Abstract
There are many results in the literature concerning linear combinations of factorials among terms of linear recurrence sequences. Recently, Grossman and Luca provided effective bounds for such terms of binary recurrence sequences. In this paper we show that under certain conditions, even the greatest prime divisor of \(u_n-a_1m_1!-\dots -a_km_k!\) tends to infinity, in an effective way. We give some applications of this result, as well.
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1 Introduction
An integer sequence \(\{u_n\}_{n\ge 0}=\{u_n(r,w,u_0,u_1)\}_{n\ge 0}\) is a binary linear recurrence if the recurrence relation
holds, where \(r,w\in {\mathbb {Z}}\backslash \{0\}\) and \(u_0,u_1\) are integers not both zero. The polynomial \(f(x)=x^2-rx-w\) attached to recurrence (1.1) is called the characteristic polynomial of the sequence \(\{u_n\}_{n\ge 0}\). We denote the discriminant of f by \(\Delta \) and assume that \(\Delta \ne 0\). Let \(\alpha \) and \(\beta \) be the roots of f with the convention that \(|\alpha | \ge |\beta |\). Putting
it is well-known that the formula
For later use we fix the notation
The sequence \(\{u_n\}_{n\ge 0}\) is called non-degenerate, if \(cd\alpha \beta \ne 0\) and \(\alpha /\beta \) is not a root of unity. Taking \(r=w=u_1=1,~ u_0=0\) the sequence \(\{u_n\}_{n \ge 0}\) becomes the classical Fibonacci sequence usually denoted by \(\{F_n\}_{n \ge 0}\) for which \(\alpha =(1+\sqrt{5})/2\) and \(\beta =(1-\sqrt{5})/2\). Grossman and Luca [2] showed that for fixed positive integers A and k, the Diophantine equation
implies that
where \(c_1=c_1(k,A)\) is an effectively computable constant depending only on k and A. Taking \(A=1\) and \(k=2\), it was shown in the same paper that \(F_{12}=4!+5!\) is the largest Fibonacci number which is a sum or difference of two factorials. Further, in [1], it was shown that \(F_7=1!+3!+3!\) is the largest Fibonacci number which is a sum of three factorials.
Let \(S=\{p_1, \ldots , p_l\}\) be a finite set of primes labelled \(p_1< \dots <p_l\) and \(P=p_l(=\max \{p_1, \ldots , p_l\})\). We denote by \({\mathscr {S}}\) the set of all positive integers whose prime factors are all in S. In particular, \(0 \not \in {\mathscr {S}}\) but \(1 \in {\mathscr {S}}\). In [3], the problem of representing \(u_n\) as a sum between a factorial and an element from \({\mathscr {S}}\) was considered. Namely, it was proven that for given integers A, B, the equation
implies that \(n \le c_2\) holds for all solutions n which are non-trivial (see the terminology of that paper for nontrivial; for example, when \(u_n=2^n+1\), the solution with \(m_1=1\), \(s=2^{n}\) for any n when \(S=\{2\}\) is trivial). Here, \(c_2\) is an explicit constant depending only on A, B, S and the sequence \(\textbf{u}=\{u_n\}_{n \ge 0}\). As a numerical application, in [3] it was shown that \(F_{24}=8!+2^53^37^1\) is the largest Fibonacci number of the form \(F_n = \pm m! \pm 2^a3^b5^c7^d\); thus the largest solution when \(\textbf{u}=\{F_n\}_{n\ge 0}\), \(A=B=1\) and \(S=\{2,3,5,7\}\).
2 Our Results
For an integer m denote by P(m) the greatest prime factor of m with the convention that \(P(0)=P(\pm 1)=1\). As before, \(\{u_n\}_{n \ge 0}\) is a non-degenerate binary recurrence sequence. Further, let \(k \ge 1\) and \(A \ge 1\) be fixed positive integers.
In view of Theorem 1 of Grossman and Luca [2] (see (1.5) and (1.6)) we have that \(u_n-\sum _{i=1}^{k}{a_im_i!} \ne 0\) for all integers \(a_1,\ldots ,a_k\) with \(|a_i|\le A\) for \(i=1,\ldots ,k\), whenever \(n>c_1\). Therefore, it is natural to examine the parameter
In this paper, we study the quantity (2.1) when \(\textbf{u}\) has \(\Delta >0\) and \(w=\pm 1\). Without loss of generality (or, replacing A by kA if needed) we assume that the unknowns \(m_i \ (i=1,\ldots ,k)\) satisfy
Our main result below implies that
in an effective way. Namely, we have the following result.
Theorem 2.1
Let \(\{u_n\}_{n \ge 0}\) be a non-degenerate binary sequence with \(\Delta >0\) and \(w=\pm 1\). Assume that \(|a_i| \le A\) for \(i=1,\ldots ,k\) and the unknowns \(m_i\) satisfy (2.2). Put \(c_3:=16(Y+2)\log (Y+2), c_4:=5.6 \cdot 10^{17} \log ^2(Y+2)\) and let \(n_1:=n_1(k)\) be the largest integer solution of the inequality
Then
whenever \(n > c_6\), where
and
As a direct consequence of Theorem 2.1, we have the following result.
Theorem 2.2
Let \(\{u_n\}_{n \ge 0}\) be a non-degenerate binary sequence with \(\Delta >0\) and \(w=\pm 1\) and let \(S=\{p_1,\ldots ,p_l\}\) be a finite set of primes. Put \(P:=\max \{p_1,\ldots ,p_l\}\). We denote by \({\mathscr {S}}\) the set of all rational integers whose prime factors are all in S. Further, let \(k \ge 1\) and \(A \ge 1\) be fixed positive integers. Consider the Diophantine equation
in integer unknowns \((m_1,m_2,\ldots ,m_k,s)\) satisfying \(s \in {\mathscr {S}}\) and (2.2). Then
where \(c_6\) and \(c_4\) are defined in the statement of Theorem 2.1 and
with
Finally, to show the strength of our above result we completely solve a simple equation of the above shape.
Theorem 2.3
Let \(\{F_n\}_{n\ge 0}\) denote the Fibonacci sequence and \(S:=\{2,3,5,7\}\). Denote by \({\mathscr {S}}\) the set of all positive integers which have no prime factor outside of S. Then all solutions of the equation
are given by
3 Linear Forms in p-Adic Logarithms
In this section, we shall present the p-adic version of a lower bound for linear forms in logarithms of algebraic numbers due to Kunrui Yu [10]. We begin by recalling some basic notions from algebraic number theory. For an algebraic number \(\eta \) of degree d over \({\mathbb {Q}}\), we define the absolute logarithmic height of \(\eta \) by the formula
where \(a_{0}\) is the leading coefficient of the minimal polynomial of \(\eta \) over \({\mathbb {Z}}\) and \(\eta ^{(i)}-\)s are the conjugates of \(\eta \) in the field of complex numbers.
