Suppose $A$ is a set of $2n$ positive integers { $a_{i}$ } with $a_{i} \geq 2$ which can be partitioned into at most $r \leq n$ coprime pairs. The following fact is true: For all $A$ and every $n \geq 1$, $$ \prod_{i = 1}^{2n} (a_{i} - 1) \equiv 0 \pmod{2^{r}}. $$ For example, the set {$2,3,5,6$} can be partitioned into two coprime pairs $(2,3)$ and $(5,6)$, so $2^{2} = 4$ divides $1 \cdot 2 \cdot 4 \cdot 5 = 40$. The set {$2,4,8,16$} cannot be partitioned into coprime pairs (i.e., $r = 0$), so the result above is not particularly interesting.
Question: Can this result be sharpened? References are welcome.
Thanks!
(Update) I think the above result can be generalized to the following: Suppose $A$ is a set of $2n$ positive integers { $a_{i}$ } with $a_{i} \geq 2$ which can be partitioned into at most $r \leq n$ pairs with odd greatest common divisor. Then $2^{r}$ divides $\prod_{i = 1}^{2n} (a_{i} - 1)$.