On Positive Integers $n$ with $\varphi (n)=\frac{2}{3}\cdot (n+1)$

Abstract

While solving a special case of a question of Erdős and Graham Steinerberger asks for all integers $n$ with $\varphi (n)=\frac{2}{3}\cdot (n+1)$. 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 $n\ge 10^{14}$.