Decomposable polynomials in second order linear recurrence sequences

We study elements of second order linear recurrence sequences $(G_n)_{n= 0}^{\infty}$ of polynomials in $\mathbb{C}[x]$ which are decomposable, i.e. representable as $G_n=g\circ h$ for some $g, h\in \mathbb{C}[x]$ satisfying $\operatorname{deg}g,\operatorname{deg}h>1$. Under certain assumptions, and provided that $h$ is not of particular type, we show that $\operatorname{deg}g$ may be bounded by a constant independent of $n$, depending only on the sequence.


Introduction and results
Let d ≥ 2 be an integer. We consider a sequence of polynomials (G n ) ∞ n=0 in C[x] satisfying the d-th order linear recurrence relation determined by A 0 , A 1 , . . . , A d−1 ∈ C[x] and initial terms G 0 , G 1 , . . . , G d−1 ∈ C[x]. Let G ∈ C(x)[T ] be the characteristic polynomial of the sequence and let α 1 , . . . , α t be its distinct roots in the splitting field L/C(x) of G, that is where k 1 , . . . , k t ∈ N. Then G n admits a representation of the form G n (x) = π 1 α n 1 + π 2 α n 2 + · · · + π d α n d , where π i ∈ L[n] for i = 1, 2, . . . , n. We say that the recurrence relation (1) is minimal if (G n ) ∞ n=0 does not satisfy a recurrence relation with smaller d and coefficients in C[x]. We say that (1) is non-degenerate if α i /α j / ∈ C * for all i = j. Finally, we say that (1) is simple if k 1 = · · · = k t = 1; in this case the π i 's lie in L. We also call the corresponding sequence (G n ) ∞ n=0 minimal, non-degenerate and simple, respectively. In this paper, we will be concerned with second-order minimal non-degenerate simple linear recurrences.
Many Diophantine problems involving linear recurrence sequences have been studied in the literature. For example, a famous problem is to estimate the number of zeros appearing in such a sequence, and more generally, to bound the number of solutions n ∈ N of the equation G n (x) = a, where a ∈ L is given (cf. [9] and the papers cited therein). Also, several authors studied the problem of giving bounds on m and n such that G n (x) = cG m (P(x)), c = c(n, m), where (G n ) ∞ n=0 is a linear recurrence sequence and P a fixed polynomial (cf. [8,[10][11][12]).
In this paper, we focus on decomposable polynomials in second order linear recurrence sequences. A polynomial f ∈ C[x] with deg f > 1 is said to be decomposable if it can be written as the composition f (x) = g(h(x)) with g, h ∈ C[x] and deg g, deg h > 1, and indecomposable otherwise. The possible ways of writing a polynomial as a composition of polynomials were studied by several authors, starting with Ritt in the 1920's in his classical paper [21]. Results in this area of mathematics have applications to various other fields, e.g. number theory, complex analysis, arithmetic dynamics, finite geometries, etc. For example, there are applications to Diophantine equations of type f (x) = g(y). In 2000, Bilu and Tichy [4], by building on the work of Siegel, Ritt, Fried and Schinzel, classified the polynomials f, g for which the equation f (x) = g(y) has infinitely many solutions in S-integers x, y. It turns out that such f and g must be representable as a composition of polynomials in a certain prescribed way.
In this paper we show that if (G n ) ∞ n=0 satisfies (1) with d = 2, under certain assumptions on G 0 , G 1 , A 0 and A 1 , if G n (x) = g(h(x)) and h(x) is not of particular type, then deg g may be bounded by a constant independent of n, depending only on the sequence (more precisely, it depends only on the degrees of G 0 , G 1 , A 0 , A 1 ). To describe what we mean by h being of particular type and to state our results, we introduce the following notions. We say that f, g ∈ C[x] are equivalent if there are linear 1 we say that f is cyclic if it is equivalent to a polynomial g with g(x) = x n for some n > 1, and we say that f is dihedral if it is equivalent to T n for some n > 2, where T n is a Chebychev polynomial, defined by the functional equation T n (x + 1/x) = x n + 1/x n . Cyclic and dihedral polynomials play an important role in Ritt's theory of polynomial decomposition, as will be explained in Sect. 2. To see that at least some exceptional cases have to be taken into account, consider e.g. the well-known family of Fibonacci polynomials F n , defined by It is easy to see that for all odd n ≥ 3, F n is an even polynomial of degree n − 1, and hence if n ≥ 5 is odd, F n (x) can be written as F n (x) = g(h(x)), where h(x) = x 2 and deg g = (n − 1)/2. Clearly, here the degree of g cannot be bounded independently of n. In this case, h is cyclic. Also, for Chebyshev polynomials T n , which satisfy the second order linear recurrence T 0 (x) = 1, T 1 (x) = x, T n+2 (x) = 2xT n+1 (x) − T n (x) for n ∈ N, it is well-known that T mn = T m •T n for any m, n ∈ N. Since deg T n = n, clearly one cannot bound deg g independently of n assuming T n (x) = g(h(x)) and deg h > 1. In this case, h is dihedral.
There is a third, trivial situation where it is clearly not possible to bound the degree of g independently of n assuming G n = g • h, namely when G m (x) ∈ C[h(x)] for every m ∈ N. Consider for example the sequence (F n (h(x))) ∞ n=0 , where F n is defined by (3) and h ∈ C[x]. This sequence satisfies a second order linear recurrence relation and we clearly cannot bound deg F n independently of n. It will be shown later that We now describe our strategy and results in detail. Let (G n ) ∞ n=0 be a minimal non-degenerate simple second order linear recurrence sequence given by (1) (with d = 2). Assume that G n is decomposable for some n ∈ N and write G n (x) = g(h(x)), where h is indecomposable, and thus deg h ≥ 2. By Gauss's lemma it follows that the polynomial h(X ) − h(x) ∈ C(h(x))[X ] is irreducible and since h (X ) = 0, it is also separable (find details in Sect. 2). Since deg h ≥ 2, there exists a root y = x in its splitting field over C(h(x)). Clearly, h(x) = h(y). As in (2), we have where α 1 , α 2 are distinct roots of the characteristic polynomial , and π 1 , π 2 ∈ L 1 . Indeed, there is a representation of this form since by assumption the characteristic polynomial has no multiple roots. Observe that π i α n i = 0 for all n ∈ N and i = 1, 2 by minimality. Conjugating (in some fixed