Sum of positive divisors if and only if perfect square let n be a positive odd integer, prove that the sum of the positive divisors of n is odd if and only if n is a perfect square.
I know that based on the prime factorization theory that every integer n can be written as the product of primes, if their sum is odd that means that there are equal pairs of even and odd divisors. Is this enough to conclude that n must be a perfect square?

I don't really understand your argument: What does "there are equal pairs of even and odd divisors" mean, especially given that there are no even divisors of $n$? You really need to be precise about the objects you're considering, and careful about how it's written.

Here's an approach that you might find useful. For each $d \in \mathbb{Z}$ such that $d | n$, you can pair $d$ with $n/d$ and notice that both are odd, so that
$$2 | (d + n/d).$$
This covers all the divisors of $n$, except when something special happens....
