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} $.