Abstract
In this paper, we consider the Diophantine equation for given integers b, c with , whereas varies among Lucas-Lehmer sequences of the second kind. We prove under some technical conditions that if the considered equation has at least three solutions (n, m) , then there is an upper bound on the size of the solutions as well as on the size of the coefficients in the characteristic polynomial of .
Keywords: Diophantine equations, Pillai’s Problem, Lucas sequences
Introduction
In recent times, many authors considered Pillai-type problems involving linear recurrence sequences. For an overview we refer to [5]. Let us note that these problems are inspired by a result due to S. S. Pillai [9, 10] who proved that for given, coprime integers a and b, there exists a constant , depending on a and b, such that for any the equation
1 |
has at most one solution .
Replacing the powers and by other linear recurrence sequences seems to be a challenging task which was supposedly picked up first by Ddamulira, Luca and Rakotomalala [4], where it was shown that
has at most one solution provided that
More generally, Chim, Pink and Ziegler [3] proved that for two fixed linear recurrence sequences , (with some restrictions) the equation
has at most one solution for all , except if c is in a finite and effectively computable set that depends on and .
In more recent years, several attempts were made to obtain uniform results, i.e. to allow to vary the recurrence sequences and in the result of Chim, Pink and Ziegler [3]. In particular, Batte, Ddamulira, Kasozi and Luca [1] showed that for all pairs with p a prime, the Diophantine equation
has at most four solutions with . This result was generalized by Heintze, Tichy, Vukusic and Ziegler [5]. They proved that for a given linear recurrence sequence (with irrational simple dominant root having a positive Binet coefficient a, such that and a are multiplicatively independent and such that the equation has no solutions with , and ), there exist an effectively computable bound such that for an integer the Diophantine equation
2 |
has at most two solutions with . Here is an effectively computable constant depending only on .
In this paper, we want to fix b in (2) and let vary over a given family of recurrence sequences. In particular, we consider the case where varies over the family of Lucas-Lehmer sequences of the second kind.
Notations and statement of the main results
In this paper, we consider Lucas-Lehmer sequences of the second kind, that is we consider binary recurrence sequences of the form
where and are the roots of the quadratic polynomial
with and . In the following, we will assume that is non-degenerate, i.e. that is not a root of unity. More precisely, we assume and , since this implies that and are distinct real numbers with . We will also assume that , which results in for all . Then we consider the Diophantine equation
3 |
where with are fixed.
Theorem 1
Let be a fixed real number and A, B coprime integers satisfying . Assume that as well as that the polynomial is irreducible and . Furthermore, assume that if and if . Assume that Eq. (3) has three solutions with for the bound . Then there exists an effectively computable constant , depending only on and b, such that . In particular, we can choose
Let us note that the bound in Theorem 1 ensures that is strictly increasing for (see Lemma 9 below). Let us also mention that it is essential to exclude the case that .
Although we can bound in terms of b and , our method does not provide upper bounds for A and |B|. However, in the case that we are more restrictive in the possible choice of B, we can prove also upper bounds for A and |B|.
Theorem 2
Let be a fixed real number and A, B coprime integers satisfying . Assume that as well as that the polynomial is irreducible. Furthermore, assume that if and if . Assume that Eq. (3) has three solutions with . Then there exist effectively computable constants , depending only on and b, and , depending only on b, such that . We can choose
and
A straight forward application of our bounds yields:
Corollary 3
Assume that . If the Diophantine equation
4 |
with has three solutions with , then either
or
and .
Another consequence of Theorem 2 together with the results of Chim, Pink and Ziegler [3] is that there exist only finitely many Diophantine equations of the form of (3) that admit more than two solutions, provided that . The following corollary gives a precise statement.
Corollary 4
Let be a fixed real number and a fixed integer. Then there exist at most finitely many, effectively computable 9-tuples , such that
A and B are coprime;
, and ;
is irreducible;
if , then ;
and ;
for .
