On decompositions of binary recurrent polynomials

Abstract

Polynomial decomposition expresses a polynomial f as the functional composition $f=g\circ h$ of lower degree polynomials g and h, and has various applications. In this paper, we will show that for a minimal, non-degenerate, simple, binary, linearly recurrent sequence $(G_n(x))_{n=0}^\infty$ of complex polynomials whose coefficients in the Binet form are constants, if $G_n(x)=g(h(x))$, then apart from some exceptional situations that have to be taken into account, the degree of g is bounded by a constant independent of n. We will build on a general but conditional result of this type that already exists in the literature. We will then present one Diophantine application of the main result.

1 Introduction

In the 1920’s, Ritt [12] studied the functional decomposition $f=f_1\circ \cdots \circ f_m$ of a complex polynomial f into indecomposable complex polynomials $f_1, \ldots , f_m$. A complex polynomial f with $\deg f>1$ is said to be indecomposable if it cannot be represented as a composition of two lower degree polynomials. Ritt gave a procedure for obtaining any decomposition of a complex polynomial from any other by applying certain transformations. Ritt’s results about polynomial decomposition have applications to number theory, complex analysis, arithmetic dynamics, finite geometries, etc. @ See e.g. [11] for an overview of the theory and applications. Of our interest in this paper are decomposition properties of binary recurrent sequences of polynomials. This topic has been studied in several papers [6,7,8] and relevant results will be mentioned later in the introduction. For $A_0(x),A_1(x), G_0(x),G_1(x)\in \mathbb {C}[x]$, let $(G_n(x))_{n=0}^\infty$ be a minimal, non-degenerate and simple binary linearly recurrent sequence of polynomials defined by

$$\begin{aligned} G_{n+2}(x)=A_{1}(x)G_{n+1}(x)+A_0(x)G_n(x), \ n\ge 0, \quad \text {so that}\ G_n(x)=\pi _1\alpha _1^n + \pi _2\alpha _2^n, \end{aligned}$$

(1.1)

for $\pi _1, \pi _2\in L$, where $L/\mathbb {C}(x)$ is the splitting field of the characteristic polynomial of the recurrence and $\alpha _1, \alpha _2\in L$ are its distinct roots (distinct since the recurrence is simple) such that $\alpha _1/\alpha _2\notin \mathbb {C}^{*}$ (since the recurrence is non-degenerate), and furthermore $A_1, A_0, \pi _1\alpha _1^n, \pi _2\alpha _2^n\ne 0$ (by minimality).

To state our first result, we recall the definitions of cyclic and dihedral polynomials. These polynomials play a prominent role in Ritt’s theory (more details will be given in Sect. 2). A polynomial h is said to be cyclic if $h(x)=\ell _1(x)\circ x^n \circ \ell _2(x)$ for some $n\ge 2$ and $\ell _1, \ell _2$ linear polynomials and dihedral if $h(x)=\ell _1(x)\circ T_n(x)\circ \ell _2(x)$ for some $n\ge 3$ and $\ell _1, \ell _2$ linear polynomials, where $T_n(x)$ is the n-th Chebychev polynomial of the first kind given by $T_{n+2}(x)=2xT_{n+1}(x)-T_n(x)$, $T_0(x)=1$, $T_1(x)=x$.

Theorem 1

Consider the sequence (1.1) such that $\pi _1,\pi _2\in \mathbb {C}$. Assume that for some $n\ge 0$ we have $G_n(x)=g(h(x))$ for $g(x), h(x)\in \mathbb {C}[x]$, where h is indecomposable and neither cyclic nor dihedral. Further asume that we do not have $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$. Then $\deg g\le C$ for a constant $C=C(\{A_i, G_i : i=0,1\})>0$, independent of n.

It is well known that $T_{mk}(x)=T_m(x)\circ T_k(x)$ for any $m, k\in \mathbb {N}$ and since for Chebychev polynomials of the first kind the coefficients in the Binet form are constants, already these polynomials illustrate that Theorem 1 would not hold if we allow h to be dihedral. Indeed, for dihedral $h(x)\in \mathbb {C}[x]$ of any degree there exists $n\ge 0$ such that $T_n(x)=g(h(x))$ for some $g(x)\in \mathbb {C}[x]$ whose degree depends on n. Furthermore, the sequence $T_n(h(x))$ with $h(x)\in \mathbb {C}[x]$ is of type (1.1) and its coefficients in the Binet form are constants, and we clearly cannot bound $\deg T_n$ independently of n. This illustrates why we also have to exclude the case when $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$. Also, for cyclic $h(x)\in \mathbb {C}[x]$ of any degree there exists a sequence $(G_n(x))_{n=0}^\infty$ satisfying (1.1), whose coefficients in the Binet form are constants, such that we do not have $G_m(x)\in \mathbb {C}[h(x)]$ for all m, but $G_n(x)=g(h(x))$ for some $n\ge 0$ and some $g(x)\in \mathbb {C}[x]$ whose degree depends on n. (Pick e.g. $\pi _1=\pi _2=1$, $A_0(x)=x$, $A_1(x)=x+1$. Then $G_{mk}(x)=x^{mk}+1=(x^m+1)\circ x^k$ for any $k, m\in \mathbb {N}$.)

Corollary 2

Let $p(x),q(x)\in \mathbb {C}[x]$ be such that $p(x)^n+q(x)^n=g(h(x))$ for some $n\ge 0$ and $g(x), h(x)\in \mathbb {C}[x]$, where h is indecomposable and neither cyclic nor dihedral. Then either $p(x), q(x)\in \mathbb {C}[h(x)]$ or $\deg g\le C$ for a constant $C=C(p, q)>0$, independent of n.

In relation to Corollary 2, we remark that it is well known (see Lemma 5) and follows from Ritt’s results that if the n-th power of a complex polynomial u is a composite of a complex polynomial h (more precisely, u is nonconstant and $u(x)^n\in \mathbb {C}[h(x)]$), where h is indecomposable and neither dihedral nor cyclic, then u is a composite of h. Moreover, Ritt [12] gave a description of all $g(x), h(x)\in \mathbb {C}[x]$ whose composition is an n-th power of a complex polynomial. Cohen [3] described all rational functions $g(x), h(x)\in \mathbb {C}(x)$ whose composition is an n-th power in $\mathbb {C}(x)$. It would be of interest to give a full description of all $g(x), h(x)\in \mathbb {C}(x)$ whose composition is a sum of two n-th powers in $\mathbb {C}(x)$.

