TITLE: Solve in positive integers: $n!=m(m+1)$
QUESTION [34 upvotes]: Does anybody know a solution to this problem? (Sorry, I've missed one summand in the previous post.)

REPLY [19 votes]: I thought I should share with you the results from my computations from the past two days.
All these computations could be carried out to significantly higher limits with better code and a bit more time. However, my motivation was just to play with them.
Please check the code before taking these computations as facts! I did not check the code too carefully.

Computation 1
The equation $$n! = m(m+1)$$ does not have integer solutions for $3 < n < 10^9$.
I used Mark Sapir's observation that if $n$ is a solution then $4n!+1$ is a square. I wrote a small program which calculates Legendre symbols $$\left(\frac{4n!+1}{p}\right),$$ with a set of primes $p>n$, to check if there might be a solution for $n$. Confirming that for $3 < n < 10^9$ there are no solutions took about one hour on my laptop using 40 primes. The code is available at http://users.jyu.fi/~tamaraja/temp/brogen.c for anyone interested to see it.

Computation 2
I wanted to test the slightly stronger guess in Mark Sapir's post that for $p>5$, $p^2$ does not divide any integer of the form $4n!+1$. This turned out not to be true.
I checked that for $5 < p <10^6$ there is one exception, namely $$761671^2 ~|~ 4\cdot446142!+1.$$ Here I used only brute force and the trivial observation that $$p ~|~ 4n!+1 \Rightarrow n < p.$$ Testing this range for $p$ took a couple of hours. The code is available at http://users.jyu.fi/~tamaraja/temp/sqrdiv.c

Computation 3
Byron Schmuland pointed out the similarity to Brocard's problem in his comment. Because the previous computations I found for the equation $$n! + 1 = m^2$$ were carried out over a decade ago, I decided to extend them a bit.
With the same approach which was used by Berndt and Galway in 2000 (and in Computation 1 above) I confirmed that the equation has no solutions for $7 < n < 10^{10}$. This took about one day. The code was the same as in Computation 1.

Computation 4
After the previous 3 computations I did a little bit of the "missing" fourth. The equation $$n!+1 \equiv 0 \pmod{p^2}$$ has no solutions $(p,n)$ with $p$ prime and $613 < p < 10^6$.