Some properties of algebraic difference equations of first order

Abstract

We prove that if $g(z)$ is a finite-order transcendental meromorphic solution of

$$\bigl(\triangle_{c} g(z)\bigr)^{2}=A(z)g(z)g(z+c)+B(z), $$

where $A(z)$ and $B(z)$ are polynomials such that $\deg A(z)>0$, then

$$1 \leq\rho(g)=\max\biggl\{ \lambda(g), \lambda\biggl(\frac{1}{g}\biggr) \biggr\} . $$

1 Introduction

Steinmetz [1] and Bank and Kaufman [2] proved that the equation

$$\bigl(g'\bigr)^{n}=R(z,g) $$

can be reduced into a list of six simple differential equations by a suitable Möbius transformation with polynomial coefficients, which include

$$ \bigl(g'\bigr)^{2}=p(z) \bigl(g-q(z)\bigr)^{2}(g- \zeta) (g-\eta), $$

(1.1)

where ζ, η are constant, and $p(z), q(z)$ are rational functions. Let $q(z)\in\mathbb{C}$. Then equation (1.1) can be transformed into

$$\bigl(g'\bigr)^{2}=P(z) \bigl(g^{2}-1\bigr). $$

Ishizaki and Korhonen [3] investigated meromorphic solutions of

$$ \bigl(\triangle g(z)\bigr)^{2}=P(z) \bigl(g(z)g(z+1)-Q(z)\bigr). $$

(1.2)

They proved that equation (1.2) possesses a continuous limit to the equation

$$\bigl(g'\bigr)^{2}=P(z) \bigl(g^{2}-1\bigr), $$

which extends to solutions in certain cases.

We assume that the reader is familiar with the basic notions of Nevanlinna theory (see, e.g., [4, 5]). Of late, several scholars [3, 6–14] studied the properties of finite-order meromorphic solutions of algebraic difference equations and obtained many interesting results.

For the special case of (1.2), Whittaker [15] has shown that the equation

$$g(z+1)=q(z)g(z), $$

where $q(z)$ is a meromorphic function of finite order $\rho(q)$, has a meromorphic solution g such that $\rho(q)\leq\rho(g)\leq\rho(q)+1$. Here $\rho(g)$ denotes the order of growth of the meromorphic function $g(z)$.

Chen [7] has extended this above result and proved that the Pielou logistic equation

$$g(z+1)=\frac{R(z)g(z)}{P(z)+Q(z)g(z)}, $$

where $R(z)$, $P(z)$, and $Q(z)$ are polynomials with $P(z)R(z)Q(z)\not\equiv0$, has a finite-order transcendental meromorphic solution g such that $1\leq\rho(g)$.

Replacing $g(z+1)$ with $\triangle g(z)$, Ishizaki [10] was concerned with the growth and value distributions of transcendental meromorphic solutions of the algebraic difference equation

$$\bigl(\triangle g(z)\bigr)^{2}=P(z)g(z). $$

In 2014, Liu [12] considered the Nevanlinna growth of an equation related to (1.2). It is interesting to consider some properties of (1.2), and our results will be stated in Section 2.

2 Main results

Theorem 2.1

Let $c\in\mathbb{C}\setminus\{0\}$, and let $A(z)$ and $B(z)$ be polynomials such that $\deg A(z)>0$. If $g(z)$ is a finite-order transcendental meromorphic solution of

$$ \bigl(\triangle_{c} g(z)\bigr)^{2}=A(z)g(z)g(z+c)+B(z), $$

(2.1)

then

$$1 \leq\rho(g)=\max\biggl\{ \lambda(g), \lambda\biggl(\frac{1}{g}\biggr) \biggr\} . $$

Remark

It is a curious problem to construct a transcendental meromorphic solution of (2.1) for the case $\deg A>0$.

Theorem 2.2

Let $c\in\mathbb{C}\setminus\{0\}$, and let $E(z)=\frac{D(z)}{F(z)}$ be an irreducible rational function, where $D(z)$ and $F(z)$ are polynomials with $\deg D(z)=d$ and $\deg F(z)=f$. If the equation

$$ \bigl(\triangle_{c} g(z)\bigr)^{2}=g(z)g(z+c)+E(z) $$

(2.2)

has a rational solution

$$g(z)=\frac{H(z)}{K(z)}=\frac{l_{h}z^{h}+\cdots+l_{0}}{m_{k}z^{k}+\cdots+m_{0}}, $$

where $l_{h}\ (\neq0), \dots, l_{0}, m_{k}\ (\neq0),\dots, m_{0}$ are constants, $\deg H(z)=h$, and $\deg K(z)=k$.