Theorem 1 relies on the main result of [8], which is a general but conditional result for an element of the sequence (1.1) to satisfy $G_n(x)=g(h(x))$, for $g(x), h(x)\in \mathbb {C}[x]$, where h is indecomposable and neither cyclic nor dihedral. To state this result precisely, assume that $G_n(x)=g(h(x))$, for $g(x), h(x)\in \mathbb {C}[x]$, where h is indecomposable. The polynomial $h(X)-h(x)\in \mathbb {C}(h(x))[X]$ is clearly separable and since $\deg h\ge 2$ by assumption, there exists $y\ne x$ such that $h(x)=h(y)$. Then $G_n(x)=G_n(y)$ and by equating the corresponding Binet forms, we obtain

$$\begin{aligned} \pi _1\alpha _1^n + \pi _2\alpha _2^n=\rho _1\beta _1^n+\rho _2\beta _2^n, \end{aligned}$$

(1.2)

where $\alpha _1, \alpha _2$ are distinct roots of the characteristic polynomial of the sequence $(G_n(x))_{n=0}^\infty$, $\beta _1, \beta _2$ are distinct roots of the characteristic polynomial of the sequence $(G_n(y))_{n=0}^\infty$, $\pi _1, \pi _2\in L_1$ and $\rho _1,\rho _2\in L_2$, where $L_1$ and $L_2$ are the splitting fields of the corresponding characteristic polynomials over $\mathbb {C}(x)$ and $\mathbb {C}(y)$, respectively. According to [8,  Thm. 1], there is a constant $C>0$, independent of n, with the following property: If $G_n(x)=g(h(x))$ for some $n\ge 0$ and $g(x), h(x)\in \mathbb {C}[x]$, where h is indecomposable and neither cyclic nor dihedral, and (1.2) has no vanishing subsum, then $\deg g\le C$. We say that there exists a vanishing subsum of (1.2) if there is a permutation $\sigma$ of the set $\{1, 2\}$ such that $\pi _i\alpha _i^n=\rho _{\sigma (i)}\beta _{\sigma (i)}^n$ for $i=1, 2$. With the above restrictions on h, one can show that this holds if and only if

$$\begin{aligned} \pi _1\pi _2A_0(x)^n=-\frac{G_1(x)^2-G_0(x)G_1(x)A_1(x)-A_0(x)G_0(x)^2}{A_1(x)^2+4A_0(x)}A_0(x)^n\in \mathbb {C}[h(x)]. \end{aligned}$$

(1.3)

See Sect. 2 for details. It appears to be difficult to classify all such $G_0, G_1, A_0, A_1$. In regard to this problem, we mention [10], where the authors solved the equation $g(x)f(x)^n=g(h(x))$ where f, g, h are unknown nonconstant complex polynomials, $n>1$, $\deg h\ge 2$ and g is separable. We also mention [4], where the authors completely classified binomials which have a non-trivial factor which is a composition of two polynomials of degree $>1$. We further mention that certain sufficient, but unfortunately not quite illuminating, conditions for (1.2) to have no vanishing subsum were presented in [8,  Thm. 2]. Finally, we mention that the constant C can be effectively computed; this is done in the proof of [8,  Thm. 1].

To the proof of Theorem 1, we will show that under the assumptions of the theorem, there does not exist a vanishing subsum of (1.2). We will build on the techniques from [8] and strengthen the arguments, in particular by utilizing a well known result of Fried (Theorem 6). We will complement Theorem 1 with the following result. It will be proved using a similar approach.

Proposition 3

Consider the sequence (1.1) with $A_0$ constant and any of $G_0, G_1, A_1$ constant. Assume $G_n(x)=g(h(x))$ for some $n\ge 0$ and $g(x), h(x)\in \mathbb {C}[x]$, where h is indecomposable and neither cyclic nor dihedral. Further assume that we do not have $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$. Then $\deg g\le C$ for a constant $C=C(\{A_i, G_i : i=0,1\})$.

We will then present some well-understood sequences from the literature exhibiting the decomposition property from Theorem 1 and Proposition 3. These include Chebyshev polynomials of the first kind, Lucas polynomials, Fibonacci polynomials, Fermat polynomials, Chebyshev polynomials of the second kind, etc. See Sect. 3 for precise definitions and for further applications.

We remark that Zannier [15] proved a result similar to Theorem 1 for lacunary polynomials, i.e. polynomials with a fixed number of terms. Theorem 1 relies on the main result of [8], which is obtained using techniques similar to Zannier’s. There are several applications of Zannier’s result, see e.g. [8] for more details . We conclude this paper by illustrating one Diophantine application of Theorem 1. Consider a minimal, simple and non-degenerate sequence $(G_n(x))_{n=0}^\infty$ satisfying

$$\begin{aligned} G_{n+2}(x)=A_{1}(x)G_{n+1}(x)+A_0(x)G_n(x), \quad n\ge 0, \end{aligned}$$

(1.4)

with $G_0(x), G_1(x), A_0(x), A_1(x)\in \mathbb {Q}[x]$, such that its coefficients in the Binet form are constants and that there is no $h(x)\in \mathbb {C}[x]$ such that $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$. Further consider another minimal, simple and non-degenerate sequence $(H_n(x))_{n=0}^\infty$ of the same type

$$\begin{aligned} H_{n+2}(x)=B_{1}(x)H_{n+1}(x)+B_0(x)H_n(x), \quad n\ge 0, \end{aligned}$$

(1.5)

with $H_0(x),H_1(x), B_0(x),B_1(x)\in \mathbb {Q}[x]$, such that its coefficients in the Binet form are constants and that there is no $h(x)\in \mathbb {C}[x]$ such that $H_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$. Recall that $P(x)\in \mathbb {C}[x]$ is said to be a composite of a cyclic or dihedral polynomial if it is nonconstant and $P(x)=g(h(x))$, where h is either cyclic or dihedral and $g(x)\in \mathbb {C}[x]$.

