Proof. In view of Lemma 3.4, it is easy to see that $ h_j (j=1, 2, 3) $ are of infinite order. If $ k=1, $ then we have $ b_0h_1+b_1h_1^{\prime}=-h_1\text{e}^{{2\gamma}/{3}}, $ $ b_0h_2+b_1h_2^{\prime}=-\omega^2h_2e^{{2\gamma}/{3}} $ and $ b_0h_3+b_1h_3^{\prime}=-\omega h_3\text{e}^{{2\gamma}/{3}}. $ Eliminating $ \text{e}^{{2\gamma}/{3}} $ from these three equations, we obtain

\begin{equation*} \frac{(h_1h_2h_3)^{\prime}}{h_1h_2h_3}=-\frac{3b_0}{b_1}, \end{equation*}

which shows that $ h_1h_2h_3=\displaystyle{D}\text{e}^{-3b_0z/b_1}, $ where D is a non-zero constant. We thus see that $ h_1h_2h_3=1-\text{e}^{-2\gamma} $ has no zeros, which is impossible. Hence, $ k\geqslant 2 $, and in the following, we will prove that k must be an odd integer.

Since $ h_1h_2h_3=1-\text{e}^{-2\gamma} $ and $ T\left(r,\text{e}^{2\gamma/3}\right)=S(r,h_j) $ for $ j=1, 2, 3, $ and therefore, we must have $ S(r,h_1)=S(r,h_2)=S(r,h_3)=S(r), $ say. Let

\begin{equation*} \alpha_1=h_1^{\prime}/h_1,\;\; \beta_1=h_2^{\prime}/h_2,\;\; \gamma_1=h_3^{\prime}/h_3. \end{equation*}

Since $ h_j(j=1, 2, 3) $ are infinite order entire functions, so α ₁, β ₁ and γ ₁ must be non-constants. In view of $ h_1h_2h_3=1-\text{e}^{-2\gamma} $ and $ T\left(r,\text{e}^{2\gamma/3}\right)=S(r), $ so we have $ \displaystyle N\left(r,1/h_j\right)=S(r) $ for $ j=1, 2, 3, $ and we see that

\begin{equation*} T(r,\alpha_1)=S(r),\;\; T(r,\beta_1)=S(r),\;\; T(r,\gamma_1)=S(r). \end{equation*}

Using the principle of mathematical induction, a simple computation shows that

\begin{equation*} h_1^{(j)}=\alpha_jh_1,\;\; h_2^{(j)}=\beta_jh_2,\;\; h_3^{(j)}=\gamma_jh_3, \end{equation*}

where $ \alpha_j,\; \beta_j $ and γ_j are meromorphic functions satisfying the following recurrence relations, respectively

\begin{equation*} \alpha_{j+1}=\alpha^{\prime}_j+\alpha_1\alpha_j,\;\; \beta_{j+1}=\beta^{\prime}_j+\beta_1\beta_j,\;\; \gamma_{j+1}=\gamma^{\prime}_j+\gamma_1\gamma_j, \end{equation*}

for $ j=0, 1, \ldots, $ where $ \alpha_0=\beta_0=\gamma_0=1. $ In view of our assumption, it is easy to see that

(3.6)\begin{align} \sum_{j=0}^{k}b_j\alpha_j&=-\text{e}^{2\gamma/3}, \end{align}

(3.7)\begin{align} \sum_{j=0}^{k}b_j\beta_j&=-\omega^2\text{e}^{2\gamma/3,} \end{align}

(3.8)\begin{align} \sum_{j=0}^{k}b_j\gamma_j&=-\omega \text{e}^{2\gamma/3}. \end{align}

A simple computation shows that

(3.9)\begin{align} \alpha_j&=\alpha_1^j+\frac{(j-1)j}{2}\alpha_1^{j-2}\alpha_1^{\prime}+\mathcal{P}_{j-2}(\alpha_1),\;\; j\geqslant 2, \end{align}

(3.10)\begin{align} \beta_j&=\beta_1^j+\frac{(j-1)j}{2}\beta_1^{j-2}\beta_1^{\prime}+\mathcal{P}_{j-2}(\beta_1),\;\; j\geqslant 2, \; \end{align}

(3.11)\begin{align} \gamma_j&=\gamma_1^j+\frac{(j-1)j}{2}\gamma_1^{j-2}\gamma_1^{\prime}+\mathcal{P}_{j-2}(\gamma_1),\;\; j\geqslant 2, \end{align}

where $ \mathcal{P}_{j-2}(\alpha_1), $ $ \mathcal{P}_{j-2}(\beta_1) $ and $ \mathcal{P}_{j-2}(\gamma_1) $ are differential polynomials, respectively, in α ₁, β ₁ and γ ₁, and of same degrees $ j-2, $ with constant coefficients. It follows from equations (3.6) and (3.9) that $ \text{e}^{{2\gamma}/{3}} $ is a differential polynomial in α ₁ of degree j with leading term $ b_k\alpha_1^k $. Therefore, we have

\begin{align*} T\left(r,e^{2\gamma/3}\right)&\leqslant k T(r,\alpha_1)+S(r,\alpha_1)\\ m(r,\alpha_1)&\leqslant m\left(r,\text{e}^{2\gamma/3}\right)+S\left(r,\text{e}^{\frac{2}{3}\gamma}\right). \end{align*}

On the other hand, we see that the poles of α ₁ come from the zeros of $ \text{e}^{-2\gamma}-1, $ and hence, we have $ N(r,\alpha_1)\leqslant 2T\left(r,\text{e}^{\gamma}\right)+S\left(r,\text{e}^{\gamma}\right) $. Therefore, we have

\begin{align*} T(r,\alpha_1)\leqslant 3T\left(r,\text{e}^{2\gamma/3}\right)+S\left(r,\text{e}^{2\gamma/3}\right). \end{align*}

By a similar argument, we obtain

\begin{align*} T(r,\beta_1)&\leqslant 3T\left(r,\text{e}^{2\gamma/3}\right)+S\left(r,\text{e}^{2\gamma/3}\right),\\ T(r,\gamma_1)&\leqslant 3T\left(r,\text{e}^{2\gamma/3}\right)+S\left(r,\text{e}^{2\gamma/3}\right). \end{align*}

Thus, each of $ S(r,\alpha_1) $, $ S(r,\beta_1) $ and $ S(r,\gamma_1) $ must be $ S\left(r,\text{e}^{2\gamma/3}\right) $ and vice versa.

