My solution: Part a) Let $S(n)$ the sum of divisors of $n$ and $T(n)$ the sum of the reciprocal divisors of $n$ $\Longrightarrow$ $T(n)=\frac{S(n)}{n}$ Suposse that exists $n$ such that $T(n)=T(p)$ where $p$ is prime $\Longrightarrow$ $\frac{S(n)}{n}=\frac{S(p)}{p}$ $\Longrightarrow$ $p|n$ and $p\neq n$ and $n=pk$ $\Longrightarrow$ $S(n)\geq n+k+1$ $\Longrightarrow$ $T(n)\geq \frac{pk+k+1}{pk}> \frac{p+1}{p}=T(p)$ $\Longrightarrow$ $T(n)\neq T(p)$ $\Longrightarrow$ $p$ is lonely for all prime $p$ Part b) We know that: $T(mn)=T(n).T(m)$ for all $m,n$ coprime $\Longrightarrow$ $T(n)$ is multiplicative It is easy to see that: $T(6)=2$ and $T(28)=2$ $\Longrightarrow$ $T(6k)=2T(k)=T(28k)$ for all $k$ coprime to $42$ $\Longrightarrow$ $n=6k$ is not lonely for all $k$ coprime to $42$ $\Longrightarrow$ exists infinitely integers $n=6k$ that are not lonely...