scholarly journals On the Diophantine Equation z(n) = (2 − 1/k)n Involving the Order of Appearance in the Fibonacci Sequence

Mathematics ◽  
2020 ◽  
Vol 8 (1) ◽  
pp. 124 ◽  
Author(s):  
Eva Trojovská
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Positive Integers

Let ( F n ) n ≥ 0 be the sequence of the Fibonacci numbers. The order (or rank) of appearance z ( n ) of a positive integer n is defined as the smallest positive integer m such that n divides F m . In 1975, Sallé proved that z ( n ) ≤ 2 n , for all positive integers n. In this paper, we shall solve the Diophantine equation z ( n ) = ( 2 − 1 / k ) n for positive integers n and k.

scholarly journals On the Ternary Exponential Diophantine Equation Equating a Perfect Power and Sum of Products of Consecutive Integers

Mathematics ◽  
2021 ◽  
Vol 9 (15) ◽  
pp. 1813
Author(s):  
S. Subburam ◽  
Lewis Nkenyereye ◽  
N. Anbazhagan ◽  
S. Amutha ◽  
M. Kameswari ◽  
...  
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Upper Bound ◽  
Exponential Diophantine Equation ◽  
Research Article ◽  
All Solutions ◽  
Sum Of Products ◽  
Consecutive Integers ◽  
Sums Of Products ◽  
Positive Integers

Consider the Diophantine equation yn=x+x(x+1)+⋯+x(x+1)⋯(x+k), where x, y, n, and k are integers. In 2016, a research article, entitled – ’power values of sums of products of consecutive integers’, primarily proved the inequality n= 19,736 to obtain all solutions (x,y,n) of the equation for the fixed positive integers k≤10. In this paper, we improve the bound as n≤ 10,000 for the same case k≤10, and for any fixed general positive integer k, we give an upper bound depending only on k for n.

scholarly journals On the Sum of Powers of Two k-Fibonacci Numbers which Belongs to the Sequence of k-Lucas Numbers

2016 ◽  
Vol 67 (1) ◽  
pp. 41-46
Author(s):  
Pavel Trojovský
Keyword(s):  
Recurrence Relation ◽  
Initial Conditions ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Lucas Numbers ◽  
Initial Values ◽  
Lucas Sequence ◽  
Positive Integers

Abstract Let k ≥ 1 and denote (Fk,n)n≥0, the k-Fibonacci sequence whose terms satisfy the recurrence relation Fk,n = kFk,n−1 +Fk,n−2, with initial conditions Fk,0 = 0 and Fk,1 = 1. In the same way, the k-Lucas sequence (Lk,n)n≥0 is defined by satisfying the same recurrence relation with initial values Lk,0 = 2 and Lk,1 = k. These sequences were introduced by Falcon and Plaza, who showed many of their properties, too. In particular, they proved that Fk,n+1 + Fk,n−1 = Lk,n, for all k ≥ 1 and n ≥ 0. In this paper, we shall prove that if k ≥ 1 and $F_{k,n + 1}^s + F_{k,n - 1}^s \in \left( {L_{k,m} } \right)_{m \ge 1} $ for infinitely many positive integers n, then s =1.

scholarly journals A NOTE ON THE DIOPHANTINE EQUATION

2013 ◽  
Vol 89 (2) ◽  
pp. 316-321 ◽  
Author(s):  
MOU JIE DENG
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Pythagorean Triple ◽  
Positive Integers

AbstractLet $(a, b, c)$ be a primitive Pythagorean triple satisfying ${a}^{2} + {b}^{2} = {c}^{2} . $ In 1956, Jeśmanowicz conjectured that for any given positive integer $n$ the only solution of $\mathop{(an)}\nolimits ^{x} + \mathop{(bn)}\nolimits ^{y} = \mathop{(cn)}\nolimits ^{z} $ in positive integers is $x= y= z= 2. $ In this paper, for the primitive Pythagorean triple $(a, b, c)= (4{k}^{2} - 1, 4k, 4{k}^{2} + 1)$ with $k= {2}^{s} $ for some positive integer $s\geq 0$, we prove the conjecture when $n\gt 1$ and certain divisibility conditions are satisfied.

