Abstract
Let \(r\ge 1\) be an integer and \(\mathbf{U}:=(U_{n})_{n\ge 0} \) be the Lucas sequence given by \(U_0=0\), \(U_1=1, \) and \(U_{n+2}=rU_{n+1}+U_n\), for all \( n\ge 0 \). In this paper, we show that there are no positive integers \(r\ge 3,~x\ne 2,~n\ge 1\) such that \(U_n^x+U_{n+1}^x\) is a member of \(\mathbf{U}\).
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1 Introduction
Let \(r\ge 1\) be an integer and \(\mathbf{U}:=(U_{n})_{n\ge 0} \) be the Lucas sequence given by \( U_0=0, ~ U_1=1, \) and
for all \( n\ge 0\). When \(r=1\) \(\mathbf{U}\) coincides with the Fibonacci sequence while when \(r=2\) \(\mathbf{U}\) coincides with the Pell sequence. It is well known that
In particular, the identity (2) tells us that the sum of the squares of two consecutive terms of \(\mathbf{U}\) is also a term of \(\mathbf{U}\). When \(r=1\), we even have \(U_n+U_{n+1}=U_{n+2}\) for all \(n\ge 0\) since \(\mathbf{U}\) is the Fibonacci sequence. We thus consider the Diophantine equation
in nonnegative integers (n, m, x) which by Eq. (2) has the parametric solution \(m=2n+1\) when \(x=2\) for any \(r\ge 1\) and even the parametric solution \(m=n+2\) when \(x=1\) if \(r=1\). For \(r=1\) Luca and Oyono [8] proved that Eq. (3) has no positive integer solutions (n, m, x) with \(n\ge 2 \) and \(x\ge 3 \). Rihane et al. [11] studied Eq. (3) when \(r=2\) and proved that there is no positive integer solution (n, m, x) of it with \(x\ne 2\). In the same spirit, Gómez Ruiz and Luca [5] studied Eq. (3) with \( \mathbf{U} = \{F_n^{(k)}\}_{n\ge -(k-2)} \), which is the \(k-\)generalised Fibonacci sequence of recurrence
with the initial conditions \( F_{-(k-2)}^{(k)} =F_{-(k-2)}^{(k)}=\cdots =F_{0}^{(k)} =0\) and \(F_1^{(k)}=1\). When \(k=2\), this sequence coincides with the sequence of Fibonacci numbers. They proved that Eq. (3) has no positive integer solution (k, n, m, x) with \( k\ge 3 \), \(n\ge 2\), and \(x\ge 1\). Another related result involving the balancing numbers was studied by Rihane et al. in [10].
In this paper, we study Eq. (3) in nonnegative integers (r, n, m, x) treating r as an integer parameter. We may assume that \(r\ge 3\) since the cases \(r\in \{1,2\}\) have been treated already in [8, 11], respectively. The solution with \((n,m)=(0,1)\) (for any r and x) is obvious so we omit it and suppose that n is positive. Our main result is the following.
Theorem 1
There is no positive integer solution (r, n, m, x) of Diophantine equation (3) with \(r\ge 3\) and \(x\ne 2\).
2 Preliminary results
2.1 The Lucas sequence
Let
be the roots of the characteristic equation \( x^2-rx-1=0 \) of the Lucas sequence \(\mathbf{U}=(U_n)_{n\ge 0}\). We put \(\Delta =r^2+4=(\alpha -\beta )^2\) for the discriminant of the above quadratic equation. The Binet formula for the general term of \(\mathbf{U}\) is given by
One may prove by induction that the inequality
holds for all positive integers n. It is also easy to show that the inequality
holds for all \(n\ge 2\). Indeed, it follows from \(U_{n+1}=rU_n+U_{n-1}>rU_n\) for \(n\ge 2\). At one point of the argument we will need the companion Lucas sequence \(\mathbf{V}:=\{V_n\}_{n\ge 0}\) given by \(V_0=2,~V_1=r\), and \(V_{n+2}=rV_{n+1}+V_n\) for all \(n\ge 0\). Its Binet formula is
There are many relations between members of \(\mathbf{U}\) and \(\mathbf{V}\) such as
We record another one. All such identities follow easily from the Binet formulas (4) and (7) of \(\mathbf{U}\) and \(\mathbf{V}\), respectively.
Lemma 1
If n is odd then
Both sequences \(\mathbf{U}\) and \(\mathbf{V}\) can be extended to negative indices either by allowing n to be negative in the Binet formulas (4) and (7) or simply by using the recurrence relation to extend \(\mathbf{U}\) to negative indices via \(U_{-n}=-rU_{-(n-1)}+U_{-(n-2)}\) for all \(n\ge 1\). The same applies to \(\mathbf{V}\). All the above formulas given in Eqs. (8), (9), (10), (11), and (13) hold when the indices are arbitrary integers, not necessarily nonnegative.
The following lemma is useful. For further details we refer the reader to the book of Koshy [6].
Lemma 2
Let \( \{U_n(r)\}_{n\ge 0} \subseteq \mathbf{Z}[r]\) be the sequence of polynomials defined by \( U_0(r)=0 \), \( U_1(r)=1 \), and
Then,
Note that the summation range above is only over these \(k\in [0,n]\) which have different parity than n. We record the following easy but useful consequence of the formula (14).
Lemma 3
We have \(r\mid U_n\) if n is even and \(r\mid U_{n}-1\) if n is odd.
