Problem Statement
Let $M(n,k)=[n+1,\ldots,n+k]$ be the least common multiple of $\{n+1,\ldots,n+k\}$.
Are there infinitely many $m,n$ and $k\geq 3$ with $m\geq n+k$ such that\[M(n,k)>M(m,k+1)?\]
Are there infinitely many $m,n$ and $k\geq 3$ with $m\geq n+k$ such that\[M(n,k)>M(m,k+1)?\]
Categories:
Number Theory
Progress
The referee of [Er79] (later identified in [Er92e] as Selfridge) found $M(96,7)>M(104,8)$ and $M(132,7)>M(139,8)$.The description in [Er79] is ambiguous whether the problem intended is the above, or to take a fixed $k\geq 3$ and ask whether infinitely many $m$ and $n$ exist, but this is easily disproved (see the comments). Presumably Erdős intended to ask the problem above (which indeed is how it appears less ambiguously in [Er92e]).
The answer is yes, as proved in a strong form by Cambie [Ca24], who proved that for all sufficiently large $k$ there exist $m\geq n+k$ with this property.
It is easy to see that there are infinitely many solutions to $M(n,k)>M(m,k)$. If $n_k$ is the smallest $n$ with this property (for some $m$) then are there good bounds for $n_k$? Erdős writes that he could prove $n_k/k\to \infty$, but knew of no good upper bounds.
Erdős also asked the following: If $u_k$ is minimal such that $M(u_k,k)>M(u_k+1,k)$ and $t<\min(u_k,T)$ then is it true that $M(t,k)\leq M(T,k)$? Stijn Cambie and Wouter van Doorn have observed that there are many counterexamples to this with $t=u_k-1$ and $T=u_k+1$. For example, when $k=7$ we have $u_k=7$, yet $M(6,7)=M(7,7)>M(8,7)$.
See also [677].
Source: erdosproblems.com/678 | Last verified: January 16, 2026