Correction to: X -coordinates of Pell equations as sums of two Tribonacci numbers
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Correction to: X-coordinates of Pell equations as sums of two Tribonacci numbers Eric F. Bravo1 · Carlos Alexis Gómez Ruiz1 · Florian Luca2,3,4 © Akadémiai Kiadó, Budapest, Hungary 2019
Correction to: Periodica Mathematica Hungarica https://doi.org/10.1007/s10998-017-0226-8 Abstract In this work, we correct an oversight from [1].
1 Introduction For a positive squarefree positive integer d and the Pell equation X 2 − dY 2 = ± 1,√ where X, Y ∈ √ Z+ , it is well known that all its solutions (X , Y ) have the form X + Y d = √ X k + Yk d = (X 1 + Y1 d)k for some k ∈ Z+ , where (X 1 , Y1 ) is the smallest positive integer solution. Let {Tn }n≥0 be the Tribonacci sequence given by T0 = 0, T1 = T2 = 1, Tn+3 = Tn+2 + Tn+1 + Tn for all n ≥ 0. Let U = {Tn + Tm : n ≥ m ≥ 0} be the set of non-negative integers which are sums of two Tribonacci numbers. In [1], we looked at Pell equations X 2 − dY 2 = ± 1 such that the containment X ∈ U has at least two positive integer solutions . The following result was proved. Theorem 1.1 For each squarefree integer d, there is at most one positive integer such that X ∈ U except for d ∈ {2, 3, 5, 15, 26}.
The original article can be found online at https://doi.org/10.1007/s10998-017-0226-8.
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Carlos Alexis Gómez Ruiz [email protected] Eric F. Bravo [email protected] Florian Luca [email protected]
1
Departamento de Matemáticas, Universidad del Valle, Calle 13 No 100-00, Cali, Colombia
2
School of Mathematics, University of the Witwatersrand, Private Bag X3, Wits 2050, South Africa
3
Max Planck Institute for Mathematics, Vivatsgasse 7, 53111 Bonn, Germany
4
Department of Mathematics, Faculty of Sciences, University of Ostrava, 30 Dubna 22, 701 03 Ostrava 1, Czech Republic
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E. F. Bravo et al.
Furthermore, for each d ∈ {2, 3, 5, 15, 26}, all solutions to X ∈ U were given together with the representations of these X ’s as sums of two Tribonacci numbers. Unfortunately, there was an oversight in [1], which we now correct. The following intermediate result is Lemma 4.1 in [1]. Lemma 1.2 Let (m i , n i , i ) be two solutions of Tm i + Tn i = X i , with 0 ≤ m i < n i for i = 1, 2 and 1 ≤ 1 < 2 . Then m 1 < n 1 ≤ 1535, 1 ≤ 1070 and n 2 < 2.5 · 1042 . The rest of the argument in [1] were just reductions of the above parameters. The first step of the reduction consisted in finding all the solutions to X 1 = Fn 1 + Fm 1 , 1 ∈ [1, 1070] 2 ≤ m 1 < n 1 ≤ 1535. Unfortunately, the case 1 = 1 was omitted in [1]. Here, we discuss the missed case 1 = 1. In order to reduce the above bound on n 2 from Lemma 1.2, we do not consider the equation P±1 (X 1 ) = X 1 since there is no polynomial equation to solve; instead, we consider each minimal solution δ := δ(X 1 , ) of Pell equation X 2 − dY 2 = = ± 1, for each X 1 = Tm 1 + Tn 1 , according to the bounds in Lemma 1.2. Thus, after some reductions using the Baker–Davenport method on the linear form in logarithms Γ1 and Γ2 from [1, inequalities 3.9 and 3.12], for (m, n, ) = (m 2 , n 2
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