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Characterizations of entire solutions for the system of Fermat-type binomial and trinomial shift equations in ℂn#

  • Goutam Haldar EMAIL logo and Abhijit Banerjee
Published/Copyright: December 31, 2023
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Demonstratio Mathematica
From the journal Demonstratio Mathematica Volume 56 Issue 1

Abstract

In this article, we investigate the existence and the precise form of finite-order transcendental entire solutions of some system of Fermat-type quadratic binomial and trinomial shift equations in C n . Our results are the generalizations of the results of [H. Y. Xu, S. Y. Liu, and Q. P. Li, Entire solutions for several systems of nonlinear difference and partial differential-difference equations of Fermat-type, J. Math. Anal. Appl. 483 (2020), 123641, 1–22, DOI: https://doi.org/10.1016/j.jmaa.2019.123641.] and [H. Y. Xu and Y. Y. Jiang, Results on entire and meromorphic solutions for several systems of quadratic trinomial functional equations with two complex variables, RACSAM 116 (2022), 8, DOI: https://doi.org/10.1007/s13398-021-01154-9.] to a large extent. Most interestingly, as a consequence of our main result, we have shown that the system of quadratic trinomial shift equation has no solution when it reduces to a system of quadratic trinomial difference equation. In addition, some examples relevant to the content of the article have been exhibited.

Keywords: functions of several complex variables; Fermat-type shift equations; entire solutions; Nevanlinna theory
MSC 2010: 30D35; 39A45

1 Introduction

It is well known that for m ≥ 3 , the Fermat equation x m + y m = 1 does not admit nontrivial solutions in rational numbers, but it does so for m = 2 . We refer the reader to take a glance on [1,2]. Using the Nevanlinna theory [3] as a tool, for the Fermat-type functional equation

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

Montel [4], Iyer [5], and Gross [6] established some remarkable results about the existence of entire and meromorphic solutions of equation (1.1). After that, a number of researchers paid their considerable attentions to study the existence of entire and meromorphic solutions of Fermat-type equation f n + g m = 1 , where f and g are, in general, meromorphic functions and m , n ∈ N (see [7–14]).

In 2004, for m = 2 , replacing g by f ′ in equation (1.1), Yang and Li [14] investigated to find the form of solutions of equation (1.1). They obtained that the transcendental entire solution of f ( z ) 2 + f ′ ( z ) 2 = 1 has the form

f ( z ) = 1 2 A e α z + 1 A e − α z ,

where A and α are nonzero complex constants.

The advent of the difference analogue lemma of the logarithmic derivative (see [15,16]) expedite the research activity to characterize the entire or meromorphic solutions of Fermat-type difference and differential-difference equations (see [17–21]).

With the help of the difference Nevanlinna theory for several complex variables, Cao and Korhonen [22] and Cao and Xu [23] obtained some interesting results on the characterizations of entire and meromorphic solutions for some Fermat-type difference equations and systems of difference equations, which are the extensions from one complex variable to several complex variables. Henceforth, we denote by z + w = ( z 1 + w 1 , z 2 + w 2 , … , z n + w n ) for any z = ( z 1 , z 2 , … , z n ) , w = ( w 1 , w 2 , … , w n ) , c = ( c 1 , c 2 , … , c n ) ∈ C n , the shift of f ( z ) is defined by f ( z + c ) , whereas the difference of f ( z ) is defined by Δ c f ( z ) = f ( z + c ) − f ( z ) (see [24]).

2 Solutions to the system of Fermat-type binomial shift equation in C n

In 2012, Liu et al. [25] proved that the transcendental entire solutions with finite-order of the Fermat-type difference equation f ( z ) 2 + f ( z + c ) 2 = 1 must satisfy f ( z ) = sin ( A z + B ) , where B is a constant and A = ( 4 k + 1 ) π ∕ 2 c , where k is an integer. Xu and Cao [26] have extended the above result to the case of several complex variables as follows.

Theorem A

[26] Let c = ( c 1 , c 2 , … , c n ) ∈ C n ⧹ { ( 0 , 0 , … , 0 ) } . Then, any nonconstant entire solution f : C n → P 1 ( C ) with finite-order of the Fermat-type difference equation f ( z ) 2 + f ( z + c ) 2 = 1 has the form of f ( z ) = cos ( L ( z ) + B ) , where L is a linear function of the form L ( z ) = a 1 z 1 + ⋯ + a n z n on C n such that L ( c ) = − π ∕ 2 − 2 k π ( k ∈ Z ) , and B is a constant on C .

After that, many researchers have studied some variants of the above equation and obtained some remarkable results in the literature (see [27–30]). In 2016, it was Gao [31] who first investigated the existence and form of entire solutions of the system of differential-difference equation

f 1 ′ ( z ) 2 + f 2 ( z + c ) 2 = 1 , f 2 ′ ( z ) 2 + f 1 ( z + c ) 2 = 1

in one complex variable and obtained the pair of finite-order transcendental entire solution ( f 1 ( z ) , f 2 ( z ) ) that satisfies

( f 1 ( z ) , f 2 ( z ) ) = ( sin ( z − i b ) , sin ( z − i b 1 ) ) or ( sin ( z + i b ) , sin ( z + i b 1 ) ) ,

where b and b 1 are constants, and c = k π , where k is a integer.

Inspired by the above result of Gao [31], Xu et al. [32] in 2020 first converted the above Theorem A into the Fermat-type systems of shift equations and obtained the following result.

Theorem B

[32] Let c = ( c 1 , c 2 ) be a constant in C 2 . Then, any pair of transcendental entire solutions with finite-order for the system of Fermat-type difference equations

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

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

  1. L ( c ) = 2 k π i , A 21 = − i , and A 22 = i , or L ( c ) = ( 2 k + 1 ) π i , A 21 = i , and A 22 = − i , here and below k is an integer;

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

From the stand point of Theorem B, it is natural to consider the following system of shift equation, namely the generalized binomial shift equation:

(2.2) f 1 ( z ) 2 + P ( z ) 2 f 2 ( z + c ) 2 = Q ( z ) , f 2 ( z ) 2 + P ( z ) 2 f 1 ( z + c ) 2 = Q ( z ) ,

where f j : C n → P 1 ( C ) be entire for j = 1 , 2 , P ( z ) , and Q ( z ) are nonzero polynomials in C n , and c = ( c 1 , c 2 , … , c n ) ∈ C n .

Before we state the main results of this article, let us set the following:

(2.3) Ψ ( z ) = ∑ i 1 = 1 n C 2 H i 1 2 ( s i 1 2 ) + ∑ i 2 = 1 n C 3 H i 2 3 ( s i 2 3 ) + ∑ i 3 = 1 n C 4 H i 3 4 ( s i 3 4 ) + ⋯ + ∑ i n − 2 = 1 n C n − 1 H i n − 2 n − 1 ( s i n − 2 n − 1 ) + H n − 1 n ( s n − 1 n ) ,

where H i 1 2 is a polynomial in s i 1 2 ≔ d i 1 j 1 2 z j 1 + d i 1 j 2 2 z j 2 with d i 1 j 1 2 c j 1 + d i 1 j 2 2 c j 2 = 0 , 1 ≤ i 1 ≤ C 2 n , and 1 ≤ j 1 < j 2 ≤ n ; H i 2 3 is a polynomial in s i 2 3 ≔ d i 2 j 1 3 z j 1 + d i 2 j 2 3 z j 2 + d i 2 j 3 3 z j 3 with d i 2 j 1 3 c j 1 + d i 2 j 2 3 c j 2 + d i 2 j 3 3 z 3 = 0 , 1 ≤ i 2 ≤ C 3 n , and 1 ≤ j 1 < j 2 < j 3 ≤ n … ; H i n − 2 n − 1 is a polynomial in s i n − 2 n − 1 ≔ d i n − 2 j 1 z j 1 + d i n − 2 j 2 z j 2 + ⋯ + d i n − 2 j n − 1 z j n − 1 with d i n − 2 j 1 c j 1 + d i n − 2 j 2 c j 2 + ⋯ + d i n − 2 j n − 1 c j n − 1 = 0 , 1 ≤ i n − 2 ≤ C n − 1 n , and 1 ≤ j 1 < j 2 < ⋯ < j n − 1 ≤ n ; and H n − 1 n is a polynomial in s n − 1 n ≔ d i n − 1 1 z 1 + d i n − 1 2 z 2 + ⋯ + d i n − 1 n z n with d i n − 1 1 c 1 + d i n − 1 2 c 2 + ⋯ + d i n − 1 n c n = 0 , where for each k , the representation of s i k in terms of the conditions of j 1 , j 2 , … , j k is unique.