2.2 Logarithmic height
Let \(\eta \) be an algebraic number of degree d with minimal primitive polynomial over the integers
where the leading coefficient \(a_0\) is positive and the \(\eta ^{(i)}\)’s are the conjugates of \(\eta \). The logarithmic height of \(\eta \) is given by
In particular, if \(\eta =p/q\) is a rational number with \(\gcd (p,q)=1\) and \(q>0\), then \(h(\eta )=\log \max \{|p|, q\}\). The following are some of the properties of the logarithmic height function \(h(\cdot )\), which will be used in the next sections of this paper without reference:
2.3 Linear forms in logarithms and continued fractions
The following result on linear forms in three logarithms is due to Mignotte [9]. The result is more general (i.e. the conditions on the parameters involved are somewhat more general), but we will only quote it in the form that we need.
Theorem 2
Consider three algebraic numbers \( \gamma _1, \gamma _2 \), and \( \gamma _3 \), which are all real, greater than 1 and multiplicatively independent. Put
Let \( b_1, b_2, b_3 \) be coprime positive integers and consider
Put
Let \( A_1, A_2 \), and \( A_3 \) be real numbers such that
Put
Then, either
or one of the following conditions holds:
-
(i)
there exist two positive integers \( r_0 \) and \( s_0 \) such that
$$\begin{aligned} r_0b_2=s_0b_1 \end{aligned}$$with
$$\begin{aligned} r_0\le 5.61A_2 (\mathcal {D}\log \mathcal {D})^{\frac{1}{3}} \quad \text {and} \quad s_0\le 5.61A_1 (\mathcal {D}\log \mathcal {D})^{\frac{1}{3}}; \end{aligned}$$ -
(ii)
there exist integers \( r_1, s_1, t_1, \) and \( t_2 \), with \( r_1s_1\ne 0 \), such that
$$\begin{aligned} (t_1b_1+r_1b_3)s_1 = r_1b_2t_2, \quad \gcd (r_1, t_1)=\gcd (s_1, t_2) =1 \end{aligned}$$which also satisfy
$$\begin{aligned}&|r_1s_1|\le 5.61\delta A_3 (\mathcal {D}\log \mathcal {D})^{\frac{1}{3}}, \quad |s_1t_1|\le 5.61\delta A_1 (\mathcal {D}\log \mathcal {D})^{\frac{1}{3}}, \\&|r_1t_2| \le 5.61\delta A_2 (\mathcal {D}\log \mathcal {D})^{\frac{1}{3}}, \end{aligned}$$where
$$\begin{aligned} \delta := \gcd (r_1, s_1). \end{aligned}$$
Moreover, when \( t_1=0 \) we can take \( r_1=1 \), and when \( t_2=0 \) we can take \( s_1 =1\).
At some point we will need to treat linear forms in two logarithms of algebraic numbers. To set the stage, let \(\gamma _1\) and \(\gamma _2\) be real algebraic numbers which are positive and let \({\mathcal {D}}:=[{\mathbb {Q}}(\gamma _1,\gamma _2):{\mathbb {Q}}]\). Let \(b_1\), \(b_2\) be nonzero integers, let \(B_1\) and \(B_2\) be real numbers larger than 1 such that
and put
Let
The following result of Laurent et al. is Corollary 2 in [7].
Theorem 3
With the above notations assuming furthermore that \(\gamma _{1} \) and \(\gamma _{2}\) are multiplicatively independent we have
Note that the fact that \(\Gamma \ne 0\) is already guaranteed by the condition that \(\gamma _1\) and \(\gamma _2\) are multiplicatively independent together with the fact that \(b_1,~b_2\) are nonzero integers.
During the calculations we get upper bounds on our variables which are too large, thus we need to reduce them. To do so we use some results from the theory of continued fractions.
For the treatment of linear forms homogeneous in two integer variables we use a well-known classical result in the theory of Diophantine approximation due to Legendre.
Lemma 4
Let \(\tau \) be an irrational number, \( {p_0}/{q_0}, {p_1}/{q_1}, {p_2}/{q_2}, \ldots \) be the sequence of convergents of the continued fraction expansion of \( \tau \) and M be a positive integer. Let N be a nonnegative integer such that \( q_N> M \). Putting \( a(M):=\max \{a_{i}: i=0, 1, 2, \ldots , N\} \) the inequality
holds for all pairs (u, v) of positive integers with \( 0<v<M \). Furthermore, if
then \(u/v=p_k/q_k\) for some \(k\ge 0\).
For a nonhomogeneous linear form in two integer variables we use a slight variation of a result due to Dujella and Pethő (see [4, Lemma 5a]). For a real number X, we write \(\Vert X \Vert := \min \{|X-n|: n\in \mathbb {Z}\}\) for the distance from X to the nearest integer.
