Does anybody know a solution to this problem? (Sorry, I've missed one summand in the [previous post](https://mathoverflow.net/questions/39207/solve-in-positive-integers-n-m2).) 



Solve in positive integers: $n!=m(m+1)$
I'm pretty sure this is open. As suggested from Brocard's problem, it is interesting to investigate the Diophantine equations $$n!=P(m)$$ for polynomials $P$. You can see the paper ["On polynomial-factorial diophantine equations"](http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/home.html), by D. Berend, J.E. Harmse where they make some advances on the problem and prove that this equation has finitely many solutions for many classes of polynomials (irreducible, with an irreducible factor of large degree or with an irreducible factor to a large power).


So by their results it is known that the equation $n!=m^r(m+1)$ has finitely many solutions if $r\geq 4$. But for $r\in \{1,2,3\}$ the problem is open.