By equations (3.6) and (3.9), we obtain

\begin{equation*} b_k\frac{\alpha_1}{\text{e}^{2\gamma}-1}\alpha_1^{k-1}+\frac{1}{\text{e}^{2\gamma}-1}\mathcal{P}(\alpha_1)=-\frac{\text{e}^{2\gamma/3}}{\text{e}^{2\gamma}-1}, \end{equation*}

where $ \mathcal{P}(\alpha_1) $ is a differential polynomial in α ₁ of degree k − 1 with constant coefficients. We also see that

\begin{equation*} m\left(r,\frac{1}{\text{e}^{2\gamma}-1}\right)=S\left(r,\text{e}^{{2\gamma}/{3}}\right)=m\left(r,\frac{\text{e}^{2\gamma/3}}{\text{e}^{2\gamma}-1}\right). \end{equation*}

By Lemma 3.3, we have $ m\left(r,{\alpha_1}/{(\text{e}^{2\gamma}-1)}\right)=S\left(r,\text{e}^{2\gamma/3}\right). $ Similarly, from equations (3.7) and (3.10), it follows that

\begin{equation*} b_k\frac{\beta_1}{\text{e}^{2\gamma}-1}\beta_1^{k-1}+\frac{1}{\text{e}^{2\gamma}-1}\mathcal{P}(\beta_1)=-\frac{\text{e}^{2\gamma/3}}{\text{e}^{2\gamma}-1}, \end{equation*}

and hence, we can show that $ m\left(r,{\beta_1}/{(\text{e}^{2\gamma}-1)}\right)=S\left(r,\text{e}^{2\gamma/3}\right). $ Combining the above two, we have

(3.12)\begin{equation} m\left(r,\frac{\alpha_1\pm \beta_1}{\text{e}^{2\gamma}-1}\right)=S\left(r,\text{e}^{2\gamma/3}\right). \end{equation}

Adding equations (3.6), (3.7) and (3.8), we obtain $ \sum_{j=0}^{k}b_j(\alpha_j+\beta_j+\gamma_j)=0 $. From this, using equations (3.9), (3.10) and (3.11), we obtain

(3.13)\begin{align}\;\;\; & b_k\bigg(\alpha_1^k+\frac{(k-1)k}{2}\alpha_1^{k-2}\alpha_1^{\prime}+\mathcal{P}_{k-2}(\alpha_1)+\beta_1^k+\frac{(k-1)k}{2}\beta_1^{k-2}\beta_1^{\prime}+\mathcal{P}_{k-2}(\beta_1)+ \gamma_1^k \\& +\frac{(k-1)k}{2}\gamma_1^{k-2}\gamma_1^{\prime}+\mathcal{P}_{k-2}(\gamma_1)\bigg)+b_{k-1}\bigg(\alpha_1^{k-1}+\frac{(k-2)(k-1)}{2}\alpha_1^{k-3}\alpha_1^{\prime}+\mathcal{P}_{k-3}(\alpha_1) \nonumber \\ &+\beta_1^{k-1}+\frac{(k-2)(k-1)}{2}\beta_1^{k-3}\beta_1^{\prime}+\mathcal{P}_{k-3}(\beta_1)+\nonumber \gamma_1^{k-1}+\frac{(k-2)(k-1)}{2}\gamma_1^{k-3}\gamma_1^{\prime} \\ &\nonumber+\mathcal{P}_{k-3}(\gamma_1)\bigg)+\sum_{j=0}^{k-2}b_j(\alpha_j+\beta_j+\gamma_j)=0. \end{align}

Moreover, we have

\begin{equation*} \alpha_1+\beta_1+\gamma_1=\frac{(h_1h_2h_3)^{\prime}}{h_1h_2h_3}=\frac{2\gamma^{\prime}}{e^{2\gamma}-1}, \end{equation*}

which yields $ \gamma_1={2\gamma^{\prime}}/{(\text{e}^{2\gamma}-1)}-(\alpha_1+\beta_1). $ Substituting this γ ₁ into equation (3.13) and using equation (3.12), we deduce that

(3.14)\begin{align} & b_k\left(1+(-1)^k\right)\alpha_1^k+b_{k-1}\left(1-(-1)^k\right)\frac{(k-1)k}{2}\alpha_1^{k-2}\alpha_1^{\prime}\\ &\quad+b_{k-1}\left(1-(-1)^k\right)\alpha_1^{k-1}\in \mathcal{D}_{k-2}(\alpha_1)\nonumber \end{align}

and

(3.15)\begin{align} & b_k\left(1+(-1)^k\right)\beta_1^k+b_{k-1}\left(1-(-1)^k\right)\frac{(k-1)k}{2}\beta_1^{k-2}\beta_1^{\prime}\\ &\quad+b_{k-1}\left(1-(-1)^k\right)\beta_1^{k-1}\in \mathcal{D}_{k-2}(\beta_1),\nonumber \end{align}

where $ \mathcal{D}_{k-2}(\alpha_1),\; \mathcal{D}_{k-2}(\beta_1) $ are, respectively, differential polynomials in α ₁, β ₁ of same degree k − 2, with coefficients being functions of small proximity relative to $ e^{\frac{2}{3}\gamma}. $ If k is even, then from equations (3.14) and (4.12), we notice $ 2b_k\alpha_1^k\in\mathcal{D}_{k-2}(\alpha_1) $ and $ 2b_k\beta_1^k\in\mathcal{D}_{k-2}(\beta_1). $

By Lemma 3.3, we have $ m(r,\alpha_1)=S\left(r,e^{2\gamma/3}\right) $ and $ m(r,\beta_1)=S\left(r,e^{2\gamma/3}\right) $. With these and in view of equations (3.6) and (3.7), one can now deduce that $ m\left(r,e^{{2\gamma}/{3}}\right)=S\left(r,\text{e}^{2\gamma/3}\right) $, which leads to a contradiction. Hence, k must be an odd integer and $ k\geqslant 3. $

Proof of Theorem 2.1

Let $ g={L}_k[f] $. Since $ E_f(\mathcal{S})=E_g(\mathcal{S}) $, where $ \mathcal{S}=\{a, b, c\} $, then there exists an entire function γ such that

(4.1)\begin{equation} \frac{(g-a)(g-b)(g-c)}{(f-a)(f-b)(f-c)}=\text{e}^{2\gamma}. \end{equation}

By the Second Fundamental Theorem, we obtain

\begin{align*} T(r,f)&\leqslant \bar{N}\left(r,\frac{1}{f-a}\right)+\bar{N}\left(r,\frac{1}{f-b}\right)+\bar{N}\left(r,\frac{1}{f-c}\right)+S(r,f)\\ &\leqslant \bar{N}\left(r,\frac{1}{g-a}\right)+\bar{N}\left(r,\frac{1}{g-b}\right)+\bar{N}\left(r,\frac{1}{g-c}\right)+S(r,f)\\ &\leqslant 3T(r,g)+S(r,f). \end{align*}

With g being a linear differential polynomial, we have $ T(r,g)\leqslant mT(r,f)+S(r,f), $ for some $ m\in\mathbb{N} $. Therefore $ S(r,f) $ can be replaced by $ S(r,g) $ and vice versa. For the sake of simplicity, we assume $ S(r):=S(r,f)=S(r,g). $ Hence equation (4.1) yields $ T(r,\gamma)=S(r) $.

