Is it true that every odd perfect number can be written in the form $\frac{r\sigma(r)}{2r - \sigma(r)}$? (Note:  This question has been cross-posted to MO.)
Is it true that every odd perfect number $N = q^k n^2$ can be written in the form $$N = \frac{r\sigma(r)}{2r - \sigma(r)}?$$
(Here, $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.)
In particular, setting
$$r = ({q^{(k-1)/2}}n)^2$$
appears to work.
The proof is contained in this paper.
MY ATTEMPT
Setting
$$r = ({q^{(k-1)/2}}n)^2$$
in
$$\frac{r\sigma(r)}{2r-\sigma(r)}=\frac{q^{k-1}n^2\sigma(q^{k-1}n^2)}{2q^{k-1}n^2 - \sigma(q^{k-1}n^2)},$$
we shall show that this last quantity is equal to $q^k n^2 = N$.
We assume (?) that equality indeed holds and work our way backwards.  (That is, we show that the steps are reversible.)
$$\frac{q^{k-1}n^2\sigma(q^{k-1}n^2)}{2q^{k-1}n^2 - \sigma(q^{k-1}n^2)}={q^k}{n^2}.$$
Cancelling ${q^{k-1}}{n^2}$ from both sides of the last equation, we get
$$\frac{\sigma(q^{k-1}n^2)}{2q^{k-1}n^2 - \sigma(q^{k-1}n^2)}=q.$$
$$\sigma(q^{k-1}n^2) = 2q^k n^2 - q\sigma(q^{k-1}n^2)$$
$$(q+1)\sigma(q^{k-1})\sigma(n^2) = 2q^k n^2.$$
Alas, here is where I get stuck.  I do not know how to rewrite
$$(q+1)\sigma(q^{k-1})$$
as
$$\sigma(q^k).$$
 A: As pointed out by Pace Nielsen via MO, the value of $r$ should be
$$r = q^{(k-1)/2} n^2$$
as considered in the paper.
Setting
$$r = q^{(k-1)/2} n^2$$
in
$$\frac{r\sigma(r)}{2r-\sigma(r)}=\frac{q^{(k-1)/2}n^2\sigma(q^{(k-1)/2}n^2)}{2q^{(k-1)/2}n^2 - \sigma(q^{(k-1)/2}n^2)},$$
we shall show that this last quantity is equal to $q^k n^2 = N$.
We assume (?) that equality indeed holds and work our way backwards.  (That is, we show that the steps are reversible.)
$$\frac{q^{(k-1)/2}n^2\sigma(q^{(k-1)/2}n^2)}{2q^{(k-1)/2}n^2 - \sigma(q^{(k-1)/2}n^2)}={q^k}{n^2}.$$
Cancelling ${q^{(k-1)/2}}{n^2}$ from both sides of the last equation, we get
$$\frac{\sigma(q^{(k-1)/2}n^2)}{2q^{(k-1)/2}n^2 - \sigma(q^{(k-1)/2}n^2)}=q^{(k+1)/2}.$$
$$\sigma(q^{(k-1)/2}n^2) = 2q^k n^2 - q^{(k+1)/2}\sigma(q^{(k-1)/2}n^2)$$
$$(q^{(k+1)/2}+1)\sigma(q^{(k-1)/2})\sigma(n^2) = 2q^k n^2.$$
Alas, here is where I get stuck.  I do not know how to rewrite
$$(q^{(k+1)/2}+1)\sigma(q^{(k-1)/2})$$
as
$$\sigma(q^k).$$
Update (March 1, 2018)
But, of course!
$$\sigma(q^k)=\frac{q^{k+1}-1}{q-1}=\frac{\bigg(q^{(k+1)/2}+1\bigg)\bigg(q^{(k+1)/2}-1\bigg)}{q-1}=\bigg(q^{(k+1)/2}+1\bigg)\sigma(q^{(k-1)/2}),$$
where we have used the fact that $k+1 \equiv 2 \pmod 4$ (i.e., $k+1$ is even).
