A study of solutions for several classes of systems of complex nonlinear partial differential difference equations in ℂ2

Xin Jiao; Yu Bin Chen; Hong Yan Xu

doi:10.1515/dema-2025-0119

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A study of solutions for several classes of systems of complex nonlinear partial differential difference equations in ℂ²

Xin Jiao , Yu Bin Chen and Hong Yan Xu

Published/Copyright: May 22, 2025

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From the journal Demonstratio Mathematica Volume 58 Issue 1

Abstract

This article is devoted to describing the entire solutions of several systems of the first-order nonlinear partial differential difference equations (PDDEs). Using the Nevanlinna theory and the Hadamard factorization theory of meromorphic functions, we establish some interesting results to reveal the existence and the forms of the finite-order transcendental entire solutions of several systems of the first-order nonlinear PDDEs:

f ( z 1 + c 1 , z 2 + c 2 ) ( a 1 g z 1 + a 2 g z 2 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) ( a 3 f z 1 + a 4 f z 2 ) = m 2 , f ( z 1 + c 1 , z 2 + c 2 ) ( a 1 f z 1 + a 2 g z 1 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) ( a 3 f z 2 + a 4 g z 2 ) = m 2 ,

and

f ( z 1 + c 1 , z 2 + c 2 ) ( a 1 f z 2 + a 2 g z 1 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) ( a 3 f z 1 + a 4 g z 2 ) = m 2 ,

where a 1 , a 2 , a 3 , a 4 , c 1 , c 2 ∈ C , m 1 , m 2 ∈ C − { 0 } . Moreover, some examples are given to explain that there are significant differences in the forms of solutions from some previous systems of functional equations.

Keywords: Nevanlinna theory; entire function; system of functional equations; several complex variables

MSC 2010: 30D35; 35M30

1 Introduction

In this article, our purpose is to study the entire solutions of several systems of the first-order nonlinear partial differential difference equations (PDDEs) with two variables. Gross [1] pointed out that the Fermat-type functional equation (which can also be called the Pythagorean functional equation)

(1.1) f ( z ) m + g ( z ) m = 1

has no transcendental meromorphic solutions when m ≥ 4 . Montel [2] proved that (1.1) has no transcendental entire solutions when m ≥ 3 . They had proved that equation (1.1) has entire solutions and pointed out that for m = 2 , the equation has non-constant entire solutions f = cos p ( z ) , g = sin p ( z ) , where p ( z ) is any non-constant entire function. The Pythagorean functional equation is also defined as the eikonal equation u z 1 2 + u z 2 2 = 1 .

In 1995, Khavinson [3] pointed out that any entire solution of the partial differential equations (PDEs)

(1.2) ∂ u ∂ z 1 2 + ∂ u ∂ z 2 2 = 1

in C 2 is necessarily linear. Later, Saleeby [4] extended the result by exploring the solutions of the Fermat-type functional equations (1.2) and obtained the following:

Theorem A

(see [4]). The entire solution of equation (1.2) must satisfy u ( z 1 , z 2 ) = c 1 z 1 + c 2 z 2 + c , where c , c 1 , c 2 ∈ C , and c 1 2 + c 2 2 = 1 .

Then, Li et al. discussed equation (1.2) with more general form such as ( ∂ f ∂ z 1 ) 2 + ( ∂ f ∂ z 2 ) 2 = f n , ( ∂ f ∂ z 1 ) 2 + ( ∂ f ∂ z 2 ) 2 = p , ( ∂ f ∂ z 1 ) 2 + ( ∂ f ∂ z 2 ) 2 = e g , where n ∈ N + , p , and g are the polynomials in C 2 [5–8]. For example, Li (2005) further investigated the functional equation of Fermat-type

(1.3) ∂ u ∂ z 1 2 + ∂ u ∂ z 2 2 = e g

and obtained.

Theorem B

(see [7]). If equation (1.3) admits an entire solution of f ( z ) with finite-order in C 2 , where g is a polynomial, then u is an entire solution of (1.3) if and only if

u = f ( c 1 z 1 + c 2 z 2 ) or
u = ϕ 1 ( z 1 + i z 2 ) + ϕ 2 ( z 1 − i z 2 ) ,
where f is an entire solution and f ′ ( c 1 z 1 + c 2 z 2 ) = ± e 1 2 g ( z ) , c 1 and c 2 are two constants satisfying c 1 2 + c 2 2 = 1 , and ϕ 1 ″ ( z 1 + i z 2 ) + ϕ 2 ′ ( z 1 − i z 2 ) = 1 4 e g ( z ) .

As Khavinson [3] and Li [5] noted, under the linear transformation x = z 1 + i z 2 and y = z 1 − i z 2 , equation (1.3) can be reduced to

(1.4) U x U y = P ,

where U ( x , y ) = u ( z 1 , z 2 ) , and P ( x , y ) = e g . By a simple calculation, one can obtain

(1.5) A ( U x x U y y − U x y 2 ) + B U x y + C = 0 ,

where A = U x U y , B = U x P y + U y P x , and C = P x P y . Obviously, equation (1.5) can be seen as a non-degenerate Monge-Ampère equation, which is widely used in differential geometry, variational methods, optimization problems, and transmission problems. There is an enormous literature dedicated to the study of Monge-Ampère equations. In view of this remark of Khavinson [3] and Li [5], there are some references focusing on the solutions of some first-order nonlinear PDEs with product form (can be found in [9–13]).

Theorem C

(see [11, Theorem 1]). Let g be a polynomial in C 2 , and let m be a non-negative integer. Then, u is an entire solution of the PDE u x u y = x m e g in C 2 if and only if the following assertions hold:

u = ϕ 1 ( x ) + ϕ 2 ( x ) , where ϕ 1 ″ ( x ) = x m e α ( x ) and ϕ 2 ′ ( y ) = e β ( y ) satisfying α ( x ) + β ( y ) = g ( x , y ) ;
u = F ( y + A x m + 1 ) , where A is a non-zero constant and ( m + 1 ) A F ′ 2 ( y + A x m + 1 ) = e g ;
u = ( x k + 1 ⁄ ( k + 1 ) ) e a y + b + C , where ( a ⁄ ( k + 1 ) ) e 2 ( a y + b ) = e g , m = 2 k + 1 , and a ( ≠ 0 ) , b , C are constant.