algebraic closure of C(x) containing α 1 , α 2 ) over C(h(x)) via x → y, we get a sequence (G n (y)) ∞ n=0 with G n (y) ∈ C[y], which satisfies the same minimal non-degenerate simple recurrence relation as (G n (x)) ∞ n=0 with x replaced by y. We conclude that where β 1 , β 2 are distinct roots of the characteristic polynomial G 2 (T ) = T 2 − A 1 (y)T d−1 − A 0 (y) in its splitting field L 2 /C(y), and ρ 1 , ρ 2 ∈ L 2 . Again we have that ρ i β n i = 0 for all n ∈ N and i = 1, 2. Since h(x) = h(y), we get G n (x) = G n (y), that is We view this last equation as an S-unit equation in function fields and seek to apply a result of Brownawell and Masser (see Theorem 3 below) to bound the height of G n and consequently the degree of g. However, this theorem can be applied directly only to equations in which no proper subsum vanishes. We will show in Sect. 4 that if h is not cyclic, then equation (4) has a proper vanishing subsum if and only if π 1 π 2 A 0 (x) n ∈ C(h(x)).
In particular, the existence of a proper vanishing subsum of (4) does not depend on the choice of the conjugate y of x over C(h(x)). However, (4) cleary depends on n and h for which G n (x) = g(h(x)) which are not known a priori. Note that if h is not cyclic and A 0 (x) = a 0 ∈ C, π 1 π 2 = π ∈ C, then there exists a vanishing subsum of (4) and one cannot apply the theorem in question; for example, this is the case for Chebyshev polynomials T n . We now state our main result.
and (G n ) ∞ n=0 be a sequence of polynomials defined by the minimal non-degenerate simple linear recurrence There is a positive real constant C = C({A i , G i : i = 1, 2}) with the following property. If for some n we have G n (x) = g(h(x)), where h is indecomposable and neither dihedral nor cyclic, and if (4) has no proper vanishing subsum, then it holds that deg g ≤ C.
We mention that the constant C in Theorem 1 can be effectively computed; this is done in the proof of the theorem. Since the bound is not very illuminating, we have not stated it above. Also note that in the theorem the situation that G m (x) ∈ C[h(x)] for all m is not excluded explicitly. It will be shown (see Lemma 9) that in this case either h is cyclic or equation (4) has a proper vanishing subsum.
Theorem 1 resembles a result of Zannier [26], who showed that if f is a polynomial with non-constant terms and f (x) = g(h(x)), where h is not of type ax k +b, a = 0, then deg g ≤ 2 ( − 1). Our proof, like Zannier's proof, involves applying Brownawell and Masser's theorem [5]. The application of this theorem in our proof requires a different approach and the technical details are more challenging. We remark that Zannier's result was one of the main ingredients of the proof of a conjecture of Schinzel [27] by the same author, which states that for f ∈ C[x] with non-constant terms, satisfying f = g • h for some g, h ∈ C[x], the number of terms of h is bounded above by B( ), where B is an explicitly computable function. Zannier's result was then used in [15,16] to study Diophantine equations of type f (x) = g(y), where f and g are arbitrary polynomials with a fixed number of non-constant terms, via the criterion of Bilu and Tichy. We remark that likewise, using our results, one may study Diophantine equations of this type where f and/or g are elements of a second order linear recurrence sequence of polynomials. We further mention that some special cases of the latter problem have already been studied in the literature, see [6,14].
In order to apply Theorem 1 one has to exclude that (4) has a proper vanishing subsum. This depends on n and h which are not known a priori. One therefore has to show that for all n ∈ N and for all h ∈ C[x], deg h > 1, h not cyclic we have that π 1 π 2 A 0 (x) n / ∈ C(h(x)) holds. Verification of this turns out to be quite non-trivial and one might suspect that it is not possible to verify it at all. We therefore complement Theorem 1 by detecting some explicit cases for which there does not exist such a vanishing subsum with the motivation to convince the reader that Theorem 1 contains useful information and is applicable. In fact we believe that no vanishing subsum exists at all unless we are in one of the exceptional cases already mentioned in the theorem. It would be very interesting to see a proof or a counterexample of this.
To detect cases when there does not exist a vanishing subsum of (4), we apply several tools. We follow a Galois-theoretic approach to decomposition questions, which originated in Ritt's work [21], and apply some recent results on polynomial decomposition from [1] and [20]. We show that the following holds.
n=0 be a sequence of polynomials defined by the minimal non-degenerate simple linear recurrence Assume that for some n we have G n (x) = g(h(x)), where h is indecomposable. If h is neither dihedral nor cyclic, and it does not hold that G m (x) ∈ C[h(x)] for all m ∈ N, then (4) has no proper vanishing subsum if A 0 (x) is constant and any of the following holds: If h is not cyclic, and it does not hold that G m (x) ∈ C[h(x)] for all m ∈ N, then (4) has no proper vanishing subsum if any of the following holds: We mention that the condition A 1 (x) 2 + 4 A 0 (x) ∈ C[x] means that the roots α 1 , α 2 of the corresponding characteristic polynomial are in C[x]. As clarified in the theorem, the condition 2G 1 (x) = G 0 (x)A 1 (x) is equivalent to the condition π 1 = π 2 . Furthermore, we mention that if G 0 (x) = A 1 (x), and either G 1 (x) = 2 A 0 (x) + G 0 (x) 2 or G 1 (x) = −2 A 0 (x), then either π 1 = α 1 and π 2 = α 2 , or π 1 = α 2 and π 2 = α 1 (see Lemma 10 and Lemma 11 for more details).
The paper is organized as follows. In Sect. 2 we shall collect some facts about polynomial decomposition; here Galois-theoretic arguments play an important role. In Sect. 3 we collect auxiliary results concerning heights in function fields, state some well-known theorems from the literature, and prove three lemmas which will be used to prove our main results. In Sect. 4 we give a proof of Theorem 2 using results from the previous two sections. In Sect. 5 we give a proof of Theorem 1. As already mentioned above, our proof of Theorem 1 involves applying the theory of S-unit equations over function fields.

