# A consequence of assuming the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers

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?