Research article

An application of $ p $-adic Baker method to a special case of Jeśmanowicz' conjecture

  • Received: 06 January 2023 Revised: 06 March 2023 Accepted: 12 March 2023 Published: 16 March 2023
  • MSC : 11D61, 11J86

  • In 1956, Jeśmanowicz conjectured that, for any positive integer $ n $, the Diophantine equation $ \left((f^{2}-g^{2})n\right)^{x}+\left((2fg)n\right)^{y} = \left((f^{2}+g^{2})n\right)^z $ has only the positive integral solution $ (x, y, z) = (2, 2, 2) $, where $ f $ and $ g $ are positive integers with $ f > g $, gcd$ (f, g) = 1 $, and $ f\not\equiv g\pmod {2} $. Let $ r = 6k+2 $, $ k \in \mathbb{N} $, $ k\geq25 $. In this paper, combining $ p $-adic form of Baker method with some detailed computation, we prove that if $ n $ satisfies $ n\equiv 0, 6, 9\pmod{12} $, $ f = g+1 $ and $ g = 2^{r}-1 $, then the conjecture is true.

    Citation: Ziyu Dong, Zhengjun Zhao. An application of $ p $-adic Baker method to a special case of Jeśmanowicz' conjecture[J]. AIMS Mathematics, 2023, 8(5): 11617-11628. doi: 10.3934/math.2023588

    Related Papers:

  • In 1956, Jeśmanowicz conjectured that, for any positive integer $ n $, the Diophantine equation $ \left((f^{2}-g^{2})n\right)^{x}+\left((2fg)n\right)^{y} = \left((f^{2}+g^{2})n\right)^z $ has only the positive integral solution $ (x, y, z) = (2, 2, 2) $, where $ f $ and $ g $ are positive integers with $ f > g $, gcd$ (f, g) = 1 $, and $ f\not\equiv g\pmod {2} $. Let $ r = 6k+2 $, $ k \in \mathbb{N} $, $ k\geq25 $. In this paper, combining $ p $-adic form of Baker method with some detailed computation, we prove that if $ n $ satisfies $ n\equiv 0, 6, 9\pmod{12} $, $ f = g+1 $ and $ g = 2^{r}-1 $, then the conjecture is true.



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