Polynomial decomposition via Galois theory
Recall that a polynomial f ∈ C[x] with deg f > 1 is called indecomposable if it cannot be written as the composition f (x) = g(h(x)) with g, h ∈ C[x], deg g > 1 and deg h > 1. Otherwise, f is said to be decomposable. Any representation of f as a functional composition of polynomials of degree > 1 is said to be a decomposition By comparison of degrees one sees that no such polynomial exists when deg μ > 1.
A lot of information about the polynomial f is encoded into its monodromy group. By Gauss's lemma it follows that Lüroth's theorem (see [22, p. 13]) states that for a field K satisfying . This theorem provides a dictionary between decompositions of f ∈ C[x] and fields between C( f (x)) and Since f is a polynomial, h can be chosen to be a polynomial by [22, p. 16]. Then f = g(h(x)) for some g ∈ C[x]. The fields between C( f (x)) and C(x) clearly correspond to groups between the two associated Galois groups -Gal(L/C( f (x))) = Mon( f ) =: G and Gal(L/C(x)) =: H (the stabilizer of x in Mon( f )). In this way, the study of ways to represent a polynomial f as a composition of lower degree polynomials reduces to a study of subgroups of the monodromy group of f , and more precisely to the study of groups between H and G. Furthermore, it can be shown that G has a transitive cyclic subgroup, that is that G = H I for some cyclic group I (I can be chosen to be the inertia group at any place of the splitting field of f (x)−t which lies over the infinite place of C(t)); see also [17,Lemma 3.4] or [24,Lemma 3.3]. In this way, the study of ways to represent a complex polynomial f as a composition of lower degree polynomials reduces to a study of subgroups of the cyclic group I .
The interested reader is referred to [17] and [20] to find out more about the Galois-theoretic setup for addressing decomposition questions which originated in Ritt's work [21]. Ritt in [21] also showed that any complete decomposition of a complex polynomial 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. He then solved the equation a • b = c • d in indecomposable complex polynomials, showing that the only solutions, up to composing with linear polynomials, are the trivial one a • b = a • b and the non-trivial solutions where h ∈ C[x], n, k, m ∈ N and T n is the n-th Chebyshew polynomial defined in the introduction. We now record two results on the topic that we will repeatedly use in the sequel.
Recall that for f ∈ C[x], we say that f is cyclic if it is equivalent to x n for some n > 1, and we say that f is dihedral if it is equivalent to T n for some n > 2. Proposition 1 is Theorem 1.3 in [20]. See also [17,Thm. 5.1]. Proposition 2 is Lemma 3.6 in [20]. See also Theorem 3.8 in [3]. We record the following corollary.

