If q(z) is a constant or r(z)\equiv 0, then the conclusion follows from Theorem B. It remains to consider the case q(z) is a non-constant polynomial and r(z)\not\equiv 0. Assume f(z) is a transcendental entire solution of (4), which is finite order, not of period *c*. Differentiating (4) and eliminating {e}^{q(z)}, we have

\begin{array}{c}f{(z)}^{n-1}(n{f}^{\prime}(z)-({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z))\hfill \\ \phantom{\rule{1em}{0ex}}=({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})p(z){({\mathrm{\Delta}}_{c}f)}^{m}-{p}^{\prime}(z){({\mathrm{\Delta}}_{c}f)}^{m}-mp(z){({\mathrm{\Delta}}_{c}f)}^{m-1}{({\mathrm{\Delta}}_{c}f)}^{\prime}.\hfill \end{array}

(7)

If n{f}^{\prime}(z)-({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z)\equiv 0, then we have f{(z)}^{n}=Ar(z){e}^{q(z)}. Writing f(z)=h(z){e}^{\frac{q(z)}{n}}, where h(z) satisfies h{(z)}^{n}=Ar(z) and *A* is a non-zero constant. Substituting f(z) into (4), we get

(A-1)r(z){e}^{q(z)}+p(z){({\mathrm{\Delta}}_{c}f)}^{m}\equiv 0.

(8)

Clearly, if A=1, then f(z) is a period function with period *c*, which contradicts the assumption. Hence, A\ne 1. Let g={e}^{\frac{q(z)}{n}}, then {({\mathrm{\Delta}}_{c}f)}^{m} can be expressed as {\sum}_{i=0}^{m}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(-1)}^{i}h{(z)}^{i}h{(z+c)}^{m-i}g{(z)}^{i}g{(z+c)}^{m-i}. Furthermore, from Lemma 2.1, we have

\begin{array}{rl}T(r,{({\mathrm{\Delta}}_{c}f)}^{m})& =T(r,\sum _{i=0}^{m}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(-1)}^{i}h{(z)}^{i}h{(z+c)}^{m-i}g{(z)}^{i}g{(z+c)}^{m-i})\\ =m(r,\sum _{i=0}^{m}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(-1)}^{i}h{(z)}^{i}h{(z+c)}^{m-i}g{(z)}^{i}g{(z+c)}^{m-i})\\ \le m(r,\frac{{\mathrm{\Sigma}}_{i=0}^{m}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(-1)}^{i}g{(z+c)}^{m-i}g{(z)}^{i}}{g{(z)}^{m}})+m(r,g{(z)}^{m})+S(r,g)\\ \le T(r,g{(z)}^{m})+S(r,g)=mT(r,g)+S(r,g).\end{array}

(9)

From (8) and (9), we get

nT(r,g)\le mT(r,g)+S(r,g),

which contradicts the condition that n>m. Therefore, we conclude that n{f}^{\prime}(z)-({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z)\not\equiv 0. We discuss the following two cases.

Case 1. n>m+1. Rewrite (7) in the following forms:

\begin{array}{c}f{(z)}^{n-m-1}(n{f}^{\prime}(z)-({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z))\hfill \\ \phantom{\rule{1em}{0ex}}=({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})p(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m}}{{f}^{m}}-{p}^{\prime}(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m}}{{f}^{m}}-mp(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m-1}{({\mathrm{\Delta}}_{c}f)}^{\prime}}{{f}^{m}}\hfill \end{array}

and

\begin{array}{c}f{(z)}^{n-m-2}\left(f(n{f}^{\prime}(z)-({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z))\right)\hfill \\ \phantom{\rule{1em}{0ex}}=({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})p(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m}}{{f}^{m}}-{p}^{\prime}(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m}}{{f}^{m}}-mp(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m-1}{({\mathrm{\Delta}}_{c}f)}^{\prime}}{{f}^{m}}.\hfill \end{array}

Applying Lemma 2.1, Lemma 2.2, and the lemma on the logarithmic derivative to the above two equations, we obtain

T(r,n{f}^{\prime}-({q}^{\prime}+\frac{{r}^{\prime}}{r})f)=m(r,n{f}^{\prime}-({q}^{\prime}+\frac{{r}^{\prime}}{r})f)+S(r,f)=S(r,f)

and

T(r,f(n{f}^{\prime}-({q}^{\prime}+\frac{{r}^{\prime}}{r})f))=m(r,f(n{f}^{\prime}-({q}^{\prime}+\frac{{r}^{\prime}}{r})f))+S(r,f)=S(r,f).

