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Q: If $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1) n^2$, does it follow that $q n^2$ is perfect?
Let $q$ be an (odd) prime, and let $\gcd(q,n)=1$.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
A number $N$ is said to be perfect if $\sigma(N)=2N$.
Here is my question:
If $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1) n^2$, does it follow that...