Proof
For any fixed value of c, we apply Theorem 2 and get an upper bound for A, independent of c. So there remain only finitely many, effectively computable possibilities for A and B. Thus we have finitely many possibilities for the recurrence . Chim, Pink and Ziegler [3] proved that, for two in absolute values strictly increasing linear recurrences and with multiplicatively independent dominant roots, there exists only finitely many, effectively computable integers c such that has more than one solution. First, we have to ensure that and b are multiplicatively independent. However, and b are multiplicatively dependent if and only if there exist integers x, y not both zero such that . But cannot be a rational number, unless , since by our earlier assumptions and is not a perfect square. Second, we note that is strictly increasing for by Lemma 10 since implies . Therefore we can apply their result in our situation and obtain that there are only finitely many, effectively computable values for c. Finally, we use Theorem 1 with to bound the . Clearly, the can now be calculated.
Let us give a quick outline for the rest of the paper. In the next section we establish several lemmas concerning properties of Lucas sequences under the restrictions that and , respectively, that we will frequently use throughout the paper. The main tool for the proofs of the main theorems are lower bounds for linear forms in logarithms of algebraic numbers. In Sect. 6 we establish bounds for , which still depend on , following the usual approach (cf. [3]). In Sect. 7, under the assumption that three solution exist, we obtain a system of inequalities involving linear forms in logarithms which contain . Combining these inequalities we obtain a linear form in logarithms which does no longer contain . Thus we obtain that Theorems 1 and 2 hold or one of the following two cases occurs:
;
.
That each of these cases implies Theorems 1 and 2 is shown in the subsequent Sects. 8 and 9.
Auxiliary results on Lucas sequences
Let and be the roots of . By our assumptions, and are distinct real numbers . Throughout the rest of the paper, we will assume that is the larger one, i.e. , which we can do since, by our assumptions, we have . Therefore we obtain
First, note that can be bounded in terms of A:
Lemma 5
Assuming that , we have .
At some point in our proofs of Theorems 1 and 2, we will use that cannot be to close to 1. In particular, the following lemma will be needed.
Lemma 6
Let be as above, in particular irreducible with real roots. Assume that . Then we have
Proof
First, let us note that the function is strictly increasing for all . Moreover, we have and . By our assumption that is irreducible, a choice for B such that is not admissible. Therefore we have
We compute
provided that , which certainly holds for . Similar computations yield
provided that .
Next, we show that under our assumptions is not to large.
Lemma 7
Assume that and that . Then we have .
Proof
First we note that
From the observation
then it follows
Lemma 8
Assume that and that . Then we have .
Proof
We have
Let us now consider the function . This function is strictly increasing with x since obviously the numerator is strictly increasing with x while the denominator is strictly decreasing with x. Therefore we obtain
since .
Now let us take a look at the recurrence sequence .
Lemma 9
Assume that and . Then is strictly increasing for .
Proof
First, we note that
certainly holds if
i.e.
5 |
Note that , by Lemma 7, and, by Lemma 5,
Moreover, note that the smallest possible value for is . Therefore (5) is certainly fulfilled if
Thus is increasing if
or equivalently
Since the rational function is strictly increasing if , strictly decreasing if , and constant if , we obtain that for
the Lucas sequence is strictly increasing.
Lemma 10
Assume that and . Then is strictly increasing for .
Proof
By (5) we know that is increasing for all if . By Lemma 5, we have and, using Lemma 8, also . Therefore we get
Remark 1
Note that assuming and implies with . Therefore all results that are proven under the assumption with also hold under the assumption and .
In view of this remark and in view of the proofs of the following lemmas, we will assume for the rest of the paper that one of the following two assumptions holds:
, and ,
, and .
Remark 2
Let us note that the bound and in assumption A2 results in the useful fact that assumption A2 implies assumption A1 with , but with instead of . The assumption is mainly used in the proofs of the Lemmas 12 and 13. Moreover, assumptions A1 and A2 both imply .
