Solution of the Diophantine equation $x^2 + p^k=y^n$
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arXiv
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| Formato: | Preprint |
| Publicado: |
2024
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| _version_ | 1866911808628981760 |
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| author | Ghosh, Arkabrata |
| author_facet | Ghosh, Arkabrata |
| contents | The main aim of this article is to find all solutions of the Diophantine equation $x^2 + p^k=y^n$ where $p \equiv 1 \pmod 4$, $\frac{p-1}{3}$ is a perfect square and the class number of $\mathbb{Z}[\sqrt{-p}]$ is $2$. In this article, I used a method involving prime factorization and class numbers which is different from using congruent number argument which is widely used in this type of problem. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2402_19445 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | Solution of the Diophantine equation $x^2 + p^k=y^n$ Ghosh, Arkabrata Number Theory 11D41, 11D61 The main aim of this article is to find all solutions of the Diophantine equation $x^2 + p^k=y^n$ where $p \equiv 1 \pmod 4$, $\frac{p-1}{3}$ is a perfect square and the class number of $\mathbb{Z}[\sqrt{-p}]$ is $2$. In this article, I used a method involving prime factorization and class numbers which is different from using congruent number argument which is widely used in this type of problem. |
| title | Solution of the Diophantine equation $x^2 + p^k=y^n$ |
| topic | Number Theory 11D41, 11D61 |
| url | https://arxiv.org/abs/2402.19445 |