Lemma 5
Let M be a positive integer, p/q be a convergent of the continued fraction expansion of the irrational number \(\tau \) such that \(q>6M\), and \(A,B,\mu \) be some real numbers with \(A>0\) and \(B>1\). Furthermore, let \(\varepsilon : = \Vert \mu q \Vert -M\Vert \tau q\Vert \). If \( \varepsilon > 0 \), then there is no solution to the inequality
in positive integers u, v, and w with
3 Proof of Theorem 1
3.1 The cases \(n=1\) or \(x=1\)
We assume that \(n\ge 1\), as the solution with \( n=0 \) is trivial. Since \( U_{n+1}< U_{n+1}+U_n< U_{n+2} \), it follows that the Diophantine equation (3) has no solution with \(x=1\). Let us assume that \(n=1\). We then get that
In particular, \(U_m\equiv 1\pmod r\). Lemma 3 shows that \(m\equiv 1\pmod 2\) and now Lemma 1 shows that
We now recall the Primitive Divisor Theorem of Carmichael [3] (see [1] for the most general statement) for the sequence \(\mathbf{U}\). It states that if \(\ell >12\), there is a prime factor p of \(U_{\ell }\) which is primitive in the sense that \(p\not \mid U_{k}\) for any positive \(k<\ell \). So, assume \(m+\delta \ge 14\). Since \(m+\delta \) is even we have \(U_{m+\delta }=U_{(m+\delta )/2}V_{(m+\delta )/2}\) by the formula (8). Since \(U_{m+\delta }\) has a primitive prime factor p, the primitive prime p must be a divisor of \(V_{(m+\delta )/2}\), which in turn must divide \(r=U_2\), a contradiction. Thus, \(m+\delta \le 12\), therefore \(m\le 13\). It thus follows that
so, if \(x\ge 13\), then
which gives \(x\le 14\). Thus, \(m\le 13,~x\le 14\). For each choice of the pair (m, x) with the components in the above ranges, the equation \(U_m(r)-1-r^x=0\) is a polynomial equation in r. After a simple computer search, we found no other solutions to Eq. (18) apart from the solution \( (n,m,x) = (1,3,2) \) which has \(x=2\) so it is part of the parametric family of solutions indicated at Eq. (2).
So, from now on we assume that \(n\ge 2\) and \(x\ge 3\).
3.2 Calculations when \(n\in [2,100]\) and \(x\in [3,100]\)
Using the Eq. (3) and the inequality (5), we get
and
From the above inequalities, we get the following result which we record for future reference.
Lemma 6
The inequalities
hold for all \(x\ge 3\) and \(n\ge 2\).
We next consider Eq. (14) given in Lemma 2. We write Eq. (3) as
Assume first that n is even. Then Eq. (20) becomes
which is equivalent to
The above relation implies that
Similarly, when n is odd, one is led to the analogous divisibility relation
So, fixing \(n\in [2,100]\) and \(x\in [3,100]\), inequalities (19) give some range for m. For each (n, x, m), divisibility relations (22) and (23) (according to whether n is even or odd) give us some possibilities for \(r\ge 3\) and now one checks whether relation (3) holds for this candidate (n, x, m, r). A computer search with Mathematica in this range for n and x which ran for a few hours found no solutions. For the search we didn’t actually checked that formula (20) holds but we checked that Eq. (20) does not hold modulo T, where T is the product of the first 20 primes. The Mathematica function powermod allowed us to compute the powers of r modulo T arising from the binomial formula rather quickly.
From now on, we assume that \(n\ge 2,~x\ge 3\) and \(\max \{n,x\}> 100\).
3.3 A small linear form in three logs
We rewrite Eq. (3) as
Dividing both sides of the above equation by \( U_{n+1}^x \) and using the inequality (6), we obtain
Put
We observe that \( \Lambda =e^{\Gamma }-1 \), where \( \Lambda \) and \( \Gamma \) are given by (26). Since \( |\Lambda |\le 2/27 \), we have that \( e^{|\Gamma |}\le 27/25 \) and using the inequality (25) we obtain
We record the above inequality for future reference.
Lemma 7
With \(\Gamma \) given by formula (26), inequality (27) holds.
We want to apply Theorem 2 with the following data:
We need to check that \( \gamma _1 \), \( \gamma _2 \), and \( \gamma _3 \) are multiplicatively independent. This we do in the next subsection.
3.4 Checking that \(\gamma _1,~\gamma _2,~\gamma _3\) are multiplicatively independent
Well, assume they are not and let i, j, k be integers not all zero such that
Squaring and rearranging the above relation we get \(\gamma _2^{2j}=(\gamma _1^2)^{-i} \gamma _3^{-2k}\in {\mathbb {Q}}\). However, \(\gamma _2^{2j}\) is also a unit, so an algebraic integer whose reciprocal is also an algebraic integer, and it is also positive, so it must be 1. Thus, \(j=0\). It now follows that i and k are both nonzero (since if one of them is, so is the other one) and further \(\gamma _3=\gamma _1^{-i/k}\). In particular, all prime factors of \(U_{n+1}\) are prime factors of \(\Delta :=r^2+4\). But this is also contemplated by the Primitive Divisor Theorem of Carmichael since primes dividing \(\Delta \) are not considered primitive. In particular, \(U_{n+1}\) does not have primitive prime factors so \(n+1\le 12\). In fact, Theorem C in [1] together with Table 1 there show that either \(n+1\in \{2,3,4,6\}\) or \(n+1\in \{5,12\}\) but in this last case, the only such Lucas sequence \(\mathbf{U}\) for which either one of \(U_5\) or \(U_{12}\) does not have primitive prime factors is the sequence of Fibonacci numbers, which is not our case. Thus, \(n+1\in \{2,3,6\}\). Further, for each prime p let z(p) be the index of appearance of p in \(\mathbf{U}\) defined as the smallest positive integer k such that \(p\mid U_k\). This always exists for our sequence \(\mathbf{U}\) since \(\alpha \) is a quadratic unit. It has the additional property that if \(\ell \) is a positive integer then \(p\mid U_{\ell }\) if and only if \(z(p)\mid \ell \). It is also well-known and easy to prove that if \(p\mid \Delta \), then \(z(p)=p\). Since also \(z(p)\mid n+1\) and \(n+1\in \{2,3,6\}\), it follows that the only possibilities for p are \(p=2,3\). Hence, \(r^2+4=2^a 3^b\). However, 3 cannot divide \(r^2+4\) for any positive integer r (because \(-4\) is not a quasartic residue modulo 3), so \(b=0\) and \(2^a=r^2+4\). Thus, \(r=2r_0\) is even, \(a\ge 3\) and the equation simplifies to \(2^{a-2}=r_0^2+1\). Hence, \(r_0\) is odd, so \(r_0^2\equiv 1\pmod 8\), therefore \(2\Vert r_0^2+1\), which leads to \(a=3,~r_0=1\), which gives \(r=2\), is not our case. Hence, indeed \(\gamma _1,~\gamma _2,~\gamma _3\) are multiplicatively independent.
