On the Diophantine Equation $F_n = F_l^k (F_l^m-1)$
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arXiv
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| Format: | Preprint |
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2024
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| _version_ | 1866917870294794240 |
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| author | Ibrahimov, Seyran S. Mahmudov, Nazim I. |
| author_facet | Ibrahimov, Seyran S. Mahmudov, Nazim I. |
| contents | 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. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2409_02047 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | On the Diophantine Equation $F_n = F_l^k (F_l^m-1)$ Ibrahimov, Seyran S. Mahmudov, Nazim I. Number Theory 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. |
| title | On the Diophantine Equation $F_n = F_l^k (F_l^m-1)$ |
| topic | Number Theory |
| url | https://arxiv.org/abs/2409.02047 |