Abstract
Let \(d \ge 2\) be an integer which is not a square. We show that if \((F_n)_{n\ge 0}\) is the Fibonacci sequence and \((X_m, Y_m)_{m\ge 1}\) is the mth solution of the Pell equation \(X^2 -dY^2 = \pm 1\), then the equation \(Y_m = F_n\) has at most two positive integer solutions (m, n) except for \(d=2\) when it has three solutions \((m,n)=(1,2),(2,3),(3,5)\).
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1 Introduction
Let \(d\ge 2\) be an integer which is not a square. Consider the Pell equation
All its positive integers solutions (X, Y) are given by \((X,Y)=(X_m,Y_m)\) for some positive integer m where
and \((X_1,Y_1)\) stands for the smallest positive integer solution. In particular, both sequences \((X_m)_{m\ge 1}\) and \((Y_m)_{m\ge 1}\) are binary recurrent of characteristic polynomial \(x^2-(2X_1)x+\lambda \), where \(\lambda =X_1^2-dY_1^2\in \{\pm 1\}\).
Several recent papers have investigated the following problem. Assume that \(\textbf{U}:= (U_n)_{n\ge 0}\) is some interesting sequence of positive integers. What can one say about the number of solutions of the containment \(X_m \in \textbf{U}\) for a generic d? What about the number of solutions of the containment \(Y_m\in \textbf{U}\)? For the sequence of X-coordinates, the answer is that for most binary recurrent sequences (Fibonacci numbers [16], repdigits [7, 10, 13] in some given integer base \(b\ge 2\), etc.), the equation \(X_m \in \textbf{U}\) has at most one positive integer solution m for any given d except for a few (finitely many) values of d, which have been explicitly calculated for the above examples \(\textbf{U}\). For the sequence of Y-coordinates, it was shown in [12] that if \(\textbf{U}\) is a fixed binary recurrent sequence of integers whose characteristic equation has real roots, then the containment \(Y_m \in \textbf{U}\) has at most two solutions m provided d exceeds some effectively computable bound depending on \(\textbf{U}.\) It has exactly two solutions for infinitely many d’s if \(\textbf{U}\) contains 1 and infinitely many even integers. The cases of the specific binary recurrent sequences with general terms \(U_n=2^n-1\) and \(U_n=L_n\), the nth Lucas number, respectively, were treated in an elementary way in [11], and following the non-elementary procedure described in [12] in [9], respectively. Here, we take \(\textbf{U}:=(F_n)_{n\ge 0}\) to be the Fibonacci sequence of initial values \(F_0=0,~F_1=1\) and recurrence \(F_{n+2}=F_{n+1}+F_n\) for all \(n\ge 0\) and study the equation \(Y_m=F_n\) in integers (m, n) with \(m\ge 1,~n\ge 1\). Our result is the following.
Theorem 1
Let \(d\ge 2\) be an integer which is not a square, \((X_m,Y_m)_{m\ge 1}\) be the mth positive integer solution to the Pell equation (1). The Diophantine equation
has at most two positive solutions (m, n) with \(m\ge 1,~n\ge 1\) except for \(d=2\) when it has three solutions \((m,n)\in \left\{ (1,2),(2,3),(3,5) \right\} \).
Our method of proof follows closely the general approach from [12] with some details borrowed from [9].
2 Preliminaries
2.1 Pell equation and Fibonacci sequence
We set \((X_1, Y_1)\) to be the minimal solution of (1). In this section, we recall a couple of facts about Pell equations and Fibonacci numbers. Let
Then
Note that \(\delta =\lambda \gamma ^{-1}\), where \(\lambda \in \{\pm 1\}\). We put \({{\mathbb {L}}}:= {{\mathbb {Q}} }(\gamma )\).
The Fibonacci sequence \(\left\{ F_n \right\} _{n\ge 0}\) is given by \(F_{n+2}=F_{n+1}+F_n\) for all \(n\ge 0\) with the initial values \(F_0=0\) and \(F_1=1.\) By setting \((\alpha ,\beta ):=((1+{\sqrt{5}})/2,(1-{\sqrt{5}})/2),\) the Binet’s formula for the Fibonacci sequence is
We note that \(\beta =-\alpha ^ {-1}\).
Lemma 1
The inequalities
hold for all \(m\ge 1\). Further, the inequalities
hold for all \(n\ge 1.\)
In the above, the inequalities involving \(Y_m\) are from Lemma 1 in [9] and the inequalities involving \(F_n\) are well–known.
2.2 Linear forms in logarithms
For an algebraic number \(\lambda \) of minimal polynomial over \({{\mathbb {Z}}}\) with positive leading coefficient
(here, \(\lambda ^{(1)}=\lambda \)), we put
for the logarithmic height of \(\lambda \). The following result is referred to in the literature as Baker’s lower bound for a non-zero linear form in logarithms (see [3, 17]).
Theorem 2
(Matveev’s theorem). Assume that \(\lambda _1,\ldots , \lambda _t\) are positive real algebraic numbers in a number field \(\mathbb {F}\) of degree D, \(b_1,\ldots ,b_t\) are rational integers, and
is not zero. Then
where
and
When \(t=2\) and \(\lambda _1,~\lambda _2\) are positive and multiplicatively independent, we can do better. Namely, let in this case let \(B_1,~B_2\) be real numbers larger than 1 such that
and
Put
Note that \(\Lambda \ne 0\) since \(\lambda _1\) and \(\lambda _2\) are multiplicatively independent. The following inequality is Corollary 2 in [15].
Theorem 3
With the above notations and conventions, assuming that \(\lambda _1,~\lambda _2\) are positive algebraic numbers which are multiplicatively independent, then
When \(t=3\) a better bound than the general one given by Theorem 2 in some special case is due to Mignotte [18, Proposition 5.2]; see also [19].
Theorem 4
Let \(\Lambda :=b_2\log \gamma _2-b_1\log \gamma _1-b_3\log \gamma _3\ne 0\) with \(b_1,b_2,b_3\) positive integers with \(\gcd (b_1,b_2,b_3)=1\) and \(\gamma _1,\gamma _2,\gamma _3\) positive real algebraic numbers \(>1\) which are multiplicatively independent in a field \(\mathbb {K}\) of degree D. Let
Let \(a_1,a_2,a_3\) be real numbers such that
Put
Then one of the following holds:
-
(i)
$$\begin{aligned} \log |\Lambda |>-790.95\Omega D^2 (\log {{\mathcal {B}}})^2; \end{aligned}$$
-
(ii)
there exist nonzero integers \(r_0,s_0\) with \(r_0b_2=s_0b_1\) satisfying the inequalities
$$\begin{aligned} |r_0|<5.61 (D\log {{\mathcal {B}}})^{1/3} a_2\quad {and}\quad |s_0|<5.61 (D \log {{\mathcal {B}}})^{1/3} a_1; \end{aligned}$$ -
(iii)
there exist integers \(r_1\ne 0,s_1\ne 0,t_1,t_2\) satisfying
$$\begin{aligned} \gcd (r_1,t_1)=\gcd (s_1,t_2)=1,\qquad (t_1b_1+r_1b_3)s_1=r_1b_2t_2, \end{aligned}$$and also
$$\begin{aligned} |r_1s_1|< & {} 5.61\delta (D\log {{\mathcal {B}}})^{1/3} a_3,\\ |s_1t_1|< & {} 5.61\delta (D\log {{\mathcal {B}}})^{1/3} a_1,\\ |r_1t_2|< & {} 5.61\delta (D \log {{\mathcal {B}}})^{1/3} a_2, \end{aligned}$$where \(\delta :=\gcd (r_1,s_1)\). If \(t_1=0\), we can take \(r_1=1\) and if \(t_2=0\) we can take \(s_1=1\).
