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. 2024 May 13;10(2):51. doi: 10.1007/s40993-024-00534-5

On Pillai’s Problem involving Lucas sequences of the second kind

Sebastian Heintze 1, Volker Ziegler 2,
PMCID: PMC11090840  PMID: 38751821

Abstract

In this paper, we consider the Diophantine equation Vn-bm=c for given integers bc with b2, whereas Vn 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 (nm) , 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 Vn.

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 c0(a,b), depending on a and b, such that for any c>c0(a,b) the equation

an-bm=c 1

has at most one solution (n,m)Z>02.

Replacing the powers an and bm 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

Fn-2m=c

has at most one solution (n,m)Z>02 provided that

c{0,1,-1,-3,5,-11,-30,85}.

More generally, Chim, Pink and Ziegler [3] proved that for two fixed linear recurrence sequences (Un)nN, (Vn)nN (with some restrictions) the equation

Un-Vm=c

has at most one solution (n,m)Z>02 for all cZ, except if c is in a finite and effectively computable set CZ that depends on (Un)nN and (Vn)nN.

In more recent years, several attempts were made to obtain uniform results, i.e. to allow to vary the recurrence sequences (Un)nN and (Vn)nN in the result of Chim, Pink and Ziegler [3]. In particular, Batte, Ddamulira, Kasozi and Luca [1] showed that for all pairs (p,c)P×Z with p a prime, the Diophantine equation

Fn-pm=c

has at most four solutions (n,m)N2 with n2. This result was generalized by Heintze, Tichy, Vukusic and Ziegler [5]. They proved that for a given linear recurrence sequence (Un)nN (with irrational simple dominant root α>1 having a positive Binet coefficient a, such that α and a are multiplicatively independent and such that the equation αz-1=axαy has no solutions with zN, x,yQ and -1<x<0), there exist an effectively computable bound B2 such that for an integer b>B the Diophantine equation

Un-bm=c 2

has at most two solutions (n,m)N2 with nN0. Here N0 is an effectively computable constant depending only on (Un)nN.

In this paper, we want to fix b in (2) and let (Un)nN vary over a given family of recurrence sequences. In particular, we consider the case where (Un)nN 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

Vn(A,B)=Vn=αn+βn,

where α and β are the roots of the quadratic polynomial

X2-AX+B

with A24B and gcd(A,B)=1. In the following, we will assume that Vn is non-degenerate, i.e. that α/β is not a root of unity. More precisely, we assume A2-4B>0 and A0, since this implies that α and β are distinct real numbers with |α||β|. We will also assume that A>0, which results in Vn>0 for all nN. Then we consider the Diophantine equation

Vn-bm=c, 3

where b,cZ with b>1 are fixed.

Theorem 1

Let 0<ϵ<1 be a fixed real number and AB coprime integers satisfying A2-4B>0. Assume that |B|<A2-ϵ as well as that the polynomial X2-AX+B is irreducible and A321/ϵ. Furthermore, assume that b2 if c0 and b3 if c<0. Assume that Eq. (3) has three solutions (n1,m1),(n2,m2),(n3,m3)N2 with n1>n2>n3N0(ϵ) for the bound N0(ϵ)=32ϵ. Then there exists an effectively computable constant C1=C1(ϵ,b), depending only on ϵ and b, such that n1<C1. In particular, we can choose

C1=4.83·1032(logb)4ϵ2log5.56·1036(logb)4ϵ24.

Let us note that the bound N0=N0(ϵ) in Theorem 1 ensures that Vn is strictly increasing for nN0 (see Lemma 9 below). Let us also mention that it is essential to exclude the case that Vn2=Vn3.

Although we can bound n1 in terms of b and ϵ, 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 κ1 be a fixed real number and AB coprime integers satisfying A2-4B>0. Assume that |B|<κA as well as that the polynomial X2-AX+B is irreducible. Furthermore, assume that b2 if c0 and b3 if c<0. Assume that Eq. (3) has three solutions (n1,m1),(n2,m2),(n3,m3)N2 with n1>n2>n31. Then there exist effectively computable constants C2=C2(κ,b), depending only on κ and b, and C3=C3(b), depending only on b, such that logA<max{C2,C3}. We can choose

C2=4.35·1010log(4κ)logblog5.98·1011logb

and

C3=9.41·109log1.4·1036(logb)4log2.23·1037(logb)44logb2.

A straight forward application of our bounds yields:

Corollary 3

Assume that |B|<A. If the Diophantine equation

Vn-2m=c 4

with c0 has three solutions (n1,m1),(n2,m2),(n3,m3)N2 with n1>n2>n31, then either

  • A<1024 or

  • n1<2.3·1040 and logA<4.48·1013.

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 |B|<κA. The following corollary gives a precise statement.

Corollary 4

Let κ1 be a fixed real number and b2 a fixed integer. Then there exist at most finitely many, effectively computable 9-tuples (A,B,c,n1,m1,n2,m2, n3,m3)Z9 such that

  • A and B are coprime;

  • Amax{κ2,1024}, |B|<κA and A2-4B>0;

  • X2-AX+B is irreducible;

  • if b=2, then c0;

  • n1>n2>n31 and m1,m2,m31;

  • Vni(A,B)-bmi=c for i=1,2,3.

