Diophantine properties of linear recursive sequences II.

Attila Pethő^{1} Institut for Mathematics and Informatics University of Debrecen H-4010 Debrecen, PO Box 12, Hungary

e-mail: pethoe@math.klte.hu

Dedicated to the 60th birthday of Professor Árpád Varecza

2000 Mathematics Subject Classification: 11D61, 11D25

1 Introduction

Let G_{n} be a k-th (k ≥ 2) order linear recursive sequence, in the sequel LRS, of integers defined by initial terms G_{0}, …, G_{k−1} ∈ ℤ and by the relation

G_{n+k} = A_{1}G_{n+k−1} + … + A_{k}G_{n}

for n ≥ 0, where A_{1}, …, A_{k} ∈ ℤ, A_{k} ≠ 0. We assume that at least one of the initial terms of G_{n} is non-zero. In the present paper we continue our survey on results concerning the mixed exponential-polynomial diophantine equation

G_{n} = P (y),  (1)

where P (y) ∈ ℤ[y] denotes a polynomial of degree d ≥ 2. In the first part [P6] we concentrated on results proved by using elementary and algebraic tools. Here we are dealing with applications of lower bounds for linear forms in logarithms of algebraic numbers.

Most of the text of the paper is four years old. In 1996 I delivered an invited talk at the Seventh International Research Conference on Fibonacci Numbers and Their Applications. The edited version of the first part of that talk appeared in [P6], while the second part remains up to now unpublished. I extended this manuscript with some new results.

Although the paper is a survey it includes new results too. You find an explicit bound for the exponent of perfect powers appearing in the classical Fibonacci and Lucas sequences. Another new result is Theorem 6, where we prove by the combination of effective and non-effective methods that, under certain condition, there exit only finitely many perfect powers in cubic recurrences.

1This paper was written when the author was a visiting professor at the Mathematical Institute of the Technical University of Graz, Austria.

Research partially supported by Hungarian National Foundation for Scientific Research Grant No 16741 and 25157.

2 Applications of lower bounds for linear forms in logarithms of algebraic numbers

The most powerful tools for the solution of equation (1) are certainly lower bounds for linear forms in logarithms of algebraic numbers. One can prove not only general finiteness theorems, but combining with numerical diophantine approximation techniques one can completely solve single and even parametrized families of equations. In this section we collect some result proved on this way.

The first general lower bound for linear forms in logarithms of algebraic numbers was proved by A. Baker [Ba1]. Since them several improvements and refinements appeared in the literature. It has a lot of applications. Analogous results were proved for linear forms in p-adic, in elliptic and in p-adic elliptic logarithms of algebraic numbers as well. Although they have also important applications in the theory of recurrences, see [ShT], we are dealing here only with the most classical case. For a collection of results proved by Baker’s method and further references we refer to the monographs [Ba2, ShT, Sm].

Here we cite three recent results of this kind. All three are very sharp, but under certain circumstances they give different results (sometimes nothing), therefore it is worth to choose the best appropriate one in a concrete case. We need some further definition. Let the algebraic number β be a zero of the irreducible polynomial p(y) = a_{t}x^{t}+…+a_{0} ∈ ℤ[y], where (a_{t}, …, a_{0}) = 1. Denote β = β_{1}, …, β_{t} the zeros of p(y). The absolute logarithmic height of β is defined by

h(β) = \frac{1}{t} log (∏_{i=1}^{t} max\{1, |β_{i}|\}) .

In the sequel let α_{1}, …, α_{m} be non-zero algebraic numbers and b_{1}, …, b_{m} rational integers. For i = 1, …, m, let log α_{i} be a determination of the logarithm of α_{i} and

Λ = b_{1} log α_{1} + … + b_{m} log α_{m}.

Assume further that Λ ≠ 0.

The first theorem is due to Baker and Wüstholz [BW].

Theorem 1 Let D = [ℚ(α_{1}, …, α_{m}) : ℚ] and B = max{|b_{1}|, …, |b_{m}|, e}. Then

log |Λ| ≥ −18(m + 1)!m^{m+1}(32D)^{m+2}h(α_{1}) ⋯ h(α_{m}) log B.

This theorem is very easy to apply, but it makes not possible for example to prove a bound for exponents for perfect powers in recursive sequences. For such purpose we shall apply the following result of Waldschmidt [W].

Theorem 2 Put

D = [ℚ(α_{1}, …, α_{m}) : ℚ] and g = [ℝ(log α_{1}, …, log α_{m}) : ℝ].

Let h_{1}, …, h_{m}, E and f be positive real numbers such that

log h_{i} ≥ h(α_{i}), (1 ≤ i ≤ m), h = max{h_{1}, …, h_{m}}

and

e ≤ E ≤ min \{h_{1}^{D}, …, h_{m}^{D}, \frac{mD}{f} (∑_{i=1}^{m} \frac{| log α_{i}|}{log h_{i}})^{−1}\} .

Assume that b_{m} ≠ 0 and put

B = max_{1≤j≤m} \{\frac{|b_{m}|}{log h_{j}} + \frac{|b_{j}|}{log h_{m}} \} ,

Z_{0} = max \{7 + 3 log m, \frac{g}{D} log E, log (\frac{D}{log E} )\} , H_{0} = max\{4mZ_{0}, log B\}

and

U_{0} = max{D^{2} log h, D^{m+2}H_{0}Z_{0} log h_{1} ⋯ log h_{m}(log E)^{−m−1}}.

Then the linear form Λ satisfies

|Λ| ≥ exp \{−1500g^{−m−2}2^{2m}m^{3m+5} (1 + \frac{g}{f} )^{m} U_{0}\} .

In solving equations related to second order recurrences we often meet cases, when only two algebraic numbers are involved, i.e. m = 2. Then there are even sharper lower bounds available. We cite here Corrollaire 1 of Laurent, Mignotte and Nesterenko [LMN].