2.3 Reduction method
The following result is Lemma 5 (a) in [8], which is a slight variation of a result of Baker and Davenport (see [2]).
Lemma 2
Let M be a positive integer, let p/q be a convergent of the continued fraction of the irrational \(\tau \) such that \(q > 6M\), and let \(A,B,\mu \) be some real numbers with \(A > 0\) and \(B > 1\). Let
where \(||\cdot ||\) denotes the distance from the nearest integer. If \(\varepsilon >0\), then there is no solution of the inequality
in positive integers m, n and k with
The following result is criterion of Legendre (see Theorem 8.2.4 in [20]).
Lemma 3
(Legendre) Let \(\tau \) be an irrational real number and x, y be integers.
-
(i)
If
$$\begin{aligned} \left| \tau -\frac{x}{y}\right| <\frac{1}{2y^2}, \end{aligned}$$then \(x/y=p_k/q_k\) is a convergent of \(\tau \). Furthermore,
$$\begin{aligned} \left| \tau -\frac{x}{y}\right| \ge \frac{1}{(a_{k+1}+2)y^2}, \end{aligned}$$where \([a_0,\ldots ,a_k,\ldots ]\) is the continued fraction expansion of \(\tau \).
-
(ii)
If \(y<q_{k+1}\), then
$$\begin{aligned} \frac{1}{(A+2)y^2}\le \left| \tau -\frac{x}{y}\right| , \end{aligned}$$where \(A:=\max \{a_j: j\le k+1\}\).
2.4 An elementary inequality
The following appears as Lemma 7 in [14].
Lemma 4
If \(s \ge 1,~ T > (4s^2)^s\) and \(T > x/(\log x)^s\), then
The statement of the above lemma is formulated in [14] only for an integer parameter s but a close look at its proof there shows that it works for any real parameter \(s\ge 1\).
3 The proof of Theorem 1
As in [9, 11], we put \({{\mathbb {M}}}:={{\mathbb {K}}}{{\mathbb {L}}},\) where \({{\mathbb {K}}}\) is the number field \({{\mathbb {Q}}}(\alpha ).\) Just like for the Lucas sequence \((L_n)_{n\ge 0}\), there are infinitely many d such that \(Y_m = F_n\) has two positive integer solutions (m, n) since \(F_1=F_2=1\) and \(F_n\) is even whenever \(3 \mid n\) (thus there are infinitely many even Fibonacci numbers). Assume \(Y_m = F_n\) has 3 positive integer solutions \((m_i, n_i)\) for \(i=1,2,3.\) Suppose further that \(m_1<m_2<m_3\). Then we have that \(n_1<n_2<n_3.\) We also assume \(n \ge 2\) since the instance \(n=1\) produces the same Fibonacci number as the instance \(n=2.\)
3.1 \(\alpha \) and \(\gamma \) are multiplicatively dependent
In this case, \({{\mathbb {K}}}={{\mathbb {L}}}\) and the fundamental unit in the ring of algebraic integers, \({{\mathcal {O}}}_{{\mathbb {L}}}\), of \({{\mathbb {L}}}\) is \(\alpha \). Thus, \(\gamma =\left( \left( 1+\sqrt{5} \right) /2 \right) ^ m\) for some positive integer m. We thus have that \(d=5\cdot d^ 2_1\) for some positive integer \(d_1\ge 1\). We rewrite equation (1) as
It is well-known that the only positive integer solutions (X, Y) of the equation \(X^2-5Y^2=\pm 4\) are of the form \((X,Y)=(L_t,F_t)\) for some integer \(t\ge 1\). Further, the sign in the right-hand side equals \((-1)^t\). Thus, the relations
hold with some positive integers \(t_i\) for \(i=1,2,3\).
It is therefore the case that \( 2d_1F_{n_i} = F_{t_i} \) for \(i = 1, 2, 3\) and \(t_1< t_2 < t_3\). The above equation gives in particular that \(n_3 \mid t_3\) with the instance \(n_3 = t_3\) being clearly impossible. So, only the instance \(n_3 < t_3\) is possible. If we assume that \(d_1\ge 2\), then the index of appearance, \(z(2d_1)\), of \(2d_1\) in the Fibonacci sequence (that is, the smallest positive integer k such that \(2d_1 \mid F_k\)) is at least 5 and it is a multiple of 3, so it is at least 6. Since \(2d_1 \mid F_{t_i}\), it is also the case that \(z(2d_1) \mid t_i\). In particular, \(t_3 = k_3z(2d_1)\) for some positive integer \(k_3 \ge 3\). We thus see that \(t_3 \ge 18\). From \(2d_1F_{n_i} = F_{t_i}\), we have that \(F_{t_3}F_{n_2} = F_{t_2}F_{n_3}\). The factor \(F_{t_3}\) appearing in the left hand side of the last equation has, by Carmichael’s Primitive Divisor Theorem (see [4, 5]), a primitive divisor, that is a prime factor \(p\mid F_{t_3}\) such that \(p\not \mid F_k\) for any \(k<t_3\). In particular, the factor p does not appear in the right–hand side of the above equation since both inequalities \(t_2 < t_3\) and \(n_3 < t_3\) hold. The case \(d_1 = 1\) yields the equation \(2F_{n_3} = F_{t_3}\), so again \(F_{n_3}\) does not have a primitive prime factor. Since \(t_3\ge 4\) (because \(t_3>n_3\)), the only possibilities are \(t_3=6,12\), which do not lead to solutions since none of \(F_{6}/2=4\) or \(F_{12}/2=72\) is a Fibonacci number. Thus, no solution exists in this case.
3.2 A useful inequality
We proceed with the proof of Theorem 1 by supposing now that \(\gamma \) and \(\alpha \) are multiplicatively independent. From the equation \(Y_m=F_n,\) we have that
This yields
Equation (3) gives that
So,
Taking the absolute value in (4), we have that
We thus record
We note that since 4d/5 is not a unit but \(\gamma ^{2m}\alpha ^{-2n}\) is a unit, the equality \(4d/5=\gamma ^{2m}\alpha ^{-2n}\) does not hold. This ensures that the left–hand side of (6) is not zero. We next suppose that \(\gamma >10^6\) and conclude that \(d\ge 94\) (see Sect. 3.2 in [9]). From the first inequality of (5), we have that
Taking absolute values in (4), we obtain
The rightmost quantity in the above inequality is less than 1/2. We use the fact that if \(|e^x-1|<y<1/2\) for real x, then \(|x|<2y\) to obtain
Next, we rewrite (3) as
We multiply through by \(\gamma ^m\) to obtain
This yields
We rule out
since it is less than 1. We thus have
Taking into account the fact that \(d\ge 94\), we can bound x as
In the case \(x<0,\) we denote \(f(x)=\sqrt{1+x}\) and note by the Mean Value Theorem that
Similarly, the case of \(x>0\) yields
So, in either case, we record
The fact that \(|x|<0.31\) yields \(|1+\zeta |>0.69,\) so \(f'(\zeta )<0.61\). Thus,
where throughout the rest of this proof we use the Landau symbol \(O_C(|x|)\) with the understanding that the constant implied by it can be taken to be C. The last equality follows from the fact that
Thus,
Hence,
To justify the last estimate put y such that \(|y|\le 1.2/\alpha ^{2n}\) and note that
and
Since \(|y|<1.2\alpha ^{-2n}\) and
we get the desired estimate. We now use this in equation (4) to get
We put
By (7), we have that \( |z|<2.08/\alpha ^{2n}<1.\) We see that
This shows that
Hence, we get
which implies that
The left–hand side of the above inequality is non-zero since if it were zero, then the exponential of the non-zero algebraic number \(\left( (-1)^{n+1}+{5\lambda ^m}/{4d}\right) \alpha ^{-2n}\) is an algebraic number. This contradicts Baker’s reformulation of the Lindermann–Weierstrass theorem. We summarise the main results of this section in the following result which is the analog of Lemma 5 in [9]. Recall that \(\lambda =X_1^2-dY_1^2\in \{\pm 1\}\).
