Problem Statement
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 $\lvert D_6\cap \{1,\ldots,n\}\rvert \gg n$?
Categories:
Number Theory
Progress
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 $k>6$.
Source: erdosproblems.com/374 | Last verified: January 14, 2026