Theorem 3 Let α_{1}, α_{2} be multiplicatively independent algebraic numbers and b_{1}, b_{2} be integers. Put D = [ℚ(α_{1}, α_{2}) : ℚ]/[ℝ(α_{1}, α_{2}) : ℝ] and

Λ = b_{1} log α_{1} − b_{2} log α_{2}.

Let h_{1}, h_{2} > 1 real numbers satisfying

log h_{i} ≥ max \{h(α_{i}), \frac{| log α_{i}|}{D}, \frac{1}{D} \} , i = 1, 2

and put

b = \frac{b_{1}}{D log h_{2}} + \frac{b_{2}}{D log h_{1}} .

Then

log |Λ| ≥ −30.9D^{4} (max \{log b, \frac{21}{D}, \frac{1}{2}\})^{2} log h_{1} log h_{2}.

In the applications of the above lower bounds the following simple lemmata are useful. For their proofs we refer to [PW].

Lemma 1 Let a ∈ ℝ. If 0 < a < 1 and 0 < |x| < a then

| log(1 + x)| < \frac{− log(1 − a)}{a}|x|.

Lemma 2 Let a ≥ 0, h ≥ 1, b > (e^{2}/h)^{h}, and let x ∈ ℝ, x > 1 satisfy

x ≤ a + b log^{h} x.

Then

x < 2^{h} (a^{1/h} + b^{1/h} log(h^{h}b))^{h} .

2.1 Perfect powers in LRS. Bounding the exponents.

Let G_{n} be a LRS with characteristic polynomial C_{G}(y) = y^{k} − A_{1}y^{k−1} − … − A_{k}. Let α_{1}, …, α_{h} denote the distinct zeros of C_{G} with multiplicities m_{1}, …, m_{h} respectively. Then there exist polynomials g_{i}(y) ∈ ℚ(α_{1}, …, α_{h})[y] of degree less than m_{i}, i = 1, …, h, such that

G_{n} = g_{1}(n)α_{1}^{n} + … + g_{h}(n)α_{h}^{n}  (2)

holds for any integer n ≥ 0.

In this section we are dealing with the solutions n, y, q ∈ ℤ of the equation

G_{n} = y^{q}.  (3)

The letters c_{1}, c_{2}, … will denote effectively computable constants depending only on the initial terms and on the coefficients of the characteristic polynomial of G_{n}.

For binary recurrences Shorey and Stewart [ShSt1] and independently the author of the present notes [P1] proved that max{n, |y|, q} < c_{1}. We come back to this result in section 2.3. For higher order recurrences much less is known. There are results only in that case, when G_{n} has a dominating characteristic zero. Under this assumption Shorey and Stewart [ShSt1] proved that q < c_{2} provided |y| > 1. Moreover Nemes and Pethő [NP1] showed that if T (y) ∈ ℤ[y] is a fixed polynomial, then q < c_{3} holds for any solutions n, y, q ∈ ℤ of the equation

G_{n} = y^{q} + T (y)  (4)

as well, provided |y| > 1. The constant c_{3} depends naturally on the coefficients of T (y) too. Recently Kiss [K3] succeeded to refine in some cases this result. He also proved [K1] the lower bound |G_{n} − y^{q}| > e^{cn}, provided α_{1} > |α_{2}| > |α_{i}|, i > 2. Shorey and Stewart [ShSt2] established under the same assumption that there exist only finitely many solution n, y, q with n ≥ 0, |y| > 1 and

q > max \{\frac{k log α_{1}}{log(α_{1}/ max\{1, |α_{2}|\})}, [ℚ(α_{1}) : ℚ] + deg T \} .

We also mention that this result is not effective, we do not know any upper bound for max{n, |y|}. A collection of the most general results of this kind you find in [ShT].

Remark that for fixed q the last equation may have infinitely many integer solution. Indeed, let H_{n} be a LRS and T (y) ∈ ℤ[y] a polynomial. Then G_{n} = T (H_{n}) is an LRS as well and the equation

G_{n} = T (y)  (5)

has infinitely many solutions. It is widely believed that the converse is also true, i.e. the equation G_{n} = T (y) has infinitely many solution only when G_{n} has a subsequence of the form T (H_{n}), with an LRS H_{n}. Positive answer for this problem was known until very recently again only for binary recurrences and only in the case when |A_{2}| = 1. (See [NP2, P4].) The situation changed with the papers [CZ1] and [CZ2]. In that papers Corvaja and Zannier proved similar results under quite general conditions. Let G_{n} be an LRS such that all of the zeros of its characteristic polynomial are simple and such that α_{1} is dominating, i.e. 1 ≠ |α_{1}| > |α_{i}|, i = 2, …, k. Let T (y) ∈ ℤ[y] be a monic polynomial.

If the equation (5) has infinitely many n, y ∈ ℤ, n ≥ 0 solutions then there exists an LRS H_{n} of algebraic numbers and an arithmetical progression P such that G_{n} = T (H_{n}) holds for all n ∈ P. Especially, let q ≥ 2 be fixed and let G_{n} be a q-th power for infinitely many n, then G_{n} = H_{n}^{q} holds for all n belonging to an arithmetical progression. In fact the theorem of Corvaja and Zannier is much more general.

In the above mentioned papers the dependence on the constants on the parameters was never examined. In this section, following essentialy the earlier line, I intend to prove an explicit upper bound for the exponents of possible perfect powers in G_{n}.

Theorem 4 Let G_{n} be a LRS such that all of the zeros of its characteristic polynomial are simple and α := α_{1} > |α_{2}| ≥ |α_{i}|, i = 3, …, k. Put

δ := \frac{log |α_{2}|}{log α},  c_{1} := ∑_{i=2}^{k} \frac{|g_{i}|}{|g_{1}|},  a_{1} = exp\{max\{h(g_{1}), e\}\}

and

c_{2} := max \{\frac{1}{log 2} (\frac{log(4c_{1})}{1 − δ} + log(c_{1} + 1) + log |g_{1}|) , \frac{log α}{2}(k!)^{14}, 3 log (\frac{5|g_{1}|}{4}) +
\frac{2 log(1.15c_{1})}{1 − δ} + \frac{1}{1 − δ} 4 · 10^{13}(k!)^{7} log a_{1} log α (217 log(k!) log log a_{1} − log(1 − δ))\} .