Lemma 5
Assume that \(Y_m=F_n\) for \(m\ge 1\) and \(n\ge 2\). The following hold
-
(i)
$$\begin{aligned} 0<\left| (2{\sqrt{d}})^{-1} \sqrt{5} \gamma ^m \alpha ^{-n}-1\right| \le \frac{2.5}{\alpha ^{2n}}. \end{aligned}$$(10)
-
(ii)
If \(\gamma >10^6\), then
$$\begin{aligned} 0<|m\log \gamma -n\log \alpha -\log (2{\sqrt{d}})+\log \sqrt{5} | \le \frac{2.08}{\alpha ^{2n}}\quad {\textrm{ for}}\quad n\ge 2. \end{aligned}$$(11) -
(iii)
If \(\gamma >10^6\), then
$$\begin{aligned} 0{} & {} <\left| m\log \gamma -n\log \alpha -\log (2{\sqrt{d}})+\log (\sqrt{5})-\left( (-1)^{n+1}+\frac{5\lambda ^m}{4d}\right) \alpha ^{-2n}\right| \nonumber \\{} & {} <\frac{3.6}{\alpha ^{4n}}. \end{aligned}$$(12)
3.3 Bounds on n and m in terms of \(\gamma \)
From Lemma 1, we have
We apply Theorem 2 on the left-hand side of (10) with the following choice of parameters; \(t:= 3\) and
We take \(b_1:= -1\), \(b_2:= -n\) and \(b_3:= m\). We assume \(n\ge 3\). Here, \(\mathbb {M}:= \mathbb {Q}(\sqrt{d},\alpha )\) contains \(\gamma _1,\gamma _2,\gamma _3\) and has \(D:= [\mathbb {M}: \mathbb {Q}] = 4\). We have
The last inequality holds as \(\gamma =X_1+{\sqrt{d}}Y_1\ge {\sqrt{d-1}}+{\sqrt{d}}\) and \(({\sqrt{d-1}}+{\sqrt{d}})^3>4d\) holds for all \(d\ge 2\). Thus, \(h(\gamma _1)<1.5\log \gamma \). Next, we have \(h(\gamma _2)=(\log \alpha )/2\) and \(h(\gamma _3)=(\log \gamma )/2\). Therefore, we can take
We note that
So, we have that \(\gamma ^{m-2}<Y_m<\alpha ^{n-1}\). This implies \(m\le n\) since \(\gamma >\alpha .\) We thus choose \(B:=n.\) We set
With the above observations, the calculations preceding Lemma 6 in [9] apply to our situation and yield the following bounds.
Lemma 6
All integer solutions (m, n) with \(m\ge 1\) and \(n\ge 2\) of the equation \(Y_m=F_n\) satisfy
-
(i)
\((m-1) \log \gamma <(n+1)\log \alpha +\log {\sqrt{2}}\);
-
(ii)
\(n < 8\cdot 10^{15} \max \{1,(\log \gamma )^3 \}\);
-
(iii)
\(m<8\cdot 10^{15} \max \{1,(\log \gamma )^2 \}\).
3.4 The case \(\gamma <10^2\)
In this case, the relation \(\gamma =X_1+\sqrt{X^2_1-\lambda }\) gives that \(X_1\le 50.\) We also use Lemma 6 to obtain the bounds \(m< 3.5\cdot 10^{19}\) and \(n<1.61 \cdot 10^{20}.\) We refer to the right-hand side of (i) in Lemma 5 and note that \(4.7/\alpha ^{2n}<1/2\) if \(n\ge 3.\) Thus, we apply (i) in Lemma 5 in the case of \(n=n_3\) and \(m=m_3\) and pass to logarithmic form to obtain the inequality
We apply Lemma 2 to sharpen the bound on \(n_3.\) To that end, we rewrite the above inequality as
with the following data:
We also set \(M:=10^{20}>m_3.\) For each value of \(\tau \) resulting from the pair \((X_1,\lambda )\) (discarding the values for which \(\tau \) is rational), we see that the denominator of the 149th convergent of \(\tau \), \(q_{149}\), in all cases satisfy \(q_{149}\ge F_{149}>6\cdot 10^{30}.\) We look for a uniform lower bound on \( \varepsilon :=||\mu q||-M||\tau q||\) for all pairs \((X_1,\lambda ).\) The inequality
holds viewing \(p_{149}/q_{149} \) as a convergent of \(\tau .\) Therefore,
Across all pairs \((X_1,\lambda )\), we found that \(\Vert q\mu \Vert >1.66\cdot 10^{-9}.\) We thus used \(\varepsilon > 1.61\cdot 10^{-9}\) and obtained the bound \(n_3<115.\)
By means of the relation \(\gamma ^{m-1}/\sqrt{2}<F_n,\) which follows from Lemma 1, we obtained a good bound on m. To search for solutions to \(Y_m=F_n,\) we went back for all pairs \((X_1,\lambda )\) and generated all values of the candidates d such that \(X^2_1-\lambda =dY^2_1\) for some integer \(Y_1.\) For each such \((d,Y_1),\) we set \(Y_2=2X_1Y_1\) and generated the sequence \(\left\{ Y_m \right\} _{m\ge 1}\) where \(Y_{m+2}=2X_1Y_{m+1}-\lambda Y_m\) with m satisfying \(\gamma ^{m-1}/\sqrt{2}<F_{115}.\) We sought to find those values of d for which the set
has cardinality 3. We found only the value \(d=2\) for which \(Y_1=F_2,Y_2=F_3,Y_3=F_5.\)
3.5 The case \(10^2 \le \gamma <10^6\)
The bound on \(\gamma \) yields \(X_1\le 5\cdot 10^5.\) We also apply Lemma 6 to obtain \(m<3.2\cdot 10^{20}.\) We set M of Lemma 2 to be \(3.3\cdot 10^{20}.\) Since \(q_{149}\ge F_{149}>6\cdot 10^{30}>6M,\) we again work with the 149th convergent of \(\tau \) for each choice of the pair \((X_1,\lambda ).\) For all choices, we found again that \(\Vert q\mu \Vert >1.66\cdot 10^{-9}.\) The inequality
also holds. The value of \(\varepsilon \) was thus uniformly chosen as \(\varepsilon > 1.61\cdot 10^{-9}\) which yields \(n_3<115.\) A search was conducted for solutions in this range over all pairs \((X_1,\lambda )\) to no fruition. From now on, we assume \(\gamma >10^6.\)
3.6 Inequalities among solutions
Suppose (m, n) and \(m',n'\) are solutions to \(Y_m=F_n.\) We can thus write
This gives us
The right-hand side follows from the fact that \(\alpha ^{n-2}\le F_n \le \alpha ^{n-1}.\) We note that
and the inequality
therefore holds. We record this as follows. This is an analogue of Lemma 7 in [9].