Differentiating both sides of equation (4.1), we obtain

(4.2)\begin{align} & \left(3g^2-2(a+b+c)g+(ab+bc+ca)\right)g^{\prime} \nonumber\\ &= \left(3f^2f^{\prime}-2(a+b+c)ff^{\prime}+(ab+bc+ca)f^{\prime}+2\gamma^{\prime}(f-a)(f-b)(f-c)\right)\text{e}^{2\gamma}. \end{align}

If z ₀ be a common zero of $ (g-a)(g-b)(g-c) $ and $ (f-a)(f-b)(f-c) $, then $ g(\alpha)\in\mathcal{S} $ and $ f(z_0)\in\mathcal{S}. $ Next, our aim is to show that z ₀ is a zero of $ [g^{\prime}(z)]^3-[f^{\prime}(z)\text{e}^{2\gamma(z)}]^3 $. In order to show this, we need to discuss several cases. Henceforth, we define

\begin{align*} \rho_0&:=\begin{pmatrix} a & b & c \\ a & b & c\\ \end{pmatrix},\; \rho_1:=\begin{pmatrix} a & b & c \\ b & c & a\\ \end{pmatrix},\;\rho_2:=\begin{pmatrix} a & b & c \\ c & a & b\\ \end{pmatrix},\\ \rho_3&:=\begin{pmatrix} a & b & c \\ a & c & b\\ \end{pmatrix},\; \rho_4:=\begin{pmatrix} a & b & c \\ c & b & a\\ \end{pmatrix},\;\rho_5:=\begin{pmatrix} a & b & c \\ b & a & c\\ \end{pmatrix}. \end{align*}

For the set $ \mathcal{S}=\{a, b, c\} $, let us introduce the notation $ f\Leftrightarrow_{\rho} g $ by which we understand that $ f(a)=g(\alpha) $, $ f(b)=g(\beta) $ and $ f(c)=g(\gamma) $, where $ \alpha, \beta, \gamma\in\mathcal{S} $ when ρ is defined by

\begin{align*} \rho&=\begin{pmatrix} a & b & c \\ \alpha & \beta & \gamma\\ \end{pmatrix}. \end{align*}

Since $ \mathcal{S} $ consists of three elements and by the assumption of the result as $ E_f(\mathcal{S})= E_g(\mathcal{S})$, we have exactly $ 3!=6 $ possible cases, which are $ f\Leftrightarrow_{\rho_j} g $ $ (j=0, 1, 2, \ldots, 5) $. We now discuss them one by one.

1. Case 1. If $ f\Leftrightarrow_{\rho_0} g $, then we see that $ f(a)=g(a) $, $ f(b)=g(b) $ and $ f(c)=g(c) $. Hence, for these three situations, from equation (4.2), a simple computation shows that, in each case, we obtain $ g^{\prime}(z_0)=f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}. $ It is clear that $ [g^{\prime}(z_0)]^3=[f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}]^3 $, which shows that z ₀ is a zero of $ [g^{\prime}(z)]^3-[f^{\prime}(z)\text{e}^{2\gamma(z)}]^3 $.

2. Case 2. If $ f\Leftrightarrow_{\rho_1} g $, then we see that $ f(a)=g(b) $, $ f(b)=g(c) $ and $ f(c)=g(a) $. Consequently, by employing equation (4.2), a simple calculation indicates that $ (a-c)g^{\prime}(z_0)=(c-b)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $, $ (b-a)g^{\prime}(z_0)=(a-c)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}. $ and $ (c-b)g^{\prime}(z_0)=(b-a)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}, $ respectively. Clearly, we have

   \begin{align*} (a-c)(b-a)(c-b)[g^{\prime}(z_0)]^3=(c-b)(a-c)(b-a)[f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}]^3, \end{align*}

   and a fact that becomes apparent upon dividing out the factor $ (a-c)(b-a)(c-b) $ shows that $ [g^{\prime}(z_0)]^3=[f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}]^3 $; hence, z ₀ is a zero of $ [g^{\prime}(z)]^3-[f^{\prime}(z)\text{e}^{2\gamma(z)}]^3 $.

3. Case 3. If $ f\Leftrightarrow_{\rho_2} g $, then we see that $ f(a)=g(c) $, $ f(b)=g(a) $ and $ f(c)=g(b) $. Henceforth, with the use of equation (4.2), a basic computation illustrates that $ (a-b)g^{\prime}(z_0)=(b-c)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $, $ (b-c)g^{\prime}(z_0)=(c-a)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $ and $ (c-a)g^{\prime}(z_0)=(a-b)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}, $ respectively. It is evident that

   \begin{align*} (a-b)(b-c)(c-a)[g^{\prime}(z_0)]^3=(b-c)(c-a)(a-b)[f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}]^3, \end{align*}

   which shows upon eliminating the factor $(a-b)(b-c)(c-a)$ that z ₀ is a zero of $ [g^{\prime}(z)]^3-[f^{\prime}(z)e^{2\gamma(z)}]^3 $.

4. Case 4. If $ f\Leftrightarrow_{\rho_3} g $, then we see that $ f(a)=g(a) $, $ f(b)=g(c) $ and $ f(c)=g(b) $. Hence, from equation (4.2), a simple computation shows that $ g^{\prime}(z_0)=f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $, $ (b-a)g^{\prime}(z_0)=(a-c)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $ and $ (c-a)g^{\prime}(z_0)=(a-b)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}, $ respectively. It is clear that

   \begin{align*} (b-a)(c-a)[g^{\prime}(z_0)]^3=(a-c)(a-b)[f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}]^3, \end{align*}

   which shows upon canceling the factor $(b-a)(c-a)$ that z ₀ is a zero of $ [g^{\prime}(z)]^3-[f^{\prime}(z)\text{e}^{2\gamma(z)}]^3 $.

5. Case 5. If $ f\Leftrightarrow_{\rho_4} g $, then we see that $ f(a)=g(c) $, $ f(b)=g(b) $ and $ f(c)=g(a) $. Therefore, by referencing equation (4.2), a simple analysis demonstrates that $ (a-b)g^{\prime}(z_0)=(b-c)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $, $ g^{\prime}(z_0)=f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $ and $ (c-b)g^{\prime}(z_0)=(b-a)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}, $ respectively. It is evident that

   \begin{align*} (a-b)(c-b)[g^{\prime}(z_0)]^3=(b-c)(b-a)[f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}]^3, \end{align*}

   which indicates upon removing the factor $(a-b)(c-b)$ that z ₀ is a zero of $ [g^{\prime}(z)]^3-[f^{\prime}(z)\text{e}^{2\gamma(z)}]^3 $.

6. Case 6. If $ f\Leftrightarrow_{\rho_5} g $, then we see that $ f(a)=g(b) $, $ f(b)=g(a) $ and $ f(c)=g(c) $. Consequently, derived from equation (4.2), a simple calculation shows that $ (a-c)g^{\prime}(z_0)=(c-b)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $, $ (b-c)g^{\prime}(z_0)=(c-a)f^{\prime}(z_0)\text{e}^{2\gamma(z_0)} $ and $ g^{\prime}(z_0)=f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}, $ respectively. Certainly, we have

   \begin{align*} (a-c)(b-c)[g^{\prime}(z_0)]^3=(c-b)(c-a)[f^{\prime}(z_0)\text{e}^{2\gamma(z_0)}]^3, \end{align*}

   which shows upon canceling the factor $(a-c)(b-c)$ that z ₀ is a zero of $ [g^{\prime}(z)]^3-[f^{\prime}(z)\text{e}^{2\gamma(z)}]^3 $.