In view of the assumptions stated above, an important consequence of Lemmas 9 and 10 is the following:
Corollary 11
Let assumption A1 or A2 be in force and let us assume that (3) has two solutions and with . Then .
Lemma 12
Let assumption A1 or A2 be in force and . Then we have .
Proof
Assume that A1 holds. Then, by Lemma 7, we have
If A2 holds, by Lemma 8, we have
Therefore in any case we get
Another lemma that will be used frequently is the following:
Lemma 13
Let assumption A1 or A2 be in force and assume that Eq. (3) has two solutions and with . Then we have
Proof
Assuming the existence of two different solutions implies by an application of Lemma 12 the inequality
which proves the first inequality.
For the second inequality, we apply again Lemma 12 to obtain
Since we assume in any case that , we get and thus
which yields the second inequality.
Finally, let us remind Carmichael’s theorem [2, Theorem XXIV]:
Lemma 14
Let A, B be coprime integers with and . For any there exists a prime1p such that , but for all , except for the case that and .
This lemma can be used to prove the following result:
Lemma 15
Assume that and . Then the Diophantine Eq. (3) has at most one solution (n, m) with .
Proof
Assume that (3) has two solutions with . Then by Carmichael’s primitive divisor theorem (Lemma 14) we deduce that and . Since and we obtain the system of equations
which yields
That is . Since we assume that and , we deduce and . We also conclude that . By considering 3-adic valuations, this yields since would imply . Hence we have .
Note that we also assume which implies that
holds. But this is only possible for and we conclude that and . Since is not irreducible, this is not an admissible case. Therefore there exists at most one solution to (3).
Lower bounds for linear forms in logarithms
The main tool in proving our main theorems are lower bounds for linear forms in logarithms of algebraic numbers. In particular, we will use Matveev’s lower bound proven in [7]. Therefore let be an algebraic number of degree d and let
be the minimal polynomial of . Then the absolute logarithmic Weil height is defined by
With this notation, the following result due to Matveev [7] holds:
Theorem 16
(Theorem 2.2 with in [7]) Denote by algebraic numbers, neither 0 nor 1, by determinations of their logarithms, by D the degree over of the number field , and by rational integers with . Furthermore let if K is real and otherwise. For all integers j with choose
and set
Assume that
Then
with
In our applications, we will be in the situation and , i.e. we have and . In this special case Matveev’s lower bounds take the following form:
Corollary 17
Let the notations and assumptions of Theorem 16 be in force. Furthermore, assume that K is a real quadratic number field and that . Then we have
6 |
Remark 3
Let us note that the form of E is essential in our proof to obtain an absolute bound for in Theorem 1. Let us also note that in the case of one could use the results of Laurent [6] to obtain numerically better values but with an term instead. This would lead to numerically smaller upper bounds for concrete applications of our theorems. However, we refrain from the application of these results to keep our long and technical proof more concise.
In order to apply Matveev’s lower bounds, we provide some height computations. First, we note the following well known properties of the absolute logarithmic height (see for example [12, Chapter 3] for a detailed reference):
Lemma 18
For any and we have
Remark 4
Note that for a positive integer b we have and, with and from Sect. 3, we have
This together with the above mentioned properties yields for the bound
One other important aspect in applying Matveev’s result (Theorem 16) is that the linear form should not vanish. We will resolve this issue with the following lemma:
Lemma 19
Assume that the Diophantine Eq. (3) has three solutions , , with . Then we have
Proof
Assume that for some . But implies which results in view of (3) in
Since is irreducible, and are Galois conjugates. Therefore, by applying the non-trivial automorphism of to the equation , we obtain since . But this implies , hence , a contradiction to our assumptions.
Now, let us assume that
This implies which results in view of (3) in . But then or is a root of unity. Both cases contradict our assumption that is irreducible and .
Finally, we want to record three further elementary lemmas that will be helpful. The first lemma is a standard fact from real analysis.
Lemma 20
If , then we have and
Proof
A direct application of Taylor’s theorem with a Cauchy and Lagrange remainder, respectively.