Lemma 7
Assume \(\gamma >10^6\). If \((m,n), (m',n')\) satisfy \(Y_m=F_n\) and \(Y_{m'}=F_{n'}\), we then have
3.7 Bounding \(n_1\)
Here, we prove the following lemma.
Lemma 8
Assume \(\gamma >10^6\) and \((n,m)\in \{(n_1,m_1),(n_2,m_2),(n_3,m_3)\}\) are all solutions of the equation \(Y_m=L_n\) with \(1\le m_1<m_2<m_3\). Then the inequality
holds.
Proof
We follow the proof of Lemma 8 in [9]. Consider the matrix
Assume first that \(\textrm{det} A \ne 0\). Writing Lemma 5(ii) for \((m,n):=(m_\ell ,n_\ell )\), for \(\ell =1,2,3\), subtracting the one for \(\ell =1\) from the ones for \(\ell \in \{2,3\}\) and using the absolute value inequality, we get
Next, we consider an application of the absolute value inequality to a suitable combination of the above inequality to obtain;
In the above inequality, \(I_\ell \) for \(\ell =2,3\) represents the expression in the absolute value sign of (16) for \(\ell =2\) and 3 respectively. This yields
where
So, by Lemma 6(iii), we get
We thus get
which proves the desired inequality when det \(A\ne 0\). Now suppose that det \(A= 0.\) Then A has rank less than 3. Let \(l_1,l_2,l_3\) be the rows of the above matrix. Note that A has rank 2 since otherwise \(l_i\) and \(l_j\) should be multiples of each other, which is not the case since their third component is equal to 1 but their first components are different. Let u, v be rational numbers such that \(l_1= ul_2+v l_3\). The numbers u, v solve the system
whose solution is \((u,v)=\left( (n_3-n_1)/(n_3-n_2), (n_1-n_2)/(n_3-n_2)\right) \). Since u, v also solve the system
we also have \((u,v)=\left( (m_3-m_1)/(m_3-m_2), (m_1-m_2)/(m_3-m_2)\right) \). We further have that \(uv\ne 0\). We next go to Lemma 5(iii) which is
We multiply the above estimate for \(i=2\) with u, for \(i=3\) with v, and subtract the one for \(i=1\), we get
Multiplying across by \(\alpha ^{2n_1}\), we get
In the left-hand side of (18) we have 3 terms. Assume first that the first is \(\ge 1/6\) in absolute value. We then have that
Hence, by Lemma 7, we get
If \(u\le 2\), then the right-hand side is less than \(104< \gamma \), a contradiction. Hence, \(u\ge 2\). Thus, \(m_3-m_1\ge 2(m_3-m_2)\), so \(m_2-m_1\ge m_3-m_2\). In particular, \(m_2-m_1\ge (m_3-m_1)/2\). Then the above inequality becomes
so \(\gamma ^{k-1}<k\) with \(k:=m_3-m_1\ge 2\), a contradiction. If the second term in the left-hand side of (18) is \(\ge 1/6\) in absolute value, we then get by a similar argument that
so
and we get again the same contradiction as in the previous case. If both the first and second are less than 1/6 in absolute value, then the left-hand side of (18) exceeds
so we get that
Instance (19) together with Lemma 6(ii) gives
which is what we wanted. \(\square \)
3.8 The case \(m_1>1\)
In this case, by Lemma 6(i), we have
which gives \(\log \gamma <31\), so \(\gamma <2.91\cdot 10^{13}\). This gives \(n_1<62\). Hence,
so \(m_1=2 \) or 3. In the case \(m_1=2\), we have \(Y_2=2X_1Y_1=F_{n_1}\), which shows that \(F_{n_1}\) is even so \(3\mid n_1\). Now \(Y_2=2X_1Y_1=F_{n_1}\). Hence, \(X_1\mid F_{n_1}\). For each \(n_1\le 61\) which is a multiple of 3 we took \(X_1\) to be a divisor of \(F_{n_1}/2\) with \(5\cdot 10^5<X_1<1.5\cdot 10^{13}\). For each one of these choices we generated \(\gamma :=X_1+{\sqrt{X_1^2-\lambda }}\) for \(\lambda \in \{\pm 1\}\) and applied the method from Sect. 3.4. We obtained \(\varepsilon >1.10\cdot 10^{-6}\) and \(n<252\). To search for solutions to \(Y_m=F_n,\) we again went back for all pairs \((X_1,\lambda )\) satisfying the above condition and generated all values of candidate d such that \(X^2_1-\lambda =dY^2_1\) for some integer \(Y_1.\) For each such \((d,Y_1),\) we set \(Y_2=2X_1Y_1\) and generated the sequence \(\left\{ Y_m \right\} _{m\ge 1}\) where \(Y_{m+2}=2X_1Y_{m+1}-\lambda Y_m\) with m satisfying \(\gamma ^{m-1}/\sqrt{2}<F_{252}.\) We sought to find those values of d for which the set
has cardinality 3. This took a few minutes and produced no solution. There is therefore no solution in this case. In the case \(m_1=3,\) we write \(Y_3=(4X_1^2-\lambda )Y_1=F_{n_1}.\) That is, \((2X_1)^2=F_{n_1}/Y_1+\lambda .\) For each \(n_1\in [1,61]\) and choice of \(\lambda \in \{\pm 1\},\) we asked Mathematica to list all divisors \(Y_1\) of \(F_{n_1}\) for which \(F_{n_1}/Y_1+\lambda \) is a square greater than \(10^{12}.\) No such value of \(Y_1\) found. So again, there is no solution in this case. From now on, we assume that \(m_1=1\).
3.9 The case \(\gamma \le 10^{10}\)
Then \(X_1\le 5\times 10^9\). We write inequalities (11) for \((m,n)=(m_i,n_i)\) and \(i=2,3\), and combine them to get
We divide across by \((m_3-m_2)\log \alpha \) to get
There are 6 values of \(\gamma \) in the range \(\gamma <10^{10}\) for which \(\log \gamma /\log \alpha \) is rational (in fact, an integer) and they appear in Sect. 3.8 in [9]. We ignore such values since for us we know that \(\gamma \) and \(\alpha \) are multiplicatively independent. Since \(\gamma <10^{10}\), Lemma 6 shows that
For all \(X_1\in [5\cdot 10^5,5\cdot 10^9]\), we generated the first 100 partial quotients \([a_0,a_1,\ldots ,a_{99}]\) of each \((\log \gamma )/\log \alpha \). The maximum A of the \(a_k\) all \(k\in [0,99]\) satisfies \(A<10^{20}\). This follows from a calculation explained in Sect. 3.8 in [9].
