(This question is an offshoot of this earlier one.)
In the paper titled Improving the Chen and Chen result for odd perfect numbers (Lemma 8, page 7), Broughan et al. show that if $$\frac{\sigma(n^2)}{q^k}$$ is a square, where $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$ and $q^k n^2$ is an odd perfect number with special/Euler prime $q$, then $k=1$. (That $q$ is the special/Euler prime means that $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, which means that $q \geq 5$.)
It is fairly easy to show that, in general, we have $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$, so that assuming $\sigma(n^2)/q^k$ is a square, we have $k=1$ by Broughan et al.'s result, and so we obtain $$\frac{\sigma(n^2)}{q}=\frac{n^2}{(q+1)/2}=D(n^2) \text{ is a square }.$$ Hence, $(q+1)/2$ is also a square. Using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$ where $B \neq 0$, $D \neq 0$, and $B \neq D$, we get $$\frac{\sigma(n^2) - n^2}{(q-1)/2}=D(n^2) \text{ is a square }.$$
Here is my question:
Does it follow that $\sigma(n^2) - n^2$ and $(q-1)/2$ are also squares?
MY ATTEMPT
Since $$\frac{q+1}{2} - \frac{q-1}{2} = 1$$ and $q \equiv 1 \pmod 4$, then $(q-1)/2$ and $(q+1)/2$ are consecutive integers. If they were both squares, then $$\bigg(\frac{q-1}{2} = 0\bigg) \land \bigg(\frac{q+1}{2} = 1\bigg)$$ which implies that $q=1$. This contradicts $q \geq 5$.
Thus, $(q-1)/2$, and therefore $\sigma(n^2) - n^2$, are not squares.
Follow-Up Question
Does this proof suffice?
If we have that $$\frac{q+1}{2}$$ and $$\frac{q-1}{2}$$ are both perfect squares and $\ q\ $ is odd, then in fact we have two consecutive integers, which can only be $\ 0\ $ and $\ 1\ $.
Hence, indeed we can conclude $\ q=1\ $ which is a contradiction.
So, your proof is valid.