The estimates in the above cases show that z ₀ is also a zero of $ \left(g^{\prime}\right)^3-\left(f^{\prime}\text{e}^{2\gamma}\right)^3. $ If z ₀ is a zero of $ (g-a)(g-b)(g-c) $ of multiplicity $ p\geqslant 2 $, then z ₀ must be a zero of $ \left(g^{\prime}\right)^3-\left(f^{\prime}\text{e}^{2\gamma}\right)^3 $ of multiplicity $ 3(p-1). $ Thus, z ₀ is not a pole of the function ψ, where ψ is defined by

(4.3)\begin{align} \psi :=\frac{\left(f^{\prime}\text{e}^{2\gamma}\right)^3-\left(g^{\prime}\right)^3 }{(g-a)(g-b)(g-c)}= \frac{\left(f^{\prime}\text{e}^{2\gamma}-g^{\prime}\right)\left(f^{\prime}\text{e}^{2\gamma}-\omega g^{\prime}\right)\left(f^{\prime}\text{e}^{2\gamma}-\omega^2 g^{\prime}\right)}{(g-a)(g-b)(g-c)}, \end{align}

where ω is a non-real cube root of unity. Clearly, ψ is an entire function. Now, we can write

\begin{align*} \frac{f^{\prime}e^{2\gamma}-g^{\prime}}{g-a}=\frac{(g-b)(g-c)f^{\prime}}{(f-a)(f-b)(f-c)}-\frac{g^{\prime}}{g-a}. \end{align*}

By the lemma of logarithmic derivative and Lemma 3.1, we obtain

\begin{align*} m\left(r,\frac{f^{\prime}\text{e}^{2\gamma}-g^{\prime}}{g-a}\right)=S(r). \end{align*}

Similarly, we obtain

\begin{align*} m\left(r,\frac{f^{\prime}\text{e}^{2\gamma}-\omega g^{\prime}}{g-b}\right)=S(r)=m\left(r,\frac{f^{\prime}\text{e}^{2\gamma}-\omega^2g^{\prime}}{g-c}\right). \end{align*}

Thus, we have $ T(r,\psi)=S(r) $.

Next, we distinguish two cases as below.

Case A. Suppose $ T\left(r,\text{e}^{2\gamma}\right)\neq S(r) $. If $ \psi=0, $ then we can write

\begin{align*} \text{e}^{2\gamma}=\omega\left(b_0+b_1\frac{f^{\prime\prime}}{f^{\prime}}+b_2\frac{f^{\prime\prime\prime}}{f^{\prime}}+\ldots+b_k\frac{f^{(k+1)}}{f^{\prime}}\right). \end{align*}

It follows from the lemma of the logarithmic derivative that $ m\left(r,\text{e}^{2\gamma}\right)=S(r), $ which contradicts the assumption.

Hence, $ \psi\not\equiv 0 $, and from equation (4.3), we can write

\begin{align*} \begin{cases} \displaystyle\frac{1}{\text{e}^{2\gamma}f^{\prime}-g^{\prime}}\displaystyle=\frac{1}{\psi}\frac{(\text{e}^{2\gamma}f^{\prime}-\omega g^{\prime})(\text{e}^{2\gamma}f^{\prime}-\omega^2g^{\prime})}{(g-a)(g-b)(g-c)};\\ \displaystyle\frac{1}{\text{e}^{2\gamma}f^{\prime}-\omega g^{\prime}}=\frac{1}{\psi}\frac{(\text{e}^{2\gamma}f^{\prime}- g^{\prime})(\text{e}^{2\gamma}f^{\prime}-\omega^2g^{\prime})}{(g-a)(g-b)(g-c)};\\ \displaystyle\frac{1}{\text{e}^{2\gamma}f^{\prime}-\omega^2g^{\prime}}=\frac{1}{\psi}\frac{(\text{e}^{2\gamma}f^{\prime}- g^{\prime})(\text{e}^{2\gamma}f^{\prime}-\omega g^{\prime})}{(g-a)(g-b)(g-c)}. \end{cases} \end{align*}

It is easy to see that

(4.4)\begin{align} & m\left(\frac{1}{\left(\text{e}^{2\gamma}f^{\prime}-g^{\prime}\right)(g-a)(g-b)(g-c)}\right) \\ &\leqslant m\left(r,\frac{1}{\psi}\right)+m\left(r,\frac{\text{e}^{2\gamma}f^{\prime}-\omega g^{\prime}}{(g-a)(g-b)(g-c)}\right)+m\left(r,\frac{e^{2\gamma}f^{\prime}-\omega^2 g^{\prime}}{(g-a)(g-b)(g-c)}\right) \nonumber\\ &\leqslant m\left(r,\frac{1}{\psi}\right)+m\left(r,\frac{f^{\prime}}{(f-a)(f-b)(f-c)}-\frac{\omega g^{\prime}}{(g-a)(g-b)(g-c)}\right) \nonumber\\ & +m\left(r,\frac{f^{\prime}}{(f-a)(f-b)(f-c)}-\frac{\omega^2 g^{\prime}}{(g-a)(g-b)(g-c)}\right) \nonumber\\ &\leqslant m\left(r,\frac{1}{\psi}\right)+2\;m\left(r,\frac{f^{\prime}}{(f-a)(f-b)(f-c)}\right)+2\;m\left(r,\frac{g^{\prime}}{(g-a)(g-b)(g-c)}\right) \nonumber\\ &\leqslant S(r).\nonumber \end{align}

Similarly, we can obtain

(4.5)\begin{align} m\left(\frac{1}{\left(\text{e}^{2\gamma}f^{\prime}-\omega g^{\prime}\right)(g-a)(g-b)(g-c)}\right)&\leqslant S(r), \end{align}

(4.6)\begin{align} m\left(\frac{1}{\left(\text{e}^{2\gamma}f^{\prime}-\omega^2g^{\prime}\right)(g-a)(g-b)(g-c)}\right)&\leqslant S(r). \end{align}

Therefore, in view of equations (4.3), (4.4), (4.5) and (4.6), we obtain

\begin{align*} & m\left(r,\frac{1}{(g-a)^4(g-b)^4(g-c)^4}\right)\\ &= m\left(r,\frac{\psi}{\left(\text{e}^{2\gamma}f^{\prime}-g^{\prime}\right)\left(\text{e}^{2\gamma}f^{\prime}-\omega g^{\prime}\right)\left(\text{e}^{2\gamma}f^{\prime}-\omega^2g^{\prime}\right)(g-a)^3(g-b)^3(g-c)^3}\right)\\ &\leqslant m\left(\frac{1}{\left(\text{e}^{2\gamma}f^{\prime}-g^{\prime}\right)(g-a)(g-b)(g-c)}\right)+m\left(\frac{1}{\left(\text{e}^{2\gamma}f^{\prime}-\omega g^{\prime}\right)(g-a)(g-b)(g-c)}\right)\\ &\quad\quad+m\left(\frac{1}{\left(\text{e}^{2\gamma}f^{\prime}-\omega^2g^{\prime}\right)(g-a)(g-b)(g-c)}\right)+m(r,\psi)\\ &\leqslant S(r), \end{align*}

which yields

(4.7)\begin{equation} m\left(r,\frac{1}{(g-a)(g-b)(g-c)}\right)=S(r). \end{equation}

It follows from equation (4.7) that

\begin{equation*} m\left(r,\frac{1}{g-a}\right)=S(r),\;\; m\left(r,\frac{1}{g-b}\right)=S(r)\;\; \text{and}\;\; m\left(r,\frac{1}{g-c}\right)=S(r). \end{equation*}

Therefore, we obtain

(4.8)\begin{equation} 3T(r,g)=N\left(r,\frac{1}{g-a}\right)+N\left(r,\frac{1}{g-b}\right)+N\left(r,\frac{1}{g-c}\right)+S(r). \end{equation}