Hence

which is a contradiction.

Case 2. n=m+1. For this case, (4) and (7) now take the following forms:

{f}^{m+1}+p(z){({\mathrm{\Delta}}_{c}f)}^{m}=r(z){e}^{q(z)}

(10)

and

\begin{array}{c}f{(z)}^{m}((m+1){f}^{\prime}(z)-({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z))\hfill \\ \phantom{\rule{1em}{0ex}}=({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})p(z){({\mathrm{\Delta}}_{c}f)}^{m}-{p}^{\prime}(z){({\mathrm{\Delta}}_{c}f)}^{m}-mp(z){({\mathrm{\Delta}}_{c}f)}^{m-1}{({\mathrm{\Delta}}_{c}f)}^{\prime}.\hfill \end{array}

(11)

Set H(z)=(m+1){f}^{\prime}(z)-({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z). Dividing (11) by f{(z)}^{m}, we have

H(z)=({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})p(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m}}{{f}^{m}}-{p}^{\prime}(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m}}{{f}^{m}}-mp(z)\frac{{({\mathrm{\Delta}}_{c}f)}^{m-1}{({\mathrm{\Delta}}_{c}f)}^{\prime}}{{f}^{m}}.

Since f(z) is entire, applying Lemma 2.1 and the lemma on the logarithmic derivative to the above equation, we conclude that

T(r,H)=m(r,H)+S(r,f)=S(r,f).

(12)

Differentiating H(z), we have

\begin{array}{c}(m+1){f}^{\u2033}-({q}^{\prime}+\frac{{r}^{\prime}}{r}){f}^{\prime}-{({q}^{\prime}+\frac{{r}^{\prime}}{r})}^{\prime}f\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{H}^{\prime}}{H}\cdot H=\frac{{H}^{\prime}}{H}((m+1){f}^{\prime}-({q}^{\prime}+\frac{{r}^{\prime}}{r})f),\hfill \end{array}

that is

(m+1){f}^{\u2033}-({q}^{\prime}+\frac{{r}^{\prime}}{r}+(m+1)\frac{{H}^{\prime}}{H}){f}^{\prime}-({q}^{\u2033}-{q}^{\prime}\frac{{H}^{\prime}}{H}+{\left(\frac{{r}^{\prime}}{r}\right)}^{\prime}-\frac{{H}^{\prime}}{H}\cdot \frac{{r}^{\prime}}{r})f=0.

The above equation can be rewritten in the following form:

\begin{array}{c}(m+1)({\left(\frac{{f}^{\prime}}{f}\right)}^{\prime}+{\left(\frac{{f}^{\prime}}{f}\right)}^{2})-({q}^{\prime}+\frac{{r}^{\prime}}{r}+(m+1)\frac{{H}^{\prime}}{H})\frac{{f}^{\prime}}{f}\hfill \\ \phantom{\rule{1em}{0ex}}-({q}^{\u2033}-{q}^{\prime}\frac{{H}^{\prime}}{H}+{\left(\frac{{r}^{\prime}}{r}\right)}^{\prime}-\frac{{H}^{\prime}}{H}\cdot \frac{{r}^{\prime}}{r})=0.\hfill \end{array}

(13)

Suppose {z}_{0} is a zero of f(z) with multiplicity *k*. If {z}_{0} is a zero of r(z) as well, then the contribution to N(r,\frac{1}{f}) is S(r,f). Assuming that {z}_{0} is not a zero of r(z), we will discuss the two subcases:

Subcase 1. Suppose {z}_{0} is a zero of H(z) with multiplicity *t*. From (13), by simple calculations, we know that k=1+t\le 2t, which means that {z}_{0} is a contribution of S(r,f) to N(r,\frac{1}{f}) by (12).

