Similarly, we can prove that the others are also non-zero. In view of Lemma 3.1, from (3.39) we obtain that(a3ek − ia1 ia2 − a4ek ) ep1(z)−p1(z+c) = 1. (3.42) In view of (3.42) and (3.39), we have − (a3 + ia1e k a4 + ia2ek ) e−p1(z)+p1(z+c) = 1. (3.43) Again, in view of Lemma 3.1, from (3.40) we obtain that(a3e−k − ia1 ia2 − a4e−k ) ep1(z)−p1(z+c) = 1. (3.44) Using (3.44) in (3.40), we obtain that − (a3ek + ia1 a4ek + ia2 ) e−p1(z)+p1(z+c) = 1. (3.45) 16 H. Y. XU, G. HALDAR EJDE-2024/26 In view of the fact that D ̸= 0, from (3.42) and (3.44), we obtain e2k = 1. Multi- plying (3.42) and (3.43), we obtain( (a22 + a24)− (a21 + a23) ) ek + i(a2a4 − a1a3)(e 2k − 1) = 0. As e2k = 1, the above equation yields a22 + a24 = a21 + a23. In view of (3.42), we conclude that p1(z)−p1(z+c) is constant. Thus, we can write p1(z) = L(z) + Φ(z) +A, where L(z) = ∑n j=1 αjzj , αj , A ∈ C, j = 1, 2, . . . , n, and Φ(z) is a polynomial defined in in Theorem 1.6(i). Therefore, from (3.42), (3.43), (3.44). and (3.45), we obtain eL(c) = a3e k − ia1 ia2 − a4ek = a3e −k − ia1 ia2 − a4e−k = −(a4e k + ia2) a3 + ia1ek = −(a4e k + ia2) a3ek + ia1 . Case 2. Let p2(z)− p1(z) be non-constant. We consider the following subcases: Subcase 2.1 Suppose p2(z) + p1(z) = k, where k is a constant in C. Then from (3.37) and (3.38), we obtain( a3e k − ia1 ) e−p1(z)+p1(z+c) + ek ( a3 + ia1e k ) e−(p1(z)+p1(z+c)) + ek ( a4 + ia2e k ) e−2p1(z) = ( ia2 − a4e k ) (3.46) and ( a3e k − ia1 ) ep1(z)−p1(z+c) + ( a3e −k + ia1 ) ep1(z)+p1(z+c) + ( ia2 + a4e −k ) e2p1(z) = ( ia2 − a4e k ) . (3.47) In a similar manner as in Case 1, we can prove that a3e k−ia1, a3+ia1e k, a4+ia2e k, ia2 − a4e k, a3e −k + ia1, and ia2 + a4e −k are non-constant. As p1(z) is a non-constant polynomial in C2, in view of Lemma 3.1, and equations (3.46) and (3.47), we obtain that( a3e k − ia1 ) e−p1(z)+p1(z+c) = ia2 − a4e k, (a3e k − ia1)e p1(z)−p1(z+c) = ia2 − a4e k. (3.48) In view of (3.46), (3.47), and (3.48), we obtain that − ( a3 + ia1e k ) ep1(z)−p1(z+c) = a4 + ia2e k, − ( a3e −k + ia1 ) ep1(z+c)−p1(z) = ia2 + ia4e −k. (3.49) Since p1(z) is a non-constant polynomial in C2, it follows from (3.48) that p1(z + c)− p1(z) = k, a constant in C. This implies that p1(z) = L(z) + Φ(z) + ξ, where L(z) = ∑n j=1 αjzj , Φ(z) is a polynomial defined in (i) of Theorem 1.6, ξ, αj are in EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 17 C, j = 1, 2, . . . , n. In view of the form of p1(z), we obtain from (3.48) and (3.49) that ( a3e k − ia1 ) eL(c) = ia2 − a4e k, (a3e k − ia1)e −L(c) = ia2 − a4e k, − ( a3 + ia1e k ) e−L(c) = a4 + ia2e k, − ( a3e −k + ia1 ) eL(c) = ia2 + ia4e −k. (3.50) Now, in view of equations of (3.50), we can easily obtain that e2L(c) = 1, e2k = −1, a21 + a23 = a22 + a24, a1a3 = a2a4. Thus, in view of (3.34) and (3.36), it follows that f1(z1, z2) = −1 2D [ (a3 − ia1e −k)eL(z)+Φ(z)+ξ + (a3 + ia1e