Theorem 4

Consider sequences $(G_n(x))_{n=0}^\infty$ and $(H_n(x))_{n=0}^\infty$ satisfying (1.4) and (1.5), respectively. Then there exists a constant $C=C(\{A_i, G_i : i=0,1\})>0$ with the following property. If $G_n(x)$ and $H_m(x)$ with $\deg G_n\ge \deg H_m>C$ are not composites of either cylic or dihedral polynomials, and the equation $G_n(x)=H_m(y)$ has infinitely many integer solutions x, y, then $G_n(x)=H_m(\ell (x))$ for a linear $\ell (x)\in \mathbb {C}[x]$.

Theorem 4 is an almost immediate consequence of Theorem 1 and the main result of [2]. The latter result is a criterion for the finiteness of integer solutions of Diophantine equations of type $f(x)=g(y)$, where $f(x), g(x)\in \mathbb {Q}[x]$ are nonconstant polynomials. All details will be given in Sect. 5.

2 Auxiliary results

We now recall some basic facts about complex polynomial decomposition. For $f(x)\in \mathbb {C}[x]$ with $\deg f>1$, we say that two decompositions $f=f_1\circ \dots \circ f_m$ and $f=g_1\circ \dots \circ g_n$ of f are equivalent if $m=n$ and there are linear $\mu _1,\dots ,\mu _{m-1}\in \mathbb {C}[x]$ such that $f_{i}\circ \mu _i=g_{i}$ and $\mu _i^{(-1)}\circ f_{i+1}=g_{i+1}$, $i=1, 2, \ldots , m-1$. Here $\mu _i^{(-1)}$ denotes the inverse of $\mu$ with respect to functional composition which clearly exists exactly when $\mu$ is a linear polynomial. For a given polynomial f there may exist several complete decompositions, that is decompositions into indecomposable polynomials, but they are all related in the following way: any complete decomposition of f can be obtained from any other through a sequence of steps, each of which involves replacing two adjacent indecomposables by two others with the same composition. The only solutions of the equation $a\circ b=c\circ d$ in indecomposable complex polynomials, up to composing with linear polynomials, are the trivial $a\circ b=a\circ b$ and the non-trivial solutions $x^m\circ x^rP(x^m)=x^rP(x)^m\circ x^m$, $T_m(x)\circ T_n(x)=T_n(x)\circ T_m(x)$ where $P(x)\in \mathbb {C}[x]$, $r, m, n\in \mathbb {N}$ and $T_n$ is the n-th Chebyshev polynomial defined in the introduction. These results are due to Ritt [12]. There are many interesting results on various topics (see e.g. [11] for an overview of such results) relying on Ritt’s findings. In particular, the following corollary of one such more recent result will be repeatedly used in this paper, [1,  Thm. 5.1] which, roughly speaking, states that ‘most’ pairs of complex polynomials have no common composite; f and h with $\deg f, \deg h>1$ have a common composite if there are nonconstant u, v such that $u(f(x))=v(h(x))$.

Lemma 5

Assume that for some $f(x), g(x)\in \mathbb {C}[x]$ where f is either cyclic or dihedral, we have $f(g(x))\in \mathbb {C}[h(x)]$ for an indecomposable $h(x)\in \mathbb {C}[x]$ which is neither cyclic nor dihedral. Then $g(x)\in \mathbb {C}[h(x)]$.

Proof

By [1,  Thm. 5.1], it follows that if $g(x), h(x)\in \mathbb {C}[x]$ satisfy $\deg g>1$, and h is indecomposable and neither cyclic nor dihedral, then g and h have a common composite if and only if either $g(x)\in \mathbb {C}[h(x)]$ or there are linear polynomials $\ell _1(x), \ell _2(x), \ell _3(x)\in \mathbb {C}[x]$ such that

$$\begin{aligned} g(x)=\ell _1(x)\circ x^m\circ \ell _3(x), \quad h(x)=\ell _2(x)\circ x^rP(x^m)\circ \ell _3(x), \end{aligned}$$

where $r, m\in \mathbb {N}$, $P(x)\in \mathbb {C}[x]$, $\gcd (\deg g, \deg h)=1$. In particular, if $\deg g>1$, then g is either cyclic or $g(x)\in \mathbb {C}[h(x)]$. However, for a cyclic polynomial there is clearly a complete decomposition consisting only of cyclic polynomials (since $x^{mn}=x^m\circ x^n$ for any m, n), and for a dihedral polynomial there is clearly a complete decomposition consisting only of dihedral polynomials and possibly cyclic polynomials of degree 2 (since $T_{mn}(x)=T_m(x)\circ T_n(x)$ for any m, n, and $T_2(x)=2x^2-1$ is cyclic). Moreover, if one complete decomposition of a complex polynomial consists only of cyclic and dihedral polynomials, then all complete decompositions consist only of cyclic and dihedral polynomials (see e.g. [11,  Thm. 1.3, Lemma 3.6]). Thus, any complete decomposition of the polynomial f(g(x)), where f is either cyclic or dihedral and g is either cyclic or $\deg g=1$, consists only of cyclic or dihedral polynomials. This implies that unless $g(x)\in \mathbb {C}[h(x)]$ or $\deg g=0$, h must be either cyclic or dihedral, a contradiction. If g is constant, the statement trivially holds. $\square$

Another famous result on the topic of polynomial decomposition that we will make use of in this paper is the following theorem due to Fried [5].

Theorem 6

For $h(x)\in \mathbb {C}[x]$ the following assertions are equivalent.

• (i) $(h(x)-h(y))/(x-y)$ is irreducible in $\mathbb {C}[x, y]$,

• (ii) h is indecomposable and if $n:=\deg h$ is an odd prime then $h(x)\ne \alpha D_n(x+b, a)+c$ with $\alpha , a, b, c\in \mathbb {C}$, with $a=0$ if $n=3$, where $D_n(x, a)$ is the n-th Dickson polynomial with parameter a satisfying

  $$\begin{aligned} D_n(x, 0)=x^n, \quad D_n(2ax, a^2)=2a^nT_n(x), \ a\ne 0 \end{aligned}$$

  (2.1)

  where $T_n$ denotes, as usual, the n-th Chebyshev polynomial of the first kind.

Thus, for an indecomposable $h(x)\in \mathbb {C}[x]$ which is neither cyclic nor dihedral, we have that $(h(X)-h(Y)/(X-Y))$ is an irreducible polynomial in $\mathbb {C}[X, Y]$. A detailed exposition of Fried’s proof of Theorem 6 can be found in [14], along with various properties of Dickson polynomials.

