Vtyjkyuk

Let $sigma(x)$ denote the sum of the divisors of the positive integer $x$.

Denote the deficiency of $x$ by
$$D(x)=2x-sigma(x).$$

I am interested in solutions to the equation
$$gcd(n^2, sigma(n^2)) = 2n^2 - sigma(n^2).$$

Case 1: $n$ is even.

In this case, it is known that $n=2^r$ are solutions for $r geq 1$.

Indeed, we have
$$gcd(2^2r, sigma(2^2r)) = 1 = 2(2^2r) - sigma(2^2r)$$
where the second equation follows from the fact that powers of two are almost perfect.

Case 2: $n$ is odd.

Added September 5, 2018 Note that $n^2 = 1$ is a trivial (odd) solution, as $1$ satisfies
$$gcd(1, sigma(1)) = 1 = 2(1) - sigma(1).$$

In this case, the only known solution (with $n>1$) is
$$n^2 = bigg(3cdot7cdot11cdot13bigg)^2 = 9018009.$$

Indeed, we obtain
$$gcd(9018009,sigma(9018009)) = 819 = 2(9018009) - sigma(9018009).$$

Here is my question:

Are there any other solutions?

My Attempt

Looking at the OEIS page for deficient-perfect numbers, it appears that no other solutions are known. Otherwise, if there existed another odd solution, then we will be able to produce an odd perfect or a spoof odd perfect number whose special prime/quasi-Euler prime has exponent $1$.

Can anybody confirm if my impressions are correct?

Reference

Conditions equivalent to the Descartes–Frenicle–Sorli Conjecture on odd perfect numbers, Notes on Number Theory and Discrete Mathematics, Volume 23, 2017, Number 2, Pages 12—20