The resolution of the Diophantine equation $$m! = n(n+1)$$ was asked on M.SE. My intuition says that this cannot be solved by elementary means - apologies if I am mistaken.

I felt that the following path was promising, consider $$21! = 2^{18}\cdot 3^9\cdot 5^4\cdot 7^3\cdot 11\cdot 13\cdot 17\cdot 19$$ where the exponents are known in general by a formula of Legendre ($\sum_r \lfloor\frac{m}{p^r} \rfloor$). Since $n$ and $n+1$ are coprime we may consider how close deleting any of these prime powers will take us to $\sqrt{m!}$ (for example $2^{18}\cdot 3^9$ is closest here, within $1.9\times 10^9$ of the square root) - a lower bound on this distance would show that the Diophantine equation has no solution.

So are there analytic tools which might be able to get a lower bound here? What are they? Or is this approach not productive?


Here is another comment, which is not really an answer, but too long for a comment. If we have a plane algebraic curve $C$, given by, say $F(x,y)=0$, then we are interested in the set $C(\mathbb{Z})$ of integral points on $C$. This set by itself does not have any useful additional structure, but we might embed the curve $C $ into another object $J$, which is not a curve, but a surface. This can be constructed for any curve and is called the Jacobian variety $J$ of the curve. The point is, that $J$ is a group. Now Weil has proved in $1928$ the following:

If $J$ is the Jacobian variety of a curve, then the abelian group $J(\mathbb{Z})$ is finitely generated.

This means that we can (in principle) get an explicit description of the group $J(\mathbb{Z})$ in terms of generators and relations. If we have that, we may be able to use the group structure and the geometry in some way to describe the elements of $J(\mathbb{Z})$ which are in the image of $C$ - these are exactly the integral points on $C$.

For details see the article of Bugeaud, Y.; Mignotte, M.; Siksek, S.; Stoll, M.; Tengely, Sz.: Integral points on hyperelliptic curves. They succesfully treat the equation $$ n(n+1)=\frac{m!}{60(m-5)!} $$ In fact, it follows that $m\in \lbrace{0, 1, 2, 3, 4, 5, 6, 7, 15, 19\rbrace}$. Of course, this does not mean we can treat $n(n+1)=m!$ this way, but at least there is hope for equations of type $\binom{n}{\ell}=\binom{m}{k}$.


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