If is a constant or , then the conclusion follows from Theorem B. It remains to consider the case is a non-constant polynomial and . Assume is a transcendental entire solution of (4), which is finite order, not of period c. Differentiating (4) and eliminating , we have
If , then we have . Writing , where satisfies and A is a non-zero constant. Substituting into (4), we get
Clearly, if , then is a period function with period c, which contradicts the assumption. Hence, . Let , then can be expressed as . Furthermore, from Lemma 2.1, we have
From (8) and (9), we get
which contradicts the condition that . Therefore, we conclude that . We discuss the following two cases.
Case 1. . Rewrite (7) in the following forms:
Applying Lemma 2.1, Lemma 2.2, and the lemma on the logarithmic derivative to the above two equations, we obtain
which is a contradiction.
Case 2. . For this case, (4) and (7) now take the following forms:
Set . Dividing (11) by , we have
Since is entire, applying Lemma 2.1 and the lemma on the logarithmic derivative to the above equation, we conclude that
Differentiating , we have
The above equation can be rewritten in the following form:
Suppose is a zero of with multiplicity k. If is a zero of as well, then the contribution to is . Assuming that is not a zero of , we will discuss the two subcases:
Subcase 1. Suppose is a zero of with multiplicity t. From (13), by simple calculations, we know that , which means that is a contribution of to by (12).
Subcase 2. Suppose is not a zero of . By (13), we get , then such a zero of must be simple and we find that must vanish at . That implies that makes a contribution of to by (12).
In a word, . Therefore, we can assume to be of the form , where is a non-constant polynomial, and is an entire function satisfying by the Hadamard factorization theorem. Substituting this expression into (10), we get
which may be written in the form
From Lemma 2.3, we have either
If , then we get
Plugging the above equation into (10), similar to (9), we get . Comparing both sides of (9), we get a contradiction. If , then we get , which means that , and this contradicts the assumption.
Therefore, from the discussions above, we find that a transcendental finite order entire function which is not of period c, cannot be a solution of (4). The proof of Theorem 1.1 is finished completely.