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 Vn. Chim, Pink and Ziegler [3] proved that, for two in absolute values strictly increasing linear recurrences (Xn)nN0 and (Ym)mM0 with multiplicatively independent dominant roots, there exists only finitely many, effectively computable integers c such that Xn-Ym=c has more than one solution. First, we have to ensure that α and b are multiplicatively independent. However, α and b are multiplicatively dependent if and only if there exist integers xy not both zero such that αx=by. But αx cannot be a rational number, unless x=0, since by our earlier assumptions A>0 and A2-4B>0 is not a perfect square. Second, we note that Vn is strictly increasing for n1 by Lemma 10 since Amax{κ2,1024} implies A4κ+4. 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 ϵ=12 to bound the ni. Clearly, the mi 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 Vn under the restrictions that |B|<A2-ϵ and |B|<κA, 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 n1, which still depend on logα, 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 logα. Combining these inequalities we obtain a linear form in logarithms which does no longer contain logα. Thus we obtain that Theorems 1 and 2 hold or one of the following two cases occurs:

  • n1m2-n2m1=0;

  • m2-m3logn1.

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 α and β be the roots of X2-AX+B. By our assumptions, α and β are distinct real numbers Q. Throughout the rest of the paper, we will assume that α is the larger one, i.e. |α|>|β|, which we can do since, by our assumptions, we have A>0. Therefore we obtain

α=A+A2-4B2andβ=A-A2-4B2.

First, note that α>1 can be bounded in terms of A:

Lemma 5

Assuming that |B|<A2, we have A2<α<2A.

At some point in our proofs of Theorems 1 and 2, we will use that β cannot be to close to 1. In particular, the following lemma will be needed.

Lemma 6

Let A,B,α,β be as above, in particular X2-AX+B irreducible with real roots. Assume that A4. Then we have

1-|β|22A+5.

Proof

First, let us note that the function f(x)=A-A2-4x2 is strictly increasing for all x<A2/4. Moreover, we have f(A-1)=1 and f(-A-1)=-1. By our assumption that X2-AX+B is irreducible, a choice for B such that |β|=1 is not admissible. Therefore we have

1-|β|min{1-f(A-2),f(A)-1,f(-A)+1,-1-f(-A-2)}.

We compute

1-f(A-2)=-(A-2)+A2-4(A-2)2=42(A-2)+A2-4(A-2)2A-2+A2-2A+1=22A-3,

provided that A2-2A+1A2-4A+8, which certainly holds for A7/2. Similar computations yield

f(A)-122A-4,f(-A)+122A+4,-1-f(-A-2)22A+5,

provided that A4.

Next, we show that under our assumptions |β| is not to large.

Lemma 7

Assume that |B|<A2-ϵ and that Aϵ8. Then we have α|β|>12Aϵ.

Proof

First we note that

αβ=A+A2-4BA-A2-4B=2A2-4B+2AA2-4B4B2A2-4|B|+2AA2-4|B|4|B|=A22|B|-1+A22|B|1-4|B|A2.

From the observation

A22|B|1-4|B|A2>Aϵ21-4Aϵ22>1,

then it follows

α|β|>A22|B|>Aϵ2.

Lemma 8

Assume that |B|<κA and that A4κ. Then we have |β|<2κ.

Proof

We have

|β|=A-A2-4B22|B|A+A2-4|B|.

Let us now consider the function f(x)=2xA+A2-4x. 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

|β|<2κAA+A2-4κA=κ21+1-4κA2κ

since A4κ.

Now let us take a look at the recurrence sequence Vn.

Lemma 9

Assume that |B|<A2-ϵ and Aϵ8. Then Vn is strictly increasing for n32ϵ.

Proof

First, we note that

Vn+1-Vn=αn+1+βn+1-αn-βn=αn(α-1)+βn(β-1)>0

certainly holds if

αn(α-1)>|β|n(|β|+1),

i.e.

α|β|n>|β|+1α-1. 5

Note that α|β|>12Aϵ, by Lemma 7, and, by Lemma 5,

|β|=|B|α<2A2-ϵA=2A1-ϵ.

Moreover, note that the smallest possible value for α is 1+52. Therefore (5) is certainly fulfilled if

12Aϵn>6A1-ϵ>3A1-ϵ-1+522A1-ϵ+11+52-1.

Thus Vn is increasing if

nϵlogA-log2>log6+(1-ϵ)logA

or equivalently

n>(1-ϵ)logA+log6ϵlogA-log2.

Since the rational function f(x)=ax+bcx+d is strictly increasing if ad-bc>0, strictly decreasing if ad-bc<0, and constant if ad-bc=0, we obtain that for

n32ϵ>1-ϵϵlog8+log6log8-log2

the Lucas sequence is strictly increasing.

Lemma 10

Assume that |B|<κA and A4κ+4. Then Vn is strictly increasing for n0.

Proof

By (5) we know that Vn is increasing for all n0 if 1>|β|+1α-1. By Lemma 5, we have α>A22κ+2>2 and, using Lemma 8, also |β|<2κ. Therefore we get

|β|+1α-1<2κ+1A2-11.