A simple computation using equations (4.1) and (4.2) shows that

(4.9)\begin{equation} \frac{\phi(g)g^{\prime}}{(g-a)(g-b)(g-c)}= \frac{\phi(f)f^{\prime}}{(f-a)(f-b)(f-c)}+2\gamma^{\prime}, \end{equation}

where $ \phi(h)=3h^2-2(a+b+c)h+(ab+bc+ca). $ It follows from equation (4.9) that

(4.10)\begin{align} & \frac{\phi^3(g)\left(g^{\prime}\right)^3}{(g-a)^3(g-b)^3(g-c)^3}\\ &=\frac{\phi^3(f)\left(f^{\prime}\right)^3}{(f-a)^3(f-b)^3(f-c)^3}+\frac{6\phi^2(f)\left(f^{\prime}\right)^2\gamma^{\prime}}{(f-a)^2(f-b)^2(f-c)^2} +\frac{12\phi(f)f^{\prime}\left(\gamma^{\prime}\right)^2}{(f-a)(f-b)(f-c)}\nonumber\\ &\quad\quad+8\left(\gamma^{\prime}\right)^3.\nonumber \end{align}

By equation (4.1) and eliminating γ, we obtain from equation (4.3) that

\begin{align*} \left(g^{\prime}\right)^3=\frac{(g-a)^3(g-b)^3(g-c)^3}{(f-a)^3(f-b)^3(f-c)^3}\left(f^{\prime}\right)^3-\psi (g-a)(g-b)(g-c) \end{align*}

from which we obtain

(4.11)\begin{align} & \frac{\left(g^{\prime}\right)^3}{(g-a)^3(g-b)^3(g-c)^3}\\ &=\frac{\left(f^{\prime}\right)^3}{(f-a)^3(f-b)^3(f-c)^3}-\frac{\psi}{(g-a)^2(g-b)^2(g-c)^2}\nonumber \end{align}

By equations (4.10) and (4.11), we deduce that

(4.12)\begin{align} & \frac{\psi\phi^3(g)}{(g-a)^2(g-b)^2(g-c)^2}\\ &=\frac{[\phi(g)-\phi(f)][\phi(g)-\omega\phi(f)][\phi(g)-\omega^2\phi(f)]\left(f^{\prime}\right)^3}{(f-a)^3(f-b)^3(f-c)^3}-\frac{6\phi^2(f)\left(f^{\prime}\right)^2\gamma^{\prime}}{(f-a)^2(f-b)^2(f-c)^2}\nonumber\\ &\quad\quad-\frac{12\phi(f)f^{\prime}\left(\gamma^{\prime}\right)^2}{(f-a)(f-b)(f-c)}-8\left(\gamma^{\prime}\right)^3. \end{align}

Since $ E_f(\mathcal{S})=E_g(\mathcal{S}) $, i.e. the functions fand g share the set $ \mathcal{S}=\{a, b, c\} $ CM, and hence, all the zeros of $ (f-a)(f-b)(f-c) $ must be zeros of $ (g-a)(g-b)(g-c) $. Let z ₁ be such a zero of $ (f-a)(f-b)(f-c) $ and $ (g-a)(g-b)(g-c) $. In view of equation (4.1), we see that equation (4.12) can be written as

(4.13)\begin{align} \psi\phi^3(g)(g-a)(g-b)(g-c)&=[\phi(g)-\phi(f)][\phi(g)-\omega\phi(f)][\phi(g)-\omega^2\phi(f)]\left(f^{\prime}\right)^3\left(e^{2\gamma}\right)^3 \nonumber\\ &\quad-6\phi^2(f)\left(f^{\prime}\right)^2\gamma^{\prime}\left(e^{2\gamma}\right)^2(g-a)(g-b)(g-c) \nonumber\\ &\quad-12\phi(f)f^{\prime}\left(\gamma^{\prime}\right)^2e^{2\gamma}(g-a)^2(g-b)^2(g-c)^2\nonumber\\ &\quad-8\left(\gamma^{\prime}\right)^3(g-a)^3(g-b)^3(g-c)^3. \end{align}

Considering equation (4.13) and noticing that the term associated with $(g-a)(g-b)(g-c)$ will vanish at z ₁, it becomes evident that z ₁ is a zero of $[\phi(g)-\phi(f)][\phi(g)-\omega\phi(f)][\phi(g)-\omega^2\phi(f)]\left(f^{\prime}\right)^3\left(\text{e}^{2\gamma}\right)^3$. Consequently, this implies that z ₁ must be a zero of $\left(\phi(g)-\phi(f)\right)\left(\phi(g)-\omega\phi(f)\right)\left(\phi(g)-\omega^2\phi(f)\right)\left(f^{\prime}\right)^3$. We observe that the poles of the right-hand side of equation (4.12) are all simple. Hence the zeros of $ (g-a)(g-b)(g-c) $ are simple as long as they are not the zeros of ψ. Therefore, we deduce that

\begin{align*} \begin{cases} \displaystyle N\left(r,\frac{1}{g-a}\right)=\bar{N}\left(r,\frac{1}{g-a}\right)+S(r);\\ \displaystyle N\left(r,\frac{1}{g-b}\right)=\bar{N}\left(r,\frac{1}{g-b}\right)+S(r);\\ \displaystyle N\left(r,\frac{1}{g-c}\right)=\bar{N}\left(r,\frac{1}{g-c}\right)+S(r). \end{cases} \end{align*}

Differentiating both sides of equation (4.3) and then eliminating $ \text{e}^{2\gamma} $, we obtain

(4.14)\begin{align} & \left(3\psi(f^{\prime\prime}+2f^{\prime}\gamma^{\prime})-f^{\prime}\psi^{\prime}\right)(g-a)(g-b)(g-c)\\ &=\left(\psi\phi(g)f^{\prime}-(6f^{\prime}\gamma^{\prime}+3f^{\prime\prime})\left(g^{\prime}\right)^2+3f^{\prime}g^{\prime}g^{\prime\prime}\right)g^{\prime}\nonumber \end{align}

Using equation (4.14), it is easy to see that any simple zero of $ (g-a)(g-b)(g-c) $ must be a zero of

\begin{equation*} \psi\phi(g)f^{\prime}-(6f^{\prime}\gamma^{\prime}+3f^{\prime\prime})\left(g^{\prime}\right)^2+3f^{\prime}g^{\prime}g^{\prime\prime}. \end{equation*}

Set

(4.15)\begin{align} \psi_1 : =\frac{\psi\phi(g)f^{\prime}-(6f^{\prime}\gamma^{\prime}+3f^{\prime\prime})\left(g^{\prime}\right)^2+3f^{\prime}g^{\prime}g^{\prime\prime}}{(f-a)(f-b)(f-c)}. \end{align}

Note that the functions f and g share the set $ \{a, b, c\} $ CM and almost all zeros of $ (g-a)(g-b)(g-c) $ are simple; hence, we have $ N(r,\psi_1)=S(r). $ Again by the lemma of logarithmic derivative, we have $ m(r,\psi_1)=S(r) $. Therefore, it is easy to see that $ T(r,\psi_1)=N(r,\psi_1)+m(r,\psi_1)=S(r). $

If $ \psi_1\not\equiv 0 $, then from equation (4.15), we obtain

\begin{align*} 3T(r,f)&\leqslant m\left(r,(f-a)(f-b)(f-c)\psi_1\right)+S(r)\\ &\leqslant m\left(r,f^{\prime}\right)+2m(r,g)+S(r)\\ &\leqslant T(r,f)+2T(r,g)+S(r)\\ &\leqslant 3T(r,f)+S(r). \end{align*}

It follows that $ T(r,f)=T(r,g)+S(r) $. In view of equation (4.8) and the estimates just before equation (4.14), a simple computation shows that

\begin{align*} 3T(r,f)=\bar{N}\left(r,\frac{1}{f-a}\right)+\bar{N}\left(r,\frac{1}{f-b}\right)+\bar{N}\left(r,\frac{1}{f-c}\right)+S(r), \end{align*}

which implies that

\begin{align*} m\left(r,\frac{1}{f-a}\right)=S(r),\;\; m\left(r,\frac{1}{f-b}\right)=S(r)\;\; \text{and} \;\;m\left(r,\frac{1}{f-c}\right)=S(r). \end{align*}

Therefore, from equation (4.1) and in view of the lemma of the logarithmic derivative, we must have $ T\left(r,\text{e}^{2\gamma}\right)=S(r,f) $, which contradicts Case A.