k)e−(L(z)+Φ(z)+ξ) ] and f2(z1, z2) = −1 2D [ (ia2 − a4e k)e−(L(z)+Φ(z)+ξ) + (ia2 + a4e −k)eL(z)+Φ(z)+ξ ] . Subcase 2.2 Let p2(z) + p1(z) be non-constant. Subcase 2.2.1 Let p1(z + c) + p2(z) = k ∈ C. Then, clearly −p1(z + c) + p2(z) is non-constant. Otherwise, we obtain that p2(z) is constant, a contradiction. It follows from (3.37) that a3e −p1(z+c)+p2(z) + ia1e p2(z+c)+p2(z) − ia1e p2(z)−p2(z+c) + a4e p1(z)+p2(z) + a4e p2(z)−p1(z) + ia2e 2p2(z) = ia2 − a3e k. (3.51) Subcase 2.2.1.1 Let ia2 − a3e k = 0. Then equation (3.51) reduces to ia1e p2(z+c)−p2(z) − ia1e −[p2(z+c)+p2(z)] + a4e p1(z)−p2(z) + a4e −[p1(z)+p2(z)] = −(ia2 + a3e −k). (3.52) Subcase 2.2.1.1.1 Let ia2 + a3e −k = 0. Then (3.52) becomes ia1e p2(z+c)+p1(z) − ia1e p1(z)−p2(z+c) + a4e 2p1(z) = −a4. (3.53) In view of Lemma 3.6, it follows from (3.53) that either ia1e p2(z+c)+p1(z) = −a4, (3.54) or −ia1e −p2(z+c)+p1(z) = −a4. (3.55) First, we assume that (3.54) holds. Since p1(z), p2(z) are non-constant polynomials in C2, it follows from (3.54) that p2(z + c) + p1(z) = k1, a constant in C. As p1(z + c) + p2(z) = k, it follows that p1(z + 2c) − p1(z) = k − k1. Then, we may assume that p1(z) = L(z) +Φ(z) + ξ, where L(z),Φ(z) are defined in the Theorem 1.6(i). Hence, p2(z) = −[L(z)+Φ(z)+ξ]+k−L(c). Thus, p1(z)+p2(z) = k−L(c), a constant in C, which contradicts to our assumption. In a similar manner we can obtain a contradiction for the case (3.55). Subcase 2.2.1.1.2 Let ia2 + a3e −k ̸= 0. Then in view of Lemma 3.1, it follows from (3.32) that ia1e p2(z+c)−p2(z) = −(ia2 + a3e −k). (3.56) 18 H. Y. XU, G. HALDAR EJDE-2024/26 Therefore, in view of (3.32) and (3.56), we obtain that −ia1e −p2(z+c)+p1(z) + a4e 2p1(z) = −a4. (3.57) In view of (3.57), we observe that N ( r, 1 e2p1(z) + 1 ) = N ( r, 1 e−p2(z+c)+p1(z) ) = S ( r, e−p2(z+c)+p1(z)) ) . Now, by the second fundamental theorem of Nevanlinna for several complex vari- ables, we obtain that T ( r, e2p1(z) ) ≤ N ( r, e2p1(z) ) +N ( r, 1 e2p1(z) ) +N ( r, 1 e2p1(z) + 1 ) + S ( r, e2p1(z) ) ≤ S ( r, e−p2(z+c)+p1(z)) ) + S ( r, e2p1(z) ) . This implies that p1(z) is constant in C, which is a contradiction. Subcase 2.2.1.2 Let ia2 − a3e k ̸= 0. Then, in view of Lemma 3.1, we obtain from (3.51) that −ia1e −p2(z+c)+p2(z) = ia2−a3e k. This implies that −p2(z+c)+p2(z) = k1, a constant in C. As p1(z + c) + p2(z) = k, it follows that p1(z + c) − p2(z) = p1(z)− p2(z) = k + k1, which is a contradiction. Subcase 2.2.2 Let p1(z + c) + p2(z) be non-constant. Subcase 2.2.2.1 Let −p1(z + c) + p2(z) = k, a constant in C. Then, from (3.37), we obtain that a3e p1(z+c)+p2(z) + ia1e p2(z+c)+p2(z) − ia1e p2(z)−p2(z+c) + a4e p1(z)+p2(z) + a4e p2(z)−p1(z) + ia2e 2p2(z) = ia2 − a3e k. Then by an argument similar one used in Subcase 2.2.1.1 and Subcase 2.2.1.2, we can easily obtain a contradiction. Subcase 2.2.2.2 Let −p1(z+ c)+p2(z) be non-constant. Then, in