In 2023, Xu et al. [14] studied the solutions of the system of the product-type PDEs with more general terms, and obtained.

Theorem D

(see [14, Corollary 2.1]). Let g be a nonconstant polynomial in C 2 , and ( u , v ) be the pair of finite-order transcendental entire solutions for system

(1.6) u z 1 v z 2 = e g , v z 1 u z 2 = e g .

If g can be represented as a polynomial in z 2 + A − 1 z 1 , then we have
( u , v ) = ( A F 1 ( z 2 + A − 1 z 1 ) , G 1 ( z 2 + A − 1 z 1 ) ) ,
where A ∈ C − { 0 } , F 1 ″ ( t ) = e φ 1 ( t ) , G 1 ″ ( t ) = e φ 2 ( t ) , and φ 1 ( t ) , φ 2 ( t ) are the nonconstant polynomials in C satisfying g = φ 1 ( z 2 + A − 1 z 1 ) + φ 2 ( z 2 + A − 1 z 1 ) ;
If g can be represented as the sum of two polynomials in C , which one is only involving in z 1 and the other is only involving z 2 , then we have
( u , v ) = ( F 2 ( z 1 ) + G 3 ( z 2 ) , G 2 ( z 1 ) + F 3 ( z 2 ) ) ,
where
F 2 ′ ( z 1 ) = e ϕ 1 ( z 1 ) , G 2 ′ ( z 1 ) = e ψ 2 ( z 1 ) , F 3 ′ ( z 2 ) = e ψ 1 ( z 2 ) , G 3 ′ ( z 2 ) = e ϕ 2 ( z 2 ) ,
and ϕ 1 ( z 1 ) , ϕ 2 ( z 2 ) , ψ 1 ( z 2 ) , ψ 2 ( z 1 ) are the polynomials in C and
g = ϕ 1 ( z 1 ) + ψ 1 ( z 2 ) = ϕ 2 ( z 2 ) + ψ 2 ( z 1 ) .

In 2018–2020, Cao and Xu [15], Cao and Xu [16,17] investigated the entire and meromorphic solutions of several types of complex differential difference equations using the difference analogs of Nevanlinna theory in C n [15–19] and obtained some theorems, which are the extensions and improvements of the previous results (including [20–27]).

Theorem E

(see [16]). Suppose f is a transcendental entire solution of equation

(1.7) ∂ f ( z 1 , z 2 ) ∂ z 1 2 + f ( z 1 + c 1 , z 2 + c 2 ) 2 = 1 ,

then f must satisfy f ( z 1 , z 2 ) = sin ( A z 1 + B ) , where A , B ∈ C and A e i A c 1 = 1 ; in the special case whenever c 1 = 0 , we have f ( z 1 , z 2 ) = sin ( z 1 + B ) .

In 2024, Xu et al. [28] extended Theorem E when (1.7) is turned into the system of complex PDEs, and obtained

Theorem F

(see [28, Theorem 1.3]). Any pair of transcendental entire solutions with finite-order for the system of Fermat-type PDEs

(1.8) ∂ f 1 ( z 1 , z 2 ) ∂ z 1 2 + f 2 ( z 1 + c 1 , z 2 + c 2 ) 2 = 1 , ∂ f 2 ( z 1 , z 2 ) ∂ z 1 2 + f 1 ( z 1 + c 1 , z 2 + c 2 ) 2 = 1 ,

have the following forms:

( f 1 ( z ) , f 2 ( z ) ) = e L ( z ) + B 1 + e − ( L ( z ) + B 1 ) 2 , η A 21 e L ( z ) + B 1 + A 22 e − ( L ( z ) + B 1 ) 2 ,

where L ( z ) = a 1 z 1 + a 2 z 2 , B 1 is a constant in C , and a 1 , c , A 21 , and A 22 satisfy one of the following cases:

A 21 = − i , A 22 = i , and a 1 = i , L ( c ) = ( 2 k + 1 2 ) π i , or a 1 = − i , L ( c ) = ( 2 k − 1 2 ) π i ;
A 21 = i , A 22 = − i , and a 1 = i , L ( c ) = ( 2 k − 1 2 ) π i , or a 1 = − i , L ( c ) = ( 2 k + 1 2 ) π i ;
A 21 = 1 , A 22 = 1 , and a 1 = i , L ( c ) = 2 k π i , or a 1 = − i , L ( c ) = ( 2 k + 1 ) π i ;
A 21 = − 1 , A 22 = − 1 , and a 1 = i , L ( c ) = ( 2 k + 1 ) π i , or a 1 = − i , L ( c ) = 2 k π i .

Recently, Liu et al. [29] studied the solutions of product-type PDDEs and obtained the following results.

Theorem G

(see [29, Theorem 2.1]). Let c = ( c 1 , c 2 ) ∈ C 2 , c 1 , c 2 ∈ C , and D ≔ β 1 γ 2 − β 2 γ 1 , and assume that ( u , v ) is a pair of finite-order transcendental entire solutions of system

(1.9) u ( z + c ) [ α 1 u ( z ) + β 1 u z 1 + γ 1 u z 2 + α 2 v ( z ) + β 2 v z 1 + γ 2 v z 2 ] = 1 , v ( z + c ) [ α 1 v ( z ) + β 1 v z 1 + γ 1 v z 2 + α 2 u ( z ) + β 2 u z 1 + γ 2 u z 2 ] = 1 .

If α 1 ≠ 0 , then (1.9) has no any pair of transcendental entire solutions with finite-order;
If α 1 = 0 and D ≠ 0 , then ( u , v ) must be the form of
( u , v ) = 1 α 2 + β 2 A 1 + γ 2 A 2 e A 1 z 1 + A 2 z 2 − B 2 , 1 α 2 − β 2 A 1 − γ 2 A 2 e − A 1 z 1 + A 2 z 2 − B 1 ,
where A 1 , A 2 , B 1 , B 2 ∈ C satisfy β 1 A 1 + γ 1 A 2 = 0 and
(1.10) e 2 ( A 1 c 1 + A 2 c 2 ) = α 2 + β 2 A 1 + γ 2 A 2 α 2 − β 2 A 1 − γ 2 A 2 , e 2 ( B 1 + B 2 ) = 1 α 2 2 − ( β 2 A 1 + γ 2 A 2 ) 2 .