Assume that n, y, q ∈ ℤ is a solution of (3) such that |y| > 1. Then

q < c_{2}.

Proof: Let 𝕂 = ℚ(α_{1}, …, α_{k}). First we remark, that as the zeros of C_{G}(y) are simple, the polynomials g_{i}(y) ∈ 𝕂[y], i = 1, …, k are actualy constants, thus the definition of a_{1}, c_{1} and c_{2} is correct.

Let n, y, q ∈ ℤ be a solution of (3) such that |y| > 1 and

q ≥ \frac{1}{log 2} (\frac{log(4c_{1})}{1 − δ} + log(c_{1} + 1) + log |g_{1}|) .

Then

|g_{1}|(c_{1} + 1)(4c_{1})^{1/(1−δ)} ≤ 2^{q} ≤ |y|^{q} < |g_{1}|(c_{1} + 1)α^{n},

thus α^{n(1−δ)} > 4c_{1}. We now obtain the following important inequality

|\frac{|y|^{q}}{|g_{1}|α^{n}} − 1| ≤ ∑_{i=2}^{k} |\frac{g_{i}}{g_{1}} | |\frac{α_{i}}{α} |^{n} ≤ c_{1}α^{−n(1−δ)} < \frac{1}{4}.

From this we obtain on the one hand

\frac{q}{log α} − \frac{log(5|g_{1}|/4)}{log α log |y|} < \frac{n}{log |y|} < \frac{q}{log α} − \frac{log(3|g_{1}|/4)}{log α log |y|}  (6)

and

|Λ_{1}| := | log |g_{1}| + n log α − q log |y|| < 1.15 · c_{1}α^{−n(1−δ)}  (7)

on the other hand. The last inequality follows from Lemma 1. The expression staying on the left hand side of (7) is a linear form in logarithms of algebraic numbers and we have just proved a sharp upper bound for it.

A useful lower bound can be derived in this case only from Theorem 2. We have in the present case m = 3, g = 1, b_{1} = 1, b_{2} = n, b_{3} = q and D ≤ k!, but actually we need only an upper bound for the degree of 𝕂 over ℚ, therefore we may take D = k!. We choose E = e, f = 1, h_{1} = a_{1}, log h_{2} = k log α and h_{3} = |y|. These parameters satisfy the assumptions of Theorem 2.

Now we are able to compute B. Indeed, we have obviously

B = max \{\frac{q}{log a_{1}} + \frac{1}{log |y|}, \frac{q}{k log α} + \frac{n}{log |y|}\}
= \frac{q}{k log α} + \frac{n}{log |y|} < \frac{2q}{log α} − \frac{log(3|g_{1}|/4)}{log α log |y|}.

Further

Z_{0} = max{7 + 3 log 3, log k!} ≤ 14 log(k!).

Now we have to distinguish two cases according to the value of H_{0}. Assume first that B ≤ (k!)^{14}. Then H_{0} = 4mZ_{0} = 168 log(k!), which yields

U_{0} < 2352(k!)^{7} log(k!) log a_{1} log α log |y|,

hence

log |Λ_{1}| > −8.64 · 10^{15}(k!)^{7} log(k!) log a_{1} log α log |y|.

Comparing this inequality with (7), dividing by (1 − δ) log |y| and using (6) we obtain the upper bound

q < \frac{1}{1 − δ} 8.64 · 10^{15}(k!)^{7} log(k!) log a_{1} log α + \frac{log(1.15c_{1})}{(1 − δ) log 2},

which is obviously smaller than c_{2}. Hence the assertion is proved in this case.

Now we turn to the other case B > (k!)^{14}, when H_{0} = log B. We have in this case

U_{0} < (k!)^{7} log a_{1} log α log |y| log B,

which yields

log |Λ_{1}| > −3.68 · 10^{12}(k!)^{7} log a_{1} log α log |y| log B.

Comparing this inequality with (7), dividing by (1−δ) log α log |y| and using (6) we obtain the inequality

X < \frac{1}{1 − δ} 7.36 · 10^{12}(k!)^{7} log a_{1} log X + \frac{2 log(1.15c_{1})}{(1 − δ) log α} + \frac{log(25|g_{1}|/12)}{log α log |y|},

where

X = \frac{2q}{log α} − \frac{3|g_{1}|/4}{log α log |y|}.

Application of Lemma 2 yields the upper bound

X < \frac{4 log(1.15c_{1})}{(1 − δ) log α} + \frac{1}{1 − δ} 14.72 · 10^{12}(k!)^{7} log a_{1} ·
(217 log(k!) log log a_{1} − log(1 − δ)) + \frac{2 log(25|g_{1}|/12)}{log α log |y|},

which implies

q < \frac{2 log(1.15c_{1})}{1 − δ} + \frac{1}{1 − δ} 7.36 · 10^{12}(k!)^{7} log a_{1} log α ·
(217 log(k!) log log a_{1} − log(1 − δ)) + 3 log(5|g_{1}|/4).

The number on the right hand side is again smaller than c_{2}, hence the theorem is proved. □

To have an impression how large is c_{2} we computed it for the Fibonacci sequence.

Corollary 1 If F_{n} = y^{q} for some n, y, q ∈ ℤ with |y|, q > 1 then

q < 5.1 · 10^{17}.

Proof: It is well known that

F_{n} = \frac{1}{\sqrt{5}} [(\frac{1 + \sqrt{5}}{2})^{n} − (\frac{1 − \sqrt{5}}{2})^{n}] .