scholarly journals Some Diophantine Problems Related to k-Fibonacci Numbers

Mathematics ◽  
2020 ◽  
Vol 8 (7) ◽  
pp. 1047
Author(s):  
Pavel Trojovský ◽  
Štěpán Hubálovský
Keyword(s):  
Recurrence Relation ◽  
Diophantine Equations ◽  
Initial Conditions ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Recursive Relation ◽  
Initial Values ◽  
Lucas Sequence ◽  
Diophantine Problems ◽  
Positive Integers

Let k ≥ 1 be an integer and denote ( F k , n ) n as the k-Fibonacci sequence whose terms satisfy the recurrence relation F k , n = k F k , n − 1 + F k , n − 2 , with initial conditions F k , 0 = 0 and F k , 1 = 1 . In the same way, the k-Lucas sequence ( L k , n ) n is defined by satisfying the same recursive relation with initial values L k , 0 = 2 and L k , 1 = k . The sequences ( F k , n ) n ≥ 0 and ( L k , n ) n ≥ 0 were introduced by Falcon and Plaza, who derived many of their properties. In particular, they proved that F k , n 2 + F k , n + 1 2 = F k , 2 n + 1 and F k , n + 1 2 − F k , n − 1 2 = k F k , 2 n , for all k ≥ 1 and n ≥ 0 . In this paper, we shall prove that if k > 1 and F k , n s + F k , n + 1 s ∈ ( F k , m ) m ≥ 1 for infinitely many positive integers n, then s = 2 . Similarly, that if F k , n + 1 s − F k , n − 1 s ∈ ( k F k , m ) m ≥ 1 holds for infinitely many positive integers n, then s = 1 or s = 2 . This generalizes a Marques and Togbé result related to the case k = 1 . Furthermore, we shall solve the Diophantine equations F k , n = L k , m , F k , n = F n , k and L k , n = L n , k .

scholarly journals ON THE NUMBER OF SOLUTIONS OF THE DIOPHANTINE EQUATION axm−byn=c

2010 ◽  
Vol 81 (2) ◽  
pp. 177-185 ◽  
Author(s):  
BO HE ◽  
ALAIN TOGBÉ
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Number Of Solutions ◽  
Integer Solutions ◽  
Image Position ◽  
Positive Integers

AbstractLet a, b, c, x and y be positive integers. In this paper we sharpen a result of Le by showing that the Diophantine equation has at most two positive integer solutions (m,n) satisfying min (m,n)>1.

scholarly journals The Proof of a Conjecture on the Density of Sets Related to Divisibility Properties of z(n)

Mathematics ◽  
2021 ◽  
Vol 9 (22) ◽  
pp. 2912
Author(s):  
Eva Trojovská ◽  
Venkatachalam Kandasamy
Keyword(s):  
Positive Integer ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Natural Density ◽  
Almost All ◽  
Divisibility Properties

Let (Fn)n be the sequence of Fibonacci numbers. The order of appearance (in the Fibonacci sequence) of a positive integer n is defined as z(n)=min{k≥1:n∣Fk}. Very recently, Trojovská and Venkatachalam proved that, for any k≥1, the number z(n) is divisible by 2k, for almost all integers n≥1 (in the sense of natural density). Moreover, they posed a conjecture that implies that the same is true upon replacing 2k by any integer m≥1. In this paper, in particular, we prove this conjecture.

ON THE DIOPHANTINE EQUATION axy + byz + czx = 0

2012 ◽  
Vol 08 (03) ◽  
pp. 813-821 ◽  
Author(s):  
ZHONGFENG ZHANG ◽  
PINGZHI YUAN
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solutions ◽  
Positive Integers

Let a, b, c be integers. In this paper, we prove the integer solutions of the equation axy + byz + czx = 0 satisfy max {|x|, |y|, |z|} ≤ 2 max {a, b, c} when a, b, c are odd positive integers, and when a = b = 1, c = -1, the positive integer solutions of the equation satisfy max {x, y, z} < exp ( exp ( exp (5))).