By observing Theorems D–H, one can suggest the following questions naturedly:

Question 1.1

How to describe the solutions if system (1.8) is changed into system of the product-type partial differential difference equations in C 2 ?

Question 1.2

What had happened about the solutions if the constant 1 is replaced by two distinct constants m 1 and m 2 in Theorem H?

2 Results and examples

Our main motive in writing this article is to discuss Questions 1.1 and 1.2. More precisely, we are concerning with the description of the transcendental entire solutions for the system of the following first-order nonlinear PDDEs:

(2.1) f ( z 1 + c 1 , z 2 + c 2 ) ( a 1 g z 1 + a 2 g z 2 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) ( a 3 f z 1 + a 4 f z 2 ) = m 2 ,

(2.2) f ( z 1 + c 1 , z 2 + c 2 ) ( a 1 f z 1 + a 2 g z 1 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) ( a 3 f z 2 + a 4 g z 2 ) = m 2 ,

and

(2.3) f ( z 1 + c 1 , z 2 + c 2 ) ( a 1 f z 2 + a 2 g z 1 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) ( a 3 f z 1 + a 4 g z 2 ) = m 2 ,

in C 2 , where a 1 , a 2 , a 3 , a 4 , c 1 , c 2 ∈ C and m 1 , m 2 ∈ C − { 0 } . The method of this article mainly comes from [9,10,14,29–31]. We obtain

Theorem 2.1

Let D 1 = a 1 c 2 − a 2 c 1 , D 2 = a 3 c 2 − a 4 c 1 , η = ( c 1 , c 2 ) ∈ C 2 and m 1 , m 2 , a 1 , a 2 , a 3 , a 4 ∈ C − { 0 } . Let ( f , g ) be a pair of finite-order transcendental entire solutions of system (2.1).

If D 1 ≠ 0 and D 2 ≠ 0 , then we have
( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) − L ( η ) − B 2 , m 22 e − L ( z ) + L ( η ) − B 1 ) ,
here and there L ( z ) = A 1 z 1 + A 2 z 2 , L ( η ) = c 1 A 1 + c 2 A 2 , H ( Z ) is a polynomial in Z ≔ c 2 z 1 − c 1 z 2 , A 1 , A 2 , B 1 , B 2 ∈ C , m 11 m 12 = m 1 , m 21 m 22 = m 2 satisfy
(2.4) e 2 ( B 1 + B 2 ) = − ( m 12 m 22 ) 2 K 1 K 2 m 1 m 2 , e 2 L ( η ) = − m 1 K 2 m 2 K 1 , e L ( η ) + B 1 + B 2 = m 12 m 22 K 2 m 2 ,
where K 1 = a 1 A 1 + a 2 A 2 , K 2 = a 3 A 1 + a 4 A 2 .
If D 1 = 0 and D 2 ≠ 0 or D 1 ≠ 0 and D 2 = 0 , then the conclusion (i) of Theorem 2.1 holds.
If D 1 = 0 and D 2 = 0 , then we have
( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) + H ( Z ) − L ( η ) − B 2 , m 22 e − L ( z ) − H ( Z ) + L ( η ) − B 1 ) ,
where L ( z ) , L ( η ) , A 1 , A 2 , B 1 , B 2 , H ( Z ) , m 12 , and m 22 satisfy (2.4).

The following examples show that every case in the forms of solutions in Theorem 2.1 can be occurred.

Example 2.1

Let

( f ( z ) , g ( z ) ) = ( − e 2 z 1 − z 2 , − i e − ( 2 z 1 − z 2 ) ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.1) with c 1 = π i , c 2 = 3 π 2 i , a 1 = a 2 = a 3 = a 4 = 2 , and m 1 = m 2 = 2 . This corresponds to the case of D 1 ≠ 0 and D 2 ≠ 0 in Theorem 2.1 (i). This shows that system (2.1) has a pair of finite-order transcendental entire solutions.

Example 2.2

Let

( f ( z ) , g ( z ) ) = − 2 e z 1 + z 2 , − 1 3 i e − ( z 1 + z 2 ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.1) with c 1 = π 6 i , c 2 = π 3 i , a 1 = a 4 = 1 , a 2 = a 3 = 2 , and m 1 = m 2 = 2 . This corresponds to the case of D 1 = 0 and D 2 ≠ 0 in Theorem 2.1 (ii). This shows that system (2.1) has a pair of finite-order transcendental entire solutions.

Example 2.3

Let

( f ( z ) , g ( z ) ) = − 2 e z 1 + z 2 , − 1 3 i e − ( z 1 + z 2 ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.1) with c 1 = π 6 i , c 2 = π 3 i , a 1 = a 4 = 2 , a 2 = a 3 = 1 , and m 1 = m 2 = 2 . This corresponds to the case of D 1 ≠ 0 and D 2 = 0 in Theorem 2.1 (iii). This shows that system (2.1) has a pair of finite-order transcendental entire solutions.

Example 2.4

Let

( f ( z ) , g ( z ) ) = ( − e 2 z 1 − z 2 + ( z 2 − z 1 ) 3 , − i e − [ 2 z 1 − z 2 + ( z 2 − z 1 ) 3 ] ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.1) with c 1 = c 2 = π 2 i , a 1 = a 2 = a 3 = a 4 = 2 and m 1 = m 2 = 2 . This corresponds to the case of D 1 = 0 and D 2 = 0 in Theorem 2.1 (ii). Obviously, we have ρ ( ( f , g ) ) = 3 . This shows that system (2.1) admits transcendental entire solution with growth order ≥ 1 .

Remark 2.1

By observing Examples 2.3 and 2.4, we find that system (2.1) has the solution when a 1 ≠ a 3 and a 2 ≠ a 4 . This differs significantly from the results in [29].

