Write the odd perfect number $m=p^k a^2$ as a product of primes $$m = p^k {p_1}^{2a_1} \cdots {p_v}^{2a_v}.$$ (Note that it is known that $v \geq 9$ by work of Nielsen.) Let $N(m)$ be the number of subscripts $i$ for which there is a subscript $j$ such that $$\gcd\Bigg(\sigma({p_i}^{2a_i} {p_j}^{2a_j}), p_i p_j \Bigg) > 1.$$
In Theorem 2 in Dandapat, Hunsucker, and Pomerance's Some new results on odd perfect numbers (1975), it is proved that
THEOREM 2 If $m$ is an odd perfect number, then $$v + 1 - \bigg(\log(v+1)/\log(2)\bigg) \leq N(m) \leq v.$$
Theorem 2 shows that $N(m)$ is not even close to $0$, but more nearly $v$.
Here is my:
INQUIRY: Does Theorem 2 imply that $\gcd(\sigma(p^k),\sigma(a^2)) > 1$?
It should be possible to prove this implication, but I am not seeing it.