• (i) If $d\geq f$ and $d-f$ is zero or an even number, then

  $$h-k=\frac{d-f}{2}. $$

• (ii) If $d< f$, then $h-k=\frac{d-f}{2}$.

Further, Example 2.3 shows that there exist rational solutions satisfying Theorem 2.2(i), and Example 2.4 shows that there exist rational solutions satisfying Theorem 2.2(ii).

Example 2.3

The equation

$$\bigl(g(z+c)-g(z)\bigr)^{2}=g(z+c)g(z)+c^{2}-z^{2}-(4+c)z-2c-4 $$

has a rational solution $g(z)=z+2$, where $d=2$, $f=0$, and $h-k=1=\frac{d-f}{2}$.

Example 2.4

The equation

$$\bigl(g(z+c)-g(z)\bigr)^{2}=g(z+c)g(z)+\frac{c^{2}-z(z+c)}{z^{2}(z+c)^{2}} $$

has a rational solution $g(z)=\frac{1}{z}$, where $d=2$, $f=4$, and $h-k=-1=\frac{d-f}{2}$.

3 Proof of Theorem 2.1

Lemma 3.1

([11])

Let $w(z)$ be a transcendental meromorphic solution of finite order of the difference equation

$$P(z, w)=0, $$

where $P(z,w)$ is a difference polynomial in $w(z)$ and its shift. If $P(z,a)\not\equiv0$ for a slowly moving target function a, that is, $T(r, a)=S(r, w)$, then

$$m\biggl(r,\frac{1}{w-a}\biggr)=S(r,w). $$

The following result obtained by Chiang and Feng [16] and Halburd and Korhonen [9, 17] independently. We state here the form stated in [16, Theorem 8.2(b)].

Lemma 3.2

([16])

Let $c_{1}$, $c_{2}$ be two arbitrary complex numbers, and let $w(z)$ be a meromorphic function of finite order ρ. Let $\varepsilon>0$ be given. Then there exists a subset $E\subset(1, \infty)$ of finite logarithmic measure such that, for all $|z|=r\notin E\cup[0, 1]$, we have

$$\exp\bigl(-r^{(\rho-1+\varepsilon)}\bigr)\leq\bigg|\frac{w(z+c_{1})}{w(z+c_{2})}\bigg|\leq\exp \bigl(r^{(\rho-1+\varepsilon)}\bigr). $$

Firstly, we prove that $\rho(g)=\rho\geq1$. We consider the following two cases separately.

Case 1.1. If $g(z)$ has infinitely many poles, we can pick a pole $z_{0}$ of $g(z)$ such that $g(z_{0})=\infty^{\pi}$, where $\pi\geq1$, then we deduce by (2.1) that $g(z_{0}+c)=\infty^{\pi_{1}}$, where $\pi _{1}\geq m$. Substituting $z_{0}+c$ for z into (2.1), we have

$$ \bigl(g(z+2c)-g(z+c)\bigr)^{2}=A(z+c)g(z+2c)g(z+c)+B(z+c). $$

(3.1)

Then (3.1) implies that $z_{0}+2c$ is a pole of g of multiplicity $\pi _{2}\geq\pi_{1} \geq\pi$.

Since $g(z)$ has infinitely many poles, following the previous steps, we pick a pole $z_{0}$ of $g(z)$ such that

$$g(z_{0}+nc)=f(\xi_{n})=\infty^{\pi_{n}}, $$

where $\pi_{n}\geq\pi$ for all $n \in \mathbb{N}^{0}$. Hence, we can choose a sequence $\{\xi_{n}=z_{0}+nc, n \in \mathbb{N}^{0}\}$ of poles of $g(z)$, the multiplicity of which is $\pi_{n}\geq\pi$, so we obtain $\lambda(\frac{1}{g})\geq1$, and therefore $\rho(g)\geq \lambda(\frac{1}{g})\geq1$.

