On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)

Yahui Yu; Jiayuan Hu; Yahui Yu; Jiayuan Hu

doi:10.3934/math.2021615

AIMS Mathematics

2021, Volume 6, Issue 10: 10596-10601. doi: 10.3934/math.2021615

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On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)

Yahui Yu ¹,
Jiayuan Hu ^{2
,
,}

1.
Department of Mathematics and Physics, Luoyang Institute of Science and Technology, Luoyang, Henan, 471023, China
2.
Department of Mathematics and Computer, Hetao College, Bayannur, Inner Mongolia, 015000, China

Received: 01 March 2021 Accepted: 19 July 2021 Published: 22 July 2021
MSC : 11D61

Let $ k $ be a fixed positive integer with $ k > 1 $. In 2014, N. Terai ^[6] conjectured that the equation $ x^2+(2k-1)^y = k^z $ has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. This is still an unsolved problem as yet. For any positive integer $ n $, let $ Q(n) $ denote the squarefree part of $ n $. In this paper, using some elementary methods, we prove that if $ k\equiv 3 $ (mod 4) and $ Q(k-1)\ge 2.11 $ log $ k $, then the equation has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. It can thus be seen that Terai's conjecture is true for almost all positive integers $ k $ with $ k\equiv 3 $(mod 4).
- polynomial-exponential diophantine equation,
- generalized Ramanujan-Nagell equation
Citation: Yahui Yu, Jiayuan Hu. On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)[J]. AIMS Mathematics, 2021, 6(10): 10596-10601. doi: 10.3934/math.2021615

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Abstract

Let $ k $ be a fixed positive integer with $ k > 1 $. In 2014, N. Terai ^[6] conjectured that the equation $ x^2+(2k-1)^y = k^z $ has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. This is still an unsolved problem as yet. For any positive integer $ n $, let $ Q(n) $ denote the squarefree part of $ n $. In this paper, using some elementary methods, we prove that if $ k\equiv 3 $ (mod 4) and $ Q(k-1)\ge 2.11 $ log $ k $, then the equation has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. It can thus be seen that Terai's conjecture is true for almost all positive integers $ k $ with $ k\equiv 3 $(mod 4).

References

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