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Slowak showed in 1999 that every odd perfect number $N = q^k n^2$ can be written in the form $$N = \dfrac{{q^k}\sigma(q^k)}{2}\cdot{D}$$ where $D>1$.

From this result, it follows that every odd perfect number $N = q^k n^2$ is a nontrivial multiple of a triangular number, more specifically, $$T(q) = \dfrac{q(q+1)}{2},$$ where $q$ is the Euler prime of $N$.

Here is my question:

Does there exist an integer $r \neq q$ such that $N = q^k n^2 = \frac{r(r+1)}{2}\cdot{d}$ is an odd perfect number with $d>1$, where $q$ is the Euler prime of $N$?

Update (April 29 2017) Trivially, $r=1$ answers my original question, so I hereby amend my question as follows:

Updated Question

Does there exist an integer $r \neq q$ (satisfying $r>1$) such that $N = q^k n^2 = \frac{r(r+1)}{2}\cdot{d}$ is an odd perfect number with $d>1$, where $q$ is the Euler prime of $N$?

Some Considerations Since $\gcd(r, (r+1)/2)=1$ and $\gcd(q^k,n^2)=1$, then either we have $$\bigg(r \mid q^k\bigg) \land \bigg((r+1)/2 \mid n^2\bigg)$$ or $$\bigg(r \mid n^2\bigg) \land \bigg((r+1)/2 \mid q^k\bigg).$$

In the first case, since $r \neq q$, $r \neq 1$, and $q$ is prime, then $q^2 \mid r$.

In the second case, since $r \neq 1$, $q \mid (r+1)/2$.

Some Additional Considerations

The equation $$N = q^k n^2 = \frac{r(r+1)}{2}\cdot{d}$$ can be rewritten as a quadratic in $r$, $$r^2 + r - \frac{2N}{d} = 0,$$ with discriminant $$1 + \frac{8N}{d} = c^2$$ for some $c \in \mathbb{N}$. (Trivially, when $d=N$, then $c=3$, in which case we have $r=1$. This is ruled out by the condition $r>1$.)

Alas, here is where I get stuck.

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