Now, we state our first result as follows.

Theorem 2.1

Let c = ( c 1 , c 2 , … , c n ) ∈ C n ⧹ { ( 0 , 0 , … , 0 ) } . If ( f 1 ( z ) , f 2 ( z ) ) be a pair of finite-order transcendental entire solution for the system (2.2), then P ( z ) reduces to a constant, say B with B 2 = 1 ; Q ( z ) = L 1 ( z ) + Ψ ( z ) + ξ 1 , where Ψ ( z ) is a polynomial as defined in equation (2.3); L 1 ( z ) = ∑ = 0 n e j z j with L 1 ( c ) = 0 , e j ∈ C , j = 1 , 2 , … , n , and ( f 1 ( z ) , f 2 ( z ) ) takes one of the following forms:

  1. f 1 ( z ) = 1 2 ( h 1 ( z ) e i ( L ( z ) + Ψ ( z ) + ξ ) + h 3 ( z ) e − i ( L ( z ) + Ψ ( z ) + ξ ) ) f 2 ( z ) = 1 2 A h 1 ( z ) e i ( L ( z ) + Ψ ( z ) + ξ ′ ) + 1 A h 3 ( z ) e − i ( L ( z ) + Ψ ( z ) + ξ ′ ) ,

    where h 1 , and h 3 are polynomials in s 1 with h 1 h 3 = Q , L ( z ) = ∑ j = 1 n α j z j such that e 2 i L ( c ) = − 1 , and e 2 i ( ξ − ξ ′ ) = A 2 and Φ ( z ) is a polynomial defined as in equation (2.3), A ( ≠ 0 ) , ξ , ξ ′ , α j ∈ C .

  2. f 1 ( z ) = 1 2 ( h 1 ( z ) e i ( L ( z ) + Ψ ( z ) + ξ ) + h 3 ( z ) e − i ( L ( z ) + Ψ ( z ) + ξ ) ) f 2 ( z ) = 1 2 1 A h 1 ( z ) e i ( L ( z ) + Ψ ( z ) − ξ ′ ) + A h 3 ( z ) e − i ( L ( z ) + Ψ ( z ) − ξ ′ ) ,

    where h 1 and h 3 are polynomials in s 1 with h 1 h 3 = Q , L ( z ) = ∑ j = 1 n α j z j such that e 2 i L ( c ) = 1 and e 2 i ( ξ + ξ ′ ) = − 1 ∕ A 2 and Φ ( z ) is a polynomial defined as in equation (2.3), A ( ≠ 0 ) , ξ , ξ ′ , α j ∈ C .

Remark 2.2

In Theorem 2.1, let n = 2 , P ( z ) = Q ( z ) = 1 , A = 1 , H ( s ) ≡ 0 , and L ( z ) = i L 1 ( z ) with L 1 ( z ) = a 1 z 1 + a 2 z 2 , ξ = i ξ 1 , and ξ ′ = i ξ 1 ′ , where a 1 , a 2 , ξ 1 , and ξ 1 ′ are all constants in C . Then, we easily obtain Theorem B from Theorem 2.1. Hence, our result is more general than Theorem B.

Remark 2.3

Let Q ( z ) = k , where k is a nonzero constant in C . Then, from Theorem 2.1, it follows that h 1 and h 2 are both constants in C .

Now, we exhibit two examples showing that our results are precise.

Example 2.4

Let L ( z ) = z 1 + z 2 , P ( z ) ≡ 1 , Q ( z ) = ( 2 z 1 + 3 z 2 ) 10 , c = ( c 1 , c 2 ) = ( 2 k + 1 ) 3 π 2 , − ( 2 k + 1 ) π , and k being an integer. In addition, let Ψ ( z ) = z 1 + 3 2 z 2 3 , and ξ and ξ ′ in C such that ξ − ξ ′ = k 1 π , k 1 ∈ C . Then, one can easily verify that ( f 1 , f 2 ) , where

f 1 ( z ) = 1 2 ( ( 2 z 1 + 3 z 2 ) 4 e i ( L ( z ) + Ψ ( z ) + ξ ) + ( 2 z 1 + 3 z 2 ) 6 e − i ( L ( z ) + Ψ ( z ) + ξ ) ) , f 2 ( z ) = 1 2 ( ( 2 z 1 + 3 z 2 ) 4 e i ( L ( z ) + Ψ ( z ) + ξ ′ ) + ( 2 z 1 + 3 z 2 ) 6 e − i ( L ( z ) + Ψ ( z ) + ξ ′ ) ) ,

is a solution to the system (2.2).

Example 2.5

Let Q ( z ) = ( 2 z 1 − 3 z 2 + z 3 ) 7 , Ψ ( z ) = ( z 1 + z 2 − z 3 ) 3 , P ( z ) ≡ 1 , and c = ( c 1 , c 2 , c 3 ) = ( 2 , 3 , 5 ) . Let L ( z ) = α 1 z 1 + α 2 z 2 + α 3 z 3 be such that 2 α 1 + 3 α 2 + 5 α 3 = k π , and choose ξ , ξ ′ in C such that ξ + ξ ′ = ( 2 k 1 + 1 ) π ∕ 2 , k , and k 1 being integers and α 1 , α 2 , and α 3 are constants in C . Then, one can easily verify that ( f 1 , f 2 ) , with

f 1 ( z ) = 1 2 [ ( 2 z 1 − 3 z 2 + z 3 ) 3 e i ( L ( z ) + Ψ ( z ) + ξ ) + ( 2 z 1 − 3 z 2 + z 3 ) 4 e − i ( L ( z ) + Ψ ( z ) + ξ ) ] , f 2 ( z ) = 1 2 [ ( 2 z 1 − 3 z 2 + z 3 ) 3 e i ( L ( z ) + Ψ ( z ) − ξ ′ ) + ( 2 z 1 − 3 z 2 + z 3 ) 4 e − i ( L ( z ) + Ψ ( z ) − ξ ′ ) ] ,

is a solution to the system (2.2).

3 Existence of solutions of quadratic trinomial shift equation in C n

Let us recall another quadratic trinomial function equation

(3.1) f ( z ) 2 + 2 α f ( z ) g ( z ) + g ( z ) 2 = 1 ,

where α is a constant in C . Note that when α = 0 , equation (3.1) is exactly the equation (1.1) with m = 2 . So, it will be interesting to investigate the existence and forms of entire and meromorphic solutions of equation (3.1) when α ≠ 0 . In this direction, Saleeby [33] investigated the entire and meromorphic solutions of equation (3.1) on C n and discovered that the transcendental entire solutions of equation (3.1), α 2 ≠ 1 , must be of the form f ( z ) = 1 2 cos ( h ( z ) ) 1 + α + sin ( h ( z ) ) 1 − α and g ( z ) = 1 2 cos ( h ( z ) ) 1 + α − sin ( h ( z ) ) 1 − α , where h is entire C n . The meromorphic solutions of equation (3.1) must be of the form f ( z ) = α 1 − α 2 β ( z ) 2 ( α 1 − α 2 ) β ( z ) and g ( z ) = 1 − β ( z ) 2 ( α 1 − α 2 ) β ( z ) , where β ( z ) is meromorphic in C n and α 1 = − α + α 2 − 1 , α 2 = − α − α 2 − 1 .

