Problem Statement
Show that the equation\[n! = a_1!a_2!\cdots a_k!,\]with $n-1>a_1\geq a_2\geq \cdots \geq a_k\geq 2$, has only finitely many solutions.
Categories:
Number Theory Factorials
Progress
This would follow if $P(n(n+1))/\log n\to \infty$, where $P(m)$ denotes the largest prime factor of $m$ (see Problem [368]). Erdős [Er76d] proved that this problem would also follow from showing that $P(n(n-1))>4\log n$.The condition $a_1<n-1$ is necessary to rule out the trivial solutions when $n=a_2!\cdots a_k!$.
Surányi was the first to conjecture that the only non-trivial solution to $a!b!=n!$ is $6!7!=10!$. More generally, Hickerson (as reported in [Er76d]) conjectured that the only non-trivial solutions to the equation in the problem statement are\[9!=2!3!3!7!,\]\[10!=6!7!,\]\[10!=3!5!7!,\]and\[16!=14!5!2!.\]Luca [Lu07b] has shown that there are only finitely many solutions, conditional on the ABC conjecture, and proved unconditionally that the number of $n\leq x$ which admit a non-trivial solution is\[\leq \exp \bigg(f(x)\frac{\log (x)}{\log\log (x)}\bigg)\]for any function $f(x)$ which tends to infinity.
This is discussed in problem B23 of Guy's collection [Gu04].
In the case when $k=2$, Erdős [Er93] proved that if $n!=a_1!a_2!$ with $n-1>a_1\geq a_2$ then\[a_1\geq n-5\log\log n,\]and says it 'would be nice' to prove $a_1\geq n-o(\log\log n)$. Bhat and Ramachandra [BhRa10] replace the $5$ with $(1+o(1))\frac{1}{\log 2}$, and also prove that the same bound holds for arbitrary $k\geq 2$.
Numerical investigations on solutions to $n!=a_1!a_2!$ have been carried out by Caldwell [Ca94] and Habsieger [Ha], and it is known that there are no solutions aside from $10!=6!7!$ for $n\leq 10^{3000}$.
Source: erdosproblems.com/373 | Last verified: January 14, 2026