Hence, by Lemma 3(ii), the left-hand side above exceeds
where \(Q=q_k\) is the maximum denominator of some convergent to \(\log \gamma /\log \alpha \) which is smaller than \(m_3-m_2<8\cdot 10^{15} (\log \gamma )^2<4.3\cdot 10^{18}\) (see Lemma 6 (iii)) since \(\gamma \le 10^{10}\). Thus, we get that
This gives, \(n_2\le 50\). Further, from Lemma 6(i), we see that
Thus, \(m_2=2\). In the case \(m_2=2\), we have \(Y_2=2X_1Y_1=F_{n_2}\), which shows that \(F_{n_2}\) is even so \(3\mid n_2\). Now \(Y_2=2X_1Y_1=F_{n_2}\). Hence, \(X_1\mid F_{n_2}\). For each \(n_2\le 50\) which is a multiple of 3 we took \(X_1\) to be a divisor of \(F_{n_2}/2\) with \(5\cdot 10^5<X_1\). For each one of these choices we generated \(\gamma :=X_1+{\sqrt{X_1^2-\lambda }}\) for \(\lambda \in \{\pm 1\}\) and applied the method from Sect. 3.4. We obtained \(n<230\). To search for solutions to \(Y_m=F_n,\) we again went back for all pairs \((X_1,\lambda )\) satisfying the above condition and generated all values of candidate d such that \(X^2_1-\lambda =dY^2_1\) for some integer \(Y_1.\) For each such \((d,Y_1),\) we set \(Y_2=2X_1Y_1\) and generated the sequence \(\left\{ Y_m \right\} _{m\ge 1}\) where \(Y_{m+2}=2X_1Y_{m+1}-\lambda Y_m\) with m satisfying \(\gamma ^{m-1}/\sqrt{2}<F_{230}.\) We sought to find those values of d for which the set
has cardinality 3. This took a few minutes and produced no solution. There is therefore no solution in this case.
3.10 Better bounds on \(m_3\)
We refer to the left-hand side of (11) which we rewrite as
for the purpose of seeking a better bound on \(m_3\). We note that \(4d>5.\) We apply Mignotte’s Theorem 4 to bound the absolute value above. We have
Here, we consider the equation \(Y_m=F_n\) with \(m>1\). Since \(\gamma \) is large, we have \(m<n\). We have \(D=[{{\mathbb {Q}}}({\sqrt{5}},{\sqrt{d}}):{{\mathbb {Q}}}]=4\).
We can take
Further,
Since
and \({\sqrt{d}}>\alpha ^2\) (in fact \(d\ge 94\)), we get that \(n/\log \gamma <m/\log \alpha \). Thus,
Next, assuming \(m>1000\), we have
Thus,
Since \(e^{0.882} \cdot 0.031<0.075\), we can take
In case (i) of Theorem 4, we get
Thus,
so
giving
In case the maximum involved in \(\log {{\mathcal {B}}}\) is 2.5, we get \(0.075m^2/\log \gamma \le e^{2.5}\), so
In the other case \(\log (0.075\,m^2/\log \gamma )<2\log (0.28\,m/{\sqrt{\log \gamma }})):=\log {{\mathcal {B}}}\), we get that
We get with \(y:=0.28 m/{\sqrt{\log \gamma }}\) that
Hence,
Letting T be the right-hand side above, we have that \(T>(4\cdot 2^2)^2=256\), so we can apply Lemma 4 with \(s=2\) to get
Hence,
Between (21), (22) and (23), the inequality (23) always holds.
We now study Case (ii) of Theorem 4. We may assume that \(r_0\) and \(s_0\) are coprime if not we simply cancel their greatest common divisor. Since \(b_1=1\), we get that \(|r_0|=1,~|s_0|=b_2=m\). Hence,
If \({{\mathcal {B}}}\le e\), then \(m < 110\log \gamma \), so inequality (23) holds, and if \({{\mathcal {B}}}>e\), then
so estimate (21) holds and in particular estimate (23) holds.
We now study Case (iii) of Theorem 4. In this case, we have a relation
In particular, \(r_1\) divides \(s_1t_1b_1\). Since \(b_1=1\) and \(\gcd (r_1,t_1)=1\), we get \(r_1\mid s_1\). Thus, \(\delta =|r_1|\). Putting \(s_1=:r_1s_1'\), we get
where
If \(t_2=0\), we take \(s_1=1\). In particular, \(m<n\le |t_1|<155 (\log \gamma ) (\log {{\mathcal {B}}})^{1/3}\). Again estimate (21) holds and in particular estimate (23) holds. Assume next that we are in the case \(t_2\ne 0\). We multiply both sides of inequality (11) by \(|r_1s_1'|\) and get
Replacing \((r_1s_1')n\) by \(t_2m-(t_1s_1')\), the left-hand side becomes
This has become a linear form in two logarithms to which we can apply Theorem 3. Here,
They are multiplicatively independent (since \(\gamma \) and \(\alpha \) are so), and they satisfy
and
Thus, we can take
and
Since \(m>1000\), the above is bounded by
Thus,
Assume first that the right-hand side above is smaller than \(21/D=21/4\). Then
In particular, estimate (21) holds and in particular estimate (23) holds. So, we assume that the right-hand side of (26) exceeds 21/D. We then get by setting
that
Comparing this with the upper bound in (25), we get
We recall that
and so
Since the left-hand side exceeds \((m-1)\log \gamma -\log {\sqrt{2}}\), we get
so
We can assume \(2\log (0.28\,m/{\sqrt{\log \gamma }})):=\log {{\mathcal {B}}};\) otherwise (23) holds.
Further,
Hence, we get
which gives again with \(y=0.28m/{\sqrt{\log \gamma }}\), that
Hence,
Denoting by T right-hand side above, we have \(T>(4\cdot (8/3)^2)^{8/3}\), so we can apply Lemma 4 with \(s=8/3\) getting
Hence,
This is a bit worse than (23) (by a factor of \(25/1.8\approx 14\)), but we have the additional linear equation (24). Recording that
we can record the following conclusion.
Lemma 9
Let \(\gamma >10^{10}\) and assume that (m, n) is a positive integer solution to \(Y_m=F_n\). Then one of the following holds:
-
(i)
$$\begin{aligned} m<1.8\cdot 10^9 (\log \gamma ) (\log \log \gamma )^2; \end{aligned}$$
or
-
(ii)
$$\begin{aligned} m<2.5\cdot 10^{10} (\log \gamma )(\log \log \gamma )^{8/3}, \end{aligned}$$
but additionally
$$\begin{aligned} a+bn+cm=0, \end{aligned}$$for some integers a, b, c all nonzero (except in the case \(t_1=0\) which has \(a=0\)) with
$$\begin{aligned} |a|\le 110 \log \gamma (\log {{\mathcal {B}}})^{1/3},\quad |b|\le 36 (\log {{\mathcal {B}}})^{1/3},\quad |c|<74(\log \gamma ) (\log {{\mathcal {B}}})^{1/3}, \end{aligned}$$where \(\log {{\mathcal {B}}}\) is bounded above as shown in (28).
In the above, we used \((a,b,c):=(t_1s_1',r_1s_1',-t_2)\).