In 2016, Liu and Yang [34] investigated the existence and the form of entire solutions of some quadratic trinomial functional equations in the complex plane C , and obtained the following results.

Theorem C

[34] If α ≠ 0 , ± 1 , then equation

f ( z ) 2 + 2 α f ( z ) f ′ ( z ) + f ′ ( z ) 2 = 1

has no transcendental meromorphic solutions.

Theorem D

[34] If α ≠ 0 , ± 1 , then the transcendental entire solutions with finite-order of the equation

(3.2) f ( z ) 2 + 2 α f ( z ) f ( z + c ) + f ( z + c ) 2 = 1

must be of order 1.

Recently, corresponding to equation (3.2), Xu and his co-authors [35,36] have extended Theorems C and D from the quadratic trinomial shift equation to systems of trinomial difference equations in one as well as several complex variables. We list a result corresponding to several complex variables below.

Theorem E

[36] Let c = ( c 1 , c 2 ) ∈ C 2 . Then, any pair of finite-order transcendental entire solutions for the system of trinomial difference equations

(3.3) f 1 ( z ) 2 + 2 α f 1 ( z ) f 2 ( z + c ) + f 2 ( z + c ) 2 = 1 , f 2 ( z ) 2 + 2 α f 2 ( z ) f 1 ( z + c ) + f 1 ( z + c ) 2 = 1

must be one of the following forms:

  1. f 1 ( z ) = 1 2 cos ( γ ( z ) + b 1 ) 1 + α + sin ( γ ( z ) + b 1 ) 1 − α ,

    f 2 ( z ) = 1 2 cos ( γ ( z ) + b 2 ) 1 + α + sin ( γ ( z ) + b 2 ) 1 − α ,

    where γ ( z ) = L ( z ) + H ( s ) , L ( z ) = a 1 z 1 + a 2 z 2 , L ( c ) ≔ a 1 c 1 + a 2 c 2 , a 1 , a 2 , b 1 , b 2 ∈ C , H ( s ) is a polynomial in s ≔ c 2 z 1 − c 1 z 2 , and L ( z ) , b 1 , and b 2 satisfy

    e 2 i L ( c ) = − α + α 2 − 1 − α − α 2 − 1 , e 2 i ( b 1 − b 2 ) = 1 .

  2. f 1 ( z ) = 1 2 cos ( γ ( z ) + b 1 ) 1 + α + sin ( γ ( z ) + b 1 ) 1 − α ,

    f 2 ( z ) = 1 2 cos ( γ ( z ) + b 2 ) 1 + α − sin ( γ ( z ) + b 2 ) 1 − α ,

    where γ ( z ) is stated as in I, and L ( z ) , b 1 , and b 2 satisfy

    e 2 i L ( c ) = 1 , e 2 i ( b 1 − b 2 ) = 1 .

From the above discussion, it is natural to ask the following questions.

Question 3.1

Can we extend Theorem E from C 2 to C n for any positive integer n ?

We also note that the all the authors in this specific field used the term “difference” equations where as all equations are mainly governed by the shift of the function and there is hardly any presence of the difference. Hence, the next question is most relevant.

Question 3.2

What can be said about the pair of entire solutions of the trinomial difference equation

(3.4) f 1 ( z ) 2 + 2 α f 1 ( z ) Δ c f 2 + ( Δ c f 2 ) 2 = 1 f 2 ( z ) 2 + 2 α f 2 ( z ) Δ c f 1 + ( Δ c f 1 ) 2 = 1 ?

Let L ˜ ( f ) = a 1 f ( z + c ) + a 2 f ( z ) and c = ( c 1 , c 2 , … , c n ) ∈ C n , where a 1 ( ≠ 0 ) , a , b , h , a 2 ∈ C . Motivated by the above questions, let us consider the following system of quadratic trinomial shift equation:

(3.5) a f 1 ( z ) 2 + 2 h f 1 ( z ) L ˜ ( f 2 ) + b L ˜ ( f 2 ) 2 = 1 a f 2 ( z ) 2 + 2 h f 2 ( z ) L ˜ ( f 1 ) + b L ˜ ( f 1 ) 2 = 1 .

Motivated by Theorem E, we consider equation (3.5), which is more general setting than equation (3.3) to obtain our main result, which provides the answers of Questions 3.1 and 3.2 in a compact and convenient manner.

Before stating the result, let us first set that

A 1 ≔ 1 1 + α + i 1 − α and A 2 ≔ 1 1 + α − i 1 − α ,

where α = h ∕ a b ( ≠ 0 , ± 1 ) is a constant in C . Now, we state our result as follows.

Theorem 3.3

Let a , b , and h be constants in C such that a b ≠ 0 and h ≠ 0 , ± a b . Then, any pair of finite-order transcendental entire solutions ( f 1 ( z ) , f 2 ( z ) ) of the system (3.5) must be one of the following forms:

  1. f 1 ( z ) = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) ) a b + h + sin ( L ( z ) + Ψ ( z ) ) a b − h , f 2 ( z ) = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + η ) a b + h + sin ( L ( z ) + Ψ ( z ) + η ) a b − h ,

    where L ( z ) = ∑ j = 1 n α j z j , and Ψ ( z ) is a polynomial defined as in equation (2.3), α j ∈ C such that

    e 2 i η = 1 , ( a + b a 2 2 − b a 1 2 ) e i η + 2 a 2 h = 0 ,

    η ∈ C , and L ( z ) satisfies the relation

    e i L ( c ) = b a 1 A 2 e i η a A 1 − b a 2 A 2 e i η = b a 1 A 2 e − i η a A 1 − b a 2 A 2 e − i η = a A 2 e i η − b a 2 A 1 b a 1 A 1 = a A 2 − b a 2 A 1 e i η b a 1 A 1 e i η .

  2. f 1 ( z ) = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + ξ ) a b + h + sin ( L ( z ) + Ψ ( z ) + ξ ) a b − h , f 2 ( z ) = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + ξ − η ) a b + h + sin ( L ( z ) + Ψ ( z ) + ξ − η ) a b − h ,

    where L ( z ) = ∑ j = 1 n α j z j , and Ψ ( z ) is a polynomial defined as in equation (2.3), α j ∈ C such that

    e 2 i η = 1 , a + b a 2 2 − b a 1 2 = 2 a b a 2 e i η ,

    ξ , η ∈ C , and L ( z ) satisfies the relation

    e i L ( c ) = a − b a 2 e i η b a 1 e i η = b a 1 e i η a − b a 2 e i η = b a 1 e − i η a − b a 2 e − i η = a − b a 2 e − i η b a 1 e − i η .

Remark 3.4

In Theorem 2.1, let n = 2 , a = b = 1 , a 1 = 1 , a 2 = 0 , and c = ( c 1 , c 2 ) ∈ C 2 with d 1 = c 2 and d 2 = c 1 . Choose η ∈ C be such that e 2 i η = 1 . Then, one can easily obtain Theorem E from Theorem 3.3. Therefore, Theorem 3.3 is more general than Theorem E.

Remark 3.5

Let a = b = 1 , a 1 = 1 , a 2 = − 1 , and α 2 ≠ 0 , 1 , where α ∈ C . Then, equation (3.5) becomes equation (3.4). Then, one can easily obtain the following Corollary.

Corollary 3.6

The system of trinomial difference equation (3.4) does not possess any pair of finite-order transcendental entire solutions.

The following examples show that our result is precise.