Next we recall a few auxiliary results recorded in [8] that we will use to prove Theorem 1. Consider the sequence (1.1) and assume $G_n(x)=g(h(x))$ for $g(x), h(x)\in \mathbb {C}[x]$, where h is indecomposable and neither cyclic nor dihedral. Let $y\ne x$ be a root of $h(X)-h(x)\in \mathbb {C}(x)[X]$ so that $h(x)=h(y)$ and consequently $G_n(x)=G_n(y)$. Then (1.2) holds. In [8], we showed that then either $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(x)$ and h is cyclic, or $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$. Indeed, since $h(x)=h(y)$, we have $\mathbb {C}(h(x))\subseteq \mathbb {C}(x)\cap \mathbb {C}(y)\subseteq \mathbb {C}(x)$. By Lüroth’s theorem ([13,  p. 13]) it follows that $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(r(x))$ for some $r\in \mathbb {C}(x)$. Moreover, since h is a polynomial, r can be chosen to be a polynomial as well by [13,  p. 16]. Then $h(x)\in \mathbb {C}[r(x)]$. Since h is indecomposable, it follows that either $\deg r=\deg h$ or $\deg r=1$, i.e. @ either $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$ or $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(x)$. If $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(x)$, then $\nu (y)=x$ for some $\nu (x)\in \mathbb {C}(x)$ and hence $h(\nu (y))=h(x)=h(y)$. We deduce that $\nu (x)\in \mathbb {C}[x]$ and $\deg \nu =1$. One can show that such h must be cyclic (see [8,  Lemma 4]), so that if h is not cyclic, then

$$\begin{aligned} \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x)). \end{aligned}$$

(2.2)

In [8] we then deduced that if h is not cyclic, there exists a vanishing subsum of (1.2) if and only if

$$\begin{aligned} \pi _1\pi _2A_0(x)^n\in \mathbb {C}(h(x)). \end{aligned}$$

(2.3)

This fact will be used repeatedly in the following section. Note that

$$\begin{aligned} \pi _1\pi _2=-\frac{G_1(x)^2-G_0(x)G_1(x)A_1(x)-A_0(x)G_0(x)^2}{A_1(x)^2+4A_0(x)}\in \mathbb {C}(x). \end{aligned}$$

(2.4)

3 Proofs of main results

Proof of Theorem 1

Let $y\ne x$ be a root of $h(X)-h(x)\in \mathbb {C}(x)[X]$ so that $h(x)=h(y)$ and consequently $G_n(x)=G_n(y)$, so that (1.2) holds. By assumption we have $\pi _1, \pi _2, \rho _1, \rho _2\in \mathbb {C}$.

Note that $\alpha _1+\alpha _2=A_1(x)$, $\alpha _1\alpha _2=-A_0(x)$, $\beta _1+\beta _2=A_1(y)$ and $\beta _1\beta _2=-A_0(y)$ by Vieta’s formulae. Further note that $\alpha _1, \alpha _2$ are not necessarily polynomials, but $\alpha _1^m+\alpha _2^m$ is a polynomial for any $m\ge 0$. Moreover, we have

$$\begin{aligned} \alpha _1^m+\alpha _2^m=D_m(A_1(x), -A_0(x))=\sum _{i=0}^{\left\lfloor {\frac{m}{2}}\right\rfloor } \frac{m}{m-j}{m-j \atopwithdelims ()j} A_0(x)^jA_1(x)^{m-2j}, \end{aligned}$$

(3.1)

where $D_m(X+Y, XY)=X^m+Y^m$ for $m\ge 0$ defines the m-th Dickson polynomial $D_m(x, a)$ with parameter a, encountered already in Theorem 6. Likewise,

$$\begin{aligned} \beta _1^m+\beta _2^m=D_m(A_1(y), -A_0(y))\in \mathbb {C}[y], \ m\ge 0. \end{aligned}$$

(3.2)

By [8,  Thm. 1] it suffices to show that (1.2), that is $\pi _1\alpha _1^n + \pi _2\alpha _2^n=\rho _1\beta _1^n+\rho _2\beta _2^n$, has no vanishing subsum. Assume the contrary. Since h is by assumption not cyclic, we can further assume that $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$, as proved in [8,  Lemma 4] and recalled in (2.2).

Since $\pi _1, \pi _2\in \mathbb {C}$, note that either $\rho _1=\pi _1$ and $\rho _2=\pi _2$, or $\rho _1=\pi _2$ and $\rho _2=\pi _1$. Indeed,

$$\begin{aligned} G_0(x)=\pi _1+\pi _2=:c_1\in \mathbb {C}, \quad \frac{2G_1(x)-G_0(x)A_1(x)}{A_1(x)^2+4A_0(x)}=(\pi _1-\pi _2)^2=:c_2\in \mathbb {C}. \end{aligned}$$

Then $G_0(y)=\rho _1+\rho _2=c_1$ and $\frac{2G_1(y)-G_0(y)A_1(y)}{A_1(y)^2+4A_0(y)}=(\rho _1-\rho _2)^2=c_2$ via $x\mapsto y$, and we easily deduce the claim. Assume without loss of generality that $\rho _1=\pi _1$ and $\rho _2=\pi _2$. We next show that the existence of a vanishing subsum of (1.2) implies $\pi _1=\pi _2$.

We have that either $\pi _1\alpha _1^n = \pi _1\beta _1^n$ and $\pi _2\alpha _2^n=\pi _2\beta _2^n$, or $\pi _1\alpha _1^n = \pi _2\beta _2^n$ and $\pi _2\alpha _2^n=\pi _1\beta _1^n$. Assume first that the former holds. Then $\alpha _1^n+\alpha _2^n = \beta _1^n+\beta _2^n$, and thus by (3.1) and (3.2) we have that $\alpha _1^n+\alpha _2^n\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$, and moreover clearly $\alpha _1^n+\alpha _2^n\in \mathbb {C}[h(x)]$. Since by assumption $G_n(x)=\pi _1\alpha _1^n+\pi _2\alpha _2^n\in \mathbb {C}[h(x)]$, it follows that $(\pi _1-\pi _2)\alpha _1^n\in \mathbb {C}[h(x)]$ and $(\pi _1-\pi _2)\alpha _2^n\in \mathbb {C}[h(x)]$. We conclude that either $\pi _1=\pi _2$ or $\alpha _1^n, \alpha _2^n\in \mathbb {C}[h(x)]$. In the latter case we easily check that if $n>0$, then we must have $\alpha _1, \alpha _2\in \mathbb {C}[x]$. Then $\alpha _1, \alpha _2\in \mathbb {C}[h(x)]$ by Lemma 5 and hence $G_m(x)\in \mathbb {C}[h(x)]$ for any $m\ge 0$, a contradiction. If $n=0$, the theorem trivially holds. Now assume $\pi _1\alpha _1^n = \pi _2\beta _2^n$ and $\pi _2\alpha _2^n=\pi _1\beta _1^n$. Then