3.5 Applying Theorem 2
Since \( \gamma _1, \gamma _2, \gamma _3 \in \mathbb {Q}(\alpha ) \), we have \( \mathcal {D}=2 \). We also have
So, we bound the heights of \(\alpha \) and \({\sqrt{r^2+4}}\) in terms of \(\log (r+1)\). Since
and
we can take
Thus \(\Omega : = A_1A_2A_3>433 n (\log (r+1))^3 > 100\) since \(n\ge 2\) and \(r\ge 3\). Then,
In the above chain of inequalities, we used the fact that \(m<(n+1)x+2<2n x\) [see inequality (19)] since \(n\ge 2\) and \(x\ge 3\). Further, putting
we have that either the inequality
holds, or the other possibilities (i), (ii) from Theorem 2 hold. We treat (i) and (ii) later and deal with the above inequality (29) at this stage. If \(\log \mathcal {B}=5\), then
On the other hand, if \(\log \mathcal {B}>5\), then
where in the above inequality we used the fact that \(e^{0.882}\times 0.12<0.3\). Thus, we get that
Comparing this inequality with Eq. (27), we get that
which implies, via the inequality
that
Using the inequality (19), we know that \( m< (n+1)x+2 <(n+1)(x+1)\) (because \(n\ge 2\)), and substituting this in (33) we get that
We now turn our attention to the possibilities (i) and (ii). In case (i), there are positive integers \(r_0,s_0\), which may be assumed to be coprime, such that \(r_0b_2=s_0b_1\). So, we get \(r_0m=s_0\) and since \(r_0,s_0\) are coprime, we take \(r_0=1, s_0=m\), and we get
Since \(m>(n-1)x+1>x\), this situation gives
This was in situation (i). In situation (ii), we have integers \(r_1,s_1,t_1,t_2\) with \(r_1s_1\ne 0\) and
Thus, for us, we have
Reducing the above equation modulo \(r_1\) we get \(t_1s_1\equiv 0\pmod {r_1}\) and since \(\gcd (t_1,r_1)=1\), we get that \(r_1\mid s_1\). So, we put \(s_1=r_1s_1'\) and simplify both sides of the above equation by \(r_1\) to get
Consequently, for us \(\delta =\gcd (r_1,s_1)=r_1\). Hence,
Assume first that \(t_2=0\). Then
which is the same as (35). Assume next that \(t_2\ne 0\). We return to inequality (27) and multiply both sides by \(t_2\) and get
We substitute \(mt_2\) by \(t_1s_1'+(r_1s_1')x\) inside the left-hand side above and then the left-hand side above becomes
Inequality (37) is of the form
and
We check that \(\eta _1\) and \(\eta _2\) are multiplicatively independent. If not, there are integers i, j not both zero such that \(\eta _1^i\eta _2^j=1\). This gives
If \(i\ne 0\), this gives a multiplicative dependence among \(\gamma _1,\gamma _2,\gamma _3\) with the exponent of \(\gamma _1\) being the nonzero integer \(-t_2i\), a contradiction with the main result of Sect. 3.4. Thus, \(i=0\), so \(j\ne 0\), and we get again a multiplicative relation among \(\gamma _2,\gamma _3\) (the exponent of \(\gamma _3\) being the nonzero integer \(-t_2j\)), which is the same contradiction. Thus, indeed \(\eta _1\) and \(\eta _2\) are multiplicatively independent and they are also positive. So, we are in position to apply Theorem 3 to the left–hand side of inequality (38). We compute \(\log B_i\) for these choices. We have, by the properties (15),
Since \(|\log \gamma _i|/2\le h(\gamma _i)\) holds for \(i=1,2,3\), it follows, by the absolute value inequality, that the same inequalities are satisfied by the numbers \(|\log \eta _i|/2\) for \(i=1,2\). Thus, since \({\mathcal {D}}=2\), we can take
We bound
Hence, we take
Now Theorem 3 gives
where \(M:=\max \{\log b'+0.4,10.5\}\). In case \(M=10.5\), we get
which gives
Finally, suppose that
Comparing inequality (39) with inequality (38), we get
Since \(2.2|t_2|<88\log (r+1)\), we get that
The first summand in the right–hand side is \(<5\) for all \(r\ge 3\). Using inequality (32), we get that
To summarise, either we are in the first situation of Theorem 2 and \(\log \mathcal {B}=5\), in which case inequality (30) holds, or \(\log \mathcal {B}>5\) in which case inequality (34) holds, or we are in the exceptional case (i) for which inequality (35) holds, which is contained in inequality (30), or we are in the exceptional situation (ii) in which case either inequality (40) holds, or inequality (63) holds. Since inequality (30) is contained in inequality (40), it follows, using the inequality \(1/88<0.12\), that we proved the following result.
