A positive integer is special if it has at least two proper divisors and is a multiple of all possible differences between two of them. 
The proper divisors of a positive integer n are all positive integers other than 1 and n which divide $n$. A positive integer is special if it has at least two proper divisors and is a multiple of all possible differences between two of them. Determine all positive integers that are special.

I determined that $6$,$8$ and $12$ are special. And that any power of $2$ greater that $8$ is not special. But I don't know how to proceed, I would appreciate if you could help me.
 A: To summarize the discussion in the comments:
Let's argue that there are no special numbers $>12$.  Toward that end, suppose that $n$ is one.
Elementary considerations show that $n$ is divisible by (at least) $2^3\times 3^2\times 5\times 7$.
(Pf:  $n $ can't be odd since the difference of two odd factors would be even. Easy to rule out powers of $2$ greater than $8$, as they would be divisible by $8-2=6$.  Thus $n$ is divisible by an odd prime.  Let $p$ be the least such.  Then $p-2$ is odd and must be $1$, else there's a smaller odd prime dividing $n$.  Hence $p=3$ so $3\,|\,n$.  It follows that $6\,|\,n$.  Hence $6-2=4$ divides $n$.  Hence $4\times 3=12\,|\,n$.  Hence $12-3=9$ divides $n$, as does $12-4=8$.  Furthermore, $8-3=5, 9-2=7$ divides $n$.  And we are done).
Let $p_1, \cdots, p_k$ be the distinct primes which divide $n$ and let $P=\prod_{i=1}^{k}p_i$ be their product.  Then, of course $P\,|\,n$.  Now, $P\neq n$ since $4\,\nmid P$.  Thus, the fact that $n$ is special means that $(P-4)\,|\,n$.  But $P-4>2$ and it is not divisible by $4$, hence it is divisible by some odd prime $q$. Then $q\,|\,n$ but $q\not \in \{p_i\}$, hence we have a contradiction.  And we are done.
