>On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)
Let $ k $ be a fixed positive integer with $ k > 1 $. In 2014, N. Terai <sup>[<xref ref-type="bibr" rid="b6">6</xref>]</sup> conjectured that the equation $ x^2+(2k-1)^y = k^z $ has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. Thi...
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doaj-2c1ee476e8234765a0e6a258647acc652021-08-03T01:21:35ZengAIMS PressAIMS Mathematics2473-69882021-07-01610105961060110.3934/math.2021615>On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)Yahui Yu0Jiayuan Hu 11. Department of Mathematics and Physics, Luoyang Institute of Science and Technology, Luoyang, Henan, 471023, China2. Department of Mathematics and Computer, Hetao College, Bayannur, Inner Mongolia, 015000, ChinaLet $ k $ be a fixed positive integer with $ k > 1 $. In 2014, N. Terai <sup>[<xref ref-type="bibr" rid="b6">6</xref>]</sup> 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). https://www.aimspress.com/article/doi/10.3934/math.2021615?viewType=HTMLpolynomial-exponential diophantine equationgeneralized ramanujan-nagell equation |
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English |
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Article |
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DOAJ |
author |
Yahui Yu Jiayuan Hu |
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Yahui Yu Jiayuan Hu >On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4) AIMS Mathematics polynomial-exponential diophantine equation generalized ramanujan-nagell equation |
author_facet |
Yahui Yu Jiayuan Hu |
author_sort |
Yahui Yu |
title |
>On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4) |
title_short |
>On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4) |
title_full |
>On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4) |
title_fullStr |
>On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4) |
title_full_unstemmed |
>On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4) |
title_sort |
>on the generalized ramanujan-nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4) |
publisher |
AIMS Press |
series |
AIMS Mathematics |
issn |
2473-6988 |
publishDate |
2021-07-01 |
description |
Let $ k $ be a fixed positive integer with $ k > 1 $. In 2014, N. Terai <sup>[<xref ref-type="bibr" rid="b6">6</xref>]</sup> 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). |
topic |
polynomial-exponential diophantine equation generalized ramanujan-nagell equation |
url |
https://www.aimspress.com/article/doi/10.3934/math.2021615?viewType=HTML |
work_keys_str_mv |
AT yahuiyu onthegeneralizedramanujannagellequationx22k1ykzwithkequiv3mod4 AT jiayuanhu onthegeneralizedramanujannagellequationx22k1ykzwithkequiv3mod4 |
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