Lemma 8
One of the following inequalities holds:
3.6 More inequalities in terms of n and x
We put
Later we shall show that \(\kappa \) is positive except possibly if \(r=3\). In this section, assuming that it is positive, we show how it gives some lower bounds for x in terms of n.
Lemma 9
The following holds :
-
(i)
\(\kappa \ne 1\),
-
(ii)
If \(\kappa =2\) and \(n\ge 3\) then \(x\ge r^{\max \{2,n-3\}}\),
-
(iii)
If \(\kappa \ge 3\), then \(\kappa \ge n/2\).
Proof
(i). If \(\kappa =1\), then \(m=nx\). So Eq. (3) becomes
If p is prime dividing \(U_n\) (which exists since \(n>1\)), then \(p\mid U_n^x\) and \(p\mid U_n\mid U_{nx}\), so from the above equation we get \(p\mid U_{n+1}\), a contradiction since \(\gcd (U_n,U_{n+1})=U_{\gcd (n,n+1)}=1\) by relation (12).
(ii) In this case \(m=nx-1\) so Eq. (3) becomes
In particular, \(U_{nx-1}-U_{n+1}^x\equiv 0\pmod {U_n^2}\). We study this congruence. In what follows for three algebraic integers a, b, c, we write \(a\equiv b\pmod c\) if \((a-b)/c\) is an algebraic integer. Write
Then
Thus,
On the other hand,
Thus,
In the last step above, we used the fact that \(\alpha ^{-1}-\alpha =-\beta -\alpha =-r\). Since the expression \(U_{nx-1}-U_{n+1}^x\) is divisible by \(U_n^2\), we get that \(U_n^2\mid \beta ^{n(x-1)} xr U_n\). Since \(\beta \) is a unit, we get that \(U_n\mid xr\). For \(n=2\), this gives us nothing since \(U_2=r\). For \(n=3\), \(U_3=r^2+1\) is coprime to r, so \(U_3\mid x\), which gives \(x\ge r^2+1>r^2\). For \(n=4\), we have that \(U_4=r(r^2+2)\) divides rx, so \(r^2+2\mid x\) giving \(x\ge r^2+2>r^2\). Finally, for \(n\ge 5\), we have that \(U_n>\alpha ^{n-2}>r^{n-2}\) and so \(x\ge U_n/r\ge r^{n-3}\). This proves (ii).
(iii) We may assume that \(n\ge 7\), otherwise the conclusion is trivial. Recall that \(V_n=\alpha ^n+\beta ^n\). Relations Eqs. (9) and (10) are
In particular, \(U_{n+1}^2\equiv (-1)^n\pmod {U_n}\) and \(V_n^2\equiv 4(-1)^n\pmod {U_n}\). We also use the fact that \(U_{-m}=(-1)^{m-1}U_m\), \(V_{-m}=(-1)^mV_m\) and relation (11) which is
Armed with these facts, writing \(m=nx-(\kappa -1)\) and
we multiply both sides of the above equation by 2 and write
We square both sides of the above equation and reduce it modulo \(U_n\) taking into account that \(U_n\mid U_{nx}\) and \(V_{nx}^2\equiv 4(-1)^{nx}\pmod {U_{nx}}\equiv 4(-1)^{nx}\pmod {U_n}\), and get
Thus, \(U_n\mid 4(U_{\kappa -1}^2-1)\). The right-hand side is nonzero since \(\kappa >2\). If \(\kappa -1\) is odd, then
by relation (9) (with \(n+1:=\kappa -1\)). Since \(n\ge 7\) and \(\mathbf{U}\) is not the Fibonacci sequence, it follows that \(U_n\) has a primitive divisor p, which must divide one of \(U_{\kappa -2}\) or \(U_{\kappa }\). Thus, \(z(p)=n\) divides one of \(\kappa -2\) or \(\kappa \), so we get \(\kappa \ge n\), which is a better conclusion than the one desired. If \(\kappa -1\) is even and \(U_n\mid 2(U_{\kappa -1}^2-1)\), we then get
so \(2\kappa -2\ge n-2\), therefore \(\kappa \ge n/2\). The above argument was based on the fact that \(r\ge 2\). In particular, if \(r\ge 4\), then the same argument gives again that
so \(\kappa \ge n/2\). So, the only case when the above arguments fail are when \(r=3\) and \(4\mid U_n\). It then follows that n is even (in fact, n is a multiple of 6, but we shall not need that), so \(r\mid U_n\mid 4(U_{\kappa -1}^2-1)\). But \(U_{\kappa -1}\) is a multiple of r (since \(\kappa -1\) is even), so \(U_{\kappa -1}^2-1\) is coprime to r. Thus, \(r\mid 4\), which is false. This finishes the proof of (iii). \(\square \)
Corollary 4
If \(\kappa >0\), then \(x>n/2\).