view of Lemma 3.1, it follows from (3.37) that −ia1e p2(z)−p2(z+c) = ia2. (3.58) As p2(z) is a non-constant polynomials in C2, it follows from (3.58) that p2(z) − p2(z + c) is a constant in C. Thus, p2(z + c) + p1(z) and −p2(z + c) + p1(z) both are non-constant. In view of Lemma 3.1, it follows from (3.38) that −ia1e −p1(z+c)+p1(z) = ia2. (3.59) Substituting (3.58) in (3.37), we obtain a3e p1(z+c)−p2(z) + a3e −(p1(z+c)+p2(z)) + ia1e p2(z+c)−p2(z) + a4e p1(z)−p2(z) + a4e −(p1(z)+p2(z)) = −ia2. (3.60) Substituting (3.59) in (3.38), we obtain a3e p2(z+c)−p1(z) + a3e −(p2(z+c)+p1(z)) + ia1e p1(z+c)−p1(z) + a4e p2(z)−p1(z) + a4e −(p1(z)+p2(z)) = −ia2. (3.61) Again, in view of Lemma 3.1, it follows from (3.60) that ia1e p2(z+c)−p2(z) = −ia2. (3.62) EJDE-2024/26 FERMAT-TYPE DIFFERENCE EQUATIONS IN Cn 19 Substituting (3.62) in (3.60), we obtain a3e p1(z+c)+p1(z) + a3e p1(z)−p1(z+c) + a4e 2p1(z) = −a4. (3.63) In view of Lemma 3.1, we obtain from (3.61) that ia1e p1(z+c)−p1(z) = −ia2. (3.64) In view of Lemma 3.1, we obtain from (3.63) that a3e p1(z)−p1(z+c) = −a4. (3.65) Substituting (3.65) in (3.63), we obtain that a3e p1(z+c)−p1(z) = −a4. (3.66) In view of (3.59), we conclude that −p1(z+ c)+ p1(z) must be constant in C. This implies that p1(z) = L(z) + Φ(z) + ξ, where L(z),Φ(z) are defined in Theorem 1.6(i). Therefore, in view of (3.59), (3.64), (3.65) and (3.66), we obtain that −ia1e −L(c) = ia2, −ia1e L(c) = ia2, a3e −L(c) = −a4, a3e L(c) = −a4. From the above fours equations, we can easily obtain that D = 0, which contradicts to our assumption. □ Proof of Theorem 2.2. Using Lemmas 3.4 and 3.5, the proof of this theorem can be carried out with arguments similar to those in the proof of [52, Theorem 1.1]. □ Concluding remark and an open question. Observe that if p(z) = L(z) + Φ(z) + ξ, where L(z) = ∑n j=1 ajzj and Φ(z) is defined as in the conclusion (i) of Theorem 1.6, then p(z+c)−p(z) must be a constant in C, c = (c1, c2, . . . , cn) ∈ Cn, ξ, aj ∈ C, j = 1, 2, . . . , n. But, we are still unable to prove the converse part. Therefore, we pose the following open problem. What will be the exact form of the polynomial p(z) : Cn → P1(C) if it satisfies the relation p(z+c)−p(z) = ξ, where c = (c1, c2, . . . , cn) ∈ Cn and ξ ∈ C? Acknowledgments. 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Hong Yan Xu School of Arts and Sciences, Suqian University, Suqian, Jiangsu 223800, China Email address: xhyhhh@126.com Goutam Haldar Department of Mathematics, Ghani Khan Choudhury Institute of Engineering and Tech- nology, Narayanpur, Malda - 732141, West Bengal, India Email address: goutamiitm@gmail.com, goutamiit1986@gmail.com 1. Introduction and main results 2. Solutions of Fermat-type partial differential-difference equations in several complex variables 3. Proof of main results Concluding remark and an open question Acknowledgments References