The topic of odd perfect numbers likely needs no introduction.
In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Let $$D(x) = 2x - \sigma(x)$$ denote the deficiency of $x$, and let $$s(x) = \sigma(x) - x$$ denote the sum of aliquot/proper divisors of $x$.
Euler proved that a hypothetical odd perfect number must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $N = q^k n^2$ is (odd) perfect, it follows that $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2,$$ from which we have $$\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}$$ since $\gcd(q^k,\sigma(q^k))=1$.
Letting $N_i$ denote the numerators and $D_i$ denote the denominators of the fractions in the equation above, we get $$N_0 = \sigma(n^2), D_0 = q^k$$ $$N_1 = 2n^2, D_1 = \sigma(q^k)$$ $$N_2 = N_1 - N_0 = D(n^2), D_2 = D_1 - D_0 = s(q^k)$$ $$N_3 = N_0 - N_2 = 2s(n^2), D_3 = D_0 - D_2 = D(q^k).$$
Note that, we obtain $$N_4 = N_1 - N_3 = 2D(n^2), D_4 = D_1 - D_3 = 2s(q^k).$$
Here is my question:
In general, if $i \neq j$ are any positive integers (which are both bigger than $1$), then is it guaranteed that $$\frac{N_k}{D_k} = \frac{N_i - N_j}{D_i - D_j}$$ is always an integer, if $N_0$ and $N_1$ are defined as above?
We can say that $\dfrac{N_i-N_j}{D_i-D_j}\in\mathbb Z$ under the condition that $D_i-D_j\not=0$ instead of $i\not=j$ (see the end of this answer) since we have that if$$\frac BA=\frac DC\in\mathbb Z\quad\text{and}\quad A\not=C$$ then $$\frac{B-D}{A-C}=\frac{B-\frac BAC}{A-C}=\frac{A(B-\frac BAC)}{A(A-C)}=\frac{B(A-C)}{A(A-C)}=\frac BA\in\mathbb Z$$
Note that $D_i=D_j$ can happen even when $i\not=j$.
For example, if we define $D_5,N_5,D_6$ and $N_6$ as $$\begin{align}D_5&:=D_0-D_4,\qquad N_5:=N_0-N_4, \\\\D_6&:=D_3-D_2,\qquad N_6:=N_3-N_2\end{align}$$ then we get $D_5=D_6$, so $\dfrac{N_5-N_6}{D_5-D_6}$ is not defined.