Proof
By Lemma 9, if \(\kappa =2\), then \(x\ge r^{\min \{2,n-3\}}\ge 3^{\min \{2,n-3\}}>n/2\) for any \(n\ge 2\). If \(\kappa \ge 3\), then
which leads to \(1+nx-n/2\ge m\). Comparing this with the lower bound \(m>(n-1)x+1\) given by inequality (19), we get \(x>n/2\). \(\square \)
3.7 Another inequality among r, n, m, x
In this section, we return to inequality (27) and rewrite it in order to deduce a good approximation of \(\log r\) by a rational number whose denominator is a multiple of \(r^2\). Let’s get to work. We need approximations of \(\log \alpha \) and \(\log {\sqrt{r^2+4}}\) in terms of \(\log r\).
Lemma 10
For \(r\ge 3\), we have the following approximations:
Proof
We have
With \(z:=4/r^2\), we have \(|z|\le 4/9\) and
The expression in parenthesis above is smaller than
Hence,
For \(\alpha \), we write
Note that \(|z_1|\le 1/r^2\). Thus,
where by the previous arguments,
It remains to expand \(z_1\). For this, we have
Since \(|\left( {\begin{array}{c}k\\ 1/2\end{array}}\right) |\le 1/4\) for all \(k\ge 1\), it follows that
Hence,
\(\square \)
The following estimate is the main result of this section.
Lemma 11
If \(r\ge 4\), then \(\kappa >0\). Furthermore,
Proof
We shall use the approximations given in Lemma 10 but we also need an approximation of \(\log U_{n+1}\). We have
where
Since \(\beta =-\alpha ^{-1}\), it follows that \(|\beta /\alpha |=1/\alpha ^2\). Thus,
where for the last inequality we used the fact that \(\alpha >r\ge 3\) and \(n\ge 2\). Inserting estimates (46) and (47) together with Eq. (49) into inequality (27), we get
We recognise the coefficient of \(\log r\) as the number we denoted \(-\kappa \) in (45). Using inequality (27), we get
Inequality (19) shows that \(m-x(n+1)\in [-2x+2,2]\). In particular, \(|m-x(n+1)|\le 2(x-1)\). We thus get, by estimates (46) and (47) together with Eq. (49), that
Since \(x\ge 3\) and \(r\ge 3\), we get that the last term satisfies
Hence,
Assume that \(\kappa \le 0\). We then get that
which implies that
If \(r\ge 5\), the right-hand side is
a contradiction. Similarly, if \(r\ge 4\) and \(x\ge 4\), then the right side is
Thus, if \(r\ge 5\) or \(r=4\) and \(x\ge 4\), then \(\kappa >0\) and now Lemma 9 applies. We will show at the end of this proof that \(\kappa > 0\) for \((r,x)=(4,3)\) as well. Multiplying both sides of estimate (51) by \(r^2\), we get
Hence,
which gives estimate (48). It remains to treat the case \((r,x)=(4,3)\). By inequality (19), we have \(m<3(n+1)+1=3n+4\) so \(\kappa =3n+1-m\ge -2\). So, the only instances in which \(\kappa \le 0\) is possible are when \(m=3n+3,~m=3n+2,~3n+1\). Well, let us show that this is not possible by proving that
Using the Binet formula (4), this is implied by
with \(\alpha =2+{\sqrt{5}}\) and \(\Delta =20\). The function of n in the left is decreasing and the function with n in the right is increasing, and the inequality holds at \(n=1\) (the left–hand side there is \(<18.5\) and the right side is \(>19.5\)), so it holds for all \(n\ge 1\). Thus, \(\kappa \ge 2\) for \(r=4\) as well. \(\square \)
3.8 The case \(n\le 100\)
We first seek bounds on r. Having the bounds in r and n, we get bounds on x using Lemma 8. Finally, for a fixed r we use Baker–Davenport on estimate Eq. (27) to lower x. The hope is that in all cases \(x\le 100\), a case which has already been treated.
We prove the following result.
Lemma 12
When \(n\le 100\), we have \(r\le 1.5\times 10^6\).
Proof
We assume \(r>10^6\). Then \(x>\kappa r^2\log r/1.01\ge nr^2\log r/2.02\) by Lemmas 9 and 11. We go through the three possibilities of Lemma 8. In case (i), we get
which gives \(r<5500\), a contradiction. Assume we are in case (ii). Then
The factor 1.001 is an upper bound on the factor \(1+1/(r\log r)\) which is valid since r is large. Now
where the last inequality holds since \(r>10^6\). Put \(y:=(x+1)/(n(\log (r+1))\). Then the above inequality is \(y>r^2/2.03\). Inequality (52) can be rewritten in terms of y as
We look at the function
Its derivative is
so our function f(y) is increasing. Since \(f(y)<1.39\times 10^6\log (r+1)\), and \(y>r^2/2.03\), it follows that \(f(r^2/2.03)<1.39\times 10^6 \log (r+1)\). This gives
which gives \(r<370000\), a contradiction. Assume we are in case (iii). We use the same substitution \(y:=(x+1)/(n(\log (r+1))\). We then get
The function \(g(y):=y/(2\log y+\log (12))^2\) is also increasing, so we deduce that
and this gives \(r<1.5\times 10^6\). \(\square \)
Having bounds on n and r, inequalities (i), (ii) and (iii) from Lemma 8 become
Any one of these inequalities implies that \(x<3\times 10^{15}\). Now we do Baker–Davenport on estimates (27) for \(n\in [2,100]\), \(r\in [3,1{,}500{,}000]\), and \(x<3\times 10^{15}\). This also gives \(m<3\times 10^{17}\) via inequality (19). We return to inequality (27) and rewrite it as follows:
Then, we apply Lemma 5 on Eq. (53) with the data:
A computer search in Mathematica reveals that \( x\le 81 \), which is a contradiction. This computation lasted 16 hours on a cluster of four 16 GB RAM computers.
