Distinctness of consecutive-block lcms: is $M(n,k)\neq M(m,k)$ whenever $m\ge n+k$? (Erdős #677)
Statement
For integers $n\geq 0$ and $k\geq 1$, let $M(n,k)=\mathrm{lcm}(n+1,n+2,\ldots,n+k)$ be the least common multiple of the block of $k$ consecutive integers starting after $n$. Is it true that $$M(n,k)\neq M(m,k)$$ for all $m\geq n+k$ — that is, once two length-$k$ blocks are disjoint, their least common multiples are always different?
Acceptance. FULLY RESOLVES: prove that $M(n,k)\neq M(m,k)$ for all $k\geq1$ and all $m\geq n+k$ (a complete proof, possibly reducing to the Thue–Siegel finiteness and then eliminating the finitely many candidate coincidences for every $k$), OR exhibit an explicit counterexample — integers $n,k$ and $m\geq n+k$ with $M(n,k)=M(m,k)$, verified by computing both least common multiples. ADVANCES: an exhaustive verified search certifying that no coincidence $M(n,k)=M(m,k)$ with $m\geq n+k$ occurs within a stated range of $n,m,k$ (code plus range plus exhaustiveness certificate); a complete classification of the mixed-length coincidences $M(n,k)=M(m,l)$ with $l<k$ inside a stated range, extending the two known solutions $M(4,3)=M(13,2)$ and $M(3,4)=M(19,2)$; or a proof of Erdős's stronger same-prime-support conjecture for a new value of $k$. Deliver the proof or counterexample, or the search code plus certified range and data.
Background
Posed by Erdős [Er79], [Er79d], [ErGr80]; listed as open on erdosproblems.com/677 (fetched 2026-07-21, status 'open'). By the Thue–Siegel theorem, for each fixed $k$ there are only finitely many pairs $m,n$ with $m\geq n+k$ and $M(n,k)=M(m,k)$, so the content is whether that finite set is in fact empty for every $k$. Erdős asked more generally how many solutions $M(n,k)=M(m,l)$ exist with $m\geq n+k$ and $l>1$, expecting very few, and none at all once $l\geq k$; the only coincidences he knew are $M(4,3)=M(13,2)$ and $M(3,4)=M(19,2)$ (both with $l<k$). In [Er79d] he conjectured the stronger statement that, apart from finitely many exceptions, if $k>2$ and $m\geq n+k$ then $\prod_{i\leq k}(n+i)$ and $\prod_{i\leq k}(m+i)$ cannot even have the same set of prime factors. The problem is discussed as B35 in Guy's collection [Gu04] and is linked to Erdős #678 and #686, and to Erdős #850 (erdosproblems.com/850, the matching-prime-support problem, also on this venue). The attacker's tool: large-scale computational search for lcm coincidences $M(n,k)=M(m,l)$ across ranges of $n,m,k,l$, backed by Thue–Siegel / Diophantine-approximation arguments to convert the known finiteness into an exact count.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #677 (T. F. Bloom) | website |
| REF-02 | Erdős Problem #850 (T. F. Bloom) — related matching-prime-support problem | website |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.