Remark 1

Note that assuming |B|<κA and Aκ2 implies |B|<A3/2=A2-ϵ with ϵ=1/2. Therefore all results that are proven under the assumption |B|<A2-ϵ with ϵ=1/2 also hold under the assumption |B|<κA and Aκ2.

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:

  1. |B|<A2-ϵ, Aϵ32 and N0=32ϵ,

  2. |B|<κA, Amax{κ2,16κ+12,1024} and N0=1.

Remark 2

Let us note that the bound Aκ2 and A1024 in assumption A2 results in the useful fact that assumption A2 implies assumption A1 with ϵ=12, but with N0=1 instead of 32ϵ. The assumption A16κ+12 is mainly used in the proofs of the Lemmas 12 and 13. Moreover, assumptions A1 and A2 both imply |B|<A2.

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 (n1,m1) and (n2,m2) with n1>n2N0. Then m1>m2.

Lemma 12

Let assumption A1 or A2 be in force and nN0. Then we have 54αn>Vn>34αn.

Proof

Assume that A1 holds. Then, by Lemma 7, we have

|β|αn<2Aϵn<14.

If A2 holds, by Lemma 8, we have

|β|αn<4κAn<14.

Therefore in any case we get

34αn<αn1-|β|αnαn+βn=Vnαn1+|β|αn<54αn.

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 (n1,m1) and (n2,m2) with n1>n2N0. Then we have

52>bm1αn1>38.

Proof

Assuming the existence of two different solutions implies by an application of Lemma 12 the inequality

54αn1>Vn1>Vn1-Vn2=bm1-bm2=bm11-1bm1-m212bm1,

which proves the first inequality.

For the second inequality, we apply again Lemma 12 to obtain

Vn1-Vn2>34αn11-53αn1-n2.

Since we assume in any case that α>A2>4, we get Vn1-Vn2>38αn1 and thus

38αn1<Vn1-Vn2=bm1-bm2<bm1,

which yields the second inequality.

Finally, let us remind Carmichael’s theorem [2, Theorem XXIV]:

Lemma 14

Let AB be coprime integers with A>0 and A2-4B>0. For any n1,3 there exists a prime1p such that pVn, but pVm for all 1m<n, except for the case that n=6 and (A,B)=(1,-1).

This lemma can be used to prove the following result:

Lemma 15

Assume that c=0 and A>1. Then the Diophantine Eq. (3) has at most one solution (nm) with n1.

Proof

Assume that (3) has two solutions (n1,m1),(n2,m2) with n1>n21. Then by Carmichael’s primitive divisor theorem (Lemma 14) we deduce that n1=3 and n2=1. Since V1=A and V3=A3-3AB we obtain the system of equations

A=bm2,A3-3AB=bm1

which yields

b2m2-3B=bm1-m2.

That is b3B. Since we assume that gcd(A,B)=1 and A>1, we deduce b3 and b=3. We also conclude that 3B. By considering 3-adic valuations, this yields m1-m2=1 since m2=0 would imply A=1. Hence we have B=32m2-1-1.

Note that we also assume A2-4B>0 which implies that

32m2-4(32m2-1-1)=4-32m2-1>0

holds. But this is only possible for m2=1 and we conclude that A=3 and B=2. Since X2-3X+2=(X-2)(X-1) 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 η0 be an algebraic number of degree d and let

adX-η(1)X-η(d)Z[X]

be the minimal polynomial of η. Then the absolute logarithmic Weil height is defined by

h(η)=1dlog|ad|+i=1dmax{0,log|η(i)|}.

With this notation, the following result due to Matveev [7] holds:

Theorem 16

(Theorem 2.2 with r=1 in [7]) Denote by η1,,ηN algebraic numbers, neither 0 nor 1, by logη1,,logηN determinations of their logarithms, by D the degree over Q of the number field K=Q(η1,,ηN), and by b1,,bN rational integers with bN0. Furthermore let κ=1 if K is real and κ=2 otherwise. For all integers j with 1jN choose

Ajmax{Dh(ηj),|logηj|,0.16},

and set

E=max|bj|AjAN:1jN.

Assume that

Λ:=b1logη1++bNlogηN0.

Then

log|Λ|-C(N,κ)max{1,N/6}C0W0D2Ω

with

Ω=A1AN,C(N,κ)=16N!κeN(2N+1+2κ)(N+2)(4(N+1))N+112eNκ,C0=loge4.4N+7N5.5D2log(eD),W0=log(1.5eEDlog(eD)).

In our applications, we will be in the situation N{2,3} and K=Q(α)R, i.e. we have D=2 and κ=1. 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 Λ0. Then we have

log|Λ|-7.26·1010log(13.81E)ΩforN=3,log|Λ|-6.7·108log(13.81E)ΩforN=2. 6

Remark 3

Let us note that the form of E is essential in our proof to obtain an absolute bound for n1 in Theorem 1. Let us also note that in the case of N=2 one could use the results of Laurent [6] to obtain numerically better values but with an log(E)2 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 η,γQ¯ and lQ we have

h(η±γ)h(η)+h(γ)+log2,h(ηγ)h(η)+h(γ),h(ηl)=|l|h(η).

Remark 4

Note that for a positive integer b we have h(b)=logb and, with α and β from Sect. 3, we have

h(α)=12max{0,logα}+max{0,log|β|}}logα.