3.9 The case \(n> 100\)
Estimate (27) together with estimate (49) give
which implies, via estimate (50), that
If \(r\ge 4\) then \(\kappa >0\), so by Lemma 9, we have \(\kappa \ge 2\). If \(\kappa \ge 3\), then
The same conclusion holds for \(\kappa =2\) since then \(x\ge r^{\min \{2,n-3\}}\ge 4^{\min \{2,n-3\}}\ge 2n+2\) for all \(n\ge 2\). We thus get
We prove the following lemma.
Lemma 13
For \(n>100\), we have \(r^{n-2}>x\).
Note that Lemmas 9 and 13 show that the case \(\kappa =2\) cannot occur for \(r\ge 4\) provided that \(n> 100\).
Proof
Assume \(x\ge r^{n-2}\). We use the bounds given by Lemma 8 on x. In case (i), we get
The function \(y\mapsto y^{99}/(\log (y+1))^2\) is increasing for all \(y\ge 3\) as one can check by computing its derivative. Thus, if the above inequality holds for r, it should hold \(r=3\) as well, which is false. In case (ii), we have
The expression \(1+1/(r\log r)\) is smaller than 1.304 at \(r=3\). Since \(1.38\times 1.304<1.8\), we get
If \(x\le n\), then we get \(r^{n-2}\le n\) for \(r\ge 3\) and \(n\ge 101\), which is false. Thus, \(n<x\), so we may use \(0.3(n+1)<n<x+1\) to get
The function \((x+1)/(\log (x+1))^3\) is increasing for \(x+1>e^3\), which is the case for us since \(x\ge r^{99}\ge 3^{99}\). Hence, the above inequality should hold for \(x+1\) replaced by \(r^{n-2}\), and that gives
Since \(r^{n/3}\ge 3^{n/3}>n\) holds for \(n>100\), we get that
so
The function \(y\mapsto y^{95}/(\log (y+1))^6\) is increasing for \(y\ge 3\), so the last inequality should hold also for r replaced by 3, which is false. A similar argument works if x is in case (iii) of Lemma 8. We don’t give further details. \(\square \)
From Lemma 13 we conclude that if \(r\ge 4\), then inequality (55) leads to
We put
We apply Theorem 3 to find a lower bound on \(\log |\Gamma _2|\) with the data:
Since \( \gamma _1, \gamma _2 \in \mathbb {Q}(\alpha ) \), we take again \( \mathbb {K}:= \mathbb {Q}(\alpha )\) with degree \( {\mathcal {D}}:=2 \). The fact that \( \gamma _1 \) and \( \gamma _2 \) are multiplicatively independent has already been checked.
We take
and
Thus,
where we used the fact that \(m-(n+1)x\in [-2x+2,2]\). By Theorem 3, we get that
We want an upper bound on r. So, assume \(r\ge 10^6\). If \( \log ({(2.5x)}/{\log (r+1)}) < 10.5\), then
Since \(x>nr^2\log r/(2+10/r)\ge 101 r^2\log r/2.02=50r^2\log r\), we get
so
so \(r\le 17\), a contradiction.
Assume next that \(\log (2.5 x/\log (r+1))>10.5\). We then get
Comparing the above inequality with estimate (56), we get
so
where we used the fact that \(r\ge 10^6\) so \(\log (r+1)/\log r<1.0001\).
We now use the bounds on x given by Lemma 8. In case (i), we have
so
so \(r<670\), a contradiction. In case (ii), we have
In case \((x+1)/(\log (r+1))\le n\), we have
So, putting \(y:=x/\log (r+1)\), we get \(y<196\log (r+1) \log (2.5 y)^2\). Note that
In the above inequalities we used the fact that \(r\ge 10^6\). The function \(y\mapsto y/\log (2.5 y)^2\) is increasing for \(y>e^2/2.5\), which is our case. Hence, the inequality \(y<196\log (r+1) \log (2.5y)^2\) should hold with y replaced by \(50r^2\), which yields \(50r^2<196 \log (r+1) \log (2.5\times 50 r^2)^2\) and gives \(r<50\), a contradiction. Thus, \(n<(x+1)/\log (r+1)\). Since \(0.3(n+1)<n\), we conclude that
Putting again \(y:=(x+1)/\log (r+1)\), we get
The function \(y\mapsto y/(\log (2.5y)^4\) is increasing for our range for \(y>50r^2>50\times 10^{16}\), so we get that the above inequality should hold by replacing y by \(50r^2\). Thus,
so \(r<2.6\times 10^8\). A similar argument holds when x is in case (iii). In that case, we may again suppose that \(n<(x+1)/\log (r+1)\). We get
so
Imposing that the above inequality holds for y replaced by \(50 r^2\), we get
so \(r<4.3\times 10^8\).
To summarise, we have proved the following.