If $ \psi_1\equiv 0 $, then from equations (4.14) and (4.15), we obtain

\begin{equation*} \frac{\psi^{\prime}}{\psi}=3\left(2\gamma^{\prime}+\frac{f^{\prime\prime}}{f^{\prime},}\right) \end{equation*}

which shows that

(4.16)\begin{equation} \left(\text{e}^{2\gamma}f^{\prime}\right)^3=\eta\psi, \end{equation}

where η is a non-zero complex number. Using equations (4.3) and (4.16), a simple computation shows that

(4.17)\begin{equation} \left(g^{\prime}\right)^3=-\psi((g-a)(g-b)(g-c)-\eta)=-\psi(g-\sigma_1)(g-\sigma_2)(g-\sigma_3), \end{equation}

where σ ₁, σ ₂ and σ ₃ are three complex numbers.

If $ \sigma_1\neq \sigma_2$, then it is easy to see that

(4.18)\begin{equation} m\left(r,\frac{1}{g^{\prime}}\right)=m\left(r,\frac{-\left(g^{\prime}\right)^2}{\psi(g-\sigma_1)(g-\sigma_2)(g-\sigma_3)}\right)=S(r,g)=S(r). \end{equation}

Again, since g is a differential polynomial in f, it follows that

\begin{align*} m\left(r,\frac{1}{(f-a)(f-b)(f-c)}\right)&\leqslant m\left(r,\frac{g^{\prime}}{(f-a)(f-b)(f-c)}\right)+m\left(r,\frac{1}{g^{\prime}}\right)+S(r)\\ &\leqslant S(r). \end{align*}

By equation (4.1), we obtain

\begin{equation*} m\left(r,\text{e}^{2\gamma}\right)=m\left(r,\frac{1}{(f-a)(f-b)(f-c)}\right)+S(r)=S(r), \end{equation*}

which shows that $ T\left(r,\text{e}^{2\gamma}\right)=m\left(r,\text{e}^{2\gamma}\right)=S(r), $ and this contradicts the assumption of Case A. Hence, we must have $ \sigma_1=\sigma_2 $. Similarly, we can prove that $ \sigma_2=\sigma_3$ and $ \sigma_3=\sigma_1=\zeta (\text{say}) $, by following the above calculation. Therefore, it is easy to see that $ \sigma_1=\sigma_2=\sigma_3. $ Thus, we have

(4.19)\begin{equation} \left(g^{\prime}\right)^3=-\psi(g-\zeta)^3. \end{equation}

Combining equations (4.1), (4.16) and (4.17), we deduce that

(4.20)\begin{equation} (g-a)(g-b)(g-c)(g-\zeta)=c_1\psi_2(f-a)(f-b)(f-c), \end{equation}

where c ₁ is a non-zero constant satisfying $ c_1^3=-\eta $ and $ \psi_2=g^{\prime}/f^{\prime}. $ From equation (4.16), we see that $\displaystyle N\left(r,1/f^{\prime}\right)=S(r). $ Thus, we have $ N(r,\psi_2)=S(r). $ By the lemma of the logarithmic derivative, we obtain $ m(r,\psi_2)=S(r) $, and hence, $ T(r,\psi_2)=S(r). $ Since g cannot be a constant, we obtain $ \psi_2\not\equiv 0. $ It follows from equation (4.20) that

(4.21)\begin{equation} 4T(r,g)=3T(r,f)+S(r). \end{equation}

We define

(4.22)\begin{equation} \psi_3 : =\frac{g^{\prime}}{g-\zeta}. \end{equation}

Hence from equation (4.19), we have $ \psi_3^3=-\psi, $ which implies that $ T\left(r,\psi_3\right)=S(r) $ and $ \psi_3\not\equiv 0. $ By equations (4.1) and (4.2), we have

(4.23)\begin{align} \phi(f)f^{\prime}\text{e}^{2\gamma}= \phi(g)(g-\zeta)\psi_3-2\gamma^{\prime}(g-a)(g-b)(g-c). \end{align}

By equation (4.16), it is easy to see that $ T\left(f^{\prime}e^{2\gamma}\right)=S(r). $ Therefore, it follows from equation (4.23) that

(4.24)\begin{equation} 2T(r,f)=2T(r,g)+S(r)\;\;\; \text{or}\;\;\; 2 T(r,f)=3T(r,g)+S(r) \end{equation}

according to $ \psi_3=\displaystyle2\gamma^{\prime}/3 $ or $ \psi_3=\gamma^{\prime} $. Combining equations (4.21) and (4.24), we obtain $ T(r,f) =S(r)$, which contradicts the assumption in Case A.

Case B. Let $ T\left(r,e^{2\gamma}\right)=S(r). $ If $ \text{e}^{2\gamma}=1, $ then by equation (4.1), we see that

\begin{equation*} (f-a)(f-b)(f-c)\equiv (g-a)(g-b)(g-c), \end{equation*}

which implies that either $ f\equiv g $ or

(4.25)\begin{align} f^2+fg+g^2-(a+b+c)(f+g)+(ab+bc+ca)=0. \end{align}

If $ f\equiv g,$ then it can be easily solved because it is an ordinary differential equation with constant coefficients. A simple computation shows that the general entire solution is $ f(z)=\sum_{j=1}^{k}a_j\text{e}^{\lambda_j z}, $ where $ \lambda_j\; (j=1, \ldots, k) $ are distinct roots of the equation $ b_kz^k+\cdots+b_1z+b_0=1. $

In the next case, since equation (4.25) is of second degree, in order to obtain f and g explicitly, we require some transformations to reduce the equation in its canonical form. By comparing equation (4.25) with the standard second-degree equation $ Af^2+2Hfg+Bg^2+2Gf+2Fg+C=0, $ we have $ A=1=B,\; H=1/2,\; G=-\displaystyle(a+b+c)/2=F $ and $ C=ab+bc+ca. $ We need to calculate Δ and D to find the nature of equation (4.25), where

\begin{align*} \Delta &= \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} =ABC+2FGH-AF^2-BG^2-CH^2,\;\;\; D=AB-H^2. \end{align*}

It is easy to see that $ \Delta=-\left((a-b)^2+(b-c)^2+(c-a)^2\right)/8 $ and $ D=\displaystyle 3/4. $ Since $ a,\; b $ and $\; c $ are distinct constants, it is easy to see that $ \Delta\neq 0 $ and D > 0. Therefore, equation (4.25) represents an ellipse. To obtain the canonical form of equation (4.25), we first translate the origin $ (0,0) $ to the point (h, k). Then the transformation formulae are $ f(z)=\mathcal{F}(z)+h $ and $ g(z)=\mathcal{G}(z)+k $, where (h, k) satisfies the equations $ Ah+Bk+G=0 $ and $ Hh+Bk+F=0 $. Solving these two equations, we obtain

\begin{equation*} (h,k)=\left(\frac{HF-BG}{AB-H^2},\;\frac{HG-AF}{AB-H^2}\right)=\left(\frac{a+b+c}{4},\; \frac{a+b+c}{4}\right). \end{equation*}

