(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.
Best Answer
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, 109 (2002), no. 7, 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 $n \equiv 17 \pmod{72}$.