Thus the conditions of Theorem 4 are fulfilled and we have to compute c_{2} with the following values of the parameters: k = 2, α = (1 + \sqrt{5})/2, δ = −1, c_{1} = 1, a_{1} = e^{e} and g_{1} = 1/\sqrt{5}, which yields the upper bound 5.1 · 10^{17} for q. □

The bound for the exponent of possible perfect powers in the Fibonacci sequence is fairly large. Remark that applying Theorem 2 directly to the linear form (7) coming from the Fibonacci sequence we could be able to improve slightly the bound for q, but even this bound would not be smaller then 10^{14}. The reason is the dependence of Waldschmidt’s bound on the parameter m.

If, however g_{1} = 1, which appears for example in the Lucas sequence, we have only two terms in (7) and may use Theorem 3, which yields a much better bound. We do this more generaly for the sequences V_{n}(A_{1}, A_{2}) which is defined by V_{0}(A_{1}, A_{2}) = 2, V_{1}(A_{1}, A_{2}) = A_{1} and by the recursion V_{n+2}(A_{1}, A_{2}) = A_{1}V_{n+1}(A_{1}, A_{2}) + A_{2}V_{n}(A_{1}, A_{2}) for n ≥ 0.

Theorem 5 Let A_{1}, A_{2} be integers such that A_{1} > 0, |A_{2}| = 1 and D := A_{1}^{2} + 4A_{2} > 0. Let further α = (A_{1} + \sqrt{D})/2. If the integers n, y, q with |y|, q > 1 solve the equation

V_{n}(A_{1}, A_{2}) = y^{q}  (8)

then

q < 55125 log α.

Moreover, if (A_{1}, A_{2}) = (1, −1) or (2, −1) then q < 13222.

Proof: Putting β = (A_{1} − \sqrt{D})/2, then V_{n} = α^{n} + β^{n} holds for all n. Assume that the integers n, y, q with |y|, q > 1 are solutions of (8). As the assertion of the theorem trivialy holds for n ≤ 10, we assume n > 10 in the sequel and obtain

|\frac{y^{q}}{α^{n}} − 1| = α^{−2n} < 0.01.

This implies the inequalities

q log y − 0.01 < n log α < q log y + 0.01
|Λ| := |q log y − n log α| < 1.1α^{−2n}.

We are ready to apply Theorem 3 to get a lower bound for |Λ|.

Set b_{1} = q, b_{2} = n, h_{1} = y and

h_{2} = \{ \begin{matrix}(log α)/2 & , if & A_{1} ≥ 3 \\ 1/2 & , if & A_{2} = 1 and A_{1} = 1, 2. \end{matrix}

Then we have in the general case

b = \frac{q}{log α} + \frac{n}{2 log y} ≤ \frac{3q}{2 log α} + 0.15

and in the special cases b < 2.04q + 0.015.

Assuming q > 55000 log α (q > 13000 in the special cases) we have

max {log b, 10.5, 0.5} = log b.

In the rest we are dealing with the general case, the proof for the special cases is completely similar. Now Theorem 3 yields

log |Λ| ≥ 247.2 log^{2} (\frac{1.725q}{log α} + 0.018) log y log α.

Comparing this with the upper bound we obtain

2q log y − 0.1 < 247.2 log^{2} (\frac{1.725q}{log α} + 0.018) log y log α.

Dividing this inequality by 2 log y log α/1.725 and putting X := \frac{1.725q}{log α} + 0.018 we have

X < 426.5 log^{2} X + 0.24,

which implies X < 95100 by Lemma 1 and the assertion follows from the definition of X. □

Even the bound 13222 is too large to try to compute with known techniques all perfect powers in the Lucas sequence. The only recursive sequence for which we know all perfect powers is the Pell sequence U_{n}(2, 1). Pethő [P5] and independently Cohn [Co] proved that the equation

P_{n} = y^{q}

has solutions only for q = 2 and these are (n, y) = (0, 0), (1, 1) and (7, 13). Pethő ruled out the odd exponents by elementary means. For q = 2 and n odd Wolfskill [Wo] proved that n < 469, and below this bound is very easy to find all squares. The statement for even index squares follows from divisibility properties of the Pell numbers.

2.2 Perfect powers in high order recurrences

If G_{n} is a LRS satisfying the conditions of Theorem 4 then there are no non-trivial (from 0 and 1 different) q-th power in G_{n} provided q is large enough. You have seen, that this bound is usualy large. But this is not so bad, because the constants in the linear form bounds are continously improved and I expect in the future for a wide class of LRS at least such a good bound as for the Lucas sequence.

Four years ago I continued with the following sentences: Annoying is that we have generaly no information about small powers. The tribonacci sequence defined in the Problem of [P6] satisfies the condition of Theorem 4, but we do not know how many squares, third powers, etc. appear in it. We are not able to prove even that their number is finite.

Similarly, we know nothing about the squares or higher powers in the sequence (2^{n} − 1)(3^{n} − 1). This seems hard for squares and a bit easier for higher powers.

The situation changed considerably in the recent years. L. Szalay [Sz] established all squares in the sequence (2^{n} − 1)(3^{n} − 1). Hajdu and Szalay [HSz] did the same for the sequences (2^{n} − 1)(6^{n} − 1) and (a^{n} − 1)(a^{kn} − 1). In both papers there are used only elementary methods.

Much more interesting are the results of Corvaja and Zannier [CZ1] and [CZ2]. They considered in [CZ1] LRS with integer characteristic roots and proved: Let

G_{n} = g_{1}a_{1}^{n} + … + g_{k}a_{k}^{n},

where k ≥ 2, g_{1}, …, g_{k} are non-zero algebraic numbers and a_{1} > a_{2} > … > a_{k} > 0 are integers with a_{1}, a_{2} coprime. Then for every integer d ≥ 2 equation (3) has only finitely many solutions n, y ∈ lN. Remark that they used in the proof W.M. Schmidt’s [Schm] celebrated subspace theorem, hence the result is not effective.

Combining this result with Theorem 4 one obtains under the same assumptions as above that there are only finitely many perfect powers in the sequence G_{n}.