Example 3.7

In Theorem 3.3, let a = b = 1 , h = 2 , a 1 = 6 , a 2 = 1 , and e 2 i η = 1 . Suppose L ( z ) = z 1 + 2 z 2 + 3 z 3 and Ψ ( z ) = ( z 1 − z 2 + z 3 ) 10 . Choose c = ( c 1 , c 2 , c 3 ) ∈ C 3 be such that

c 1 + 2 c 2 + 3 c 3 = − i log 3 + 1 − 2 c 1 − c 2 + c 3 = 0 .

Then, one can easily verify that ( f 1 , f 2 ) is a pair of transcendental entire solution of equation (3.5), where

f 1 ( z ) = 1 2 cos ( L ( z ) + Ψ ( z ) ) 3 − i sin ( L ( z ) + Ψ ( z ) )

and

f 2 ( z ) = 1 2 cos ( L ( z ) + Ψ ( z ) + η ) 3 − i sin ( L ( z ) + Ψ ( z ) + η ) .

Example 3.8

In Theorem 3.3, let a = 2 , b = 1 , h = 2 , a 1 = 2 , a 2 = 4 , and e 2 i η = − 1 . Suppose L ( z ) = 2 z 1 + 3 z 2 + 4 z 3 and Ψ ( z ) = ( c 2 c 3 z 1 − 2 c 3 c 1 z 2 + c 1 c 2 z 3 ) 11 . Choose c = ( c 1 , c 2 , c 3 ) ∈ C 3 be such that

e i L ( c ) = − 2 + 1 + 2 − 1 ( 2 2 + 1 ) 2 − 1 + ( 2 2 − 1 ) 2 + 1 .

Then, one can easily verify that ( f 1 , f 2 ) is a pair of transcendental entire solution of equation (3.5), where

f 1 ( z ) = 1 2 1 2 1 4 cos ( L ( z ) + Ψ ( z ) ) 2 + 2 − i sin ( L ( z ) + Ψ ( z ) ) 2 − 2

and

f 2 ( z ) = − 1 2 1 2 1 4 cos ( L ( z ) + Ψ ( z ) ) 2 + 2 − i sin ( L ( z ) + Ψ ( z ) ) 2 − 2 .

Remark 3.9

If h = ± a b , then equation (3.5) reduces to [ a f 1 ( z ) ± b L ˜ ( f 2 ) ] 2 = 1 and [ a f 2 ( z ) − b L ˜ ( f 1 ) ] 2 = 1 . If a f 1 ( z ) − b L ˜ ( f 2 ) = 1 and a f 2 ( z ) ± b L ˜ ( f 1 ) = 1 , then we must have

( a − b a 2 2 ) f 1 ( z ) = a + b ( a 1 + a 2 ) + 2 b a 1 a 2 f 1 ( z + c ) + b a 1 2 f 1 ( z + 2 c ) , ( a − b a 2 2 ) f 2 ( z ) = a + b ( a 1 + a 2 ) + 2 b a 1 a 2 f 2 ( z + c ) + b a 1 2 f 2 ( z + 2 c ) .

Then, it is easy to find the transcendental entire solutions with finite or infinite-order when a 2 = 0 . For example,

( f 1 ( z ) , f 2 ( z ) ) = b a 1 2 a ( e ( z 1 − 2 z 2 ) π i + z 1 + z 2 ) , b a 1 2 a ( − e ( z 1 − 2 z 2 ) π i + z 1 + z 2 ) , ( f 1 ( z ) , f 2 ( z ) ) = b a 1 2 a ( sin ( e ( z 1 − 2 z 2 ) π i ) + z 1 + z 2 ) , b a 1 2 a ( − sin ( e ( z 1 − 2 z 2 ) π i ) + z 1 + z 2 )

are the solutions of equation (3.5).

4 Proofs of the main results

Before we starting the proof of the main results, we present here some necessary lemmas which will play key role to prove the main results of this article.

Lemma 4.1

[37] Let f j ≢ 0 ( j = 1 , 2 … , m ; m ≥ 3 ) be meromorphic functions on C n such that f 1 , … , f m − 1 are not constants and f 1 + f 2 + ⋯ + f m = 1 , as well as

∑ j = 1 m N n − 1 r , 1 f j + ( m − 1 ) N ¯ ( r , f j ) < λ T ( r , f j ) + O ( log + T ( r , f j ) )

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, f m = 1 .

Lemma 4.2

[37] Let f j ≢ 0   ( j = 1 , 2 , 3 ) be meromorphic functions on C n such that f 1 are not constant and f 1 + f 2 + f 3 = 1 , as well as

∑ j = 1 3 N 2 r , 1 f j + 2 N ¯ ( r , f j ) < λ T ( r , f j ) + O ( log + T ( r , f j ) )

holds for all r outside possibly a set with finite logarithmic measure, where λ < 1 is a positive number. Then, either f 2 = 1 or f 3 = 1 .

Lemma 4.3

[38–40] 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.

Lemma 4.4

[41] 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:

  1. the internal function h is a polynomial and the external function g is of finite-order; or

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

Proof of Theorem 2.1

Let ( f 1 , f 2 ) be a pair of finite-order transcendental entire solution of system (2.2). First, we write equation (2.2) as follows:

(4.1) ( f 1 ( z ) + i P ( z ) f 2 ( z + c ) ) ( f 1 ( z ) − i P ( z ) f 2 ( z + c ) ) = Q ( z ) ( f 2 ( z ) + i P ( z ) f 1 ( z + c ) ) ( f 2 ( z ) − i P ( z ) f 1 ( z + c ) ) = Q ( z ) .

In view of equation (4.1), Lemmas 4.3 and 4.4, it follows that there exist nonconstant polynomials h 1 ( z ) and h 2 ( z ) , and any nonzero polynomials h 11 ( z ) , h 12 ( z ) , h 21 ( z ) , and h 22 ( z ) in C n with h 11 ( z ) h 12 ( z ) = h 21 ( z ) h 22 ( z ) = Q ( z ) such that

f 1 ( z ) + i P ( z ) f 2 ( z + c ) = h 11 ( z ) e i h 1 ( z ) f 1 ( z ) − i P ( z ) f 2 ( z + c ) = h 12 ( z ) e − i h 1 ( z ) f 2 ( z ) + i P ( z ) f 1 ( z + c ) = h 21 ( z ) e i h 2 ( z ) f 2 ( z ) − i P ( z ) f 1 ( z + c ) = h 22 ( z ) e − i h 2 ( z ) ,

which yield that

(4.2) f 1 ( z ) = 1 2 [ h 11 ( z ) e i h 1 ( z ) + h 12 ( z ) e − i h 1 ( z ) ] ; f 2 ( z + c ) = 1 2 i P ( z ) [ h 11 ( z ) e i h 1 ( z ) − h 12 ( z ) e − i h 1 ( z ) ] ; f 2 ( z ) = 1 2 [ h 21 ( z ) e i h 2 ( z ) + h 22 ( z ) e − i h 2 ( z ) ] ; f 1 ( z + c ) = 1 2 i P ( z ) [ h 21 ( z ) e i h 2 ( z ) − h 22 ( z ) e − i h 2 ( z ) ] .

Now, after some easy calculations, we obtain from equation (4.2) that

(4.3) i P ( z ) h 11 ( z + c ) − h 22 ( z ) e i ( h 1 ( z + c ) + h 2 ( z ) ) + i P ( z ) h 12 ( z + c ) − h 22 ( z ) e − i ( h 1 ( z + c ) − h 2 ( z ) ) + h 21 ( z ) h 22 ( z ) e 2 i h 2 ( z ) = 1 ,

and

(4.4) i P ( z ) h 21 ( z + c ) − h 12 ( z ) e i ( h 2 ( z + c ) + h 1 ( z ) ) + i P ( z ) h 22 ( z + c ) − h 12 ( z ) e − i ( h 2 ( z + c ) − h 1 ( z ) ) + h 11 ( z ) h 12 ( z ) e 2 i h 1 ( z ) = 1 .