Under this transformation, equation (4.25) transformed to $ \mathcal{F}^2+\mathcal{FG}+\mathcal{G}^2+\tau=0, $ where

\begin{equation*} \tau=\frac{\Delta}{D}=-\frac{(a-b)^2+(b-c)^2+(c-a)^2}{6}. \end{equation*}

To remove the product term $ \mathcal{FG} $, we take the transformation, namely, rotation through an angle θ, and the transformation formulae are

\begin{align*} \mathcal{F}&=\mathcal{U}\cos\theta-\mathcal{V}\sin\theta,\\ \mathcal{G}&=\mathcal{U}\cos\theta+\mathcal{V}\sin\theta, \end{align*}

where $ \theta=({1}/{2})\tan^{-1}\left(2H/(A-B)\right)=\pi/4. $ Thus, we have $ \mathcal{F}=\displaystyle(\mathcal{U}-\mathcal{V})/\sqrt{2} $ and $ \mathcal{G}=\displaystyle(\mathcal{U}+\mathcal{V})/\sqrt{2}. $ Applying this transformation, the equation takes the form $ \alpha\mathcal{U}^2+\beta\mathcal{V}^2+\tau=0 $, and by the invariance of this transformation, we must have $ \alpha+\beta=A+B=2 $ and $ \alpha\beta-0^2=AB-H^2=\displaystyle 3/4 $. A simple computation shows that $ (\alpha, \beta)=\left(3/2,\; 1/2\right) $ or $ (\alpha, \beta)=\left(1/2,\; 3/2\right) $. Therefore, the transformed equation is one of the following

\begin{align*} \frac{3\mathcal{U}^2}{2}+\frac{\mathcal{V}^2}{2} &=\frac{1}{6}\left((a-b)^2+(b-c)^2+(c-a)^2\right),\\ \frac{\mathcal{U}^2}{2}+\frac{3\mathcal{V}^2}{2}&=\frac{1}{6}\left((a-b)^2+(b-c)^2+(c-a)^2\right). \end{align*}

Denoting η by $ \eta=\left((a-b)^2+(b-c)^2+(c-a)^2\right) $, we discuss the following two cases.

Subcase 1. If $ 3\mathcal{U}^2/2+\mathcal{V}^2/2 =\left((a-b)^2+(b-c)^2+(c-a)^2\right)/6, $ then we have

(4.27)\begin{align} \left(\frac{\mathcal{U}}{\pm\displaystyle\frac{\sqrt{\eta}}{3}}\right)^2+\left(\frac{\mathcal{V}}{\pm\sqrt{\displaystyle\frac{\eta}{3}}}\right)^2=1. \end{align}

This is a Fermat-type equation of the form $ F^2(z)+G^2(z)=1 $, and hence, by the result of Gross [Reference Gross4], the only non-constant entire solutions are $ F(z)=\sin(h(z)) $ and $ G(z)=\cos(h(z)) $, where h(z) is an entire function. Therefore, by equation (4.27), we obtain $ \mathcal{U}(z)=\pm\displaystyle({\sqrt{\eta}}/{3})\sin(h(z)) $ and $ \mathcal{V}(z)=\pm{\displaystyle({\sqrt{\eta/{3}}})}\cos(h(z)). $ Hence, f and g can be expressed as

\begin{align*} f(z)&=\mathcal{F}(z)+h\\ &=(\mathcal{U}-\mathcal{V})/\sqrt{2}+(a+b+c)/4\\ &=\pm \sqrt{\eta/54}\left(\sqrt{3}\sin(h(z))-3\cos(h(z))\right)+(a+b+c)/4 \end{align*}

and

\begin{align*} g(z)&=\mathcal{G}(z)+k\\ &=(\mathcal{U}+\mathcal{V})/\sqrt{2}+(a+b+c)/4\\ &=\pm \sqrt{\eta/54}\left(\sqrt{3}\sin(h(z))+3\cos(h(z))\right)+(a+b+c)/4. \end{align*}

Case 2. If $\mathcal{U}^2/2+3\mathcal{V}^2/2=\left((a-b)^2+(b-c)^2+(c-a)^2\right)/6, $ then we have

(4.28)\begin{align} \left(\frac{\mathcal{U}}{\pm\sqrt{\displaystyle\frac{\eta}{3}}}\right)^2+\left(\frac{\mathcal{V}}{\pm\displaystyle\frac{\sqrt{\eta}}{3}}\right)^2=1. \end{align}

By the similar argument as in Case 1, we have

\begin{align*} \mathcal{U}(z)=\pm\sqrt{\displaystyle\frac{\eta}{3}}\sin(h(z))\;\; \text{and}\;\; \mathcal{V}(z)=\pm\displaystyle\frac{\sqrt{\eta}}{3}\cos(h(z)). \end{align*}

Therefore, f and g can be expressed as

\begin{align*} f(z)&=\mathcal{F}(z)+h\\ &=(\mathcal{U}-\mathcal{V})/\sqrt{2}+(a+b+c)/4\\ &=\pm \sqrt{\eta/54}\left(3\sin(h(z))-\sqrt{3}\cos(h(z))\right)+(a+b+c)/4 \end{align*}

and

\begin{align*} g(z)&=\mathcal{G}(z)+k\\ &=(\mathcal{U}+\mathcal{V})/\sqrt{2}+(a+b+c)/4\\ &=\pm \sqrt{\eta/54}\left(3\sin(h(z))+\sqrt{3}\cos(h(z))\right)+(a+b+c)/4. \end{align*}

We now assume that $ e^{2\gamma}\neq 1. $

Under the transformation $ f(z)=\mathfrak{F}(z)+\displaystyle (a+b+c)/3 $and$ g(z)=\mathfrak{G}(z)+\displaystyle (a+b+c)/3 $, equation (4.1) is transformed to

(4.29)\begin{align} (\mathfrak{F}-\mathcal{B})(\mathfrak{F}-\omega\mathcal{B})(\mathfrak{F}-\omega^2\mathcal{B})\text{e}^{2\gamma}=(\mathfrak{G}-\mathcal{B})(\mathfrak{G}-\omega \mathcal{B})(\mathfrak{G}-\omega^2\mathcal{B}), \end{align}

where $\displaystyle \mathcal{B}=(2a-b-c)/3 $, $ \mathcal{B}\omega \displaystyle=(2b-c-a)/3 $ and $ \mathcal{B}\omega^2\displaystyle =(2c-a-b)/3, $ and so we have

\begin{equation*} \omega^2(2a-b-c)=\omega (2b-c-a)= 2c-a-b. \end{equation*}

We set three functions $ h_j(z)\;(j=1, 2 3) $ as follows:

(4.30)\begin{align} h_1(z)&=\mathfrak{F}(z)-\text{e}^{-\frac{2}{3}\gamma}\mathfrak{G}(z), \end{align}

(4.31)\begin{align}h_2(z)&=\mathfrak{F}(z)-\omega \text{e}^{-\frac{2}{3}\gamma}\mathfrak{G}(z), \end{align}

(4.32)\begin{align} h_3(z)&=\mathfrak{F}(z)-\omega^2\text{e}^{-\frac{2}{3}\gamma}\mathfrak{G}(z). \end{align}

Then, we have

(4.33)\begin{align} h_1h_2h_3=\mathcal{B}^3(1-\text{e}^{-2\gamma})\neq 0. \end{align}