Next, we want to state another estimate from real analysis:
Lemma 21
Let and such that and . Then we have
Proof
Since the case is trivial, we may assume . From the mean value theorem, we then get
The third lemma is due to Pethő and de Weger [8].
Lemma 22
Let , and be the largest solution of . If , then
and if , then
A proof of this lemma can be found in [11, Appendix B].
A lower bound for |c| in terms of and
The purpose of this section is to prove a lower bound for |c|. In particular, we prove the following proposition:
Proposition 23
Assume that assumption A1 or A2 holds and assume that Diophantine Eq. (3) has two solutions and with . Then we have
with .
Proof
Let us first take a look at the case
Here we get immediately
and are done. Now we can assume
We consider equation
and obtain
which yields
The goal is to apply Matveev’s theorem (Theorem 16) with . Note that with and we choose and , in view of Remark 4, and obtain
Due to Lemma 13 we have
Therefore we obtain by Corollary 17 that
which implies the content of the proposition.
Bounds for in terms of
In this section, we will assume that assumption A1 or A2 holds. However, in the proofs we will mainly consider the case that assumption A1 holds. Note that this is not a real restriction since assumption A2 implies assumption A1 with and instead of (cf. Remark 2). Also assume that Diophantine Eq. (3) has three solutions , , with . In this section we follow the approach of Chim et al. [3] and prove upper bounds for in terms of . To obtain explicit bounds and to keep track of the dependence on and of the bounds we repeat their proof. This section also delivers the set up for the later sections which provide proofs of our main theorems. Moreover, note that the assumption that three solutions exist, simplifies the proof of Chim et al. [3].
The main result of this section is the following statement:
Proposition 24
Assume that assumption A1 or A2 holds and that Diophantine Eq. (3) has three solutions , , with . Then we have
where we choose in case that assumption A2 holds.
Since we assume the existence of two solutions, we have
and therefore obtain
7 |
Let us write . Note that we have , by Lemma 7 and Lemma 5. With this notation we get the inequality
Note that, depending on whether or , we have
Therefore, using Lemma 13, we obtain
8 |
First, let us assume that the maximum in (8) is . Under our assumptions, we have and , which implies . Thus taking logarithms and applying Lemma 20 yields
We apply Matveev’s theorem (Theorem 16) with . Note that with and we choose and , in view of Remark 4, to obtain
Note that, due to Lemma 13,
Therefore we obtain from Corollary 17 that
where for the last inequality we used . Thus we have
which, using Lemma 22, proves Proposition 24 in this case. Note that this also proves, in this specific case, Theorem 1.
Now we assume that the maximum in (8) is . By our assumptions on A we have and obtain, by Lemma 20,
As computed before, an application of Matveev’s theorem yields
and therefore
9 |
For the rest of the proof of Proposition 24, we will assume that (9) holds. Since we assume that a third solution exists, the statement of Lemma 13 also holds for and instead of and . In particular we have
which yields
10 |
Let us rewrite Eq. (7) again to obtain the inequality
As previously noted we have and therefore we obtain
![]() |
We aim to apply Matveev’s theorem to with . Note that, due to Remark 4 and the properties of heights, we obtain
Thus we obtain as before, and from Matveev’s theorem
which yields
11 |
If we put and apply Lemma 22 to (11), then we end up with
which yields the content of Proposition 24.
Combining linear forms of logarithms
As done before, let us assume that Diophantine Eq. (3) has three solutions , , with . Again we assume that assumption A1 or A2 holds.
