PROVED (LEAN)
This has been solved in the affirmative and the proof verified in Lean.
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\[M(n,k)>M(m,k+1)?\]
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].
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This page was last edited 11 January 2026.
Additional thanks to: Boris Alexeev, Stijn Cambie, and Wouter van Doorn
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 #678, https://www.erdosproblems.com/678, accessed 2026-01-18