This together with the above mentioned properties yields for tZ>0 the bound

h(αt±1)tlogα+log2.

One other important aspect in applying Matveev’s result (Theorem 16) is that the linear form Λ should not vanish. We will resolve this issue with the following lemma:

Lemma 19

Assume that the Diophantine Eq. (3) has three solutions (n1,m1), (n2,m2), (n3,m3)N2 with n1>n2>n3>0. Then we have

Λi:=nilogα-milogb0,fori=1,2,3Λ:=n2logα-m2logb+logαn1-n2-1bm1-m2-10,

Proof

Assume that Λi=nilogα-milogb=0 for some i{1,2,3}. But Λi=0 implies αni-bmi=0 which results in view of (3) in

αni+βni-bmi=βni=c.

Since X2-AX+B is irreducible, α and β are Galois conjugates. Therefore, by applying the non-trivial automorphism of K=Q(α) to the equation βni=c, we obtain αni=c since cQ. But this implies βni=αni, hence |α|=|β|, a contradiction to our assumptions.

Now, let us assume that

Λ=n2logα-m2logb+logαn1-n2-1bm1-m2-1=0.

This implies αn1-αn2=bm1-bm2 which results in view of (3) in βn1=βn2. But then β=0 or β is a root of unity. Both cases contradict our assumption that X2-AX+B is irreducible and βR.

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 |x|12, then we have log(1+x)2|x| and

29x2x-log(1+x)2x2.

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 xR and nN such that |2nx|<12 and n1. Then we have

(1+x)n-1e1/4n|x|.

Proof

Since the case x=0 is trivial, we may assume x0. From the mean value theorem, we then get

(1+x)n-1|x|n(1+|x|)n-1e1/4n.

The third lemma is due to Pethő and de Weger [8].

Lemma 22

Let a,b0, h1 and xR be the largest solution of x=a+b(logx)h. If b>(e2/h)h, then

x<2ha1/h+b1/hlog(hhb)h,

and if b(e2/h)h, then

x2ha1/h+2e2h.

A proof of this lemma can be found in [11, Appendix B].

A lower bound for |c| in terms of n1 and α

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 (n1,m1) and (n2,m2) with n1>n2N0. Then we have

|c|>αn1-K0log(27.62n1)-|β|n1

with K0=2.69·109logb.

Proof

Let us first take a look at the case

|c|+|β|n1αn112.

Here we get immediately

|c|+|β|n112αn1>αn1-1

and are done. Now we can assume

|c|+|β|n1αn1<12.

We consider equation

αn1+βn1-bm1=c

and obtain

bm1αn1-1|c|+|β|n1αn1<12

which yields

m1logb-n1logα2|c|+|β|n1αn1.

The goal is to apply Matveev’s theorem (Theorem 16) with N=2. Note that with η1=b and η2=α we choose A1=2logb and A2=2logα, in view of Remark 4, and obtain

E=maxm1logblogα,n1.

Due to Lemma 13 we have

m1logblogα<n1+log(5/2)logα<2n1.

Therefore we obtain by Corollary 17 that

2.68·109logblogαlog(27.62n1)-logm1logb-n1logαn1logα-log(|c|+|β|n1)-log2

which implies the content of the proposition.

Bounds for n1 in terms of logα

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 ϵ=12 and N0=1 instead of N0=32ϵ (cf. Remark 2). Also assume that Diophantine Eq. (3) has three solutions (n1,m1), (n2,m2), (n3,m3) with n1>n2>n3N0. In this section we follow the approach of Chim et al. [3] and prove upper bounds for n1 in terms of α. To obtain explicit bounds and to keep track of the dependence on logb and logα 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 (n1,m1), (n2,m2), (n3,m3) with n1>n2>n3N0. Then we have

n1<2.58·1022logαϵ(logb)2log7.12·1023logαϵ(logb)22,

where we choose ϵ=1/2 in case that assumption A2 holds.

Since we assume the existence of two solutions, we have

Vn1-bm1=c=Vn2-bm2

and therefore obtain

αn1+βn1-αn2-βn2=bm1-bm2. 7

Let us write γ:=min{α,α/|β|}. Note that we have γ>12Aϵ>14αϵ, by Lemma 7 and Lemma 5. With this notation we get the inequality

bm1αn1-1αn2-n1+bm2αn1+|β|n1+|β|n2αn1.

Note that, depending on whether |β|>1 or |β|1, we have

|β|n1+|β|n2αn12αn12γ-n1if|β|12·|β|n1αn12γ-n1if|β|>1.

Therefore, using Lemma 13, we obtain

bm1αn1-1αn2-n1+bm2αn1+2γ-n12max72αn2-n1,2γ-n17maxαn2-n1,γ-n1. 8

First, let us assume that the maximum in (8) is γ-n1. Under our assumptions, we have Aϵ32 and n13, which implies 7γ-n1<12. Thus taking logarithms and applying Lemma 20 yields

|m1logb-n1logα=:Λ|14γ-n1.

We apply Matveev’s theorem (Theorem 16) with N=2. Note that with η1=b and η2=α we choose A1=2logb and A2=2logα, in view of Remark 4, to obtain

E=maxm1logblogα,n1.