Case 1.2. If $g(z)$ is a transcendental meromorphic function with finitely many poles, then we can rewrite $g(z)$ as

$$ g(z)=\frac{g_{1}(z)}{P(z)}, $$

(3.2)

where $g_{1}(z)$ is a transcendental entire function, and $P(z)$ is a polynomial. Substituting (3.2) into (2.1), we have

$$ \biggl(\frac{g_{1}(z+c)}{P(z+c)}-\frac{g_{1}(z)}{P(z)}\biggr)^{2}=A(z) \frac {g_{1}(z+c)}{P(z+c)}\frac{g_{1}(z)}{P(z)} +B(z). $$

(3.3)

By computing (3.3) we have

$$ \frac{P(z)}{P(z+c)}\frac{g_{1}(z+c)}{g_{1}(z)}+\frac{P(z+c)}{P(z)}\frac {g_{1}(z)}{g_{1}(z+c)} =2+A(z)+\frac{B(z)P(z)P(z+c)}{g_{1}(z)g_{1}(z+c)}. $$

(3.4)

We prove that $\rho(g)=\rho(g_{1})=\rho\geq1$. Suppose, on the contrary to the assertion, that $\rho(g)=\rho(g_{1})=\rho<1$. For any given ε ($0<\varepsilon<\frac{1-\rho(g_{1})}{2}$), by Lemma 3.2 we obtain

$$\begin{gathered} \bigg|\frac{g_{1}(z+c)}{g_{1}(z)}\bigg|\leq\exp\bigl(r^{\rho(g_{1})-1+\varepsilon}\bigr)=\exp \bigl(o(1)\bigr),\\ \bigg|\frac{g_{1}(z)}{g_{1}(z+c)}\bigg|\leq\exp\bigl(r^{\rho(g_{1})-1+\varepsilon}\bigr)=\exp \bigl(o(1) \bigr)\end{gathered} $$

(3.5)

outside a finite logarithmic measure E. As $z_{k}$ satisfies $|g_{1}(z_{k})|=M(r_{k}, g_{1})$, $|z_{k}|=r_{k} \notin E$, $r_{k}\rightarrow\infty$, we deduce by (3.4) and (3.5) that

$$\begin{aligned} \big|A(z_{k})\big| ={}&\bigg|\frac{P(z_{k})}{P(z_{k}+c)}\frac{g_{1}(z_{k}+c)}{g_{1}(z_{k})}+\frac {P(z_{k}+c)}{P(z_{k})}\frac{g_{1}(z_{k})}{g_{1}(z_{k}+c)} - \frac{B(z_{k})P(z_{k})P(z_{k}+c)}{g_{1}(z_{k})g_{1}(z_{k}+c)}-2\bigg| \\ \leq{}& \bigg|\frac{P(z_{k})}{P(z_{k}+c)}\frac{g_{1}(z_{k}+c)}{g_{1}(z_{k})}\bigg|+\bigg|\frac {P(z_{k}+c)}{P(z_{k})} \frac{g_{1}(z_{k})}{g_{1}(z_{k}+c)}\bigg| \\ &+\bigg|\frac{B(z_{k})P(z_{k})P(z_{k}+c)}{M(r_{k}, g_{1})^{2}}\frac {g_{1}(z_{k})}{g_{1}(z_{k}+c)}\bigg|+2\\ \leq{}& M,\end{aligned} $$

where M is some finite constant, a contradiction, since $\deg A(z)>0$. Hence we have ${\rho(g)\geq1}$.

Next, we prove that $\max\{\lambda(g), \lambda(\frac{1}{g})\}=\rho(g)$. If $B(z)\not\equiv0$, then we set

$$P(z, g)=\bigl(g(z+c)-g(z)\bigr)^{2}-A(z)g(z+c)g(z) -B(z). $$

Since $P(z, 0)=-B(z)\not\equiv0$, by Lemma 3.1 we deduce that

$$N\biggl(r, \frac{1}{g}\biggr)=T(r, g)+S(r, f). $$

Hence $\lambda(g)=\rho(g)$.

If $B(z)\equiv0$, then (2.1) can be reduced to

$$\bigl(g(z+c)-g(z)\bigr)^{2}=A(z)g(z+c)g(z). $$

Next, we prove that $\max\{\lambda(g), \lambda(\frac{1}{g})\}=\rho(g)$. Suppose, on the contrary to the assertion, that $\max\{\lambda(g), \lambda(\frac{1}{g})\}=\alpha<\rho(g)$. We next divide the proof into the following two cases.