Let us reconsider inequality (8) with replaced by , respectively. Then we obtain
12 |
Let us assume for the next paragraphs that
13 |
Then, by applying Lemma 20 to (8) and (12), we obtain the system of inequalities
We eliminate the term from these inequalities by considering and obtain the inequality
14 |
Let us write M for the maximum on the right hand side of (14). If , we obtain the inequality . Since we will study the case that in Sect. 8, we will assume for the rest of this section that , i.e. we have . Therefore we have to consider five different cases. In each case we want to find an upper bound for if possible:
- The case In this case we get
which yields
since we assume . - The case In this case we obtain
which yields
To obtain from this inequality a bound for is not straight forward and we will deal with this case in Sect. 9. - The case This case implies
and we obtain - The case By a similar computation as in the first case, we obtain in this case the inequality
- The case Almost the same computations as in the case that lead to
In the case that (13) does not hold, i.e. that , we obtain by similar computations in each of the five possibilities the following inequalities:
The case ;
The case ;
The case ;
The case ;
The case .
Let us recap what we have proven so far in the following lemma:
Lemma 25
Assume that assumption A1 or A2 holds and assume that Diophantine Eq. (3) has three solutions , , with . Then one of the following three possibilities holds:
-
(i)
;
-
(ii)
;
-
(iii)
Since we will deal with the first and second possiblity in the next sections, we close this section by proving that the last possibility implies Theorems 1 and 2. Therefore let us plug in the upper bound for into inequality (11) to obtain
Writing , this inequality turns into
and an application of Lemma 22 implies
Thus Theorem 1 is proven in this case.
Now let us assume that assumption A2 holds. By Remark 2 we get the bound
If we insert our upper bound for into the upper bound for , we obtain
This proves Theorem 2 in the current case.
The case
We distinguish between the cases and .
The case – bound for
In this case we have
15 |
Furthermore it holds
Since
holds under our assumptions, we may apply Lemma 20 to get the two inequalities
Multiplying the first inequality by and the second one by as well as forming the difference afterwards yields
16 |
Let us note that , and therefore we obtain
Together with the estimate
this implies
17 |
Let us assume for the moment that the maximum is . Then we obtain
which implies and thus Theorem 2. Plugging in in Proposition 24 yields the content of Theorem 1 in this case.
Therefore we assume now . By Proposition 23 we obtain
which yields
Note that, due to our assumptions, we have the bound
which implies . Thus we get
and
As previously, solving this inequality with the help of Lemma 22 yields
18 |
which proves Theorem 1 in this case.
So we may now assume that assumption A2 holds and the bound for with is valid. This yields
19 |
The case bound for
For it remains to prove the bound for stated in our second theorem. We can already use the above proven bound (19) for since we assume assumption A2. Note that under assumption A2 we have , by Lemma 8. Hence the quantities are bounded by effectively computable constants depending only on and b.
Let us consider the case . Note that by the usual Galois conjugation argument. If , then (16) gives us
which yields
As (see inequality (15)) we obtain
which yields .
If , then (16) gives us
which implies
and hence .
Thus we may now assume . Let us note that under the assumption we can deduce
20 |
and otherwise we would get the constant in Theorem 2 (cf. the calculations below). Then, by Lemma 20, we get
Recalling from the beginning of Sect. 7 the bound
we can again eliminate the term from these inequalities by considering the form and obtain the inequality
21 |
If , then we have
which yields , and together with the bound (19) this gives us constant in Theorem 2.
Hence we can assume and replace in the discussion above by as well as by . Since by (20) we have
we may apply Lemma 20 also to this expression and get an analogous version of (16) with replaced by . The consideration of yields in the case , in the same way as above, , and in the case we apply the analogous version of (16), as done above, with the conclusion
a contradiction stating that this case is not possible. For this reason we may now assume . Thus altogether we obtain
22 |
From (22) we deduce that one of the two factors of the left hand side is smaller than . By thinking of constant in Theorem 2, we may assume . So we have . Let us first assume that
This implies and further
Therefore we have . But Lemma 15 states that there cannot be three solutions for .
Now we may assume
Here we get the further bound
Assuming , we obtain by an application of Lemma 21 that
This together with our assumption gives us
provided that
Thus we may assume provided that is large enough. But this also implies
If , we get
Therefore we may assume is positive. Since for any real numbers and we have , we obtain from Lemma 6 together with Lemma 5
Hence we get
in this case.
