Entire Solutions of the Second-Order Fermat-Type Differential-Difference Equation

In this work, we assume that the readers are familiar with general definitions and fundamental theories of Nevanlinna theory [1–3]. f is a meromorphic function which means f is meromorphic in the finite complex plane C. If f has no poles, we call f is an entire function. We denote by S(r, f), any function satisfyingS(r, f) � o(T(r, f)), r⟶∞, outside of a possible exceptional set of finite logarithmic measure. For a meromorphic function f(z), we define its shift by fc(z) � f(z + c) and its difference operators by


Introduction
In this work, we assume that the readers are familiar with general definitions and fundamental theories of Nevanlinna theory [1][2][3]. f is a meromorphic function which means f is meromorphic in the finite complex plane C. If f has no poles, we call f is an entire function. We denote by S(r, f), any function satisfyingS(r, f) � o(T(r, f)), r ⟶ ∞, outside of a possible exceptional set of finite logarithmic measure. For a meromorphic function f(z), we define its shift by f c (z) � f(z + c) and its difference operators by e classic Fermat-type functional equation has been intensively studied in recent years. If n ≥ 4, equation (2) has no transcendental meromorphic solutions [4]. If n ≥ 3, equation (2) has no transcendental entire solutions [4]. If n � 2, all entire solutions are the forms of f(z) � sin(h(z)) and g(z) � cos(h(z)), where h(z) is any entire function [5]. Yang [6] studied the following generalized Fermat-type functional equation: where n and m are positive integers and obtained the following theorem. Theorem 1. If (1/n) + (1/m) < 1, then equation (3) has no nonconstant entire solutions f(z) and g(z).
For further research solutions of the Fermat-type functional equation, Yang and Li [7] considered the following special Fermat-type functional equation: and they obtained the following theorem.
In the following, c is a nonzero constant, unless otherwise specified.
Liu [8] investigated the following special Fermat-type functional equation: and h 1 (z)h 2 (z) � 1. Liu et al. [9] proved that the nonconstant finite-order entire solutions of (5) must have order one. en, Liu et al. [10] considered the entire solutions of the following difference equations: In 2019, Dang and Chen [12] obtained the meromorphic solutions of the following special Fermat-type functional equation: In the following, we will consider the entire solutions with finite order of the Fermat-type differential-difference equation and obtain the following result.
Theorem 6. f(z) be entire solutions of finite order of differential-difference equation (12) if and only if f(z) be the following forms: where a, b, d, and e are constants, e ac ≠ 1, Obviously, from eorem 6, we immediately get the following example.

Example 1.
e transcendental entire function solutions with finite order of the differential-difference equation must satisfy where a, b, d, and e are constants, e ac ≠ 1, e 2ac + 1 � 0, and a 2 � i(e ac − 1) 2 ; f(z) � az 2 + bz + d, where 4a 2 c 4 + 4a 2 � 1. en, we will study the following system of differentialdifference equations: and obtain the next result.
en, the transcendental entire function solutions with finite order of the system of differential-difference equations must be in the following forms: where e kac � (− 1) k , e ac ≠ 1, and where e kac + (− 1) k � 0, e ac ≠ 1, and e b 1 − b 2 � − 1.
Lemma 2 (see [1]). Suppose that f 1 (z), f 2 (z), . . . , f n (z)(n ≥ 2) are meromorphic functions and g 1 (z), g (z) , . . . , g n (z) are entire functions satisfying the following conditions: (1) n j�1 f j (z)e g j (z) ≡ 0 (2) e orders of f j are less than those of e g k − g l for Lemma 3 (Hadamard's factorization theorem; see [1]). Let f be an entire function of finite order ρ(f) with zeros where P(z) is the canonical product of f formed with nonnull of f and Q(z) is a polynomial of degree less than ρ(f).

Proof of Theorem 6
Suppose that f(z) is an entire solution with finite order which satisfies (12). We rewrite (12) as follows: It It follows from (27) that en, Combining (28) and (30), we get If m ≥ 2, then for 0 ≤ j < l ≤ k, we have and by (33) and Lemma 2, we obtain (− 1) k− k C k k � 0, which is contradicting. Hence en, by (33), we have Obviously, (41) and Lemma 2, we obtain that By (42), it is easy to get e ac ≠ 1 and (e ac ) k + (− 1) k � 0. It follows from (42) that a 2 � i(e ac − 1) k . us, it follows from (26) that Case 2. Suppose that f is a polynomial. en, it follows from (26) that p(z) is a constant, and f(z) � az 2 + bz + d.
It follows from (47) and (49)  where e i and d i are constants. Combining (79) and e b 1 − b 2 � ± 1, we get that f 2 (z) � ± f 1 (z) + ez + d, where e and d are constants.