By Lemma 4.1, it follows from equations (4.3) and (4.4) that either

i P ( z ) h 11 ( z + c ) − h 22 ( z ) e i ( h 1 ( z + c ) + h 2 ( z ) ) = 1 or i P ( z ) h 12 ( z + c ) − h 22 ( z ) e − i ( h 1 ( z + c ) − h 2 ( z ) ) = 1 ,

and

i P ( z ) h 21 ( z + c ) − h 12 ( z ) e i ( h 2 ( z + c ) + h 1 ( z ) ) = 1 or i P ( z ) h 22 ( z + c ) − h 12 ( z ) e − i ( h 2 ( z + c ) − h 1 ( z ) ) = 1 .

Now, we consider the following four possible cases:

Case 1. Let

(4.5) i P ( z ) h 11 ( z + c ) − h 22 ( z ) e i ( h 1 ( z + c ) + h 2 ( z ) ) = 1 , i P ( z ) h 21 ( z + c ) − h 12 ( z ) e i ( h 2 ( z + c ) + h 1 ( z ) ) = 1 .

Next equations (4.3), (4.4), and (4.5) yield

(4.6) i P ( z ) h 12 ( z + c ) h 21 ( z ) e − i ( h 1 ( z + c ) + h 2 ( z ) ) = 1 , i P ( z ) h 22 ( z + c ) h 11 ( z ) e − i ( h 2 ( z + c ) + h 1 ( z ) ) = 1 .

Since h 1 ( z ) and h 2 ( z ) are two nonconstant polynomials, it yields from equation (4.5) that h 1 ( z + c ) + h 2 ( z ) = η 1 and h 2 ( z + c ) + h 1 ( z ) = η 2 , where η 1 and η 2 are constants in C . Thus, we assume that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ and h 2 = − ( L ( z ) + Ψ ( z ) ) + ξ ′ , where L ( z ) = ∑ j = 1 n α j z j , and Ψ ( z ) is a polynomial defined as in equation (2.3), and α j , ξ , ξ ′ are all constants in C .

From equations (4.5) and (4.6), we obtain P ( z ) 2 Q ( z + c ) = Q ( z ) . This implies that P ( z ) must be constant, say B such that B 2 = 1 . Thus, we must have Q ( z + c ) = Q ( z ) , which implies that Q ( z ) = L 1 ( z ) + Ψ ( z ) + ξ 1 , where Ψ ( z ) is a polynomial as defined in equation (2.3), L 1 ( z ) = ∑ = 0 n e j z j with L 1 ( c ) = 0 and e j ∈ C , j = 1 , 2 , … , n . As Q ( z ) is a periodic function of period c , and Q ( z ) = h 11 ( z ) h 12 ( z ) = h 21 ( z ) h 22 ( z ) , we have h 11 , h 12 , h 21 , and h 22 as c -periodic functions.

Therefore, equations (4.5) and (4.6) give

(4.7) − i B K 1 e i ( L ( c ) + ξ + ξ ′ ) = 1 ; − i B K 2 e i ( − L ( c ) + ξ + ξ ′ ) = 1 ; i B K 2 e − i ( L ( c ) + ξ + ξ ′ ) = 1 ; i B K 1 e − i ( − L ( c ) + ξ + ξ ′ ) = 1 ,

where K 1 = h 11 ( z ) ∕ h 22 ( z ) and K 2 = h 21 ( z ) ∕ h 12 ( z ) , which implies that K 1 = K 2 = A , a nonzero constant in C . Now, from equation (4.7), we can easily obtain the following.

e 2 i ( ξ + ξ ′ ) = − 1 ∕ A 2 , say , and e 2 i L ( c ) = 1 .

Thus, from equation (4.2), we obtain

f 1 = 1 2 ( h 1 ( z ) e i ( L ( z ) + Ψ ( z ) + ξ ) + h 3 ( z ) e − i ( L ( z ) + Ψ ( z ) + ξ ) ) ,

and

f 2 = 1 2 1 A h 1 ( z ) e i ( L ( z ) + Ψ ( z ) − ξ ′ ) + A h 3 ( z ) e − i ( L ( z ) + Ψ ( z ) − ξ ′ ) .

Case 2. Let

(4.8) i P ( z ) h 11 ( z + c ) − h 22 ( z ) e i ( h 1 ( z + c ) + h 2 ( z ) ) = 1 , i P ( z ) h 22 ( z + c ) − h 12 ( z ) e − i ( h 2 ( z + c ) − h 1 ( z ) ) = 1 .

As h 1 ( z ) and h 2 ( z ) are polynomials in C n , from equation (4.8), we have h 1 ( z + c ) + h 2 ( z ) = η 1 and − h 2 ( z + c ) + h 1 ( z ) = η 2 , where η 1 and η 2 are constants in C . This implies that h 1 ( z + 2 c ) + h 1 ( z ) = η 1 + η 2 , which yields that h 1 ( z ) is constant, a contradiction.

Case 3. Let

i P ( z ) h 12 ( z + c ) − h 22 ( z ) e i ( − h 1 ( z + c ) + h 2 ( z ) ) = 1 , i P ( z ) h 21 ( z + c ) − h 12 ( z ) e i ( h 2 ( z + c ) + h 1 ( z ) ) = 1 .

Then, by similar arguments as in Case 2, we obtain a contradiction.

Case 4. Let

(4.9) i P ( z ) h 12 ( z + c ) − h 22 ( z ) e i ( − h 1 ( z + c ) + h 2 ( z ) ) = 1 , i P ( z ) h 22 ( z + c ) − h 12 ( z ) e i ( − h 2 ( z + c ) + h 1 ( z ) ) = 1 .

From equation (4.3), (4.4), and (4.9), we obtain

(4.10) i P ( z ) h 11 ( z + c ) h 21 ( z ) e i ( h 1 ( z + c ) − h 2 ( z ) ) = 1 , i P ( z ) h 21 ( z + c ) h 11 ( z ) e i ( h 2 ( z + c ) − h 1 ( z ) ) = 1 .

Since h 1 ( z ) and h 2 ( z ) are two nonconstant polynomials in C n , equation (4.9) yields that − h 1 ( z + c ) + h 2 ( z ) = η 1 and − h 2 ( z + c ) + h 1 ( z ) = η 2 , where η 1 and η 2 are constants in C . This implies that − h 1 ( z + 2 c ) + h 1 ( z ) = η 1 + η 2 and − h 2 ( z + 2 c ) + h 2 ( z ) = η 1 + η 2 . Thus, we assume that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ and h 2 = L ( z ) + Ψ ( z ) + ξ ′ , where L ( z ) = ∑ j = 1 n α j z j and Ψ ( z ) is a polynomial defined as in equation (2.3), α j , ξ , ξ ′ ∈ C . Then, by similar arguments as in Case 1 of Theorem 2.1, we obtain the conclusion II of Theorem 2.1.□

Proof of Theorem 3.3

Let us make a transformation

(4.11) a f 1 ( z ) = U , b L ( f 2 ) ( z ) = V .

Then, the first equation of equation (3.5) reduces to

(4.12) U 2 + 2 α U V + V 2 = 1 ,

where α = h ∕ a b . Let

(4.13) U = 1 2 ( U 1 − V 1 ) , V = 1 2 ( U 1 + V 1 ) .

Then, equation (4.12) becomes

(4.14) ( 1 + α ) U 1 2 + ( 1 − α ) V 1 2 = 1 .

Now, in view of Lemmas 4.3 and 4.4, we obtain from (4.14) that

1 + α U 1 + i 1 − α V 1 = e i h 1 ( z ) 1 + α U 1 − i 1 − α V 1 = e − i h 1 ( z ) ,

where h 1 ( z ) is a nonconstant polynomial, from which we can obtain that

(4.15) U 1 = e i h 1 ( z ) + e − i h 1 ( z ) 2 1 + α , V 1 = e i h 1 ( z ) − e − i h 1 ( z ) 2 i 1 − α .

