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Let $x$ be a positive integer.

Denote the sum of divisors of $x$ by $$\sigma(x) = \sum_{d \mid x}{d},$$ and the deficiency of $x$ by $$D(x) = 2x - \sigma(x).$$

A number $N$ is said to be perfect if $\sigma(N)=2N$. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

The Descartes-Frenicle-Sorli Conjecture on odd perfect numbers predicts that $k=1$.

In the paper titled Conditions equivalent to the Descartes–Frenicle–Sorli Conjecture on odd perfect numbers – Part II, the following results are proved:

If $N = q^k n^2$ is an odd perfect number given in Eulerian form, then $k=1$ if and only if $$N = \bigg(\frac{q(q+1)}{2}\bigg)\cdot{D(n^2)}.$$

If $N = q^k n^2$ is an odd perfect number given in Eulerian form, then $k=1$ if and only if $$N = \frac{n^2 \sigma(n^2)}{D(n^2)}.$$

Putting these two results together, we have that

If $N = q^k n^2$ is an odd perfect number given in Eulerian form, then $k=1$ if and only if $$\bigg(\frac{q(q+1)}{2}\bigg)\cdot{D(n^2)} = N = \frac{n^2 \sigma(n^2)}{D(n^2)}.$$

This last equation implies that $k=1$ if and only if $$\bigg(D(n^2)\bigg)^2 = \frac{n^2}{(q+1)/2}\cdot\frac{\sigma(n^2)}{q}.$$

Here is my question:

Does the equation $$\bigg(D(n^2)\bigg)^2 = \frac{n^2}{(q+1)/2}\cdot\frac{\sigma(n^2)}{q}$$ imply that $(q+1)/2$ and $\sigma(n^2)/q$ are both squares?

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