Let \({{\mathbb {L}}}\) be an algebraic number field of degree \(d_{\mathbb L}\) and denote by \({\mathcal {O}}_{{\mathbb {L}}}\) the ring of integers of \({\mathbb {L}}\). Let \(\pi \) be a prime ideal in \({\mathcal {O}}_{{\mathbb {L}}}\) and denote by \(e_{\pi }\) the ramification index of \(\pi \), and by \(f_{\pi }\) the residue class degree of \(\pi \). For the unique prime number \(p\in {\mathbb {Z}}\) such that \(\pi \mid p{{\mathcal {O}}}_{{\mathbb {L}}}\), we say that \(\pi \) lies above p. Further, it is well known that
where \(\pi _1,\ldots ,\pi _g\) are prime ideals of \({\mathcal {O}}_{{\mathbb {L}}}\). The prime ideal \(\pi \) is one of the primes \(\pi _i\), say \(\pi _1\), and its \(e_{\pi }\) equals \(e_1\). The number \(f_{\pi }\) is the dimension of the finite field \({\mathcal O}_{{\mathbb {L}}}/\pi \) over its prime field \({\mathbb {Z}}/p{\mathbb {Z}}\), or, equivalently, can be computed via the formula \(\#\left( {\mathcal O}_{{\mathbb {L}}}/\pi \right) =p^{f_{\pi }}\). In the special case \({{\mathbb {L}}}={{\mathbb {Q}}}\) we have \(\pi =p\) and \(d_{\mathbb L}=e_{\pi }=f_{\pi }=1\).
For a non-zero algebraic number \(\gamma \in {\mathbb {L}}\) we write \(\nu _\pi (\gamma )\) for the exponent of \(\pi \) in the factorization in prime ideals of the principal fractional ideal \(\gamma {\mathcal O}_{{\mathbb {L}}}\). It is well known that for every non-zero integer j and prime ideal \(\pi \) of \({\mathcal {O}}_{{\mathbb {L}}}\) lying above the rational prime p we have
Let \(\eta _1,\ldots ,\eta _l\) be non-zero algebraic numbers in \({{\mathbb {L}}}\) and let
where \(d_1,\ldots ,d_l \in {\mathbb {Z}}\).
With the above definitions and notations, Yu [10] proved the following result.
Lemma 3.1
Let \(\pi \) be a prime ideal in \({\mathcal {O}}_{{\mathbb {L}}}\) lying above the rational prime p with the convention that \(\pi =p\) and \(d_{{\mathbb {L}}}=e_{\pi }=f_{\pi }=1\) if \({\mathbb {L}}={\mathbb {Q}}\). Consider the linear form \(\varLambda \) defined by (3.2) and let
and
If \(\Lambda \ne 0\), then
Following Lenstra, Lenstra and Lovász [5], we recall the definition of an LLL-reduced basis of a lattice \({\mathcal {L}}\subset {\mathbb {R}}^n\). For a basis \(\{b_1,b_2,\ldots ,b_n\}\) of the lattice \({\mathcal {L}}\) the Gram-Schmidt procedure provides an orthogonal basis \(\{b_1^{*}, b_2^{*},\ldots ,b_n^{*}\}\) of \({\mathcal {L}}\) with respect to the inner product \(\langle \cdot , \cdot \rangle \) of \({\mathbb {R}}^n\) given inductively by
We call a basis \(\{b_1,b_2,\ldots ,b_n\}\) for a lattice \({\mathcal {L}}\) LLL-reduced if
and
where \(\Vert .\Vert \) denotes the ordinary Euclidean length.
To reduce the initial upper bounds for the parameters, we shall also need the following three standard lemmas.
Lemma 3.2
Let \(b_1, \dots ,b_n\) be an LLL-reduced basis of a lattice \({\mathcal {L}}\subset {\mathbb {R}}^n\). Then \(c_6:=||b_1||^2/2^{n-1}\) is a lower bound for the length of the shortest vector of \({\mathcal {L}}\).
Proof
This is a simplified version of Theorem 5.9 of [9]. \(\square \)
Lemma 3.3
Let \(\alpha _1, \dots , \alpha _n\in {\mathbb {R}}\) be real numbers and \(x_1, \dots , x_n \in {\mathbb {Z}}\) with \(|x_i| \le X_i\). Put \(X_0:=\max X_i\), \(S:=\sum _{i=1}^{n-1} X_i^2\), \(T:=0.5+0.5\cdot \sum _{i=1}^n X_i\) and assume that
holds for some positive real constants \(c_2, c_5, H\) and positive integer q. Let \(C\ge (nX_0)^n\) and let \({\mathcal {L}}\) denote the lattice of \({\mathbb {R}}^n\) generated by the columns of the matrix
Let \(c_6\) denote a lower bound on the length of the shortest non-zero vector of the lattice \({\mathcal {L}}\). If \(c_6^2 > T^2+S\) then we have either
or
Proof
This is Lemma VI.I of [9]. \(\square \)
Lemma 3.4
Let \(z \in {\mathbb {C}}\) with \(|z-1|\le a\in (0,1)\). Then
Proof
This is Lemma B.2 of [9]. \(\square \)
4 Preliminary Results on Binary Recurrence Sequences
The next lemma contains several known results which are very useful for the estimates needed in the paper.
Lemma 4.1
Let \(\{u_n\}_{n \ge 0}\) be a non-degenerate binary recurrence sequence given by (1.3) and let Y defined by (1.4). Then the following hold:
-
(i)
\(\max \{h(\alpha ),h(\beta ), h(\alpha /\beta ), h(c), h(d), h(c/d) \} < 8\log (Y+2).\)
-
(ii)
If \(n>c_3:=16(Y+2)\log (Y+2)\) then \(u_n \ne 0\).
-
(iii)
If \(n>c_3\) then \(|u_n|>|\alpha |^{n-c_{10}\log {n}}\), where \(c_{10}:=4 \cdot 10^{11}\log (Y + 2)\).
Proof
(i) is Lemma 8, (ii) is Lemma 9 and (iii) is Lemma 11 of [3]. \(\square \)
The following lemma is also well-known (see for instance Lemma 1 of [2]) and it provides a lower bound for the p-adic valuation of factorials.
