On the Diophantine Equation $F_n = F_l^k (F_l^m-1)$

TL;DR

This paper solves a specific Fibonacci Diophantine equation by applying advanced number theory techniques, proving that only one particular quadruple of positive integers satisfies the equation.

Contribution

The paper uniquely determines all solutions to the Fibonacci-based Diophantine equation using Matveev's theorem and a modified Baker-Davenport method.

Findings

Only the quadruple (6, 3, 3, 1) satisfies the equation.

Applied Matveev's theorem to bound solutions.

Used divisibility properties of Fibonacci numbers.

Abstract

In this paper, we examine the Diophantine problem given by the equation $F_n = F_l^k (F_l^m - 1)$, where $n, l, m \geq 1$ and $k \geq 3$. Here, $\{ F_t \}_{t=0}^{\infty} $ denotes the Fibonacci numbers, defined by the recurrence relation $F_0 = 0$, $F_1 = 1$, and $F_t = F_{t-1} + F_{t-2}$ for $t \geq 2$. By applying Matveev's theorem, which provides lower bounds for linear forms in logarithms of algebraic numbers, along with a modified Baker-Davenport reduction method and a divisibility property of Fibonacci numbers, we show that $(n, l, k, m) = (6, 3, 3, 1)$ is the only positive integer quadruple that satisfies this equation.