What are all the integral solutions of $n!=m(m^2-1)$?

Question

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 $n<20$.

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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...

Answer 1

In Richard Guy's third edition of Unsolved Problems in Number Theory, he writes on p.301 in D25 Equations involving factorial $n$. that

Simmons notes that $n!=(m-1)m(m+1)$ for $(m,n)=(2,3),(3,4),(5,5),$ and $(9,6)$ and asks if there are other solutions. More generally he asks if there are any other solutions of $n!+x=x^k.$

This Simmons is in reference to:

Gustavus J. Simmons, A factorial conjecture, J. Recreational Math., 1(1968) 38.

So it would seem that this problem was open in 2004 when Guy wrote the third edition of his book, but I don't know if it has been solved in the last 17 years.

Answer 2

Interestingly enough, there might be something even deeper going on. Let us define a family of elliptic curves given by the Weierstrass equations

$$E_n: y^2=x^3-x-n!$$

Your question is asking whether $E_n$ has any integer solutions $(x,0)$, which by the Nagell-Lutz theorem is equivalent to asking if $E_n$ has any points of 2-torsion. A natural generalization of this conjecture is to ask whether or not the torsion group $E_n(\mathbb{Q})$ is trivial for any $n\neq3,4,5,6$.

Using Sage one can compute the torsion groups of elliptic curves, and computationally it checks out that $E_n(\mathbb{Q})$ is trivial for any $n\leq10,000$, except for $n=3,4,5,6$ where we have that the only torsion points were the points of 2-torsion.

It turns out, in fact, that this generalized problem is $\mathbf{equivalent}$ to your conjecture. This is true because $x^3-x$ is always a multiple of $3$ (check by hand of by Fermat's Little Theorem) so any non $2$-torsion solution $y^2=x^3-x-n!$ will also be a multiple of $3$. By Nagell Lutz this would imply that the discriminant $\Delta_{E_n/\mathbb{Q}}=-4(27n!^2-4)$ would be a multiple of $3$ which as we can clearly see it is not.

Answer 3

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 $2$.