The proof utilizes the ideas of the papers [11, 15], although some details are different. Suppose contrary to the assertion that both \(f(z)F_{1}(z,f)-q(z)\) and \(f(z)F_{2}(z,f)-q(z)\) have finitely many zeros. Since \(f(z)\) is of finite order, by the Hadamard factorization theorem we can write

$$ f(z)F_{1}(z,f)-q(z)=p_{1}(z)e^{\alpha(z)} $$

(3.1)

and

$$ f(z)F_{2}(z,f)-q(z)=p_{2}(z)e^{\beta(z)}, $$

(3.2)

where \(\alpha(z)\), \(\beta(z)\), \(p_{1}(z)\), \(p_{2}(z)\) are polynomials. From Remark 5 we know that \(T(r,F_{1}(z,f))\leq T(r,f)+S(r,f)\) and \(T(r,F_{2}(z,f))\leq T(r,f)+S(r,f)\). First, if \(T(r,F_{1}(z,f))=S(r,f)\) and \(T(r,F_{2}(z,f))=S(r,f)\), then from the second main theorem for three small functions [8, Theorem 2.5] we know that

$$\begin{aligned} T(r,f) \leq& \overline{N}(r,f)+\overline{N} \biggl(r,\frac{1}{f(z)-\frac {q(z)}{F_{1}(z,f)}} \biggr)+ \overline{N} \biggl(r,\frac{1}{f(z)-\frac {q(z)}{F_{2}(z,f)}} \biggr)+S(r,f) \\ =&\overline{N} \biggl(r,\frac{1}{\frac{p_{1}(z)}{F_{1}(z,f)}} \biggr)+\overline{N} \biggl(r, \frac{1}{\frac{p_{2}(z)}{F_{2}(z,f)}} \biggr)+S(r,f)=S(r,f), \end{aligned}$$

which is impossible.

Second, suppose that \(T(r,F_{1}(z,f))\neq S(r,f)\) and \(T(r,F_{2}(z,f))\neq S(r,f)\). We affirm that \(e^{\alpha}\), \(e^{\beta}\), and \(e^{\alpha+\beta}\) are not small functions with respect to \(f(z)\). Otherwise, if \(e^{\alpha}\) is a small function with respect to \(f(z)\), we have \(f(z)F_{1}(z,f)=t(z)\) from (3.1), where \(t(z)=q(z)+p_{1}(z)e^{\alpha(z)}\) is a small function with respect to \(f(z)\). Lemma 2.7 and Lemma 2.3 imply that

$$ T(r,F_{1})=m(r,F_{1})=S(r,f), $$

and we get \(T(r,f)=S(r,f)\), a contradiction. We can use a similar method to obtain that \(e^{\beta}\) is not a small function of \(f(z)\). From (3.1) and (3.2) we have

$$f(z)^{2}F_{1}(z,f)F_{2}(z,f)-q(z)f(z) \bigl[F_{1}(z,f)+F_{2}(z,f)\bigr]+q(z)^{2}=p_{1}(z)p_{2}(z)e^{\alpha +\beta}. $$

If \(e^{\alpha+\beta}\) is a small function, by Lemma 2.7 and Lemma 2.3 we have \(T(r, F_{1}F_{2})=S(r,f)\) and \(T(r, \frac {F_{1}F_{2}}{f})=S(r,f)\), and hence \(T(r,f)=S(r,f)\), which is impossible.

Since *α* is a polynomial, *α* and \(\alpha'\) are small functions of *f*. Differentiating (3.1) and eliminating \(e^{\alpha }\), we obtain

$$ a_{1}fF_{1}-f'F_{1}-fF'_{1}=b_{1}, $$

(3.3)

where \(a_{1}=\frac{p_{1}'}{p_{1}}+\alpha'\) and \(b_{1}=(\frac {p_{1}'}{p_{1}}+\alpha')q-q'\). Now, we will show that \(a_{1}\not\equiv 0\). Indeed, if \(a_{1}\equiv0\), then by a simple integration there exists a nonzero constant \(C_{1}\) such that \(C_{1}=p_{1}e^{\alpha}\), which implies a contradiction to the fact that \(e^{\alpha}\) is not a small function with respect to \(f(z)\). Similarly, we have \(b_{1}\not \equiv0\).

By the same arguments as before, (3.2) gives

$$ a_{2}fF_{2}-f'F_{2}-fF'_{2}=b_{2}, $$

(3.4)

where \(a_{2}=\frac{p_{2}'}{p_{2}}+\beta'\) and \(b_{2}=(\frac {p_{2}'}{p_{2}}+\beta')q-q'\). We also obtain \(a_{2}\not\equiv0\) and \(b_{2}\not\equiv0\).

In view of Lemma 2.8 and Remark 6, we assume that \(z_{0}\) is a simple zero of *f* such that \(z_{0}\) is not a zero or pole of \(b_{i}\)
\((i=1,2)\). Equations (3.3) and (3.4) imply that

$$ \bigl(f'F_{1}+b_{1}\bigr) (z_{0})=0 $$

(3.5)

and

$$ \bigl(f'F_{2}+b_{2}\bigr) (z_{0})=0. $$

(3.6)

Hence we have

$$ (b_{2}F_{1}-b_{1}F_{2}) (z_{0})=0. $$

(3.7)

We will consider two cases depending on whether \(b_{2}F_{1}-b_{1}F_{2}\equiv0\) or not.

*Case* 1. \(b_{2}F_{1}-b_{1}F_{2}\not\equiv0\). Set

$$ h=\frac{b_{2}F_{1}-b_{1}F_{2}}{f}. $$

(3.8)

From Lemma 2.3 we have \(m(r,h)=S(r,f)\). On the other hand, from (3.7) and Lemma 2.8 we have

$$\begin{aligned} N(r,h) =&N \biggl(r,\frac{b_{2}F_{1}-b_{1}F_{2}}{f} \biggr) \\ =&N_{1} \biggl(r,\frac{b_{2}F_{1}-b_{1}F_{2}}{f} \biggr)+S(r,f)=S(r,f). \end{aligned}$$

(3.9)

Thus \(T(r,h)=S(r,f)\). Rewrite (3.8) in the form

$$ F_{1}=\frac{h}{b_{2}}f+\frac{b_{1}}{b_{2}}F_{2}. $$

(3.10)

By differentiating (3.10) we have

$$ F'_{1}= \biggl(\frac{h}{b_{2}} \biggr)'f+\frac{h}{b_{2}}f'+ \biggl( \frac {b_{1}}{b_{2}} \biggr)'F_{2}+\frac{b_{1}}{b_{2}}F'_{2}. $$

(3.11)

Substituting (3.10) and (3.11) into (3.3), we get

$$ \begin{aligned} \biggl[a_{1} \frac{h}{b_{2}}- \biggl(\frac{h}{b_{2}} \biggr)' \biggr]f^{2}-2\frac{h}{b_{2}}ff' + \biggl[a_{1}\frac{b_{1}}{b_{2}}- \biggl(\frac{b_{1}}{b_{2}} \biggr)' \biggr]fF_{2} &-\frac{b_{1}}{b_{2}}f'F_{2}- \frac{b_{1}}{b_{2}}fF'_{2}=b_{1}. \end{aligned} $$

(3.12)

In addition, equation (3.4) can be written as

$$ a_{2}\frac{b_{1}}{b_{2}}fF_{2}- \frac{b_{1}}{b_{2}}f'F_{2}-\frac {b_{1}}{b_{2}}fF'_{2}=b_{1}. $$

(3.13)

Combining (3.12) and (3.13), we get

$$ \biggl[a_{1}\frac{h}{b_{2}}- \biggl( \frac{h}{b_{2}} \biggr)' \biggr]f-2\frac {h}{b_{2}}f'+ \biggl[a_{1}\frac{b_{1}}{b_{2}}- \biggl(\frac{b_{1}}{b_{2}} \biggr)'-a_{2}\frac{b_{1}}{b_{2}} \biggr]F_{2}=0. $$

(3.14)

Note that \(-2\frac{h}{b_{2}}\not\equiv0\). We proceed to prove that

$$ a_{1}\frac{h}{b_{2}}- \biggl(\frac{h}{b_{2}} \biggr)'\not\equiv0\quad \mbox{and} \quad a_{1} \frac{b_{1}}{b_{2}}- \biggl(\frac {b_{1}}{b_{2}} \biggr)'-a_{2} \frac{b_{1}}{b_{2}}\not\equiv0. $$

(3.15)

Otherwise, if \(a_{1}\frac{h}{b_{2}}-(\frac{h}{b_{2}})'\equiv0\), then by the definition of \(a_{1}\) and a simple integration we have

$$ p_{1}e^{\alpha}=C_{2} \frac{h}{b_{2}}, $$

(3.16)

where \(C_{2}\) is a nonzero constant. Since \(T (r,\frac {h}{b_{2}} )=S(r,f)\), \(p_{1}e^{\alpha}\) is a small function of *f*, a contradiction. If \(a_{1}\frac{b_{1}}{b_{2}}-(\frac {b_{1}}{b_{2}})'-a_{2}\frac{b_{1}}{b_{2}}\equiv0\), then we have \(a_{1}-a_{2}\equiv\frac{b_{1}'}{b_{1}}-\frac{b_{2}'}{b_{2}}\). A simple integration yields that

$$ \frac{p_{1}}{p_{2}}e^{\alpha-\beta}=C_{3}\frac{b_{1}}{b_{2}}:=\gamma, $$

where \(C_{3}\) is a nonzero constant, and *γ* is a small function of *f*. From (3.1) and (3.2) we have

$$ f(F_{1}-\gamma F_{2})=q(1-\gamma). $$

(3.17)

If \(\gamma\not\equiv1\), then by Lemma 2.7 we get

$$ m(r,F_{1}-\gamma F_{2})+S(r,f)=T(r,F_{1}-\gamma F_{2})=S(r,f). $$

From this and from (3.17) we have

$$ T(r,f)=T \biggl(r,\frac{q(1-\gamma)}{F_{1}-\gamma F_{2}} \biggr)=S(r,f), $$

a contradiction. If \(\gamma\equiv1\), then we have \(F_{1}\equiv F_{2}\), which contradicts the hypothesis of Theorem 1.1. Thus \(a_{1}\frac {b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'-a_{2}\frac{b_{1}}{b_{2}}\not\equiv0\).

From these discussions we can rewrite (3.14) as

$$ F_{2}=mf+nf', $$

(3.18)

where

$$m=\frac{(\frac{h}{b_{2}})'-a_{1}\frac{h}{b_{2}}}{(a_{1}-a_{2})\frac {b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'},\qquad n=\frac{\frac {2h}{b_{2}}}{(a_{1}-a_{2})\frac{b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'}. $$

Furthermore, (3.10) and (3.18) give

$$ F_{1}=sf+tf', $$

(3.19)

where

$$ s=\frac{b_{1}m+h}{b_{2}},\qquad t=\frac{b_{1}}{b_{2}}n. $$

(3.20)

Differentiating (3.18), we have

$$ F'_{2}=m'f+ \bigl(m+n'\bigr)f'+nf''. $$

(3.21)

Substituting (3.18) and (3.21) into (3.4), we obtain

$$ \bigl(a_{2}m-m'\bigr)f^{2}+ \bigl(a_{2}n-2m-n'\bigr)ff'-n \bigl(f'\bigr)^{2}-nff''=b_{2}. $$

(3.22)

Differentiating (3.22), we have

$$ \begin{aligned}[b] & \bigl(a_{2}m-m' \bigr)'f^{2}+ \bigl[2\bigl(a_{2}m-m' \bigr)+\bigl(a_{2}n-2m-n'\bigr) \bigr]ff' \\ &\quad {}+\bigl(a_{2}n-2m-2n'\bigr) \bigl[\bigl(f' \bigr)^{2}+ff'' \bigr]-n\bigl(3f'f''+ff''' \bigr)=b_{2}'. \end{aligned} $$

(3.23)

Since we have supposed that \(f(z_{0})=0\), \(f'(z_{0})\neq0\), and \(b_{2}(z_{0})\neq0 \), from (3.22) and (3.23), respectively, we obtain

$$ \bigl(n\bigl(f'\bigr)^{2}+b_{2} \bigr) (z_{0})=0 $$

and

$$ \bigl[\bigl(a_{2}n-2m-2n'\bigr) \bigl(f' \bigr)^{2}-3nf'f''-b_{2}' \bigr](z_{0})=0, $$

which leads to

$$ \bigl(\bigl[b_{2}\bigl(a_{2}n-2m-2n' \bigr)+b_{2}'n\bigr]f'-3b_{2}nf'' \bigr) (z_{0})=0. $$

Let

$$ H=\frac{[b_{2}(a_{2}n-2m-2n')+b_{2}'n]f'-3b_{2}nf''}{f}. $$

It is obvious that *H* is a small function of *f* and

$$ f''=\frac{b_{2}(a_{2}n-2m-2n')+b_{2}'n}{3b_{2}n}f'- \frac{H}{3b_{2}n}f. $$

(3.24)

Substituting (3.24) into (3.22), we have

$$ g_{1}f^{2}+g_{2}ff'+g_{3} \bigl(f'\bigr)^{2}=b_{2}, $$

(3.25)

where

$$\begin{aligned} &g_{1}=a_{2}m-m'+\frac{H}{3b_{2}}, \\ &g_{2}=\frac{1}{3} \biggl(2a_{2}-\frac{b_{2}'}{b_{2}} \biggr)n-\frac {4}{3}m-\frac{1}{3}n',\qquad g_{3}=-n. \end{aligned}$$

It is easy to see that \(g_{3}\not\equiv0\). We proceed to prove \(g_{2}\not\equiv0\). Otherwise, we get

$$ \frac{g_{2}}{g_{3}}=\frac{2}{3}\frac{h'}{h}+ \frac{1}{3}\frac {n'}{n}-\frac{1}{3}\frac{b_{2}'}{b_{2}}- \frac{2}{3}(a_{1}+a_{2})=0, $$

(3.26)

since, by the definition of \(a_{1}\) and \(a_{2}\),

$$ \alpha'+\beta'+\frac{p_{1}'}{p_{1}}+\frac{p_{2}'}{p_{2}}= \frac {h'}{h}+\frac{1}{2} \biggl(\frac{n'}{n}-\frac{b_{2}'}{b_{2}} \biggr). $$

By integration we have

$$ p_{1}p_{2}e^{\alpha+\beta}=C_{3}h\biggl( \frac{n}{b_{2}}\biggr)^{\frac{1}{2}}, $$

where \(C_{3}\) is a nonzero constant. Since \(T (r,C_{3}h(\frac {n}{b_{2}})^{\frac{1}{2}} )=S(r,f)\), we can deduce that \(e^{\alpha +\beta}\) is a small function of *f*, a contradiction. Thus \(g_{2}\not \equiv0\). Differentiating (3.25), we have

$$ g_{1}'f^{2}+ \bigl(2g_{1}+g_{2}'\bigr)ff'+ \bigl(g_{2}+g_{3}'\bigr) \bigl(f' \bigr)^{2}+g_{2}ff''+2g_{3}f'f''=b_{2}'. $$

(3.27)

By the same method used to deal with (3.22) and (3.23) we have

$$ f''=\frac{b_{2}'g_{3}-b_{2}(g_{2}+g_{3}')}{2b_{2}g_{3}}f'+ \frac {R}{2b_{2}g_{3}}f, $$

(3.28)

where

$$ R=\frac{[b_{2}(g_{2}+g_{3}')-b_{2}'g_{3}]f'+2b_{2}g_{3}f''}{f} \quad \mbox{and}\quad T(r,R)=S(r,f). $$

Substituting (3.28) into (3.27), we get

$$ \begin{aligned}[b] &\biggl(g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}} \biggr)f^{2}+ \biggl(2g_{1}+g_{2}'+\frac{1}{2} \frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2}\bigl(g_{2}+g_{3}' \bigr)\frac{g_{2}}{g_{3}} +\frac{R}{b_{2}} \biggr)f'f \\ &\quad {}+\frac{b_{2}'g_{3}}{b_{2}}\bigl(f'\bigr)^{2}=b_{2}'. \end{aligned} $$

(3.29)

Combining (3.29) and (3.25), we have

$$ \biggl(g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}}- \frac{b_{2}}{b_{2}}'g_{1} \biggr)f+ \biggl(2g_{1}+g_{2}'- \frac{1}{2}\frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2} \bigl(g_{2}+g_{3}'\bigr)\frac{g_{2}}{g_{3}}+ \frac{R}{b_{2}} \biggr)f'=0. $$

(3.30)

Let

$$ q_{1}=g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}}- \frac{b_{2}}{b_{2}}'g_{1} $$

and

$$ q_{2}=2g_{1}+g_{2}'- \frac{1}{2}\frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2} \bigl(g_{2}+g_{3}'\bigr)\frac{g_{2}}{g_{3}}+ \frac{R}{b_{2}}. $$

Now we claim that \(q_{1}\) and \(q_{2}\) vanish identically. If \(q_{2}\not \equiv0\), then by the previous analysis we can get that \(q_{2}\) is a small function of *f*. From (3.30) we have \(q_{2}(z_{0})=0\). Thus

$$ T(r,f)=N_{1} \biggl(r,\frac{1}{f} \biggr)+S(r,f)\leq N \biggl(r,\frac {1}{q_{2}} \biggr)+S(r,f)=S(r,f), $$

a contradiction. Again by (3.30) we get \(q_{1}\equiv0\). Eliminating *R* from \(q_{1}\equiv0\) and \(q_{2}\equiv0\), we have

$$ \begin{aligned}[b] &g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)\frac {b_{2}'}{b_{2}}+g_{2}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)+2g_{2}g_{2}'g_{3}-g_{2}^{2}g_{3}'-4g_{1}'g_{3}^{2} \\ &\quad =g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)\frac {b_{2}'}{b_{2}}+g_{2}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)-g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)'+g_{3}'\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)=0. \end{aligned} $$

(3.31)

We continue by discussing two subcases depending on whether \(4g_{1}g_{3}-g_{2}^{2}\) vanishes identically or not.

*Subcase* 1. If \(4g_{1}g_{3}-g_{2}^{2}\not\equiv0\), then from (3.31) we have

$$ \frac{g_{2}}{g_{3}}=\frac {(4g_{1}g_{3}-g_{2}^{2})'}{4g_{1}g_{3}-g_{2}^{2}}-\frac {b_{2}'}{b_{2}}-\frac{g_{3}'}{g_{3}}. $$

Combining this equation with (3.26), we get that

$$ 2(a_{1}+a_{2})=2\frac{h'}{h}+\frac{n'}{n}+2 \frac{b_{2}'}{b_{2}}-3\frac {(4g_{1}g_{3}-g_{2}^{2})'}{4g_{1}g_{3}-g_{2}^{2}}. $$

Similarly as in the proof of \(g_{2}\not\equiv0\), we can deduce that \(e^{\alpha+\beta}\) is a small function of *f*, a contradiction.

*Subcase* 2. If \(4g_{1}g_{3}-g_{2}^{2}\equiv0\), then, on the one hand, by using \(q_{1}\equiv0\) we can rewrite equation (3.28) as

$$ f''= \biggl(\frac{b_{2}'g_{1}}{b_{2}g_{2}}-\frac{g_{1}'}{g_{2}} \biggr)f +\frac{1}{2} \biggl(\frac{b_{2}'}{b_{2}}-\frac{g_{2}}{g_{3}}- \frac {g_{3}'}{g_{3}} \biggr)f', $$

and then from this equation and from (3.24) we have

$$ \frac{g_{1}'}{g_{2}}-\frac{b_{2}'}{b_{2}}\frac{g_{1}}{g_{2}}=\frac{H}{3b_{2}n}. $$

On the other hand, by the definitions of \(g_{1}\) and \(g_{3}\) we have

$$ \frac{ g_{1}}{g_{3}}=\frac{m'}{n}-a_{2}\frac{m}{n}- \frac{H}{3b_{2}n}. $$

Therefore we get

$$ \frac{g_{1}'}{g_{2}}-\frac{b_{2}'}{b_{2}}\frac{g_{1}}{g_{2}}+ \frac{ g_{1}}{g_{3}}-\frac{m'}{n}+a_{2}\frac{m}{n}=0. $$

(3.32)

For brevity, we denote

$$ D=(a_{1}-a_{2})\frac{b_{1}}{b_{2}}-\biggl( \frac{b_{1}}{b_{2}}\biggr)'. $$

By calculation we have

$$ \frac{ g_{1}}{g_{3}}=\frac{1}{4} \biggl(\frac{g_{2}}{g_{3}} \biggr)^{2},\qquad \frac{g_{1}}{g_{2}}=\frac{1}{4} \frac{g_{2}}{g_{3}},\qquad \frac{ g_{1}'}{g_{2}}=\frac{1}{2} \biggl( \frac{g_{2}}{g_{3}} \biggr)'+\frac{1}{4}\frac{g_{2}}{g_{3}} \frac{g_{3}'}{g_{3}}, $$

and

$$ \frac{m'}{n}= \biggl(\frac{m}{n} \biggr)'+ \frac{m}{n}\frac{n'}{n},\qquad \frac{m}{n}=\frac{1}{2} \biggl(\frac{h'}{h}-\frac {b_{2}'}{b_{2}}-a_{1} \biggr),\qquad \frac{n'}{n}=\frac{h'}{h}-\frac {b_{2}'}{b_{2}}-\frac{D'}{D}. $$

From (3.26) we have

$$ \frac{g_{2}}{g_{3}}=\frac{h'}{h}-\frac{2}{3}\frac{b_{2}'}{b_{2}}- \frac {1}{3}\frac{D'}{D}-\frac{2}{3}(a_{1}+a_{2}). $$

Substituting these identities into (3.32), we have

$$ \begin{aligned} &\frac{1}{4} \biggl(\frac{h'}{h}- \frac{2}{3}\frac{b_{2}'}{b_{2}}-\frac {1}{3}\frac{D'}{D}- \frac{2}{3}(a_{1}+a_{2}) \biggr) \biggl( \frac{8}{3}\frac{b_{2}'}{b_{2}}-2\frac{h'}{h}+\frac{4}{3} \frac {D'}{D}+\frac{2}{3}(a_{1}+a_{2}) \biggr) \\ &\quad {}+\frac{1}{2} \biggl(\biggl(\frac{h'}{h}\biggr)'- \biggl(\frac{b_{2}'}{b_{2}}\biggr)'-a_{1}' \biggr)+ \frac{1}{2} \biggl(\frac{h'}{h}-\frac{b_{2}'}{b_{2}}- \frac {D'}{D}-a_{2} \biggr) \biggl(\frac{h'}{h}- \frac{b_{2}'}{b_{2}}-a_{1} \biggr) \\ &\quad {}-\frac{1}{2} \biggl(\biggl(\frac{h'}{h}\biggr)'- \frac{2}{3}\biggl(\frac {b_{2}'}{b_{2}}\biggr)'- \frac{1}{3} \biggl(\frac{D'}{D} \biggr)'- \frac {2}{3}(a_{1}+a_{2})' \biggr)=0, \end{aligned} $$

which leads to

$$\begin{aligned} &\frac{7}{18}\frac{b_{2}'}{b_{2}}\frac{D'}{D}+\frac{1}{6} \biggl(\frac {D'}{D} \biggr)' -\frac{5}{18}(a_{1}+a_{2}) \frac{D'}{D}-\frac{1}{6}(a_{1}+a_{2}) \frac {h'}{h}+\frac{1}{18} \biggl(\frac{b_{2}'}{b_{2}} \biggr)^{2} \\ &\qquad {}-\frac{1}{18}(a_{1}+a_{2})\frac{b_{2}'}{b_{2}}+ \frac{1}{2}a_{1}\frac{D'}{D} +\frac{1}{2}a_{1}a_{2}- \frac{1}{6}\frac{b_{2}'}{b_{2}}-\frac {1}{2}a_{1}'+ \frac{1}{3}(a_{1}+a_{2})' \\ &\quad =\frac{1}{9}(a_{1}+a_{2})^{2}. \end{aligned}$$

Since

$$\lim_{z\rightarrow\infty}\frac{R'(z)}{R(z)}=0 $$

if \(R(z)\) is a nonzero rational function, dividing both sides of the above equation by \(\frac{(a_{1}+a_{2})^{2}}{2}\) and taking the limit, we have

$$ \lim_{z\rightarrow\infty}\frac{a_{1}a_{2}}{(a_{1}+a_{2})^{2}}=\frac {2}{9}+\lim _{z\rightarrow\infty}\frac{1}{3}\frac{h'}{h}\frac{1}{(a_{1}+a_{2})}. $$

It follows by Lemma 2.4 that

$$ \lim_{z\rightarrow\infty}\frac{a_{1}a_{2}}{(a_{1}+a_{2})^{2}}=\lim_{z\rightarrow\infty} \frac{\alpha'\beta'}{(\alpha'+\beta')^{2}}=\frac{2}{9}. $$

By setting \(\alpha(z)=a_{m}z^{m}+\cdots+a_{0}\) and \(\beta (z)=b_{m}z^{m}+\cdots+b_{0}\) we have

$$ \lim_{z\rightarrow\infty}\frac{\alpha'\beta'}{(\alpha'+\beta ')^{2}}=\frac{a_{m}b_{m}}{(a_{m}+b_{m})^{2}}= \frac{2}{9}, $$

which implies that \(\frac{a_{m}}{b_{m}}=2\) or \(\frac{a_{m}}{b_{m}}=\frac {1}{2}\). We first consider the case \(\frac{a_{m}}{b_{m}}=2\). From (3.1), (3.2), (3.18), and (3.19) we have

$$ sf^{2}+tff'-q=A(z)e^{2b_{m}z^{m}} $$

and

$$ mf^{2}+nff'-q=B(z)e^{b_{m}z^{m}}, $$

where

$$ A(z)=p_{1}e^{a_{m-1}z^{m-1}+\cdots+a_{0}} \quad \mbox{and} \quad B(z)=p_{2}e^{b_{m-1}z^{m-1}+\cdots+b_{0}}. $$

Then

$$ mf^{2}+nff'=q+B \biggl(\frac{sf^{2}+tff'-q}{A} \biggr)^{\frac{1}{2}}. $$

Combining (2.2) with the expressions of *A* and *B*, we obtain

$$ \begin{aligned}T(r,F_{2})&=T\bigl(r, mf+nf' \bigr) \\ &=m \biggl(r,\frac{q}{f}+B \biggl(\frac{sf^{2}+tff'-q}{Af^{2}} \biggr)^{\frac{1}{2}} \biggr)+S(r,f) \\ &\leq m\biggl(r,\frac{q}{f}\biggr)+m(r,B)+\frac{1}{2}m \biggl(r, \frac {sf^{2}+tff'-q}{Af^{2}} \biggr)+S(r,f) \\ &\leq m\biggl(r,\frac{s}{A}\biggr)+m\biggl(r,\frac{t}{A}\biggr)+m \biggl(r,\frac{f'}{f}\biggr)+m\biggl(r,\frac {q}{Af^{2}}\biggr)+S(r,f) \\ &=S(r,f). \end{aligned} $$

Hence we get \(T(r,F_{2})=S(r,f)\), a contradiction. For the case \(\frac{a_{m}}{b_{m}}=\frac{1}{2}\), by the same argument we can also deduce \(T(r,F_{1})=S(r,f)\), a contradiction.

*Case* 2. \(b_{2}F_{1}-b_{1}F_{2}\equiv0\).

Using the same arguments as in the proof of (3.14), we have

$$ a_{1}\frac{b_{1}}{b_{2}}-\biggl(\frac{b_{1}}{b_{2}} \biggr)'-a_{2}\frac {b_{1}}{b_{2}}\equiv0, $$

which leads to

$$ \frac{p_{1}}{p_{2}}e^{\alpha-\beta}=C_{3}\frac{b_{1}}{b_{2}}, $$

where \(C_{3}\) is a nonzero constant. Then by (3.1) and (3.2), if \(\frac{C_{3}b_{1}}{b_{2}}\equiv1\), then \(F_{1}\equiv F_{2}\), which contradicts the hypothesis of Theorem 1.1. If \(\frac{C_{3}b_{1}}{b_{2}}\not\equiv1\), then from

$$fF_{1}-\frac{C_{3}b_{1}}{b_{2}}fF_{2}=q-q\frac{C_{3}b_{1}}{b_{2}} $$

we have \(T(r,f)=S(r,f)\), which is impossible.

Thus we obtain that at least one of \(f(z)F_{1}(z,f)-q(z)\) and \(f(z)F_{2}(z,f)-q(z)\) has infinitely many zeros.