On Positive Integers $n$ with $ϕ(n)=\frac{2}{3} \cdot (n+1)$
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arXiv
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| Formato: | Preprint |
| Publicado: |
2025
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| _version_ | 1866912350595973120 |
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| author | Hercher, Christian |
| author_facet | Hercher, Christian |
| contents | While solving a special case of a question of Erdős and Graham Steinerberger asks for all integers $n$ with $ϕ(n)=\frac{2}{3} \cdot (n+1)$. He discovered the solutions $n\in\{5, 5 \cdot 7, 5\cdot 7\cdot 37, 5\cdot 7\cdot 37\cdot 1297\}$ and found that any additional solution must be greater than $10^{10}$. He conjectured that there are no such additional solutions to this problem.
We analyze this problem and prove:
*) Every solution $n$ must be square-free.
*) If $p$ and $q$ are prime factors of a solution $n$ then $p\nmid (q-1)$.
*) Any solution additional to the set given by Steinerberger has to have at least 7 prime factors.
*) For any additional solution it holds $n\geq 10^{14}$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2504_19915 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | On Positive Integers $n$ with $ϕ(n)=\frac{2}{3} \cdot (n+1)$ Hercher, Christian Number Theory While solving a special case of a question of Erdős and Graham Steinerberger asks for all integers $n$ with $ϕ(n)=\frac{2}{3} \cdot (n+1)$. He discovered the solutions $n\in\{5, 5 \cdot 7, 5\cdot 7\cdot 37, 5\cdot 7\cdot 37\cdot 1297\}$ and found that any additional solution must be greater than $10^{10}$. He conjectured that there are no such additional solutions to this problem. We analyze this problem and prove: *) Every solution $n$ must be square-free. *) If $p$ and $q$ are prime factors of a solution $n$ then $p\nmid (q-1)$. *) Any solution additional to the set given by Steinerberger has to have at least 7 prime factors. *) For any additional solution it holds $n\geq 10^{14}$. |
| title | On Positive Integers $n$ with $ϕ(n)=\frac{2}{3} \cdot (n+1)$ |
| topic | Number Theory |
| url | https://arxiv.org/abs/2504.19915 |