$$\begin{aligned} \alpha _1^n+\alpha _2^n&=\frac{(\pi _1+\pi _2)(\pi _1\beta _1^n+\pi _2\beta _2^n) -\pi _1\pi _2(\beta _1^n+\beta _2^n)}{\pi _1\pi _2}\\&=\frac{(\pi _1+\pi _2)G_n(y)-\pi _1\pi _2(\beta _1^n+\beta _2^n)}{\pi _1\pi _2}. \end{aligned}$$

Since $\pi _1, \pi _2\in \mathbb {C}$, by (3.1) and (3.2) we have $\alpha _1^n+\alpha _2^n\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$, and moreover $\alpha _1^n+\alpha _2^n\in \mathbb {C}[h(x)]$. Since also $\pi _1\alpha _1^n+\pi _2\alpha _2^n\in \mathbb {C}[h(x)]$, it follows that either $\pi _1=\pi _2$ or $\alpha _1^n, \alpha _2^n\in \mathbb {C}[h(x)]$, as in the former case. The latter possibility we exclude as before. It remains to consider the case $\pi _1=\pi _2$.

If $\pi _1=\pi _2=:\pi$, then $G_n(x)=\pi (\alpha _1^n+\alpha _2^n)=\pi (\beta _1^n+\beta _2^n)=G_n(y)$. By assumption there exists a vanishing subsum of this sum, and we may assume without loss of generality that $\alpha _1^n=\beta _1^n$ and $\alpha _2^n=\beta _2^n$. Thus, $\alpha _1=\zeta \beta _1$ and $\alpha _2=\mu \beta _2$ for $\zeta , \mu \in \mathbb {C}$ such that $\zeta ^n=\mu ^n=1$. Since then $A_0(x)=-\alpha _1\alpha _2=-\zeta \beta _1\mu \beta _2=\zeta \mu A_0(y)$, it follows that $A_0(x)\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$. Moreover, clearly $A_0(x)\in \mathbb {C}[h(x)]$. Since then $A_0(x)=A_0(y)$, it follows that $\zeta \mu =1$. A short calculation shows that

$$\begin{aligned} A_1(x)=\frac{\left( \zeta +\frac{1}{\zeta }\right) A_1(y) + \left( \zeta -\frac{1}{\zeta }\right) \sqrt{A_1(y)^2+4A_0(y)}}{2}, \end{aligned}$$

(3.3)

and hence

$$\begin{aligned} A_1(x)^2-\left( \zeta +\frac{1}{\zeta }\right) A_1(x)A_1(y) +A_1(y)^2-\left( \zeta -\frac{1}{\zeta }\right) ^2A_0(x)=0. \end{aligned}$$

Denote

$$\begin{aligned} H_1(X, Y):&=A_1(X)^2-\left( \zeta +\frac{1}{\zeta }\right) A_1(X)A_1(Y)+A_1(Y)^2\nonumber \\&\quad -\left( \zeta -\frac{1}{\zeta }\right) ^2A_0(X)\in \mathbb {C}[X, Y]. \end{aligned}$$

(3.4)

Recall that by assumption $h(x)=h(y)$. Since h is neither cyclic nor dihedral, by Theorem 6 it follows that $H(X,Y)=(h(X)-h(Y))/(X-Y)\in \mathbb {C}[X,Y]$ is irreducible. Since $H_1(x, y)=0$, it follows that $H(X, Y)\mid H_1(X,Y)$. (We clearly also have

$$\begin{aligned} H(X, Y)\mid A_1(X)^2-\left( \zeta +\frac{1}{\zeta }\right) A_1(X)A_1(Y) +A_1(Y)^2-\left( \zeta -\frac{1}{\zeta }\right) ^2A_0(Y), \end{aligned}$$

but this follows from (3.4) and $A_0(x)\in \mathbb {C}[h(x)]$, which is what we will use instead.) It follows that the highest homogenous part of H(X, Y) divides the highest homogenous part of $H_1(X, Y)$. (A similar argument appeared in [15,  Lemma 3].) If $\deg A_0>2\deg A_1$, then

$$\begin{aligned} \frac{X^{\deg h}-Y^{\deg h}}{X-Y}\mid X^{\deg A_0}, \end{aligned}$$

which is clearly a contradiction. If $2\deg A_1>\deg A_0$, then

$$\begin{aligned} \frac{X^{\deg h}-Y^{\deg h}}{X-Y}\mid X^{2 \deg A_1} -\left( \zeta +\frac{1}{\zeta }\right) X^{\deg A_1}Y^{\deg A_1}+Y^{2 \deg A_1}. \end{aligned}$$

It follows that $\delta ^{2\deg A_1}-\left( \zeta +\frac{1}{\zeta }\right) \delta ^{\deg A_1}+1=0$ for all $\delta$ satisfying $\delta \ne 1$ and $\delta ^{\deg h}=1$. Thus for all such $\delta$ we have that either $\delta ^{\deg A_1}=\zeta$ or $\delta ^{\deg A_1}=\zeta ^{-1}$. Then either $\zeta =\pm 1$ or $\{1,\zeta , 1/\zeta \}$ is a cyclic subgroup of order 3 of the group of k-th roots of unity, where $k=\deg h$. If $\zeta =\pm 1$, then $A_1(x)=\pm A_1(y)$ by (3.3), and hence $A_1(x)\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$, and moreover clearly $A_1(x)\in \mathbb {C}[h(x)]$. Since then $A_0(x), A_1(x), G_0(x), G_1(x)\in \mathbb {C}[h(x)]$ from the recurrence relation it follows that $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$, a contradiction. In the latter case we have $\zeta ^3=1$, and hence $\alpha _1^3+\alpha _2^3=\beta _1^3+\beta _2^3$ from $\alpha _1=\zeta \beta _1$ and $\alpha _2=\frac{1}{\zeta } \beta _2$. By (3.1) and (3.2) we conclude that

$$\begin{aligned} A_1(x)^3+3A_0(x)A_1(x)=\alpha _1^3+\alpha _2^3\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x)). \end{aligned}$$