Case 1. Suppose that $\rho(g)=1$. Then we obtain

$$ g(z)=m(z)\exp^{qz+p}, $$

(3.6)

where $q\neq0$ and p are constants, and $m(z)$ is a meromorphic function such that $\rho(m)=\alpha<1$. Substituting (3.6) into (2.1), we obtain

$$ \bigl(m(z+c)\exp^{q(z+c)+p}-m(z)\exp^{qz+p}\bigr)^{2}=A(z)m(z+c) \exp ^{q(z+c)+p}m(z)\exp^{qz+p}. $$

(3.7)

By computing (3.7) we obtain

$$\begin{aligned}[b] &m^{2}(z+c)\exp^{2qc+2p}\exp^{2qz}+m^{2}(z) \exp^{2p}\exp ^{2qz}\\&\quad=\bigl(A(z)+2\bigr)m(z)m(z+c) \exp^{qc+2p}\exp^{2qz},\end{aligned} $$

(3.8)

that is,

$$ \bigl(A(z)+2\bigr)\exp^{qc+2p}\exp^{2qz}=\frac{m(z+c)}{m(z)} \exp^{2qc+2p}\exp ^{2qz}+\frac{m(z)}{m(z+c)}\exp^{2p} \exp^{2qz}. $$

(3.9)

By Lemma 3.2 we obtain

$$\begin{gathered} \bigg|\frac{m(z+c)}{m(z)}\bigg|\leq\exp\bigl(r^{\rho(m)-1+\varepsilon}\bigr)=\exp\bigl(o(1)\bigr),\\ \bigg|\frac{m(z)}{m(z+c)}\bigg|\leq\exp\bigl(r^{\rho(m)-1+\varepsilon}\bigr)=\exp \bigl(o(1)\bigr) \end{gathered} $$

(3.10)

outside a finite logarithmic measure. By (3.9) and (3.10), as $|z|\rightarrow\infty$, we obtain

$$\begin{aligned} \big|\bigl(A(z)+2\bigr)\exp^{qc+2p}\big|&=\bigg|\frac{m(z+c)}{m(z)} \exp^{2qc+2p}+\frac {m(z)}{m(z+c)}\exp^{2p}\bigg| \\ &\leq\bigg|\frac{m(z+c)}{m(z)}\exp^{2qc+2p}\bigg|+\bigg|\frac{m(z)}{m(z+c)}\exp ^{2p}\bigg|\leq M_{1}\end{aligned} $$

outside a finite logarithmic measure, where $M_{1}$ is a finite constant. This is impossible, since $\deg A(z)>0$.

Case 2. Suppose that $\rho(g)>1$. Then

$$ g(z)=m(z)\exp^{l(z)}, $$

(3.11)

where $l(z)$ is a polynomial such that $\rho(g)=\deg l(z)>1$, and $m(z)$ is a meromorphic function such that $\rho(m)<\rho(g)$. Substituting (3.11) into (2.1), we obtain

$$ \bigl(m(z+c)\exp^{l(z+c)}-m(z)\exp^{l(z)}\bigr)^{2}=A(z)m(z+c) \exp ^{l(z+c)}m(z)\exp^{l(z)}. $$

(3.12)

Let

$$l(z)=p_{k}z^{k}+p_{k-1}z^{k-1}+ \cdots+p_{1}z+p_{0}, $$

where $p_{k}\neq0$. Then

$$\begin{aligned}& l(z+c)=p_{k}z^{k}+(ckp_{k}+p_{k-1})z^{k-1}+Q(z), \end{aligned}$$

(3.13)

$$\begin{aligned}& l(z+c)-l(z)=(ckp_{k})z^{k-1}+Q_{1}(z), \end{aligned}$$

(3.14)

where $Q(z)$ and $Q_{1}(z)$ are polynomials of degree at most $k-2$. Equalities (3.12) and (3.14) imply that

$$m^{2}(z+c)\exp^{2ckp_{k}z^{k-1}+2Q_{1}(z)}+m^{2}(z)=\bigl(A(z)+2 \bigr)m(z+c)m(z)\exp ^{ckp_{k}z^{k-1}+Q_{1}(z)}, $$

that is,

$$ \begin{aligned}[b] \big|\exp^{2ckp_{k}z^{k-1}+2Q_{1}(z)}\big|&=\bigg|{-}\frac {m^{2}(z)}{m^{2}(z+c)}+\bigl(A(z)+2\bigr) \frac{m(z)}{m(z+c)}\exp^{ckp_{k}z^{k-1}+Q_{1}(z)}\bigg| \\ &\leq\bigg|\frac{m^{2}(z)}{m^{2}(z+c)}\bigg|+\bigg|\bigl(A(z)+2\bigr)\frac{m(z)}{m(z+c)}\exp ^{ckp_{k}z^{k-1}+Q_{1}(z)}\bigg|.\end{aligned}$$

