TITLE: What are all the integral solutions of $n!=m(m^2-1)$?
QUESTION [17 upvotes]: Observe that:

$3!=2(2^2-1)$
$4!=3(3^2-1)$
$5!=5(5^2-1)$
$6!=9(9^2-1)$

Question: What are all the integral solutions of $n!=m(m^2-1)$?
I guess it is just $(n,m) = (3,2),(4,3),(5,5),(6,9)$, but how to prove that there is no other one?
I checked that there is no other one for $n<20$.

Let reformulate the problem using Cardano's formula:    consider the cubic equation $$x^3+px+q=0,$$
its discriminant is $\Delta = -(4p^3+27p^2)$. Our case corresponds to $(p,q) = (-1,-n!)$, so $\Delta = 4-27(n!)^2 <0$. Thus the cubic has one real root and two non-real complex conjugate roots, and by Cardano's formula, the real root is $$ \left(-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}} \right)^{1/3} +\left(-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}\right)^{1/3} $$
So the problem "reduces" to ask whether there exists an integer $n>6$ such that
$$ \left(\frac{n!}{2} +\sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3} +\left(\frac{n!}{2} - \sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3}  $$
is also an integer. It is not clear that it helps...

REPLY [2 votes]: Variations of your problem are very well studied. For example, Brocard's problem asks about whether or not there are any solutions $(n,m)$ to $n!=m^2-1$ other than $(4,5)$, $(5,11)$, and $(7,71)$. This conjecture dates back to Erdos.
An interesting generalization is asking how many solutions $n!=P(m)$ has for any integer polynomial. Under the abc conjecture, there are only ever finitely many solutions if $P(m)$ has degree at least $2$.