then for any complete decomposition of f the collection of monodromy groups of the indecomposable polynomials consists only of dihedral groups. Furthermore, if f is cyclic, then for any complete decomposition of f the collection of monodromy groups of the indecomposable polynomials consists only of cyclic groups.
Proof. By Proposition 2, it suffices to prove the statement in the cases f (x) = T m (x) and f (x) = x m for m ∈ N, respectively. Note that since T mn (x) = T m (T n (x)) for any m, n ∈ N and Mon( f ) is dihedral if and only if f is dihedral, for any m ∈ N there exists a complete decomposition of T m (x) such that the collection of monodromy groups of the indecomposable polynomials consists only of dihedral groups. By Proposition 1, for any complete decomposition of T m (x) the collection of monodromy groups of the indecomposable polynomials consists only of dihedral groups. By the same argument, for any complete decomposition of x m the collection of monodromy groups of the indecomposable polynomials consists only of cyclic groups.
In the literature, quite often Ritt's and related results are expressed in terms of Dickson polynomials D n (x, a) (with parameter a), as they satisfy We refer to Turnwald's paper [24] for various properties of Chebyshev and Dickson polynomials. We now list some that will be of importance to us in this paper.
Proposition 3. All of the following holds: x, a), a n ) for any m, n ∈ N.
• D n (x + y, x y) = x n + y n .
• Let n ≥ 2 and let ζ n ∈ C be a primitive n-th root of unity. Put γ k = ζ k n + ζ −k when n is odd and when n is even.
For the proof of Theorem 2 we will also need the following result about polynomials with a common composite, which can be deduced from a result of Beals, Wetherell and Zieve [1,Thm. 5 then f 1 and f 2 are said to have a common composite. 'Most' pairs of complex polynomials have no common composite (this follows to the most part already from Ritt's results, see [1] for the details). The following fact will be repeatedly used in our proof of Theorem 2.