(3.15)

By Lemma 3.2 we obtain

$$ \bigg|\frac{m(z)}{m(z+c)}\bigg|\leq\exp\bigl(r^{\rho(m)-1+\varepsilon}\bigr) $$

(3.16)

outside a possible set of finite logarithmic measure E. As $|z|=r\notin E\cup[0, 1]$, and $r\rightarrow\infty$, we deduce by (3.15) and (3.16) that

$$ \begin{aligned}[b] & \big|\exp^{2ckp_{k}z^{k-1}+2Q_{1}(z)}\big| \\ &\quad\leq\exp\bigl(2r^{\rho(m)-1+\varepsilon}\bigr)+\bigl|r^{M}\exp\bigl(r^{\rho (m)-1+\varepsilon} \bigr)\exp^{ckp_{k}z^{k-1}+Q_{1}(z)}\bigr|,\end{aligned}$$

(3.17)

where M is a positive constant.

We can find a sequence $\{z_{k}\}$ ($|z_{k}|\rightarrow\infty$) such that $|z_{k}|=r_{k}\notin E\cup[0, 1]$, and $cp_{k} z_{k}^{k-1}=|cp_{k}|r_{k}^{k-1}$ as $r_{k}\rightarrow\infty$. We obtain

$$ \big|\exp^{2ckp_{k}z_{k}^{k-1}+2Q_{1}(z_{k})}\big| =\exp^{2k|cp_{k}|r_{k}^{k-1}}\big|\exp^{Q_{1}(z_{k})}\big|\geq \exp^{\frac {3}{2}k|cp_{k}|r_{k}^{k-1}}. $$

(3.18)

By (3.17) and (3.18), for any given ε ($0<\varepsilon<\frac {k-\rho(m)}{2}$), we obtain

$$\exp^{\frac{3}{2}k|cp_{k}|r_{k}^{k-1}}\leq\exp\bigl(2r_{k}^{\rho(m)-1+\varepsilon}\bigr) +r_{k}^{M}\exp\bigl(r_{k}^{\rho(m)-1+\varepsilon}\bigr) \exp^{ckp_{k}r_{k}^{k-1}} \leq\exp^{\frac{6}{5}k|cp_{k}|r_{k}^{k-1}}, $$

which is impossible. Hence we proved that $\max\{\lambda(g), \lambda (\frac{1}{g})\}=\rho(g)$. Theorem 2.1 is thus proved.

4 Proof of Theorem 2.2

Suppose that (2.2) has a rational solution $g(z)$ and has poles $l_{1}, l_{2},\dots,l_{k}$. Hence $g(z)$ can be represented as

$$ g(z)=\frac{H(z)}{K(z)} =\sum^{k}_{j=1} \biggl[\frac{t_{js_{j}}}{(z-l_{j})^{s_{j}}} +\cdots+\frac{t_{j_{1}}}{(z-l_{j})}\biggr]+b_{0}+b_{1}z+ \cdots+b_{r}z^{r}, $$

(4.1)

where $b_{0},\dots,b_{r}, t_{js_{j}},\dots,t_{j1}$ are constants.

(i) If $d>f$ and $d-f$ is an even number, then (2.2) and (4.1) imply that

$$ \biggl(\frac{H(z+c)}{K(z+c)}-\frac{H(z)}{K(z)}\biggr)^{2}- \frac{H(z+c)}{K(z+c)}\frac {H(z)}{K(z)}=\frac{D(z)}{F(z)}. $$

(4.2)

Let $\deg H(z)=h$ and $\deg K(z)=k$. Suppose $h< k$. Then

$$ \lim_{z\rightarrow\infty}\frac{H(z+c)}{K(z+c)}=0,\qquad \lim _{z\rightarrow\infty}\frac{H(z)}{K(z)}= 0. $$

(4.3)

From (4.2) with (4.3) we obtain

$$\frac{D(z)}{F(z)}=\biggl(\frac{H(z+c)}{K(z+c)}-\frac{H(z)}{K(z)} \biggr)^{2} -\frac{H(z+c)}{K(z+c)}\frac{H(z)}{K(z)}\rightarrow0. $$

This is impossible, since $\frac{D(z)}{F(z)}\rightarrow\infty$ as $z\rightarrow\infty$.