3.11 A bound on \(m_2\)
We write inequalities (11) for \((m,n)=(m_i,n_i)\) and \(i=2,3\), and combine them to get
We apply Theorem 3 with
We have \(D=4\), \(\log B_1=0.5\log \gamma ,~\log B_2=0.5\log \alpha \) and
Now
Hence,
where the third inequality above was obtained from the second inequality of Lemma (1). Putting \(\Lambda _1\) for the left-hand side of (29) we get
Comparing with (29), we get
Hence,
Thus,
If the maximum is 21/4, we then get \(\log (2.42m_3)<21/4 = 5.25\), so \(m_2<m_3<e^{5.25}/(2.42)<80\). If the maximum is \(\log (2.42m_3)\), we then get
Since \(m_3\ge 3\) and \(376(\log (2.42\cdot 3))^2 > 80\), inequality (30) always holds. Let us record it.
Lemma 10
Let \(\gamma >10^{10}\) and assume that (m, n) is a positive integer solution to \(Y_m=F_n.\) Then
3.12 Bounding \(\gamma \)
We return to (11) which we rewrite as
Now
where
Note that \(|\log (1+\zeta )|<1.01|\zeta |\) for \(|\zeta |<10^{-6}.\)
We also have that
and so \(\alpha ^{n_2}>2.2\gamma \), which gives \(2.08/\alpha ^{2n_2}<{0.42}/{\gamma ^2}\). Thus,
We write the above inequality for \((m,n)=(m_i,n_i)\) with \(i=2,3\), multiply the one for \(i=2\) with \(m_3-1\) and the one for \(i=3\) with \(m_2-1\) and apply the absolute value inequality to get
We rewrite the above as
and treat it as a linear form in two logarithms.
The right-hand side in (31) is \(<6m_3/\gamma ^2\). Assume that
By Lemma 9, we get
which implies \(\gamma <1.5\cdot 10^7\), a case already treated (see Section (3.9)). Thus, we may assume that \(\gamma ^{2-1/10}>6m_3\), so the right-hand side of (31) is smaller that \(1/\gamma ^{1/10}\). Since \(\sqrt{5}F_{n_1}\) and \(\alpha \) are multiplicatively independent (this is true since no rational power of \(\alpha \) is a positive integer larger than 1), we can apply Theorem 3 to bound the left-hand side of (31). Here, we have \(\lambda _1:=\alpha ,~\lambda _2:=\sqrt{5}F_{n_1}\). Thus, \(D=2\), \(\log B_1=1/2.\) Since
We choose \(~\log B_2=(n_1+2)\log \alpha > \log \sqrt{5}F_{n_1}\). Further,
Note that
where we used the fact that \(1/\gamma ^{1/10}<1/10 < \log 2\le \log F_{n_1}\) together with the fact that \(m_2\ge 2\). We recall \((n_1+2)\log \alpha > \log \sqrt{5}F_{n_1}\) to obtain
Thus, Theorem 3 together with (31) give that
Hence,
If the maximum above is at 10.5, we then get
If the maximum above is at \(0.14+\log b'\), then
which gives \(\log \gamma <2.5\cdot 10^8\). Hence, by Lemma 9 again we get
This bound is about \(10^{10}\) times sharper than what Lemma (6) gives on \(m_3.\) Between estimates (32) and (33), we conclude that estimate (33) always holds. Thus, by Lemma 10, we have
Further, by Lemma 8, we get
so \(n_1\le 81\). Let us record what we have proved so far.
Lemma 11
Assuming \(\gamma >10^{10}\) and \(m_1=1,\), we have
We need to reduce the above bounds. Returning to (31), we get
Assume first that \(m_3\le 2m_2\). Then \(m_3\le 2.2\cdot 10^6\) and the right-hand side above is at most \(6\cdot (3m_2)/\gamma ^2<2\cdot 10^7/\gamma ^2\). The quantity on the left-hand side in (34) exceeds \(10^{-68}\) in this range. To compute it, we used the fact that if \(\tau \) is irrational and \(m<x\), then
according to Lemma 3(i), where \([a_0,\ldots ,a_k,\ldots ]\) is the continued fraction expansion of \(\tau \), \(p_k/q_k=[a_0,\ldots ,a_k]\) is the kth convergent to \(\tau \) and k is maximal such that \(q_k<x\). In a matter of seconds, Mathematica generated for each \(n_1\in [2,100]\) (the bound on \(n_1\) is 81 in fact) the largest k such that \(q_k<2.2\cdot 10^6\) for \(\tau =\log \sqrt{5} F_{n_1}/\log \alpha \). The smallest value of the left-hand side in (34) is larger than \(10^{-68}\). We thus get
which gives \(\gamma <6\cdot 10^{37}\). Assume next that \(m_3\ge 2m_2\). In this case, we have that \((m_3+m_2)/(m_3-m_2)\le 3\), and the right-hand side in (34) is at most \(18/\gamma ^2\). A similar calculation as before (with x replaced by \(2\cdot 10^{22}\)) gives that the left-hand side of (34) exceeds \(10^{-100}\). Hence, we get
so \(\gamma <4.3\cdot 10^{50}\). Lemma 9 shows that
Thus,
Further,
Hence, \(n_1\le 51\). We perform another reduction cycle with the already improved bounds. If \(m_3\le 2m_2\), then \(m_3<857180\), the right-hand side in (34) is at most \(1.1\cdot 10^7/\gamma ^2\) and the left-hand side exceeds \(10^{-41}\). Thus, we get
so \(\gamma <1.1\cdot 10^{25}\). Next, assuming \(m_3>2m_2\), the left-hand side in (34) is at most \(18/\gamma ^2\). We calculate again a lower bound on the left-hand sides of (34) (now with \(x=1.9\cdot 10^{14}\)), getting that it exceeds \(3\cdot 10^{-57}\). We thus get
so \(\gamma <5.5\cdot 10^{28}\). Thus,
and
Further, by Lemma 8, we have
so \(n_1\le 50\). After another cycle of reduction, we only obtained the following bounds
Let us summarise what we have proved.
Lemma 12
Assuming \(\gamma >10^{10}\), \(m_1=1\), we have
The only inequality to still justify is the one about \(n_2\) but it follows from
where \(R=\log {\sqrt{2}} +\log 1.1\cdot 10^{13}.\) This implies that \(n_2<5.4\cdot 10^7\).
3.13 The case \(m_2=2\)
Here, \(2X_1=Y_2/Y_1=F_{n_2}/F_{n_1}>\alpha ^{n_2-n_1-1}\)(see Lemma (1)). Since \(2X_1<\gamma +1\le 6.06\cdot 10^{27}\), we get that \(n_2-n_1\le 133\). We get \(n_2\le 195\). Since \(F_{n_2}\) is even, we have that \(n_2\) is a multiple of 3.
For each pair of integers \(n_1\le 50,n_2\le 195\) such that \(X_1=F_{n_2}/(2F_{n_1})\) is an integer satisfying \(5\cdot 10^8<X_1<3.1\cdot 10^{27},\) we generated \(\gamma :=X_1+{\sqrt{X_1^2-\lambda }}\) for \(\lambda \in \{\pm 1\}\) and applied the method from Sect. 3.4. We obtained and \(n_3<240\). To search for solutions to \(Y_m=F_n,\) we again went back for all pairs \((X_1,\lambda )\) satisfying the above condition and generated all values of possible candidate d such that \(X^2_1-\lambda =dY^2_1\) for some integer \(Y_1.\) For each such \((d,Y_1),\) we set \(Y_2=2X_1Y_1\) and generated the sequence \(\left\{ Y_m \right\} _{m\ge 1}\) where \(Y_{m+2}=2X_1Y_{m+1}-\lambda Y_m\) with m satisfying \(\gamma ^{m-1}/\sqrt{2}<F_{240}.\) We sought to find those values of d for which the set
has cardinality 3. This took a few minutes and produced no solution. There is therefore no solution in this case.