From Theorem 2.1, we obtain the following corollary easily.

Corollary 2.1

Let η = ( c 1 , c 2 ) ∈ C 2 and m 1 , m 2 ∈ C − { 0 } . If ( f , g ) is a pair of finite-order transcendental entire solutions for the system

(2.5) f ( z 1 + c 1 , z 2 + c 2 ) g z 1 ( z 1 , z 2 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) f z 1 ( z 1 , z 2 ) = m 2 ,

If c 2 ≠ 0 , then we have
( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) − L ( η ) − B 2 , m 22 e − L ( z ) + L ( η ) − B 1 ) ;
If c 2 = 0 , then we have
( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) + H ( z 2 ) − L ( η ) − B 2 , m 22 e − L ( z ) − H ( z 2 ) + L ( η ) − B 1 ) ,
where L ( z ) = A 1 z 1 + A 2 z 2 , L ( η ) = c 1 A 1 + c 2 A 2 , and H ( z 2 ) is a polynomial in z 2 , A 1 , A 2 , B 1 , B 2 ∈ C , m 11 m 12 = m 1 , m 21 m 22 = m 2 satisfy
e 2 ( B 1 + B 2 ) = − ( m 12 m 22 A 1 ) 2 m 1 m 2 , e 2 L ( η ) = − m 1 m 2 , e L ( η ) + B 1 + B 2 = m 12 m 22 A 1 m 2 .

The following examples show that the forms of solutions of system (2.5) are precise.

Example 2.5

Let

( f ( z ) , g ( z ) ) = ( e − z 1 + 2 z 2 + 1 , 2 e z 1 − 2 z 2 − 1 ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.5) with c 1 = c 2 = log 2 and m 1 = 4 , m 2 = − 1 . This corresponds to the case of c 2 ≠ 0 in Corollary 2.1 (i). This shows that system (2.5) has a pair of finite-order transcendental entire solutions.

Example 2.6

Let

( f ( z ) , g ( z ) ) = ( − i e z 1 + A z 2 + z 2 3 , − 2 i e − ( z 1 + A z 2 + z 2 3 ) ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.5) with c 1 = log 2 , c 2 = 0 , and m 1 = 4 , m 2 = − 1 . This corresponds to the case of c 2 = 0 , H ( z 2 ) = z 2 3 in Corollary 2.1 (ii). Obviously, we have ρ ( ( f , g ) ) = 3 . This shows that system (2.5) admits transcendental entire solution with growth order ≥ 1 .

Remark 2.2

Examples 2.5 and 2.6 show that system (2.1) has the solution when m 1 ≠ m 2 . This is an extension of the results in [29].

Corollary 2.2

Let η = ( c 1 , c 2 ) ∈ C 2 and m 1 , m 2 ∈ C − { 0 } . Let ( f , g ) be a pair of finite-order transcendental entire solutions for the system

(2.6) f ( z 1 + c 1 , z 2 + c 2 ) ( g z 1 + g z 2 ) = m 1 , g ( z 1 + c 1 , z 2 + c 2 ) ( f z 1 + f z 2 ) = m 2 .

If c 1 ≠ c 2 , then we have
( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) − L ( η ) − B 2 , m 22 e − L ( z ) + L ( η ) − B 1 ) ;
If c 1 = c 2 , then we have
( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) + H ( z 1 − z 2 ) − L ( η ) − B 2 , m 22 e − L ( z ) − H ( z 1 − z 2 ) + L ( η ) − B 1 ) ,
where L ( z ) = A 1 z 1 + A 2 z 2 , L ( η ) = c 1 A 1 + c 2 A 2 , and H ( Z ) is a polynomial in Z ≔ z 1 − z 2 , A 1 , A 2 , B 1 , B 2 ∈ C , m 11 m 12 = m 1 , and m 21 m 22 = m 2 satisfy
e 2 ( B 1 + B 2 ) = − [ m 12 m 22 ( A 1 + A 2 ) ] 2 m 1 m 2 , e 2 L ( η ) = − m 1 m 2 , e L ( η ) + B 1 + B 2 = m 12 m 22 ( A 1 + A 2 ) m 2 .

The following examples show that the forms of solutions of system (2.6) are precise.

Example 2.7

Let

( f ( z ) , g ( z ) ) = ( e z 1 + z 2 , i e − ( z 1 + z 2 ) ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.6) with c 1 = π i , c 2 = 3 2 π i and m 1 = m 2 = 2 . This corresponds to the case of c 1 ≠ c 2 in Corollary 2.2 (i). This shows that system (2.6) has a pair of finite-order transcendental entire solutions.

Example 2.8

Let

( f ( z ) , g ( z ) ) = ( e 2 z 1 − z 2 + ( z 1 − z 2 ) 3 log 2 + 1 , − 2 e − ( 2 z 1 − z 2 + ( z 1 − z 2 ) 3 log 2 + 1 ) ) .

Then, ( f , g ) is a pair of transcendental entire solutions of system (2.6) with c 1 = c 2 = log 2 , and m 1 = 4 , m 2 = − 1 . This corresponds to the case of c 1 = c 2 and H ( z 1 − z 2 ) = ( z 1 − z 2 ) 3 log 2 in Corollary 2.2(ii). Obviously, we have ρ ( ( f , g ) ) = 3 . This shows that system (2.6) admits transcendental entire solution with growth order ≥ 1 .

Theorem 2.2

Let η = ( c 1 , c 2 ) ∈ C 2 and a 1 , a 2 , a 3 , a 4 , m 1 , m 2 ∈ C − { 0 } . Then, the system of (2.2) has no any pair of finite-order transcendental entire solutions.

Theorem 2.3

Let η = ( c 1 , c 2 ) ∈ C 2 and a 1 , a 2 , a 3 , a 4 , m 1 , m 2 ∈ C − { 0 } . Then, the system of (2.3) has no any pair of finite-order transcendental entire solutions.

3 Proofs of Theorem 2.1 and Corollaries 2.1–2.2

We first start the following lemmas that play key roles in proving our main theorems.