Moreover combining Theorem 4 with Theorem 2 of [CZ2] we are able to prove the following theorem.

Theorem 6 Let G_{n} be a third order LRS. Assume that the characteristic polynomial of G_{n} is irreducible and has a dominating root. Then there are only finitely many perfect powers in G_{n}.

Remark 1 The tribonacci numbers are defined by the initial terms T_{0} = T_{1} = 0, T_{2} = 1 and by the recursion T_{n+3} = T_{n+2} + T_{n+1} + T_{n} for n ≥ 0. In Part I [P6] I asked whether T_{0} = T_{1} = 0, T_{2} = T_{3} = 1, T_{5} = 4, T_{10} = 81, T_{16} = 3136 = 56^{2} and T_{18} = 10609 = 103^{2} are the only perfect squares among the tribonacci numbers. It follows from Theorem 6 that there are only finitely many perfect powers in T_{n}. This is a partial answer to my question. Remark that Theorem 6 is not effective, hence we have presently no algorithm for computing all powers in the sequence T_{n}.

Proof of Theorem 6. The assumptions of Theorem 4 fulfill, hence if G_{n} = y^{q} and |y| > 1 then q < c_{2} with an effectively computable constant c_{2}.

In the rest of the proof let 2 ≤ q < c_{2} be fixed. Assume that (3) has for this q infinitely many solutions in n, y ∈ lN, y > 1. Then there exist by [CZ2] (Theorem 2) non-zero algebraic numbers d_{1}, …, d_{r}, δ_{1}, …, δ_{r} such that

G_{n} = g_{1}α_{1}^{n} + g_{2}α_{2}^{n} + g_{3}α_{3}^{n} = (d_{1}δ_{1}^{n} + … + d_{r}δ_{r}^{n})^{q}  (9)

holds for all elements n of an arithmetical progression P. We may assume without loss of generality that |δ_{1}| ≥ … ≥ |δ_{r}|.

There exist, by a theorem of van der Poorten and Schlickewei [PSch], for any ε > 0 constants c_{4} and c_{5}(ε) such that

c_{5}(ε)|δ_{1}|^{n(1−ε)} ≤ |∑_{i=1}^{r} d_{i}δ_{i}^{n}| ≤ c_{4}|δ_{1}|^{n}

holds for all n > n(ε). On the other hand

|g_{1}||α_{1}|^{n}/2 ≤ |G_{n}| ≤ 2|g_{1}||α_{1}|^{n}

is true, provided n is large enough.

Hence on one hand

|\frac{δ_{1}^{q}}{α_{1}} |^{n} ≥ \frac{|g_{1}|}{2c_{4}^{q}}

for all sufficiently large n, which implies

|α_{1}| ≤ |δ_{1}|^{q}.

On the other hand we have

(\frac{|δ_{1}^{q}|^{1−ε}}{|α_{1}|} )^{n} ≥ \frac{2|g_{1}|}{(c_{5}(ε))^{q}}

for any fixed ε > 0 and all n ≥ n(ε). This implies

|α_{1}| ≥ |δ_{1}|^{q(1−ε)}

for all ε > 0. The lower and upper bounds for |α_{1}| imply |δ_{1}| = |α_{1}|^{1/q}.

Assume that |δ_{1}| = … = |δ_{s}| = |α_{1}|^{1/q} > |δ_{s+1}| ≥ |δ_{r}|. Define D_{1}(n) = ∑_{i=1}^{s} d_{i}δ_{i}^{n} and D_{2}(n) = ∑_{i=s+1}^{r} d_{i}δ_{i}^{n} and rewrite (9) as

g_{1}α_{1}^{n} + g_{2}α_{2}^{n} + g_{3}α_{3}^{n} = (D_{1}(n) + D_{2}(n))^{q} = D_{1}(n)^{q} + ∑_{j=1}^{q} \binom{q}{j}D_{1}(n)^{q−j}D_{2}(n)^{j}.

This implies

g_{1}α_{1}^{n} − D_{1}(n)^{q} = ∑_{j=1}^{q} \binom{q}{j}D_{1}(n)^{q−j}D_{2}(n)^{j} − g_{2}α_{2}^{n} − g_{3}α_{3}^{n}.  (10)

There is staying on the left hand side a finite power sum. If it is non-zero then for any ε > 0 its absolute value is bounded below by c_{6}(ε)|α_{1}|^{n(1−ε)}. In contrast the right hand side is obviously bounded above by c_{7}|α_{1}|^{n(q−1)/q} < c_{7}|α_{1}|^{n(1−η)} with a fixed η > 0. This is a contradiction, hence

g_{1}α_{1}^{n} = D_{1}(n)^{q},

which can be written as

g_{1}^{1/q}(α_{1}^{1/q})^{n} − ∑_{i=1}^{s} d_{i}δ_{i}^{n} = 0.

This is an S-unit equation in the field ℚ(g_{1}^{1/q}, α_{1}^{1/q}, d_{1}, …, d_{s}, δ_{1}, …, δ_{s}), which has infinitely many solutions in n ∈ lN. By the theorem of Evertse [Ev1] there exists a 1 ≤ i ≤ s such that g_{1}^{1/q}(α_{1}^{1/q})^{n} = d_{i}δ_{i}^{n} holds infinitely often. We may assume i = 1 without loss of generality, i.e. we have

g_{1}^{1/q}(α_{1}^{1/q})^{n} = d_{1}δ_{1}^{n} and ∑_{i=2}^{s} d_{i}δ_{i}^{n} = 0

for infinitely many n ∈ lN.