Lemma 4.2
Let p be a prime number and let m be a positive integer. If \(m \ge p\), then \(\nu _p(m!)>\frac{m}{2p}\).
The following lemma is an elementary result due to Pethő and de Weger [7]. It will be used in the proof of Theorem 2.2. For a proof of Lemma 4.3 we refer to Appendix B of [9].
Lemma 4.3
Let \(u,v \ge 0, h \ge 1\) and \(x \in {\mathbb {R}}\) be the largest solution of \(x=u+v(\log {x})^h\). Then
In the proof of Theorem 2.1 we need an upper bound for the quantity of the form \(\nu _p(u_n-t)\), where \(t\in {\mathbb {Z}}\).
Lemma 4.4
Let \(\{u_n\}_{n \ge 0}\) be a non-degenerate sequence given by (1.3) with \(\Delta >0\) and \(w=\pm 1\). Further, let Y defined by (1.4). If \(n>c_3\) then for prime p and integer t with \(u_n \ne t\) we have
where
Proof
We have the representation
where
with \(cd\alpha \beta \ne 0\) and \(\alpha /\beta \) is not a root of unity. Let \({\mathbb {L}}={{\mathbb {Q}}}({\alpha })\). Let p be an arbitrary but fixed prime and denote by \(\pi \) a prime ideal dividing p in \({\mathbb {L}}\). Write \(e_{\pi }\) and \(f_{\pi }\) for the ramification index and for the residue class degree of \(\pi \), respectively. Since \(\alpha \beta =\pm 1\), it is clear that both of \(\alpha \) and \(\beta \) are units in \({\mathbb {L}}={{\mathbb {Q}}}({\alpha })\), and therefore \(\nu _{\pi }(\alpha )=\nu _{\pi }(\beta )=0\).
Suppose first that \(t=0\).
Since \(\nu _{\pi }(\alpha )=0\), we have by (4.3), (3.1), \(e_{\pi } \ge 1\) and the additive property of the function \(\nu _{\pi }\), that
where
Let us bound the quantities of the right hand side of (4.5). Denote by \({\mathcal {N}}({\mathcal {I}})\) the norm of the ideal \({\mathcal {I}}\). By (4.4), we clearly have
and therefore
So,
We next bound \(\nu _{\pi }(\Lambda )\) from above. If \(\Lambda =0\) then \(u_n=0\) also holds, which by our assumption \(n>c_3\) and (ii) of Lemma 4.1 leads to a contradiction. Thus, we may suppose that \(\Lambda \ne 0\). We apply Lemma 3.1 to bound \(\nu _{\pi }(\Lambda )\) on choosing
By (i) of Lemma 4.1, we can take
Applying Lemma 3.1, we get
If \(\max \{\log {p},8\log (Y+2)\}=\log {p}\) we obtain by (4.5), (4.7), (4.8), the fact that \(p \ge 2, Y \ge 1, n>c_3\) and some routine calculations that
while if \(\max \{\log {p},8\log (Y+2)\}=8\log (Y+2)\) we get
leading to a sharper upper bound than stated in the case \(t=0\).
Assume next that \(t \ne 0\). By \(\beta =w\alpha ^{-1}=\pm \alpha ^{-1}\) and (4.3), an easy calculation gives
where
Recall that \({{\mathbb {L}}}={\mathbb {Q}}(\alpha )\) and fix \(w \in \{-1,1\}\) as well as the parity of n. Define the number field \({\mathbb {K}}\) by
It is clear that in both cases \(d_{{\mathbb {K}}}=[{\mathbb {K}}:{\mathbb {Q}}] \le 4\). Let p be a prime and let \(\mathfrak {p}\) be a prime ideal in \({\mathbb {K}}\) dividing p. Write \(e_{\mathfrak {p}}\) and \(f_{\mathfrak {p}}\) for the ramification index and for the residue class degree of \(\mathfrak {p}\), respectively. Using (3.1), Eq. (4.9) gives
Since \(\alpha \) is a unit also in \({\mathbb {K}}\) and \(e_{\mathfrak {p}} \ge 1\), by combining (4.12) with the additivity of the function \(\nu _{\mathfrak {p}}\) we get
Putting \(\Lambda _1:=\alpha ^{-n}z_1-1\) and \(\Lambda _2:=\alpha ^{-n}z_2-1\) inequality (4.13) can be rewritten as
Since \([{\mathbb {K}}:{\mathbb {L}}] \le 2\) and \(\nu _{\mathfrak {p}}(c) \le \nu _{\mathfrak {p}}(u_1-u_0\beta )\) we have
so
We next bound \(\nu _{\mathfrak {p}}(\Lambda _i)\) for \(i=1,2\). It is clear that \(\Lambda _i=0 \ (i=1,2)\) holds if and only if \(\alpha ^n=z_i \ (i=1,2)\), which by (4.9) is equivalent with \(u_n=t\). However, this is excluded. Therefore \(\Lambda _i \ne 0\) for \(i=1,2\). Note, that if \(d_{{\mathbb {K}}}=4\), the number field \({\mathbb {K}}\) is a biquadratic number field. It is well known that in this case \(f_{\mathfrak {p}} \le 2\). Hence, it is clear that for \(\mathfrak {p} \in {\mathbb {K}}\) we are in one of the following cases
In order to bound \(\nu _{\mathfrak {p}}(\Lambda _i)\) for \(i=1,2\) from above we use twice Lemma 3.1 with the parameters
to bound \(\nu _{\mathfrak {p}}(\Lambda _1)\) and with
to bound \(\nu _{\mathfrak {p}}(\Lambda _2)\). By (i) of Lemma 4.1, we have \(h(\alpha )<8\log (Y+2)\) and therefore we may choose in both cases
Further, by combining (4.10) and (i) of Lemma 4.1 with some well-known properties of the absolute logarithmic height function h(.), we may write for \(i=1,2\) that
where \(\gamma =\sqrt{t^2-4 w^n cd}\). Since
we obtain
Thus, a straightforward calculation leads to
Since \(\Delta >0\), we have \(|\alpha |>|\beta |\), which together with \(w=\pm 1=\alpha \beta \) implies that \(|\beta |<1\). Further, since \(\alpha =r-\beta \), we have \(|\alpha | \le |r|+|\beta |<Y+1\), which together with \(\Delta =(\alpha -\beta )^2\) leads to
Now, the combination of (4.19), (1.4) and (4.20) gives
which since \(Y<Y+2,Y \ge 1,|t| \ge 1\) implies that
Since
we get by (4.18), (4.21), \(Y \ge 1\) and \(|t| \ge 1\) that
Thus, (4.22) shows that we may choose in both cases
By (4.16), we may write
Applying Lemma 3.1 we get by (4.14), (4.15) and (4.24) that
which together with \(p \ge 2, Y \ge 1, |t| \ge 1\) and \(n > c_3 \ge 48\log {3}\) leads to the desired upper bound. The proof of Lemma 4.4 is complete. \(\square \)
The next lemma deals with sums of factorials in binary recurrence sequences. This was originally proved by Grossman and Luca (see Theorem 1 of [2]). For our purposes, we need a totally explicit version of Theorem 1 of [2] in the case where \(\Delta >0\) and \(w=\pm 1\).