ON THE EXPONENTIAL DIOPHANTINE EQUATION

2014 ◽  
Vol 90 (1) ◽  
pp. 9-19 ◽  
Author(s):  
TAKAFUMI MIYAZAKI ◽  
NOBUHIRO TERAI
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Baker’S Method ◽  
Elementary Methods ◽  
Positive Integers

AbstractLet $m$, $a$, $c$ be positive integers with $a\equiv 3, 5~({\rm mod} \hspace{0.334em} 8)$. We show that when $1+ c= {a}^{2} $, the exponential Diophantine equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$ under the condition $m\equiv \pm 1~({\rm mod} \hspace{0.334em} a)$, except for the case $(m, a, c)= (1, 3, 8)$, where there are only two solutions: $(x, y, z)= (1, 1, 2), ~(5, 2, 4). $ In particular, when $a= 3$, the equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(8{m}^{2} - 1)}\nolimits ^{y} = \mathop{(3m)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$, except if $m= 1$. The proof is based on elementary methods and Baker’s method.

scholarly journals COUNTING THE NUMBER OF SOLUTIONS TO THE ERDŐS–STRAUS EQUATION ON UNIT FRACTIONS

2013 ◽  
Vol 94 (1) ◽  
pp. 50-105 ◽  
Author(s):  
CHRISTIAN ELSHOLTZ ◽  
TERENCE TAO
Keyword(s):  
Positive Integer ◽  
Lower Bounds ◽  
Diophantine Equation ◽  
Upper And Lower Bounds ◽  
Natural Numbers ◽  
Number Of Solutions ◽  
Waring’S Problem ◽  
Unit Fractions ◽  
Image Position ◽  
Positive Integers

AbstractFor any positive integer $n$, let $f(n)$ denote the number of solutions to the Diophantine equation $$\begin{eqnarray*}\frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\end{eqnarray*}$$ with $x, y, z$ positive integers. The Erdős–Straus conjecture asserts that $f(n)\gt 0$ for every $n\geq 2$. In this paper we obtain a number of upper and lower bounds for $f(n)$ or $f(p)$ for typical values of natural numbers $n$ and primes $p$. For instance, we establish that $$\begin{eqnarray*}N\hspace{0.167em} {\mathop{\log }\nolimits }^{2} N\ll \displaystyle \sum _{p\leq N}f(p)\ll N\hspace{0.167em} {\mathop{\log }\nolimits }^{2} N\log \log N.\end{eqnarray*}$$ These upper and lower bounds show that a typical prime has a small number of solutions to the Erdős–Straus Diophantine equation; small, when compared with other additive problems, like Waring’s problem.

scholarly journals ON THE DIOPHANTINE EQUATION x2 + C = 2yn

2009 ◽  
Vol 05 (06) ◽  
pp. 1117-1128 ◽  
Author(s):  
FADWA S. ABU MURIEFAH ◽  
FLORIAN LUCA ◽  
SAMIR SIKSEK ◽  
SZABOLCS TENGELY
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Main Tool ◽  
Sharp Bound ◽  
Primitive Divisors ◽  
Positive Integers

In this paper, we study the Diophantine equation x2 + C = 2yn in positive integers x,y with gcd (x,y) = 1, where n ≥ 3 and C is a positive integer. If C ≡ 1 (mod 4), we give a very sharp bound for prime values of the exponent n; our main tool here is the result on existence of primitive divisors in Lehmer sequences due to Bilu, Hanrot and Voutier. We illustrate our approach by solving completely the equations x2 + 17a1 = 2yn, x2 + 5a113a2 = 2yn and x2 + 3a111a2 = 2yn.

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