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Refik Keskin

Publications and source records attributed to Refik Keskin.

8 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↗
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↗
arXiv2018

A Note On the Exponential Diophantine Equation (a^n-1)(b^n-1)=x^2

In 2002, F. Luca and G. Walsh solved the Diophantine equation in the title for all pairs (a,b) such that 1 4 with 2|n. Also, we solved (a^n-1)(b^n-1)=x^2 for the pairs (a,b)=(2,50),(4,49),(12,45),(13,76),(20,77),(28,49), and (45,100). Lastly, we show that when b is even, the equation (a^n-1)(b^(2n)a^n-1)=x^2 has no solutions n,x.

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↗
arXiv2013

Positive Integer Solutions of the Pell Equation $x^{2}-dy^{2}=N,$ $% d\in \left\{k^{2}\pm 4,\text{}k^{2}\pm 1\right\} $ and $N\in \left\{\pm 1,\pm 4\right\}

Let $\ k$ be a natural number and $d=k^{2}\pm 4$ or $k^{2}\pm 1$. In this paper, by using continued fraction expansion of $\sqrt{d},$ we find fundamental solution of the equations $x^{2}-dy^{2}=\pm 1$ and we get all positive integer solutions of the equations $x^{2}-dy^{2}=\pm 1$ in terms of generalized Fibonacci and Lucas sequences. Moreover, we find all positive integer solutions of the equations $x^{2}-dy^{2}=\pm 4$ in terms of generalized Fibonacci and Lucas sequences. Although some of the results are well known, we think our method is elementary and different from the others.

math.NT↗
arXiv2012

Generalized Fibonacci and Lucas Numbers of the form $5x^{2}$

Let $(U_{n}(P,Q) $ and $(V_{n}(P,Q) $ denote the generalized Fibonacci and Lucas sequence, respectively. In this study, we assume that $Q=1.$ We determine all indices $n$ such that $U_{n}=5\square $ and $U_{n}=5U_{m}\square $ under some assumptions on $P.$ We show that the equation $V_{n}=5\square $ has the solution only if $n=1$ for the case when $% P$ is odd. Moreover, we show that the equation $V_{n}=5V_{m}\square $ has no solutions.

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