Note that, due to Lemma 13,

m1logblogα<n1+log(5/2)logα<2n1.

Therefore we obtain from Corollary 17 that

2.68·109logblogαlog(27.62n1)-log|Λ|n1logγ-log14n1(ϵlogα-log4)-log14n1ϵ2logα-log14,

where for the last inequality we used Aϵ32. Thus we have

5.38·109logbϵlog(27.62n1)>n1,

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 αn2-n1. By our assumptions on A we have 7αn2-n1<12 and obtain, by Lemma 20,

|m1logb-n1logα=:Λ|14αn2-n1.

As computed before, an application of Matveev’s theorem yields

2.68·109logblogαlog(27.62n1)-log|Λ|(n1-n2)logα-log14

and therefore

2.69·109logblog(27.62n1)>n1-n2. 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 m2 and n2 instead of m1 and n1. In particular we have

m1<n1logαlogb+log52logbn1logαlogb+2log52m2>n2logαlogb+log38logbn2logαlogb+2log38

which yields

m1-m2<logαlogb(n1-n2)+log49<2.7·109logαlog(27.62n1). 10

Let us rewrite Eq. (7) again to obtain the inequality

bm2αn2·bm1-m2-1αn1-n2-1-1|β|n1+|β|n2αn1-αn24γ-n1.

As previously noted we have 4γ-n1<12 and therefore we obtain

graphic file with name 40993_2024_534_Equ184_HTML.gif

We aim to apply Matveev’s theorem to Λ with η3=bm1-m2-1αn1-n2-1. Note that, due to Remark 4 and the properties of heights, we obtain

max{Dh(η3),logη3,0.16}2(m1-m2+1)logb+2(n1-n2+1)logα1.1·1010logblogαlog(27.62n1)=:A3.

Thus we obtain E2n1 as before, and from Matveev’s theorem

3.2·1021(logblogαlog(27.62n1))2n1logγ-log8n1ϵ2logα-log8,

which yields

6.42·1021logαϵ(logblog(27.62n1))2>n1. 11

If we put n=27.62n1 and apply Lemma 22 to (11), then we end up with

n<7.12·1023logαϵ(logb)2log7.12·1023logαϵ(logb)22

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 (n1,m1), (n2,m2), (n3,m3) with n1>n2>n3N0. Again we assume that assumption A1 or A2 holds.

Let us reconsider inequality (8) with n1,m1,n2,m2 replaced by n2,m2,n3,m3, respectively. Then we obtain

bm2αn2-1αn3-n2+bm3αn2+2γ-n23maxαn3-n2,52bm3-m2,2γ-n2. 12

Let us assume for the next paragraphs that

M0:=max3αn3-n2,152bm3-m2,6γ-n2,7αn2-n1,4γ-n1<12. 13

Then, by applying Lemma 20 to (8) and (12), we obtain the system of inequalities

|m1logb-n1logα=:Λ1|max14αn2-n1,8γ-n1,|m2logb-n2logα=:Λ2|max6αn3-n2,15bm3-m2,12γ-n2.

We eliminate the term logα from these inequalities by considering Λ0=n2Λ1-n1Λ2 and obtain the inequality

|Λ0|=|(n2m1-n1m2)logb|max12n1αn3-n2,30n1bm3-m2,24n1γ-n2,28n2αn2-n1,16n2γ-n1. 14

Let us write M for the maximum on the right hand side of (14). If n2m1-n1m20, we obtain the inequality logbM. Since we will study the case that n2m1-n1m2=0 in Sect. 8, we will assume for the rest of this section that n2m1-n1m20, i.e. we have logbM. Therefore we have to consider five different cases. In each case we want to find an upper bound for logα if possible:

  • The case M=12n1αn3-n2: In this case we get
    loglogblog(12n1)-(n2-n3)logα
    which yields
    logαlog12n1/logb<log(17.4n1),
    since we assume b2.
  • The case M=30n1bm3-m2: In this case we obtain
    loglogblog(30n1)-(m2-m3)logb
    which yields
    m2-m3log(30n1/logb)logb<1.45log(43.3n1).
    To obtain from this inequality a bound for logα is not straight forward and we will deal with this case in Sect. 9.
  • The case M=24n1γ-n2: This case implies
    loglogblog(24n1)-n2logγlog(24n1)-n2ϵ2logα
    and we obtain
    logα2log24n1/logbϵ<2log(34.7n1)ϵ.
  • The case M=28n2αn2-n1: By a similar computation as in the first case, we obtain in this case the inequality
    logα<log(40.4n2)<log(40.4n1).
  • The case M=16n2γ-n1: Almost the same computations as in the case that M=24n1γ-n2 lead to
    logα<2log(23.1n1)ϵ.

In the case that (13) does not hold, i.e. that M01/2, we obtain by similar computations in each of the five possibilities the following inequalities:

  • The case M0=3αn3-n2: logαlog6;

  • The case M0=152bm3-m2: m2-m33;

  • The case M0=6γ-n2: logαlog144ϵ;

  • The case M0=7αn2-n1: logαlog14;

  • The case M0=4γ-n1: logαlog64ϵ.