Subcase 2. Suppose {z}_{0} is not a zero of H(z). By (13), we get {k}^{2}-k=0, then such a zero of f(z) must be simple and we find that {q}^{\prime}+\frac{{r}^{\prime}}{r}+(m+1)\frac{{H}^{\prime}}{H} must vanish at {z}_{0}. That implies that {z}_{0} makes a contribution of S(r,f) to N(r,\frac{1}{f}) by (12).

In a word, N(r,\frac{1}{f})=S(r,f). Therefore, we can assume f(z) to be of the form f(z)=G(z){e}^{s(z)}, where s(z) is a non-constant polynomial, and G(z) is an entire function satisfying N(r,\frac{1}{G})=T(r,G)=S(r,f) by the Hadamard factorization theorem. Substituting this expression into (10), we get

G{(z)}^{m+1}{e}^{(m+1)s(z)}+p(z){(G(z+c){e}^{s(z+c)}-G(z){e}^{s(z)})}^{m}=r(z){e}^{q(z)},

and so

\begin{array}{c}G{(z)}^{m+1}{e}^{(m+1)s(z)}+p(z)\left(\sum _{i=0}^{m-1}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(-1)}^{i}G{(z+c)}^{m-i}G{(z)}^{i}{e}^{(m-i)s(z+c)+is(z)}\right)\hfill \\ \phantom{\rule{1em}{0ex}}+p(z){(-1)}^{m}G{(z)}^{m}{e}^{ms(z)}=r(z){e}^{q(z)},\hfill \end{array}

which may be written in the form

\begin{array}{c}\frac{G(z){e}^{s(z)}}{p(z){(-1)}^{m-1}}+\frac{r(z){e}^{q(z)}}{p(z){(-1)}^{m}G{(z)}^{m}{e}^{ms(z)}}\hfill \\ \phantom{\rule{1em}{0ex}}+\frac{{\sum}_{i=0}^{m-1}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(-1)}^{i}G{(z+c)}^{m-i}G{(z)}^{i}{e}^{(m-i)s(z+c)+is(z)}}{{(-1)}^{m-1}G{(z)}^{m}{e}^{ms(z)}}=1.\hfill \end{array}

(14)

From Lemma 2.3, we have either

{f}_{2}(z)=\frac{r(z){e}^{q(z)}}{p(z){(-1)}^{m}G{(z)}^{m}{e}^{ms(z)}}\equiv 1

or

{f}_{3}(z)=\frac{{\sum}_{i=0}^{m-1}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(-1)}^{i}G{(z+c)}^{m-i}G{(z)}^{i}{e}^{(m-i)s(z+c)+is(z)}}{{(-1)}^{m-1}G{(z)}^{m}{e}^{ms(z)}}\equiv 1.

If {f}_{2}(z)\equiv 1, then we get

r(z){e}^{q(z)}=p(z){(-1)}^{m}G{(z)}^{m}{e}^{ms(z)}=p(z){(-1)}^{m}f{(z)}^{m}.

Plugging the above equation into (10), similar to (9), we get T(r,r{e}^{q}-p{({\mathrm{\Delta}}_{c}f)}^{m})\le mT(r,f)+S(r,f). Comparing both sides of (9), we get a contradiction. If {f}_{3}(z)\equiv 1, then we get f{(z)}^{m+1}\equiv r(z){e}^{q(z)}, which means that {\mathrm{\Delta}}_{c}f\equiv 0, and this contradicts the assumption.

Therefore, from the discussions above, we find that a transcendental finite order entire function f(z) which is not of period *c*, cannot be a solution of (4). The proof of Theorem 1.1 is finished completely.