Sum of consecutive powers as a perfect power
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arXiv
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| Hauptverfasser: | , |
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| Format: | Preprint |
| Veröffentlicht: |
2026
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| _version_ | 1866911694116093952 |
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| author | Koutsianas, Angelos Tzanakis, Nikos |
| author_facet | Koutsianas, Angelos Tzanakis, Nikos |
| contents | In this paper we study the equation $$
x^k + (x+1)^k = y^n,\quad n\geq 3, $$ when $k\equiv 2\pmod{4}$. We prove that the only solutions are for $x=0, -1$ when $6\leq k\leq 100$ or for a $k$ with odd prime factors congruent to $3\pmod{4}$. We use linear forms in logarithms, the modular method and the resolution of Thue equations. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2605_18348 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | Sum of consecutive powers as a perfect power Koutsianas, Angelos Tzanakis, Nikos Number Theory 1D41, 11G10 In this paper we study the equation $$ x^k + (x+1)^k = y^n,\quad n\geq 3, $$ when $k\equiv 2\pmod{4}$. We prove that the only solutions are for $x=0, -1$ when $6\leq k\leq 100$ or for a $k$ with odd prime factors congruent to $3\pmod{4}$. We use linear forms in logarithms, the modular method and the resolution of Thue equations. |
| title | Sum of consecutive powers as a perfect power |
| topic | Number Theory 1D41, 11G10 |
| url | https://arxiv.org/abs/2605.18348 |