Lemma 14
If \(r\ge 4\), then \(r<4.3\times 10^8\).
Having bounds for r it is easy to find bounds for x. For example,
Next, if x is in case (i), then
If x is in case (ii), then
which gives \(x<5\times 10^{19}\). Finally, if x is as in case (iii), then
so \(x<3\times 10^{20}\).
Thus, \(r<4.3\times 10^8\) and \(x<3\times 10^{20}\). Inequality (56) gives that
where for the last inequality we used that \(r^n=r^2 r^{n-2}\ge 16x\) by Lemma 13. In particular, the ratio \((x(n+1)-1)/(x-1)\) is a convergent of \(\log {\sqrt{r^2+4}}/\log \alpha \). Since \(x<3\times 10^{20}< F_{100}\), it follows that \((x(n+1)-1)/(x-1)=p_k/q_k\) for some \(k\in [0,99]\). So, we apply Lemma 4 on Eq. (59) with the data:
With the help of a computer search in Mathematica, we checked all these possibilities over all the values for \(4\le r\le 4.3\times 10^8\) and found that \( n\le 46 \), which is a contradiction. This computation lasted 6 hours on a cluster of four 16 GB RAM computers.
3.10 The case \(r=3\)
The case \(r=3\) is special since there we don’t know that \(\kappa >0\) so some of the inequalities used for the case \(r\ge 4\) do not apply. In the case \(n\le 100\), Lemma 8 gives
or
which gives \(x<1.7\times 10^{14}\), or
which gives \(x<3.3\times 10^{11}\). Now we can do Baker–Davenport on estimate (27) and get better bounds on x. In case \(n>100\), estimate (54) together with Lemma 13 hold and give
We keep the notation r and \(\alpha \) although this section only applies to \(r=3\) for which \(\alpha =(3+{\sqrt{13}})/2\). Put \(\ell :=\min \{n-1,x-1\}\). The lower bound estimate (58) still applies and gives
When \(\ell =n-1\), we get
In case the maximum above is 10.5 we get \(x<16{,}000\) and \(n<40{,}000\). If the maximum above is not at 10.5, we then get
Going via the possibilities (i), (ii), (iii), we get
which gives \(x<2\times 10^{16}\), or finally
which gives \(x<3.1\times 10^{15}\). So, in all instances, \(x<2\times 10^{16}\), and now
Since Lemma 13 still applies, it follows that estimate (60) gives
so again \((x(n+1)-m)/(x-1)=p_k/q_k\) is a convergent of \(\log {\sqrt{r^2+4}}/\log \alpha \) with \(x<2\times 10^{16}<F_{80}\), so \(k\in [0,\ldots ,16]\). So, everything works fine if \(\ell =n-1\).
In case \(\ell =x-1\), one gets
which gives \(x<5\times 10^4\). And one wonders how one should finish it off. We can expand another linear form in logarithms which is small, or we may recall the following main theorem from [2].
Theorem 5
Assume that \(s\not \in \{1,2,4\}\) is minimal such that \(U_m\mid U_{n+1}^s-U_n^s\). Then \(m<20{,}000s^2\).
In our instance, since \(U_m=U_n^x+U_{n+1}^x\), one checks that the minimal s is exactly 2x. Thus, \(m<80{,}000x^2\), and since \(m>(n-1)x\) by estimate (19), we get \(n\le 80{,}000x<4\times 10^9\). Thus, \(n<4\times 10^9\) and \(x<5\times 10^4\). It is still a large range and we need to reduce it.
We consider the element
Lemma 13 together with the fact that \(\alpha >r\) implies that
where the last inequality holds for all \( n> 100 \). Now, we write
and
If n is odd, then
because y is very small. On the other hand, if n is even, then
because y is very small. Thus, the following inequalities hold in both cases,
and
Now, we return to (3) and rewrite it as
or
Multiplying both sides of the above inequality by \( \alpha ^{-(n+1)x}(r^2+4)^{x/2} \), we obtain that
so we may divide both sides of it by \(1+\alpha ^{-x}\) and get
Since \(n>100\), the left-hand side is small (say smaller than 1/2), so we can pass to a logarithmic form and get
For us, the parameter x is in [3, 50000]. Given x, we have \(m<2\times 10^{14}\). So, this is a suitable inequality to apply Baker-Davenport to. To do so, we rewrite Eq. (63) as
Then, we apply Lemma 5 on Eq. (64) with the data:
A computer search in Mathematica reveals that \( n\le 94 \), which is the final contradiction. This computations lasted a few hours on a cluster four 16 GB RAM computers. \(\square \)
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Acknowledgements
This paper was written when both authors visited the Max Planck Institute for Mathematics Bonn, in 2019. They thank this institution for the hospitality and the fruitful working environment.
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Open access funding provided by Austrian Science Fund (FWF). M. D. was supported by the FWF Projects: F5510-N26—Part of the special research program (SFB), “Quasi Monte Carlo Methods: Theory and Applications” and W1230—“Doctoral Program Discrete Mathematics”. F. L. was supported by Grant RTNUM19 from CoEMaSS, Wits, South Africa.
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Ddamulira, M., Luca, F. On the exponential Diophantine equation related to powers of two consecutive terms of Lucas sequences. Ramanujan J 56, 651–684 (2021). https://doi.org/10.1007/s11139-020-00278-7
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DOI: https://doi.org/10.1007/s11139-020-00278-7