For such integers n we obtain

∑_{j=1}^{q} \binom{q}{j}d_{1}^{q−j}δ_{1}^{(q−j)n}D_{2}(n)^{j} − g_{2}α_{2}^{n} − g_{3}α_{3}^{n} = 0  (11)

from equation (10). As α_{2}/α_{3} is not a root of unity the sum g_{2}α_{2}^{n} + g_{3}α_{3}^{n} never vanish if n is large enough, hence D_{2}(n) ≠ 0 for such n’s. The function \binom{q}{j}d_{1}^{q−j}δ_{1}^{(q−j)n}D_{2}(n)^{j} is for fixed n strictly decreasing in j, hence writing it as a sum of exponential functions it has at least q summands such that no proper subsums of their sum vanishes for infinitely many n. From these one can be equal to g_{2}α_{2}^{n} and an other equal to g_{3}α_{3}^{n}, hence if q > 2 then (11) cannot hold for infinitely many n.

Finally let q = 2. Then (11) simplifies to

2d_{1}δ_{1}^{n} (∑_{i=2}^{r} d_{i}δ_{i}^{n}) + (∑_{i=2}^{r} d_{i}δ_{i}^{n})^{2} − g_{2}α_{2}^{n} − g_{3}α_{3}^{n} = 0.  (12)

This equation has infinitely many solutions in n. As ∑_{i=2}^{r} d_{i}δ_{i}^{n} ≠ 0 for all large enough n, there exists a minimal index 2 ≤ i_{0} ≤ r and J ⊆ R = {2, …, r} such that i_{0} ∈ J,

∑_{j∈R\J} d_{j}δ_{j}^{n} = 0 and ∑_{j∈J_{1}⊆J} d_{j}δ_{j}^{n} ≠ 0

for infinitely many solutions of (12) and for all ∅ ≠ J_{1} ⊆ J. Therefore |J| = 1 and

2d_{1}δ_{1}^{n}d_{i_{0}}δ_{i_{0}}^{n} + d_{i_{0}}^{2}δ_{i_{0}}^{2}n − g_{2}α_{2}^{n} − g_{3}α_{3}^{n} = 0.

Hence the system of equations

2d_{1}δ_{1}^{n}d_{i_{0}}δ_{i_{0}}^{n} = g_{2}α_{2}^{n}
d_{i_{0}}^{2}δ_{i_{0}}^{2}n = g_{3}α_{3}^{n}

or the analogous system of equations, where the right hand sizes are interchanged, has infinitely many solutions in n.

These implies δ_{1}δ_{i_{0}} = α_{2}ζ_{2} and δ_{i_{0}}^{2} = α_{3}ζ_{3}, with some roots of unity ζ_{2}, ζ_{3}. We also have δ_{1}^{2} = α_{1}ζ_{1} with a root of unity ζ_{1}. These three relations imply

\frac{α_{2}^{2}}{α_{1}α_{3}} = \frac{ζ_{1}ζ_{3}}{ζ_{2}^{2}} .

Using finally that α_{1}α_{2}α_{3} is an integer A, we conclude that α_{2}^{3} = Aζ, where ζ is a root of unity. Taking conjugates we conclude that |α_{1}| = |α_{2}| = |α_{3}|, which contradicts the assumption that one of the roots is dominating. The Theorem is proved. □

2.3 Perfect powers in second order recurrences

For second order recurrences we can bound effectively not only the exponents of the perfect powers appearing in the sequence, but also the largest index for which a term of the sequence can be a perfect power. This was proved by Shorey and Stewart [ShSt1] and independently by me [P1].

Theorem 7 Let G_{n} be a non-degenerated second order LRS and d ∈ ℤ. Then there exist effectively computable positive constants c_{1}, c_{2} depending only on G_{0}, G_{1}, A_{1}, A_{2} and on d such that if for the integers n, y, q such that q > 1 the equation

G_{n} = y^{q}  (13)

holds, then:

(a) If |y| > 1, then max{|y|, n, q} < c_{1},

(b) if |y| ≤ 1, then n < c_{2}.

Proof: We may assume by Theorem 4, that if |y| > 1, then q is bounded by an effectively computable positive constants c_{3} depending only on G_{0}, G_{1}, A_{1}, A_{2} and on d. Let fix q such that 1 < q < c_{3}.

Write

G_{n} = g_{1}α_{1}^{n} + g_{2}α_{2}^{n}

and let

H_{n} = \frac{g_{1}α_{1}^{n} − g_{2}α_{2}^{n}}{α_{1} − α_{2}} .

It is easy to check that H_{n} is also an LRS of integers, G_{n} and H_{n} have the same characteristic polynomial and

G_{n}^{2} − (α_{1} − α_{2})^{2}H_{n}^{2} = 4(α_{1}α_{2})^{n}.

Remark that (α_{1} − α_{2})^{2} = A_{1}^{2} + 4A_{2} and α_{1}α_{2} = −A_{2} are integers. Let S be the set of all prime divisors of A_{2} and n, y be a solution of (13) for our fixed q with |y| > 1. There

exist integers r, s such that n = s(2q) + r and 0 ≤ r < 2q. The rational numbers \frac{G_{n}}{(A_{2})^{qs}}

and \frac{y}{s} are S-integral solutions of the hyperelliptic equation

z^{2} = (A_{1}^{2} + 4A_{2})u^{2q} + 4(−A_{2})^{r}.

The polynomial staying on the right hand side of this equation has for all 0 ≤ r < 2q at least three simple zeros, hence it has only finitely many effectively computable S-integer solutions.

As there exists only finitely many possibilities for q and for r the number of the hyperelliptic equations to be considered is finite, thus the number of solutions is finite and effectively computable.

The rest of the proof is easy, you find details in [P1, ShSt1] or in [ShT]. □

2.4 A method for determining squares in second order LRS

The above proof is simple and is very well for theoretical purposes. Its drawback is that it is hard to find power values in given LRS following its line, because we transform the problem to finitely many hyperelliptic equations and their solution is not at all easy.

Mignotte and Pethő [MP2] proposed a more direct way to solve (13) in case q = 2 and |A_{2}| = 1. We present now their method in this special form. They remarked that the method can be generalized to arbitrary A_{2} and for arbitrary q.