Therefore, equations (4.11), (4.13) and (4.15) together imply

(4.16) f 1 ( z ) = 1 2 2 a ( A 1 e i h 1 ( z ) + A 2 e − i h 1 ( z ) ) , L ˜ ( f 2 ) = 1 2 2 b ( A 2 e i h 1 ( z ) + A 1 e − i h 1 ( z ) ) .

Similarly, from the second equation of equation (4.1), we can obtain

(4.17) f 2 ( z ) = 1 2 2 a ( A 1 e i h 2 ( z ) + A 2 e − i h 2 ( z ) ) L ˜ ( f 1 ) = 1 2 2 b ( A 2 e i h 2 ( z ) + A 1 e − i h 2 ( z ) ) ,

where h 2 ( z ) is a nonconstant polynomial in C n .

After simple computations, from equations (4.16) and (4.17), we obtain that

(4.18) b a 1 A 1 e i ( h 1 ( z + c ) + h 2 ( z ) ) + b a 1 A 2 e i ( − h 1 ( z + c ) + h 2 ( z ) ) + b a 2 A 1 e i ( h 2 ( z ) + h 1 ( z ) ) + b a 2 A 2 e i ( h 2 ( z ) − h 1 ( z ) ) − a A 2 e 2 i h 2 ( z ) = a A 1

and

(4.19) b a 1 A 1 e i ( h 2 ( z + c ) + h 1 ( z ) ) + b a 1 A 2 e i ( − h 2 ( z + c ) + h 1 ( z ) ) + b a 2 A 1 e i ( h 1 ( z ) + h 2 ( z ) ) + b a 2 A 2 e i ( h 1 ( z ) − h 2 ( z ) ) − a A 2 e 2 i h 1 ( z ) = a A 1 .

Now, we discuss two possible cases.

Case 1. Let a 2 ≠ 0

Subcase 1.1. Let h 2 ( z ) − h 1 ( z ) = η , where η is a constant in C . Then, it can be easily seen that h 1 ( z ) + h 2 ( z ) and h 2 ( z ) + h 1 ( z + c ) are nonconstant polynomials in C n .

Therefore, equations (4.18) and (4.19), respectively, can be rewritten as follows:

(4.20) b a 1 A 1 e i η e i ( h 1 ( z + c ) + h 1 ( z ) ) + b a 1 A 2 e i η e i ( h 1 ( z ) − h 1 ( z + c ) ) + ( b a 2 A 1 − a A 2 e i η ) e i η e 2 i h 1 ( z ) = a A 1 − b a 2 A 2 e i η

and

(4.21) b a 1 A 1 e i η e i ( h 1 ( z + c ) + h 1 ( z ) ) + b a 1 A 2 e − i η e i ( h 1 ( z ) − h 1 ( z + c ) ) + ( b a 2 A 1 e i η − a A 2 ) e 2 i h 1 ( z ) = a A 1 − b a 2 A 2 e − i η .

Note that equation (4.20) implies b a 2 A 1 − a A 2 e i η and a A 1 − b a 2 A 2 e i η cannot be zero at the same time. Otherwise, from equation (4.20), we obtain that A 1 e 2 i h 1 ( z + c ) = − A 2 , which implies that h 1 ( z + c ) , and hence h 1 ( z ) is constant, a contradiction.

Let b a 2 A 1 − a A 2 e i η ≠ 0 and a A 1 − b a 2 A 2 e i η = 0 . Then, we can write equation (4.20) as

b a 1 A 1 e i ( h 1 ( z + c ) − h 1 ( z ) ) + b a 1 A 2 e − i ( h 1 ( z + c ) + h 1 ( z ) ) = b a 2 A 1 − a A 2 e i η .

By the second main theorem of Nevanlinna for several complex variables, we obtain

T ( r , e − i ( h 1 ( z + c ) + h 1 ( z ) ) ) ≤ N ¯ ( r , e − i ( h 1 ( z + c ) + h 1 ( z ) ) ) + N ¯ r , 1 e − i ( h 1 ( z + c ) + h 1 ( z ) ) + N ¯ r , 1 e − i ( h 1 ( z + c ) + h 1 ( z ) ) − w + S ( r , e − i ( h 1 ( z + c ) + h 1 ( z ) ) ) ≤ N ¯ r , 1 e i ( h 1 ( z + c ) − h 1 ( z ) ) + S ( r , e − i ( h 1 ( z + c ) + h 1 ( z ) ) ) ≤ S ( r , e − i ( h 1 ( z + c ) + h 1 ( z ) ) ) + S ( r , e − i ( h 1 ( z + c ) − h 1 ( z ) ) ) ,

where w = ( b a 2 A 1 − a A 2 e i η ) ∕ b a 1 A 2 . This implies that h 1 ( z + c ) + h 1 ( z ) , and hence h 1 ( z ) is constant, a contradiction.

Similarly, we can obtain a contradiction for the case b a 2 A 1 − a A 2 e i η = 0 and a A 1 − b a 2 A 2 e i η ≠ 0 .

Therefore, b a 2 A 1 − a A 2 e i η ≠ 0 and a A 1 − b a 2 A 2 e i η ≠ 0 . By similar arguments, we obtain that b a 2 A 1 e i η − a A 2 ≠ 0 and a A 1 − b a 2 A 2 e − i η ≠ 0 .

By Lemma 4.1, equations (4.20) and (4.21) yield

(4.22) b a 1 A 2 e i η e − i h 1 ( z + c ) + i h 1 ( z ) = a A 1 − b a 2 A 2 e i η b a 1 A 2 e − i η e − i h 1 ( z + c ) + i h 1 ( z ) = a A 1 − b a 2 A 2 e − i η .

Taking into account equations (4.20), (4.21), and (4.22), we obtain

(4.23) b a 1 A 1 e i h 1 ( z + c ) − i h 1 ( z ) = a A 2 e i η − b a 2 A 1 b a 1 A 1 e i η e i h 1 ( z + c ) − i h 1 ( z ) = a A 2 − b a 2 A 1 e i η .

From (4.22), we conclude that h 1 ( z + c ) − h 1 ( z ) is constant, and hence we assume that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ , where L ( z ) = ∑ j = 1 n α j z j and Ψ ( z ) is a polynomial as defined in equation (2.3), α j , ξ ∈ C for j = 1 , 2 , … , n . Therefore, in view of equations (4.22) and (4.23), we easily obtain

(4.24) b a 1 A 2 e i η e − i L ( c ) = a A 1 − b a 2 A 2 e i η b a 1 A 2 e − i η e − i L ( c ) = a A 1 − b a 2 A 2 e − i η b a 1 A 1 e i L ( c ) = a A 2 e i η − b a 2 A 1 b a 1 A 1 e i η e i L ( c ) = a A 2 − b a 2 A 1 e i η .

Therefore, it follows from equation (4.24) that

e i L ( c ) = b a 1 A 2 e i η a A 1 − b a 2 A 2 e i η = b a 1 A 2 e − i η a A 1 − b a 2 A 2 e − i η = a A 2 e i η − b a 2 A 1 b a 1 A 1 = a A 2 − b a 2 A 1 e i η b a 1 A 1 e i η .

From the last equation, we can obtain that

e 2 i η = 1 , ( a + b a 2 2 − b a 1 2 ) e i η + 2 a 2 h = 0 .

Hence, from equations (4.16) and (4.17), we obtain

f 1 = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) ) a b + h + sin ( L ( z ) + Ψ ( z ) ) a b − h , f 2 = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + η ) a b + h + sin ( L ( z ) + Ψ ( z ) + η ) a b − h .