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 (n1,m1), (n2,m2), (n3,m3) with n1>n2>n3N0. Then one of the following three possibilities holds:

  • (i)

    n2m1-n1m2=0;

  • (ii)

    m2-m3<1.45log(43.3n1);

  • (iii)

    logα<2log(34.7n1)ϵ.

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 logα into inequality (11) to obtain

1.29·1022logbϵ2(log(34.7n1))3>n1.

Writing n=34.7n1, this inequality turns into

4.48·1023logbϵ2(log(n))3>n

and an application of Lemma 22 implies

n1<1.04·1023logbϵ2log1.21·1025logbϵ23.

Thus Theorem 1 is proven in this case.

Now let us assume that assumption A2 holds. By Remark 2 we get the bound

n1<4.16·1023logb2log4.84·1025logb23.

If we insert our upper bound for n1 into the upper bound for logα, we obtain

logA<logα+log25log(34.7n1)5log1.45·1025logb2log4.84·1025logb23.

This proves Theorem 2 in the current case.

The case n1m2-n2m1=0

We distinguish between the cases c0 and c<0.

The case c0 – bound for n1

In this case we have

0c=Vn3-bm3<αn3+βn3<2αn3. 15

Furthermore it holds

bm1αn1=1+βn1-cαn1as well asbm2αn2=1+βn2-cαn2.

Since

βn2-cαn2|β|αn2+2αn3-n2<12

holds under our assumptions, we may apply Lemma 20 to get the two inequalities

29βn1-cαn12βn1-cαn1-m1logb-n1logα2βn1-cαn12,29βn2-cαn22βn2-cαn2-m2logb-n2logα2βn2-cαn22.

Multiplying the first inequality by n2 and the second one by n1 as well as forming the difference afterwards yields

2n29βn1-cαn12-2n1βn2-cαn22n2βn1-cαn1-n1βn2-cαn22n2βn1-cαn12-2n19βn2-cαn22. 16

Let us note that (a+b)24max{|a|2,|b|2}, and therefore we obtain

cn1αn2-n2αn1n2|β|n1αn1+n1|β|n2αn2+8n2max|β|2n1α2n1,c2α2n110n1max|β|n2αn2,c2α2n2.

Together with the estimate

n1αn2-n2αn1>n1αn21-1α>78·n1αn2

this implies

c<12max|β|n2,c2αn2. 17

Let us assume for the moment that the maximum is c2αn2. Then we obtain

αn2<12c<24αn3

which implies α<24 and thus Theorem 2. Plugging in α<24 in Proposition 24 yields the content of Theorem 1 in this case.

Therefore we assume now c<12|β|n2. By Proposition 23 we obtain

αn1-K0log(27.62n1)-|β|n1<|c|<12|β|n2

which yields

n1-K0log(27.62n1)logα<log13+n1max{log|β|,0}.

Note that, due to our assumptions, we have the bound

α1-ϵ48α1-ϵ>4A1-ϵ>2αAϵ>|β|

which implies (1-ϵ4)logα>log|β|. Thus we get

n1ϵ4logα<K0logαlog(27.62n1)+log13

and

n1<1.08·1010logbϵlog(27.62n1).

As previously, solving this inequality with the help of Lemma 22 yields

n1<2.17·1010logbϵlog2.99·1011logbϵ 18

which proves Theorem 1 in this case.

So we may now assume that assumption A2 holds and the bound for n1 with ϵ=12 is valid. This yields

n1<4.34·1010logblog5.98·1011logb. 19

The case c0- bound for logA

For c0 it remains to prove the bound for logA stated in our second theorem. We can already use the above proven bound (19) for n1 since we assume assumption A2. Note that under assumption A2 we have |β|<2κ, by Lemma 8. Hence the quantities |c|,|β|n1,|β|n2 are bounded by effectively computable constants depending only on κ and b.

Let us consider the case |c-βn2|2(|c|+|β|n1)αn2-n1. Note that βn2c by the usual Galois conjugation argument. If c>βn2, then (16) gives us

(2n1-n2)|c|+|β|n1αn1n2βn1-cαn1-n1βn2-cαn22n2βn1-cαn12-2n19βn2-cαn222n1|β|n1+|c|αn12

which yields

αn12(|β|n1+|c|).

As c<2αn3 (see inequality (15)) we obtain

0.2αn1<αn1-2c2|β|n1<2(2κ)n1

which yields α<20κ.

If c<βn2, then (16) gives us

n1-n22βn2-cαn2n1βn2-cαn2-n2βn1-cαn12n1βn2-cαn22-2n29βn1-cαn122n1βn2-cαn22

which implies

14·βn2-cαn24·|β|n2+|c|αn252·|β|n2αn252·|β|α<104κα-1

and hence α<104κ.

Thus we may now assume |c-βn2|<2(|c|+|β|n1)αn2-n1. Let us note that under the assumption α>4(|c|+max{1,|β|}n1) we can deduce

bm3αn3-1|c|+|β|n3αn3<14, 20

and otherwise we would get the constant C2 in Theorem 2 (cf. the calculations below). Then, by Lemma 20, we get

|m3logb-n3logα=:Λ3|2|c|+2|β|n3αn324|β|n2+2|β|n3αn326(2κ)n1αn3.

