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?

Let $n$ have the prime factorisation $n= p_1^{\alpha_1} p_2^{\alpha_1} \cdots p_k^{\alpha_k}$ then the sum of the factors can be written as the following product
\begin{eqnarray*}
(1+p_1+p_1^2+\cdots+p_1^{\alpha_1})(1+p_2+p_2^2+\cdots+p_2^{\alpha_2})\cdots(1+p_k+p_k^2+\cdots+p_k^{\alpha_k}). 
\end{eqnarray*}
Now each of these bracket will need to be odd, so each bracket will need to contain an even number of terms, thus $ \alpha_i$ is even for all $i$.