Lemma 3.1

[32,33] For an entire function F on C n , F ( 0 ) ≠ 0 and put ρ ( n F ) = ρ < ∞ . Then, there exist a canonical function f F and a function g F ∈ C n such that F ( z ) = f F ( z ) e g F ( z ) . For the special case n = 1 , f F is the canonical product of Weierstrass.

Remark 3.1

Here, ρ ( n F ) denotes the order of the counting function of zeros of F .

Lemma 3.2

[34] If g and h are entire functions and g ( h ) is an entire function of finite-order, then there are only two possible cases: either

the internal function h is a polynomial and the external function g is of finite-order; or else
the internal function h is not a polynomial but a function of finite-order, and the external function g is of zero order.

Lemma 3.3

[30, Lemma 3] and [29, Lemma 3.3] Let g ( u ) = g ( x , y ) be a polynomial in C 2 , and u 0 = ( x 0 , y 0 ) , x 0 , y 0 ∈ C be two constants. If g ( u + u 0 ) − g ( u ) = g ( x + x 0 , y + y 0 ) − g ( x , y ) is a constant, then g ( u ) can be represented as the form of

g ( x , y ) = L ( u ) + H ( s ) ,

where L ( u ) = α x + β y , α , β are the constants, and H ( s ) is a polynomial in s in C , s ≔ y 0 x − x 0 y .

3.1 Proof of Theorem 2.1

Let ( f , g ) be a pair of transcendental entire solutions of system (2.1). By observing the form of system (2.1), it is easy to obtain that a 3 f z 1 + a 4 f z 2 , a 1 g z 1 + a 2 g z 2 have no zeros and poles. Thus, by Lemmas 3.1 and 3.2, there exist two polynomials p and q in C 2 such that

(3.1) a 1 g z 1 + a 2 g z 2 = m 11 e q , f ( z + η ) = m 12 e − q ,

and

(3.2) a 3 f z 1 + a 4 f z 2 = m 21 e p , g ( z + η ) = m 22 e − p .

Obviously, p and q are nonconstant polynomials; otherwise, f ( z + η ) or g ( z + η ) is constant. This is a contradiction with the assumption of f and g being transcendental functions. Thus, it yields from (3.1) and (3.2) that

m 21 e p ( z + η ) = a 3 f z 1 ( z + η ) + a 4 f z 2 ( z + η ) = − m 12 ( a 3 q z 1 + a 4 q z 2 ) e − q , m 11 e q ( z + η ) = a 1 g z 1 ( z + η ) + a 2 g z 2 ( z + η ) = − m 22 ( a 1 p z 1 + a 2 p z 2 ) e − p ,

i.e.,

(3.3) m 21 m 12 e p ( z + η ) + q ( z ) = − ( a 3 q z 1 + a 4 q z 2 ) , m 11 m 22 e q ( z + η ) + p ( z ) = − ( a 1 p z 1 + a 2 p z 2 ) .

This implies that p ( z + η ) + q ( z ) = ξ 1 and q ( z + η ) + p ( z ) = ξ 2 , where ξ 1 and ξ 2 are the constants in C . Otherwise, we can obtain a contradiction that the left-hand side of both equations in (3.3) are transcendental and the right-hand side are polynomials since p and q are polynomials. Hence, we have that p ( z + 2 η ) − p ( z ) = ξ 1 − ξ 2 and q ( z + 2 η ) − q ( z ) = ξ 2 − ξ 1 . By Lemma 3.3, we have p ( z ) = L ( z ) + H ( Z ) + B 1 and q ( z ) = − L ( z ) − H ( Z ) + B 2 , where L ( z ) = A 1 z 1 + A 2 z 2 , H ( Z ) is a polynomial in Z ≔ c 2 z 1 − c 1 z 2 , and A 1 , A 2 , B 1 , B 2 are the constants in C . Substituting these into (3.3), we have

(3.4) m 21 m 12 e L ( η ) + B 1 + B 2 = ( a 3 A 1 + a 4 A 2 ) + ( a 3 c 2 − a 4 c 1 ) H ′ ( Z ) ,

and

(3.5) m 11 m 22 e − L ( η ) + B 1 + B 2 = − ( a 1 A 1 + a 2 A 2 ) − ( a 1 c 2 − a 2 c 1 ) H ′ ( Z ) ,

where L ( η ) = A 1 c 1 + A 2 c 2 , m 11 m 12 = m 1 , m 21 m 22 = m 2 , a 1 , a 2 , a 3 , a 4 ∈ C − { 0 } . This implies that ( a 3 c 2 − a 4 c 1 ) H ′ ( Z ) , ( a 1 c 2 − a 2 c 1 ) H ′ ( Z ) are the constants. Set D 1 = a 1 c 2 − a 2 c 1 , D 2 = a 3 c 2 − a 4 c 1 .

Now, we will consider three cases as follows.

(i) If D 1 ≠ 0 and D 2 ≠ 0 , then H ′ ( Z ) ≡ ϑ 1 , ϑ 1 ∈ C . Thus, it follows that H ( Z ) = ϑ 1 ( c 2 z 1 − c 1 z 2 ) + ϑ 0 , where ϑ 0 ∈ C . Then, L ( z ) + H ( Z ) is still a linear form of z 1 and z 2 . Hence, we still denote L ( z ) + H ( Z ) as L ( z ) . Thus, we have p ( z ) = L ( z ) + B 1 and q ( z ) = − L ( z ) + B 2 , where L ( z ) = A 1 z 1 + A 2 z 2 , and A 1 , A 2 , B 1 , and B 2 are the constants in C . By combining with (3.1), (3.2), (3.4), (3.5), we obtain

(3.6) ( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) − L ( η ) − B 2 , m 22 e − L ( z ) + L ( η ) − B 1 ) ,

here and there

(3.7) e 2 ( B 1 + B 2 ) = − ( m 12 m 22 ) 2 K 1 K 2 m 1 m 2 , e 2 L ( η ) = − m 1 K 2 m 2 K 1 , e L ( η ) + B 1 + B 2 = m 12 m 22 K 2 m 2 ,

where K 1 = a 1 A 1 + a 2 A 2 , K 2 = a 3 A 1 + a 4 A 2 .