In view of equation (4.33), we note that $ h_j(z)\; (j=1, 2, 3) $ all are non-constant entire functions satisfying $\displaystyle N\left(r,1/h_j(z)\right)=S(r) $ for $ j=1, 2, 3. $ Let $ \alpha_1=h_1^{\prime}/h_1 $, $ \beta_1=h_2^{\prime}/h_2 $ and $ \gamma_1=h_3^{\prime}/h_3. $ Then, $ \alpha_1,\; \beta_1 $ and γ ₁ are non-constant meromorphic functions, and employing the lemma of the logarithmic derivative, we obtain that $ T(r,\alpha_1)=S(r), $ $ T(r,\beta_1)=S(r) $ and $ T(r,\gamma_1)=S(r). $ By the principal of the mathematical induction, it can be seen that

\begin{equation*} h_1^{(j)}=\alpha_jh_1,\;\; h_2^{(j)}=\beta_jh_2\;\; \text{and}\;\; h_3^{(j)}=\gamma_jh_3, \end{equation*}

where α_j, β_j and γ_j are meromorphic functions satisfying $ T(r,\alpha_j)=T(r,\beta_j)=T(r,\gamma_j)=S(r) $, for $ j=1, 2, \ldots $, and the following recurrence formulas, respectively,

\begin{align*} \alpha_{j+1}=\alpha^{\prime}_j+\alpha_j\alpha_1,\; \beta_{j+1}=\beta^{\prime}_j+\beta_j\beta_1\; \text{and}\; \gamma_{j+1}=\gamma^{\prime}_j+\gamma_j\gamma_1, \end{align*}

for $ j=0, 1, 2, \ldots $ and $ \alpha_0=\beta_0=\gamma_0=1. $

From equations (4.30), (4.31) and (4.32), we obtain

\begin{align*} \mathfrak{F}(z)&=\frac{1}{3}\left(h_1(z)+h_2(z)+h_3(z)\right)\\ \mathfrak{G}(z)&=-\frac{1}{3}\left(h_1(z)\text{e}^{2\gamma/3}+\omega^2h_2(z)\text{e}^{2\gamma/3}+\omega h_3(z)\text{e}^{2\gamma/3}\right). \end{align*}

Thus, we have

(4.34)\begin{align} \begin{cases} f(z)&=\dfrac{1}{3}\left(h_1(z)+h_2(z)+h_3(z)+(a+b+c)\right)\\ g(z)&=-\dfrac{1}{3}\left(\left(h_1(z)+\omega^2h_2(z)+\omega h_3(z)\right)\text{e}^{2\gamma/3}-(a+b+c)\right). \end{cases} \end{align}

It is easy to see that

\begin{equation*} T(r,h_j)=\displaystyle\frac{1}{3}T(r,f)+S(r)\;\;\mbox{for}\;\; j=1, 2, 3. \end{equation*}

By equation (4.34) and in view of the definition of g, we deduce that

(4.35)\begin{align} \sum_{j=0}^{k}\frac{b_j}{3}&\bigg(h_1^{(j)}(z)+h_2^{(j)}(z)+h_3^{(j)}(z)\bigg)+\frac{b_0}{3}(a+b+c)\\ &=-\frac{\left(h_1(z)+\omega^2h_2(z)+\omega h_3(z)\right)\text{e}^{2\gamma/3}-(a+b+c)}{3}.\nonumber \end{align}

Comparing both sides of equation (4.35) and then employing Lemma 3.2, we obtain the following set of equations:

\begin{align*} \begin{cases} \displaystyle\sum_{j=0}^{k}b_jh_1^{(j)}(z)=-h_1(z)\text{e}^{2\gamma/3},\\ \displaystyle\sum_{j=0}^{k}b_jh_2^{(j)}(z)=-\omega^2h_2(z)\text{e}^{2\gamma/3},\\ \displaystyle\sum_{j=0}^{k}b_jh_3^{(j)}(z)=-\omega h_3(z)\text{e}^{2\gamma/3}, \end{cases} \end{align*}

with $ 3b_{-1}+(b_0-1)(a+b+c)=0. $

Proof of Theorem 2.6

Following the proof of Theorem 2.1, we notice that $ h_1, h_2 $ and h ₃ are entire functions. Let $ h_j^{*}=3h_j/(2a-b-c) $, $ j=1, 2, 3. $ Then we have $ h_1^{*}h_2^{*}h_3^{*}=1-\text{e}^{-2\gamma} $ and $ T\left(r,\text{e}^{2\gamma/3}\right)=S(r,h_j^{*}) $ for $ j=1,\; 2,\; 3. $ We note that the functions $ h_j^{*} $ $ j=1, 2, 3 $ are still satisfied with the equations in equation (2.1).

If k = 1 or k is any even integer, then by Lemma 3.5, we see that γ must be a constant. Let $ \text{e}^{2\gamma}=1, $ then by the same argument as in Case B, we have $ f\equiv g $ or $ f^2+fg+g^2-(a+b+c)(f+g)+ab+bc+ca=0, $ and hence, conclusions (1) and (2), respectively, hold in Theorem 2.5.

Suppose that $ \text{e}^{2\gamma}\neq 1, $ then by equations (2.1) and (4.33), there exist non-zero constants A ₁, A ₂, A ₃ and λ such that $ h_1(z)={A}_1e^{\lambda z}, $ and hence, we obtain $ h_2(z)={A}_2\text{e}^{\omega^2\lambda z} $ and $ h_3(z)={A}_3 \text{e}^{\omega\lambda z} $, where $ {A}_1{A}_2{A}_3=\left(\displaystyle (2a-b-c)/3\right)^3\left(1-\text{e}^{-2\gamma}\right). $ Assuming $ \rho=e^{2\gamma/3} $, we have

\begin{align*} {A}_1{A}_2{A}_3=\left(\displaystyle (2a-b-c)/3\right)^3\left(\displaystyle(\rho^3-1)/\rho^3\right), \end{align*}

and a simple computation shows that

\begin{align*} \sum_{j=0}^{k}b_j\lambda^j=-\rho,\;\;\; \sum_{j=0}^{k}b_j\omega^{2j}\lambda^j=-\omega^2\rho,\;\;\; \sum_{j=0}^{k}b_j\omega\lambda^j=-\omega\rho. \end{align*}

In view of the functional forms of f and g, respectively, in equation (4.34), we obtain

\begin{align*} f(z)&=\;\;\;\frac{1}{3}\left({A}_1\text{e}^{\lambda z}+{A}_2\text{e}^{\omega^2\lambda z}+{A}_3 \text{e}^{\omega\lambda z}+(a+b+c)\right),\\ g(z)&=-\frac{1}{3}\left(\rho{A}_1\text{e}^{\lambda z}+\rho\omega^2{A}_2\text{e}^{\omega^2\lambda z}+\rho\omega{A}_3 \text{e}^{\omega\lambda z}-(a+b+c)\right). \end{align*}

If f is a finite order entire function, then g is also of finite order because f ^ʹ, $ f'' $, $ \ldots $, $ f^{(k)} $ are all entire functions of finite order. Using equations (4.1), (4.30), (4.31) and (4.32), we observe that h ₁, h ₂ and h ₃ are also functions of finite order. Consequently, by utilizing equation (2.1) and Lemma 3.4, we can deduce that γ must be a constant. As a result, it becomes evident that f and g still satisfy one of the cases outlined in Theorem 2.5. This completes the proof.