Suppose $h=k$. Then

$$ \lim_{z\rightarrow\infty}\frac{H(z+c)}{K(z+c)}= \beta, \qquad \lim _{z\rightarrow\infty}\frac{H(z)}{K(z)}=\beta, $$

(4.4)

where $\beta\in\mathbb{C}\setminus\{0\}$. Relations (4.2) and (4.4) yield that

$$\frac{D(z)}{F(z)}=\biggl(\frac{H(z+c)}{K(z+c)}-\frac{H(z)}{K(z)} \biggr)^{2} -\frac{H(z+c)}{K(z+c)}\frac{H(z)}{K(z)}\rightarrow- \beta^{2}, $$

a contradiction, since $\frac{D(z)}{F(z)}\rightarrow\infty$ as $z\rightarrow\infty$. Hence, we obtain that $h>k$. So $b_{r}\neq0$ ($r\geq1$). As $z\rightarrow\infty$, we have

$$\begin{gathered} g(z)=b_{r} z^{s}\bigl(1+o(1)\bigr),\qquad g(z+c)=b_{r}z^{s} \bigl(1+o(1)\bigr), \\ \frac{D(z)}{F(z)}=\alpha z^{d-f}\bigl(1+o(1)\bigr), \end{gathered} $$

(4.5)

where $\alpha\in\mathbb{C}\setminus\{0\}$. As $z\rightarrow\infty$, by (4.2) and (4.5) we can deduce

$$ -b_{r}^{2} z^{2r}\bigl(1+o(1)\bigr)=\alpha z^{d-f}\bigl(1+o(1)\bigr). $$

(4.6)

Relation (4.6) implies that

$$h-k=r=\frac{d-f}{2}. $$

If $d=f$, then, as $z\rightarrow\infty$, we obtain

$$\frac{D(z)}{F(z)}=\alpha\bigl(1+o(1)\bigr), $$

where $\alpha\in\mathbb{C}\setminus\{0\}$. If $h< k$, then using the similar method as before, we can obtain a contradiction. If $h>k$, then $b_{r}\neq0$ ($r\geq1$). By (4.2), as $z\rightarrow\infty$, we obtain

$$ -b_{r}^{2} z^{2r}\bigl(1+o(1)\bigr)=\alpha z^{d-f}\bigl(1+o(1)\bigr)=\alpha\bigl(1+o(1)\bigr), $$

(4.7)

a contradiction. Hence $h=k$, that is,

$$h-k=0=\frac{d-f}{2}. $$

(ii) We next consider the case $d< f$. Suppose that $h>k$. Then $b_{r}\neq0$ ($r\geq1$). Using the similar method as before, as $z\rightarrow\infty$, by (4.2) we obtain that

$$3b_{r}^{2} z^{2r}\bigl(1+o(1)\bigr)=0, $$

a contradiction.

If $h=k$, then using the similar method as before, we obtain

$$\frac{D(z)}{F(z)}=\biggl(\frac{H(z+c)}{K(z+c)}-\frac{H(z)}{K(z)} \biggr)^{2} -\frac{H(z+c)}{K(z+c)}\frac{H(z)}{K(z)}\rightarrow \beta^{2}\neq0 \quad \mbox{as } z\rightarrow\infty, $$

which is a contradiction, since $\frac{D(z)}{F(z)}\rightarrow0$ as $z\rightarrow\infty$. Hence $h< k$. Substituting $g(z)=\frac{H(z)}{K(z)}$ into (2.2), we have

$$\begin{aligned}[b] &F(z)H^{2}(z+c)K^{2}(z)-3F(z)H(z)H(z+c)K(z)K(z+c)+F(z)H^{2}(z)K^{2}(z+c) \\ &\quad =D(z)K^{2}(z)K^{2}(z+c). \end{aligned}$$

(4.8)

Since

$$\begin{gathered} \deg\bigl(F(z)H^{2}(z+c)K^{2}(z)-3F(z)H(z)H(z+c)K(z)K(z+c)+F(z)H^{2}(z)K^{2}(z+c)\bigr) \\ \quad =f+2h+2k, \\ \deg\bigl(D(z)K^{2}(z)K^{2}(z+c)\bigr)=d+4k.\end{gathered} $$

From this and from (4.8) we have

$$h-k=\frac{d-f}{2}. $$