(ii) If D 1 = 0 and D 2 ≠ 0 or D 1 ≠ 0 and D 2 = 0 , then we can still obtain H ′ ( Z ) ≡ ϑ 1 , ϑ 1 ∈ C . Thus, it follows that H ( Z ) = ϑ 1 ( c 2 z 1 − c 1 z 2 ) + ϑ 0 , where ϑ 0 ∈ C . Then, L ( z ) + H ( Z ) is still a linear form of z 1 and z 2 . Hence, we still denote L ( z ) + H ( Z ) as L ( z ) . Thus, we have p ( z ) = L ( z ) + B 1 and q ( z ) = − L ( z ) + B 2 , where L ( z ) = A 1 z 1 + A 2 z 2 , A 1 , A 2 , B 1 , and B 2 are the constants in C . Using the same argument as in Case (i), we can obtain the conclusion (ii) of Theorem 2.1.

(iii) If D 1 = 0 and D 2 = 0 , then it follows from (3.4) and (3.5) that

(3.8) m 21 m 12 e L ( η ) + B 1 + B 2 = a 3 A 1 + a 4 A 2 , m 11 m 22 e − L ( η ) + B 1 + B 2 = − ( a 1 A 1 + a 2 A 2 ) .

By combining with (3.1), (3.2), and (3.8), we obtain

(3.9) ( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) + H ( Z ) − L ( η ) − B 2 , m 22 e − L ( z ) − H ( Z ) + L ( η ) − B 1 ) ,

where L ( z ) , L ( η ) , B 1 , and B 2 satisfy (3.7).

Therefore, we complete the proof of Theorem 2.1.

3.2 Proof of Corollary 2.1

Let ( f , g ) be a pair of transcendental entire solutions of system (2.5). By observing the form of system (2.5), it is easy to obtain that f z 1 ( z 1 , z 2 ) and g z 1 ( z 1 , z 2 ) have no zeros and poles. Thus, by Lemmas 3.1 and 3.2, there exist two polynomials p and q in C 2 such that

(3.10) g z 1 ( z 1 , z 2 ) = m 11 e q , f ( z + η ) = m 12 e − q ,

and

(3.11) f z 1 ( z 1 , z 2 ) = m 21 e p , g ( z + η ) = m 22 e − p .

Obviously, p and q are nonconstant polynomials; otherwise, f ( z + η ) or g ( z + η ) is a constant. This is a contradiction with the assumption of f and g being transcendental functions. Thus, it yields from (3.10) and (3.11) that

m 21 e p ( z + η ) = f z 1 ( z + η ) = − m 12 q z 1 e − q ,

m 11 e q ( z + η ) = g z 1 ( z + η ) = − m 22 p z 1 e − p ,

i.e.,

(3.12) m 21 m 12 e p ( z + η ) + q ( z ) = − q z 1 , m 11 m 22 e q ( z + η ) + p ( z ) = − p z 1 .

This implies that p ( z + η ) + q ( z ) = ξ 1 and q ( z + η ) + p ( z ) = ξ 2 , where ξ 1 and ξ 2 are the constants in C . Otherwise, we can obtain a contradiction that the left-hand side of both equations in (3.12) are transcendental and the right-hand side are polynomials since p and q are polynomials. Hence, we have that p ( z + 2 η ) − p ( z ) = ξ 1 − ξ 2 and q ( z + 2 η ) − q ( z ) = ξ 2 − ξ 1 . By Lemma 3.3, we have p ( z ) = L ( z ) + H ( Z ) + B 1 and q ( z ) = − L ( z ) − H ( Z ) + B 2 , where L ( z ) = A 1 z 1 + A 2 z 2 , H ( Z ) is a polynomial in Z ≔ c 2 z 1 − c 1 z 2 , and A 1 , A 2 , B 1 , B 2 are the constants in C . Substituting these into (3.12), we have

(3.13) m 21 m 12 e L ( η ) + B 1 + B 2 = A 1 + c 2 H ′ ( Z ) , m 11 m 22 e − L ( η ) + B 1 + B 2 = − A 1 − c 2 H ′ ( Z ) ,

where L ( η ) = A 1 c 1 + A 2 c 2 , m 11 m 12 = m 1 , m 21 m 22 = m 2 . This implies that c 2 H ′ ( Z ) is a constant. Now, two cases can be considered in the following:

(i) If c 2 ≠ 0 , then H ′ ( Z ) ≡ ϑ 1 , ϑ 1 ∈ C . Thus, it follows that H ( Z ) = ϑ 1 ( c 2 z 1 − c 1 z 2 ) + ϑ 0 , where ϑ 0 ∈ C . Then, L ( z ) + H ( Z ) is still a linear form of z 1 and z 2 . Hence, we still denote L ( z ) + H ( Z ) as L ( z ) . Thus, we have p ( z ) = L ( z ) + B 1 and q ( z ) = − L ( z ) + B 2 , where L ( z ) = A 1 z 1 + A 2 z 2 , and A 1 , A 2 , B 1 , and B 2 are the constants in C . By combining with (3.10), (3.11), and (3.13), we obtain

(3.14) ( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) − L ( η ) − B 2 , m 22 e − L ( z ) + L ( η ) − B 1 ) ,

where

(3.15) e 2 ( B 1 + B 2 ) = − ( m 12 m 22 A 1 ) 2 m 1 m 2 , e 2 L ( η ) = − m 1 m 2 , e L ( η ) + B 1 + B 2 = m 12 m 22 A 1 m 2 .

(ii) If c 2 = 0 , then it follows from (3.13) that

(3.16) m 21 m 12 e L ( η ) + B 1 + B 2 = A 1 , m 11 m 22 e − L ( η ) + B 1 + B 2 = − A 1 .

By combining with (3.10), (3.11), and (3.16), we obtain

(3.17) ( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) + H ( z 2 ) − L ( η ) − B 2 , m 22 e − L ( z ) − H ( z 2 ) + L ( η ) − B 1 ) ,

where L ( z ) , L ( η ) , B 1 , and B 2 satisfy (3.15).

