scholarly journals On Two Problems Related to Divisibility Properties of z(n)

Mathematics ◽  
2021 ◽  
Vol 9 (24) ◽  
pp. 3273
Author(s):  
Pavel Trojovský

The order of appearance (in the Fibonacci sequence) function z:Z≥1→Z≥1 is an arithmetic function defined for a positive integer n as z(n)=min{k≥1:Fk≡0(modn)}. A topic of great interest is to study the Diophantine properties of this function. In 1992, Sun and Sun showed that Fermat’s Last Theorem is related to the solubility of the functional equation z(n)=z(n2), where n is a prime number. In addition, in 2014, Luca and Pomerance proved that z(n)=z(n+1) has infinitely many solutions. In this paper, we provide some results related to these facts. In particular, we prove that limsupn→∞(z(n+1)−z(n))/(logn)2−ϵ=∞, for all ϵ∈(0,2).

2020 ◽  
Vol 2020 ◽  
pp. 1-7
Author(s):  
Long Chen ◽  
Kaimin Cheng ◽  
Tingting Wang

Let p be an odd prime number and n be a positive integer. Let vpn, N∗, and Q+ denote the p-adic valuation of the integer n, the set of positive integers, and the set of positive rational numbers, respectively. In this paper, we introduce an arithmetic function fp:N∗⟶Q+ defined by fpn≔n/pvpn1−vpn for any positive integer n. We show several interesting arithmetic properties about that function and then use them to establish some curious results involving the p-adic valuation. Some of these results extend Farhi’s results from the case of even prime to that of odd prime.


2014 ◽  
Vol 10 (04) ◽  
pp. 915-933 ◽  
Author(s):  
Florian Luca ◽  
Carl Pomerance

Let z(N) be the order of appearance of N in the Fibonacci sequence. This is the smallest positive integer k such that N divides the k th Fibonacci number. We show that each of the six total possible orderings among z(N), z(N + 1), z(N + 2) appears infinitely often. We also show that for each nonzero even integer c and many odd integers c the equation z(N) = z(N + c) has infinitely many solutions N, but the set of solutions has asymptotic density zero. The proofs use a result of Corvaja and Zannier on the height of a rational function at 𝒮-unit points as well as sieve methods.


Mathematics ◽  
2021 ◽  
Vol 9 (22) ◽  
pp. 2912
Author(s):  
Eva Trojovská ◽  
Venkatachalam Kandasamy

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.


Mathematics ◽  
2020 ◽  
Vol 8 (6) ◽  
pp. 876 ◽  
Author(s):  
Pavel Trojovský

The order of appearance z : Z > 0 → Z > 0 is an arithmetic function related to the Fibonacci sequence ( F n ) n . This function is defined as the smallest positive integer solution of the congruence F k ≡ 0 ( mod n ) . In this paper, we shall provide lower and upper bounds for the functions ∑ n ≤ x z ( n ) / n , ∑ p ≤ x z ( p ) and ∑ p r ≤ x z ( p r ) .


2020 ◽  
Vol 14 (1) ◽  
pp. 139-142
Author(s):  
Sudhangshu B. Karmakar

AbstractAn elementary proof that the equation x2n + y2n = z2n can not have any non-zero positive integer solutions when n is an integer ≥ 2 is presented. To prove that the equation has no integer solutions it is first hypothesized that the equation has integer solutions. The absence of any integer solutions of the equation is justified by contradicting the hypothesis.


2015 ◽  
Vol 11 (08) ◽  
pp. 2341-2351
Author(s):  
Alain Kraus

Soient K un corps de nombres et p un nombre premier ≥ 5. Notons μp le groupe des racines p-ièmes de l'unité. On définit p comme étant K-régulier si p ne divise pas le nombre de classes du corps K(μp). Sous l'hypothèse que p est K-régulier et inerte dans K, on établit le second cas du théorème de Fermat sur K pour l'exposant p. On utilise pour cela des arguments classiques, ainsi que le théorème de Faltings selon lequel une courbe de genre au moins deux sur K n'a qu'un nombre fini de points K-rationnels. De plus, si K est un corps quadratique imaginaire, distinct de [Formula: see text], on en déduit un énoncé permettant souvent en pratique de démontrer le théorème de Fermat sur K pour un exposant K-régulier donné. Mots-clés: Théorème de Fermat; corps de nombres. Let K be a number field and p a prime number ≥5. Let us denote by μp the group of the pth roots of unity. We define p to be K-regular if p does not divide the class number of the field K(μp). Under the assumption that p is K-regular and inert in K, we establish the second case of Fermat's Last Theorem over K for the exponent p. We use in the proof classical arguments, as well as Faltings' theorem stating that a curve of genus at least two over K has a finite number of K-rational points. Moreover, if K is an imaginary quadratic field, other than [Formula: see text], we deduce a statement which allows often in practice to prove Fermat's Last Theorem over K for a given K-regular exponent.


Author(s):  
Benson Schaeffer

In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem. AMS 2020 subject classification: 11A99, 11D41 Diophantine equations, Fermat’s equation ∗The corresponding author. E-mail: [email protected] 1 1 Introduction To prove Fermat’s Last Theorem, it suffices to show that the equation A p + B p = C p (1In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem.


Mathematics ◽  
2021 ◽  
Vol 9 (20) ◽  
pp. 2638
Author(s):  
Eva Trojovská ◽  
Kandasamy Venkatachalam

The order of appearance of n (in the Fibonacci sequence) z(n) is defined as the smallest positive integer k for which n divides the k—the Fibonacci number Fk. Very recently, Trojovský proved that z(n) is an even number for almost all positive integers n (in the natural density sense). Moreover, he conjectured that the same is valid for the set of integers n ≥ 1 for which the integer 4 divides z(n). In this paper, among other things, we prove that for any k ≥ 1, the number z(n) is divisible by 2k for almost all positive integers n (in particular, we confirm Trojovský’s conjecture).


2007 ◽  
Vol 76 (1) ◽  
pp. 133-136 ◽  
Author(s):  
Andrzej Dąbrowski

Let p be a prime number ≥ 5, and n a positive integer > 1. This note is concerned with the diophantine equation x4 − y4 = nzp. We prove that, under certain conditions on n, this equation has no non-trivial solution in Z if p ≥ C(n), where C(n) is an effective constant.


1986 ◽  
Vol 59 (2) ◽  
pp. 76 ◽  
Author(s):  
Jonathan P. Dowling

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