Lemma 4.5
Let \(\{u_n\}_{n \ge 0}\) be the non-degenerate binary recurrence sequence given by (1.3) with \(\Delta >0\) and \(w=\pm 1\). Further, let \(k \ge 1\) and \(A \ge 1\) be fixed positive integers. Consider the equation
in integer unknowns \((n,m_1,\ldots ,m_k)\) with
Then we have \(n \le \min (c_3,n_0)\), where \(n_0:=n_0(k)\) is the largest positive integer solution of the inequality
Proof
If \(n\le c_3\), then the statement is automatic. So throughout the proof we shall assume that \(n>c_3\). Consider the Eq. (4.25) satisfying assumption (4.26). We may assume that there is no vanishing subsum on the right hand side of (4.25); that is, that we have
for each non-empty subset \(I \subset \{1,2,\ldots ,k\}\). Note, that (4.27) can be assumed without loss of generality, since otherwise we obtain an equation similar to (4.25) with fewer terms.
For \(j=1,\ldots ,k\) put
We show by induction that
For \(j=1\) we have \(|4N_j|=|4a_km_k!|\). Further, since \(n>c_3\), by applying Lemma 4.4 with \(t=0\) and \(p=2\), we obtain
Further, by (4.26) it is clear that
If \(m_k \ge 4\cdot (4c_4\log {n})\) then Lemma 4.2 yields \(\nu _2(m_k!) > 4c_4\log {n}\), contradicting (4.30). Thus, \(m_k < 4\cdot (4c_4\log {n})=16c_4\log {n}\), whence
We may assume that \(n>16c_4\), since otherwise we obtain \(n \le 16c_4\), which is better than the stated inequality. Since for \(n>c_3 (\ge 48\log {3})\) one has \(\log {\log {n}}/\log n<0.35\) we may write by (4.31) that
Suppose now, that (4.29) holds for some \(1 \le j < k\). By rewriting (4.25) in the form
we easily see by (4.27) that the right hand side of (4.33) is nonzero and hence \(u_n \ne N_j\). Further, by (4.27) \(N_j \ne 0\) also holds. Therefore, we may apply Lemma 4.4 with \(p=2\) and \(t=N_j\). We obtain that
Further, by (4.26) it is clear that
If \(m_{k-j} \ge 4 \cdot 4c_4\log (|4N_j|)\log n\) then Lemma 4.2 yields
contradicting (4.34). Thus, \(m_{k-j} < 16c_4\log (|4N_j|)\log {n}\), whence
We may assume that \(n > 16c_4\log (|4N_j|)\). Indeed, if \(n \le 16c_4\log (|4N_j|)\), we then get by (4.29) that
Further, since \(j \le k-1\) the above inequality implies that
which is better than the stated bound for n. Since for \(n>c_3 (\ge 48\log {3})\) one has \(\log {\log {n}}/\log n<0.35\) we may write by \(n > 16c_4\log (|4N_j|)\) and (4.35) that
It is clear that
and hence
Thus,
which is equivalent to
Now, (4.37) implies that
which leads to
Since \(A \ge 1, |N_j| \ge 1\) and \(21.6c_4(\log ^2{n})-1 \ge 1\) one has
which by (4.38) yields
The combination of (4.29) and (4.39) gives
Finally, since
we obtain by (4.40) that
finishing the induction.
Recall that by assumption \(n>c_3\), which guarantees that \(u_n \ne 0\). The above inductive argument together with (iii) of Lemma 4.1 shows that
whence
which together with \(n>c_3, \log (4A)>0, |\alpha | \ge (1+\sqrt{5})/2, k \ge 1\) and the definitions of \(c_{10}\) and \(c_4\) implies
The lemma is proved. \(\square \)
5 Proof of Theorem 2.1
Proof of Theorem 2.1
Suppose that
Since \(n_1 \ge n_0\), Lemma 4.5 implies that
Thus, we may write
where \(s \ne 0\) is some integer, \(|a_i| \le A\) and
We let \(P:=P(s)\). By employing an inductive argument similar to the one applied in Lemma 4.5, we derive an explicit upper bound for n in terms of P, A, k and Y in equation (5.2), leading to an explicit lower bound for P and therefore also for \(P(u_n-(a_1m_1!+\cdots +a_km_k!))\).
We may assume without loss of generality that there is no vanishing subsum on the right hand side of (5.2), that is that
holds for each non-empty subset \(I \subset \{1,2,\ldots ,k\}\) and each \(\delta \in \{0,1\}\). Indeed, if there is an index set \(I \subset \{1,2,\ldots ,k\}\) and \(\delta \in \{0,1\}\) such that
then (5.2) implies that
Now, (5.5) shows that for \(\delta =1\) we obtain an equation similar to (4.25) which for \(n>n_0\) cannot happen, while for \(\delta =0\) we get an equation similar to (5.2) with fewer terms.
If \(|s|=m_i!\) for some \(i=1,\ldots ,k\), then (5.2) leads to an equation of the form
which by Lemma 4.5 gives \(n \le n_1\), which is a contradiction in view of (5.1).
Put \(m_{k+1}:=0\) and let \(m_0\) be such that \(\max \{|s|,m_1!\}<m_0!\). By (5.1), there exists an index \(0 \le i_0 \le k\) such that
and for \(i=1,\ldots ,k+1\) we put
For \(j=1,\ldots ,k+1\), we set
We show by induction on j that for \(1 \le j \le k+1\)
For \(j=1\) we easily see that
Case 1. \(|s|<m_k!\).