Thus, this completes the proof of Corollary 2.1.

3.3 Proof of Corollary 2.2

Let ( f , g ) be a pair of transcendental entire solutions of system (2.6). By observing the form of system (2.6), it is easy to obtain that f z 1 + f z 2 , g z 1 + g z 2 have no zeros and poles. Thus, by Lemmas 3.1 and 3.2, there exist two polynomials p and q in C 2 such that

(3.18) g z 1 + g z 2 = m 11 e q , f ( z + η ) = m 12 e − q ,

and

(3.19) f z 1 + f z 2 = m 21 e p , g ( z + η ) = m 22 e − p .

m 21 e p ( z + η ) = f z 1 ( z + η ) + f z 2 ( z + η ) = − m 12 ( q z 1 + q z 2 ) e − q , m 11 e q ( z + η ) = g z 1 ( z + η ) + g z 2 ( z + η ) = − m 22 ( p z 1 + p z 2 ) e − p ,

i.e.,

(3.20) m 21 m 12 e p ( z + η ) + q ( z ) = − ( q z 1 + q z 2 ) , m 11 m 22 e q ( z + η ) + p ( z ) = − ( p z 1 + p z 2 ) .

This implies that p ( z + η ) + q ( z ) = ξ 1 and q ( z + η ) + p ( z ) = ξ 2 , where ξ 1 and ξ 2 are the constants in C . Otherwise, we can obtain a contradiction that the left-hand side of both equations in (3.20) are transcendental and the right-hand side are polynomials since p and q are polynomials. Hence, we have that p ( z + 2 η ) − p ( z ) = ξ 1 − ξ 2 and q ( z + 2 η ) − q ( z ) = ξ 2 − ξ 1 . By Lemma 3.3, we have p ( z ) = L ( z ) + H ( Z ) + B 1 and q ( z ) = − L ( z ) − H ( Z ) + B 2 , where L ( z ) = A 1 z 1 + A 2 z 2 , H ( Z ) is a polynomial in Z ≔ c 2 z 1 − c 1 z 2 , and A 1 , A 2 , B 1 , and B 2 are the constants in C . Substituting these into (3.20), we have

(3.21) m 21 m 12 e L ( η ) + B 1 + B 2 = ( A 1 + A 2 ) + ( c 2 − c 1 ) H ′ ( Z )

and

(3.22) m 11 m 22 e − L ( η ) + B 1 + B 2 = − ( A 1 + A 2 ) − ( c 2 − c 1 ) H ′ ( Z ) ,

where L ( η ) = A 1 c 1 + A 2 c 2 , m 11 m 12 = m 1 , m 21 m 22 = m 2 . This implies that ( c 2 − c 1 ) H ′ ( Z ) is a constant. Now, two cases can be considered in the following:

(i) If c 2 ≠ c 1 , then H ′ ( Z ) ≡ ϑ 1 , ϑ 1 ∈ C . Thus, it follows that H ( Z ) = ϑ 1 ( c 2 z 1 − c 1 z 2 ) + ϑ 0 , where ϑ 0 ∈ C . Then, L ( z ) + H ( Z ) is still a linear form of z 1 and z 2 . Hence, we still denote L ( z ) + H ( Z ) as L ( z ) . Thus, we have p ( z ) = L ( z ) + B 1 and q ( z ) = − L ( z ) + B 2 , where L ( z ) = A 1 z 1 + A 2 z 2 , and A 1 , A 2 , B 1 , and B 2 are the constants in C . By combining with (3.18), (3.19), (3.21), and (3.22), we obtain

(3.23) ( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) − L ( η ) − B 2 , m 22 e − L ( z ) + L ( η ) − B 1 ) ,

where

(3.24) e 2 ( B 1 + B 2 ) = − [ m 12 m 22 ( A 1 + A 2 ) ] 2 m 1 m 2 , e 2 L ( η ) = − m 1 m 2 , e L ( η ) + B 1 + B 2 = m 12 m 22 ( A 1 + A 2 ) m 2 .

(ii) If c 2 = c 1 , then it follows from (3.21) and (3.22) that

(3.25) m 21 m 12 e L ( η ) + B 1 + B 2 = A 1 + A 2 , m 11 m 22 e − L ( η ) + B 1 + B 2 = − ( A 1 + A 2 ) .

By combining with (3.18), (3.19), and (3.25), we obtain

(3.26) ( f ( z ) , g ( z ) ) = ( m 12 e L ( z ) + H ( z 1 − z 2 ) − L ( η ) − B 2 , m 22 e − L ( z ) − H ( z 1 − z 2 ) + L ( η ) − B 1 ) ,

where L ( z ) , L ( η ) , B 1 , and B 2 satisfy (3.24).

Thus, this completes the proof of Corollary 2.2.

4 Proof of Theorem 2.2

Let ( f , g ) be a pair of finite-order transcendental entire solutions of system (2.2). By observing the form of system (2.2), it is easy to obtain that a 1 f z 1 + a 2 g z 1 , a 3 f z 2 + a 4 g z 2 have no zeros and poles. Thus, by Lemmas 3.1 and 3.2, there exist two polynomials p and q in C 2 such that

(4.1) a 1 f z 1 + a 2 g z 1 = m 11 e p , f ( z + η ) = m 12 e − p ,

and

(4.2) a 3 f z 2 + a 4 g z 2 = m 21 e q , g ( z + η ) = m 22 e − q .

m 11 e p ( z + η ) = a 1 f z 1 ( z + η ) + a 2 g z 1 ( z + η ) = − a 1 m 12 p z 1 e − p − a 2 m 22 q z 1 e − q , m 21 e q ( z + η ) = a 3 f z 2 ( z + η ) + a 4 g z 2 ( z + η ) = − a 3 m 12 p z 2 e − p − a 4 m 22 q z 2 e − q ,

i.e.,

(4.3) e p ( z + η ) + p ( z ) = − a 1 m 12 m 11 p z 1 − a 2 m 22 m 11 q z 1 e p − q ,

(4.4) e q ( z + η ) + q ( z ) = − a 4 m 22 m 21 q z 2 − a 3 m 12 m 21 p z 2 e q − p .

