Research article

On a conjecture concerning the exponential Diophantine equation $ (an^{2}+1)^{x}+(bn^{2}-1)^{y} = (cn)^{z} $

  • Received: 16 April 2024 Revised: 14 June 2024 Accepted: 20 June 2024 Published: 26 June 2024
  • Let $ a $, $ b $, $ c $, and $ n $ be positive integers such that $ a+b = c^{2} $, $ 2\nmid c $ and $ n > 1 $. In this paper, we prove that if $ \gcd(c, n) = 1 $ and $ n\geq 117.14c $, then the equation $ (an^{2}+1)^{x}+(bn^{2}-1)^{y} = (cn)^{z} $ has only the positive integer solution $ (x, y, z) = (1, 1, 2) $ under the assumption $ \gcd(an^{2}+1, bn^{2}-1) = 1 $. Thus, we affirm that the conjecture proposed by Fujita and Le is true in this case. Moreover, combining the above result with some existing results and a computer search, we show that, for any positive integer $ n $, if $ (a, b, c) = (12, 13, 5) $, $ (18, 7, 5) $, or $ (44, 5, 7) $, then this equation has only the solution $ (x, y, z) = (1, 1, 2) $. This result extends the theorem of Terai and Hibino gotten in 2015, that of Alan obtained in 2018, and Hasanalizade's theorem attained recently.

    Citation: Shuanglin Fei, Guangyan Zhu, Rongjun Wu. On a conjecture concerning the exponential Diophantine equation $ (an^{2}+1)^{x}+(bn^{2}-1)^{y} = (cn)^{z} $[J]. Electronic Research Archive, 2024, 32(6): 4096-4107. doi: 10.3934/era.2024184

    Related Papers:

  • Let $ a $, $ b $, $ c $, and $ n $ be positive integers such that $ a+b = c^{2} $, $ 2\nmid c $ and $ n > 1 $. In this paper, we prove that if $ \gcd(c, n) = 1 $ and $ n\geq 117.14c $, then the equation $ (an^{2}+1)^{x}+(bn^{2}-1)^{y} = (cn)^{z} $ has only the positive integer solution $ (x, y, z) = (1, 1, 2) $ under the assumption $ \gcd(an^{2}+1, bn^{2}-1) = 1 $. Thus, we affirm that the conjecture proposed by Fujita and Le is true in this case. Moreover, combining the above result with some existing results and a computer search, we show that, for any positive integer $ n $, if $ (a, b, c) = (12, 13, 5) $, $ (18, 7, 5) $, or $ (44, 5, 7) $, then this equation has only the solution $ (x, y, z) = (1, 1, 2) $. This result extends the theorem of Terai and Hibino gotten in 2015, that of Alan obtained in 2018, and Hasanalizade's theorem attained recently.


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