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

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

  • $\begingroup$ If the answer is yes, I think that nobody can give you an example. $\endgroup$
    – ajotatxe
    Jan 2, 2017 at 11:33
  • $\begingroup$ @ajotatxe, how about if the answer is NO? $\endgroup$ Jan 2, 2017 at 11:56
  • 1
    $\begingroup$ 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. $\endgroup$
    – user243301
    Jan 2, 2017 at 12:05
  • $\begingroup$ Thank you for your observations, @user243301! It is good to see more people getting interested in research on (odd) perfect numbers, ehh? =) $\endgroup$ Jan 2, 2017 at 12:07
  • $\begingroup$ 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 $=1mod4$. $\endgroup$
    – user243301
    Jan 27, 2017 at 17:24

1 Answer 1


The answer turns out to be YES.


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 $d>1$.


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 $d'' = ({q^{k-1}}{\sigma(q^k)}{d'})/(q+1) > 1$.

  • $\begingroup$ In fact, it is easy to show that, if $k=1$, then $$d = D(n^2) = 2n^2 - \sigma(n^2)$$ where $D(x)$ is the deficiency of the positive integer $x$. $\endgroup$ May 2, 2020 at 14:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .