arXiv subjects

Zafer Şiar

Publications and source records attributed to Zafer Şiar.

6 recordsLinked to original sources

arXiv2022

On Perfect Powers in k-Generalized Pell-Lucas Sequence

Let k>=2 and let (Q_{n}^{(k)})_{n>=2-k} be the k-generalized Pell sequence defined by Q_{n}^{(k)}=2Q_{n-1}^{(k)}+Q_{n-2}^{(k)}+...+Q_{n-k}^{(k)} for n>=2 with initial conditions Q_{-(k-2)}^{(k)}=Q_{-(k-3)}^{(k)}=...=Q_{-1}^{(k)}=0, Q_{0}^{(k)}=2,Q_{1}^{(k)}=2. In this paper, we solve the Diophantine equation Q_{n}^{(k)}=y^{m} in positive integers n,m,y,k with m,y,k>=2. We show that all solutions (n,m,y) of this equation in positive integers n,m,y,k such that 2<=y<=100 are given by (n,m,y)=(3,2,4),(3,4,2) for k>=3. Namely, Q_{3}^{(k)}=16=2^4=4^2 for k>=3.

math.NT↗

arXiv2021

A Note on Terai's Conjecture Concerning the Exponential Diophantine Equation x^{2}+b^{y}=c^{z}

Let (a,b,c) be a primitive Pythagorean triple, i.e., a^{2}+b^{2}=c^{2} with gcd(a,b,c)=1, a even and b odd. Terai's conjecture says that the Diophantine equation x^{2}+b^{y}=c^{z} has only the positive integer solutions (x,y,z)=(a,2,2). In this study, we prove that Terai's conjecture is true when b is a product of two primes and c=5(mod8).

math.NT↗

arXiv2020

Repdigits in k-generalized Pell sequence

Let $k\geq 2$ and let $(P_{n}^{(k)})_{n\geq 2-k}$ be $k$-generalized Pell sequence defined by \begin{equation*}P_{n}^{(k)}=2P_{n-1}^{(k)}+P_{n-2}^{(k)}+...+P_{n-k}^{(k)}\end{equation*} for $n\geq 2$ with initial conditions \begin{equation*}P_{-(k-2)}^{(k)}=P_{-(k-3)}^{(k)}=\cdot \cdot \cdot =P_{-1}^{(k)}=P_{0}^{(k)}=0,P_{1}^{(k)}=1. \end{equation*} In this paper, we deal with the Diophantine equation \begin{equation*}P_{n}^{(k)}=d\left( \frac{10^{m}-1}{9}\right)\end{equation*} in positive integers $n,m,k,d$ with $k\geq 2,$ $m\geq 2$ and $1\leq d\leq 9$. We will show that repdigits with at least two digits in the sequence $\left( P_{n}^{(k)}\right)_{n\geq 2-k}$ are the numbers\ $P_{5}^{(3)}=33$ and $P_{6}^{(4)}=88.$

math.NT↗

arXiv2020

An exponential Diophantine equation related to the difference of powers of two Fibonacci numbers

In this paper, we prove that there is no x>=4 such that the difference of x-th powers of two consecutive Fibonacci numbers greater than 0 is a Lucas number.

math.NT↗

arXiv2018

On the Exponential Diophantine Equation $(a^2-2)(b^2-2)=x^2$

In this paper, we consider the equation $(a^n-2^{m})(b^n-2^{m})=x^2$. By assuming the abc conjecture is true, in [8], Luca and Walsh gave a theorem, which implies that the above equation has only finitely many solutions $n,x$ if a and b are different fixed positive integers. We solve the above equation when $m=1$ and $(a,b)=(2,10),(4,100),(10,58),(3,45)$. Moreover, we show that $(a^2-2)(b^2-2)=x^2$ has no solution n,x if 2|n and gcd$(a,b)=1$. We also give a conjecture which says that the equation $(2^2-2)((2P_k)^n-2)=x^2$ has only the solution $(n,x)=(2,Q_k)$, where $k>3$ is odd and $P_k,Q_k$ are Pell and Pell Lucas numbers, respectively. We also conjecture that if the equation $(a^2-2)(b^2-2)=x^2$ has a solution $n,x$, then $n<7$, where $2<a<b$.

math.NT↗

arXiv2017

On the Diophantine equation F_{n}-F_{m}=2^{a}

In this paper, we solve Diophantine equation in the tittle in nonnegative integers m,n, and a. In order to prove our result, we use lower bounds for linear forms in logarithms and and a version of the Baker-Davenport reduction method in diophantine approximation.

math.NT↗