I. Diouf

most citedOn solutions of the Diophantine equation

L_{n} + L_{m} = 3^{a}

4 citations · 4 across the 2 of their papers we have counts for

collaborators

Showing math.NTShow all

4 papers · 1 filter

math.NT2023

The Diophantine equation $a (\frac{b ^{k} - 1}{b - 1}) = U_{n} - U_{m}$

P. Tiebekabe, I. Diouf, A. Tall +1

Here, we find all positive integer solutions of the Diophantine equation in the title, where $(U_{n})_{n ⩾ 0}$ is the generalized Lucas sequence $\mathcal{U}_0=0, \…

math.NT2022

On the Diophantine equation $k = 1 \sum 5 F_{n_{k}} = 2^{a}$

Pagdame Tiebekabe, Ismaïla Diouf

Let $(F_{n})_{n \geq 0}$ be the Fibonacci sequence given by $F_{0} = 0, F_{1} = 1$ and $F_{n + 2} = F_{n + 1} + F_{n}$ for $n \geq 0$ . In this paper, we have determined all the powers of 2 which…

math.NT2022★ 4 cited

On solutions of the Diophantine equation $L_{n} + L_{m} = 3^{a}$

Pagdame Tiebekabe, Ismaila Diouf

Let $(L_{n})_{n \geq 0}$ be the Lucas sequence given by $L_{0} = 2, L_{1} = 1$ and $L_{n + 2} = L_{n + 1} + L_{n}$ for $n \geq 0$ . In this paper, we are interested in finding all powers of three…

math.NT2022

On solutions of the Diophantine equation $F_{n_{1}} + F_{n_{2}} + F_{n_{3}} + F_{n_{4}} = 2^{a}$

Pagdame Tiebekabe, Ismaïla Diouf

Let $(F_{n})_{n \geq 0}$ be the Fibonacci sequence given by $F_{0} = 0, F_{1} = 1$ and $F_{n + 2} = F_{n + 1} + F_{n}$ for $n \geq 0$ . In this paper, we solve all powers of two which are sums of…

The Diophantine equation a(b−1bk−1​)=Un​−Um​

On the Diophantine equation k=1∑5​Fnk​​=2a

On solutions of the Diophantine equation Ln​+Lm​=3a

On solutions of the Diophantine equation Fn1​​+Fn2​​+Fn3​​+Fn4​​=2a

The Diophantine equation $a (\frac{b ^{k} - 1}{b - 1}) = U_{n} - U_{m}$

On the Diophantine equation $k = 1 \sum 5 F_{n_{k}} = 2^{a}$

On solutions of the Diophantine equation $L_{n} + L_{m} = 3^{a}$

On solutions of the Diophantine equation $F_{n_{1}} + F_{n_{2}} + F_{n_{3}} + F_{n_{4}} = 2^{a}$