Let G_{n} be a second order recursive sequence with |A_{2}| = 1, d ∈ ℤ and consider the equation

G_{n} = g_{1}α_{1}^{n} + g_{2}α_{2}^{n} = dy^{2}.

As |A_{2}| = 1 the numbers α_{1}, α_{2} are real quadratic units. Put 𝕂 = ℚ(α_{1}). Let γ′ denotes the conjugate of γ∈𝕂. Then α_{2} = α_{1}′. We may assume without loss of generality that α_{1} > |α_{2}|. Adjusting our equation appropriately (changing g_{1} and g_{2} with α_{1}g_{1} and α_{2}g_{2} respectively if n is odd; multiplying the equation with the square free part of g_{2} and with −1 if necessary) we see that it is enough to deal with the equation

aα_{1}^{2m} − b^{2}α_{2}^{2m} = cy^{2},

where a, b and c∈ℤ_{𝕂} and m, y≥0 are integers. Our aim is to prove an upper bound for m.

Let 𝕃 = 𝕂(\sqrt{−c}) and assume that 𝕃 is a quadratic extension of 𝕂, i.e. [𝕃 : ℚ] = 4. Then our equation implies

N_{𝕃/ℚ}(bα_{2}^{m} + \sqrt{−c}y) = N_{𝕃/ℚ}(a) = A,  (14)

with an integer A. Choose in ℤ_{𝕃}, in the ring of integers of 𝕃, units η_{2}, …, η_{r}; r = 1, 2 or 3 such that the group U generated by η_{1} = α, η_{2}, …, η_{r} has finite index in the group of units of ℤ_{𝕃}. There exists in ℤ_{𝕃} a maximal finite set of, with respect to U, non–associated elements of norm A. This set will be denoted by A. Then there exist for all m, x∈ℤ with (14) a γ∈A and ε∈U such that

bα_{2}^{m} + \sqrt{−c}y = γε.  (15)

Let order the conjugates 𝕃^{(i)}, i = 1, 2, 3, 4 of 𝕃 according the following ordering of the conjugates of \sqrt{−c} : \sqrt{−c}, −\sqrt{−c}, \sqrt{−c′}, −\sqrt{−c′}. Let R denote the regulator of U, i.e. the absolute value of the determinant of the matrix (log |η_{i}^{(j)}|)_{1≤i,j≤r} and finally, for δ ∈ ℤ_{𝕂} let

\overline{|δ|} := max\{|δ^{(i)}, i = 1, …, 4\}.

It is easy to see that if m > m_{0} then

\frac{1}{2} \frac{\sqrt{|a|}}{|γ^{(i)}|}α_{1}^{m} < |ε^{(i)}| < \frac{2\sqrt{|a|}}{|γ^{(i)}|}α_{1}^{m}  (16)

for i = 1, 2; and if b′ > 0, which we may assume without loss of generality, then

\frac{b′}{2|γ^{(3)}|}α_{1}^{m} < |ε^{(3)}| < 2\frac{b′}{|γ^{(3)}|}α_{1}^{m}  (17)

and

\frac{|a′|}{2b′|γ^{(4)}|}α_{1}^{−3m} < |ε^{(4)}| < \frac{2|a′|}{b′|γ^{(4)}|}α_{1}^{−3m}  (18)

hold. We remark that if b′ < 0 then only the role of ε^{(3)} and ε^{(4)} changes.

The last inequalities imply that if c′ > 0 then (14) has only finitely many solutions and they are very easy to compute. In fact ε^{(3)} and ε^{(4)} are in this case conjugate complex numbers, hence

\frac{b′}{2|γ^{(3)}|}α_{1}^{m} < |ε^{(3)}| = |ε^{(4)}| < \frac{2|a′|}{b′|γ^{(4)}|}α_{1}^{−3m},

i.e m < \frac{1}{4} log |\frac{4a′γ^{(3)}}{b^{′2}γ^{(4)}} | .

The situation is more interesting when c′ < 0. Then ε^{(3)} and ε^{(4)} are real numbers and we will use estimations on linear forms in logarithms of algebraic numbers to establish an upper bound for m.

Let first c > 0 (and c′ < 0). Then 𝕃 has two nonreal and two real conjugates, and there exist u_{1}, u_{2} ∈ ℤ with ε = η_{1}^{u_{1}}η_{2}^{u_{2}}. The estimations (16) with i = 2 and (17) yield

|u_{1}| < \frac{2m log α_{1}| log |η_{2}^{(3)}| − log |η_{2}^{(2)}||}{R} + c_{1}

and

|u_{2}| < \frac{4m log^{2} α_{1}}{R} + c_{1},

where

c_{1} = 2 log (3\overline{|a|}\overline{|b^{2}|}\overline{|\frac{1}{γ} |}) max\{log α_{1}, log \overline{|η_{2}|}\}/R.

We have

γ^{(1)}ε^{(1)} + γ^{(2)}ε^{(2)} = 2bα_{2}^{m},

hence

|1 + \frac{γ^{(2)}}{γ^{(1)}} (\frac{η_{2}^{(2)}}{η_{2}^{(1)}} )^{u_{2}} | < \frac{4b}{\sqrt{|a|}}α_{1}^{−2m}.

If m > m_{0}, then \frac{4b}{\sqrt{|a|}}α_{1}^{−2m} < \frac{1}{2} and so

|Λ_{1}| = |log (−\frac{γ^{(2)}}{γ^{(1)}} ) + u_{2} log (\frac{η_{2}^{(2)}}{η_{2}^{(1)}} ) + u_{0}π| < \frac{4.1|b|}{\sqrt{|a|}}α_{1}^{−2m},

where u_{0}∈ℤ and we take the principal value of the complex logarithm function, i.e. −π ≤ log(z) ≤ π for every z ∈ ℂ. The last inequality yields |u_{0}| < |u_{2}| + 2.

