Problem Statement
Prove the following proposition.
If $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 – q^k$ is not a square.
Motivation
Let $q^k n^2$ be an odd perfect number with special prime $q$. Then $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
By Pomerance, et al., we know that $q^k < n^2$, so that $n^2 – q^k$ is a positive integer. Also, since $n^2$ is a square and $q \equiv 1 \pmod 4$, then
$$n^2 – q^k \equiv 1 – 1 \equiv 0 \pmod 4.$$
My Attempt
Suppose that $q^k n^2$ is an odd perfect number with special prime $q$, and that $n^2 – q^k = s^2$, for some $s \geq 2$.
Then
$$n^2 – s^2 = q^k = (n + s)(n – s)$$
so that we obtain
$$\begin{cases}
{q^{k-v} = n + s \\
q^v = n – s}
\end{cases}$$
where $v$ is a positive integer satisfying $0 \leq v \leq (k-1)/2$.
It follows that we have the system
$$\begin{cases}
{q^{k-v} + q^v = q^v (q^{k-2v} + 1) = 2n \\
q^{k-v} – q^v = q^v (q^{k-2v} – 1) = 2s}
\end{cases}$$
Since $q$ is a prime satisfying $q \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, from the first equation it follows that $v=0$, so that we obtain
$$\begin{cases}
{q^k + 1 = 2n \\
q^k – 1 = 2s}
\end{cases}$$
which yields
$$n = \frac{q^k + 1}{2} < q^k.$$
Lastly, note that the inequality $q<n$ has been proved by Brown (2016), Dris (2017), and Starni (2018), so that we are faced with the inequality
$$q < n < q^k.$$
This implies that $k>1$.
Finally, notice that $k>1$ contradicts the Descartes-Frenicle-Sorli Conjecture, while $n<q^k$ contradicts the Dris Conjecture.
Question
Is it possible to remove the reliance of this proof on the truth of either the Descartes-Frenicle-Sorli Conjecture or the Dris Conjecture?
Best Answer
Here's a way to finish the proof without appealing to any conjecture.
If $q^k n^2$ is a perfect number with $\operatorname{gcd}(q,n)=1$, we have $$ \sigma(q^k) \sigma(n^2) = 2 q^k n^2. $$ We know that $\sigma(q^k) = (q^{k+1}-1)/(q-1)$ and you've shown that $n = (q^k + 1)/2$, so we can conclude that $$ 2(q^{k+1}-1) \sigma(n^2) = (q-1) q^k (q^k + 1)^2.\tag{$*$} $$ Consider the GCD of $q^{k+1}-1$ with the right-hand side: $$ \operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)\operatorname{gcd}(q^{k+1}-1,q^k+1)^2, $$ since $q^k$ is coprime to $q^{k+1} - 1$.
Noticing that $q^{k+1} - 1$ = $q(q^k + 1) - (q + 1)$, we find $\operatorname{gcd}(q^{k+1}-1,q^k+1) = \operatorname{gcd}(q+1,q^k+1)$, which is $q+1$ because $k$ is odd.
Thus $$ \operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)(q+1)^2. $$ Since $k\equiv 1 \pmod 4$ and you have shown $k \gt 1$, we have $k \ge 5$. If $(*)$ holds, the left-hand side of the inequality must be $q^{k+1}-1$, which is then greater than $q^5$. But the right-hand side is less than $q^4$, so this is impossible.