Now recall that h is indecomposable and thus from $\mathbb {C}(h(x))\subseteq \mathbb {C}(A_1(x), h(x))\subseteq \mathbb {C}(x)$ by Lüroth’s theorem ([13,  p. 13]) it follows that there are no intermediate fields between $\mathbb {C}(h(x))$ and $\mathbb {C}(x)$. Thus, either $\mathbb {C}(h(x))= \mathbb {C}(A_1(x), h(x))$ or $\mathbb {C}(A_1(x), h(x))=\mathbb {C}(x)$. Since $A_1(x)^3+3A_0(x)A_1(x)\in \mathbb {C}(h(x))$ and $A_0(x)\in \mathbb {C}(h(x))$ we have that $A_1(x)$ is a root of a cubic polynomial over $\mathbb {C}(h(x))$, and hence either $\mathbb {C}(A_1(x), h(x))=\mathbb {C}(h(x))$ or $\deg h=[\mathbb {C}(x):\mathbb {C}(h(x)]=[\mathbb {C}(A_1(x), h(x)):\mathbb {C}(h(x))]\le 3$. However, any polynomial of degree 2 is cyclic and of degree 3 either cyclic or dihedral ($ax^3+bx+c+d\in \mathbb {C}[x]$ with $a\ne 0$ is cylic if $b^2=3ac$, and dihedral otherwise). If $\mathbb {C}(A_1(x), h(x))=\mathbb {C}(h(x))$, then clearly $A_1(x)\in \mathbb {C}[h(x)]$ and then $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$, a contradiction.

If $2\deg A_1= \deg A_0$, then

$$\begin{aligned}&\frac{X^{\deg h}-Y^{\deg h}}{X-Y}\mid a_1^2\left( X^{2 \deg A_1} -\left( \zeta +\frac{1}{\zeta }\right) X^{\deg A_1}Y^{\deg A_1}+Y^{2 \deg A_1}\right) \\&\quad -a_0\left( \zeta -\frac{1}{\zeta }\right) ^2 Y^{2\deg A_1}, \end{aligned}$$

where $a_1$ is the leading coefficient of $A_1$ and $a_0$ is the leading coefficient of $A_0$. It follows that for any $\delta \ne 1$ such that $\delta ^{\deg h}=1$, we have

$$\begin{aligned} a_1^2\left( \delta ^{2\deg A_1}-\left( \zeta +\frac{1}{\zeta }\right) \delta ^{\deg A_1}+1\right) -a_0\left( \zeta -\frac{1}{\zeta }\right) ^2=0. \end{aligned}$$

(3.5)

Since $A_0(x)\in \mathbb {C}[h(x)]$ and $\deg A_0=2\deg A_1>0$ ($A_0$ and $A_1$ are nonconstant since otherwise the sequence is constant, which would violate the minimality assumption), it follows that $\deg h\mid 2\deg A_1$. Therefore $\delta ^{2\deg A_1}= 1$ and hence $\delta ^{\deg A_1}=\pm 1$. We deduce

$$\begin{aligned} -a_1^2\left( \zeta +\frac{1}{\zeta }\pm 2\right) =a_0\left( \zeta +\frac{1}{\zeta }-2\right) \left( \zeta +\frac{1}{\zeta }+2\right) \end{aligned}$$

Recall that $\zeta$ is an n-th root of unity. It follows that either $\zeta =\pm 1$, or $\zeta ^3=1$ (and $a_1$ and $a_0$ are related in a certain way). In the former case, $A_1(x)=\pm A_1(y)$ by (3.3), and hence $A_1(x)\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$ and moreover $A_1(x)\in \mathbb {C}[h(x)]$, so as before we conclude that $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$, a contradiction. In the latter case, we conclude that $A_1(x)^3+3A_0(x)A_1(x)\in \mathbb {C}(h(x))$. We eliminate this possibility using the same argument as in the case $2\deg A_1> \deg A_0$. $\square$

Proof of Corollary 2

If $p(q)=\pm q(x)$, or one of p and q is a zero polynomial, or $p(x)^m=q(x)^m$ for some $m\ge 1$, then the statement follows from Lemma 4, since $p(x)^n+q(x)^n$ is a constant times $p(x)^n$. Otherwise, consider the sequence (1.1) with $A_0(x)=-p(x)q(x)$, $A_1(x)=p(x)+q(x)$, $G_0(x)=2$ and $G_1(x)=A_1(x)$, which is minimal, non-degenerate and simple, and $G_m(x)=p(x)^m+q(x)^m$ for all $m\ge 0$. By Theorem 1 it follows that either $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$ or $\deg g\le C(p, q)$. If the former holds, then $p(x)+q(x)\in \mathbb {C}[h(x)]$ and $p(x)^2+q(x)^2\in \mathbb {C}[h(x)]$, so also $p(x)q(x)\in \mathbb {C}[h(x)]$, and thus $A_0(x), A_1(x)\in \mathbb {C}[h(x)]$. Furthermore, clearly either

$$\begin{aligned} p(x)=\frac{A_1(x) + \sqrt{A_1(x)^2+4A_0(x)}}{2}, \quad q(x)=\frac{A_1(x) - \sqrt{A_1(x)^2+4A_0(x)}}{2}, \end{aligned}$$

or vice versa. Since p and q are polynomials, we have that $A_1(x)^2+4A_0(x)=D(x)^2$ for some $D(x)\in \mathbb {C}[x]$. It follows that $D(x)^2\in \mathbb {C}[h(x)]$. By Lemma 5 we have $D(x)\in \mathbb {C}[h(x)]$, and hence $p(x), q(x)\in \mathbb {C}[h(x)]$. $\square$

Proof of Proposition 3

As in the proof of Theorem 1, let $y\ne x$ be a root of $h(X)-h(x)\in \mathbb {C}(x)[X]$ so that $h(x)=h(y)$ and consequently $G_n(x)=G_n(y)$. Then (1.2) holds. Also as in the proof of Theorem 1, we may assume that $\mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$ since h is not cyclic. As before, by [8,  Thm. 1] it suffices to show that (1.2) has no vanishing subsum. We assume the contrary.