So it remains to consider the situation
With (19) and Lemma 5 this implies
and Theorem 2 is proven in that case.
The case
The case can be treated with similar arguments. Therefore we will only point out the differences.
We start with the observation
and write again
Note that, using Lemma 13,
holds under our assumptions if in addition . The case is included in the next section. Thus we get again the inequality chain (16) and furthermore the bound
If the maximum is , then we have
which implies . This will be handled in Sect. 9. Therefore we may now again assume . In the same way as in the case we obtain again the upper bound (18) proving Theorem 1 also in the case . Moreover, we get under assumption A2 the same upper bound (19) for . In particular, the quantities are bounded by effectively computable constants depending only on and b.
The reader might already have noticed that, in the case , we sometimes have written |c| and sometimes c. We did this in order to reuse these calculations now for the case . The only adaptions we need for when going through the previous subsection are the following: First, the special case
now, by Lemma 13, yields
and thus . Second, we have to consider instead of , which implies
If , we get
and if is negative, we obtain from Lemma 6 together with Lemma 5
and again
The other steps work as above. Hence Theorem 2 is proven in this case as well.
Let us summarize what we have proven so far:
Lemma 26
Assume that assumption A1 or A2 holds and assume that Diophantine Eq. (3) has three solutions , , with . Then at least one of the following three possibilities holds:
- assumption A1 holds and
- assumption A2 holds and
or .
The case
In view of Theorems 1 and 2 and Lemma 26 we may assume that assumption A1 or A2 holds and that .
First, we reconsider inequality (8) and note that implies either or . In the first cases we have an upper bound for and by Proposition 24 also an upper bound for ; the second case contradicts assumption A1 and A2 respectively. Thus Theorems 1 and 2 are shown in those situations. Now we may apply Lemma 20 and obtain, as in Sect. 6, the inequality
23 |
Next, let us consider the inequality
In particular, note that by Lemma 13. Assuming that yields which implies by Proposition 24 Theorems 1 and 2. Similarly, using , the assumption gives us either or and we are done as well. Thus we may apply Lemma 20 and obtain
24 |
Eliminating the term from the linear forms and by considering yields together with (23) and (24) the bound
25 |
where
Now we have to distinguish between the cases and .
The case
Since this case can only occur if b and are multiplicatively dependent. This is only possible if and , i.e. if . Note that our assumptions imply if . Therefore we obtain
which implies .
From Lemma 13 we know that . Hence, using the facts and , we get the inequality
which implies and thus . An application of Proposition 24 yields Theorems 1 and 2.
Remark 5
Let us note that in the case the argument above does not work. This is the reason why we exclude if .
The case
Here we may apply Matveev’s theorem, Theorem 16, to . Note that the case can be excluded by the previous subsection.
First, let us find an upper bound for . We deduce from (25) the bound
which implies
Furthermore, using Lemma 18, we have
Therefore we may choose in Theorem 16.
Now we obtain by Matveev’s theorem
and then
Together with the upper bound for this yields
and thus
26 |
Similar as in Sect. 7 we plug in this upper bound for into (11) and obtain the inequality
Writing and applying Lemma 22 gives us an upper bound for and in the sequel for , namely
This concludes the proof of Theorem 1.
Now let us assume that assumption A2 holds. Then we put and get
in particular for (cf. Corollary 3). If we insert this upper bound into (26) with , then we obtain
which finally proves Theorem 2.
Acknowledgements
We want to thank the anonymous referees for carefully reading our manuscript and their many helpful suggestions which improved the quality of the paper significantly. This research was funded in whole or in part by the Austrian Science Fund (FWF) [10.55776/I4406]. For open access purposes, the author has applied a CC BY public copyright license to any author accepted manuscript version arising from this submission.
Funding
Open Access funding provided by the Paris Lodron University of Salzburg.
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Footnotes
This prime p is called a primitive divisor.
This work was supported by the Austrian Science Fund (FWF) under the project I4406
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