Preliminaries and auxiliary results
Our strategy involves the use of height functions in function fields. In what follows, let L be a finite extension of the rational function field C(x). For a ∈ C define the valuation ν a as follows.
. These are all (normalized) discrete valuations on C(x). All of them can be extended in at most [L : C(x)] ways to a discrete valuation on L and again in this way one obtains all discrete valuations on L. Furthermore, for f ∈ L * the sum formula ν( f ) = 0 holds, where the sum is taken over all discrete valuations on L. We just mention that there are different equivalent descriptions of the notion of discrete valuations as e.g. places or the rational points on a(ny) nonsingular complete curve over C with function field L. Now, define the projective height H of u 1 , . . . , u n ∈ L/C(x), where n ≥ 2 and not all u i zero, via Also, for a single element f ∈ L * , we set In both cases the sum is taken over all discrete valuations ν on L. Note that ν( f ) = 0 only for a finite number of valuations ν and that We state some basic properties of the projective height.

Lemma 2. Denote as above by H the projective height on L/C(x)
. Then for f, g ∈ L * the following properties hold: Proof. H( f ) ≥ 0 clearly holds by definition. To show that H( f + g) ≤ H( f ) + H(g), note that min(0, ν( f + g)) ≥ min(0, ν( f )) + min(0, ν(g)). Namely, if min(0, ν( f + g)) = 0, this clearly holds. Otherwise, by the definition of discrete valuations we have ν( f + g) ≥ min(ν( f ), ν(g)) and it follows that min(0, ν( f + g)) = ν( f + g) ≥ min(0, ν( f )) + min(0, ν(g)). Hence, To see that (6) holds, observe that by (2) and (3), it follows that if a ∈ C, then  H(a f ) = H( f + a) = H( f ). We argue by induction on n = deg A. The statement holds for n = 0 since in this case H (A( f ) Let us now assume that deg A = n + 1 and that the statement is true for lower-degree polynomials. If A(T ) = aT n+1 + b, with a, b ∈ C, the claimed equality clearly holds. Otherwise, let m > 0 be the unique integer such so that we can apply the induction hypothesis to A 1 . We claim that Indeed, if ν( f m ) > 0 then ν( f ) > 0, and by the strict triangle inequality for valuations it follows that ν(A 1 ( f )) = 0 for A 1 (0) = 0. On the other hand, if ν( f m ) < 0, and consequently ν( f ) < 0, then (again by the strict triangle inequality) we have ν(A 1 ( f )) < 0. So the claimed equality holds in any case. We conclude We use the following result due to Brownawell and Masser taken from [13] (more precisely, this is a direct consequence of [5, Thm. B and Cor. 1]), which gives an upper bound for the height of S-units, which arise as a solution of certain S-unitequations. Recall that for a set S of discrete valuations, we call an element of L an S-unit, if it has poles and zeros only at places in S, or equivalently, the set of S-units in L is Furthermore, we use the following classical estimates for the genus of a compositum of function fields, which are taken from [23, p. 130, p. 132].

Theorem 5. (Riemann's Inequality) Suppose that F = C(x, y). Then we have the following estimate for the genus g of F/C:
We now prove three lemmas that we will need in the proofs of our main results. Y ). Then the highest homogeneous part of H 1 (X, Y ) divides the highest homogeneous part of H (X, Y ), which is a constant multiple of

Lemma 3. Let h ∈ C[x] be indecomposable and let y = x be a root of h(X ) − h(x) ∈ C(x)[X ]. If h is neither cyclic nor dihedral, then
Therefore

Lemma 4. Let h ∈ C[x] be indecomposable and let y = x be a root of h(X
. Then either C(x)∩C(y) = C(x) and h is cyclic or C(x)∩C(y) = C(h(x)).