In this case \(N_1=s\). Recall that \(P=\max \{p: p\mid s\}\). By (5.1), we have that \(u_n \ne 0\) and therefore by applying Lemma 4.4 with \(t=0\) and with each prime factor \(p \mid s\), we obtain \(\nu _p(u_n)<c_4p^2\log {n}\), which yields
On using (5.3) for every prime p we infer that
If \(m_k \ge 2P(c_4P^2\log {n})=2c_4P^3\log {n}\), then Lemma 4.2 and \(p \le P\) show that
holds for every \(p \mid s\), which by (5.2), (5.10) and (5.11) forces
to hold for every \(p \mid s\). Thus, (5.12) and (5.10) imply
Since \(\pi (P)<{2P}/{\log {P}}\) (see Corollary 1 in [8]), the above inequality leads to
Suppose now that \(m_k<2c_4P^3\log {n}\). Then since \(m_k! \le m_k^{m_k}\), we may write
If \(n<2c_4P^3\), then \(P>n^{1/3}(2c_4)^{-1/3}\), which is a sharper lower bound for P than stated. Therefore, we may assume that \(n \ge 2c_4P^3\) which combined with (5.14) and with \(\log {\log {n}}/\log {n}<0.35\) which holds for \(n>n_1\), we get
Since \(\log (|4N_1|)= \log (|4s|)\), we obtain from \(|s|<m_k!\) and (5.15) that
Case 2. \(|s|>m_k!\).
In this case, \(N_1=a_km_k!\). If \(m_k \ge 2c_4P^3\log {n}\) then using the same argument as in Case 1, we get that \(\log (4|s|)<\log {4}+2c_4P^3\log {n}\), which together with \(m_k!<|s|\) and \(|a_k| \le A\) implies that
Assume now that \(m_k<2c_4P^3\log {n}\). Then by the argument applied in the corresponding part of Case 1, we obtain
Finally, (5.13), (5.16), (5.17) and (5.18) show that in fact the bound occurring in (5.18) is appropriate for all cases proving the assertion for \(j=1\).
Assume now that (5.9) holds for some \(1 \le j<k+1\). Rewrite (5.2) as
where \(\delta \in \{0,1\}\) and \(\ell :=\ell (\delta ,j,k)=k+1-j-\delta \). It is clear that \(N_{j+1}=N_j+t_{j+1}\), where
Further, \(N_j \ne 0\) and \(u_n-N_j \ne 0\) hold in view of (5.4). Thus, we apply Lemma 4.4 with \(t=N_j\) to obtain \(\nu _p(u_n-N_j)<c_4p^2\log (|4N_j|)\log {n}\), for every prime p. If \(p \mid s\) then \(p \le P\), so
Using (5.3), we get that
We wish to estimate \(\log {|4t_{j+1}|}\). To do so, we split the proof into three cases according to the value of \(t_{j+1}\) (see 5.20).
Assume first that \(\delta =1 \ \text {and} \ |s|<m_{\ell }!\).
Then \(t_{j+1}=s\). If \(m_{\ell } \ge 2P(c_4P^2\log (|4N_j|)\log {n})=2c_4P^3\log (|4N_j|)\log {n}\) then Lemma 4.2 and \(p \le P\) shows that
which by (5.19), (5.21) and (5.22) forces
Thus, (5.21) and \(p \le P\) imply
Since \(\pi (P)<2P/\log P\), the above inequality leads to
Suppose now that \(m_{\ell } < 2c_4P^3\log (|4N_j|)\log {n}\). Then, by the same argument as in the corresponding part of the case \(j=1\), we obtain
If \(n<2c_4P^3\log (|4N_j|)\) then (5.9); i.e., the induction hypothesis and \(j \le k\) yield
which leads to a sharper lower bound for P than stated. Therefore, we may assume that \(n \ge 2c_4P^3\log (|4N_j|)\), which by (5.25), \(|s|<m_{\ell }!\) and \(\log (\log {n})/\log {n}<0.35\) gives
Suppose now that \(\delta =1 \ \text {and} \ |s|>m_{\ell }!\).
In this case, we have \(t_{j+1}=a_{\ell }m_{\ell }!\). If \(m_{\ell } \ge 2c_4P^3\log (|4N_j|)\log {n}\), then by the same argument as in the corresponding part of the previous case, we obtain that \(\log (|4s|)\) is “small” (by “small” we mean a quantity bounded polynomially in both P and \(\log n\)), that is
which by \(|4a_{\ell }s|>|4a_{\ell }m_{\ell }!|\) gives
Assume now that \(m_{\ell }< 2c_4P^3\log (|4N_j|)\log {n}\). By the same argument as in the corresponding part of the previous case, we obtain that \(\log (|4a_{\ell }m_{\ell }!|)\) is “small”, that is
Finally, suppose that \(\delta =0\). Then it is straightforward that \(t_{j+1}=a_{\ell }m_{\ell }!\). We apply Lemma 4.4 with \(t=N_j\) and with some prime \(p_1 \mid s\). We get
By (5.3) and (5.19) (with \(\delta =0\)) it is clear that for \(p_1\) (actually for each prime \(p\mid s\)), we have
If \(m_{\ell } \ge 2c_4P^3\log (|4N_j|)\log {n}\) then Lemma 4.2 and \(p_1 \le P\) shows that
which is a contradiction in view of (5.29) and (5.30). Therefore, we may suppose that \(m_{\ell } < 2c_4P^3\log (|4N_j|)\log {n}\). By the same argument as in the corresponding part of the previous case, we obtain that \(\log (|4a_{\ell }m_{\ell }!|)\) is “small”, that is
Now (5.24), (5.26), (5.27), (5.28) and (5.29) show that in fact the bound occurring in (5.29) is appropriate for all cases proving that
Since \(N_{j+1}=N_j+t_{j+1}\), we obtain by (5.32) and the triangle inequality that
whence
Inequality (5.33) leads to
which by \(A \ge 1, |N_j| \ge 1, P \ge 2, n>n_1\) and \(c_4 \ge 5.6 \cdot 10^{17}\log ^2{3}\) yields
Further, by the combination of (5.34) with the inductive hypothesis (5.9), we infer that
Since for every \(u,v \in {\mathbb {R}}\) with \(u>1,v>1\) and every integer \(j \ge 1\) one has \(u+v(u+v)^j<(u+v)^{j+1}-1\), we get by (5.35)
whence
finishing the induction.