As the field ℚ(γ^{(2)}/γ^{(1)}, η_{2}^{(2)}/η_{2}^{(1)}, \sqrt{−1}) is obviously a subfield of ℚ(\sqrt{−c}, \sqrt{−c′}\sqrt{−1}), which is of degree at most 16 over ℚ we can set D = 16 and B = |u_{2}| + 2 in Theorem 1 which yields the lower bound

|Λ_{1}| > exp \{−1.25 · 10^{18}h (\frac{γ^{(2)}}{γ^{(1)}} ) h (\frac{η_{2}^{(2)}}{η_{2}^{(1)}} ) log(|u_{2}| + 2)\} .

Comparing the lower and upper bounds for |Λ_{1}| we conclude

2m log α_{1} − log \frac{4.1|b|}{\sqrt{|a|}} < 1.25 · 10^{18}h (\frac{γ^{(2)}}{γ^{(1)}} ) h (\frac{η_{2}^{(2)}}{η_{2}^{(1)}} ) log (\frac{4m log^{2} α}{R} + c_{1} + 8) . (19)

This inequality yields an upper bound for m, which we shall only compute knowing the actual values of the occouring parameters.

Let now c < 0 (and c′ < 0). Then all conjugates of 𝕃 are real and there exist u_{1}, u_{2}, u_{3} ∈ ℤ with ε = η_{1}^{u_{1}}η_{2}^{u_{2}}η_{3}^{u_{3}}. We recall η_{1} = α_{1}. The estimations (16) with i = 1, 2 and (17) yield

|u_{i}| < \frac{4m log^{2} α_{1} log h}{R} + c_{2}, i = 2, 3,

where

c_{2} = 3\sqrt{3} log (3\overline{|a|}\overline{|b^{2}|}\overline{|1/γ}) log α_{1} log \overline{|η_{2}|} log \overline{|η_{3}|}/R

and h = max\{\overline{|η_{2}|}, \overline{|η_{3}|}\}.

Similarly to the above case, but working with real instead of complex logarithms we get

|Λ_{2}| = |log |\frac{γ^{(2)}}{γ^{(1)}} | + u_{2} log |\frac{η_{2}^{(2)}}{η_{2}^{(1)}} | + u_{3} log |\frac{η_{3}^{(2)}}{η_{3}^{(1)}} || < \frac{5.6|b|}{\sqrt{|a|}}α_{1}^{−2m}.

The parameters in the application of Theorem 1 are the same as earlier except that D = 8 and B = max{|u_{2}|, |u_{3}|}. Hence Theorem 1 implies

2m log α_{1} −log \frac{5.6|b|}{\sqrt{|a|}} < 3.87·10^{16}h (\frac{γ^{(2)}}{γ^{(1)}} ) h (\frac{η_{2}^{(2)}}{η_{2}^{(1)}} ) h (\frac{η_{3}^{(2)}}{η_{3}^{(1)}} ) log (\frac{4m log^{2} α_{1} log h}{R} + c_{2}) .

(20)

We again do not express an upper bound for m explicitely, but rather show through an example how to use inequalities (19) and (20).

Example Let G_{0} = 0, G_{1} = 10 and G_{n+2} = 10G_{n+1} − G_{n} and consider the equation G_{2m+1} = □. It is easy to see, that if we are able to solve this equation then we can solve G_{n} = □ too. (See [MP2].)

To solve the equation G_{2m+1} = □ we intend to apply the method described above. First we compute the necessary initial data. We have α_{1} = 5 + 2\sqrt{6}, α_{2} = 5 − 2\sqrt{6} and c = −12 + 5\sqrt{6}, thus our equation has the form

α_{1}^{2m} − α_{2}^{2}α_{2}^{2m} = 4α_{2}\sqrt{6}w^{2} = 4(5\sqrt{6} − 12)w^{2}.

It is easy to see that it has only one solution (m, w) = (0, 1) in the range 0 ≤ m ≤ 10. If m > 10 then (16) is obviously true. Thus we may assume in the sequel m > 10.

The algebraic number field 𝕃 = ℚ(\sqrt{6}, \sqrt{12 − 5\sqrt{6}}), has two real and two non-real

conjugates. The units α_{1} and η_{2} = 5 − 2\sqrt{12 − 5\sqrt{6}} − 2\sqrt{6} are fundamental units in ℤ_{𝕃}, hence its regulator is R = 6.83836 and we get

|u_{2}| < 3.07398m + 8.95847.

As γ = 1 there are only two summands in Λ_{1}, actually it has the form

Λ_{1} = |u_{2} log (\frac{5 − 2\sqrt{6} + 2\sqrt{12 − 5\sqrt{6}}}{5 − 2\sqrt{6} − 2\sqrt{12 − 5\sqrt{6}}}) + u_{0}π| < 0.042α_{1}^{−2m}.  (21)

As we proved, there are generaly three logarithms in Λ_{1}, but in the actual example we have only two, therefore in the, to (14) analogous inequality we get a much better constant. More precisely we have

4.58486m + 3.17387 < 6.81595 · 10^{11} log(12.4m + 40),

which implies m < 5 · 10^{12} and |u_{2}| < 1.55 · 10^{13}. Dividing the inequality for Λ_{1} by u_{2}π we see that, as m > 10, u_{0}/u_{2} is a convergent of

log (\frac{η_{2}^{(2)}}{η_{2}^{(1)}} ) /π = δ = .93557845273700309088141600367180617252445255312155...

The denominator of the 26-th convergent of δ, \frac{51706546491839}{55266927472061}, is larger than 10^{14}, hence by the well known extremality property of the convergents of real numbers we obtain

|u_{0} − u_{2}δ| ≥ |51706546491839 − 55266927472061 · δ| > 0.16132 · 10^{−13}.

Comparing this inequality with (21) we obtain

0.16132 · 10^{−13} < 0.0134α_{1}^{−2m},

which implies m ≤ 5. Thus our equation has only the trivial solution (m, w) = (0, 1).

Aknowledgement. The author thanks very much for the carefull work of the referee.

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