Proof. By assumption, h(x) = h(y).
Note that thus C(h(x)) ⊆ C(x) ∩ C(y) ⊆ C(x). By Lüroth's theorem (see [22, p. 13]) it follows that C(x) ∩ C(y) = C(r (x)) for some r ∈ C(x). Moreover, since h is a polynomial, r can be chosen to be a polynomial as well by [22, p. 16] Let Aut(h) denote the group of linear polynomials ∈ C[x] such that h • = h. It follows that ν ∈ Aut(h) and since ν(y) = x = y, it follows that Aut(h) is a non-trivial group. Recall that h is by assumption indecomposable. We now show that Mon(h) is cyclic, and hence that h is cyclic. This has been shown in Remark 2.14 in [20], as well as in Corollary 6.6 in [17]. For the sake of completeness we recall the proof.
First recall from Sect. 2 that if L is the splitting field of h(X ) − t over C(t) and x is such that h(x) = t, then G := Mon(h) = Gal(L/C(h(x))), and if we set H = Gal(L/C(x)), then G = H I for some cyclic group I . Now note that Aut(h) ∼ = N G (H )/H . Since h is indecomposable, there are no intermediate fields between C(h(x)) and C(x), and thus no proper subgroups between H and G, so either N G (H ) = G or N G (H ) = H . In the latter case, Aut(h) is trivial, a contradiction. Thus H G. Since H contains no nontrivial normal subgroups of G (because L is the normal closure of C(x)/C(h(x))), we must have H = 1, and G = H I = I , so G is cyclic. By Proposition 2 it follows that h is cyclic. Zannier [26,Lemma 3] showed that for an arbitrary h ∈ C[x] with deg h ≥ 1, there exists a conjugate y of x over C(h(x)) with the above properties: (1) then states that for q ∈ C[x], we have q(x) = q(y) if and only if q ∈ C[h(x)], while (2) and (3) are the same as above. Note that in Lemma 5, we put some conditions on h, but y is an arbitrary conjugate of x (such that y = x).

Lemma 5. Let h ∈ C[x] be indecomposable and let y = x be a root of h(X ) − h(x)∈ C(x)[X ]. Then the following hold. (1) For q ∈ C[h(x)] we have q(x) = q(y). Furthermore, if h is not cyclic and q(x) = q(y) for some q ∈ C[x], then q ∈ C[h(x)].
Proof of Lemma 5. The first statement follows from h(x) = h(y). Assume now that h is not cyclic and that q(x) = q(y) for some q ∈ C[x]. By Lemma 4 it follows that C(x) ∩ C(y) = C(h(x)). Since q(x) = q(y), it follows that q(x) ∈ C(x) ∩ C(y) = C(h(x)). Furthermore, since h, q ∈ C[x], we have q(x) ∈ C[h(x)]. This completes the proof of (1). We prove the other two statements completely analogously to the proof of Lemma 3 from [26]. By setting H (X, H 1 (x, y) = 0, then one shows by the same argument as in the proof of Lemma 3 that deg (3) is a consequence of the fact that the genus of a plane curve of degree ≤ d is bounded by (d − 1)(d − 2)/2.

Proof of Theorem 2
In this section we prove Theorem 2 using results from the previous two sections.

Recall that
n=0 is a sequence of polynomials defined by the minimal non-degenerate simple linear recurrence We are assuming that for some n we have G n = g • h, where h is indecomposable, and that x and y, which define equation 4, are such that h(x) = h(y) and x = y. We will use this notation throughout this section. In this notation, we have the following characterization of the existence of a proper vanishing subsum of (4) in the case when C(x) ∩ C(y) = C(h(x)). Note that by Lemma 4, either this holds or h is cyclic. Lemma 6. If C(x) ∩ C(y) = C(h(x)), then there exists a proper vanishing subsum of (4) if and only if π 1 π 2 A 0 (x) n ∈ C(h(x)).

Note that by Lemma 4 and Lemma 6 it follows that if
then either h is cyclic or there exists a proper vanishing subsum of (4). On the other hand, we have the following.
Proof. By π 1 π 2 = π A 0 (x) m and by Lemma 4 and Lemma 6, it follows that if there exists a proper vanishing subsum of (4), then either h is cyclic or A 0 (x) m+n ∈ C[h(x)]. Assuming the latter, by Lemma 5 we have A 0 (x) = ζ A 0 (y) for some (m+n)-th root of unity ζ . Then A 0 (x) ∈ C(x)∩C(y) = C(h(x)). Since deg A 0 = 1 and deg h ≥ 2, we have a contradiction.
In Theorem 2 we are assuming that we do not have G m (x) ∈ C[h(x)] for all m ∈ N. We have the following characterization of this situation.