Recall that \(n>c_6=\max \{n_1,c_3\}\), which guarantees that \(u_n \ne 0\). The above inductive argument together with (iii) of Lemma 4.1 shows that
leading to
which since \(n>c_3, \log (4A)>0, |\alpha | \ge (1+\sqrt{5})/2, k \ge 1\) and the definitions of \(c_{10}\) and \(c_4\) implies that
Finally, (5.37) and \(P \ge 2\) yield
which is equivalent to
The theorem is proved. \(\square \)
Remark
For the equation
where \(F_n\) is the Fibonacci sequence, we may use (5.38) (or (5.39)) with \(k=2, A=1, P=7, Y=1, c_4=5.6 \cdot 10^{17} \log ^2(Y+2)=5.6 \cdot 10^{17}\log ^2{3}\) to obtain
6 Proof of Theorem 2.2
Proof
It is enough to show that assumption \(n>c_6\) implies \(n < c_8\) yielding the desired upper bound (2.5); i.e., \(n \le \max \{c_6,c_8\}=c_7\).
Suppose that \(n>c_6\) and rewrite Eq. (2.4) in the form
We investigate the quantity \(P\left( u_n-\sum _{i=1}^{k}{a_i m_i!}\right) \); i.e., the greatest prime divisor of \(u_n-\sum _{i=1}^{k}{a_i m_i!}\). On one hand, by (6.1) we have
On the other hand, since \(n>c_6\), Theorem 2.1 gives
where \(c_5(n)\) is defined in the statement of Theorem 2.1. Now, the combination of (6.2) and (6.3) yields
By \(A \ge 1, n>c_6\) and the definition of \(c_4\) we easily see that
which together with (6.4) yields
with
Finally, by applying Lemma 4.3 to (6.5) with the parameters
we obtain that \(n < c_8\), where
The theorem is proved. \(\square \)
7 Preliminary Results on Fibonacci and Lucas Numbers
The recurrence sequence \(\{F_n\}_{n\ge 0}\) defined by
is called the Fibonacci sequence, and the elements belonging to this sequence are called Fibonacci numbers. The recurrence sequence \(L_n\) given by
is called the companion sequence of the Fibonacci sequence, and the elements belonging to this sequence are called Lucas numbers. We have \(\alpha =(1+{\sqrt{5}})/2\) and \(\beta =(1-{\sqrt{5}})/2\) for the above sequences.
In this section we collect results about Fibonacci and Lucas numbers which are needed in the proof of Theorem 2.3.
Lemma 7.1
Let \(F_n\) denote the \(n^\textrm{th}\) Fibonacci number.
-
(1)
\(2\mid F_n \iff 3 \mid n\);
-
(2)
for \(k\ge 2\) we have \(2^k\mid F_n\iff 3\cdot 2^{k-2}\mid n\);
-
(3)
\(3^k\mid F_n\iff 4\cdot 3^{k-1}\mid n\);
-
(4)
\(5^k\mid F_n\iff 5^k\mid n\);
-
(5)
\(7^k\mid F_n\iff 8\cdot 7^{k-1}\mid n\);
-
(6)
\(11^k \mid F_n \iff 10\cdot 11^{k-1} \mid n\);
-
(7)
\(13^k \mid F_n \iff 7\cdot 13^{k-1} \mid n\);
-
(8)
\(17^k \mid F_n \iff 9\cdot 17^{k-1} \mid n\);
-
(9)
\(19^k \mid F_n \iff 18\cdot 19^{k-1} \mid n\);
-
(10)
\(29^k \mid F_n \iff 14\cdot 29^{k-1} \mid n\).
Proof
This is a simple consequence of the Main Theorem, Lemma 1 and Lemma 2 of [4]. \(\square \)
Lemma 7.2
Let \(L_n\) denote the \(n^\textrm{th}\) Lucas number. Then
Proof
This is a simple consequence of Lemma 2 of [4]. \(\square \)
Lemma 7.3
Let N be a positive integer not of the form \(F_m\) for some positive integer m. Then for all positive integers \(n \ge 3\) one has
Proof
This is Lemma 1 of [1]. \(\square \)
Lemma 7.4
Let \(n\ge 0\) be an integer and \(m\in \{ 3,4,5,6,7,8,9,10, 12, 14, 18 \}\). Assume that
Then the the parity of n and m must be the same.
Proof
We have the following cases to consider:
-
if \(m=3\) then \(F_n\equiv 2 \pmod 8\) which by Lemma 7.1 implies that \(n\equiv \pm 3 \pmod {12}\);
-
if \(m=4\) then \(3 \mid F_n\) which by Lemma 7.1 implies \(4\mid n\);
-
if \(m=5\) then \(5\mid F_n\), which by Lemma 7.1 implies \(5\mid n\). If n would be even, then by \(10\mid n\) we would get \(11\mid F_n\) and since \(11\mid 30!\) this contradicts the fact \(F_m=5\), consequently n must be odd;
-
if \(m=6\) then we have \(8\mid F_n\) which by Lemma 7.1 implies \(6\mid n\);
-
if \(m=7\) then we have \(13 \mid F_n\) which by Lemma 7.1 implies \(7\mid n\) and if n would be even, then we would have \(14\mid n\) implying \(29 \mid F_n\) which together with \(F_m =13\) contradicts \(30! \mid F_n-F_m\);
-
if \(m=8\) then we have \(7\mid F_n\) which by Lemma 7.1 implies \(8\mid n\);
-
if \(m=9\) then we have \(17\mid F_n\) which by Lemma 7.1 implies \(9\mid n\) and if n would be even, then we would have \(18\mid n\) implying \(19 \mid F_n\) which together with \(F_m =34\) contradicts \(30! \mid F_n-F_m\);
-
if \(m=10\) then we have \(11 \mid F_n\) which by Lemma 7.1 implies \(10\mid n\);
-
if \(m=12\) then we have \(16 \mid F_n\) which by Lemma 7.1 implies \(12\mid n\);
-
if \(m=14\) then we have \(29 \mid F_n\) which by Lemma 7.1 implies \(14\mid n\);
-
if \(m=18\) then we have \(8 \mid F_n\) which by Lemma 7.1 implies \(6\mid n\).
Thus, we have proved that the parity of n and m must be the same. \(\square \)
Lemma 7.5
Let \(m \le n\) be two nonnegative integers such that \(m \equiv n \pmod 2\). Let \(\delta :=(-1)^{(m-n)/2}\). Then,
Proof
See Lemma 2 of [6]. \(\square \)
Lemma 7.6
Let \(F_n\) denote the Fibonacci sequence.
(i) Assume that \((p,k)\in \{ (2,267), (3,168), (5,114), (7,95) \}\) and let \(m_2\) be an integer with \(1\le m_2 \le 600\). Then the congruence
has no solutions in integers \(4\le n \le 10^{77}\).
