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(Note: This question has been cross-posted to MO.)

Can an odd perfect number be a nontrivial multiple of a triangular number?

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    If the answer is yes, I think that nobody can give you an example. – 2017-01-02
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    @ajotatxe, how about if the answer is *NO*? – 2017-01-02
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    This is only to do feedback about it. We assume that there exists $m>1$ an integer such that, using the Eulerian form of an odd perfect number $N=P^{4\lambda+1}M^2$, then $$2P^{4\lambda+1}M^2=mn(n+1),$$ that is our non trivial triangle number is $m\frac{n(n+1)}{2}$. Then I believe that one can try analyse if it is possible deduce something, because one has $(P,M)=(n,n+1)=1$. If my calculations were rigths we can deduce also that $\sigma(mn(n+1))=3\cdot mn(n+1)$. On the other hand also one has Touchard's theorem to try do deductions ( if are feasibles in this question), dividing. Thanks. – 2017-01-02
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    Thank you for your observations, @user243301! It is good to see more people getting interested in research on (odd) perfect numbers, ehh? =) – 2017-01-02
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    I was thinking that if it ins't in the literature you can to study also the similar statement of this Steven Kahan, *Perfectly Odd Cubes*, Mathematics Magazine, Vol. 71, No. 2 (1998). I presume that you can read it with your MyJstor. What am I saying? That the first identity in the proof, here where the integer in RHS is your odd perfect number, seems that there is no problem when one do a comparison with Euler's theorem for OPN since being the $n$ in the identity odd, one can to take the second factor in such RHS as your Euler factor, and it is $=1mod4$, also our RHS is $=1mod4$. – 2017-01-27
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    I don't know nothing about the comparison of sizes between both factors in RHS, if we assume as was said that RHS is an OPN. The second factor also can has exponents $=1 mod 4$, see for example the toy example $97=97^1$. Thus I am saying that if it is interesting you can do the similar question that tell us the paper, but now for odd perfect numbers for an unknown integer $n$, and try to deduce if it is possible, that is if such assumption has mathematical meaning. Good luck. – 2017-01-27
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    @user243301, but is it even known that an odd perfect number is a sum of cubes? – 2017-01-28
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    I don't know nothing about it, if is well known or maybe is it's possible explain that it is interesting. What I am saying is that from my viewpoint is a similar question than this in your post, because you are asking if make sense that an OPN be a triangle number (a nontrivial multiple of a triangle number). I tried deduce an absurd by contradicition using Euler's theorem for OPN, and choosing the second factor as your Euler factor $q^{4λ+1}$ and thus the other factor as $m^2$ I can't deduce a contradition with the simple claims that I said, that is dividing by $4$, for an unknown odd $n$. – 2017-01-28

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The answer turns out to be YES.

Claim

If $N = q^k n^2$ is an odd perfect number, then $N$ can be written in the form $$N = \dfrac{q(q+1)}{2}\cdot{d}$$ for some integer $d>1$.

Proof

By Slowak's 1999 result, every odd perfect number must have the form $$\dfrac{{q^k}\sigma(q^k)}{2}\cdot{d'}$$ for some integer $d' > 1$.

If $k=1$, then our claim readily follows from Slowak's result.

Otherwise, if $k>1$, since $q \mid q^k$ and $(q + 1) \mid \sigma(q^k)$ for all $k \equiv 1 \pmod 4$, then we have that every odd perfect number must have the form $$\dfrac{q(q+1)}{2}\cdot{d''}$$ for some integer $d'' = ({q^{k-1}}{\sigma(q^k)}{d'})/(q+1) > 1$.