Recalling from the beginning of Sect. 7 the bound

|m1logb-n1logα=:Λ1|max14αn2-n1,8γ-n1,

we can again eliminate the term logα from these inequalities by considering the form Λ0=n3Λ1-n1Λ3 and obtain the inequality

|Λ0|=|(n3m1-n1m3)logb|max52n1(2κ)n1α-n3,28n3αn2-n1,16n3γ-n1max52n1(2κ)n1α-n3,28n3αn2-n1,16n3α-n1/4. 21

If n3m1-n1m30, then we have

logbmax52n1(2κ)n1α-n3,28n3αn2-n1,16n3α-n1/4

which yields logα5+logn1+n1log(4κ), and together with the bound (19) this gives us constant C2 in Theorem 2.

Hence we can assume n3m1-n1m3=0 and replace in the discussion above m2 by m3 as well as n2 by n3. Since by (20) we have

βn3-cαn3=bm3αn3-1<14,

we may apply Lemma 20 also to this expression and get an analogous version of (16) with n2 replaced by n3. The consideration of |c-βn3|2(|c|+|β|n1)αn3-n1 yields in the case c>βn3, in the same way as above, α<20κ, and in the case c<βn3 we apply the analogous version of (16), as done above, with the conclusion

14·βn3-cαn34·βn3-cαn3<1,

a contradiction stating that this case is not possible. For this reason we may now assume |c-βn3|<2(|c|+|β|n1)αn3-n1. Thus altogether we obtain

|β|n3βn2-n3-1=βn2-βn3<4(|c|+|β|n1)αn2-n1. 22

From (22) we deduce that one of the two factors of the left hand side is smaller than 2αn2-n12|c|+|β|n1. By thinking of constant C2 in Theorem 2, we may assume α>64(|c|+|β|n1). So we have 2αn2-n12|c|+|β|n1<14. Let us first assume that

|β|n3<2αn2-n12|c|+|β|n1.

This implies |β|n3<14 and further

|c|<|β|n3+2(|c|+|β|n1)αn3-n1<14+18<1.

Therefore we have c=0. But Lemma 15 states that there cannot be three solutions for c=0.

Now we may assume

βn2-n3-1<2αn2-n12|c|+|β|n1.

Here we get the further bound

||β|-1||β|n2-n3-1βn2-n3-1<2αn2-n12|c|+|β|n1.

Assuming α>64n22(|c|+|β|n1), we obtain by an application of Lemma 21 that

||β|n2-1|2.6n2αn2-n12|c|+|β|n1.

This together with our assumption ||c|-|β|n2||c-βn2|<2(|c|+|β|n1)αn2-n1 gives us

||c|-1|<2.6n2αn2-n12|c|+|β|n1+2(|c|+|β|n1)αn2-n1<12

provided that

α109n22(|c|+|β|n1).

Thus we may assume c=1 provided that α is large enough. But this also implies

|1-βn3|<2(1+|β|n1)αn3-n12(1+|β|n1)α-2.

If βn3<0, we get

α<2(1+|β|n1)2(|c|+|β|n1).

Therefore we may assume βn3=|β|n3 is positive. Since for any real numbers x>0 and n1 we have |1-x||1-xn|, we obtain from Lemma 6 together with Lemma 5

24α+5<22A+5|1-|β|||1-βn3|<2(|c|+|β|n1)α-2.

Hence we get

α<9(|c|+|β|n1)

in this case.

So it remains to consider the situation

α<109n22(|c|+|β|n1)<1417n12(2κ)n1.

With (19) and Lemma 5 this implies

logA<n1log(4κ)+2logn1+log2834<4.35·1010log(4κ)logblog5.98·1011logb

and Theorem 2 is proven in that case.

The case c<0

The case c<0 can be treated with similar arguments. Therefore we will only point out the differences.

We start with the observation

0<|c|=-c=bm3-Vn3<bm3

and write again

bm1αn1=1+βn1-cαn1as well asbm2αn2=1+βn2-cαn2.

Note that, using Lemma 13,

βn2-cαn2|β|αn2+|c|αn2<116+bm3αn2<116+52bm3-m2<12

holds under our assumptions if in addition m2-m32. The case m2-m3=1 is included in the next section. Thus we get again the inequality chain (16) and furthermore the bound

|c|<12max|β|n2,|c|2αn2.

If the maximum is |c|2αn2, then we have

25bm2<αn2<12|c|<12bm3

which implies m2-m33. This will be handled in Sect. 9. Therefore we may now again assume |c|<12|β|n2. In the same way as in the case c0 we obtain again the upper bound (18) proving Theorem 1 also in the case c<0. Moreover, we get under assumption A2 the same upper bound (19) for n1. In particular, the quantities |c|,|β|n1,|β|n2 are bounded by effectively computable constants depending only on κ and b.

The reader might already have noticed that, in the case c0, we sometimes have written |c| and sometimes c. We did this in order to reuse these calculations now for the case c<0. The only adaptions we need for c<0 when going through the previous subsection are the following: First, the special case

αn12(|β|n1+|c|)

now, by Lemma 13, yields

115αn1<845bm1<αn1-2bm3<αn1-2|c|2|β|n1<2(2κ)n1

and thus α<60κ. Second, we have to consider c=-1 instead of c=1, which implies

|1+βn3|<2(1+|β|n1)αn3-n12(1+|β|n1)α-2.