(ii) Assume that \((p,k)\in \{ (2,56), (3,36), (5,26), (7,21) \}\) and let \(m_2\) be an integer with \(1\le m_2 \le 600\). Then the congruence
has no solutions in integers \(4\le n \le 10^{15}\).
Proof
-
(i)
The problem is finite since all parameters and unknowns in the congruence are bounded. However, a direct computation is not possible due to the size of the range of n. Thus, we used the constructive method indicated below. For given \(m_2\) and p we first we solved the congruence
$$\begin{aligned} F_n\equiv m_2! \pmod p \end{aligned}$$by checking all values of \(0\le n \le \pi (p)\), where \(\pi (p)\) denotes the pth Pisano-period. Then we worked inductively. If the solutions of the congruence
$$\begin{aligned} F_n\equiv m_2! \pmod {p^u} \end{aligned}$$are \(s_1, \dots , s_t\) modulo \(\pi (p)p^{u-1}\) then the solutions of the congruence
$$\begin{aligned} F_n\equiv m_2! \pmod {p^{u+1}} \end{aligned}$$(7.1)must be among \(s_i+j\pi (p)p^{u-1}\) (\(i=1, \dots , t\), \(j=0,1,\dots , p-1\)) modulo \(\pi (p)p^u\). Here one must be careful again, since computing the Fibonacci number of index \(s_i+j\pi (p)p^{u-1}\) after a while is not possible, so instead we computed recursively the values \(\alpha ^{s_i+j\pi (p)p^{u}} \pmod {I_u}\) and \(\beta ^{s_i+j\pi (p)p^{u}} \pmod {I_u}\) where \(\alpha \) and \(\beta \) are the roots of the companion polynomial of the Fibonacci sequence and \(I_u\) denotes the ideal of the ring of integers of \({\mathbb {Q}}(\alpha )\) generated by \(p^u\) for \(u=1,\dots , k\). Clearly as \(\alpha ^s_i \pmod {I_u}\) and \(\alpha ^{j\pi (p)p^{u-1}} \pmod {I_u}\) were already computed in the previous step, we only raised \(\alpha ^{j\pi (p)p^{u-1}} \pmod {I_u}\) to power p and multiplied the result by \(\alpha ^s_i \pmod {I_u}\) to obtain \(\alpha ^{s_i+j\pi (p)p^{u}} \pmod {I_u}\), and did the same for \(\beta \). This procedure worked fast, and we could check the congruence
$$\begin{aligned} \alpha ^{s_i+j\pi (p)p^{u}}-\beta ^{s_i+j\pi (p)p^{u}} \equiv (\alpha -\beta )m_2! \pmod {I_{u+1}} \end{aligned}$$to decide whether \(s_i+j\pi (p)p^{u-1}\) is a solution of (7.1) or not. The above algorithm programmed in Magma proved our assertion for given \(m_2, p, k\) in under a few seconds.
-
(ii)
The very same algorithm proves this statement in even less running time. \(\square \)
8 Proof of Theorem 2.3
Proof
By (5.40) we infer that for any solution of the equation (2.6) we must have
We will split the analysis into cases.
Case I. Assume \(m_1!\ge \sqrt{F_n}\).
Then we have
and we further split our treatment of Case I. into subcases:
Case I(1). Assume \(m_1\le 10^4\).
Then we have
and by (8.1) we obtain
By Lemma 7.1, we have
Case I(1)(i). Assume \(m_2\ge 49\).
Then we clearly have
Thus, Eq. (2.6) implies
Now we compute the list \({\mathcal {L}}\) of all values
and we check for each \(1\le n \le 3.828 \cdot 10^5\) whether \(F_n-s \in {\mathcal {L}}\), where
Since the size of \({\mathcal {L}}\) and the number of values for \(F_n\) is large, and also the values with which we need to do arithmetic are too large, instead of checking equality we check congruences
for \(p=20011, 20021, 20023\). Denote the list \({\mathcal {L}}\) mod p by \({\mathcal {L}}_p\). First for every \(u=0,1,\ldots ,20010\) we collected all indices i such that \({\mathcal {L}}_{20011}[i]=u\) in a list \({\mathcal {J}}_u\). Then for the smallest positive residue \(u\equiv F_n-s\pmod {20011}\) and for all indices j in \({\mathcal {J}}[u]\) we checked if \({\mathcal {L}}_{20021}[j] \equiv F_n-s \pmod {20021}\) and \({\mathcal {L}}_{20023}[j] \equiv F_n-s \pmod {20023}\) holds. If for all j in \({\mathcal {J}}[u]\) one of the above congruences was false, then we excluded n from the list of possible solutions (at least in this case). The computation took 1085 s on an Intel Xeon W-2245 3.90GHz CPU processor and the only values for n which were not excluded by this procedure were \(n= 198489, 228652, 375659\). Then, as explained above, \(s=2^{\nu _2(F_n)}3^{\nu _3(F_n)}5^{\nu _5(F_n)}7^{\nu _7(F_n)}\) is fixed and we computed the value \(F_n-s\). If \(F_n-s=m_1!+m_2!\) with \(m_1>m_2\), then \(m_1!\) is the largest factorial which is smaller than \(F_n-s\), and we checked that \(F_n-s-m_1!\) is not a factorial, thus excluding that value of n, too. In the three remaining cases we obtained the following data:
n | s | \(m_1\) |
---|---|---|
198489 | 2 | 11444 |
228652 | 3 | 12987 |
375659 | 1 | 20271 |
and we conclude that none of the values \(n= 198489, 228652, 375659\) is a solution in this case.
Case I(1)(ii). Assume \(1\le m_2\le 48\) and \(m_1\ge 56\).
Then we additionally have \(m_1-m_2 \ge 8\) which clearly implies
Thus, whenever for every \(p\in S\) either \(\nu _p(m_2!)\ne \nu _p(F_n)\) or
then we must have
Thus, s is explicitly given. So, we compute \(F_n-s\) and exclude all such values of n for which \(F_n-s\) is not the sum of two factorials, as we did it in Case I(1)(i). There are 1338980 cases when the pairs \((n,m_2)\) do not fulfill the above conditions. For each such pair \((n,m_2)\) we compute for each \(p\in S\) the value \(\nu _p(F_n-m_2!)\) and we see that
which implies that
Thus, also in these cases s is explicitly given, and then we compute \(F_n-s\) and exclude all such values of n for which \(F_n-s\) is not the sum of two factorials, as we did it in Case I(1)(i). There are only 3 cases where the above procedure does not work, namely \((n,m_2)=(1,1),(2,1),(3,2)\), when we do have \(F_n=m_2!\), which clearly cannot lead to a solution. The computation of this case took 2018 s on a Intel Xeon W-2245 3.90 GHz CPU processor.