Subcase 1.2. Let h 2 ( z ) − h 1 ( z ) be nonconstant. Observe that h 2 ( z ) − h 1 ( z + c ) is also nonconstant.

Subcase 1.2.1. Let h 2 ( z ) + h 1 ( z ) = η , where η is a constant in C .

Therefore, equations (4.18) and (4.19), respectively, yield

(4.25) b a 1 A 1 e i ( h 1 ( z + c ) + h 2 ( z ) ) + b a 1 A 2 e i ( − h 1 ( z + c ) + h 2 ( z ) ) + b a 2 A 2 e i ( h 2 ( z ) − h 1 ( z ) ) − a A 2 e 2 i h 2 ( z ) = a A 1 − b a 2 A 1 e i η

and

(4.26) b a 1 A 1 e i ( h 2 ( z + c ) + h 1 ( z ) ) + b a 1 A 2 e i ( − h 2 ( z + c ) + h 1 ( z ) ) + b a 2 A 2 e i ( h 1 ( z ) − h 2 ( z ) ) − a A 2 e 2 i h 1 ( z ) = a A 1 − b a 2 A 1 e i η .

Let a − b a 2 e i η = 0 . From equation (4.25), we obtain

(4.27) b a 1 A 1 e i ( h 1 ( z + c ) − h 2 ( z ) ) + b a 1 A 2 e − i ( h 1 ( z + c ) + h 2 ( z ) ) = a A 2 − b a 2 A 2 e − i η .

Now, from equation (4.27), we claim a A 2 − b a 2 A 2 e − i η ≠ 0 . Otherwise, we must obtain e 2 i h 1 ( z + c ) = − A 2 ∕ A 1 , which implies that h 1 ( z + c ) , and hence h 1 ( z ) is constant, which is impossible. Now, by second main theorem of Nevanlinna for several complex variables, we obtain

T ( r , e i ( h 1 ( z + c ) + h 1 ( z ) ) ) ≤ N ¯ ( r , e i ( h 1 ( z + c ) + h 1 ( z ) ) ) + N ¯ r , 1 e i ( h 1 ( z + c ) + h 1 ( z ) ) + N ¯ r , 1 e i ( h 1 ( z + c ) + h 1 ( z ) ) − w 1 + S ( r , e i ( h 1 ( z + c ) + h 1 ( z ) ) ) ≤ N ¯ r , 1 e i ( h 1 ( z ) − h 1 ( z + c ) ) + S ( r , e i ( h 1 ( z + c ) + h 1 ( z ) ) ) + S r , 1 e i ( h 1 ( z ) − h 1 ( z + c ) ) + S ( r , e i ( h 1 ( z + c ) + h 1 ( z ) ) ) ,

where w 1 = ( a A 2 e i η − b a 2 A 2 ) / b a 1 A 2 . This implies that h 1 ( z + c ) + h 1 ( z ) , and hence h 1 ( z ) is constant, a contradiction.

Hence, a − b a 2 e i η ≠ 0 . Therefore, Lemma 4.1 together with equations (4.25) and (4.26) give

(4.28) b a 1 A 1 e i ( h 1 ( z + c ) + h 2 ( z ) ) = a A 1 − b a 2 A 1 e i η b a 1 A 1 e i ( h 2 ( z + c ) + h 1 ( z ) ) = a A 1 − b a 2 A 1 e i η .

Next, equations (4.25), (4.26), and (4.28) yield

(4.29) b a 1 e − i ( h 1 ( z + c ) + h 2 ( z ) ) = a − b a 2 e − i η b a 1 e − i ( h 2 ( z + c ) + h 1 ( z ) ) = a − b a 2 e − i η .

In view of equation (4.28), we have h 1 ( z + c ) + h 2 ( z ) = η 1 and h 2 ( z + c ) + h 1 ( z ) = η 2 , where η 1 and η 2 are constants in C . Thus, we assume that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ and h 2 ( z ) = − ( L ( z ) + Ψ ( z ) ) + ξ ′ , where L ( z ) = ∑ = 1 n α j z j and Ψ ( z ) is a polynomial as defined in equation (2.3), α j , ξ , ξ ′ ∈ C and η = ξ + ξ ′ .

Therefore, from equations (4.28) and (4.29), we obtain

(4.30) b a 1 e i ( L ( c ) + η ) = a − b a 2 e i η b a 1 e i ( − L ( c ) + η ) = a − b a 2 e i η b a 1 e − i ( L ( c ) + η ) = a − b a 2 e − i η b a 1 e − i ( − L ( c ) + η ) = a − b a 2 e − i η ,

which yield

e i L ( c ) = a − b a 2 e i η b a 1 e i η = b a 1 e i η a − b a 2 e i η = b a 1 e − i η a − b a 2 e − i η = a − b a 2 e − i η b a 1 e − i η ,

e 2 i η = 1 , 2 a b a 2 e i η = a − b a 1 2 + b a 2 2 .

Again, from equations (4.16) and (4.17), we obtain

f 1 = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + ξ ) a b + h + sin ( L ( z ) + Ψ ( z ) + ξ ) a b − h ,

f 2 = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + ξ − η ) a b + h − sin ( L ( z ) + Ψ ( z ) + ξ − η ) a b − h .

Subcase 1.2.2. Let h 2 ( z ) + h 1 ( z ) be nonconstant.

If h 1 ( z + c ) + h 2 ( z ) is nonconstant, then by Lemma 4.1, we obtain from equation (4.18)

b a 1 A 2 e i ( − h 1 ( z + c ) + h 2 ( z ) ) = a A 1 ,

which implies that − h 1 ( z + c ) + h 2 ( z ) is a constant. Let − h 1 ( z + c ) + h 2 ( z ) = η 1 , where η 1 ∈ C . If h 2 ( z + c ) + h 1 ( z ) = η 2 , a complex constant in C , then we must have h 2 ( z + 2 c ) + h 2 ( z ) = η 1 + η 2 , which implies that h 2 ( z ) is constant, a contradiction. Thus, h 2 ( z + c ) + h 1 ( z ) is nonconstant. Then, by Lemma 4.1, we obtain from equation (4.19)

b a 1 A 2 e i ( − h 2 ( z + c ) + h 1 ( z ) ) = a A 1 .

This implies that − h 2 ( z + c ) + h 1 ( z ) = const. = η 3 . But, we must have − h 1 ( z + c ) + h 1 ( z ) = η 1 + η 3 . Thus, we conclude that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ , where L ( z ) = ∑ j = 1 n α j z j , and Ψ ( z ) is a polynomial as defined in equation (2.3), α j , and ξ are all constants in C , and therefore, h 2 ( z ) = L ( z ) + Ψ ( z ) + ξ + η 1 + L ( c ) . This implies h 2 ( z ) − h 1 ( z ) = η 1 + L ( c ) = constant , a contradiction. Hence, h 1 ( z + c ) + h 2 ( z ) = η 4 , a constant in C . Now, if h 2 ( z + c ) + h 1 ( z ) = η 5 , a constant in C , then we must have h 1 ( z + 2 c ) − h 1 ( z ) = η 4 − η 5 =  constant. This implies that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ , where L ( z ) , Ψ ( z ) , and ξ are just defined above. Therefore, h 1 ( z ) = − ( L ( z ) + Ψ ( z ) + ξ ) + η 4 − L ( c ) = constant. Then, h 2 ( z ) + h 1 ( z ) = η 4 − L ( c ) = constant, a contradiction. Thus, h 2 ( z + c ) + h 1 ( z ) is nonconstant.

Then, by Lemma 4.1 and from equation (4.19), we obtain

b a 1 A 2 e i ( − h 2 ( z + c ) + h 1 ( z ) ) = a A 1 ,

which yields that − h 2 ( z + c ) + h 1 ( z ) = constant = η 6 . Then, we must have h 1 ( z + 2 c ) + h 1 ( z ) = η 4 + η 6 = constant. This implies that h 1 ( z ) is constant, which is not possible.