Suppose that p − q is a constant. Observing the form of (4.3), it is easy to obtain that p ( z + η ) + p ( z ) is a constant. But according to (4.1), we know that p cannot be a constant. Thus, this is a contradiction.
Suppose that p − q is a nonconstant. We can easily know that q − p is also a nonconstant. Now three cases will be considered below.

Case 1. Suppose that q z 1 ≡ 0 . By combining with (4.3), we obtain

e p ( z + η ) + p ( z ) = − a 1 m 12 m 11 p z 1 ,

which means that p ( z + η ) + p ( z ) and − a 1 m 12 m 11 p z 1 are all constants. This is a contradiction with the assumption of f and g being transcendental functions.

Case 2. Suppose that p z 2 ≡ 0 . By combining with (4.4), we obtain

e q ( z + η ) + q ( z ) = − a 4 m 22 m 21 q z 2 ,

which means that q ( z + η ) + q ( z ) and − a 4 m 22 m 21 q z 2 are all constants. This is a contradiction with the assumption of f and g being transcendental functions.

Case 3. Suppose that q z 1 ≢ 0 and p z 2 ≢ 0 . By combining with (4.3) and (4.4), we obtain

deg p ⩾ deg q , deg q ⩾ deg p .

i.e.,

deg p = deg q .

We make G = e p ( z + η ) + p ( z ) . Using the Nevanlinna second fundamental theorem, we have from (4.3) that

T ( r , G ) ⩽ N ( r , G ) + N r , 1 G + N r , 1 G + a 1 m 12 m 11 p z 1 + S ( r , G ) ⩽ N r , m 11 − a 2 m 22 q z 1 e p − q + S ( r , G ) ⩽ O ( log r ) + S ( r , G ) = o ( T ( r , G ) ) ,

which is impossible.

Therefore, we complete the proof of Theorem 2.2.

5 Proof of Theorem 2.3

Let ( f , g ) be a pair of finite-order transcendental entire solutions of system (2.3). By observing the form of system (2.3), it is easy to obtain that a 1 f z 2 + a 2 g z 1 , a 3 f z 1 + a 4 g z 2 have no zeros and poles. Thus, by Lemmas 3.1 and 3.2, there exist two polynomials p and q in C 2 such that

(5.1) a 1 f z 2 + a 2 g z 1 = m 11 e p , f ( z + η ) = m 12 e − p ,

and

(5.2) a 3 f z 1 + a 4 g z 2 = m 21 e q , g ( z + η ) = m 22 e − q .

m 11 e p ( z + η ) = a 1 f z 2 ( z + η ) + a 2 g z 1 ( z + η ) = − a 1 m 12 p z 2 e − p − a 2 m 22 q z 1 e − q , m 21 e q ( z + η ) = a 3 f z 1 ( z + η ) + a 4 g z 2 ( z + η ) = − a 3 m 12 p z 1 e − p − a 4 m 22 q z 2 e − q ,

i.e.,

(5.3) e p ( z + η ) + p ( z ) = − a 1 m 12 m 11 p z 2 − a 2 m 22 m 11 q z 1 e p ( z ) − q ( z ) ,

(5.4) e q ( z + η ) + q ( z ) = − a 4 m 22 m 21 q z 2 − a 3 m 12 m 21 p z 1 e q ( z ) − p ( z ) .

Suppose that p − q is a constant. By observing the form of (5.3), it is easy to obtain that p ( z + η ) + p ( z ) is a constant. But according to (5.1), we know that p cannot be a constant. Thus, this is a contradiction.
Suppose that p − q is a nonconstant. We can easily know that q − p is also a nonconstant.

Now, three cases will be considered in the following:

Case 1. If q z 1 ≡ 0 . By combining with (5.3), we obtain

e p ( z + η ) + p ( z ) = − a 1 m 12 m 11 p z 2 ,

which means that p ( z + η ) + p ( z ) and − a 1 m 12 m 11 p z 2 are all constants. This is a contradiction with the assumption of f and g being transcendental functions.

Case 2. If p z 1 ≡ 0 . By combining with (5.4), we obtain

e q ( z + η ) + q ( z ) = − a 4 m 22 m 21 q z 2 ,

which means that q ( z + η ) + q ( z ) and − a 4 m 22 m 21 q z 2 are all constants. This is a contradiction with the assumption of f and g being transcendental functions.

Case 3. If q z 1 ≢ 0 and p z 1 ≢ 0 . By combining with (5.3) and (5.4), we obtain

deg p ⩾ deg q , deg q ⩾ deg p ,

i.e.,

deg p = deg q .

Let G = e p ( z + η ) + p ( z ) . Using the Nevanlinna second fundamental theorem, we have from (5.3) that

T ( r , G ) < N ( r , G ) + N r , 1 G + N r , 1 G + a 1 m 12 m 11 p z 2 + S ( r , G ) < N r , m 11 − a 2 m 22 q z 1 e p − q + S ( r , G ) < O ( log r ) + S ( r , G ) = o ( T ( r , G ) ) ,

which is impossible.

Therefore, we complete the proof of Theorem 2.3.

Acknowledgment

We express our sincere thanks to the anonymous reviewers for their valuable comments and suggestions.

Funding information: This work was supported by NSF Grant 12161074, the Talent Introduction Research Foundation of Suqian University (106-CK00042/028), Suqian Sci & Tech Program (Grant No. M202206), Qing Lan Project of Jiangsu Province and Suqian Talent Xiongying Plan of Suqian, and the College Student Innovation and Entrepreneurship Training Program Project (No. 202414160013Z).
Author contributions: XJ and HYX completed the main part of this article. XJ, YBC, and HYX corrected the main theorems. All authors gave the final approval for publication.
Conflict of interest: The authors declare that none of the authors have any competing interests in the manuscript.
Ethical approval: The conducted research is not related to either human or animal use.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2024-04-28

Revised: 2024-12-11

Accepted: 2025-03-17

Published Online: 2025-05-22

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Keywords for this article

Nevanlinna theory; entire function; system of functional equations; several complex variables

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