Case I(1)(iii). Assume \(1\le m_2\le 48\) and \(m_1\le 55\).
Then
and by (8.1) we obtain
Now for \(n<920\), \(1\le m_2\le 48\) and \(m_2<m_1\le 55\) we check whether
and if yes, then we have found a solution of our equation. Altogether, we found the solutions listed in Theorem 2.3. This case had a running time of a few seconds. (Clearly, one could also check for the condition \(F_n\le (m_1!)^2\) if interested only on the solutions belonging to Case I.)
Case I(2). Assume \(m_1>10^4\).
In this case we still have (8.1) (i.e. \(F_n\le (m_1!)^2\)) since we are in a subcase of Case I. Further, recall that by (5.40) all solutions of the Eq. (2.6) have
This together with Lemma 7.1 shows that
Case I(2)(i). Assume \(m_1>10^4\) and \(m_2\ge 600\).
Then we have
This proves that we again have (8.2) implying that
Using Lemma 7.3, we obtain
This gives
Thus, \(m_2\le \frac{8}{7} \cdot 10^{10.9}<10^{11}\) and
Hence,
Now we use again Lemma 7.3 to obtain
and consequently
We get \( m_1< 2\cdot 10^{20.45}<10^{20.8}\) and this implies
Now we get
so
We repeat the above procedure. Using Lemma 7.1, this shows that
By the assumption \(m_2\ge 600\) we get again (8.4) so equations (8.2) hold implying that
Now using a short computer program we consider the equation
modulo primes between 100 and 600. For each such prime p we have
and the computer search shows that this congruence is fulfilled simultaneously for all primes between 100 and 600 if and only if
That is, we must have
Now our Eq. (2.6) takes the form
with \(m=1,2,3,4,5,6,8,12\). By Lemma 7.4, in Eq. (8.6) the parity of n and m must be the same. So, we can use Lemma 7.5 and we obtain
where \(\delta =\pm 1\). Recall that since \(m_2\ge 600\), we have
and since \(\nu _2(L_k)\le 2\) (see Lemma 7.2), we obtain that
However, this shows that
which contradicts (8.3). So, we have shown that in Case I(2)(i) our equation has no solution.
Case I(2)(ii) Assume that \(m_1>10^4\) and \(m_2 < 600\).
Now we show that in this case
For if not assume for example that \(\nu _2(s)\ge 267\). Consider the Eq. (2.6) as a congruence modulo \(2^{267}\). Thus we obtain that for any solution of (2.6) fulfilling the conditions of this subcase we have
However, by Lemma 7.6 (i), this has no solutions with \(4\le n\le 10^{77}\). But solutions with \(n>10^{77}\) do not exist at all, and since \(m_1\) is large \(n<4\) also cannot happen in this case. So we conclude that if there exists a solution in the present subcase, then it must have \(\nu _2(s)< 267\). A similar reasoning proves the other inequalities of (8.7).
Now since \(s\le 2^{267}3^{168}5^{114}7^{95}\) we may use again the ideas implemented in Case I(2)(i). We have
and using again Lemma 7.3 we obtain
Consequently,
so we get \( m_1< 2\cdot 10^{11.55}<10^{12}\). This implies
Now we conclude by
so
Using Lemma 7.6 (ii) the same way as we used its statement (i) at the beginning of this case, we obtain that
Now using a short computer program as in Case I(2)(i) we considered the equation
modulo primes between 100 and 800. For each such prime p we have
and the computer search shows that this congruence is fulfilled simultaneously for all primes between 100 and 800 if and only if
That is, we must have
The running time for this computation was 112 s on a Intel Xeon W-2245 3.90GHz CPU processor.
Now our Eq. (2.6) takes the form
with \(m=3,4,5,6,7,8,9,10, 12, 14, 18\). By Lemma 7.4, in Eq. (8.9) the parity of n and m must be the same. So, we can use Lemma 7.5 and we obtain
where \(\delta =\pm 1\). Recall that by \(m_1\ge 10^4\) we have
and since by Lemma 7.2 we have \(\nu _2(L_k)\le 2\), we obtain that
However, this shows that
which contradicts (8.3). So we have shown that in Case I(2)(i) our equation has no solution.
Case II. We assume \(m_1!\le {\sqrt{F_n}}\).
Then from Eq. (2.6) with the notation \(s=2^a3^b5^c7^d\) we obtain
and by the condition of Case II we have
We clearly may assume that \(n>10\), so \(4/\alpha ^{n/2}<0.4\). Now using Lemma 3.4 we infer that
The conditions of Lemma 3.3 are fulfilled with
and
Choosing \(C=10^{400}\) and using the LLL-algorithm implemented in Magma we obtain an LLL-reduced basis of \({\mathcal {L}}\). By Lemma 3.2 we get a lower bound \(c_6\) for the length of the shortest vector of \({\mathcal {L}}\). Finally, Lemma 3.3 provides the upper bound \(H\le 3077\). Now using Lemma 3.3 with
by the above procedure we infer that \(H\le 219\). Now we use once more Lemma 3.3 with \(H=220\), \(X_0=2\cdot 220+1\), \(C=10^{24}\), and by the above procedure we get
This shows that \(n<200\) and consequently, \(m_2<m_1<36\). So to conclude the proof of our theorem for all natural numbers \(n < 200\) and \(m_2<m_1<36\) with \(F_n-m_1!-m_2!>0\) we check whether there exist \(a,b,c,d\in {\mathbb {N}}\) such that
and we get exactly the solutions listed in Theorem 2.3. (Clearly, one could also check the condition \(F_n > (m_1!)^2\) if interested only in the solutions belonging to Case II). \(\square \)
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We thank the anonymous referee for comments which improved the quality of our manuscript.
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Open access funding provided by University of the Witwatersrand. The research of Authors A.B., L.H. and I. P. was supported in part by the Eőtvős Loránd Research Network (ELKH), by the NKFIH Grants 128088 and 130909, and the Projects EFOP-3.6.1-16-2016-00022 cofinanced by the European Union and the European Social Fund.
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Bérczes, A., Hajdu, L., Luca, F. et al. Additive Diophantine Equations with Binary Recurrences, \({{\mathcal {S}}}\)-Units and Several Factorials. Results Math 78, 116 (2023). https://doi.org/10.1007/s00025-023-01871-0
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DOI: https://doi.org/10.1007/s00025-023-01871-0