If βn3>0, we get

α<2(1+|β|n1)2(|c|+|β|n1),

and if βn3=-|β|n3 is negative, we obtain from Lemma 6 together with Lemma 5

24α+5<22A+5|1-|β|||1+βn3|<2(|c|+|β|n1)α-2

and again

α<9(|c|+|β|n1).

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 (n1,m1), (n2,m2), (n3,m3) with n1>n2>n3N0. Then at least one of the following three possibilities holds:

  1. assumption A1 holds and
    n1<1.04·1023logbϵ2log1.21·1025logbϵ23;
  2. assumption A2 holds and
    logA<4.35·1010log(4κ)logblog5.98·1011logb
    or
    logA<5log1.45·1025logb2log4.84·1025logb23;
  3. m2-m3<1.45log(43.3n1).

The case m2-m3logn1

In view of Theorems 1 and 2 and Lemma 26 we may assume that assumption A1 or A2 holds and that m2-m3<1.45log(43.3n1).

First, we reconsider inequality (8) and note that 7maxαn2-n1,γ-n112 implies either α14 or Aϵ28. In the first cases we have an upper bound for α and by Proposition 24 also an upper bound for n1; 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

|n1logα-m1logb=:Λ1|14maxαn2-n1,γ-n1. 23

Next, let us consider the inequality

αn2bm2-bm3-1=αn2bm3(bm2-m3-1)-1αn3+|β|n2+|β|n3bm2-bm3<6αn3-n2+12γ-n218max{αn3-n2,γ-n2}.

In particular, note that bm2-bm312bm2>316αn2 by Lemma 13. Assuming that 18αn3-n212 yields α36 which implies by Proposition 24 Theorems 1 and 2. Similarly, using n22, the assumption 18γ-n212 gives us either α6 or Aϵ12 and we are done as well. Thus we may apply Lemma 20 and obtain

|n2logα-m3logb-log(bm2-m3-1)=:Λ2|36max{αn3-n2,γ-n2}. 24

Eliminating the term logα from the linear forms Λ1 and Λ2 by considering Λ=n1Λ2-n2Λ1 yields together with (23) and (24) the bound

|Λ|36n1max{αn3-n2,γ-n2}+14n2max{αn2-n1,γ-n1}50n1max{αn3-n2,γ-n2,αn2-n1}200n1α-ϵ, 25

where

Λ=(m1n2-m3n1)logb-n1log(bm2-m3-1).

Now we have to distinguish between the cases Λ=0 and Λ0.

The case Λ=0

Since n10 this case can only occur if b and bm2-m3-1 are multiplicatively dependent. This is only possible if b=2 and m2-m3=1, i.e. if bm2-m3-1=1. Note that our assumptions imply c0 if b=2. Therefore we obtain

0c=Vn3-bm3=αn3+βn3-bm3

which implies bm32αn3.

From Lemma 13 we know that bm2>38αn2. Hence, using the facts b=2 and m2=m3+1, we get the inequality

38αn2<bm2=2bm34αn3

which implies αn2-n3<11 and thus α<11. An application of Proposition 24 yields Theorems 1 and 2.

Remark 5

Let us note that in the case c<0 the argument above does not work. This is the reason why we exclude b=2 if c<0.

The case Λ0

Here we may apply Matveev’s theorem, Theorem 16, to Λ. Note that the case bm2-m3-1=1 can be excluded by the previous subsection.

First, let us find an upper bound for |m1n2-m3n1|. We deduce from (25) the bound

|m1n2-m3n1|logbn1log(bm2-m3-1)+200n1n1(m2-m3)logb+200n1

which implies

|m1n2-m3n1|90n1log(43.3n1).

Furthermore, using Lemma 18, we have

maxDh(bm2-m3-1),log(bm2-m3-1),0.162(m2-m3+1)logb3.5log(43.3n1)logb.

Therefore we may choose E=52n1 in Theorem 16.

Now we obtain by Matveev’s theorem

4.69·109log(719n1)log(43.3n1)(logb)2-log|Λ|

and then

4.69·109(log(719n1)logb)2-log|Λ|.

Together with the upper bound for |Λ| this yields

4.69·109(log(719n1)logb)2ϵlogα-log(200n1)

and thus

logα<4.7·1091ϵ(log(719n1)logb)2. 26

Similar as in Sect. 7 we plug in this upper bound for logα into (11) and obtain the inequality

719n1<2.17·1034ϵ-2(logblog(719n1))4.

Writing n=719n1 and applying Lemma 22 gives us an upper bound for n and in the sequel for n1, namely

n1<4.83·1032(logb)4ϵ2log5.56·1036(logb)4ϵ24.

This concludes the proof of Theorem 1.

Now let us assume that assumption A2 holds. Then we put ϵ=12 and get

n1<1.94·1033(logb)4log2.23·1037(logb)44,

in particular n1<2.3·1040 for b=2 (cf. Corollary 3). If we insert this upper bound into (26) with ϵ=12, then we obtain

logα<9.4·109log1.4·1036(logb)4log2.23·1037(logb)44logb2

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.

Data availability

This manuscript has no associated data.

Footnotes

1

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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