Case 2. Let a 2 = 0 . Then, (4.18) and (4.19), respectively, reduces to

(4.31) b a 1 A 1 e i ( h 1 ( z + c ) + h 2 ( z ) ) + b a 1 A 2 e i ( − h 1 ( z + c ) + h 2 ( z ) ) − a A 2 e 2 i h 2 ( z ) = a A 1 ,

and

(4.32) b a 1 A 1 e i ( h 2 ( z + c ) + h 1 ( z ) ) + b a 1 A 2 e i ( − h 2 ( z + c ) + h 1 ( z ) ) − a A 2 e 2 i h 1 ( z ) = a A 1 .

By Lemma 4.1, we obtain from (4.31) that either

b a 1 A 1 e i ( h 1 ( z + c ) + h 2 ( z ) ) = a A 1 or b a 1 A 1 e i ( − h 1 ( z + c ) + h 2 ( z ) ) = a A 1 .

In a similar manner, from equation (4.32), we obtain that either

b a 1 A 1 e i ( h 2 ( z + c ) + h 1 ( z ) ) = a A 1 or b a 1 A 1 e i ( − h 2 ( z + c ) + h 1 ( z ) ) = a A 1 .

Now, we consider the following four possible cases:

Subcase 2.1. Let

(4.33) b a 1 A 1 e i ( h 1 ( z + c ) + h 2 ( z ) ) = a A 1 b a 1 A 1 e i ( h 2 ( z + c ) + h 1 ( z ) ) = a A 1 .

It follows from equation (4.33) that h 1 ( z + c ) + h 2 ( z ) = η 1 and h 2 ( z + c ) + h 1 ( z ) = η 1 , where η 1 and η 2 are constants in C . This implies that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ and h 2 ( z ) = − ( L ( z ) + Ψ ( z ) ) + ξ ′ , L ( z ) , Ψ ( z ) are defined as in Theorem 3.3 and ξ and ξ ′ are constants in C .

Again, in view of equation (4.31), (4.32), and (4.33), it follows that

(4.34) b a 1 A 2 e − i ( h 1 ( z + c ) + h 2 ( z ) ) = a A 2 b a 1 A 2 e − i ( h 2 ( z + c ) + h 1 ( z ) ) = a A 2 .

From equation (4.33) and (4.34), we obtain

(4.35) b a 1 e i ( L ( c ) + ξ + ξ ′ ) = a b a 1 e i ( − L ( c ) + ξ + ξ ′ ) = a b a 1 e − i ( L ( c ) + ξ + ξ ′ ) = a b a 1 e i ( − L ( c ) + ξ + ξ ′ ) = a ,

from which we can easily deduce the following:

e 2 i L ( c ) = 1 , e 2 i η = 1 , b a 1 2 = a with η = ξ + ξ ′ .

Therefore, from equations (4.16) and (4.17), we obtain

f 1 = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + ξ ) a b + h + sin ( L ( z ) + Ψ ( z ) + ξ ) a b − h ,

f 2 = 1 2 b a 1 4 cos ( L ( z ) + Ψ ( z ) + ξ − η ) a b + h − sin ( L ( z ) + Ψ ( z ) + ξ − η ) a b − h ,

which is the conclusion II of Theorem 3.3 with a 2 = 0 .

Subcase 2.2.

Let

b a 1 A 1 e i ( h 1 ( z + c ) + h 2 ( z ) ) = a A 1 b a 1 A 2 e i ( − h 2 ( z + c ) + h 1 ( z ) ) = a A 1 .

From the above equations, we obtain h 1 ( z + c ) + h 2 ( z ) = η 1 and − h 2 ( z + c ) + h 1 ( z ) = η 2 , where η 1 and η 2 are two constants in C . This implies h 1 ( z + 2 c ) + h 1 ( z ) = η 1 + η 2 , which yields that h 1 ( z ) is constant, a contradiction.

Subcase 2.3.

Let

b a 1 A 2 e i ( − h 1 ( z + c ) + h 2 ( z ) ) = a A 1 b a 1 A 1 e i ( h 2 ( z + c ) + h 1 ( z ) ) = a A 1 .

Similar to the arguments of Subcase 2.2, we obtain a contradiction.

Subcase 2.4.

Let

(4.36) b a 1 A 2 e i ( − h 1 ( z + c ) + h 2 ( z ) ) = a A 1 b a 1 A 2 e i ( − h 2 ( z + c ) + h 1 ( z ) ) = a A 1 .

In view of equations (4.31), (4.32), and (4.36), we deduce

(4.37) b a 1 A 1 e i ( h 1 ( z + c ) − h 2 ( z ) ) = a A 2 b a 1 A 1 e i ( h 2 ( z + c ) − h 1 ( z ) ) = a A 2 .

Now, from equation (4.36), we know − h 1 ( z + c ) + h 2 ( z ) = η 1 and − h 2 ( z + c ) + h 1 ( z ) = η 2 , where η 1 , η 2 ∈ C . This implies h 1 ( z ) − h 1 ( z + 2 c ) = η 1 + η 2 = h 2 ( z ) − h 2 ( z + 2 c ) . Therefore, we assume that h 1 ( z ) = L ( z ) + Ψ ( z ) + ξ and h 2 ( z ) = L ( z ) + Ψ ( z ) + ξ ′ , where L ( z ) and Ψ ( z ) are defined as in Theorem 3.3 and ξ , ξ ′ ∈ C .

Next, from equations (4.36) and (4.37), we obtain

(4.38) b a 1 A 2 e i ( − L ( c ) + ξ ′ − ξ ) = a A 1 b a 1 A 2 e i ( − L ( c ) + ξ − ξ ′ ) = a A 1 b a 1 A 1 e i ( L ( c ) + ξ − ξ ′ ) = a A 2 b a 1 A 1 e i ( L ( c ) + ξ ′ − ξ ) = a A 2 ,

from which we can easily obtain the following:

e 2 i η = 1 , b a 1 2 = a , e 2 i L ( c ) = A 2 2 A 1 2 , where ξ − ξ ′ = η .

In this case, the form of the pair of finite-order transcendental entire solutions ( f 1 ( z ) , f 2 ( z ) ) will be same as I of Theorem 3.3 with a 2 = 0 .□

Proof of Corollary 3.6

Suppose a = b = a 1 = 1 and a 2 = − 1 . Then, from the conclusion I of Theorem 3.3, we obtain that

(4.39) e i L ( c ) = A 2 e i η A 1 + A 2 e i η = A 2 e − i η A 1 + A 2 e − i η = A 2 e i η + A 1 A 1 = A 2 + A 1 e i η A 1 e i η .

From first and second of equation (4.39), we obtain that e 2 i η = 1 . From first and fourth of equation (4.39), we obtain A 1 2 + A 2 2 = 0 , i.e., α = 0 , which is a contradiction.

Again, from the conclusion II of Theorem 3.3, we obtain that 4 = 1 , which is a contradiction.

Hence, the system of equation (3.4) does not possess any pair of finite-order transcendental entire solutions.□

# This work was completed with the support of TEX-pert.

Acknowledgment

The authors would like to thank the referee(s) for the helpful suggestions and comments to improve the exposition of the article.

  1. Funding information: The authors have not received any kind of fund for this research work.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors declare that they have no conflict of interest.

  4. Human/animals participants: The authors declare that there is no research involving human participants and/or animals in this work.

  5. Data availability statement: Data sharing is not applicable to this article as no database were generated or analyzed during the current study.

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Received: 2022-10-22
Revised: 2023-05-15
Accepted: 2023-07-05
Published Online: 2023-12-31

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2023-0104
Keywords for this article
functions of several complex variables; Fermat-type shift equations; entire solutions; Nevanlinna theory
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