Lemma 8. We have that G m (x) ∈ C[h(x)] for all m ∈ N if and only if
, then by the recurrence relation it follows that G m (x) ∈ C[h(x)] for every m ∈ N.
Conversely, assume that G m (x) ∈ C[h(x)] for all m ∈ N. If G 0 , G 1 , G 2 , G 3 (or any four consecutive elements of the sequence) satisfy G 2 1 − G 0 G 2 = 0, then the linear system (the last equality follows immediately by integrality). Since G m (x) ∈ C[h(x)] for all m ∈ N it cannot always hold that G 2 m+1 = G m G m+2 because in this case a short calculation shows that contradicting the assumption that (G n ) ∞ n=0 is a second order linear recurrence (observe that in this case necessarily Proof. Since h is not cyclic, by Lemma 4 it follows that C(x) ∩ C(y) = C(h(x)). (11) we conclude that π 1 π 2 ∈ C(h(x)) and hence π 1 π 2 A 0 (x) n ∈ C(h(x)). By Lemma 6 it follows that (4) has a proper vanishing subsum.
We complete a proof of Theorem 2 with the help of two lemmas. First note that by (10) it follows that π 1 = π 2 if and only if 2G 1

Lemma 10. If h is neither dihedral nor cyclic, and it does not hold that G m (x) ∈ C[h(x)] for all m, then (4) has no proper vanishing subsum if
Furthermore, if h is not cyclic, and it does not hold that G m (x) ∈ C[h(x)] for all m, then (4) has no proper vanishing subsum if π 1 = π 2 = π ∈ C and either Proof. Assume that h is not cyclic, and it does not hold that G m (x) ∈ C[h(x)] for all m, and that there exists a proper vanishing subsum of (4). Recall that by Lemma 4 and Lemma 6 it follows that π 1 π 2 A 0 (x) n ∈ C(h(x)).
Assume first that A 0 (x) = a 0 ∈ C and π 1 = π 2 =: π . Then G 0 (x) = 2π and since A 0 (x) = a 0 ∈ C, it follows that π 1 π 2 A 0 (x) n = a n 0 G 0 (x) 2 4 ∈ C(h(x)), Furthermore, by Proposition 3 we have it follows that for any m ∈ N we have that a contradiction with the assumption. If deg A 1 = 1, we have Obviously, D n (A 1 (x), −a 0 ) is equivalent to D n (x, −a 0 ), which is either cyclic or dihedral. By Lemma 1, Proposition 1 and Proposition 2 it follows that h is either cyclic or dihedral, a contradiction with the assumption. We conclude that deg A 1 , deg h > 1 and A 1 and h have a common composite. We now use Proposition a contradiction with the assumption. Assume thus that . By Proposition 4, since h is neither cyclic nor dihedral it follows that for some linear polynomials 1 , 2 , 3 ∈ C[x] and s, r ∈ N, s ≥ 2. In particular, A 1 is cyclic. By Proposition 1 and Lemma 1 it follows that the collection of monodromy groups in any complete decomposition of D n (A 1 (x), −a 0 ) consists only of cyclic or dihedral groups. Since D n (A 1 (x), −a 0 ) ∈ C[h(x)], by Proposition 2 it follows that h is either cyclic or dihedral, a contradiction.
We now prove the second statement. Assume that π 1 = π 2 = π ∈ C and either Recall that by Lemma 4 and Lemma 6 it again follows that It follows that A 0 (x) n ∈ C[h(x)] and thus A 0 (x) n = A 0 (y) n by Lemma 5. Then A 0 (x) = ζ A 0 (y) for some n-th root of unity ζ , so A 0 (x) ∈ C(x)∩C(y) = C(h(x)), by Lemma 5 we have D n (A 1 (x), −A 0 (x)) = D n (A 1 (y), −A 0 (y)).
Since A 0 (x) = A 0 (y) we further get Using Proposition 3 we get that either A 1 (x) = ±A 1 (y) or for γ k , δ k ∈ C given in the proposition.
a contradiction with the assumption. Thus we get that (14) holds. A short calculation shows that , a contradiction with the assumption.

