Compute a natural number $n\geq 2$ s.t. $p\mid n \Longrightarrow p^2\nmid n$ AND $p-1\mid n \Longleftrightarrow p\mid n$ for all prime divisor p of n. Question: Compute a natural number $n\geq 2$ that satisfies:  


*

*For each prime divisor $p$ of $n$, $p^2$ does not divide $n$.  

*For each prime number $p$, $p-1$ divides $n$ if and only if $p$ divides $n$.  


I'm stuck with this one like forever.
My first intuition was to find the canonical form of such $n$, which is $\prod_{i} p_i^{\alpha_i}$, where $\alpha_i$ must be equal to  $1$ for all $i\in [1,r]$ for some $r\in\mathbb{N}$.
But it got me nowhere.
Then, I started trying to find such $n$ from the second condition. Yet again, I have no clue whatsoever.
Hence, I'm here asking for help. Any hints?
Many thanks,
D.

Here's a partial answer: Suppose $n$ is a solution. Then $n(n+1)$ is a solution if $n+1$ is prime, since $n+1\mid n(n+1)$ and $n\mid n(n+1)$, and the condition is clearly satisfied for all primes less than $n$. So in addition to $2$, $3$, $6$, and $42$, we also get
\begin{align*}
  42(43) = 1806 = 2\cdot 3\cdot 7\cdot 43 .
\end{align*}
This fails at the next step since $1807=13\cdot 139$. This doesn't prove that there aren't other solutions, however.
