(Note: This question has been cross-posted from MSE.)

Euclid and Euler proved that every even perfect number is of the form $m = \frac{{M_p}\left(M_p + 1\right)}{2}$ where $M_p = 2^p - 1$ is a prime number, called a Mersenne prime. Thus, an even perfect number is triangular.

On the other hand, Euler showed that an odd perfect number, if one exists, takes the form $N = q^k n^2$, where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$. (Descartes, Frenicle and subsequently Sorli conjectured that $k = 1$ always holds.)

Here is my question:

Has it been proved that odd perfect numbers cannot be triangular?

Added March 26 2016

If $\sigma(q) = 2n^2$, then it would follow that $n < q$, which implies that $k = 1$. The odd perfect number $N = q^k n^2$ then takes the form $N = \frac{q(q + 1)}{2}$. Unfortunately, it is known that $\sigma(q^k) \leq \frac{2n^2}{3}$.

Any pointers to the existing literature containing such a proof would be most appreciated.


Curiously enough, I asked myself the same question several days ago... I couldn't settle it; yet, resorting to Jacques Touchard's theorem on the form of odd perfect numbers (cf. J. A. Holdener, "A theorem of Touchard on the form of odd perfect numbers". Amer. Math. Monthly, vol. 109, issue 7 (Aug. - Sep., 2002), pp. 661-663), one can easily establish the following result:

Proposition. If $\frac{n(n+1)}{2}$ is and odd perfect number, then $n \equiv 1 \pmod{24}$ or $n \equiv 9 \pmod{72}$ or $n \equiv 17 \pmod{72}$.

  • $\begingroup$ +1, this answer adds perspective to the question. Thanks! $\endgroup$ – Jose Arnaldo Bebita Dris Sep 16 '16 at 2:01

Not an answer, but I just want to point out some thoughts that recently occurred to me, which are related to this problem.

By this answer, we know that every odd perfect number $N = q^k n^2$ can be written in the form $$N = \dfrac{q(q+1)}{2} \cdot d$$ where $d > 1$. (That is, an odd perfect number is a nontrivial multiple of the triangular number $$T(q) = \dfrac{q(q+1)}{2},$$ where $q$ is the Euler prime of $N$.)

If $k=1$, then it is easy to show that $$d = D(n^2)$$ where $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of the non-Euler part $n^2$.

(I have asked a separate question for the form of $d'$ here, if $N = q^k n^2$ is an odd perfect number with $k > 1$, and $N = \dfrac{q(q+1)}{2} \cdot d'$ with $d' > 1$.)


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.