From now on, \(m_2>2\).
3.14 The final calculations
We return to (11) which we rewrite as
We rework \(\log (2{\sqrt{d}}F_{n_1})\) in order to get additional terms.
Since \(m_2\ge 3\), we have
so
Hence, (11) yields
which gives
We write the above estimates for \((m,n)=(m_i,n_i)\) for \(i=2,3\) and take a linear combination of them to get, via the absolute value inequality
This yields,
so the expression in the left–hand side above has sign \(-\lambda \). Furthermore,
We study the ratio \((m_3+m_2-2)/(m_3-m_2)\).
-
(i)
If \(m_3\le 2m_2\), by Lemma 12, we get \(m_3\le 2\cdot 404168<808336\). Thus,
$$\begin{aligned} \frac{m_3+m_2-2}{m_3-m_2}<2m_3<1.7\cdot 10^6 \end{aligned}$$in this case. Hence,
$$\begin{aligned} O_{1.51}\left( \frac{m_3+m_2-2}{(m_3-m_2)\gamma ^2}\right) =O_{1.51}\left( \frac{1.7\cdot 10^6}{\gamma ^2}\right) =O_{0.001}\left( \frac{1}{\gamma }\right) \end{aligned}$$in this case since \(1.51\cdot 1.7\cdot 10^6<10^7<\gamma /1000\).
-
(ii)
If \(m_3>2m_2\), then
$$\begin{aligned} \frac{m_3+m_2-2}{m_3-m_2}<\frac{1.5m_3}{m_3/2}=3, \end{aligned}$$so
$$\begin{aligned} O_{1.51}\left( \frac{m_3+m_2-2}{(m_3-m_2)\gamma ^2}\right) =O_{4.53}\left( \frac{1}{\gamma ^2}\right) =O_{1.5\cdot 10^{-9}}\left( \frac{1}{\gamma }\right) . \end{aligned}$$
Hence, in the first case, we get
We use the fact that \(|(1+x)^{1/L}-1|<1.001|x|/L\) valid for all \(x\in (-1/10^{8},1/10^{8})\) and extract square roots getting
Finally, we can take reciprocals getting
By the same argument, in the second case we get that the above estimate holds with \(O_{0.1}(1)\) replaced by \(O_{1.5\cdot 10^{-8}}\). Further, since \(\gamma =2X_1+O_1(1/\gamma )\), we get that
where \(\zeta =O_{0.2}(1)\) in the first case and \(\zeta =O_{1.6\cdot 10^{-8}}(1)\) in the second case. In both cases,
and the expression under the absolute value has the sign \(-\lambda \). It remains to find an efficient process to detect all possible fractions of the form
Well, we return to estimate (36) and note that its right-hand side is
We distinguish several cases.
Case 1. The case \(m_3\le 2m_2\).
In this case, we have \(m_3<808336, \gamma >10^{10}\), so \(\gamma >3(m_3-m_2)\). Thus, the right-hand side in (36) is smaller than
By a well-known criterion of Lagrange (see Lemma 3(i)), the fraction
is a convergent \(p_k/q_k\) of \(\log \sqrt{5} F_{n_1}/\log \alpha \). Here, k is a nonnegative integer such that \(q_k<m_3-m_2<m_3 < 808,336\). Thus, we generate all such possibilities, then we calculate \(X_1\) using formula (38). Having obtained \(X_1\), we use the method from Sect. 3.9 to find a small bound on \(n_2\). The maximum value of the number A there was at most \(A<5.18\cdot 10^{25}\). Thus, inequality (20) gives
so \(n_2<78\). This leads to \(m_2=2\) because \(\gamma > 10^{10}\), but this case was already covered with no solution found.
From now on, we may assume that \(m_3>2m_2\).
Case 2. Considerations about \(\gamma \) and \(m_3\).
If \(\gamma > 2.1m_3\), then again the right-hand side of (36) is smaller than
so Legendre’s result applies leading again to \(n_2\le 78\) and \(m_2=2\).
It remains to deal with the case when \(\gamma <2.1m_3\). Note that in this case both \(m_3>10^{10}/2.1>4.7\cdot 10^9\) and \(m_3-m_2>m_3/2>2.3\cdot 10^9\) are large making direct computations with them ineffective. We write
where a and b are coprime. Thus,
for some positive integer D. Write
Case 3. The case when \(m_3 \le 73\gamma \) and \(m_3 > 2m_2\).
Then the right-hand side in (39) above is at most
Hence, returning to estimate (36), we have
If \(D\ge 150\), then the right-hand side above is smaller than \(1/(2b^2)\), so Legendre’s criterion applies, which was already covered in Case 1. Thus, we may assume that \(D<150\). In this case
The above is a particular instance of an inequality of the form
where \(\tau :=\log \sqrt{5}F_{n_1}/\log \alpha \), \(a/b=((m_2-1)n_3-(m_3-1)n_2)/(m_3-m_2)\) and \(K=10,900\). At this stage we use the following theorem of Worley [21] for the irrational \(\tau \), which generalises Legendre’s result.
Theorem 5
Assume (41) holds. There exist r, s with \(r>0,~s\ge 0\), \(rs<2K\) and \(k\ge 1\) such that
or \(1\le rs<K\), k is such that \(a_{k+1}=1\) and
In case \(s=0\), we can take \(r=1\) (since a and b are coprime) and then we have \(a/b=p_k/q_k\). This is the only case possible when \(K\le 1/2\) since then \(2K\le 1\) so \(rs<1\) giving \(s=0\). This is Legendre’s result. Since we already did the calculation corresponding to this case, we assume that \(rs\ge 1\). Worley describes the number k. In case \(rs\ne 0\), then let \(\ell \ge -1\) be the unique positive integer such that if we write \(\tau =[a_0,a_1,\ldots ,a_\ell ,a_{\ell +1},\ldots ]\) and \(a/b=[a_0,a_1,\ldots ,a_\ell ,b_1,\ldots ,b_u]\) then \(b_1\ne a_{\ell +1}\). In Worley’s notation, let \(\beta =[b_1,\ldots ,b_{u}]\) and \(\gamma =[a_{\ell +1},\ldots ]\). Then \(\beta =b_1+\rho /d\), where \(0\le \rho <d\) and \(b_1\ne a_{\ell +1}\). Worley distinguishes between the cases \(\beta >\gamma \) and \(\beta <\gamma \). In case \(\beta >\gamma \), he shows that \(k=\ell +1\), and \((r,s,\eta )=(d,d(b_1-a_{\ell +1})+\rho ,+1)\) satisfies the required inequality \(rs<2K\) except when \(b_1=a_{\ell +1}+1\) and \(a_{\ell +2}=1\), in which case \(k=\ell +1\) and \((r,s,\eta )=(d,\rho ,+1)\) and \(b=rq_{\ell +2}+sq_{\ell }\). Thus, in this case k is an index such that \(p_k<b<2Kp_{k+1}\) except when \(a_{\ell +2}=1\), in which case the upper bound is slightly worse namely \(b<2Kp_{k+2}<4Kp_{k+1}\). In the case \(\beta \in (0,\gamma )\), Worley distinguishes cases according to the value of \(\min \{\beta ,a_{\ell +1}/2\}\). If \(\beta \le a_{\ell +1}/2\), then he writes \(\beta =m+\rho /d\) and shows that \(k=\ell \) and \((r,s)=(dm+\rho ,d)\) have the required property. This also works when \(m=0\). If \(\beta >a_{\ell +1}/2\), then he writes \(\beta =a_{\ell +1}-m-\rho /d\), and shows that \(k=\ell +1\) and \((r,s,\eta )=(dm+\rho ,d,-1)\) have the required properties. This is the only case of the negative sign. In particular, in this case \(k=\ell \) and \(k=\ell +1\) according to whether the sign is \(+1\) or \(-1\) and at any rate \(q_{\ell -1}+q_\ell /(2K)<b<q_{\ell +1}\). In practice this helps to corner k when doing the calculations. Thus, for \(n_1\in [2,50]\), we apply Worley’s Theorem 5 with \(\tau :=\log \sqrt{5} F_{n_1}/\log \alpha \) and \(K:=10,900\). We calculate the continued fraction and the convergents \(p_k/q_k, ~p_{k-1}/q_{k-1}\) of \(\tau \) such that \(k\le k_{\tau }\), where \(k_{\tau }\) is maximal such that \(q_{k_{\tau }}\le m_3\le 7.16\cdot 10^{13}\). A simple calculation reveals that \(k_{\tau } \le 32\), which is reached at \(n_1 = 4\). Then we generated the fractions
If \(a_{k+1}=1\) we must also consider the fractions of the form
In order to reduce the amount of fractions a/b we consider also the condition (40). Then we used equation (38) to generate all possible candidates \(X_1\) obtainable in this way. As we obtain them we check for the extra condition namely that in estimate (37) we have that the distance for the formula we use to round up to \(X_1\) (distance to the nearest integer \(X_1\)) is smaller than \((1.6/2)\cdot 10^{-8}<10^{-8}\). Furthermore, we are left only with those \(X_1\)’s that make \(\gamma = X_1+\sqrt{X_1^2-\lambda } > 10^{10}\), for \(\lambda \in \{\pm 1\}\). After a few hours of computation we obtained the following list of \(X_1\)’s
Only to these candidates we apply the method from Sect. 3.9. The exact quadruple \((\eta ,n_1,k, X_1)\) resulting from the code is captured in the table below.
\(\eta \) | \(n_1\) | k | \(X_1\) | \(\lambda \) | \(\max \left\{ a_i: i \le 68 \right\} \) |
|---|---|---|---|---|---|
\(-1\) | 5 | 24 | 15048438972 | −1 | 116 |
\(-1\) | 10 | 13 | 58298763342 | −1 | 91 |
\(-1\) | 14 | 8 | 727111625435 | −1 | 55 |
\(-1\) | 29 | 4 | 8581755725011 | −1 | 294 |
\(+1\) | 3 | 23 | 86632618310552 | −1 | 323 |
\(+1\) | 10 | 16 | 25042219341229 | −1 | 69 |
\(+1\) | 16 | 15 | 3375923511388 | −1 | 279 |
\(a_{k+1}=1\) | 30 | 9 | 7538130851888704 | −1 | 82 |
With the values of \(X_1\) so obtained, we used the method from Sect. 3.9 to obtain a small bound on \(n_2.\) From the first 68 partial quotients \([a_0,a_1,\ldots ,a_{67}]\) of each \((\log \gamma )/\log \alpha ,\) we obtained the maximum A of the \(a_k\) for all \(k\in [0,67]\) satisfies \(A<323\). So
By Lemma 1, the above implies
which is impossible for \(m_2 >2\) and \(\gamma >10^{10}\).
From now on, \(m_3>73\gamma \). If \(m_3 < 1.8\cdot 10^9(\log \gamma )(\log (\log \gamma ))^2\), then we get
which gives \(\gamma <5.4\cdot 10^9\), a contradiction.
Hence, we only have treat:
Case 4. The case when \(m_3 > 73\gamma \) and \(m_3 > 1.8\cdot 10^9(\log \gamma )(\log (\log \gamma ))^2\).
Then Lemma 9 shows that,
so \(\gamma < 2.1 \cdot 10^{11}\). We then get
Thus, we get that
for some nonzero integers a, b, c with
In particular, we have the system
The second equation follows from (35). Note that since \(\gamma >10^{10}\), it follows that the error term in the second equation above is \(<10^{-17}\). We solve it with Cramer’s rule for \(m_3-1\) getting
The numerator is in absolute value at most
In the above, we used the fact that \(2<\sqrt{5} F_{n_1}=\sqrt{5}Y_1<\gamma \) which is true since \(d\ge 94\) for \(\gamma \ge 10^{10}.\) Hence,
We thus get
The amount above that involves \(O_{127}\) is at most \(127\cdot 55^{1/3}\cdot m_3^{-1}<10^{-9}\) since \(m_3>73\gamma >7.3\cdot 10^{11}.\) Exponentiating we get
The second equality holds true because
and \(1.001<128/127\). Since \(m_3>73\gamma \), we have
In the above, we used that \(\gamma <2.1\cdot 10^{11}\) and \(\log {{\mathcal {B}}}<55\). Since
and \(|b|\le 36(\log {{\mathcal {B}}})^{1/3}<140, \) we get that
Thus,
This shows that \(X_1\) is determined by pairs of integers |b|, |c| with
and then \(X_1\) is one of the integers of the form \(\lfloor 0.5\alpha ^{|c|/|b|}\rceil +\ell \), where
With these values of \(X_1\), after checking that \(X_1\in [5\cdot 10^9, 1.1\cdot 10^{11}]\) (the upper bound follows since \(X_1<(\gamma +1)/2\) and \(\gamma <2.1\cdot 10^{11}\)), we used the method of Sect. 3.9. We got \(A<9\cdot 10^{21}\). This calculation took a few minutes. So,
Thus, \(n_2<100\). However, by Lemma 1
Since \(\gamma > 10^{10}\), this gives \(m_2=2\), which is a case already treated.
This finishes the proof of the theorem.
Data Availability
Data sharing not applicable to this article as no datasets were generated or analysed during the current study.
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Acknowledgements
The authors thank the anonymous referee for comments which improved the quality of the manuscript. F. L. worked on this paper during a visit at the Max Planck Institute for Software Systems in Saarbrücken, Germany in Spring 2022. He thanks the people of this Institute for hospitality and support. F. S. Z. worked on this project as part of his Ph.D. at Wits funded by an NRF-TWAS fellowship. Computations were performed using High Performance Computing infrastructure provided by the Mathematical Sciences Support unit at the University of the Witwatersrand.
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Luca, F., Zottor, F.S. On Y-coordinates of Pell equations which are Fibonacci numbers. Bol. Soc. Mat. Mex. 29, 49 (2023). https://doi.org/10.1007/s40590-023-00519-x
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DOI: https://doi.org/10.1007/s40590-023-00519-x