If $A_0$ and $A_1$ are constants, then clearly $\alpha _1, \alpha _2, \beta _1, \beta _2\in \mathbb {C}$, and consequently $\pi _1, \pi _2\in \mathbb {C}[x]$ and $\rho _1, \rho _2\in \mathbb {C}[y]$. Since there exists a vanishing subsum of (1.2), we have that either $\pi _1\alpha _1^n = \rho _1\beta _1^n$ and $\pi _2\alpha _2^n=\rho _2\beta _2^n$, or $\pi _1\alpha _1^n = \rho _2\beta _2^n$ and $\pi _2\alpha _2^n=\rho _1\beta _1^n$. In either case,

$$\begin{aligned} \pi _1\alpha _1^n , \pi _2\alpha _2^n \in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x)) \end{aligned}$$

and moreover clearly $\pi _1\alpha _1^n , \pi _2\alpha _2^n \in \mathbb {C}[h(x)]$. Thus $\pi _1,\pi _2 \in \mathbb {C}[h(x)]$ and hence $G_m(x)=\pi _1\alpha _1^m+\pi _2\alpha _2^m\in \mathbb {C}[h(x)]$ for all $m\ge 0$, a contradiction.

Now ssume $A_0(x)=a_0$ and $G_m(x)=c$ for $m\in \{0, 1\}$, with $a_0, c\in \mathbb {C}$. Then also $A_0(y)=a_0$ and $G_m(y)=c$ via $x\mapsto y$. Note that the statement of the theorem trivially follows for $n\le 3$. We may thus assume $n>3$. Since, by assumption, there exists a vanishing subsum of (1.2), by multiplication and Vieta’s formulae it follows that $\pi _1\pi _2A_0(x)^n=\rho _1\rho _2A_0(y)^n$, and hence $\pi _1\pi _2=\rho _1\rho _2$. Then also $\pi _1\pi _2\alpha _1^m\alpha _2^m=\rho _1\rho _2\beta _1^m\beta _2^m$ since $(-\alpha _1\alpha _2)^m=A_0(x)^m=a_0^m=A_0(y)^m=(-\beta _1\beta _2)^m$. Since we also have $G_m(x)=G_m(y)=c$, it follows that $\pi _1\alpha _1^m+\pi _2\alpha _2^m=\rho _1\beta _1^m+\rho _2\beta _2^m$. We conclude that either $\pi _1\alpha _1^m=\rho _1\beta _1^m$ and $\pi _2\alpha _2^m=\rho _2\beta _2^m$ or $\pi _1\alpha _1^m=\rho _2\beta _2^m$ and $\pi _2\alpha _2^m=\rho _1\beta _1^m$. Without loss of generality we may assume that the fomer holds. Since there exists a vanishing subsum of (1.2), we have that either $\pi _1\alpha _1^n = \rho _1\beta _1^n$ and $\pi _2\alpha _2^n=\rho _2\beta _2^n$, or $\pi _1\alpha _1^n = \rho _2\beta _2^n$ and $\pi _2\alpha _2^n=\rho _1\beta _1^n$. We now show that in either case

$$\begin{aligned} D_{n-m}(A_1(x), -A_0(x))=D_{n-m}(A_1(x), -a_0)=\alpha _1^{n-m}+\alpha _2^{n-m}\in \mathbb {C}[h(x)]. \end{aligned}$$

Recall that by assumption $n>3$ and thus $n-m>2$. In the former case, from $\pi _1\alpha _1^m=\rho _1\beta _1^m$ and $\pi _1\alpha _1^n = \rho _1\beta _1^n$, we conclude $\alpha _1^{n-m}= \beta _1^{n-m}$, and likewise from $\pi _2\alpha _2^m=\rho _2\beta _2^m$ and $\pi _2\alpha _2^n=\rho _2\beta _2^n$, that $\alpha _2^{n-m} =\beta _2^{n-m}$. Then $\alpha _1^{n-m}+\alpha _2^{n-m} = \beta _1^{n-m}+\beta _2^{n-m}$ and by (3.1) and (3.2) we conclude that $\alpha _1^{n-m}+\alpha _2^{n-m} \in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$, and hence $D_{n-m}(A_1(x), -A_0(x))\in \mathbb {C}[h(x)]$. If on the other hand $\pi _1\alpha _1^n = \rho _2\beta _2^n$ and $\pi _2\alpha _2^n=\rho _1\beta _1^n$, then from $\pi _1\alpha _1^m=\rho _1\beta _1^m$ and $\pi _2\alpha _2^m=\rho _2\beta _2^m$ we deduce

$$\begin{aligned} \alpha _1^{n-m}+\alpha _2^{n-m}=\frac{\rho _2\beta _2^n}{\rho _1\beta _1^{m}}+\frac{\rho _1\beta _1^n}{\rho _2\beta _2^{m}}= \frac{G_0(y)G_{m+n}(y)-\rho _1\rho _2(\beta _1^{m+n}+\beta _2^{m+n})}{\rho _1\rho _2(-A_0(y))^m}. \end{aligned}$$

By (2.4), it follows that $\rho _1\rho _2\in \mathbb {C}(y)$, and thus by (3.1) and (3.2) we conclude that $\alpha _1^{n-m}+\alpha _2^{n-m}\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$, and hence $D_{n-m}(A_1(x), -A_0(x))=\alpha _1^{n-m}+\alpha _2^{n-m}\in \mathbb {C}[h(x)]$. Now, since $A_0(x)=a_0\in \mathbb {C}$ and $n-m>2$ by assumption, from (2.1) we have that $D_{n-m}(X, -a_0)$ is either dihedral (if $a_0\ne 0$) or cyclic (if $a_0=0$). By Lemma 5, from $D_{n-m}(A_1(x), -a_0)\in \mathbb {C}[h(x)]$, it follows that $A_1(x)\in \mathbb {C}[h(x)]$. Since $\pi _1\pi _2=\rho _1\rho _2\in \mathbb {C}(x)\cap \mathbb {C}(y)=\mathbb {C}(h(x))$ and $A_0(x), A_1(x)\in \mathbb {C}[h(x)]$, by (2.4) it follows that

$$\begin{aligned} G_1(x)^2-G_0(x)G_1(x)A_1(x)-a_0G_0(x)^2\in \mathbb {C}[h(x)]. \end{aligned}$$