Lemma 11. If h is neither dihedral nor cyclic, and it does not hold that G m (x) ∈ C[h(x)] for all m, then (4) has no proper vanishing subsum if
Furthermore, if h is not cyclic, and it does not hold that G m (x) ∈ C[h(x)] for all m, then (4) has no proper vanishing subsum if G 0 (x) = A 1 (x) and any of the following holds: Proof. Assume that h is not cyclic, and it does not hold that for all m, and that there exists a proper vanishing subsum of (4). Recall that by Lemma 4 and Lemma 6 it follows that π 1 π 2 A 0 (x) n ∈ C(h(x)).
and consequently by Lemma 5 that Then either A 1 (x) = ±A 1 (y) or for any m, a contradiction with the assumption. Thus we get that (14) holds. A short calculation shows that It remains to examine the case when in addition to the assumptions stated at the beginning of the proof we have A 0 (x) = a 0 ∈ C and h is not dihedral.
By (16) (15) it follows that G m (x) ∈ C[h(x)] for any m ∈ N, a contradiction with the assumption. Assume thus that As in the proof of Lemma 10, we conclude deg A 1 > 1. By Proposition 4, since h is neither cyclic nor dihedral it follows that for some linear polynomials 1 , 2 , 3 ∈ C[x] and s, r ∈ N, s ≥ 2. In particular, A 1 is cyclic. By Proposition 1 and Lemma 1 it follows that the collection of monodromy groups in any complete decomposition of D i (A 1 (x), −a 0 ) consists only of cyclic or dihedral groups. Since D i (A 1 (x), −a 0 ) ∈ C[h(x)], by Proposition 2 it follows that h is either cyclic or dihedral, a contradiction.
Proof of Theorem 2. By Lemma 7, Lemma 10 and Lemma 11 we conclude the proof of Theorem 2.

Proof of Theorem 1
Proof of Theorem 1. Assume that G n (x) = g(h(x)), where h is indecomposable and neither cyclic nor dihedral. Recall that x and y, which define (4), are such that h(x) = h(y) and x = y. From Lemma 4 it follows that C(x) ∩ C(y) = C(h(x)). Assume further that there is no proper vanishing subsum of (4) and write it as Define and also Let F = C(x, y, α 1 , α 2 , β 1 , β 2 ) and let H be the projective height on F/C(x), defined as in Sect. 3. By Lemma 2, we find the estimate and similarly we argue for u 2 . So, for i = 1, 2, 3, we have Note that if for some i we have H (v i ) = 0, then Since (G n (x)) ∞ n=0 is non-degenerate, the same holds for the sequence (G n (y)) ∞ n=0 , i.e. β 1 /β 2 / ∈ C. It follows that H(v 2 ) = H(β 1 /β 2 ) = 0 and thus On the other hand, we find the following upper bound for the height of G n (x): Using (18), we conclude that Now consider equation (17), which by assumption has no proper vanishing subsum. Let A = {α i , π i , β i , ρ i , i = 1, 2} and put where S 0 denotes the set of finite valuations and S ∞ denotes the set of infinite valuations on F. Then by Theorem 3 it follows that where g is the genus of F/C. We now estimate the genus and |S| in terms of deg h. We start with the genus. In order to use Castelnuovo's inequality (Theorem 4), we define Note that C is the field of constants of F we have that C(x, α 1 , . Now Riemann's inequality (Theorem 5) yields , we conclude that where C 1 := max{deg A 0 , 2 deg A 1 } (it will be shown later that indeed C 1 ≥ 1, by the non-degeneracy of the sequence).

It therefore follows that
Next, we estimate the height of w 2 in a similar way: (G 1 (y)) + H(G 0 (y))) + H(β 1 ) + H(β 2 ) ≤ 2(deg G 1 + deg G 0 )H(y) + 3C 1 H(y) To give a suitable lower bound for H(G n (x)), note that since G n = g • h we have Finally, we conclude that and therefore that deg g < 2C.