Related to this
Martin Hopf coined in another multiplicative function $$\varphi_2(n)=\sum_{d\mid n} \varphi(d)\cdot \varphi(\frac{n}{d})$$ the Dirichlet convolution of $\varphi(n)$, the totient function.
Let $n\ge 2$ be an integer and $m:=\varphi_2(n)$. If $n^2\mid m^m-1$ holds, can we conclude that $n$ must be squarefree ?
What I worked out :
- $n$ must be cubefree : If $p^3\mid n$ for some prime $p$ , we have $p\mid m$ which is a contradiction to $p\mid m^m-1$.
- For every prime factor $p$ of $n$ we have $p-1\mid m$ : We have $\varphi_2(p)=2(p-1)$ and $\varphi_2(p^2)=(p-1)\cdot (3p-1)$. This could be extended, but I already ruled out $p^3\mid n$. $m$ is the product of such expressions , where $p$ runs over the prime factors of $n$.
- $p^2\mid n$ means $p^4\mid m^m-1$ which implies that for some base $b$ we have $b^{p-1}\equiv 1\mod p^4$ which is a strong requirement , unless $b\equiv \pm1\mod p$.
- No non-squarefree $n$ upto $10^{10}$ satisfies $n^2\mid m^m-1$.