OPEN
This is open, and cannot be resolved with a finite computation.
For any $m\in \mathbb{N}$, let $F(m)$ be the minimal $k\geq 2$ (if it exists) such that there are $a_1<\cdots <a_k=m$ with $a_1!\cdots a_k!$ a square. Let $D_k=\{ m : F(m)=k\}$. What is the order of growth of $\lvert D_k\cap\{1,\ldots,n\}\rvert$ for $3\leq k\leq 6$? For example, is it true that $\lvert D_6\cap \{1,\ldots,n\}\rvert \gg n$?
The open status of this problem reflects the current belief of the owner of this website. There may be literature on this problem that I am unaware of, which may partially or completely solve the stated problem. Please do your own literature search before expending significant effort on solving this problem. If you find any relevant literature not mentioned here, please add this in a comment.
Studied by Erdős and Graham
[ErGr76] (see also
[LSS14]). It is known, for example, that:
- no $D_k$ contains a prime,
- $D_2=\{ n^2 : n>1\}$,
- $\lvert D_3\cap \{1,\ldots,n\}\rvert = o(\lvert D_4\cap \{1,\ldots,n\}\rvert)$,
- the least element of $D_6$ is $527$, and
- $D_k=\emptyset$ for $k>6$.
View the LaTeX source
View history
Additional thanks to: Zachary Chase
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #374, https://www.erdosproblems.com/374, accessed 2026-08-09
1 claimed proof for this problem