If $m=0$ and thus $G_0(x)=c$, then $a_0G_0(x)^2$ is constant and $G_1(x)^2-cG_1(x)A_1(x)\in \mathbb {C}[h(x)]$. Since also $A_1(x)\in \mathbb {C}[h(x)]$, we deduce that $(2G_1(x)-cA_1(x))^2\in \mathbb {C}[h(x)]$. By Lemma 5 it follows that $2G_1(x)-cA_1(x)\in \mathbb {C}[h(x)]$ and hence $G_1(x)\in \mathbb {C}[h(x)]$. If $m=1$, we analogously conclude that $G_0(x)\in \mathbb {C}[h(x)]$. Therefore $G_0(x), G_1(x), A_0(x), A_1(x) \in \mathbb {C}[h(x)]$ in either case, so $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$, a contradiction. $\square$

4 Some remarks in relation to our main results

In Table 1 we list some well-studied binary recurrent sequences of polynomials that our main results can be applied to. All of these polynomials are generated by the Lucas polynomial sequence. Note that for each polynomial sequence in the second column we have $G_m(x)=\alpha _1^m+\alpha _2^m$ for all $m\ge 0$, where $\alpha _1, \alpha _2$ are such that $A_1(x)=\alpha _1+\alpha _2$ and $A_0(x)=-\alpha _1\alpha _2$, and therefore Theorem 1 can be applied. All the sequences in the first column have constant $G_0$ and $A_0$, and therefore Proposition 3 can be applied.

Table 1 $(G_n(x))_{n=0}^\infty$ satisfying $G_{n+2}(x)=A_{1}(x)G_{n+1}(x)+A_0(x)G_n(x), \ n\ge 0$

Furthermore, consider the sequence

$$\begin{aligned} G_n(x)=(Ax+B)G_{n-1}(x)+D G_{n-2}(x), \ n\ge 1, \quad G_{-1}(x)=0, \ G_0(x)=g_1\ne 0, \end{aligned}$$

(4.1)

where the coefficients $A, B, D\in \mathbb {C}$ satisfy $A, D\ne 0$ and do not depend on n. Fuchs, Pethő and Tichy [9] considered this sequence while studying a problem related to ours. Under certain assumptions, they gave a bound on the number of distinct $n,m\ge 0$ such that $G_n(x)=G_m(P(x))$ for a fixed nonconstant $P(x)\in \mathbb {C}[x]$. Note that $\deg G_m=m$, so Proposition 3 gives an upper bound on m and n such that $G_n(x)=G_m(P(x))$ for a fixed nonconstant polynomial $P\in \mathbb {C}[x]$ if P is not a composite of a cyclic or a dihedral polynomial, or such that $G_m(x)$ and P(x) are composites of the same polynomial of degree $>1$ for all $m\ge 0$.

5 Proof of Theorem 4

As mentioned in the introduction, Theorem 4 is an almost immediate consequence of Theorem 1 and the main result of [2]. To state the latter result we define the so called standard pairs of polynomials. In what follows $a, b\in \mathbb {Q}{\setminus }{0}$, $m, n\in \mathbb {N}$, $r\in \mathbb {N}\cup \{0\}$, $p(x) \in \mathbb {Q}[x]$ is nonzero and $D_{m} (x,a)$ is the m-th Dickson polynomial with parameter a, defined in Theorem 6 (Table 2).

Table 2 Standard pairs

Theorem 7

Let $f(x), g(x)\in \mathbb {Q}[x]$ be non-constant polynomials. Then the equation $f(x)=g(y)$ has infinitely many rational solutions with a bounded denominator if and only if $f(x)=\phi \left( f_{1}\left( \lambda (x\right) \right)$, $g(x)=\phi \left( g_{1}\left( \mu (x)\right) \right)$, where $\phi (x)\in \mathbb {Q}[x]$, $\lambda (x), \mu (x)\in \mathbb {Q}[x]$ are linear polynomials, and $\left( f_{1},g_{1}\right)$ is a standard pair over $\mathbb {Q}$ such that the equation $f_1(x)=g_1(y)$ has infinitely many rational solutions with a bounded denominator.

The proof of Theorem 7 in [2] relies on Siegel’s classical theorem on integral points on curves, and is consequently ineffective. Thus, Theorem 4 is also ineffective. We remark that among the ingredients in the proof of Theorem 7 were Ritt’s decompositions results.

Proof of Theorem 4

Assume that the equation $G_n(x)=H_m(y)$ has infinitely many solutions in integers x, y. Then $G_n(x)=\phi \left( f_{1}\left( \lambda (x\right) \right)$ and $H_m(x)=\phi \left( g_{1}\left( \mu (x)\right) \right)$, where $\phi (x)\in \mathbb {Q}[x]$, $\lambda (x), \mu (x)\in \mathbb {Q}[x]$ are linear polynomials, and $\left( f_{1},g_{1}\right)$ is a standard pair over $\mathbb {Q}$, according to Theorem 7. Note that $\phi$ is nonconstant since $G_n$ and $H_m$ are by assumption nonconstant. From the table we see that if both $\deg f_1>1$ and $\deg g_1>1$, then either $G_n$ or $H_m$ is a composite of an either cyclic or dihedral polynomial, a contradiction. Since $\deg G_n\ge \deg H_m$ by assumption, it follows that either $\deg g_1=1$ and $\deg f_1>1$, or both $f_1$ and $g_1$ are linear polynomials. If the latter holds, then clearly $\phi (x)=H_m(\ell _1(x))$ for linear $\ell _1$, and thus $G_n(x)=H_m(\ell (x))$ for linear $\ell (x)\in \mathbb {Q}[x]$. If $\deg g_1=1$ and $\deg f_1>1$, then clearly $\phi (x)=H_m(\ell (x))$ for linear $\ell (x)\in \mathbb {Q}[x]$, and hence $G_n(x)=H_m(\ell (f_1(\lambda (x)))$. Since by assumption there does not exist $h(x)\in \mathbb {C}[x]$ such that $G_m(x)\in \mathbb {C}[h(x)]$ for all $m\ge 0$ and $G_n$ is also not a composite of a cyclic or a dihedral polynomial, by Theorem 1 it follows that $\deg H_m<C(\{A_i, G_i : i=0,1\})$. $\square$