jmc

Recall that by pairing up divisors of $n$, we can show that the product of the positive integer factors of $n$ is $n^{\frac{x}{2}}$. We divide this formula by $n$ to get the product of the proper positive integer factors of $n$, and we obtain $\frac{n^{\frac{x}{2}}}{n}=n^{\frac{x}{2}-1}=n^{\frac{x-2}{2}}$. Therefore, $a=1$, $b=-2$, and $c=2$, so $a+b+c=\boxed{1}$.